question_id
string | site
string | title
string | body
string | link
string | tags
list | votes
int64 | creation_date
timestamp[s] | comments
list | comment_count
int64 | category
string | diamond
int64 |
|---|---|---|---|---|---|---|---|---|---|---|---|
201
|
mathoverflow
|
Is there a model category describing shape theory?
|
Is there a nice model structure on some category of topological spaces compatible with [shape theory](https://en.wikipedia.org/wiki/Shape_theory_\(mathematics\))? In particular, weak equivalences should induce isomorphisms on sheaf cohomology.
As an example, the [Polish Circle](https://en.wikipedia.org/wiki/Shape_theory_\(mathematics\)#Warsaw_Circle) has the same sheaf cohomology with constant coefficients as $S^1$, and both spaces also have equivalent categories of covering spaces. So one could ask if in such a model structure, the polish circle would be weakly equivalent to an ordinary one.
**Edit:** I would be happy with an answer for any nice category of topological spaces which contains the Polish circle, e.g. compactly generated weak Hausdorff spaces. Please ignore the following paragraph.
Also, sheaf theory is defined on rather general topological spaces, so it would be nice if the underlying category of topological spaces would be much more general than, say, compactly generated weak Hausdorff spaces.
|
https://mathoverflow.net/questions/345396/is-there-a-model-category-describing-shape-theory
|
[
"at.algebraic-topology",
"sheaf-theory",
"model-categories",
"shape-theory"
] | 18 | 2019-11-06T08:52:29 |
[
"@DenisNardin Thanks for the suggestion. I see the point now and did the edit.",
"@SebastianGoette Well, it is sort of the same definition, only you need to generalize it to a notion of the shape of a topos rather than just of topological spaces (you might have heard it under the name \"étale homotopy type\").The point is that the underlying space of an algebraic variety does not have much useful information (e.g. all curves over an alg. closed field are homeomorphic!),and you need more refined objects (e.g the étale topos)to recover the information present in the topology for manifolds. Still, your question is interesting, I would only remove the reference to alg. varieties.",
"@DenisNardin I was really thinking of the shape as a topological space, as in the wikipedia article. I was not even aware that there is a separate definition in algebraic geometry.",
"@SimonHenry It's not idempotent, but it becomes one if you consider the profinite shape instead (see the appendix to SAG and the Barwick-Glasman-Haine work on the stratified shape)",
"In Lurie's perspective shape theory is an adjunction between the $\\infty$-category of $\\infty$-toposes and the $\\infty$-category of pro-spaces. I always had the feeling that this adjunction is idempotent ( not sure if this is really true though, is there someone who know ?) if it is the case \"shapes\" are a localization of the category of pro-spaces, as such it should be possible to construct a model structure representing them (maybe on pro-simplicial sets...). If this works one can start thinking about whether this transfer to a model structure on some category of topological spaces...",
"@SebastianGoette I'll have to admit I am a bit confused. What do you mean with the shape of an algebraic variety then? The notion I know is not encoded in the topology.",
"@DenisNardin I don't expect to see the 'algebraic' cohomology (in whatever sense) here, because the structure sheaf seems to be an extra datum that the topological space does not have. But maybe apart from constant sheaves, we also want to (maybe have to) consider constructible sheaves. But then all information we need is still encoded in the topology.",
"@SimonHenry I must admit that I have no idea. The correct notion of equivalence seems to be part of the problem. Also I wonder whether it is possible to work on topological spaces, or whether one needs topoi.",
"What is the precise definition of 'shape equivalences' (it is not in the wikipedia link) ? I'm not expert in shape theory but I believe there has been several non-equivalent definition... is it equivalent to the definition Lurie gave for infinity toposes ?",
"Hmm... the shape of algebraic varieties does not depend on the underlying topological space, unless I'm mistaken (rather, it is the shape of the étale topos), so I doubt a naive approach would work there."
] | 10 |
Science
| 0 |
202
|
mathoverflow
|
Are simplicial finite CW complexes and simplicial finite simplicial sets equivalent?
|
**Edit** Originally the question was whether an arbitrary diagram of finite CW complexes can be approximated by a diagram of finite simplicial sets. In view of Tyler's comment, this was clearly asking for too much. I restricted the question from arbitrary diagrams to simplicial diagrams.
* * *
It is well-known that for any finite CW complex $K$ one can construct a finite simplicial set whose geometric realization is equivalent to $K$. But the construction is not functorial.
Suppose we have a simplicial object in finite CW complexes. Can one always construct a simplicial object of finite simplicial sets, whose levelwise geometric realization is connected to the original diagram by a zig-zag of levelwise weak equivalences?
One may ask a similar question in a more general context. Suppose we have an $\infty$-category $\mathcal C$. Suppose $A$ is a collection of (say compact) objects of $\mathcal C$. Let $CW^f(A)$ be the category of finite cellular objects generated by $A$. Roughly speaking, $CW^f(A)$ consists of objects that can be built as a finite homotopy colimit of objects of $A$. I am leaving the definition of $CW^f(A)$ a little vague - feel free to use any reasonable notion. Let $S_\bullet^f(A)$ be the category of simplicial objects in $\mathcal C$ that in each simplicial degree are a finite sum of elements of $A$, and are degenerate above some dimension. Geometric realization gives a functor $S_\bullet^f(A)\to CW^f(A)$. Now we have an obvious generalization of the question:
Suppose we have a simplicial object in $CW^f(A)$. Can we find a simplicial object in $S_\bullet^f(A)$ whose levelwise geometric realization is levelwise (weakly) equivalent to it? If it is not true in general, is there a reasonable sufficient assumption?
|
https://mathoverflow.net/questions/338660/are-simplicial-finite-cw-complexes-and-simplicial-finite-simplicial-sets-equival
|
[
"at.algebraic-topology",
"homotopy-theory"
] | 18 | 2019-08-19T00:48:40 |
[
"The point of the original question is that a model does not support every morphism. Tyler's example of a functor from $B\\mathbb N$ is just the choice of an object and a morphism. If you allow Kan complexes or finite CW complexes, then they are cofibrant-fibrant and thus they support arbitrary morphisms. In the purely homotopical setting, the object supports the morphism. Tyler's example was $S^1$, which is as nicely cellular as you can imagine: represent it as the constant simplicial object. Maybe having to spell out the cells allows counterexamples, but not that one.",
"@BenWieland I am feeling a little slow. Why it is not a counterexample?",
"@BenWieland yes, absolutely. you can replace each object in a diagram with any weakly equivalent object and get an equivalent diagram at the cost of working coherently.",
"If the \"more general\" version is purely homotopical, Tyler's example isn't a counterexample, is it? (a counterexample to an even more general version) And a nitpick, just a problem for objects, if a \"finite homotopy colimit\" includes retracts, then you run into Wall finiteness problems, but @R.vanDobbendeBruyn's version doesn't.",
"A possible more general finiteness condition could be the assumption that every object in the diagram has finitely many maps out of it. (This has popped up in some inductive constructions I ran into; examples include finite diagrams and semisimplicial objects.)",
"@AchimKrause That is, indeed, a great deal simpler than what I had in mind.",
"In Tyler's example, can't we just observe that for any finite simplicial set $K$, only finitely many maps on $\\pi_1$ are induced by endomorphisms $K\\to K$? This is because there are only finitely many maps $K\\to K$ in total.",
"Hi Tyler, thanks. This sounds right and suggests I should impose a local finiteness condition on the indexing diagram (such as, there can be only finitely many morphisms between any two objects). What I really wanted to know was about simplicial objects: can any simplicial object in finite CW complexes be approximated by a simplicial object in finite simplicial sets? I will edit the question.",
"Hi Greg, I suspect that the map $B\\Bbb N \\to CW$, representing $S^1$ with the endomorphisms $2^k$ for $k > 0$, can't be realized by a diagram of finite simplicial sets, but to prove it I'd want to come up with some explicit growth condition on the number of edges needed to represent elements of $\\pi_1$ for a fixed finite model. This may be a little harder for s.sets than for $\\infty$-categories because the mapping complexes aren't Kan complexes."
] | 9 |
Science
| 0 |
203
|
mathoverflow
|
What is the logical complexity of the Hodge conjecture?
|
The Hodge conjecture seems to me the most mysterious among the Millennium problems (and many others). In particular, I am not sure about its logical complexity. It is not difficult to see that the conjecture is equivalent to a $\Pi^0_2$ statement if restricted to varieties over $\overline{\mathbb{Q}}$. (Basically, it is because these varieties form a recursive set, and the statement sounds like "for any Hodge class there is an algebraic cycle".) However the conjecture may be true over $\overline{\mathbb{Q}}$ but false in general ; if I am not mistaken, this possibility has not being ruled out yet. My guess is that the Hodge conjecture may be formulated as a $\Pi^0_3$ statement but, at present, not as $\Pi^0_2$. Is it true?
Some remarks. As is well known, the Hodge conjecture may be separated into two parts:
I. Hodge classes are absolute,
II. Absolute Hodge classes are algebraic.
The second part is $\Pi^0_2$. (It follows from results of Voisin, "Hodge loci and absolute Hodge classes", Compositio Mathematica, Vol. 143 Part 4, 945-958, 2007. ) So, the logically hard part seems to be the first one. In view of Proposition 0.5 of the Voisin's paper, it boils down to whether Hodge loci are defined over $\overline{\mathbb{Q}}$ or not. I have some ideas about how to translate this into $\Pi^0_3$ but nothing close to $\Pi^0_2$. Naturally, if it is true that the Hodge conjecture can _not_ be formulated as a $\Pi^0_2$ statement for now, there is no way one can actually prove this fact. In case it is not unlikely to be so, I am only asking for an expert opinion. Also, if possible, I would appreciate a slick and sound proof of the "upper bound" $\Pi^0_3$ (although I have some ideas about it, I do not really like them).
|
https://mathoverflow.net/questions/289745/what-is-the-logical-complexity-of-the-hodge-conjecture
|
[
"ag.algebraic-geometry",
"lo.logic"
] | 18 | 2018-01-02T03:07:33 |
[
"@Libli See en.wikipedia.org/wiki/Arithmetical_hierarchy",
"what are $\\Pi_{2}^0$ and $\\Pi_{3}^0$?"
] | 2 |
Science
| 0 |
204
|
mathoverflow
|
Is this Variation of the Continuum Hypothesis Inconsistent with ZFC or ZF?
|
It is a well-known fact that the Generalized Continuum Hypothesis is undecidable from ZFC. For similar sentences $\phi$, this is simply equivalent to ZFC having a model $M$ for which $M\models\phi$.
I wanted to see if there was any way to make all $\beth$-numbers limits. At first, this seemed dubious, but now, after creating a variation on GCH, I have realized it seems completely legitimate.
The variation I created is called the **Cofinal Continuum Hypothesis** , or CFCH. It is defined as follows:
$$\forall\kappa(2^\kappa=\aleph_{\kappa^+})$$
This immediately destroys the idea of the continuum hypothesis, as with this definition $2^{\aleph_0}=\aleph_\Omega$, which is very far off from $\aleph_1$.
The reason I chose specifically this definition is that its cofinality preserving properties when compared to GCH, namely that $\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)$. This property "piggybacks" off of GCH's undecidability in a sense; most proofs that a variation on GCH is inconsistent with ZFC (or similar theories) involve some failure of an inequality with $\mathrm{cof}(2^\kappa)$ and $\mathrm{cof}(\kappa)$. With this definition, it is clear to see that this problem no longer exists because in GCH $\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)$.
Because of this construction, one could generalize it to $2^\kappa$ is the $\kappa^+$-th aleph fixed point, or $2^\kappa=\aleph_{\aleph_\kappa}$. These, however, seem even more audacious than CFCH itself.
**However** , the reason I feel this is still dubious is because of it's interesting relation to the Singular Cardinal Hypothesis. It is clear that, with this definition, the first strong limit cardinal is a singular cardinal. This becomes a problem when introduced to the Singular Cardinal Hypothesis, as, assuming and CFCH and SCH:
$$\forall\alpha(\mathrm{cof}(\beth_{\omega\cdot\alpha})\neq\beth_{\omega\cdot\alpha}\rightarrow 2^{\beth_{\omega\cdot\alpha}}=\beth_{\omega\cdot\alpha}^+)$$ $$2^{\beth_\omega}=\beth_\omega^+$$ $$2^{\beth_\omega}=\aleph_{\beth_\omega^+}$$ $$\beth_\omega^+=\aleph_{\beth_\omega^+}$$
This is a contradiction, as it implies that a successor cardinal is an $\aleph$-fixed point. Thus, CFCH is inconsistent with SCH. Proving CFCH equiconsistent to ZFC would be a great feat, as it would prove $\neg$SCH equiconsistent to ZFC as well, thus showing that the existence of a measurable cardinal with Mitchell order $\kappa^{++}$ is equiconsistent to ZFC (as a result of Gitik), and every other large cardinal axiom implied by said axiom (which has not yet been done). This is why I doubt that CFCH will be proven equiconsistent with ZFC.
**Here are some other interesting facts implied by CFCH:**
* $\beth_{\omega\cdot\alpha}$ is the $\alpha$-th $\aleph$-fixed point (and thus all strong limit cardinals are $\aleph$-fixed points). This is yet another doubt I have for CFCH.
* $\mathrm{cof}(\beth_\alpha)=\mathrm{cof}(\aleph_\alpha)$ (and thus for regular cardinals $\aleph_\alpha$, $\mathrm{cof}(\beth_\alpha)=\aleph_\alpha$ and for limit ordinals $\alpha$, $\mathrm{cof}(\beth_\alpha)=\alpha$.)
* The Konig's Theorem corollaries about cofinality seem are implied by it
($\kappa<\mathrm{cof}(2^\kappa)=\mathrm{cof}(\kappa^+)=\kappa^+$ and $2^\kappa<(2^\kappa)^{\mathrm{cof}(2^\kappa)})=2^{\kappa\cdot\mathrm{cof}(2^\kappa)}=2^{\mathrm{cof}(2^\kappa)}>2^\kappa$)
* As for other variations of the Generalized Continuum Hypothesis and Continuum Hypothesis:
* $\mathrm{CFCH}\rightarrow\neg\mathrm{GCH}$
* $\mathrm{CFCH}\rightarrow\neg\mathrm{SCH}$
* $\mathrm{CFCH}\rightarrow\neg\mathrm{CH}$
* $\mathrm{CFCH}\rightarrow\neg\mathrm{NCH}$ (The Natural Continuum Hypothesis, which is found in another MathExchange link "[A New Continuum Hypothesis](https://mathoverflow.net/questions/148756/a-new-continuum-hypothesis-revised-version)")
**In general, it would be really useful if the reader of this question reply in the comments with either a link to a website listing many interesting models of ZF and ZFC or several Forcing methods or they send me the names of the models themselves.** Of course, you could reply to this question with comments pertaining to the question as well.
|
https://mathoverflow.net/questions/283443/is-this-variation-of-the-continuum-hypothesis-inconsistent-with-zfc-or-zf
|
[
"lo.logic",
"set-theory",
"continuum-hypothesis"
] | 18 | 2017-10-13T21:31:39 |
[
"I mainly mentioned NCH because it seems like this is a \"variant\" of NCH (where cardinal exponentiation speeds up)",
"Just as a reminder, I would like to mention that despite its tempting appearance and intuitive background from natural numbers, the Natural Continuum Hypothesis (NCH) is inconsistent with ZFC due to Konig's lemma. However, it sounds interesting to me that you considered the cofinality version of CH independently.",
"Under CFCH, $2^{\\aleph_n} = \\aleph_{\\omega_{n+1}}$ and therefore $2^{\\aleph_{\\omega}} = \\aleph_{\\omega_\\omega}^\\omega = \\aleph_{\\omega_{\\omega + 1}}$. By the theory of pcf, we get that if $a$ is the set of all cardinals between $\\aleph_{\\omega}$ and $\\aleph_{\\omega_\\omega}$, the possible cofinalities of $\\prod a$ is the regular cardinals in the range between $\\aleph_{\\omega_\\omega + 1}$ and $\\aleph_{\\omega_{\\omega + 1}}$ which has cardinality $|a|^+$. A recent paper of Gitik claims that it is consistent, but it is open if one can obtain this type of pcf structure in accessible cardinals.",
"Ah, so ZFC + CFCH is even stronger.",
"Easton's theorem implies that ZFC + \"for all regular $\\kappa$, $2^\\kappa = \\aleph_{\\kappa^+}$\" is equiconsistent with ZFC. Other variants, such as \"the $\\kappa^+$-th fixpoint\" or $\\aleph_{\\aleph_{\\kappa^+}}$ are possible, too. - But it is well known that ZFC+ non-SCH is much stronger than ZFC, in terms of consistency strength."
] | 5 |
Science
| 0 |
205
|
mathoverflow
|
Is the universality of the surreal number line a weak global choice principle?
|
I'd like to consider the principle asserting that the surreal number line is universal for all class linear orders, or in other words, that every linear order (including proper-class-sized) linear orders, order-embeds into the surreal number line.
This principle is the class-analogue of the corresponding result for countable linear orders, proved by Cantor, namely, that every countable linear order embeds into the rational line $\newcommand\Q{\mathbb{Q}}\Q$. Using the "forth" part of Cantor's back-and-forth method, one enumerates a given countable linear order and then maps each new point to a rational number filling the corresponding cut in the images of the previous points.
A similar transfinite extension of this argument works with the surreal numbers. Namely, if $\langle A,\leq_A\rangle$ is any class-sized linear order and $A$ is $\newcommand\Ord{\text{Ord}}\Ord$-enumerable, then we may map each new point to the first-born surreal number filling the corresponding cut in the images of the previous elements of $A$, and thereby construct an embedding of $A$ to the surreal numbers.
What this argument shows is that, provided the elements of the linear order $A$ are $\Ord$-enumerable, then we may recursively build up an embedding of the linear order into the surreal number line. In other words, if the principle of global choice holds, which is equivalent to the assertion that every class is $\Ord$-enumerable, then every class-sized linear order embeds into the surreal line. So global AC implies that the surreals are universal for class linear orders.
Similar uses of global choice underlie many other assertions of universality for the surreal numbers, with respect to class-sized structures, as ordered fields and so on. It is fundamental in all the arguments that I know for such universality that one should be able to enumerate the elements of the domain structure in order type $\Ord$, in order to carry out the transfinite embedding procedure.
Since this use of the $\Ord$-enumerations has always struck me as essential, I have long believed that the global axiom of choice was required in order to prove that the surreal number line is universal for class linear orders.
But upon recent reflection, I realized that I don't actually have a proof that global AC is required. I simply don't have a proof that global AC is required for universality, and at the same time, I don't know how to omit it.
Hence these questions:
* Is the global axiom of choice required to prove that the surreal number line is universal for all class-sized linear orders?
* Can one prove the universality of the surreal number line merely in GB+AC, without the global axiom of choice?
* Or is the universality of $\text{No}$ an intermediate principle between AC and global AC?
To answer these questions, it would seem to be useful to settle the issue of the surreal-numbers-universality principle (which I propose we denote by $\text{UNo}$), in some of the known models where global AC fails. For example, in my answer to the question [Does ZFC prove the universe is linearly orderable?](https://mathoverflow.net/a/110823/1946), I showed that there are models of ZFC with no definable linear order of the universe. If we could settle the universality of the surreals in that model, it would answer at least one of the questions above. For example, if the model had surreal-universality UNo, then we would know that universality is not equivalent to global AC; if it is didn't have it, then we would know that UNo isn't provable in GB+AC.
(This topic is related to some ideas I'll be speaking about at my talk on [The hypnagogic digraph](http://jdh.hamkins.org/the-hypnagogic-digraph-jmm-2016/), which I am giving this Friday at the JMM in Seattle.)
|
https://mathoverflow.net/questions/227849/is-the-universality-of-the-surreal-number-line-a-weak-global-choice-principle
|
[
"lo.logic",
"set-theory",
"axiom-of-choice",
"linear-orders",
"surreal-numbers"
] | 18 | 2016-01-06T22:01:00 |
[
"Or to put it the other way round: the fact that a class-sized linear order A is embeddable in No apparently does not give a way to Ord-enumerate A. If one wants to somehow use the property `all class-sized linear orders are embeddable in No', then again, we trivially know that all class-sized linear orders isomorphic to a subclass of No are embeddable...so you would have to come up with a class-sized linear order B defined outside of No, and from B's embeddability in No deduce that No is Ord-enumerable. [Well, I hope I'm making some sense here, feel free to disregard my question/comments].",
"The question intrigues me although I have extremely little knowledge of such matters (but I'm pondering on variations of surreal numbers for constructive math). If I understand correctly, No itself is Ord-enumerable iff Global AC holds? (I don't know the first thing about this, but read it in another post). If that is correct, then intuitively I'd say: global AC is not necessary for UNo, because one can embed No in No without global AC... This may seem a strange argument, but it is meant to illustrate that one can have weaker conditions which already give embeddability.",
"Keep in mind that the surreal numbers No themselves are definable in L[G], and as elements these define every object in L[G].",
"Also, when you appeal to UNo, do you have to shout UNo when you have one line left in your proof, or else you have to take four more lines to finish?",
"Joel, what I'd try to do is to show that every linear ordering comes from a bounded part.",
"If you think about $L[G]$ in the model in my answer to your question about linear orders of $V$, then the question is whether UNo holds there. We know there is no linear order of all of $L[G]$ there, so perhaps there is a paucity of definable linear orders there (always allowing parameters). If so, then we might expect UNo to hold. But I can't quite prove it.",
"I just opened my eyes ten minutes ago, it's my pleasure to have choice related questions to think about when I wake up. :-)",
"@AsafKaragila But let me say that I'm very glad you're looking at this question. I'd appreciate any insight you might have.",
"I think of global-AC as an assertion in second-order set theory, as in Goedel-Bernays set theory, so one doesn't need to worry about definability. That is, one simply asserts that there is a proper class choice function on all nonempty sets, or equivalently, that there is an Ord-enumeration of V. In particular, global AC definitely holds after adding a Cohen real.",
"When you say global choice, does it come with parameters? Specifically, if you add a Cohen real to $L$, you have global choice with parameters, but not global choice itself. It stands to reason that in such model you might have what you were looking for.",
"I'll post my slides on Friday, and you can read all about it!",
"Hypnagogic digraph! Quite an intriguing title. Will there be forthcoming discussions of the hypnopompic digraph as well? :-)"
] | 12 |
Science
| 0 |
206
|
mathoverflow
|
Does the "holomorphic spheres-to-continuous spheres" forgetful function respect the mixed Hodge structures on homotopy groups?
|
For each smooth, projective, complex variety $X$ that is simply connected, John Morgan constructed a natural mixed Hodge structure on the homotopy group $\pi_k(X,x)\otimes \mathbb{Q}$. This was extended to (possibly singular) quasi-projective varieties independently by Hain, by Deligne, and by Navarro Aznar. For ~~any~~ the basepoint connected component of the double loop space $\Omega^2 X_0\sim [(S^2,0),(X,x)]_0$, each homotopy group $\pi_k(\Omega^2 X_0)\otimes \mathbb{Q}$ is canonically isomorphic to $\pi_{k+2}(X)\otimes \mathbb{Q}$, $k>0$. Thus these also have natural mixed Hodge structures.
Grothendieck constructed a Hom scheme $$H=\text{Hom}_{\mathbb{C}}((\mathbb{CP}^1,0),(X,x))$$ parameterizing algebraic maps (i.e., holomorphic maps in case $X$ is projective). Each quasi-compact, open and closed subset $H_\beta$ of the Hom scheme is quasi-projective. Assuming that $H_\beta$ is connected, there is a well-defined homotopy class of maps $$\Phi_{\beta}:H_\beta \to \Omega^2 X_0,$$ that associates to each holomorphic map from $\mathbb{CP}^1 = S^2$ the underlying continuous map. There are associated pushforward maps on homotopy groups, $$\Phi_{k,\beta}: \pi_k(H_\beta)\otimes \mathbb{Q} \to \pi_{k+2}(X,x)\otimes \mathbb{Q}.$$ Assume now that $H_\beta$ is simply connected so that the groups $\pi_k(H_\beta)\otimes \mathbb{Q}$ have mixed Hodge structures. (There are many cases of $X$ where the associated spaces $H_\beta$ are connected and simply connected. Also, with no hypothesis on $H_\beta$, one can also ask about compatibility with mixed Hodge structures of the induced homomorphism $H_1(H_\beta;\mathbb{Q}) \to \pi_3(X,x)\otimes \mathbb{Q}.$) Also, as Dan Petersen points out, it is necessary to form the Tate twist of the target, i.e., $\pi_{k+2}(X,x)\otimes \mathbb{Q}(1)$.
**Question**. Is the pushforward map $\Phi_{k,\beta}$ a morphism of mixed Hodge structures?
Of course there are many complex varieties $X$ for which $H$ is empty, and then the answer is trivially yes. Yet there are many other complex projective manifolds, e.g., Fano manifolds, where the maps $\Phi_{k,\beta}$ are nontrivial.
|
https://mathoverflow.net/questions/233979/does-the-holomorphic-spheres-to-continuous-spheres-forgetful-function-respect
|
[
"ag.algebraic-geometry",
"homotopy-theory",
"hodge-structure"
] | 18 | 2016-03-18T08:22:50 |
[
"@AknazarKazhymurat No, unfortunately I have not found an answer to this question.",
". . . This reduces to the smooth, projective case. In that case, the Sullivan minimal model of the De Rham complex agrees with the Sullivan minimal model of the De Rham cohomology (with zero differentials). Since the Sullivan minimal model of a dga is (roughly) a multilinear algebra operation on the dga, the work is to check that these multilinear operations respect the weight and Hodge filtrations.",
"@VivekShende. \"Is this mixed hodge structure constructed by actually making a simplicial scheme ...\" Whether it can be made that way, I do not believe that is how Morgan first constructed these mixed Hodge structures. Morgan uses a single log resolution and an algebraic construction (basically by spectral sequences) of the rational homotopy type of the open stratum in terms of those of the (projective, smooth) boundary strata. Deligne proved that this spectral sequence respects mixed Hodge structures. Morgan checks that is also respects Sullivan minimal models . . .",
"Is this mixed hodge structure constructed by actually making a simplicial scheme (which?) whose cohomology groups are the homotopy groups of the original $X$?",
"@DanPetersen. You are correct, there should be a Tate twist. I will add it now.",
"I'd expect there to be a Tate twist at least. If $k=0$ you get a map from $H_0(H_\\beta)=\\mathbf Q(0)$ to $H_2(X)$, which has weight $-2$.",
"\"Is it easy to see?\" Yes; this is part of adjointness of the space of continuous functions with the compact-open topology. Let $C$, $X$, and $H$ be topological spaces with $H$ Hausdorff. Let $F:H\\times C \\to X$ be continuous. The claim is that the induced function $\\Phi:H\\to [C,X]$ is continuous, where $[C,X]$ has the compact-open topology. For an open subset $U$ of $X$, for a compact subset $K$ of $C$, the subset $(H\\times K)\\cap \\Phi^{-1}(X\\setminus U)$ is a closed subset of $H\\times K$, hence it is proper over $H$. So it has closed image in $H$, and the complement of the image is open.",
"Yes, I did mean that topology. Is it easy to see? Are the $H_\\beta$s actually homeomorphic to subspaces of $\\Omega^2X$?",
"\"Is it easily seen that there is a continuous $\\Phi_\\beta$?\" Certainly $\\Phi_\\beta$ is not continuous for the Zariski topology on $H_\\beta$. It is continuous for the analytic topology on the associated analytic space of $H_\\beta$. Similarly, the notation $\\pi_{k+2}(X,x)$ means the homotopy group for the associated analytic space of $X$.",
"Is it easily seen that there is a continuous $\\Phi_\\beta$?",
"@TomChurch. You are correct. I am eliding the issue that the function $\\Phi_\\beta$ is only well-defined up to homotopy. There is a canonical function $\\Phi_\\beta$ to the connected component of $\\Omega^2 X$ corresponding to the homotopy class $\\beta\\in \\pi_2(X)$. The homotopy equivalence of this component with the basepoint component is only well-defined up to homotopy.",
"What does the subscript 0 mean in $\\Omega^2 X_0\\sim [(S^2,0),(X,x)]_0$? If it means \"basepoint component\", then why do you say \"for any connected component\"? If not, why is the isomorphism between $\\pi_k(\\Omega^2 X_0)\\otimes \\mathbb{Q}$ and $\\pi_{k+2}(X)\\otimes \\mathbb{Q}$ canonical?"
] | 12 |
Science
| 0 |
207
|
mathoverflow
|
Deforming a basis of a polynomial ring
|
The ring $Symm$ of symmetric functions in infinitely many variables is well-known to be a polynomial ring in the elementary symmetric functions, and has a $\mathbb Z$-basis of Schur functions $\\{S_\lambda\\}$, with structure constants $c_{\lambda\mu}^\nu$ called Littlewood-Richardson numbers.
I have been studying a commutative deformation of these structure constants, i.e. lifting $c_{\lambda\mu}^\nu$ to a 2-variable polynomial $c_{\lambda\mu}^\nu(a,b)$, for which I have a very explicit formula (much too complicated to describe here). However, since $Symm$ is a polynomial ring (i.e. free), any commutative deformation of it must be trivializable: there must be a way to deform each $S_\lambda$ to a function $S_\lambda(a,b)$ whose ordinary multiplication has my structure constants.
> Given the $\\{c_{\lambda\mu}^\nu(a,b)\\}$, how can I trivialize this family, finding such deformations $S_\lambda(a,b)$?
It seems like the answer should be nonunique (hence harder to find); it's easy to imagine deformations of the basis that don't change the structure constants at all.
> What extra conditions should one put on the deformation in order to make the trivialization unique?
|
https://mathoverflow.net/questions/199537/deforming-a-basis-of-a-polynomial-ring
|
[
"co.combinatorics",
"ac.commutative-algebra",
"symmetric-functions",
"algebraic-combinatorics",
"schur-functions"
] | 18 | 2015-03-09T18:40:56 |
[
"I think $a=0$ is triangular one way, $b=0$ triangular the other way. The one-box rule is here: mathoverflow.net/questions/88569/…",
"Is it possible that the deformation is upper-triangular (i.e. that $S_\\lambda(a,b) = S_\\lambda + \\textrm{lower order terms}$)? How complicated are the structure constants when $\\mu$ is 1 box?"
] | 2 |
Science
| 0 |
208
|
mathoverflow
|
An algebraic strengthening of the Saturation Conjecture
|
The Saturation Conjecture (proved by Knutson-Tao) asserts that $c_{n\mu,n\nu}^{n\lambda}\neq 0\Rightarrow c_{\mu,\nu}^{\lambda} \neq 0$, where $c$ denotes a Littlewood-Richardson coefficient and $n$ is a positive integer. Let $\mathfrak{o}$ be a (commutative) discrete valuation ring with a finite residue field $k=\mathfrak{o}/\mathfrak{p}$. Let $M$ be a finite $\mathfrak{o}$-module. (For instance, we can take $\mathfrak{o}$ to be the $p$-adic integers $\mathbb{Z}_p$, in which case $M$ is a finite abelian $p$-group.) Thus there is a unique partition $\lambda=(\lambda_1,\lambda_2,\dots,\lambda_r)$ such that $$ M \cong \bigoplus_{i=1}^r (\mathfrak{o}/\mathfrak{p}^{\lambda_i}). $$ We call $\lambda$ the _type_ of $M$. A fundamental result of Philip Hall is that if $M$ has type $\lambda$, then there exists a submodule $N$ of type $\mu$ and cotype $\nu$ (i.e., $M/N$ has type $\nu$) if and only if $c^{\lambda}_{\mu\nu}\neq 0$. See Macdonald, _Symmetric Functions and Hall Polynomials_ , Chapter II, (4.3).
Now suppose that $M$ has type $n\lambda$, and $N$ has type $n\mu$ and cotype $n\nu$. Is there is a submodule $L$ of $M$ of type $\lambda$, such that $L\cap N$ has type $\mu$ and cotype (with respect to $L$) $\nu$?
|
https://mathoverflow.net/questions/212368/an-algebraic-strengthening-of-the-saturation-conjecture
|
[
"co.combinatorics",
"symmetric-functions"
] | 18 | 2015-07-26T11:57:23 |
[
"@DavidSpeyer You are right about the definition of Hall algebra. I have fixed this.",
"A note on terminology (and then I will think about this interesting question). I usually understood the Hall algebra to refer to the ring whose elements were formal sums of isomorphism classes of $\\mathfrak{o}$-modules, and where multipication was given by counting submodules of given type and co-type. See, for example, arxiv.org/abs/math/0611617 . Also, as I imagine you know, Derksen and Weyman gave a far reaching generalization of the saturation conjecture in terms of this sort of Hall algebra -- see ams.org/journals/jams/2000-13-03/S0894-0347-00-00331-3",
"Is there a combinatorial interpretation for some special case (similar to Yamanouchi tableaux for the LR-coeffs)? The saturation conjecture implies the analogous statement for Kosktka and skew Kosktka coefficients, so is there a \"Kostka\"-analogue of this question, which should be easier to prove?"
] | 3 |
Science
| 0 |
209
|
mathoverflow
|
Smooth curves on smooth varieties
|
Let $X$ be a smooth, proper algebraic variety over a field $k$, of positive dimension.
Is it true that $X$ contains a smooth Zariski-closed curve?
If it is projective, this is true by Bertini. But is it true in general?
|
https://mathoverflow.net/questions/112524/smooth-curves-on-smooth-varieties
|
[
"ag.algebraic-geometry"
] | 18 | 2012-11-15T13:44:16 |
[
"@algori : yes, indeed ! The answer is positive if $X$ is separably rationally connected (over an algebraically closed field). Indeed, if $X$ is of dimension $\\leq 2$, it is projective so there is no problem. And if it is of dimension $\\geq 3$ it contains (lots of) smooth rational curves by Theorem IV 3.9 in Koll'ar's book \"Rational curves on algebraic varieties\". ",
"Olivier -- it's worse than you think: I'm not sure how to prevent the curve from intersecting itself.",
"@algori : The problem with this strategy is that it is really possible to have a divisor $E$ on $X'$, exceptional over $X$, such that every curve meeting $E$ has singular image in $X$. This happens for instance if $X$ is a smooth surface, and $X'$ is constructed by blowing up a point, then blowing up a point on the exceptional divisor, then blowing up the intersection of the two exceptional divisors, and if you choose $E$ to be the third exceptional divisor. The problem is that the differential of $X'\\to X$ is identically zero at every point of $E$. ",
"Here is a strategy for showing that the answer is positive: blow up a proper smooth variety, call it $X$, to get a projective smooth $X'$ such that $X'\\setminus D'$ is isomorphic to $X\\setminus D$ with $D\\subset X$ closed and $D'\\subset X'$ a divisor with normal crossings; then take a curve in $X'$ that intersects transversally all the components of $D'$ and project back again."
] | 4 |
Science
| 0 |
210
|
mathoverflow
|
An "exercise" on von Neumann algebra tensor product
|
The following problem appears to be an easy exercise on von Neumann algebra tensor products, but since I've been failing to find a rigorous proof, I'd like to make sure it's not that trivial. Suppose $M$ and $N$ are von Neuamann factors such that $L^\infty[0,1] \mathbin{\bar\otimes} M \cong L^\infty[0,1] \mathbin{\bar\otimes} N$. Does it follow $M \cong N$? It's true under the assumption of separability by the disintegration theory, but not so sure if $M$ (and hence $N$) does not have separable predual.
|
https://mathoverflow.net/questions/270650/an-exercise-on-von-neumann-algebra-tensor-product
|
[
"oa.operator-algebras",
"von-neumann-algebras"
] | 18 | 2017-05-25T03:32:20 |
[
"Obviously when $M$ or $N$ are Type $\\mathrm{I}$ it can done in a different way and the result is well-known, but I'm unsure how to proceed if the factors in question are Type $\\mathrm{II}_\\infty$ or Type $\\mathrm{III}$, even assuming separability.",
"This is obviously achievable if $M$ and $N$ are $\\mathrm{II}_1$ factors, with $H$ and $K$ being their standard representations. In which case $L^2(0, 1) \\otimes H$ and $L^2(0, 1) \\otimes K$ are standard representations of $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$, respectively, so a spatial isomorphism of the above form just comes from any isomorphism between $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$. But without trace can this still be done?",
"I'm late here, but I'm curious as to how the question is resolved even in the separable case. All the treatments I could find on direct integral theory always needs a spatial isomorphism between $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$. More precisely, suppose $M$ acts on $H$ and $N$ acts on $K$, it seems to always require a unitary between $L^2(0, 1) \\otimes H$ and $L^2(0, 1) \\otimes K$ that conjugates $L^\\infty(0, 1) \\bar{\\otimes} M$ and $L^\\infty(0, 1) \\bar{\\otimes} N$ onto each other."
] | 3 |
Science
| 0 |
211
|
mathoverflow
|
What is the Hochschild cohomology of the Fukaya-Seidel category?
|
Let $(Y, \omega)$ be a compact symplectic manifold and let $Fuk(X,\omega)$ be its Fukaya category. The Hochschild cohomology of this category should be given by $HH^\bullet(Fuk(Y,\omega))=H^\bullet(Y, \mathbb{C})$, at least at the level of vector spaces (the product structure should be the quantum cohomology on $H^\bullet(Y,\mathbb{C})$: see for example [Hochschild (co)homology of Fukaya categories and (quantum) (co)homology](https://mathoverflow.net/questions/11081/hochschild-cohomology-of-fukaya-categories-and-quantum-cohomology) ). As $Fuk(Y,\omega)$ should be a Calabi-Yau category (see for example [Are Fukaya categories Calabi-Yau categories?](https://mathoverflow.net/questions/13114/are-fukaya-categories-calabi-yau-categories) ), the Hochschild homology $HH_\bullet(Fuk(Y,\omega))$ coincides with the Hochschild cohomology $HH^\bullet(Fuk(Y,\omega))$ up to a shift of the grading.
Let me deform the preceding setting: let me assume that $(Y,\omega)$ is a Kähler manifold and that $W \colon Y \rightarrow \mathbb{C}$ is an holomorphic function on $Y$ (in particular $Y$ is noncompact if one wants $W$ non-constant). One should be able to define the Fukaya-Seidel category $FS((Y,\omega),W)$ (I probably need to assume that $W$ has only isolated non-degenerate critical points given the current technology). The Hochschild homology of this category should be given by:
$HH_\bullet(FS((Y,\omega),W)=H^\bullet(Y,Y_{-\infty},\mathbb{C})$,
where we take the cohomology relative to the fiber $Y_{-\infty}=W^{-1}(z)$ of a point $z\in\mathbb{C}$ with $Re(z)<<0$. But now $FS((Y,\omega),W)$ is no longer Calabi-Yau and so the Hochschild cohomology is no longer simply related to the Hochschild homology. So my question is:
What is (or should be) the Hochschild cohomology $HH^\bullet(FS((Y,\omega),W))$ of the Fukaya-Seidel category $FS((Y,\omega),W)$?
If $(Y,\omega),W)$ is the mirror of a smooth projective Fano variety $X$ then
$FS((Y,\omega),W)=D^b Coh(X)$
and one knows by the Hochschild-Kostant-Rosenberg theorem that
$HH_\bullet(D^bCoh(X))=\oplus_{p-q=\bullet}H^p(X,\Omega_X^q)$
and
$HH^\bullet(D^bCoh(X))=\oplus_{p+q=\bullet}H^p(X,\wedge^q T_X)$.
In particular, Hochschild homology and cohomology are indeed very different in general.
This question is motivated by the general principle saying that the deformations of a category are governed by $HH^2$. In particular, a subquestion could be: what are the deformations of the Fukaya-Seidel category?, and this should be equivalent to understand what $HH^2$ is.
Disclaimer: when I write that something "should be" true, it is what I think is the general expectation but as I am not an expert in symplectic geometry, I don't know the technical state of the art of the rigorous proofs. As my question is itself of the "should be" form, I hope it is fine.
|
https://mathoverflow.net/questions/218232/what-is-the-hochschild-cohomology-of-the-fukaya-seidel-category
|
[
"sg.symplectic-geometry",
"mirror-symmetry",
"hochschild-cohomology",
"fukaya-category"
] | 18 | 2015-09-13T14:06:08 |
[
"Another important paper which is related to your question is Seidel's symplectic homology as a Hochschild homology. If you denote by $\\mathscr{B}$ the Fukaya category of the fiber and $\\mathscr{A}$ the Fukaya-Seidel category, then $HH_\\ast(\\mathscr{A}\\oplus t\\mathscr{B}[[t]])$ gives the symplectic homology of the total space.",
"The deformation issues concerning fiberwise compactification of a Lefschetz fibration are discussed in II1/2. The key conceptual piece is the fixed point Floer cohomology.",
"You probably need to read Seidel's papers entitled Fukaya $A_\\infty$ structures associated to Lefschetz fibrations I, II, II1/2."
] | 3 |
Science
| 0 |
212
|
mathoverflow
|
A cohomology class associated with a complex representation of a group
|
$\newcommand\CC{\mathbb C}\newcommand\ZZ{\mathbb Z}\newcommand\ad{\mathsf{ad}}\newcommand\Ext{\operatorname{Ext}}$ Suppose that $G$ is a finite group and that it acts on a finite dimensional complex vector space $V$.
For each $g\in G$ we may consider the subspace $V_g$ of $V$ spanned by the eigenvectors of $g$ corresponding to eigenvalues different from $1$, and put $d(g)=\dim V_g$. We pick any nonzero linear map $\omega_g:\Lambda^{d(g)}V_g\to\CC$.
If $h\in G$ then the map $x\in \Lambda^{d(g)}V_{hgh^{-1}}\mapsto \omega_g(h^{-1}v)\in\CC$ makes sense, and we write it $h\cdot\omega_g$. Since it is not zero, there is a nonzero scalar $\lambda(h,g)\in\CC^\times$ such that $h\cdot\omega_g=\lambda(h,g)\omega_{hgh^{-1}}$. Associativity of the group action implies that $$\lambda(gh,k)=\lambda(h,k)\lambda(g,hkh^{-1})$$ for all $g$, $h$, $k\in G$. This means that the function $\lambda:G\times G\to\CC^\times$ is a $1$-cocycle in the complex which computes $\Ext_{\ZZ G}^\bullet(\ZZ G^\ad,\CC^\times)$, when we use a bar resolution of the $G$-module $\ZZ G^\ad$ (which is the permutation $G$-module constructed from the conjugation action of $G$ on itself) We may therefore take its class $[\lambda]$ in $\Ext^1_{\ZZ G}(\ZZ G^\ad,\CC^\times)$.
This class depends only on the representation and not on the choice of the $\omega_g$'s, so we write it $c(V)$.
The long exact sequence corresponding to the exponential sequence $0\to\ZZ\to\CC\to\CC^\times\to0$ allows us to identify that $\Ext^1$ with $\Ext^2_{\ZZ G}(\ZZ G^\ad,\ZZ)$. If $CG$ is the set of conjugacy classes and for each $c\in CG$ we let $\ZZ c$ be the permutation module corresponding to the conjugation action on $c$, we have $\ZZ G^\ad=\bigoplus_{c\in CG}\ZZ c$. If $G_c$ denotes the centralizer of an element of $c$, some form of Shapiro's lemma allows us to identify $\Ext^2_{\ZZ G}(\ZZ c,\ZZ)$ with $\Ext^2_{\ZZ G_c}(\ZZ,\ZZ)$ and then our class $[\lambda]$ is an element of $$\bigoplus_{c\in CG}H^2(G_c,\ZZ).$$ This can be written a bit more canonically as $$\left(\bigoplus_{g\in G}H^2(C_g,\ZZ)\right)^G$$ with now $C_g$ the centralizer of $g\in G$ and $G$ acting on the direct sum in the only sensible way. If $G$ is abelian, this can be identified further to $\ZZ G\otimes H^2(G,\ZZ)$.
> **Question:** What is this cohomology class $c(V)$?
This reminds me of Burghelea's description of the cyclic homology of the group algebra, so it might be some form of Chern class (if some vector bundle constructed from V on the free loop space of BG?).
This class shows up when one computes the Hochschild cohomology of the cross product algebra $S(V)\rtimes G$, but explaining how would make this post too much longer. It appears as an obstruction to making this nice :-/
**On triviality:** suppose there is a central element $z$ in $G$ which acts by multiplication by a scalar different from $1$. In that case $c=\\{z\\}$ is a conjugacy class, $G_c=G$, $\lambda(g,z)=\det g$ for all $g$, and the component of $c(V)$ in $\Ext^1_{\ZZ G}(\ZZ c,\CC^\times)=\hom(G,\CC^\times)$ is the determinant map. If the representation does not land in $SL(V)$, then this is not zero. Examples of representations like this are the geometric representation of Weyl groups which have nontrivial center, in which case the longest element is central and acts as $-1$ —see [this question and its answers](https://mathoverflow.net/questions/81858/does-i-belong-to-weyl-group) for the list of which types work— or abelian groups acting with some element not having $1$ in its spectrum.
In the other direction, if the representation respects a symplectic form $\omega$ on $V$, then all the $d(g)$ are even and we may take $\omega_g$ to be the restriction of the $\tfrac12d(g)$th power $\omega^{d(g)/2}$ to $V_g$. In this case, $\lambda(h,g)=1$ identically, so $c(V)=0$.
**Note.** If $e$ is the exponent of $G$ and $\Omega$ is any set of $e$th roots of unity, we can consider the subspace $V_g^\Omega$ spanned by eigenvectors of $g$ corresponding to eigenvalues in $\Omega$, and proceeding as above we get a class $c_\Omega(V)$ in the same direct sum. The class above corresponds to taking $\Omega$ the set of all $e$th roots of unity different from $1$. The assigment $V\mapsto\bigoplus_{g\in G}V_g^\Omega$, for $c$ a conjugacy class of $G$, is a nice functor into Yetter-Drinfeld modules, by the way.
|
https://mathoverflow.net/questions/232090/a-cohomology-class-associated-with-a-complex-representation-of-a-group
|
[
"rt.representation-theory",
"group-cohomology"
] | 18 | 2016-02-24T21:43:54 |
[
"(Here LBG is the free loop space on BG) The first Chern class of this descended bundle is an element of $H^2(LBG,\\mathbb Z)$, which is a direct sum of cohomologies of centralizers. Is that class the same as the one described? That would be an answer to \"what is c(V)?\".",
"@Andreas, well, I know that the class is a bunch of homomorphisms (which are coherent in some way): that is how it is defined! What I am asking is what the class is, in the sense of is it something? For example, from V you can construct a vector bundle $\\tilde V=EG\\times_GV$ on $BG$, which you can pull back along the evaluation map $LBG\\times S^1\\to BG$ to get a vector bundle on that product which descends onto $LBG\\times_{S^1}S^1$, which is $LBG$.",
"@MarianoSuárez-Alvarez: Your question was \"What is this cohomology class?\". Maybe I do not understand what a possible answer would be, but didn't appear the class in $H^1$ naturally from the representation. What are you looking for?",
"$\\beta$ will be a cocycle here. An easy way to prove this is the following: in the twisted group algebra $\\mathbb{C}^{\\alpha}G$ we have the equality, if $g$ and $h$ commute in $G$, $U_h^{-1}U_gU_h = \\beta(g,h)U_g$. Now it is quite easy to prove that the resulting map $C_g\\rightarrow \\mathbb{C}^{\\times}$ is a homomorphism, and thus a one cocycle, since the action on $\\mathbb{C}$ is trivial.",
"I honestly do not understand what «that is all» means in this context.",
"The identification $H^2(C_g,\\mathbb Z)=H^1(C_g,\\mathbb C^{\\times})$ makes it look more obscure than it is. For each $g$, the centralizer $C_g$ fixes the subspace spanned by non-fixed eigenvectors and hence there is a determinant homomorphism $C_g \\to \\mathbb C^{\\times}$, i.e., a class in $H^1(C_g,\\mathbb C^{\\times})$, that's all.",
"@EhudMeir, indeed, my example with the determinant was an example of something else! :-| I'll correct it. if I start with with a $2$-cocycle $\\alpha:G\\times G\\to\\CC^\\times$, fix $g\\in G$ and let $\\beta:h\\in G_g\\mapsto \\alpha(g,h)/\\alpha(h,g)\\in\\CC^\\times$, I am only being able to prove that $d\\beta=\\beta^2\\smile\\beta^2$, no that it is a $1$-cocycle on $G_g$; is that construction treated somewhere? (I can do it if $\\alpha:G\\times G^{\\ad}\\to\\CC^\\times$ is a $1$-cocycle giving an element of $\\Ext^1(\\ZZ G^\\ad,\\CC^\\times)$, though: is that what you meant?)",
"Another question: if $[\\alpha]\\in H^2(G,\\mathbb{C}^{\\times})$, then we get another element in the same group $\\oplus_{c\\in CG}H^2(G_c,\\mathbb{Z})\\cong \\oplus_{c\\in CG}H^1(G_c,\\mathbb{C}^{\\times})$, namely if $h$ commutes with $g$ then we have the one cocycle $h\\mapsto \\alpha(g,h)/\\alpha(h,g)$. Is it possible to see any sort of connection between elements arising from two cocycles and from representations?",
"Shouldn't $\\lambda(g,1_G)=1$? That is: for the identity element $e=1_G$ it holds that $d(e) = 0$ because the only eigenvalue of $e$ is 1. I guess that in this case the interpretation of $\\wedge^0 V_e$ should be $\\mathbb{C}$, and the induced map should be the identity?"
] | 9 |
Science
| 0 |
213
|
mathoverflow
|
Cohomological characterization of CM curves
|
In his 1976 classical Annals paper on $p$-adic interpolation, N. Katz uses the fact that if $E_{/K}$ is an elliptic curve with complex multiplications in the quadratic field $F$, up to a suitable tensoring the decomposition of the algebraic $H_{\rm dR}^{1}(E,K)$ in eigenspaces for the natural $F^\times$-action coincides with the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$ and (for ordinary good reduction at $p$) with the Dwork-Katz decomposition of $H_{\rm dR}^{1}(E)\otimes B$ for $p$-adic algebras $B$.
Then, he asks for a converse statement. Namely, is it true that if the Hodge decomposition of $H_{\rm dR}^{1}(E,{\Bbb C})$, where $E_{/K}$ is an elliptic curve, is induced by a splitting of the algebraic de Rham, then $E$ has complex multiplications?
The question is left unanswered in that paper. Does anyone know if the question has been answered since?
|
https://mathoverflow.net/questions/13681/cohomological-characterization-of-cm-curves
|
[
"elliptic-curves",
"cohomology",
"nt.number-theory"
] | 18 | 2010-02-01T06:36:19 |
[
"Ok, I guess I was a bit lazy myself too..... :-) $K$ is a finite extension of $\\Bbb Q$ over which the complex multiplications are defined, that is $F\\subseteq K$. The comparison between algebraic $H^1$ and the Rham over $\\Bbb C$ is allowed by choosing (and fixing) an embedding $K\\rightarrow{\\Bbb C}$.",
"What is $K$ in this question? Too lazy to find Katz' paper in the mess behind me :-) To compute the de Rham cohomology with coefficients in $C$ you'll need a map $K\\to C$ right? Or are we considering all such maps? Or have I misunderstood the question?"
] | 2 |
Science
| 0 |
214
|
mathoverflow
|
Lipschitz constant of a homotopy
|
Let $n>0$, $\mathbb S^n$ be $n$-sphere and $1\in \mathbb S^n$ be its north pole.
A am looking for an example of compact manifold $M$ with a continuous $n$-parameter family of maps $h_x\colon M\to M$, $x\in \mathbb S^n$ such that
* $h_1=\mathop{\rm id}_M$
* For any Riemannian metric $g$ on $M$ and any family of maps $h'_x\sim h_x$ (i.e., such that the maps $h,h'\colon \mathbb S^n\times M\to M$ are homotopic) there is $x\in\mathbb S^n$ such that the Lipschitz constant of $h'_x\colon (M,g)\to(M,g)$ is $\ge 10000$.
**Comments.**
* I think about this problem for few years; I do not think it is an easy one, but I want to make sure that I did not miss a well known trick. A positive answer might have interesting consequences for collapsing with lower curvature bound.
* If $n=0$, so $\mathbb S^0=\\{+1,-1\\}$ then $M=\mathbb T^k$ and $h_{-1}=$ a hyperbolic linear map with a big eigenvalue will do.
* Denote by $F$ the space of all maps $M\to M$. If $\pi_n(F,\mathop{\rm id}_M)$ is finitely generated then for every $h_x\colon (M,g)\to (M,g)$ there is $h'_x$ with a fixed Lipschitz constant $L=L(M,g)$ for all $x$.
* I know few examples of families $h_x$ for which one can not find a metric $g$ and homotopic family $h'_x$ with Lipschitz constant $=1$ (in other words the family can not be made isometric).
* I do not know an answer even if instead of 10000 I would have $1+\varepsilon$ for a fixed $\varepsilon> 0$.
|
https://mathoverflow.net/questions/99276/lipschitz-constant-of-a-homotopy
|
[
"dg.differential-geometry",
"mg.metric-geometry",
"homotopy-theory",
"at.algebraic-topology"
] | 18 | 2012-06-11T00:57:45 |
[
"If it were me, I'd try to start : Let $\\Sigma$ be an exotic $n$-sphere. By conjugacy through the round sphere, this carries a very-transitive action by homeomorphisms of $S O (n+1)$; this action cannot be by diffeomorphisms, or else one could use it to give $\\Sigma$ an $S O (n+1)$-equivariant Riemann metric... But I can't see whether this is helpful after perturbing away from the subspace of group actions; still, I do expect exotic spheres to be useful somewhere, especially since it is known that some carry obstructions to global curvature positivities.",
"@Henrik: it is almost correct, you should change \"any\" to \"some\".",
"The paper \"Homotopical effects of dilation\" by Gromov considers similar question, but I haven't checked the exact relationship. Links: projecteuclid.org/euclid.jdg/1214434601 , or (non-paywalled) seven.ihes.fr/~gromov/PDF/4%5B19%5D.pdf",
"So is the following rephrasing correct: The question is whether any element in $\\pi_k(Map(M,M))$ lies in the image of $\\pi_k(Lip_{10000}(M,g))$ for any Riemannian metric $g$ on $M$, where $Lip_C(M,g)$ denotes the space of self maps of $(M,g)$ Lipschitz constant $C$."
] | 4 |
Science
| 0 |
215
|
mathoverflow
|
local equivalence of loop group representations
|
Let $G$ be a compact, simple, connected, simply connected (cscsc) Lie group, and let its smooth loop group $LG:=C^\infty(S^1,G)$. Given an interval $I\subset S^1$, we have the **_local loop group_** $$ L_IG := \\{\gamma\in LG \ | \ \forall z\not\in I \ \gamma(z)=e\\} $$ which is a subgroup of $LG$. Let $k\ge 1$ be an integer. The level $k$ central extension of $LG$ is denoted $\mathcal{L}G_k$. It restricts to a central extension of the local loop group that we denote $\mathcal{L}_IG_k$.
A representation of $\mathcal{L}G_k$ on a Hilbert space is called _**positive energy**_ if it admits a covariant action of $S^1$ (i.e., the action should extend to $S^1\ltimes \mathcal{L}G_k$) whose infinitesimal generator has positive spectrum. Here, the center of $\mathcal{L}G_k$ is required to act by scalar multiplication.
> **Definition 1:**
> _Two level $k$ positive energy representation of the loop group are called **locally equivalent** if they become equivalent when restricted to $\mathcal{L}_IG_k$._
The follows is believed to be true:
> **Claim 2:**
> _Let $G$ be a cscsc group and let $V$ and $W$ be any two positive energy representations of $\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._
* * *
~~I know a paper that proves the following:~~
> **Theorem 3:**
> _Let $G$ be a **simply laced** cscsc group and let $V$ and $W$ be two positive energy representations of $\mathcal{L}G_k$. Then $V$ and $W$ are locally equivalent._
_**Edit:** The argument in [[GF]](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full) seems to contain a mistake (on lines -4 and -3 of page 600)_
~~The basic ingredients that are needed (see page 599 of [Gabbiani & Fröhlich _[Operator algebras and conformal field theory](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-155/issue-3/Operator-algebras-and-conformal-field-theory/cmp/1104253398.full)_] for the proof) are the following two facts about positive energy representations of simply laced loop groups:
• Every level 1 rep can be obtained from the vacuum rep by precomposing the action by an outer automorphism of $\mathcal{L}G_1$ that is the identity on $\mathcal{L}_IG_1$.
• Every level $k$ rep appears in the restriction of a level 1 rep under the map $\mathcal{L}G_k\to \mathcal{L}G_1$ induced by the $k$-fold cover of $S^1\to S^1$. ~~
There are proofs in the literature, due to A. Wassermann ([here](https://arxiv.org/abs/math/9806031) p23) and V. Toledano-Laredo ([here](https://arxiv.org/abs/math/0409044) p82) respectively, for the cases $LSU(n)$ and $LSpin(2n)$, that are based on the theory of free fermions -- actually, Toledano only treats half of the representations of $LSpin(2n)$.
* * *
Is there a proof of Claim 2 in the literature?
How does one prove Claim 2?
|
https://mathoverflow.net/questions/66859/local-equivalence-of-loop-group-representations
|
[
"loop-spaces",
"conformal-field-theory",
"von-neumann-algebras",
"rt.representation-theory",
"reference-request"
] | 18 | 2011-06-03T16:35:03 |
[
"After having a further look at his preprint, it is now my opinion that Antony Wassermann does not provide a complete proof of the desired claim. So the question is open.",
"Many thanks to the person who sent me Wassermann's preprint.",
"There seems to be a proof of Claim 2 in some unpublished notes, namely in Böckenhauer and Evans \"Modular Invariants, Graphs and $\\alpha$-induction for Nets of Subfactors. II\" on page 66 is claimed that local equivalence hold for any compact connected simple $G$ with reference to the unplished notes of Wassermann \"Subfactors arising from positive energy representations of some infinte dimensional groups\", 1999.",
"probably not really helpful, but there seems to be a paper in preparation by Toledano Laredo which maybe could shed some light about this question in the B and C case"
] | 4 |
Science
| 0 |
216
|
mathoverflow
|
Steenrod algebra at a prime power
|
Let $n=p^k$ be a prime power.
When $k=1$, the algebra of stable operations in mod $p$ cohomology is the [Steenrod algebra](http://en.wikipedia.org/wiki/Steenrod_algebra) $\mathcal{A}_p$. It has a nice description in terms of generators and relations. Its dual (as a Hopf algebra) is also well understood by work of Milnor.
What about when $k>1$? Has any work been done on trying to understand the algebra of stable operations in mod $n$ cohomology? For instance, is there a description of $\mathcal{A}_4$ in terms of generators and relations? What about the dual Hopf algebra?
Of course I am just asking about the cohomology ring $H^\ast (H\mathbb{Z}/n;\mathbb{Z}/n)$, where $H\mathbb{Z}/n$ denotes the mod $n$ Eilenberg--Mac Lane spectrum. So I fully expect the answer to be either one of "yes, this is well-known and classical" or "no, this is known to be a big mess".
|
https://mathoverflow.net/questions/106602/steenrod-algebra-at-a-prime-power
|
[
"at.algebraic-topology",
"cohomology",
"steenrod-algebra"
] | 18 | 2012-09-07T05:54:20 |
[
"@elidiot: I don't know anything about mod p^k Dyer-Lashof operations, but that's probably just my own ignorance. It seems worthy of an MO question for sure!",
"I know that this is another question but as it is related I'm taking the liberty to ask it here: has anything been written about mod p^k Dyer Lashof operations ? I would especially be interested in mod 4 ones. Please tell me if this deserves a new thread.",
"@Sean: No. Shortly after asking this question I started working on other things. I did check out Adams' Berkeley notes, but couldn't find anything therein. Tyler's answer to mathoverflow.net/questions/119604/computation-of-hz-4-hz-4 seems encouraging though.",
"@Mark: Did you figure out a satisfactory answer? I had a thought about one possible way to \"reverse engineer\" it.",
"@Tyler: Thanks for the tip. I've ordered Adams' Berkeley notes from our library, I'll let you know if I find anything!",
"@Mark: I do remember reading somewhere in one of Adams' texts - possibly the Berkeley lectures? - about looking at cohomology operations into or out of mod $2^n$ cohomology. I might start looking there. The method is to filter $H\\mathbb{Z}/p^n$ by $H\\mathbb{Z}/p^k$ and look at the two associated Bockstein sequences that you get. Sorry for the indeterminacy in references.",
"@Tyler: This sounds promising. Was this a self-made calculation? If not, is there any hope that you could remember where you are remembering it from? Thanks.",
"Err, a lot of the multiplications become zero.",
"If I remember correctly, this can be calculated because the mod-$p$ Steenrod algebra has a relatively simple structure when considered as a bimodule over the subalgebra generated by the Bockstein. (I'm less sure at odd primes.) For $p=2$: in degrees zero and one you get $\\mathbb{Z}/2^k$, generated by the Bockstein; I believe that in higher degrees it's abstractly isomorphic to the Steenrod algebra as a module except that admissible monomials starting with $Sq^1$ start with the Bockstein $\\beta$ and admissible monomials ending with $Sq^r$, $r$ odd, end with $\\beta Sq^{r-1}$. A lot of the mul",
"@Sean: I am perhaps not so interested in the answer for finite fields, but I would be interested to know why it is easier (presumably issues with the Kunneth Theorems and duality)?",
"This is not directly related, but it is not hard to compute the Steenrod algebra corresponding a finite field. I am guessing you are not interested in that though."
] | 11 |
Science
| 0 |
217
|
mathoverflow
|
Elliptic $\infty$-line bundles over $B G$
|
Theorem 5.2 in Jacob Lurie's "Survey of Elliptic Cohomology" ([pdf](http://www.math.harvard.edu/~lurie/papers/survey.pdf)) states the equivalence of two maps
$$ B G \longrightarrow B \mathrm{GL}_1(A) $$
for $A$ an $E_\infty$-ring carrying an oriented derived elliptic curve $E \to \mathrm{Spec}(A)$.
The theorem is stated for $G = \mathrm{Spin}$. That's the case of relevance in the paragraph directly following its statement. But, in view of the whole 2-equivariance story, the statement of the theorem as such would make sense for more general $G$ equipped with a homomorphism $G \to O$, as in the discussion on the preceding page.
**First question:** Might this statement still be true for more general $G\to O$? (What's the idea of the proof, anyway?)
More in detail, one of the two maps is the composite
$$ B G \to B O \stackrel{J}{\longrightarrow} B \mathrm{GL}_1(\mathbb{S}) \longrightarrow B \mathrm{GL}_1(A) $$
where $J$ denotes the $J$-homomorphism and $\mathbb{S}$ the sphere spectrum. (Here I am giving a nominally different but hopefully straightforwardly equivalent description of what is indicated on p. 38 of Lurie's note.) The other one is the restriction of the derived theta-line bundle
$$ \theta \;\colon\; \mathrm{Loc}_{\mathrm{Spin}}(E) \longrightarrow \mathbf{B} \mathbb{G}_m $$
along the global point inclusion $\mathrm{Spec}(A)\to \mathrm{Loc}_{\mathrm{Spin}}(E)$ that picks the trivial local system/flat connection (that's again my slight refomulation which I believe is straightforwardly equivalent, but check -- my $\mathrm{Loc}_{\mathrm{Spin}}(E)$ is Lurie's "$M_{\mathrm{Spin}}$").
Regarding the first map, for $G = \mathrm{Spin}$ and for the special case $A = \mathrm{tmf}$ this is discussed in section 8 of Ando-Blumberg-Gepner [arXiv:1002.3004](http://arxiv.org/abs/1002.3004) and crucially equivalent to the $\mathrm{tmf}$-line bundle which is associated via the String-orientation $\sigma$ to the "Chern-Simons 3-bundle" classified by $\tfrac{1}{2}p_1$, i.e. to the map
$$ B \mathrm{Spin} \stackrel{\tfrac{1}{2}p_1}{\longrightarrow} B^4 \mathbb{Z} \stackrel{\tilde \sigma}{\longrightarrow} B \mathrm{GL}_1(\mathrm{tmf}) \,. $$
I suppose the argument there straightforwardly generalizes from $A = \mathrm{tmf}$ to any $A$ as above. But as a sanity check:
**Second question** : is that right?
In conlusion I am wondering:
**Third question** : In which generality in $E_\infty$-rings $A$ carrying oriented derived elliptic curves and in compact simply connected simple Lie groups $G$, is it true that the restriction of the derived theta line bundle along the $\mathrm{Spec}(A)$-point inclusion is equivalent to the $A$-line bundle classified by
$$ B G \stackrel{c_2}{\longrightarrow} B^4 \mathbb{Z} \stackrel{\tilde \sigma}{\longrightarrow} B \mathrm{GL}_1(A) \,. $$
?
(Where now $c_2$ denotes the generator of $H^4(B G, \mathbb{Z}) \simeq \mathbb{Z}$.)
|
https://mathoverflow.net/questions/185721/elliptic-infty-line-bundles-over-b-g
|
[
"at.algebraic-topology",
"elliptic-curves",
"cohomology",
"derived-algebraic-geometry"
] | 18 | 2014-10-29T10:11:41 |
[] | 0 |
Science
| 0 |
218
|
mathoverflow
|
Computation of low weight Siegel modular forms
|
We have these huge tables of elliptic curves, which were generated by computing modular forms of weight $2$ and level $\Gamma_0(N)$ as N increased.
For abelian surfaces over $\mathbb{Q}$ we have very little as far as I know. The Langlands philosophy suggests that every abelian surface should be attached to a Siegel modular form of weight $(2,2)$ on $GSp_4$, but the problem is that this weight is not cohomological, which has the concrete consequence that it's going to be tough to compute such things using group cohomology. In particular one of the reasons that modular symbols work for computing elliptic curves, fails in this situation.
I guess though that one might be able to somehow use the trace formula to compute the trace of various Hecke operators on Siegel modular forms of weight $(2,2)$ and various levels, because presumably the trace formula translates the problem into some sort of "class group" (in some general sense) computation, plus some combinatorics.
[EDIT: from FC's comment, it seems that my guess is wrong.]
Has anyone ever implemented this and tabulated the results?
[NB I know that people have done computations for low level and high weight, for example there's a lovely paper of Skoruppa that outlines how to compute in level 1; my question is specifically about the weights that are tough to access]
|
https://mathoverflow.net/questions/22949/computation-of-low-weight-siegel-modular-forms
|
[
"computational-number-theory",
"siegel-modular-forms",
"nt.number-theory",
"trace-formula"
] | 18 | 2010-04-28T23:46:38 |
[
"@David: in which case my answer to your aside is then simply \"yes\".",
"@Kevin: I am aware of the low-dimensional coincidences SO(2,1)=PGL(2) and Spin(2,3)=Sp(4). I suppose n > 2 was implicit in my query. :)",
"@David: there is a low-degree coincidence: the dual group of GSp_4 is GSp_4 (think about the Dynkin diagram of Sp_4). I agree that the pattern should not continue for Sp_6 (think about the Dynkin diagram again).",
"Aside: Shouldn't the automorphic representation associated to an abelian n-fold live on SO(2n+1), rather than a symplectic group? (Look at the L-groups)",
"Thanks for the reference FC. What you say about Sp_6 is terrifying!",
"@David: I think that trying to describe the moduli space is not what I am after. I am after an algorithm to compute traces of Hecke operators on spaces of cusp forms that takes a level and a prime (or some slightly more precise data; a diagonal matrix with powers of p on the diagonal) as an input and gives a number as an output and doesn't care less about whether a certain moduli space is general type or not. The same way as modular symbols would do the job for classical modular forms. But from FC's comment it seems that the trace formula won't do it either :-/",
"@FC: I am pretty sure one can't use the trace formula to compute forms of weight 1 but I have so little understanding of the trace formula that I don't know why this is the case. Seems to me that you're saying that the same obstruction stops me computing low weight Siegel forms :-( ",
"@Kevin: Curves of genera up 6 have a very pleasant description, and you can comfortably (depends on your standard of comfort of course) live with the description of curves up to genus 15; I'd buy a curve of genus up to 15 as a moduli space any day. I have yet to see a general type three-fold with a nice description (unless you cooked it for this purpose). Disclaimer: I don't know the method involved, you may well be right, it's just my vague intuition speaking here.",
"I don't buy this. Isn't that like saying \"wouldn't computational efforts with elliptic curves of conductor 100 or more be hard because the moduli space has genus at least 2\", isn't it? The point is that you don't compute the moduli space---that's the last thing you want to do! You use the trace formula applied to a carefully-chosen function which will give you essentially the trace of Frobenius on the cohomology in terms of some much more algebraic/combinatorial/number-theoretic data and compute that instead. Well, that's my suggestion, but a lot of thought needs to go in to making it work.",
"Wouldn't any computational efforts with N > 3 be very difficult simply because the moduli space is general type ?"
] | 10 |
Science
| 0 |
219
|
mathoverflow
|
Quotients of residually finite groups by amenable normal subgroups
|
My questions are:
> Is there any group, which cannot be written as the quotient of a residually finite group by an amenable normal subgroup? Is it possible for large classes of groups?
and
> Is there a group, such that every extension by an amenable group is split?
|
https://mathoverflow.net/questions/40991/quotients-of-residually-finite-groups-by-amenable-normal-subgroups
|
[
"gr.group-theory"
] | 18 | 2010-10-03T23:19:30 |
[
"@Andreas: Me too! Can you send me the text? This sounds very interesting!",
"Have you imagined a way to answer the first question considering only the one-relator groups?",
"@Andreas: Could you please send me a copy of this paper, also? I am very interested in this.",
"@Andreas: Can you send me the text? I suggest you publish the result. Do you need advice about a journal? ",
"@Mark: I proved it and it is not published, since I do not yet know what to do with it.",
"Also, 'if the answer is \"yes\", it would provide a new way to construct groups because old ways don't seem to work' is exactly the sort of motivation I was asking for.",
"Mark - I agree that if the answer to the first question were 'no' then it would be remarkable! I was wondering what motivates Andreas to ask the question. It's always useful when the asker gives all the information that s/he has about a problem.",
"@Andreas: who proved the result of measured group theory that you mentioned, and where?",
"I meant that no motivation is required because if the answer to the first question is \"no\", then it would be the first non-trivial property of all groups (which would disprove Gromov's thesis). And if the answer is \"yes\", it would provide a new way to construct groups because old ways don't seem to work. ",
"̯@Henry: If the second question has a positive answer, then this would (most likely) help to answer the first. My motivation is the study of sofic groups. In the sense of measured group theory, every sofic group is the quotient of a residually finite group by an amenable subgroup (but that takes some time to explain). I was wondering what kind of condition one gets in the more rigid setup of ordinary group theory. Burnside groups are not known to be sofic, so Mark's comment is very interesting.",
"@Henry: None of these questions requires an additional motivation. @Andreas: I suspect that the free Burnside group of large enough exponent is an example. That would solve mathoverflow.net/questions/37344/… (see the discussion there). But that is a difficult question.",
"Could you give some motivation, and explain what (if anything) the two questions have to do with each other?"
] | 12 |
Science
| 0 |
220
|
mathoverflow
|
What is known about module categories over general monoidal categories?
|
All of the literature I have seen on module categories over monoidal categories has been in the rigid $k$-linear semisimple case, more or less in the spirit of Ostrik's paper,
* Ostrik, V. _Module categories, weak Hopf algebras and modular invariants_ , Transformation Groups **8** (2003) 177–206, doi:[10.1007/s00031-003-0515-6](https://doi.org/10.1007/s00031-003-0515-6), arXiv:[math/0111139](https://arxiv.org/abs/math/0111139).
However, the basic definition of module category makes perfect sense in general, so I am wondering if there has been any work done in more general settings.
In particular, I am interested in the following sort of question. Let $\mathcal{C}$ be a braided monoidal category (maybe ultimately with some added assumptions, such as rigidity). Any left $\mathcal{C}$-module category is automatically a $(\mathcal{C}, \mathcal{C})$-bimodule category via the braiding. For bimodule categories there should be a good notion of "tensoring over $\mathcal{C}$," which makes the 2-category of bimodules (and hence the 2-category of left modules) into a monoidal 2-category. If we extract invertible objects and morphisms at all levels, we should obtain a sort of Brauer 3-group for $\mathcal{C}$. In the fusion category setting, this object is considered in the recent preprint
* Pavel Etingof, Dmitri Nikshych, Victor Ostrik, _Fusion categories and homotopy theory_ , Quantum Topol. **1** (2010), 209–273, [journal](https://ems.press/journals/qt/articles/2876), arXiv:[0909.3140](https://arxiv.org/abs/0909.3140).
I'd like to know how this is related to the "internal" Brauer 3-group of $\mathcal{C}$, whose objects are Azumaya algebras in $\mathcal{C}$, morphisms are invertible bimodules, and 2-morphisms are invertible bimodule morphisms. In the fusion category setting, the connection is furnished by the main theorem in Ostrik's paper, which states that any semisimple indecomposable module category is equivalent to the category of modules over some algebra in $\mathcal{C}$. (I think this implies that the two notions of Brauer 3-group are equivalent in the fusion category setting.) Is any result along these lines known in more generality? It's not even clear to me that the module category of left modules over an Azumaya algebra in $\mathcal{C}$ is invertible, or even what the tensor product of two module categories of modules is.
|
https://mathoverflow.net/questions/6775/what-is-known-about-module-categories-over-general-monoidal-categories
|
[
"ct.category-theory",
"monoidal-categories"
] | 18 | 2009-11-24T23:08:19 |
[
"The \"module categories are categories of modules\" yoga is probably very special to the finite rigid setting. We tried to be careful about what exactly was needed for this here:arxiv:1406.4204. An interesting example is the following (non-rigid) tensor category C. It is finite, semisimple and has two simple objects 1 and x. The object 1 is the unit and we have $x \\otimes x = 0$. This determines the monoidal structure by linearity. If I remember right, the subcategory generated by x is a module category which is not of the form A-mod(C) for an algebra A in C.",
"The Deligne tensor product doesn't exist for arbitrary abelian categories. But if you relax \"abelian\" to instead mean \"has finite colimits\" then there is a very general tensor product that always exists called the \"Kelly tensor product\". A similar statement for locally presentable categories is in Cor 2.2.5 of (arXiv:1105.3104). I think the categorical machinery underlying these results can be adapted to the case of tensor products of module cats in the generality you are considering, but I don't know of a reference that does it.",
"An Azumaya algebra is one in which the tensor actions $X \\mapsto A \\otimes X$ and $X \\mapsto X \\otimes A$ are equivalences between $\\mathcal{C}$ and the categories of left $A \\otimes A^{\\text{op}}$-modules and right $A^{\\text{op}} \\otimes A$-modules, respectively. This definition comes from the paper by Van Oystaeyen and Zhang, \"The Brauer Group of a Braided Monoidal Category.\"",
"Could you recall the definition of Azumaya algebra in a general braided monoidal category?"
] | 4 |
Science
| 0 |
221
|
mathoverflow
|
Almost complex 4-manifolds with a "holomorphic" vector field
|
**Main question.** What is the class of smooth orientable 4-dimensional manifolds that admit an almost complex structure $J$ and a vector field v, that preserves $J$?
_The following sub question is rewritten thanks to the comment of Robert Bryant:_
Is it true that if $(M,J)$ admits a vector field that preserves $J$ then there is $J'$ on $M$ homotopic to $J$ that is preserved by an $S^1$-action on $M$?
_After three years I don't think indeed that the following is such a reasonable motivation_
POSSIBLE MOTIVATION. Claire Voisin gave a construction of the Hilbert scheme of points for every almost complex 4-fold by an analogy with the Hilbert scheme of points of a complex surface. The first calculation of the Euler charactericstics of Hilbert scheme of points of complex surfaces was done via localisation techniques, for $CP^2$. Now, if we have a "holomorphic" vector field on an almost complex manifold, this could potentially help to reduce the calculation of the Euler charachteristics of its Hilbert scheme to the study of fixed points of the manifold. So the question is how flexible this notion is... But in its nature this question seems to be more a question (maybe not a hard one) on dynamical systems.
|
https://mathoverflow.net/questions/9518/almost-complex-4-manifolds-with-a-holomorphic-vector-field
|
[
"4-manifolds",
"ds.dynamical-systems",
"gt.geometric-topology",
"complex-geometry"
] | 18 | 2009-12-21T16:14:00 |
[
"Robert, thanks for this remark, you are right that this second phrase makes no sense. I rewrote so that I it makes sense.",
"@Dmitri: Perhaps I don't understand your terminology, but it seems to me that $S^4$ admits a smooth $S^1$ action, but it doesn't admit any almost complex structure, let alone one with a (nontrivial) symmetry vector field. Thus, your second sentence is mysterious to me.",
"Yes, I refer to the flow associated to the vector field. I don't want to impose that all zeros are isolated. ",
"When you say the vector field preserves the almost-complex structure, are you referring to the flow associated to the vector field? And do you want the vector field to be everywhere nonzero? Nonzero at all but finitely many points? ",
"Tim, thanks for the comment! In fact the approach I propose is exactly to give a justification of your words:)) How do we calculate the Euler charecteristics of a manifold - count the number of zeros of a vector field. I want to say, that for a Hilbert scheme this should be a result of the same caclulation. I think it would be cool to have a \"simple\" calculation of the Euler charateristics. By the way this speculation can also be applied to DT invariants -- when you count 0-dim subschemes on 3 dimensional CY manifolds. This calculation was done in 2005 and was one of MNOP conjectures.",
"It's an intriguing question, but it seems to me a surprisingly delicate way to calculate the Euler characteristic of the Hilbert scheme, which one expects to be a universal function of the Betti numbers of the 4-manifold, as in the integrable case. Perhaps one can prove this using the Cech spectral sequence coming from the open cover of Hilb arising from a good cover of the 4-manifold? [BTW, I know you're quoting Voisin, but the following rule is worth insisting on: a 4-fold is an algebraic or analytic variety of dimension 4. A 4-manifold has real dimension 4.]",
"Thanks :) Maybe this sound cooler than it should be. He is the reference, it was published in 2002 people.math.jussieu.fr/~voisin/Articlesweb/almost.pdf",
"This sounds really cool. Is this recent work of Voisin?"
] | 8 |
Science
| 0 |
222
|
mathoverflow
|
$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian?
|
During my research I came across this question, I proposed it in the chat, but nobody could find a counterexample, so I allow myself to ask you :
$G$ a group, with $p$ a prime number, and $|G|=2^p-1$, is it abelian ?
|
https://mathoverflow.net/questions/273976/g-a-group-with-p-a-prime-number-and-g-2p-1-is-it-abelian
|
[
"nt.number-theory",
"gr.group-theory",
"finite-groups"
] | 18 | 2017-07-08T05:27:26 |
[
"@YemonChoi Well, I would wait 500 years for the math community to find a complex 600-page solution, like for Fermat.",
"If you are claiming that you can answer this question, then you can either add an answer below, or write up your work as a preprint in the normal academic way, or write it in a blog post and leave a link. Saying \"I can answer this, can you?\" is not a constructive use of this site, nor is it collegial.",
"I have rolled back the edits to this old question, which seem to be purely cosmetic and done for promotion. Yes the question is appropriate for MO, but that doesn't mean you need to add some kind of \"RL\" label",
"Indeed, your suggestion mathoverflow.net/questions/273976/… is the same as @spin's.",
"msp.org/pjm/1967/22-3/pjm-v22-n3-p15-p.pdf",
"I did a quick check for $p$ at most $800$ or so and found no examples. In the end this is a question about number theory and not group theory, and it might be difficult. For example it is an old open problem to determine whether $2^p - 1$ is squarefree for all primes $p$. If the answer to that old problem is yes, then you are asking whether every group of order $2^p - 1$ is cyclic.",
"One can use the Cunningham tables to do a quick check on the factors for more than 50 prime exponents. Gerhard \"Not Computationally Motivated This Morning\" Paseman, 2017.07.08.",
"@PéterKomjáth: No, we have then $2^p \\equiv 2$ (mod $q^3$), respectively, $2^{p-1} \\equiv 1$ (mod $q^3$).",
"@Glorfindel, it's hard to believe that never happens, but a quick test in Mathematica shows that (I screwed up in Mathematica or) it doesn't happen for the first 50 primes.",
"Yes, but probably easier to solve. I doubt there is a prime $q$ for which $q^3 | 2^p - 1$, but the other 'case', two primes $q_1, q_2$ in the factorization for which $q_1 | q_2 - 1$ might very well be possible.",
"It depends on the factorization of $2^p-1$, see groupprops.subwiki.org/wiki/…"
] | 11 |
Science
| 0 |
223
|
mathoverflow
|
Symmetries of local systems on the punctured sphere
|
Let $X=S^2\setminus D$, for $D\subset S^2$ some finite set of points, say with $|D|=n\geq 1$. The category of locally constant sheaves of $\mathbb{C}$-vector spaces on $X$ (equivalently, complex representations of $\pi_1(X)=F_{n-1}$), $\text{LocSys}(X)$, has many natural autoequivalences.
For example, if $f: S^2\to S^2$ is a homeomorphism fixing $D$, then $\mathbb{V}\mapsto f^*\mathbb{V}$ is an autoequivalence of $\text{LocSys}(X)$. Two homeomorphisms fixing $D$, homotopic through homeomorphisms fixing $D$, induce naturally equivalent autoequivalences, so one obtains an action of the mapping class group $\text{Mod}(S^2, D)$ on $\text{LocSys}(X)$. Since $\text{Mod}(S^2, D)$ is the pure spherical braid group on $n$ strands, I'll denote it by $B_n$. My question is:
> What are the auotequivalences of $\text{LocSys}(X)$ commuting (up to natural isomorphism) with the actions of $B_n$?
I know of three sources of examples. The first is invertible objects in $\text{LocSys}(X)$: if $\mathbb{L}$ is a rank one local system on $X$, the functor $\mathbb{V}\mapsto \mathbb{V}\otimes \mathbb{L}$ is $B_n$-equivariant. The second is the evident action of the group $\text{Aut}(\mathbb{C}/\mathbb{Q})$ on the structure constants of any given local system. Both of these operations preserve the rank of a local system.
The other is Katz's middle convolution operation, which does not.
Let $\Delta\subset X\times X$ be the diagonal, $\pi_1, \pi_2: X\times X\setminus \Delta\to X$ the two projections, $a: X\times X\setminus \Delta\to \mathbb{C}^\times$ the map $(x, y)\mapsto x-y$ (where here we choose an isomorphism $S^2\simeq \mathbb{CP}^1$ sending some point of $D$ to $\infty$ to coordinatize $S^2$), and $j: X\times X\setminus \Delta\hookrightarrow S^2\times X$ the evident inclusion. Then given a non-trivial rank one local system $\chi$ on $\mathbb{C}^\times$, the middle convolution $$MC_{\chi}(\mathbb{V}):=R^1\pi_{2\ast}j_{\ast}(\pi_1^{\ast}\mathbb{V}\otimes a^{\ast}\chi).$$ It is not obvious but Katz shows it is true that $MC_\chi$ is (up to equivalence) inverse to $MC_{\chi^{-1}}$. Moreover this operation (not entirely obviously) commutes with the action of $B_n$.
Of course one can compose these operations in various ways to get more autoequivalences commuting with the $B_n$-action.
> Is that everything?
I'm also curious about the analogous question with $S^2\setminus D$ replaced by a surface of higher genus.
|
https://mathoverflow.net/questions/446416/symmetries-of-local-systems-on-the-punctured-sphere
|
[
"ag.algebraic-geometry",
"rt.representation-theory",
"braid-groups",
"local-systems"
] | 17 | 2023-05-08T12:40:43 |
[] | 0 |
Science
| 0 |
224
|
mathoverflow
|
Does the Ackermann function count something?
|
Let $\mathrm{FinSet}$ be the category of finite sets.
A _finite set structure_ is a faithful functor $F\colon C\to \mathrm{FinSet}$ such that, for any $n\geq 1$, there are only finitely many isomorphism classes of objects $F$ maps to $\\{1, \dots, n\\}$.
> **Question.** Is there a natural finite set structure realizing the Ackermann function (or some other computable function growing faster than any primitive recursive function)?
|
https://mathoverflow.net/questions/392387/does-the-ackermann-function-count-something
|
[
"co.combinatorics",
"ct.category-theory"
] | 17 | 2021-05-10T07:20:52 |
[
"So, I do not know what the Ackermann function counts, but the TREE function does count something explicit and combinatorial, and the Ackermann function is tiny in comparison with TREE. mathoverflow.net/questions/93828/how-large-is-tree3",
"The inverse Ackermann function appears in some algorithms, e.g. using a \"soft heap\" you can find the minimum spanning tree in time $O(m\\alpha(m, n))$. A decade ago I probably knew why and whether something was being counted.",
"Interesting question. Maybe some nested set-partition-like structures? A stupid answer would be to take a deterministic rewrite system that takes $A\\left(n,m\\right)$ steps to terminate (I think there should be one), and just say \"the intermediate states of this rewrite system\".",
"There is a whole \"school\" of enumerative combinatorics done using Joyal's species! Relevant names are Joyal himself, Labelle, Leroux, Viennot, Mendez, Rajan...",
"Not sure why this (good) question has a category theory tag, but there is a kind of \"categorical number theory / combinatorics\" that was studied by Steve Schanuel. I don't know any more than that.",
"@SimonHenry even though the question isn't phrased in this way, I suspect it is asking about a species of structure with the Ackermann function as generating species. I have no idea what the sequence $a_n=A(n,n)$ could be counting, though, and throwing the very few terms of the sequence that can be computed at OEIS doesn't seem a viable choice: is there a combinatorial interpretation for those numbers?",
"@fosco : yes but,I am not aware of anything that forces the exponential generating function of combinatorial species to converge, so why would the convergence brings anything to the question ? But maybe you have in mind some result proving the convergence for some class of \"nice\" species ?",
"@SimonHenry yes you can associate some category but I wondered if there was a natural (in the vague sense of the word) choice. Maybe a structure arising independently elsewhere in mathematics.",
"I'm not sure how to understand this question. For any function $f:\\mathbb{N} \\to \\mathbb{N}$ I can chose a sequence of sets $P_n$ such that $\\# P_n = f(n)$ and take $C$ to be a naively defined category with an object for each $i \\in P_n$ mapping to a set with $n$ elements... so any functions can be obtained this way.",
"hi @SimonHenry ! To a combinatorial species you can associate various formal power series, and especially from the exponential series you can understand a lot about the species itself --although not everything. Ad yes, that series it's not convergent, so there is no associated function you can study. This makes the problem elusive, even provided you find a meaning for what kind of structure $A(n,n)$ is counting.",
"@fosco, surely not; $f(n) \\ge (n!)^2$ eventually.",
"@fosco : why is the convergence of the series relevant to the question ?",
"oh, fair enough; well, is the series $\\sum \\frac{f(n)}{n!}X^n$ convergent in at least a disk $B(0,\\epsilon[$?",
"@fosco I had in mind $f(n)=A(n, n)$",
"The Ackermann function $A(m,n)$ takes two inputs; apart from this, if I understand well, you're asking about the existence of a species of structure whose associated exponential series (in the variables X,Y) is $\\sum \\frac{A(m,n)}{n!m!}X^nY^m$. First of all, is this series convergent?"
] | 15 |
Science
| 0 |
225
|
mathoverflow
|
Picture of Lambert's proof that $\pi$ is irrational?
|
With a suitably generous notion of "picture proof" or "proof without words" or "geometric proof," there do exist such proofs of [the irrationality of square roots](https://www.cut-the-knot.org/proofs/GraphicalSqRoots.shtml) and even of the [irrationality of $e$](https://arxiv.org/pdf/0704.1282.pdf). However, I am not aware of anything that could plausibly be called a [geometric proof of the irrationality of $\pi$](https://math.stackexchange.com/q/1658174). Given that the original definition of $\pi$ was geometric, it seems a pity that there is no such thing. Or maybe there is?
Perhaps one can try to draw pictures to accompany [Lambert's irrationality proof](https://math.stackexchange.com/a/895728). For example, is there a way to draw a picture of the following fact? $$\tan(a/b) = {a \over\displaystyle b - {a^2 \over \displaystyle 3b - {a^2 \over \displaystyle 5b - {\displaystyle a^2 \over 7b - \cdots}}}}$$ And if so, is there any way to draw a picture of the fact that such a continued fraction is irrational when $a$ and $b$ are positive integers?
|
https://mathoverflow.net/questions/376318/picture-of-lamberts-proof-that-pi-is-irrational
|
[
"nt.number-theory",
"continued-fractions",
"transcendental-number-theory",
"irrational-numbers",
"alternative-proof"
] | 17 | 2020-11-12T11:09:09 |
[
"I don't have a specific suggestion, but of possible related interest is a collection of references I assembled 3 years ago in my answer to Irrationality of $\\pi^2$ and $\\pi^3$.",
"Correction: The above formula should have $\\tan(x^{-1})$ on the right hand side. I also think there might be a sign error somewhere.",
"I'll think more about this later but: Let $SL_2(\\mathbb{R})$ act on $\\mathbb{R}$ by Mobius transformations in the usual way. It looks to me like your continued fraction says that $\\begin{bmatrix} 0 & 1 \\\\ -1 & x \\\\ \\end{bmatrix} \\begin{bmatrix} 0 & 1 \\\\ -1 & 3x \\\\ \\end{bmatrix} \\begin{bmatrix} 0 & 1 \\\\ -1 & 5x \\\\ \\end{bmatrix} \\cdots (0) = \\tan(x)$. So maybe draw a sequence of lines given by partial products of this, and see that their slopes are approaching $\\tan(x)$ or, equivalently, their angles are approaching $x$? (Here $x = a/b$.)"
] | 3 |
Science
| 0 |
226
|
mathoverflow
|
When is the determinant an $8$-th power?
|
I am working over $\mathbb{R}$ (though most of the story goes over any field). I am looking for linear spaces of matrices such that the restriction of the determinant to this spaces can be written (non-trivially) as the power of another polynomial. Let me give some examples where such a phenomenon appears, before asking my question in more details.
* let $n = 2m$ be an even integer. Then, the restriction of the determinant to $\bigwedge^2 \mathbb{R}^n \subset \mathrm{End}(\mathbb{R}^n)$ can be written as the square of a non-zero polynomial : the pfaffian.
* let $n = 2m$ be again even. Let $\mathcal{H}_{m}(\mathbb{C})$ be the set of $m \times m$ Hermitian matrices with complex coefficients. Using the matrix representation of $i$ (square root of $-1$) as: $$ i = \begin{pmatrix} 0 & -1 \\\ 1 & 0 \\\ \end{pmatrix},$$ one can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{C})$:
$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{C})} \right)^2.$$
* let $n = 4m$. Let $\mathcal{H}_{m}(\mathbb{H})$ be the set of $m \times m$ Hermitian matrices with quaternionic coefficients. Using the $4 \times 4$ matrix representation of $i,j,k$ (square roots of $-1$ in $\mathbb{H}$), One can embed $\mathcal{H}_{m}(\mathbb{C})$ as a sub-algebra of $\mathrm{S}^{2} \mathbb{R}^{n} \subset \mathrm{End}(\mathbb{R}^{n})$. Then, it is again easily checked that the following equality holds on $\mathcal{H}_{m}(\mathbb{H})$:
$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{n})} = \left( \mathrm{det}_{\mathcal{H}_{m}(\mathbb{H})} \right)^4.$$
**Question** : Is there a similar story for $\mathcal{H}_{3}(\mathbb{O})$?
More precisely, one can define a good notion of determinant for Hermitian $3 \times 3$ matrices with octonionic coefficients. I was wondering if there is an embedding $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{S}^{2} \mathbb{R}^{24} \subset \mathrm{End}(\mathbb{R}^{24})$, such that the following hold on $\mathcal{H}_{3}(\mathbb{O})$:
$$ \mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})} = \left( \mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})} \right)^8?$$
The algebra $\mathcal{H}_{3}(\mathbb{O})$ being non-associative, it can not be embedded **as an algebra** into a matrix algebra. On the other hand, what I am asking is (probably) considerably weaker : I just want an embedding of vector spaces $\mathcal{H}_{3}(\mathbb{O}) \hookrightarrow \mathrm{End}(\mathbb{R}^{24})$ such that the restriction of $\mathrm{det}_{\mathrm{End}(\mathbb{R}^{24})}$ to $\mathcal{H}_{3}(\mathbb{O})$ is the $8$-th power of $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}$.
Of course, since $\mathcal{H}_{3}(\mathbb{O})$ can not be embedded into $\mathrm{End}(\mathbb{R}^{24})$ as an algebra, it will probably be harder to check such an equality of determinants on $\mathcal{H}_{3}(\mathbb{O})$. Indeed, we can't use the representative of the Hermitian comatrix in $\mathrm{End}(\mathbb{R}^{24})$ and just multiply it with the orginal matrix to get $\mathrm{det}_{\mathcal{H}_{3}(\mathbb{O})}.Id_{24}$. But I thought that it was perhaps possible to find another trick for $\mathcal{H}_{3}(\mathbb{O})$?
Thanks for your help.
|
https://mathoverflow.net/questions/278981/when-is-the-determinant-an-8-th-power
|
[
"ag.algebraic-geometry",
"rt.representation-theory",
"matrices",
"ra.rings-and-algebras",
"octonions"
] | 17 | 2017-08-17T13:13:33 |
[
"But I may be wrong and I would be delighted to learn that there is such an embedding!",
"@DenisSerre : Yes sure, I tried a lot of stuff related to the Jordan algebra theory. In fact I am now able to prove that there is an embedding of $\\mathcal{H}_{3}(\\mathbb{O})$ in $\\mathbb{R}^{24}$ such that $\\mathrm{det}_{\\mathbb{R}^{24}}$ is the $4$-th power of a certain sextic related to $\\mathrm{det}_{\\mathcal{H}_{3}(\\mathbb{0})}$. However, I did a lot of computations with Macaulay2, and I start to believe that there is no embedding such that the restriction of $\\mathrm{det}_{\\mathbb{R}^{24}}$ to $\\mathcal{H}_{3}(\\mathbb{O})$ is a $8$-th power.",
"Have you tried using the fact that ${\\cal H}_3({\\mathbb O})$ is the exceptional Jordan algebra, that called the Albert algebra ?",
"@MarekMitros : Thanks! Yes I have seen this question a bit after having posted mine. It is very much related. Though, I guess mine is more focused on the equation lining the two dterminants while the other one focuses more to the existence of a determinant for hermitian matrices with octonionic coefficients.",
"See also mathoverflow.net/questions/239954/…",
"The first suggestion of embedding octonions into matrices is to map octonion $x$ to $8\\times 8$ matrix $L_x$ or $R_x$ of left and right multiplication by octonion $x$. Similarly octonion matrix $2\\times 2$ can be mapped to block matrix consisting of four blocks $L_{a_{i,j}}$ where $a_{i,j}$ is octonion in original matrix. Do we have formula for determinant of block matrix ?",
"What is that first map? The multiplication table should be neither symmetric nor skew-symmetric. I see how given an $\\mathbb{R}$-linear map $f:\\mathbb{O}\\hookrightarrow\\mathrm{End}(\\mathbb{R}^8)$ satisfying $f(z)f(\\bar z)=|z|^2I_8$ one could extend to $\\mathcal{H}_2(\\mathbb{O})$ with the right determinant relation. That would require 7 anticommuting matrices $e_1,\\dots,e_7$ in $\\mathrm{End}(\\mathbb{R}^8)$ that square to $-I_8$. How are you building these out of the multiplication table?",
"@MTyson In the case of $\\mathcal{H}_{2}(\\mathbb{O})$ I think there is something going on. Let $\\mathbb{O} \\hookrightarrow \\mathrm{S}^2 \\mathbb{R}^8$ which associates to an octonion its multiplication table. This gives rise to an embedding $\\mathcal{H}_{2}(\\mathbb{O})$ in $\\mathrm{S}^2 \\mathbb{R}^{16}$. Using the block computation of a determinant for a $2*2$ block matrix (with the bottom right being a translation), one sees that $\\mathrm{det}_{\\mathrm{End}(\\mathbb{R}^{16})} = (\\mathrm{det}_{\\mathcal{H}_{2}(\\mathbb{O})})^3$",
"Do you know anything about the $\\mathcal{H}_{2}(\\mathbb{O})$ case?"
] | 9 |
Science
| 0 |
227
|
mathoverflow
|
Kan's simplicial formula for the Whitehead product
|
In his article on _Simplicial Homotopy Theory_ (Advances in Math., 6, (1971), 107 –209) Curtis quotes a formula (on page 197) for the Whitehead and Samelson products in a simplicial group $G$. The formula for the Samelson product $\langle x,y\rangle$ of elements $x\in \pi_p(G)$ and $y\in \pi_q(G)$ is in terms of an ordered product of commutators of pairs of certain degenerate elements. The order of multiplication is determined by an antilex total order on the set of $(p,q)$-shuffles. I tried, and think I succeeded in filling in the details of this sketch from Curtis. I understand that the result was due to Dan Kan but that he never published it.
My questions are
(i) is there a published proof of this formula and if so where? (As I said I have a proof that it seems to work, but do not yet have one that it actually deserves to be called the Samelson product and that the related Whitehead product corresponds to the classical form, e.g. involving a 'universal example'.)
and
(ii) I have so far failed to prove that this formula gives a bilinear pairing on homotopy groups. (It is simple to prove a related formula which incorporates an action as in the Witt-Hall identities for commutators but I do not see why the action involved should be trivial as would seem to be required for the bilinearity result to hold.)
Can anyone suggest a reference to a classical proof of bilinearity of the (topological) Whitehead product that is fairly categorical and hence adaptable to my simplicial setting. Most sources seem to leave it as an exercise and Whitehead's original approach although clear is very topological in nature.
(Edit: I note that Adams in his _Student's Guide to Algebraic Topology_ , stated: The Whitehead product is bilinear and anticommutative; this can be proved by diagram chasing with the universal example'. Unfortunately I have yet to find the diagram through which I have to chase, although I have tried several ones that initially seemed to give some hope of being good for the task in hand.)
|
https://mathoverflow.net/questions/296479/kans-simplicial-formula-for-the-whitehead-product
|
[
"at.algebraic-topology",
"homotopy-theory",
"simplicial-stuff"
] | 17 | 2018-03-29T00:19:56 |
[] | 0 |
Science
| 0 |
228
|
mathoverflow
|
Almost monochromatic point sets
|
There are many sort of equivalent theorems about monochromatic configurations in finite colorings, such as [Van der Waerden](https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem), [Hales-Jewett](https://en.wikipedia.org/wiki/Hales%E2%80%93Jewett_theorem) or [Gallai's theorem](https://arxiv.org/abs/1411.1038), the latter of which states that in a finite coloring of $\mathbb Z^d$ or $\mathbb R^d$, there is a homothetic (i.e., scaled and translated) copy of any finite configuration $S$. Motivated by [this problem](https://mathoverflow.net/questions/299616/bichromatic-pencils), I wonder if similar statements hold if instead we require that in a configuration $(S,s_0)$ all points in $S$ are monochromatic, while $s_0$ has a different color from the rest.
Obviously, we need to impose some conditions on the coloring and the configuration. About the coloring, I only want to demand that it is non-monochromatic, i.e., not all points of the space are colored with the same color. About $(S,s_0)$, I want to require that $s_0\notin conv(S)$, i.e., $s$ is not in the convex hull of some points from $S$, as then we might not have a solution if the "first" half of the space is red, while the "second" half is blue.
> > Is there always an almost monochromatic copy (homothetic or isometric) of any finite $(S,s_0)$ with $s_0\notin conv(S)$ in a non-monochromatic finite coloring of $\mathbb R^d$?
Note that the answer is no for $\mathbb Z$ if $s_0=0$ and $S=\\{1,2\\}$ as shown by coloring odd numbers red and even numbers blue. This particular configuration, however, is easy to find in $\mathbb Q$.
|
https://mathoverflow.net/questions/300604/almost-monochromatic-point-sets
|
[
"nt.number-theory",
"co.combinatorics",
"mg.metric-geometry",
"ramsey-theory",
"polymath16"
] | 17 | 2018-05-19T13:50:40 |
[
"@fedja How can I cite your comment?",
"@fedja I've just discovered that essentially the same problem was posed by some dude called Erdős and his pals in '75, and they made the same conjecture, so you might want to spell out your argument for their sake; see Conjecture 4 in old.renyi.hu/~p_erdos/1975-12.pdf.",
"Didn't I spell it out? $\\tau$ is a field automorphism of $\\mathbb C$ preserving $\\mathbb Q$.",
"@fedja I'm not sure, but it feels like your reasoning goes the other way. Let me try. We know that there is a bad coloring (of even $\\mathbb C$) for $(0,1,s_0)$ whenever $0<s_0<1$ and from this we want to show that for some $\\tau$ there is also a bad coloring (of $\\mathbb R$) for $(0,1,\\tau(s_0))$ where $1<\\tau(s)$. And then what do you require of $\\tau$?",
"If I understand things right, you can design a field automorphism $\\tau$ of $\\mathbb C$ over $\\mathbb Q$ so that $\\tau(s_0)<0$. Your configuration is then $0,1,\\tau(s_0)$ and in your coloring $\\tau(z)$ is blue if $\\Re z\\ge 0$ and red if $\\Re z<0$. The point is that the \"similar\" configurations are obtained by linear transformations and $a\\tau(s_0)+b=\\tau(\\tau^{-1}(a)s_0+\\tau^{-1}(b))$",
"@fedja Sorry, but my field automorphisms would need some polishing, though your idea certainly seems good. Could you give me an example of how you want to derive a coloring?",
"You might be interested in mathoverflow.net/q/3322/806, which was a problem in a similar spirit in $[n]$.",
"What prevents you from taking the \"convex\" configuration $(0,s_0,1)$ with some transcendental $s_0\\in(0,1)$ and the \"half-plane\" coloring and apply a field automorphism of $\\mathbb C$ over $Q$ that moves $s_0$ to a negative number to create a non-convex configuration and some crazy coloring that just encode the old ones?"
] | 8 |
Science
| 0 |
229
|
mathoverflow
|
On manifolds which do not admit (smooth) actions of finite groups
|
**Question** : Assume a smooth manifold $M$ does not admit any effective smooth group actions of finite groups $G \neq 1$, does it follow that $M$ also admits no _continuous_ effective group actions of finite groups $G \neq 1$?
Manifolds which do not admit actions of finite groups are interesting because if $M$ is compact and we choose a Riemannian metric $g$ on $M$, then $\text{Isom}(M,g)$ is a compact Lie group by Steenrod-Myers and must be trivial, because otherwise it contains a non-trivial finite group acting on $M$.
For this nice property, it is enough to consider only smooth actions.
However, there are some interesting [articles](https://arxiv.org/pdf/math/0606714v1.pdf) which construct manifolds which do not admit any _continuous_ actions of finite groups. And so I'm wondering if its enough to eliminate smooth actions if the manifold is smooth to get a manifold of this type.
|
https://mathoverflow.net/questions/253815/on-manifolds-which-do-not-admit-smooth-actions-of-finite-groups
|
[
"at.algebraic-topology",
"differential-topology",
"smooth-manifolds",
"group-actions"
] | 17 | 2016-11-03T01:45:57 |
[
"The linked article of V. Puppe seems to say \"manifold\" for \"(connected) closed manifold\". Is the same implicit assumption made in the question?",
"Restatement of the question using group theory language: If $\\mathrm{Diff}^\\infty(M)$ is torsion-free, does it follow that $\\mathrm{Homeo}(M)$ is torsion-free?"
] | 2 |
Science
| 0 |
230
|
mathoverflow
|
Jets of sections of vector bundles expressed by symmetrized iterated covariant derivatives - who did it first?
|
The (non-unique) bundle isomorphism between the bundle $J^r E$ of $r$-th order jets of sections of a _vector_ bundle $\pi:E\rightarrow M$ and the direct sum $$\bigoplus^r_{k=0}\vee^kT^*M\otimes E\rightarrow M$$ of the vector bundles of $E$-valued totally symmetric rank-$k$ covariant tensors for all $k=0,\ldots,r$ (here $\vee^1 T^*M=T^*M$ and $\vee^0 T^*M\rightarrow M=$ the trivial line bundle $\mathrm{pr}_1:M\times\mathbb{R}\rightarrow M$ of scalars, so that $\vee^0T^*M\otimes E=E$) may be chosen at the level of $r$-th order jet prolongations $j^r s$ of smooth sections $s\in\Gamma(E)=\\{s'\in C^\infty(M,E)\ |\ \pi\circ s'=\mathrm{id}_M\\}$ of $E$ by means of symmetrized iterated covariant derivatives of $s$. This is used to write linear partial differential operators between spaces of smooth sections of vector bundles in a global form (see below).
To do so, we start by fixing a covariant derivative on $E$ and a (say, torsion-free) covariant derivative on $M$, and then extend them to a covariant derivative on $E$-valued covariant tensor fields on $M$ by means of the Leibniz rule for tensor products of smooth sections. We denote this common extension by $\nabla$. Given $s\in\Gamma(E)$, we define the $k$-th order iterated covariant derivative $\nabla^k s$ of $s$ with respect to $\nabla$ recursively as $$\nabla^0s=s\ ,\,\nabla^1 s=\nabla s\ ,\,\nabla^{k+1}s=\nabla(\nabla^k s)\ ,$$ so that $\nabla^k s\in\Gamma(\otimes^k T^*M\otimes E)$. The totally symmetric part of $\nabla^k s$ (which, of course, belongs to $\Gamma(\vee^k T^*M\otimes E)$) will be denoted by $D^k_\nabla s$.
> One can show that the map $$j^r s \mapsto i_\nabla(j^r s)\doteq(s,D_\nabla s=\nabla s,\ldots,D^r_\nabla s)$$ is a $C^\infty(M)$-linear isomorphism between the $C^\infty(M)$-modules $\Gamma(J^r E)$ and $\Gamma(\oplus^r_{k=0}\vee^kT^*M\otimes E)$, thus establishing an instance of the bundle isomorphism stated in the first paragraph. Now, given two vector bundles $\pi:E\rightarrow M$, $\pi':F\rightarrow M$, a linear partial differential operator $P$ of type $E\rightarrow F$ and order $r\geq 0$ is the composite of the $r$-th order jet prolongation map $\Gamma(E)\ni s\mapsto j^r s\in\Gamma(J^r E)$ with a $C^\infty(M)$-linear map $p$ from $\Gamma(J^r E)$ into $\Gamma(F)$ (the "symbol map" of $P$). Using $i_\nabla$, one may write $$Ps=p(j^r s)=(p\circ i_\nabla^{-1})(s,D_\nabla s,\ldots,D^r_\nabla s)=\sum^r_{k=0}a_k D_\nabla^k s\ ,$$ where $a_k\in\Gamma(\vee^k T M\otimes E^*\otimes F)=$ the $k$-th order coefficient of $P$ with respect to $\nabla$ is uniquely determined by $P$ and $\nabla$ for each $k=0,\ldots,r$ (_remark:_ uniqueness of $a_k$ no longer holds if the symmetry requirement is dropped). Moreover, $a_r$ (the "principal symbol" of $P$) does not depend on $\nabla$.
The earliest reference I could find on this statement is Chapter IV, Section 9 (more precisely, the Corollary to Theorem 7, pp. 90-91) of the book edited by Richard S. Palais, _Seminar on the Atiyah-Singer Index Theorem_ (Princeton University Press, 1965). This chapter was written by Palais himself, and it has no references.
> **Question:** Is this really the first published proof of this result? As stated is it due to Palais, or does it go back even farther?
|
https://mathoverflow.net/questions/240329/jets-of-sections-of-vector-bundles-expressed-by-symmetrized-iterated-covariant-d
|
[
"reference-request",
"dg.differential-geometry",
"ho.history-overview",
"jets"
] | 17 | 2016-06-02T22:30:00 |
[
"Conversely, given $P$ as defined in the question above, it is clear that $[P,f]=$ commutator of $P$ and (multiplication by) a smooth function $f$ on $M$ is a linear differential operator of order $k-1$. (Remark: one has to add to your definition the \"initial condition\" that linear differential operators of order $-1$ are identically zero)",
"@AliTaghavi Not really - such a definition is essentially the same as the one in terms of jet bundles. To see that, it suffices to know that the latter's fiber over $p\\in M$ may be defined as the quotient of the space $E(p)$ of germs of smooth sections of $\\pi$ over $p$ modulo the subspace of such germs vanishing to order $k+1$ at $p$. As such, it can be seen by induction on $k$ that given smooth functions $f_1,\\ldots,f_{k+1}$ on $M$, one always has $D((f_1-f_1(p))\\cdots(f_{k+1}-f_{k+1}(p))s)(p)=0$ for all $p\\in M$, $s\\in\\Gamma(\\pi)$, hence $D$ is necessarily of the form stated for $P$ above.",
"I confess I did not understand your terminologies completely. But what would be happen if one define a diffm operator of order k as a linear map $D$ such that $s\\mapsto D(fs)-fD(s)$ would be of order k-1.(inductive definition). Is there a problem of GLOBAL definition of differentoal operatores in this manner?",
"Pohl's results were derived independently by Libermann (1963) (a student of Ehresmann) and Feldman (1963), but one also cannot find any explicit isomorphism there. There is a couple of references by Ehresmann (1955) and Libermann (1961) on higher-order connections I couldn't get access to (they are both in rather obscure proceedings volumes) and maybe there is something there closer to what I ask, but I really don't know.",
"@DmitriZaitsev That's true, but that's not what I am asking. I've also tracked Ehresmann's pioneer work on jets and (arbitrary-order) connections, and couldn't find any trace of the result I've stated - to wit, relating jets to iterated (first-order) covariant derivatives. The book by Jafarpour and Lewis quoted by Umberto Lupo above traces related results to a Trans. AMS paper by Pohl (1966) whose preprint actually dates back to 1963, but there (as the authors themselves state) no such formula can be found explicitly.",
"The jets are usually attributed to Ehresmann",
"That is actually a very nice reference (despite lacking a bit of historical care, as you noticed and most people using/quoting the result do), thanks!",
"I had not known that the result dates back to Palais' work. I learned it from engineering.ucsb.edu/~saber.jafarpour/time_varying.pdf (see Lemma 2.1) and apparently those authors did not come across your reference either."
] | 8 |
Science
| 0 |
231
|
mathoverflow
|
Trying to reconcile two facts about the Appell-Lerch sum learned from Polishchuk and Zwegers
|
One of the key characters in the [thesis](https://arxiv.org/abs/0807.4834) of Zwegers is the modular correction $\tilde\mu(u,v;\tau)=\mu(u,v;\tau)+\frac i2R(u-v;\tau)$ of the Lerch sum $\mu(u,v;\tau)=\frac{e^{\pi i u}}{\vartheta(v;\tau)}\sum_{n\in\mathbb Z}(-1)^n\frac{e^{\pi i(n^2+n)\tau}e^{2\pi inv}}{1-e^{2\pi in\tau}e^{2\pi iu}}$, where $$ R(u;\tau)=\sum_{\nu\in\frac12+\mathbb Z}(-1)^{\nu-\frac12}\left\\{\operatorname{sgn}(\nu)-E\left(\left(\nu+\frac{\Im(u)}{\Im(\tau)}\right)\sqrt{2\,\Im(\tau)}\right)\right\\}e^{-\pi i\nu^2\tau}e^{-2\pi i\nu u}, $$ with $E(x)=2\int_0^xe^{-\pi t^2}dt$.
In an illuminating (for me at any rate) paper "[M. P. Appell's function and Vector Bundles of Rank 2 on Elliptic Curves](https://pdfs.semanticscholar.org/4aef/7841fc86e1e2024c020094f75a1e6e9ef8f9.pdf)", Polishchuk explains that the function $\kappa_a(z;q)=\sum_{n\in\mathbb Z}\frac{q^{n^2/2}}{q^n-a}z^n$ can be uniquely characterized by the fact that $(\kappa_a(z;q),\theta(z;q))$ defines a holomorphic section of the rank 2 vector bundle on the elliptic curve $\mathbb C^*/q^{\mathbb Z}$ given by the quotient $(z,v_1,v_2)\sim(qz,av_1+v_2,q^{-1/2}z^{-1}v_2)$ of $\mathbb C^*\times\mathbb C^2$.
What confuses me is this - the way $\kappa$ is involved in the section of a bundle on the universal elliptic curve, it must have certain modularity properties; and it is clearly holomorphic (well, for $a\notin q^{\mathbb Z}$). On the other hand $\mu$, which is obviously closely related to it, although holomorphic, is not modular, while $\tilde\mu$, being modular, is not holomorphic. So seemingly $\tilde\mu$ must have a counterpart in the context considered by Polishchuk; but the latter does not mention any possible nonholomorphic modular corrections.
How exactly are $\mu$, $\tilde\mu$ and $\kappa$ related to each other?
|
https://mathoverflow.net/questions/266199/trying-to-reconcile-two-facts-about-the-appell-lerch-sum-learned-from-polishchuk
|
[
"ag.algebraic-geometry",
"elliptic-curves",
"modular-forms",
"vector-bundles"
] | 17 | 2017-04-02T11:59:10 |
[] | 0 |
Science
| 0 |
232
|
mathoverflow
|
Are there two non-equivalent exotic structures on $\mathbb{R}^4$ coming from topologically slice, non-slice knots?
|
For a knot $K \subset S^3$, which is topologically slice but not slice (in a smooth way), there's a four manifold $\mathbb{R}^4_K$, homeomorphic but not diffeomorphic to standard euclidean $\mathbb{R}^4$.
There are ways to find such knots, particularly knots with the Alexander polynomial equal to 1, are topologically slice (Freedman). Among these knots one can find non-slice knots using obstructions like $\tau$-invariant (coming from Knot Floer Homology) or Rasmussen $s$-invariant (coming from Khovanov Homology). Knots with non-zero $\tau$- or $s$- invariant are not slice.
The negative Whitehead double of left trefoil is an example of these knots.
Question. The above argument proves the existence of an exotic structure on $\mathbb{R} ^4$, but this method proves the existence of only one exotic structure on $\mathbb{R} ^4$. Can you show if there are two different knots, $K$ and $K'$ (both topologically slice, non-slice knots), which they give us two non-equivalent exotic $\mathbb{R} ^4$s? Or if all of $\mathbb{R}_K^4$ coming from these knots are the same, i.e. you can get to only one exotic structure through this method.
|
https://mathoverflow.net/questions/277368/are-there-two-non-equivalent-exotic-structures-on-mathbbr4-coming-from-top
|
[
"differential-topology",
"knot-theory",
"4-manifolds",
"heegaard-floer-homology",
"khovanov-homology"
] | 17 | 2017-07-27T03:27:14 |
[] | 0 |
Science
| 0 |
233
|
mathoverflow
|
Need explicit formula for certain "$q$-numbers" involving gcd's
|
The question is motivated by yet another possible approach to a combinatorial problem formulated previously in ["Special" meanders](https://mathoverflow.net/q/146802/41291). I'm not giving details of the connection as I believe the question is sufficiently motivated by itself.
I've got a vector space with basis $\left\\{e_n\mid n\geqslant0\right\\}$ and scalar product $$ \left\langle e_m,e_n\right\rangle=q^{\gcd(m,n)} $$ (with the convention $\gcd(0,n)=n$ for all $n\geqslant0$).
What I need is a maximally explicit expression for an orthogonal basis $\left\\{o_n\mid n\geqslant0\right\\}$ with respect to this scalar product. I do not mind if the $o_n$ are not of unit norm (and this is clearly a minor point anyway).
Here are the first few values (calculated with MAPLE). Up to arbitrary rescalings, \begin{align*} o_{{0}}&=e_{{0}}\\\ o_{{1}}&=e_{{1}}-qe_{{0}}\\\ o_{{2}}&=e_{{2}}- \left( q+1 \right) e_{{1}}+qe_{{0}}\\\ o_{{3}}&=e_{{3}}-qe_{{2}}-e_{{1}}+qe_{{0}}\\\ o_{{4}}&=\left( q+1 \right) e_{{4}}-{q}^{2}e_{{3}}- \left( {q}^{2}+q+1 \right) e_{{2}}+{q}^{2}e_{{1}}+{q}^{2}e_{{0}}\\\ o_{{5}}&=\left( {q}^{2}+q+1 \right) e_{{5}}- \left( {q}^{2}+1 \right) qe_{{4}}- \left( {q}^{2}+1 \right) qe_{{3}}+ \left( {q}^{3}-{q}^{2}-1 \right) e_{{1}}+ \left( {q}^{2}+1 \right) qe_{{0}}\\\ o_{{6}}&=\left( {q}^{3}+{q}^{2}+2\,q+1 \right) e_{{6}}- \left( {q}^{3} +q-1 \right) qe_{{5}}- \left( {q}^{3}+q-1 \right) qe_{{4}}\\\&\- \left( {q} ^{2}+1 \right) \left( {q}^{2}+q+1 \right) e_{{3}}- \left( {q}^{3}+{q} ^{2}+2\,q+1 \right) e_{{2}}+ \left( 2\,{q}^{4}+{q}^{3}+3\,{q}^{2}+1 \right) e_{{1}}+ \left( {q}^{3}+q-1 \right) qe_{{0}}\\\ o_{{7}}&=\left( {q}^{2}+1 \right) e_{{7}}- \left( {q}^{2}-q+1 \right) qe_{{6}}- \left( {q}^{2}-q+1 \right) qe_{{5}}- \left( {q}^{2}-q+1 \right) qe_{{4}}\\\&\+ \left( {q}^{2}-q+1 \right) qe_{{2}}+ \left( {q}^{3} -2\,{q}^{2}+q-1 \right) e_{{1}}+ \left( {q}^{2}-q+1 \right) qe_{{0}} \end{align*}
The inverse transformation (again up to rescalings) does not look any more enlightening (except that everything is positive):
\begin{align*} e_{{0}}&=o_{{0}}\\\ e_{{1}}&=o_{{1}}+qo_{{0}}\\\ e_{{2}}&=o_{{2}}+ \left( q+1 \right) o_{{1}}+{q}^{2}o_{{0}}\\\ e_{{3}}&=o_{{3}}+qo_{{2}}+ \left( {q}^{2}+q+1 \right) o_{{1}}+{q}^{3}o_ {{0}}\\\ e_{{4}}&=o_{{4}}+{\frac {{q}^{2}}{q+1}}o_{{3}}+ \left( {q}^{2}+1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+1 \right) o_{{1}} +{q}^{4}o_{{0}}\\\ e_{{5}}&=o_{{5}}+{\frac { \left( {q}^{2}+1 \right) q}{{q}^{2}+q+ 1}}o_{{4}}+{\frac {q \left( {q}^{2}+1 \right)}{q+1}} o_{{3}}+q \left( {q}^{2} +1 \right) o_{{2}}+ \left( {q}^{4}+{q}^{3}+{q}^{2}+q+1 \right) o_{{1}} +{q}^{5}o_{{0}}\\\ e_{{6}}&=o_{{6}}+{\frac { \left( {q}^{3}+q-1 \right) q}{{q}^{3}+ {q}^{2}+2\,q+1}}o_{{5}}+{\frac { \left( {q}^{3}+q-1 \right) q}{{q}^{2} +q+1}}o_{{4}}+{\frac { \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}} o_{{3}}\\\ &\+ \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+ 1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) o_{{1}}+{q}^{6}o_{{0}}\\\ e_{{7}}&=o_{{7}}+{\frac { \left( {q}^{2}-q+1 \right) q}{{q}^{2}+ 1}}o_{{6}}+{\frac { \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) q}{{q}^{3}+{q}^{2}+2\,q+1}}o_{{5}}+ \left( {q}^{2}-q+1 \right) qo_{{4}}\\\ &+{\frac {q \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}}o_{{3}} +q \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right) o_{{2}} \+ \left( {q}^{6}+{q}^{5}+{q}^{4}+{q}^{3}+{q}^{2}+q+1 \right) o_{{1}}+{q}^{7}o_{{0}}\\\ e_{{8}}&=o_{{8}}+{\frac { \left( {q}^{2}+1 \right) {q}^{4}}{ \left( {q}^{2}+q+1 \right) \left( {q}^{3}+q+1 \right) }}o_{{7}}+{\frac {{q} ^{4}}{{q}^{2}+q+1}}o_{{6}}+{\frac { \left( {q}^{2}+1 \right) {q}^{4}}{{q}^{3}+{q}^{2}+2\,q+1}}o_{{5}}\\\ &+{\frac { {q}^{6}+{q}^{4}+{q}^{2}+q+1 }{{q}^{2}+q+1}}o_{{4}} +{\frac {{q}^{2} \left( {q}^{2}+q+1 \right) \left( {q}^{2}-q+1 \right)}{q+1}}o_{{3}}+ \left( {q}^{2}+1\right) \left( {q}^{4}+1 \right) o_{{2}}+ \left( q+1 \right) \left( {q}^{2}+1 \right) \left( {q}^{4}+1 \right) o_{{1}}+{q}^{8}o_{ {0}} \end{align*}
There are some patterns, but I cannot even guess any statement that I could try to prove about these coefficients. Also the above was computed under the assumption that the transformation matrix is triangular, i. e. that $o_n$ only depends on $e_k$ with $k\leqslant n$. It is quite possible that there is a nicer basis which violates this assumption, but again I cannot think of any natural alternative form of the transformation matrix.
What I know is an explicit orthogonal basis for a similar scalar product "without $q$": if I would have just $\left\langle e_m,e_n\right\rangle=\gcd(m,n)$, then an orthogonal basis would be given by $e_n=\sum_{d|n}o_d$, so $o_n=\sum_{d|n}\mu\left(\frac nd\right)e_d$. Strangely enough, I obtain this from the above by substituting $q=0$, and again I have no idea why.
As @alpoge points out in a comment, if one removes $e_0$, then this actually works "with $q$" too.
PS As suggested by @Wolfgang in a comment, I've looked at polynomials resulting from the substitutions $q=\pm1$, $e_n=x^n$. Here:
$q=1$: \begin{align*} &1\\\ &x-1\\\ & \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 2\,x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 3\,{x}^{2}+x+2 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 5\,{x}^{3}+4\,{x}^{2}+8\,x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( {x}^{2}+x+1 \right) \left( 2\,{x}^{2}-x+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 9\,{x}^{5}+7\,{x}^{4}+14\,{x}^{3}+10\,{x}^ {2}+6\,x+2 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 11\,{x}^{6}+8\,{x}^{5}+16\,{x}^{4}+10\,{x} ^{3}+15\,{x}^{2}+9\,x+3 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 7\,{x}^{5}+6\,{x}^{4}+5\,{x}^{3}+4\,{x}^{2 }+10\,x+1 \right) \left( {x}^{2}+1 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 16\,{x}^{8}+11\,{x}^{7}+22\,{x}^{6}+12\,{x }^{5}+18\,{x}^{4}+3\,{x}^{3}+9\,{x}^{2}-6\,x+5 \right) \left( x-1 \right) ^{2}\\\ & \left( x+1 \right) \left( 21\,{x}^{9}+19\,{x}^{8}+38\,{x}^{7}+34\,{x }^{6}+51\,{x}^{5}+45\,{x}^{4}+39\,{x}^{3}+33\,{x}^{2}+6\,x+2 \right) \left( x-1 \right) ^{2} \end{align*} $q=-1$: \begin{align*} 0&:1\\\ 1&:1+x\\\ 2&:\left( x-1 \right) \left( 1+x \right)\\\ 3&:\left( x-1 \right) \left( 1+x \right) ^{2}\\\ 4&:- \left( x-1 \right) \left( 1+x \right) ^{2}\\\ 5&:\left( x-1 \right) \left( {x}^{2}+x+2 \right) \left( 1+x\right) ^{2}\\\ 6&:-\left( x-1 \right) \left( {x}^{3}+2\,{x}^{2}+2\,x+3 \right) \left( 1+x \right) ^{2}\\\ 7&:\left( x-1 \right) \left( {x}^{2}+x+1 \right) \left( 2\,{x}^{2}- x+3 \right) \left( 1+x \right) ^{2}\\\ 8&:-\left( x-1 \right) \left( {x}^{4}+2\,{x}^{2}+2 \right) \left( 1 +x \right) ^{3}\\\ 9&:\left( x-1 \right) \left( {x}^{5}+{x}^{4}+{x}^{3}+3\,{x}^{2}+3 \right) \left( 1+x \right) ^{3}\\\ 10&:- \left( x-1 \right) \left( {x}^{2}+1 \right) \left( {x}^{5}+2\, {x}^{4}+{x}^{3}+2\,{x}^{2}+2\,x+3 \right) \left( 1+x \right) ^{2}\\\ 11&:\left( x-1 \right) \left( 4\,{x}^{8}+{x}^{7}+8\,{x}^{6}+2\,{x}^{ 5}+12\,{x}^{4}+3\,{x}^{3}+11\,{x}^{2}+4\,x+5 \right) \left( 1+x \right) ^{2} \end{align*}
|
https://mathoverflow.net/questions/210448/need-explicit-formula-for-certain-q-numbers-involving-gcds
|
[
"nt.number-theory",
"co.combinatorics",
"orthogonal-matrices"
] | 17 | 2015-06-30T00:37:31 |
[
"@OfirGorodetsky Thanks for the link! Reading that - seems an interesting idea to give it a combinatorial interpretation in terms of the Möbius function of some poset. In fact this might be important for the original problem which is of combinatorial nature.",
"@alpoge But this is great! I was sticking to $e_0$ so much it never occurred to me that dropping it gives such a nice orthogonal basis! It is true that I still need $e_0$ essentially (it is the unit of certain algebra), but your version does almost all of it. Could you please post this as an answer? It's worth it in any case, and if nobody will find a way to incorporate $e_0$ I will go for it",
"Just a remark\\restatement, nothing fancy: Let $A_{i,j} = q^{(i,j)}$ be a $(n+1) \\times(n+1)$ matrix. Decompose $A$ as $B^T B$ (Cholesky-Decomposition; $B$ can be upper-triangular). The condition that $o_i = \\sum_{j} o_{i,j} e_j$ are orthonormal w.r.t to the given inner product is the same as requiring that $B o_i$ are orthonormal w.r.t to the ordinary inner product, so we can take $o_i = B^{-1} e_i$. Hence the problem reduces to Cholesky-decomposing $A$, and inverting. Bruce Sagan's talk here about GCD-matrices might provide useful input: users.math.msu.edu/users/sagan/Slides/mfp5.pdf",
"Formally doesn't $o_n := \\sum_{d\\vert n} \\mu(d) e_{n/d}$ for $n > 0$ and $o_0 := e_0 - \\sum_{n\\geq 1} o_n$ work? (Not that that helps much, since the expression for $o_0$ is nonsense.)",
"@Wolfgang Well $q=-1$ looks somehow less hopeless but still I have no clue :) I'll add it too",
"and I suppose the same with $q=-1$ doesn't reveal more either?",
"I had skipped $n=6 $, so that was a wrong conclusion of mine! :(.",
"@PeterMueller I've added that",
"@PeterMueller I use $\\gcd(0,0)=0$, I think it is not senseless in view of $\\gcd(0,n)=\\gcd(n,n)=n$ for all $n$.",
"@OP: What is $\\langle e_0,e_0\\rangle$? It seems to me that the problem is not well formulated, as $gcd(0,0)=\\infty$. In particular, I don't know how to interpret your assertion $0=\\langle o_0,o_1\\rangle=\\langle e_0,e_1-qe_0\\rangle=q-q\\cdot q^{gcd(0,0)}$.",
"@Wolfgang I've looked at it, what happens that seeminglu each of these polynomials (well, starting from 3) is divisible by $(x-1)^2(x+1)$, but the quotient seems impenetrable to me. I'll add it to the question though",
"@Wolfgang Many thanks for the suggestion, sounds promising! I'll definitely try it",
"Have you tried putting $q=1$ and replacing $e_i$ by $x^i$ in the $o_j$ expressions? The resulting polynomials seem to split into linear and quadratic factors, and maybe you'll find some patterns in those factors, which might give some ideas to start with."
] | 13 |
Science
| 0 |
234
|
mathoverflow
|
Bunnity of multilinear maps
|
Is there a way to compute the following nullity of multilinear maps? As it is different from any nullity I know of, I call it _bunnity_ after myself:-)) If it already has a name, it be nice to know it. References are particularly appreciated.
Let $V$ be a $d$-dimensional vector space over a field $K$, $f:V^{\otimes n}\rightarrow K$ a multilinear map. We define the bunnity by
$Bun(f) = sup \\{ \sum_{k=1}^n dim(U_k) \ | \ f(U_1\otimes U_2 \ldots \otimes U_n)=0\\}$.
If $n=2$, the bunnity is nullity plus dimension. It is more complicated if $n\geq 3$. I expect that for $n=3$ or $n=4$, there should be an algorithm because the 3-subspace problem is finite, while the 4-subspace problem is tame. I would be pleasantly surprised if there is an algorithm for a general $n$.
**EDIT** I forgot to say what $U_i$ are. They are **non-zero** vector subspaces of $V$.
|
https://mathoverflow.net/questions/206211/bunnity-of-multilinear-maps
|
[
"linear-algebra",
"multilinear-algebra"
] | 17 | 2015-05-10T07:36:12 |
[
"The dimension of the null-space: think of $f$ as a map $V\\rightarrow V^\\ast$ and take the dimension of its kernel. Similarly, the usual multinullity is defined for any $n$: use position $i$ to write $f$ as $V\\rightarrow (V^{\\otimes (n-1)})^\\ast$ and take the dimension of its kernel.",
"Could you remind what definition of \"nullity\" do you use (e.g. when $n=2$)?",
"You should offer a bounty to whoever manages to frame the answer in terms of Wascal's triangle.",
"Inequality. In the usual tensor rank/nullity, you choose each $U_i$ separately. I ask you to choose them simultaneously.",
"What's the connection to multilinear tensor rank?"
] | 5 |
Science
| 0 |
235
|
mathoverflow
|
Question about combinatorics on words
|
Let $\\{a_1,a_2,...,a_n\\}$ be an alphabet and let $\\{u_1,...,u_n\\}$ be words in this alphabet, and $a_i\mapsto u_i$ be a substitution $\phi$.
Question: Is there an algorithm to check if for some $m,k$ some prefix of the word $\phi^m(a_1)$ coincides with some suffix of the word $\phi^k(a_2)$?
This is related to a question about the R. Thompson group $F$.
|
https://mathoverflow.net/questions/250290/question-about-combinatorics-on-words
|
[
"co.combinatorics",
"gr.group-theory",
"semigroups-and-monoids",
"combinatorics-on-words"
] | 17 | 2016-09-19T20:08:39 |
[
"@domotorp The equality version is decidable, by Theorem 4 of [1]. Unfortunately, this doesn't seem very helpful for the original question, because it goes via saying that if the substitution homomorphism is not injective, then it factors through a smaller alphabet and we are happy by induction. [1] A. Ehrenfeucht, G. Rozenberg, Simplifications of homomorphisms, Information and Control, Volume 38, Issue 3, 1978 (sciencedirect.com/science/article/pii/S0019995878900955)",
"With the below simple trick we can ensure that if a prefix equals a suffix, then the whole words are equal. I'm thinking that deciding equality might be already undecidable. Trick: $\\phi(a_1)=a_{start} a_1' a_{end}$ and $\\phi(a_2)=a_{start} a_2' a_{end}$, where $\\phi(a_{start})=a_{start}$ and $\\phi(a_{end})=a_{end}$, and none of $a_1, a_2, a_{start}, a_{end}$ are in the image of any other $a_i$.",
"@NaradRampersad: You can ask him and post his answer here, if he is not on MO himself.",
"I don't have an actual solution to the problem, but I know the best person to ask about it: Juha Honkala of the University of Turku. If anyone knows the answer it would be him.",
"This is a decision problem. I need to decide whether or not the property holds. The second $n$ is a $k$.",
"Assuming the words are finite, there is the brute force method with no guarantee of termination. Are you looking for an algorithm with such a guarantee, or do you want even more? Also, is the n in the question the same as the alphabet size? (Probably not.) Gerhard \"Just Checking On Some Details\" Paseman, 2016.09.19."
] | 6 |
Science
| 0 |
236
|
mathoverflow
|
The topos for forcing in computability theory
|
My understanding is that forcing (such as Cohen forcing) can be described via a topos. For example this [nlab article on forcing](http://ncatlab.org/nlab/show/forcing) describes forcing as a "the topos of sheaves on a suitable site."
My question concerns forcing in computability theory, for example as described in Chapter 3 or these [lecture notes of Richard Shore](http://www.math.cornell.edu/~shore/papers/pdf/SingLect2NS.pdf). The idea is that the generics are those which meet all _computable_ dense sets of forcing conditions. (Computable can mean a few things. Often it is taken to mean a $\Sigma^0_1$ set of forcing conditions. Also, usually the forcing posets are countable.) Since there are only countably many such dense sets, such effective generics exist.
> Is there a known/canonical type of topos corresponding to the forcing in computability theory?
Any references would be appreciated.
_FYI: My background is in computability theory, proof theory, and computable analysis. I know little about topos theory, but I am willing to learn a bit. I am mostly asking this question because I want to compare some ideas I have about effective versions of Solovay forcing with some work by others about the topos corresponding to Solovay forcing. Also, it is always nice to learn new things._
|
https://mathoverflow.net/questions/195794/the-topos-for-forcing-in-computability-theory
|
[
"ag.algebraic-geometry",
"ct.category-theory",
"lo.logic",
"computability-theory",
"topos-theory"
] | 17 | 2015-02-05T19:55:21 |
[
"Have you, since asking the question two years ago, learned about a topos-theoretic treatment of forcing in computability theory?",
"@BjørnKjos-Hanssen, I would be happy to know the answer in that case. (The application I have in mind is a bit more general, but I think knowing what is going on in the case of a computable forcing poset is sufficient for me to get a general idea of what is going on.)",
"Did you also want the forcing partial order to be a (close to) computable relation?"
] | 3 |
Science
| 0 |
237
|
mathoverflow
|
Elements of finite fields with many powers of trace zero
|
Let $p$ be an odd prime number, $n>1$ be an integer, and $\mathrm{tr}$ be the trace map of the field extension $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$. For which pair $(p,n)$ does there exists $x\in\mathrm{GF}(p^{2n})$ such that $x^{2(1+p^n)}=1$ and $\mathrm{tr}(x^{1+p^j})=0$ for $j=0,1,\dots,n-1$?
After computing the gcd of the corresponding $n+1$ polynomials over $\mathrm{GF}(p)$ for small $p$ and $n$, I believe that the answer should be ``if and only if $p$ divides $n$" and moreover, when $p$ divides $n$, all such $x$ would form a subgroup of $\mathrm{GF}(p^{2n})^\times$ of index $2(1+p^{n/p})$.
**Update.** If $(p,n)$ is such a pair, then denoting by $X$ the $(2n)\times(2n)$ matrix with $(i,j)$th entry $x^{p^{i+j-2}}$ we deduce from the conditions that $$ X^2= \begin{pmatrix} 0&2naI_n\\\ 2naI_n&0 \end{pmatrix}, $$ where $a=x^{1+p^n}=\pm1$. This yields two necessary conditions:
(1). If $n$ is not divisible by $p$ then $n$ is odd.
_Proof._ Suppose on the contrary that $n$ is even. Then $$ |X|^2= \begin{vmatrix} 0&2naI_n\\\ 2naI_n&0 \end{vmatrix} =(-1)^n(2na)^{2n}=(2na)^{2n} $$ is a square in $\mathrm{GF}(p)$, which implies that $|X|\in\mathrm{GF}(p)$. However, changing columns shows that $-|X|=(-1)^{2n-1}|X|$ is the determinant of the matrix with $(i,j)$th entry $x^{p^{i+j-1}}$. Hence $-|X|=|X|^p$ and so $|X|\notin\mathrm{GF}(p)$, a contradiction.
(2). If $n$ is not divisible by $p$ then $x,x^p,\dots,x^{p^{2n-1}}$ is a normal basis of $\mathrm{GF}(p^{2n})/\mathrm{GF}(p)$.
_Proof._ $|X|^2=(-1)^n(2na)^{2n}\neq0$ implies that $|X|\neq0$ and so $x,x^p,\dots,x^{p^{2n-1}}$ are $\mathrm{GF}(p)$-linearly independent.
|
https://mathoverflow.net/questions/257733/elements-of-finite-fields-with-many-powers-of-trace-zero
|
[
"nt.number-theory",
"galois-theory",
"finite-fields"
] | 17 | 2016-12-20T21:45:08 |
[] | 0 |
Science
| 0 |
238
|
mathoverflow
|
Groups generated by 3 involutions
|
Let $r(m)$ denote the residue class $r+m\mathbb{Z}$, where $0 \leq r < m$. Given disjoint residue classes $r_1(m_1)$ and $r_2(m_2)$, let the class transposition $\tau_{r_1(m_1),r_2(m_2)}$ be the permutation of $\mathbb{Z}$ which interchanges $r_1+km_1$ and $r_2+km_2$ for every $k \in \mathbb{Z}$ and which fixes everything else.
1. Is it algorithmically decidable whether all orbits on $\mathbb{Z}$ under the action of a group generated by 3 given class transpositions are finite?
2. Is it algorithmically decidable whether a group generated by 3 given class transpositions acts transitively on the set of nonnegative integers in its support (i.e. set of moved points)?
**Added on Dec 11, 2013:** This question will appear as Problem 18.47 in:
Kourovka Notebook: _Unsolved Problems in Group Theory_. Editors V. D. Mazurov, E. I. Khukhro. 18th Edition, Novosibirsk 2014.
Remarks:
Ad 1.: A hard case is $G = \left\langle\tau_{0(2),1(2)}, \tau_{0(5),4(5)}, \tau_{1(4),0(6)}\right\rangle$. For example we have $|32^G| = 6296$ and $|25952^G| = 245719352$, and the largest point in the latter orbit is about $10^{5759}$. The finiteness of the orbit $173176^G$ is not known to the author so far.
EDIT: In the meantime, the length of the cycle of $g := \tau_{0(2),1(2)} \cdot \tau_{0(5),4(5)} \cdot \tau_{1(4),0(6)}$ containing 173176 has been computed by Jason B. Hill (<http://mail.gap-system.org/pipermail/forum/2012/003948.html>) -- it is 47610700792, and the largest point in the cycle is about $10^{76785}$. An earlier attempt of the same computation ended without success after one week of CPU time (<http://mail.gap-system.org/pipermail/forum/2012/003915.html>). Possibly the cycle length is also the orbit length $|173176^G|$.
Ad 2.: A hard case is $G = \left\langle\tau_{1(2),4(6)}, \tau_{1(3),2(6)}, \tau_{2(3),4(6)}\right\rangle$. As the question whether this group acts transitively on $\mathbb{N} \backslash 0(6)$ is equivalent to Collatz' $3n+1$ conjecture (cf. <http://en.wikipedia.org/wiki/Collatz_conjecture>), likely only a negative answer might be given without solving a known open problem. This will likely also not be easy -- at least unless someone has e.g. a good idea on how to encode arbitrary computations with just 3 class transpositions.
**Added on Apr 24:** An easy case is $G = \left\langle\tau_{0(2),1(2)}, \tau_{0(3),2(3)}, \tau_{1(2),2(4)}\right\rangle$. By means of computation it can be checked that this group acts at least 5-transitively on $\mathbb{N}_0$.
|
https://mathoverflow.net/questions/112527/groups-generated-by-3-involutions
|
[
"permutation-groups",
"computational-group-theory",
"gr.group-theory",
"nt.number-theory",
"computability-theory"
] | 17 | 2012-11-15T14:14:42 |
[] | 0 |
Science
| 0 |
239
|
mathoverflow
|
Is there a n/2 version of the Erdős-Hanani conjecture?
|
This question comes out of REU research from this past summer. Unfortunately weeks of thought led to only trivial observations and the conclusion that the problem is quite hard.
Fix $k,t$. Let $F$ be a set of $k$-subsets of $[n] := \\{1,\ldots,n\\}$ of minimal cardinality such that $F$ covers all $t$-subsets of $[n]$ (covers in the sense that any $t$-subset of $[n]$ is a subset of an element of $F$.) Let $\kappa_n := |F|$. The Erdős-Hanani conjecture states that
> $\kappa_n = \binom{n}{t} / \binom{k}{t}(1 + o(1))$.
Of course $\binom{n}{t} / \binom{k}{t}$ is a lower bound on $\kappa_n$, so the EH conjecture is saying that the obvious necessary condition is asymptotically sufficient. Rödl proved the EH conjecture in 1985.
This question is about what happens when $k$ and $t$ are not fixed. Specifically, take $k = \lfloor n/2 \rfloor$ and $t = \lfloor n/2\rfloor - 1$. Define $F$ and $\kappa_n$ as above. Is it true that
> $\kappa_n = \frac{1}{\lfloor n / 2 \rfloor} \binom{n}{\lfloor n/2 \rfloor}(1 + o(1))$?
# Background
The EH conjecture lead to the study of what is called "packing in a hypergraph." See <http://en.wikipedia.org/wiki/Packing_in_a_hypergraph>. Rödl's proof introduced what is now called the "Rödl nibble" and is pseudo-random in nature. Spencer gave a lovely proof using branching processes. There are a lot of results from the late 80s to 90s that say, as Kahn puts it in "Asymptotics of Hypergraph Matching, Covering and Coloring Problems", that hypergraphs are asymptotically well-behaved _as long as their edge sizes are bounded_! Unfortunately the $n/2$ version of EH involves hypergraphs of unbounded edge size and the existing methods appear useless.
# Some ideas
A straightforward application of the method of alterations (or equivalently, some easy analysis of the greedy algorithm) gives that $\kappa_n \leq \log n \frac{1}{\lfloor n / 2 \rfloor} \binom{n}{\lfloor n/2 \rfloor} (1 + o(1))$, so the whole question is whether we can eliminate this log factor.
A set of $\lfloor n/2 \rfloor$-subsets has maximum coverage of $(\lfloor n/2 \rfloor - 1)$-subsets when all its elements have pairwise symmetric difference of at least 4. So really this is a coding theory problem. The paper "Lower bounds for constant weight codes" by Graham and Sloane shows that we can find a set $H$ of $\lfloor n /2\rfloor$-subsets of $[n]$ such that $|H| \geq \frac{1}{2}\binom{n}{\lfloor n/2 \rfloor}$ and the hamming distance between elements is at least 4. Let $G$ be the set of $(\lfloor n/2 \rfloor - 1)$-subsets covered by $H$. $G$ is half the size we want it to be, but we only used half as many elements are we are allowed. So we might be optimistic that by allowing some small overlap we can cover everything we want. If we take a permutation $\sigma \in S_n$ and look at $\sigma(H)$ (i.e. apply the permutation to the elements of the elements of $H$) it covers $\sigma(G)$. Of course $|\sigma(G)| = |G|$. We could hope that a good choice of $\sigma$ gives $|G \cup \sigma(G)| \approx 2|G|$ and we have found an appropriate set $F := H \cup \sigma(H)$. I asked the question of whether such a $\sigma$ must exist before: [Size of union of a set of subsets and its permutations](https://mathoverflow.net/questions/101886/size-of-union-of-a-set-of-subsets-and-its-permutations). That question is interesting in its own right, but this EH conjecture is really why I wanted an answer.
|
https://mathoverflow.net/questions/121961/is-there-a-n-2-version-of-the-erd%c5%91s-hanani-conjecture
|
[
"co.combinatorics"
] | 17 | 2013-02-15T17:02:33 |
[
"@Gerhard Paseman: it looks like the repository has $n < 100$, $k \\leq 25$ and $t \\leq 8$. With such small values it is not possible to detect the presence or absence of a log factor.",
"Some structural results on this part of the boolean lattice can be found with the search phrases \"middle levels\" and \"boolean\" together. For example, see the references in arxiv.org/pdf/math.CO/0608485.pdf .",
"I should mention also that the Graham-Sloane construction means that the EH-conjecture for $k = \\phi(n)$ and $t = k - 1$ has a positive answer when $\\phi(n) = o(n)$, so $n/2$ is in fact the \"hard case.\"",
"@Brendan McKay: Yes, thank you. It is corrected.",
"Does your description have $t$ and $l$ mixed up? If so, please edit.",
"Have you checked out the La Jolla Covering Repository online? That might give a nice suggestion of what to expect. Gerhard \"Ask Me About System Design\" Paseman, 2013.02.15"
] | 6 |
Science
| 0 |
240
|
mathoverflow
|
Katz--Mazur for abelian varieties
|
Over $\mathbb Z$, there is a smooth DM stack $A_g$ classifying abelian varieties.
Over $\mathbb Z[\frac 1N]$, there is finite etale cover $A_g(N)_{\mathbb Z[\frac 1N]}\to A_g\otimes\mathbb Z[\frac 1N]$ classifying abelian varieties with $N$-level structure.
> Is there a finite proper map $A_g(N)\to A_g$ over $\mathbb Z$ with a _nice moduli interpretation_ , which recovers the usual notion of $N$-level structure over $\mathbb Z[\frac 1N]$?
**Remark:** In fact, I believe that such an $A_g(N)\to A_g$ is determined canonically from $A_g(N)_{\mathbb Z[\frac 1N]}\to A_g\otimes\mathbb Z[\frac 1N]$ by "normalizing in the function field", so the content of the question is really to have a good moduli interpretation for this canonical extension.
In more informal terms, I'm just asking:
> What is the "right" notion of $p$-level structure on an abelian variety in characteristic $p$?
The impression I get from reading in various places is that the answer to this is well-known, but I haven't been able to find a good reference. I don't have a specific type of level structure in mind. I'd just like to know what is known (and where to find it) for some common notions of level structure. **EDIT** : Actually, for level structures consisting of a choice of _subgroup(s)_ , there is a natural choice of moduli space over $\mathbb Z$ classifying abelian varieties with a choice of subgroup scheme(s) (or equivalently, isogenies) and this is valid in "bad" characteristics -- see, e.g. the end of [Chai](http://www.ams.org/mathscinet-getitem?mr=1124633). So, let me state my question for level structures given by choices of _point(s)_ , e.g. "a point of exact order $N$", or full level $N$ structure.
For elliptic curves, [Katz--Mazur](http://www.ams.org/mathscinet-getitem?mr=772569) showed that the answer is provided by the notion of a _Drinfeld level structure_ (first considered by [Drinfeld](http://www.ams.org/mathscinet-getitem?mr=384707)). For example, the "right" notion of a $p$-torsion point on an elliptic curve $E\to S$ (for _any_ scheme $S$) is a point $P:S\to E$ so that:
1. $pP=1$.
2. The Cartier divisor $\sum_{i=0}^{p-1}[iP]$ is a subgroup scheme.
As Katz--Mazur write, it is not clear how to generalize this notion to abelian varieties since points are no longer divisors.
|
https://mathoverflow.net/questions/152986/katz-mazur-for-abelian-varieties
|
[
"ag.algebraic-geometry",
"nt.number-theory",
"moduli-spaces",
"abelian-varieties",
"characteristic-p"
] | 17 | 2013-12-28T11:48:13 |
[
"Dear Daniel Litt: Thanks for tracking it down; that is indeed the right paper. It is also worth noting (in view of the question posed) that this paper provides a sense in which mere normalization in higher level (which doesn't have positive-dimensional fibers) is the \"wrong\" thing to consider, suggesting that perhaps the question posed about normalization is not a \"useful\" point of view for applications with $g > 1$.",
"The paper of Chai and Norman user76758 alludes to is likely this one: jstor.org/stable/2374734",
"Note that $A_g$ classifies principally polarized abelian schemes (of relative dimension $g$); the polarization aspect comes along for free uniquely when $g=1$ but not otherwise, as you undoubtedly know. I was once told that there is a paper by Chai and Norman (title I do not know) which shows in some sense that there isn't a good version of Drinfeld's idea beyond relative dimension 1. Perhaps email your question to Chai (or Kai-Wen Lan) if you don't receive a satisfactory answer here.",
"+1: very nice question!"
] | 4 |
Science
| 0 |
241
|
mathoverflow
|
Souslin trees and weakly compact cardinals
|
In [Souslin trees on the first inaccessible cardinal](https://mathoverflow.net/questions/143530/souslin-trees-on-the-first-inaccessible-cardinal) it is asked if it is consistent that there are no $\kappa-$Souslin trees at the least inaccessible cardinal $\kappa$.
In this question I would like to ask an apparently simpler question:
**Question.** Is it consistent that there are no $\kappa-$Souslin trees for some inaccessible not weakly compact cardinal $\kappa$?
**Remarks.** (A) If $\kappa$ is weakly compact, then the tree property holds at $\kappa$, in particular there are no $\kappa-$Souslin trees,
(B) If $V=L,$ then $\kappa$ is weakly compact iff there are no $\kappa-$Souslin trees.
(C) $\Diamond_\kappa+GCH$ does not imply the existence of a $\kappa-$Souslin tree, for $\kappa$ inaccessible. For example if $\kappa$ is ineffable and $GCH$ holds, then $\Diamond_\kappa+GCH$ holds and there are no $\kappa-$Souslin trees.
|
https://mathoverflow.net/questions/161868/souslin-trees-and-weakly-compact-cardinals
|
[
"lo.logic",
"set-theory",
"forcing",
"large-cardinals"
] | 17 | 2014-03-30T04:28:37 |
[
"Its about your comment above (since in the situation of your comment $S$ is non reflecting at every $\\alpha$, as witnessed by $\\text{acc }C_\\alpha$).",
"Your comment is about my above comment for a weaker sufficient condition?",
"@Mohammad: you don't actually need a $\\square (\\kappa )$ sequence, all you need is a non reflecting stationary set $S$ with diamond on it.",
"@Yair: Your last comment seems like a familiar topic all of a sudden! :-)",
"If there is a $\\kappa$-closed $\\kappa$ saturated ideal on $\\kappa$ and $\\kappa$ is not measurable then $\\kappa$ is inaccessible and there is a $\\kappa$-Suslin tree (by splitting the positive sets in a partial binary tree). Actually, in this case every normal $\\kappa$-tree contains a Suslin subtree.",
"Does the existence of a $\\kappa-$Souslin tree for $\\kappa$ inaccessible have any effects on saturated ideals on $\\kappa$ or something else? Maybe this way we can reduce the problem to something that we know more about it.",
"I think it does not give the answer. Note that a sufficient condition for having a $\\kappa-$Souslin tree for $\\kappa$ inaccessible, is square-with-built-in-diamond at $\\kappa$, as is implicit in [Sh:624]. A weaker sufficient conditions is, e.g.,: there exists a $\\square(\\kappa)-$sequence $< C_\\alpha: \\alpha<\\kappa >$, and a stationary set $S$ in $\\kappa$, which is disjoint from $acc(C_\\alpha)$ for all $\\alpha<\\kappa$, and $\\Diamond(S)$ holds. To force the failure of these principles, we need more than weakly compact cardinals.",
"What happens if we start with a weakly compact cardinal $\\kappa$ in $V$, and then force to kill the Mahloness of $\\kappa$ by adding club $C\\subset\\kappa$ avoiding the regular cardinals below $\\kappa$? Does this create a $\\kappa$-Suslin tree?"
] | 8 |
Science
| 0 |
242
|
mathoverflow
|
Maximum automorphism group for a 3-connected cubic graph
|
The following arose as a side issue in a project on graph reconstruction.
**Problem:** Let $a(n)$ be the greatest order of the automorphism group of a 3-connected cubic graph with $n$ vertices. Find a good upper bound on $a(n)$.
There is a [paper of Opstall and Veliche](http://arxiv.org/abs/math/0608645) that finds the maximum over all cubic graphs, but the maximum occurs for graphs very far from being 3-connected.
I foolishly **conjecture** : for $n\ge 16$, $a(n) < n 2^{n/4}$.
When $n$ is a multiple of 4 there is a vertex-transitive cubic graph achieving half the conjectured bound, so if true the bound is pretty sharp.
The maximum does not always occur for vertex-transitive graphs. There is no vertex-transitive counterexample to the conjecture with less than 1000 vertices.
Any bound for which the exponential part is $2^{cn}$ for $c<\frac12$ is potentially useful.
**Added:** As verret pointed out in a comment, a [paper of Potočnik, Spiga and Verret](http://arxiv.org/abs/1010.2546), together with the computation I mentioned, establishes the conjecture for vertex-transitive graphs, so the remaining problem is whether one can do better for non-transitive graphs. For 20, and all odd multiples of 2 vertices from 18 to at least 998 (but not for 4-16 or 24 vertices) the graph achieving the maximum is not vertex-transitive.
|
https://mathoverflow.net/questions/136929/maximum-automorphism-group-for-a-3-connected-cubic-graph
|
[
"co.combinatorics",
"gr.group-theory",
"graph-theory",
"finite-groups",
"automorphism-groups"
] | 17 | 2013-07-16T18:29:15 |
[
"In the cubic vertex-transitive case and n twice an odd number, it immediately follows from the same paper that you get a polynomial rather than exponential upper bound. For example Corollary 4 yields that, for large enough n, we have |G|<n^2. This is not best possible, but is not far off, at least for some values of n. If you need more precise estimates, I can show you a few more references that deal with this. (By the way, the link in your \"added\" section has a typo.)",
"@verret: Any idea what the maximum is for transitive cubic graphs of order an odd multiple of 2?",
"Great! I vaguely remembered such a paper but hadn't managed to find it today.",
"The family of vertex-transitive graphs you have in mind are in fact best possible among large enough cubic vertex-transitive graphs. (n=100 or so should already suffice.) This is shown in the paper : \"Bounding the order of the vertex-stabiliser in 3-valent vertex-transitive and 4-valent arc-transitive graphs\", arxiv.org/abs/1010.2546. Therefore, a counter-example to your conjecture will necessarily be not vertex-transitive."
] | 4 |
Science
| 0 |
243
|
mathoverflow
|
Actions on ℍⁿ generated by torsion elements
|
Let $n$ be a large integer.
I am looking for a cocompact properly discontinuous isometric action on $n$-dimensional Lobachevky space which is generated by elements of finite order.
Or equivalently, I need a cocompact properly discontinuous isometric action $\Gamma\curvearrowright\mathbb{H}^n$ such that $\mathbb{H}^n/\Gamma$ is simply connected, here $\mathbb{H}^n/\Gamma$ stays for quotient space (NOT for orbifold).
**Comments.**
* I am aware that for large $n$ these is no cocompact action generated by reflections (Vinberg, 1984).
* Consider the group of matrices with integer coefficients from $\mathbb{Q}[\sqrt{5}]$ which preserve the quadratic form $$\tfrac{1+\sqrt{5}}{2}\cdot x_0^2-x_1^2-\cdots-x_n^2.$$ This gives cocompact properly discontinuous isometric action, say $\Gamma\curvearrowright\mathbb{H}^n$. (For $n=2$, the group $\Gamma$ contains the Coxeter group of regular right-angled pentagon.)
_I would be very happy if the subgroup generated by the elements of finite order in $\Gamma$ would have finite index. Or, equivalently, if_ $$|\pi_1(\mathbb{H}^n/\Gamma)|<\infty.$$
**P.S.** It seems that examples of such actions are not known. (In addition to Agol's comment, I've got a letter from Vinberg stating this.)
|
https://mathoverflow.net/questions/102644/actions-on-%e2%84%8d%e2%81%bf-generated-by-torsion-elements
|
[
"group-actions",
"hyperbolic-geometry",
"algebraic-groups",
"open-problems"
] | 17 | 2012-07-19T04:21:58 |
[
"@ Anton: I mean he computed the reflection subgroup for $n$ such that the subgroup is finitely generated. The full group is then generated by the reflection subgroup together with the symmetries of its fundamental domain, so in some sense he computed $O(n,1;Z)$. All that I'm pointing out is that these are actually generated by involutions. ",
"@Agol; What do you mean by \"up to the dimensions that Vinberg had computed them\"? Did Vinberg gave a list of generators of $O(n,1;\\mathbb{Z})$ for small $n$?",
"At one point, I checked that all of the groups $O(n,1;Z)$ are generated by rank one involutions (reflections through planes and points), up to the dimensions that Vinberg had computed them. In fact, the even unimodular Lorentzian lattice in dimension 25 is also generated by such involutions (Daniel Allcock checked this using properties of the Leech lattice and Conway groups). I don't see why this might not be true in all dimensions, but I would have no clue how to approach proving it. Are you interested in non-uniform lattices too? ",
"I suspect this is an open question, but I'm not aware of a reference explicitly stating it. "
] | 4 |
Science
| 0 |
244
|
mathoverflow
|
Does a symplectic group act on a tensor power of a spin representation?
|
$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}$More specifically, let $S_k$ be the spin representation of $\Spin(2k+1)$. Then is there are action of $\Sp(2r-2)$ on $\bigotimes^{2r}S_k$ which commutes with the action of $\Spin(2k+1)$?
I am not asking for a duality. That is, I do not require that each isotypic vector space is an irreducible representation of $\Sp(2r-2)$.
The motivation is that the space of invariant tensors in $\bigotimes^{2r}S_k$ looks like the irreducible representation of $\Sp(2r-2)$ with highest weight $k\omega_{r-1}$. For instance, for $k=1$, both vector spaces have dimension given by the Catalan numbers $C_r = \frac{1}{r+1}\binom{2r}{r}$. More generally, for arbitrary $k$ the dimension of both of these spaces is $$ \prod_{1\leq i \leq j \leq r-1} \frac{i+j+2k}{i+j} $$ which is the number of so-called "_$k$ -fans of nested Dyck paths of semilength $r$_" (see, e.g., Section 3.1.6 of [Federico Ardila - Algebraic and geometric methods in enumerative combinatorics](https://arxiv.org/abs/1409.2562)).
|
https://mathoverflow.net/questions/57244/does-a-symplectic-group-act-on-a-tensor-power-of-a-spin-representation
|
[
"co.combinatorics",
"rt.representation-theory",
"lie-groups",
"invariant-theory"
] | 17 | 2011-03-03T03:38:18 |
[
"(For an explanation of a similar, well-known numerical \"coincidence,\" but for $SL_n$ and $GL_m$, using Howe duality, see: sbseminar.wordpress.com/2007/08/10/the-ubiquity-of-howe-duality/…)",
"As mentioned in the bounty message, I would be interested in any algebraic explanation of the numerical coincidence of the dimensions of the spaces discussed in Bruce's question. Does some version of Howe duality apply here?"
] | 2 |
Science
| 0 |
245
|
mathoverflow
|
Antichains of Cardinals in ZF Without Choice...
|
With the Axiom of Choice, the cardinals form a nice linearly ordered "set". In the absence of the Axiom of Choice, the cardinals form a partially ordered "set". Broadly, I am wondering what properties these posets can have.
A specific question I am interested in is the following.
**Is there, for each (infinite) subset $S \subseteq \mathbb{N}$ containing $1$ and not containing $0$, a model of ZF in which there is a maximal antichain of cardinals of size $n$ if and only if $n \in S$?**
Though I have a mild interest in knowing how such a model would be constructed (assuming a positive answer), my primary interest is in knowing that it is (is not) possible. Hence, if such a result exists in the literature, a citation would be all that I ask for.
|
https://mathoverflow.net/questions/82972/antichains-of-cardinals-in-zf-without-choice
|
[
"lo.logic",
"set-theory",
"axiom-of-choice"
] | 17 | 2011-12-08T07:09:36 |
[
"Joel: My motivation for requiring infinite $S$ was the initial context: How complicated can the theory of cardinals in a model of $ZF$ (in the language of posets) be? That said, now my interest goes beyond the initial context, so I am curious about sets such as $\\{1,2\\}$. Thanks for elaborating! <p> For comparison, the theory of the cardinals of a model of $ZFC$ (as a poset or under addition) is decidable. <p> Note that, so the underlying universe is a set rather than a proper class, I am restricting the \"model\" of set theory to any downward closed subset of cardinals within the \"model\".",
"Asher, suppose that $\\{\\a\\}$ was a maximal antichain (again, I'm ignoring finite sets) then $a$ is comparable with its Hartog number and thus well orderable. So everyone below $a$ can be well ordered as well. It's clear that $W_{\\aleph_0}$ implies $1\\in S$; on the other hand, if $W_{\\aleph_0}$ does not hold, pick $a$ to be a D-finite set, every cardinal above it cannot be well ordered and thus incomparable with its Hartog number. Therefore no one above $a$ can be a maximal antichain of size $1$.",
"Asher, the connection is that I believe that whenever there is an antichain of size 2, then one can make larger finite antichains, but it wasn't clear to me how to make larger finite maximal antichains. But if the answer to my question was that it always held, then one could not attain $S=\\{1,2}$, for example. Meanwhile, you had asked about infinite $S$, but it isn't clear to me why you insist on that.",
"Joel: Naively, I would expect models of ZF with finite antichains having no finite maximal antichain extension. Is there a connection between this phenomena and the primary question?",
"Trevor: Thanks, I was not even certain that (nontrivial) finite maximal antichains were consistent with ZF. Asaf and Joel: Indeed, the only reason I required $0 \\not \\in S$ and $1 \\in S$ was to avoid such trivialities with the finite cardinals. Excepting them, why is $1 \\in S$ if and only if $W_{\\aleph_0}$ holds? Naively, it seems plausible to have a set an infinite set $S$ with no countable subset, with some other cardinal (with countable subsets and size $S$ subsets) comparable to all cardinals. ",
"Asaf, I don't think your embedding fact settles the question I raise in my comment, since perhaps the finite antichain can be extended to a maximal antichain that does not lie in the copy of the poset in the cardinals. ",
"@Joel: Sure, if you allow finite cardinals then $1\\in S$ is a requirement. I just don't think it is very interesting that way. Considering only infinite cardinals, on the other hand... :-) As for your second question, I'd not think so. Jech has a theorem [Axiom of Choice, Thm 11.1] which embeds every poset into the cardinals, take a poset with only infinite maximal antichains, then its embedding will produce a model in which there are finite antichains that cannot be extended into finite maximal antichains.",
"Do we have any reason to hope that every finite antichain of cardinals can be extended to a finite maximal antichain? ",
"Asaf, I don't think what you say is correct. We need $1\\in S$ always because, for example, {7} is a maximal antichain, since every set either has fewer than 7 elements or at least 7. Similarly for {0} or {k} for any finite k. ",
"To answer a very small part too, $1\\in S$ if and only if $W_{\\aleph_0}$ holds, that is every infinite set has a countable subset.",
"To answer a very small part of your question, $\\lbrace \\omega_1, \\mathbb{R} \\rbrace$ is a maximal antichain in all known natural models of the Axiom of Determinacy. See A.E. Caicedo and R. Ketchersid, \\emph{A trichotomy theorem in natural models of AD}, available online at sites.google.com/site/richardketchersid/home/research/adcf-final.pdf."
] | 11 |
Science
| 0 |
246
|
mathoverflow
|
"extended TQFT" versus "TQFT with defects"
|
There are two ways in which higher categories appear in topological field theory: in extended TFTs and in TFTs with defects. How are these appearances related?
According to the Atiyah-Segal axioms, a d-dimensional TFT is a symmetric monoidal functor
\begin{equation*}\text{Bord}_d\rightarrow\mathcal{C}\end{equation*}
where the target category $\mathcal{C}$ is a symmetric monoidal category, typically $\text{Vect}$. The objects of $\text{Bord}_d$ are closed $(d-1)$-manifolds and map to the vector spaces of $\text{Vect}$ under the TFT. The morphisms are d-dimensional bordisms of the closed $(d-1)$-manifolds and map to the linear maps of $\text{Vect}$.
This description can be "extended down" to define a d-dimensional n-extended TFT as a symmetric monoidal $n$-functor
\begin{equation*}\text{Bord}_d^n\rightarrow\mathcal{C}\end{equation*}
where $\mathcal{C}$ is now an symmetric monoidal $n$-category. The objects of $\text{Bord}_d^n$ are $(d-n)$-manifolds (for a "fully extended" TFT, _i.e._ $n=d$, the bordism objects are points) and map to the objects of $\mathcal{C}$, which can be thought of as $(n-1)$-categories. One should really specify $\mathcal{C}$ as some extension of $\text{Vect}$, but I am ignoring this technicality by thinking of the n-category of all (small) $(n-1)$-categories. More generally, $k$-morphisms are $(d-n+k)$-dimensional bordisms between $(d-n+k-1)$-manifolds and map to $(n+k-1)$-categories. There are also technicalities surrounding how one manages "manifolds with corners" in $\text{Bord}_d^n$, but allow me to gloss over them here.
Meanwhile, one uses higher categories to decorate manifolds with defects. In this picture, $k$-dimensional manifolds are decorated with $k$-dimensional (extended?) TFTs called "$k$-defects." Lower dimensional submanifolds can be decorated with "defects within defects." The $k$-defect assigned to a $k$-dimensional "boundary" between two $(k+1)$-dimensional regions, each with attached $(k+1)$-defects, amounts to a morphism of $(k+1)$-defects. Fusion of defects and sub-defects endows the set of $k$-defects with the structure of a $k$-category. (For details, see <http://arxiv.org/abs/1002.0385>.) If we again imagine the $k$-category of (small) $(k-1)$-categories, we can understand decoration as an assignment of $(k-1)$-categories to $k$-manifolds.
This construction of TFT with defects feels "upside down" compared to extended TFT: $k$-manifolds are decorated by $(k-1)$-categories and map to $(d-k-1)$-categories under the TFT functor. I realize that decoration is not a functor from a bordism category, but is it a functor in some other sense (from some category where higher degree morphisms are lower dimensional submanifolds)? Defects may have an interpretation as inserted operators (Wilson loops, surface operators, etc); can these two formalisms be combined to compute path integrals with operator insertions in an extended theory? In general, I am curious about the relation between how higher categories are used to define an extended TFT and how they are used to characterize TFTs with defects.
Here is a particular problem from <http://arxiv.org/abs/1309.1489>. Consider a fully extended 4D TFT. The TFT functor assigns $1$-categories to Riemann surfaces $\Sigma$. Meanwhile, in the defect description, 2D TFTs form a $2$-category. Restricting ourselves to a single object leaves us with a symmetric monoidal category of $1$-dimensional domain walls (essentially boundary conditions) of the 2D theory. Is this category of boundary conditions identified with the category assigned by the TFT functor (as is claimed)? It seems that this identification is only possible since $\dim\Sigma=2=\text{codim }\Sigma$ in dimension four and does not reflect a general relation between extended TFT and TFT with defects.
|
https://mathoverflow.net/questions/184454/extended-tqft-versus-tqft-with-defects
|
[
"mp.mathematical-physics",
"higher-category-theory",
"quantum-field-theory",
"extended-tqft"
] | 17 | 2014-10-14T19:07:01 |
[
"One approach to this is discussed in detail at arxiv.org/abs/1009.5025 . See sections 2 and 6, and the remark about defects near the start of section 6.7.",
"Perhaps the following question needs to be answered first: a defect theory (an object of the k-category of defects that decorates a k-fold) is itself a TFT. How is this object realized as a TFT? In particular, given a d-dim theory Bord->C, how does one characterize the (d-1)-dim TFTs on boundaries (or domain walls)? And how are these boundary theories related to the defect decoration?",
"@KevinWalker Following your hint, the upside-down can be turned rightside-up, and the degrees and dimensions work out; however, I am still unclear about how to identify the data of the TFT with the defect data. For a d-dim theory Bord->C, the defects on a k-fold M form a k-category, the objects (defect theories) of which might be interpreted as k-morphisms of C. Endomorphisms in C of a defect theory form a (n-k-1)-category of boundary conditions (or an (n-k)-category with one object) of the defect theory. Somehow this might be identified with the (n-k-1)-category assigned to M by the TFT.",
"Forget about TQFTs for a moment and consider an $n$-category $C$. To a $k$-morphism $x$ of $C$ we can, of course, associate the $k$-morphism $x$ itself. We can also associate to $x$ the $(n-k)$-category of endomorphisms of $x$ in $C$. The \"upside-down-ness\" you are noticing is closely related to this familiar example.",
"Yes, representations of cobordisms with singularities encode TFTs with boundaries and defects. For \"pre-quantum\" field theory this is discussed in sections 3.9.14.4 to 3.9.14.6 of arxiv.org/abs/1310.7930 (improved version in preparation at ncatlab.org/schreiber/show/Local+prequantum+field+theory). For quantization of this: arxiv.org/abs/1402.7041 . This is based on discussion with Domenico Fiorenza and Alessandro Valentino that recently appeared as arxiv.org/abs/1409.5723",
"I think you want the cobordism hypothesis with singularities (Theorem 4.3.11 in arxiv.org/abs/0905.0465)."
] | 6 |
Science
| 0 |
247
|
mathoverflow
|
The Grothendieck Ring of Higher Stacks
|
The Grothendieck ring of varieties is defined to be the free abelian group spanned by isomorphism classes of varieties modulo the _cut & paste_ (or scissor) relations, which say that $[X] = [U] + [Y]$, for any closed $Y \subset X$ and where $U$ is the open complement.
Now, we can equally speak of a Grothendieck ring of schemes, or algebraic spaces, and they all turn out to be isomorphic via the obvious inclusions. (for the scheme case one sees this by noticing that $[X] = [X_\text{red}] - 0$) There is also a Grothendieck ring of Artin stacks (everything here is of finite type over a field of characteristic zero, possibly algebraically closed, and with affine diagonal). This ring turns out to be a localisation of the previous one.
My question is: what happens if we include _higher_ stacks?
I think there is work by Toën on the Grothendieck ring for derived stacks, but here I'm only asking about higher (underived) stacks. Do we have a similar phenomenon to the case of varieties VS schemes? In the sense that $[X] = [\pi_0(X)] - 0$, where $\pi_0$ of a derived stack is its underived truncation?
|
https://mathoverflow.net/questions/121984/the-grothendieck-ring-of-higher-stacks
|
[
"ag.algebraic-geometry",
"stacks"
] | 17 | 2013-02-16T05:10:36 |
[
"are there any interesting phenomena you expect to happen? or just for fun?"
] | 1 |
Science
| 0 |
248
|
mathoverflow
|
Vector Bundles on the Moduli Stack of Elliptic Curves
|
As is well known, there is classification of line bundles on the moduli stack of elliptic curves over a nearly arbitrary base scheme in the paper _The Picard group of $M_{1,1}$_ by Fulton and Olsson: every line bundle is isomorphic to a tensor power of the line bundle of differentials $\omega$ and $\omega^{12}$ is trivial.
I am now interested in a similar classification scheme for higher dimensional vector bundles (i.e. etale-locally free quasi-coherent sheaves of finite rank). I am especially interested in the prime 3, so you may assume, 2 is inverted, or even that we work over $\mathbb{Z}_{(3)}$. I found really very little in the literature on these questions. I know only of two strategies to approach the topic:
1) I think, I can prove that every vector bundle $E$ on the moduli stack over $\mathbb{Z}$ localized at $p$ for $p>2$ is an extension of the form $0\to L \to E \to F$ where $L$ is a line bundle and $F$ a vector bundle of one dimension smaller than $E$ (this may hold also for $p=2$, but I haven't checked). In a paper of Tilman Bauer ([Computation of the homotopy of the spectrum tmf](http://arxiv.org/abs/math/0311328)) Ext groups of the so called Weierstraß Hopf algebroid are computed, which should amount to a computation of the Ext groups of the line bundles on the moduli stack of elliptic curves if one inverts $\Delta$. It follows than that every vector bundle on the moduli stack of elliptic curves over $\mathbb{Z}_{(p)}$ is isomorphic to a sum of line bundles for $p>3$ if I have not made a mistake. But for $p=3$, there are many non-trivial Ext groups and I did not manage to see which of the occuring vector bundles are isomorphic.
2) One can try to find explicit examples of a non-trivial higher dimensional vector bundles. A candidate was suggested to me by M. Rapoport: for every elliptic curve $E$ over a base scheme $S$ we have an universal extension of $E$ by a vector bundle. Take the Lie algebra of this extension and we get a canonical vector bundle over $S$. As explained in the book _Universal Extensions and One Dimensional Crystalline Cohomology_ by Mazur and Messing, this is isomorphic to the deRham cohomology of $E$. This vector bundle is an extension of $\omega$ and $\omega^{-1}$ and lies in a non-trivial Ext group. But I don't know how to show that this bundle is non-trivial.
I should add that I am more a topologist than an algebraic geometer and stand not really on firm ground in this topic. I would be thankful for any comment on the two strategies or anything else concerning a possible classification scheme.
|
https://mathoverflow.net/questions/20824/vector-bundles-on-the-moduli-stack-of-elliptic-curves
|
[
"ag.algebraic-geometry"
] | 17 | 2010-04-09T03:06:40 |
[
"The fact that vector bundles on $\\mathcal M_{1,1}$ split as sums of line bundles in characteristic larger than 3 can also be seen by the classical description of $\\mathcal M_{1,1}$ as an open substack of the weighted projective stack $\\mathbb P(4,6)$, coming from the Weierstrass form. Any locally free sheaf on $\\mathcal M_{1,1}$ extends to a reflexive sheaf on $\\mathbb P(4,6)$, which is locally free, because $\\mathbb P(4,6)$ is regular of dimension 1. It is not hard to prove that any locally free sheaf on a weighted projective stack $\\mathbb P(m,n)$ splits as a direct sum of line bundles.",
"@Cardano: Sure.",
"@Charles: Good point, I should have noticed the link with the non-liftability to char. 0 for the Hasse invariant when $p = 3$ (and its square when $p=2$). Thanks. ",
"@Tyler Lawson: you use the covering $\\mathcal{M}(4) \\to \\mathcal{M}$ and some flatness criterion. I can send you a pdf file with some details if you like. ",
"You can also deduce that $Ext^1(O, \\omega^2)$ is non-trivial from the fact that there are non-trivial \"mod p\" modular forms of weight 2 when p=2 or 3; these are computed (for instance) in Deligne's note on \"formulas, after Tate\" in Antwerp 4.",
"@Brian: The computation of the Ext-groups of tensor powers of $\\omega$ on this moduli stack is written up in the Bauer paper that was linked to under part (1).",
"Since the coarse space is affine and the automorphism groups of the geometric points have order a power of 2 times a power of 3, and the Ext-group (or $\\omega^{-1}$ by $\\omega$, to be precise) is identified with degree-1 coherent cohomology (of $\\omega^2$), its nontriviality is also not visible over $\\mathbf{Z}[1/6]$. So can you also briefly indicate how you know it is nontrivial?",
"How did you prove that every vector bundle is an extension?"
] | 8 |
Science
| 0 |
249
|
mathoverflow
|
Combinatorial identity involving the Coxeter numbers of root systems
|
The setup is:
$R$ = irreducible (reduced) root system;
$D$ = connected Dynkin diagram of $R$, with nodes numbered $1,2,...,r$;
$\hat D$ = extended Dynkin diagram, nodes numbered $0,1,2,...,r$;
$\alpha_k$ = $k^{th}$ simple root ($1\le k\le r$); $\alpha_0$ = -(highest root);
label $n_k$ ($1\le k\le n$) = multiplicity of root k in the highest root; $n_0=1$;
$n^\vee_k$ = dual label = multiplicity of co-root $\alpha_k^\vee$ in the highest _short_ co-root;
$\quad\quad=n_k$ ($\alpha_k$ long) or $n_k/c$ ($\alpha_k$ short), where $c=2$ (types $B_n,C_n,F_4$) or $3$ (type $G_2$) (also $n_0^\vee=1$);
$h(D)$ = Coxeter number of D = $\sum_{k=0}^rn_k$ (for example $h(E_8)=30$);
Now, delete node $k$ from $\hat D$ where $n_k>1$. Write $\hat D-\\{k\\}$ as a union of connected Dynkin diagrams $D_1,\dots, D_s$. Let $h_1,\dots, h_s$ be the Coxeter numbers of $D_1,\dots, D_s$.
**Lemma** : If $\alpha_k$ is long:
$$ \sum_{i=1}^{s}\frac1{h_i}\sum_{j\in D_i}n_j=n_k $$ If $\alpha_k$ is short: $$ \sum_{i=1}^s\frac1{h_i}\sum_{j\in D_i}n_j^\vee=n_k^\vee $$
Note that the sum over $D_i$ involves the labels $n_j$ coming from $D$, which have no obvious relation to those coming from $D_i$. On the other hand $h_i$ is the Coxeter number for $D_i$, so this is a mix of data from $D$ and $\\{D_i\\}$.
The proof is case-by-case.
> **Question** : is there a conceptual proof of the Lemma?
**Examples:** (Bourbaki numbering)
1) $D=E_8$, $k=2$, $n_2=3$, $\hat D-\\{2\\}$ = type $A_8$, $s=1$, $h(A_8)=9$: $$ \frac19(2+4+6+5+4+3+2+1)=3=n_2 $$ (As mentioned above, the numbers $2,4,6,5,4,3,2,1$ come from $E_8$, not $A_8$.)
2) $D=E_8$, $k=4$, $n_4=6$, $\hat D-\\{6\\}=A1+A2+A5$, $s=3$, $h(A_i)=i+1$: $$ \frac12(3) + \frac13(2+4) + \frac16(5+4+3+2+1) = \frac32 + 2 + \frac{15}6 = 6 = n_4 $$ Note that the terms in the sum are not all integers.
3) $D=E_8$, $k=8$, $n_8=2$, $\hat D-\\{8\\}=E_7+A_1$, $s=2$, $h(E_7)=18$, $h(A_1)=2$: $$ \frac1{18}(2+3+4+6+5+4+3)+\frac12(1)=\frac{27}{18}+\frac12=2=n_8 $$
4) Here is an example involving $n_k^\vee$: $D=C_8$, $k=3$, $\alpha_k$ is short, $n_k=2$, $n^\vee_k=1$ (in fact all $n^\vee_j=1$). $D-\\{3\\}=C_3+C_5$, $h(C_n)=2n$: $$ \frac16(1+1+1) + \frac1{10}(1+1+1+1+1)=\frac12+\frac12=1=n_3^\vee $$
5) By popular demand, $G_2$:
a) $k=2$ (long root), $n_2=2$, $n_2^\vee=2$, $\hat D-\\{2\\}=A_1+A_1$. Note that $n_1=3,n_1^\vee=1$.
$$ \frac12(1)+\frac12(1)=1=n_1^\vee $$
b) $k=1$ (short root), $n_1=3$, $n_1^\vee=1$, $\hat D-\\{3\\}=A_2$, $h(A_2)=3$. Note that $n_2=n_2^\vee=2, n_0=n_0^\vee=1$: $$ \frac13(2+1)=1=n_1^\vee $$
**Remark:** Write $H$ for the subgroup of $G$ corresponding to $\hat D-\\{k\\}$. This identity plays a role in some computations related to embedding elliptic elements of the Weyl group of $H$ in those of $G$ (the Coxeter element is an example).
|
https://mathoverflow.net/questions/38992/combinatorial-identity-involving-the-coxeter-numbers-of-root-systems
|
[
"co.combinatorics",
"rt.representation-theory",
"lie-algebras",
"root-systems",
"weyl-group"
] | 17 | 2010-09-16T09:47:55 |
[
"@Jeff: One small nit-pick about your formulation is that the Coxeter number $h$ only depends on the Weyl group, not on $D$; for example, Lie types $B_r, C_r$ yield the same $h$. Coxeter originally found the concept in the framework of finite real groups generated by reflections, which is independent of Lie theory and root systems but of course has a lot of overlap. Your question, however, definitely comes from Lie theory.",
"Maybe a proof of this result can be furnished via Vinberg's classification of Dynkin diagrams in terms of the existence of a subadditive function? I think the coefficients $n_k$ you are describing are just the coefficients of the unique (up to scale) subadditive function...",
"This is an intriguing observation, which may not have a convincing explanation outside representation theory or the combinatorial geometry of affine Weyl groups. It might for example come from Langlands duality in some context not so visible in classical Lie theory (The tag weyl-group and/or lie-algebra might be appropriate. Also, it makes the dualization process clearer if you include the case $G_2$.) "
] | 3 |
Science
| 0 |
250
|
mathoverflow
|
How many Hecke operators span the Hecke algebra?
|
This is a generalisation of my [earlier question](https://mathoverflow.net/questions/42809/how-many-hecke-operators-span-the-level-1-hecke-algebra) about generators for the level 1 Hecke algebra.
Let $\Gamma$ be a congruence subgroup of $\operatorname{SL}_2(\mathbb{Z})$, and $k \ge 1$ an integer. There's an explicit bound $S$ (the "Sturm bound") known such that if $f \in M_k(\Gamma)$ (the space of modular forms of weight $k$ and level $\Gamma$) satisfies $a_n(f) = 0$ for $0 \le n \le S$, then $f = 0$, where $a_n(f)$ denote the coefficients of the $q$-expansion of $f$ (at the cusp $\infty$).
Since $M_k(\Gamma)$ is finite-dimensional and doesn't contain any nonzero constant functions, there must actually exist some $S'$ such that if $a_n(f) = 0$ for $1 \le n \le S'$ (but with no assumption on $a_0(f)$), then in fact $f = 0$. Can one give an effective upper bound on $S'$ (in terms of $k$ and the index $[\operatorname{SL}_2(\mathbb{Z}) : \Gamma]$)?
I'd be interested to know the answer to this even for the special case $\Gamma = \Gamma_0(N)$ or $\Gamma_1(N)$. For these groups it is equivalent to asking that the Hecke algebra acting on $M_k(\Gamma)$ is spanned by the operators $T_1, \dots, T_{S'}$ (hence the title of the question).
|
https://mathoverflow.net/questions/42815/how-many-hecke-operators-span-the-hecke-algebra
|
[
"nt.number-theory",
"modular-forms"
] | 17 | 2010-10-19T11:30:31 |
[
"Joel: see e.g. wstein.org/books/modform/modform/newforms.html#sturm-s-theorem.",
"David, do you know a good reference for sturm bounds?",
"A data point: for $\\Gamma_0(p)$ in weight 4, $p$ prime, the Sturm bound $S$ is $(p + 1)/3$, but the Eisenstein series $E_4(pz)$ has $a_n = 0$ for $1 \\le n < p$, so $S'$ must be at least $p$.",
"Julian, if you read the question you'll see that I've carefully explained why what I'm looking for is not the same as the Sturm bound.",
"Hmmm... isn't it called a \"Sturm bound\"?",
"I've decided to undelete this old question -- I did delete it because I wanted to set it as a student project, but the student didn't solve it.",
"@David: if you don't want anyone to answer the question you should delete it."
] | 7 |
Science
| 0 |
251
|
mathoverflow
|
monomorphisms and epimorphisms of local rings
|
I want to understand the structure of monomorphisms/epimorphisms in the category of local rings (with local homomorphisms), or dually in the category of local schemes. Let $LR$ denote this category.
* Every monomorphism is injective.
Proof: Let $R \to S$ be a monomorphism and $a,b \in R$ mapped to the same element. Regard $a,b$ as ring maps $R[t] \to R$. Let $\mathfrak{p},\mathfrak{q}$ the preimages of the maximal ideal $\mathfrak{m}$. Then we get two morphisms of local rings $R[t]_{\mathfrak{p}} \times_S R[t]_{\mathfrak{q}} \to R$ which agree when composed with $R \to S$, thus they are equal. Evaluating at $(t,t)$ gives $a=b$. There is a shorter proof which uses $R[t]_{(t)+\mathfrak{m}}$.
2. Is every monomorpism regular? (i.e. the equalizer of some morphism pair)
3. Is every monomorphism extremal? (i.e. $i=ep$ with $p$ epi implies $p$ iso)
4. Is every epimorphism, which is also injective, and isomorphism?
5. Is every epimorphism surjective?
Clearly we have 4 => 3 <=> 2 and 1 => 2. Since in $LR$ every morphism can be factorized as $ep$ where $p$ is surjective and $e$ is injective, we also have 3 => 4.
In the category of rings, 1,2,3,4 are false (see [this discussion](https://mathoverflow.net/questions/109/what-do-epimorphisms-of-commutative-rings-look-like)). However, the counterexamples don't carry over to $LR$. In the "category theory bible" (joy of cats!) you can find nothing about local rings. The questions are partially motivated by Anton's problem about [coequalizers](https://mathoverflow.net/questions/63/can-a-coequalizer-of-schemes-fail-to-be-surjective) in the category of (local) schemes.
In Lazards article about flat epimorphisms ([pdf link](http://archive.numdam.org/ARCHIVE/SAC/SAC_1967-1968__2_/SAC_1)), there is an example (1.6) in which 4 fails. I have not understood it yet. What is $Z$ there? Perhaps $B$ is a localization of $k[T,Z]$ and not $k[T]$? But I don't believe that $f$ is well-defined.
If the answers turn out to be "no":
* Is there a nice description for the regular monomorphisms? What about field extensions?
If $L/K$ is a finite galois extension, then $K \to L$ is a regular monomorphism iff $L/K$ is cyclic (see Brians comments).
|
https://mathoverflow.net/questions/24066/monomorphisms-and-epimorphisms-of-local-rings
|
[
"ac.commutative-algebra",
"ct.category-theory"
] | 17 | 2010-05-10T00:17:01 |
[
"Oh yes, this is the example. I'll take a look at Paul Balmer's explanation.",
"The link to Lazard's article seems to be broken; is this (mathoverflow.net/questions/19282/…) the counterexample you were referring to? If so, I don't see why $C$ maps to $D$ either.",
"@Martin: here's proof of non-existence when Gal. group G for K/k is non-cyclic. Suppose $s,t:K \\rightrightarrows A$ is equalizer pair for $k \\rightarrow K$ in category $LR$. Then $A$ is naturally a $K \\otimes_k K$-algebra, and by locality of $A$ and structure of $K \\otimes_k K$ (using $K/k$ is Galois), the algebra structure factors through projection to a factor field. Thus, $t=s \\circ g$ for some $g \\in G$. Hence, by univ. property of $A$ and inj. of $s$, any $f:B \\rightarrow K$ s.t. $g \\circ f=f$ factors through $k$. Thus, $K^g=k$, so $G=\\langle g \\rangle$. Contradiction.",
"I'm thinking about that, but it's pretty difficult to start. Even for nice small extensions ...",
"@Martin: sorry, for #1 I got disoriented by Spec and was thinking of co-equalizers. But how about a Galois extension whose Galois group doesn't admit a pair of generators? By restricting to local rings can you find an equalizer pair description for that?",
"If $K/k$ is a cyclic galois extension with generator $\\sigma$ of the galois group, then $k \\to K$ is regular since it is the equalizer of $id,\\sigma : K \\to K$.",
"1 fails for quadratic Galois ext'n of fields $K/k$ (due to non-locality of $K \\otimes_k K$ and how its isom with $K \\times K$ is defined). In category of local noetherian rings, map from ring to completion is an epimorphism (due to Krull intersection theorem, hence the noetherian condition), and in non-complete case provides counterexamples to 2, 3, 4. Doesn't show 2,3,4 fail in $LR$, but kills a reason to care, since affirmative results then seem useless. Category theory is like point-set topology: can get wrapped up in \"weird examples\"; not necessarily a good use of time. "
] | 7 |
Science
| 0 |
252
|
mathoverflow
|
Does every connected set that is not a line segment cross some dyadic square?
|
A _dyadic square_ is a subset of $R^2$ of the form $x + 2^{-n} [0,1]^2$ with $x \in 2^{-m} Z^2$, for integers $m,n \geq 0$. We say that a set $A$ _crosses_ a square $S$ if there exists a connected subset of $A \cap S$ which intersects two opposite sides of the square $S$. Clearly, the 45 degree line $\\{ (\pi + t, t) : t \in R \\}$ does not cross any dyadic square. Does every non-trivial closed, connected set that is not a line segment cross some dyadic square?
|
https://mathoverflow.net/questions/120415/does-every-connected-set-that-is-not-a-line-segment-cross-some-dyadic-square
|
[
"geometry",
"discrete-geometry",
"euclidean-geometry",
"gt.geometric-topology",
"mg.metric-geometry"
] | 17 | 2013-01-31T06:07:14 |
[
"Since you @mirko don't mind rephrasing-reposting the question I'd repost as a dimension/measure-theoretical question on curves in the plane, someone will spit some result at you really quick if you're convinced it's true. But I think it's false, you should be able to get good control of limiting behavior with a rectilinear construction and finish using basic results about direct limits of continua.",
"Really, some zero-span curve can't be constructed with just a uniform direct limit by oscillating off corners in Koch curve style with rectilinear paths by like a scaled factor of $> \\frac{1}{4^n}$? In my head it works, the argument should go like 'what's the biggest dyadic square it crosses' and then induct. Has anyone tried to write it down? In my head it works. If there is no example then a dimension theorist probably can handwave why, you should tag dimension theory and GMT.",
"@IanAgol No, I don't have a proof for the pseudoarc, though this is usually what I think of, as a totally path-disconnected continuum (i.e. path-components are points). If $C$ is say hereditarily indecomposable plane continuum I have considered (something elementary), take $\\frac1n$-neighborhood (union of all $\\frac1n$-balls with centers in $C$) this is path-connected, try to use it somehow (then let $n\\to\\infty$), but haven't had much success. Your comment reminded me that the pseudoarc is arc-like(chainable), so perhaps use this to modify the path-connected case, but I have to think about it",
"@Mirko: do you think that you can prove it for pseudo-arcs?",
"I believe I have a positive answer under the additional assumption that the set in question is path-connected. I posted a related question which eventually got a bit long, but see the definition of hsb plane continuum there, at the beginning, along with the Result after the \"Edit May 29-31, 2023\" here: mathoverflow.net/q/446092 . For completeness, here are a couple of other, related questions that I posted recently mathoverflow.net/q/446317 and math.stackexchange.com/q/4694709 . (I don't feel like posting my \"path-connected\" solution here, it is only a special case.)",
"Is it pure curiosity? In other words --- do you have some motivation for this question?",
"Hello Kevin, this is a nice question and seems difficult, though I came up with some ideas that I need to develop. Do you know if it is an open question, and if anyone has done work on it, any references? Did you find it somewhere in a book or a paper, or did you come up with it on your own? I am inclined to believe that there is no example, though I changed my mind about it several times yesterday and today. I will email a colleague whose area is closer to this question, to see if he knows anything or has any advice (just sending him the link to your question as posted here). Thank you,",
"Hi Gunter, I pointed out that there were two mistakes in this answer. A day later it was removed by somebody.",
"There used to be an interesting answer posted here. Where is it? Was it wrong?",
"Gunter Rote, I do not mean the condition $m\\leq n$. E.g. the square $[1/8,5/8]\\times [1/4,3/4]$ is a dyadic square with my definition. Perhaps you would like to call such a square a translated dyadic square.",
"I think that every path-connected component of such a non-crossing set should be a line segment.",
"I would have expected the condition $m\\le n$. Is that what you mean? Or is this another version of the problem? Every line segment that is not $\\pm45$ degrees, and every sufficiently smooth curve that is not a $\\pm45$ degree line segment must cross a dyadic square (even with the strong definition of dyadic square). So if there are counterexamples they are pretty pathological."
] | 12 |
Science
| 0 |
253
|
mathoverflow
|
Rational equivalence of smooth closed manifolds
|
All spaces below will be assumed simply connected. A continuous map is a _rational equivalence_ if it induces an isomorphism of the rational homology groups. Two spaces are _rationally equivalent_ if they can be connected by a sequence of rational equivalences (not necessarily all going in the same direction). Now suppose $X$ and $Y$ are rationally equivalent smooth compact manifolds without boundary. Can one choose a sequence of rational equivalences from $X$ to $Y$ that only passes through smooth compact manifolds of dimension $\dim X=\dim Y$?
Note: if $X$ and $Y$ are CW-complexes, or more generally, compactly generated Hausdorff spaces, then $X$ is rationally equivalent to $Y$ if and only if the $\mathbb{Q}$-localizations $X_\mathbb{Q}$ and $Y_\mathbb{Q}$ are homotopy equivalent, see e.g. Felix, Halperin, Thomas, Rational homotopy theory, Proposition 9.8. If this is the case, then there is a roof $X\to X_\mathbb{Q}\gets Y$ of rational equivalences. The middle of the roof is of course an infinite complex (except when it is a point).
|
https://mathoverflow.net/questions/429394/rational-equivalence-of-smooth-closed-manifolds
|
[
"at.algebraic-topology",
"gt.geometric-topology",
"smooth-manifolds",
"rational-homotopy-theory"
] | 16 | 2022-08-29T13:31:35 |
[
"To add to Denis T's comment, if the rational homotopy type in question is formal, then $X$ admits endomorphisms of arbitrarily divisible degree (see Infinitesimal Computations in Topology Theorem 12.2, and F. Manin's \"Positive weights and self-maps\" for a detailed discussion, Corollary 1.1 arxiv.org/abs/2108.02173v3) and so, upon precomposing with an appropriate endomorphism one can lift the map $X \\to X_{Q}$ through $Y \\to X_{Q}$ to get a direct rational equivalence $X \\to Y$.",
"...And SW classes are functorial along any maps which are $\\Bbb Z/2$ cohomology equivalence, so one can maybe just use them instead of more complicated things like cokernel of Hurewitz map for oriented bordism provided our cohomology rings are \"sturdy\" enough in sense I defined above.",
"Well, yeah, that's why I'm writing it as a comment and not an answer, but I think my non-example can be promoted. By methods of Manuel Amann I think one can produce \"sturdy\" PD rational types, i. e. ones without self-maps of degree $\\neq 1$ and retaining that property under connected sums. Then one can use some secondary obstructions (i. e. homology classes not realizable as manifolds) to distinguish two realisations of that type. Sturdiness will prevent positive degree things from killing our obstructions, which will forbid any nonzero degree maps in either direction.",
".. also, the tangent Stiefel-Whitney classes are preserved under genuine homotopy equivalences of manifolds, but not in general under rational ones.",
"Denis: this is a good example, but rational equivalences may have degree $\\neq 1$. For example, any map $S^n\\to S^n$ of degree $\\neq 0$ is a rational equivalence.",
"I think there's no way to connect $S^2 \\times S^3$ with the unique other 5-mfd with $H_2 = \\Bbb Z$ (double of nontrivial $D^3$-bundle over $S^2$) through degree 1 maps because one has trivial $w_2$ and other one does not. They are rationally equivalent, since they are both formal."
] | 6 |
Science
| 0 |
254
|
mathoverflow
|
Transcendence of sum of reciprocals of factorials
|
For $A \subseteq \mathbb{N}$, define $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$. It is easy to see that for every infinite $A$, $x_A$ is irrational.
**Question** : Is there an infinite $A \subseteq \mathbb{N}$ for which $\displaystyle x_A = \sum_{n \in A} \frac{1}{n!}$ is algebraic?
|
https://mathoverflow.net/questions/419523/transcendence-of-sum-of-reciprocals-of-factorials
|
[
"nt.number-theory",
"analytic-number-theory",
"transcendental-number-theory"
] | 16 | 2022-04-02T22:22:00 |
[
"At least we can prove that if $x_{A}$ is an algebraic number then $A$ is not a cofinite subset of $\\mathbb{N}$. :)",
"For other results along these lines, see Kumar and Vance and the references therein. I'm not familiar with them all but at first glance they all require faster growth rates than $n!$. And for an arbitrary $A$ it seems to me that there's not much you can say about $\\sum_{n\\in A} 1/n!$ other than the general rate at which the summands shrink.",
"I suspect the answer is no but that a proof is beyond current technology. The proof that $e$ is transcendental relies on special properties of the function $e^x$, not just on the rate of growth of $n!$. There are transcendence results that rely only on how fast the summands shrink but they require a faster growth rate than $n!$. For example, Nyblom proves that if for some fixed $\\lambda>2$ we have $\\liminf_{n\\to\\infty} a_{n+1}/a_n^{λ+1}>1$ then $\\sum_n 1/a_n$ is transcendental.",
"Exactly this class of numbers is proved irrational (and some examples given, which I think all happen to be transcendental) in Martin Griffiths, \"Irrational sums from reciprocals of factorials\", Math. Gaz. 2015, jstor.org/stable/24496963"
] | 4 |
Science
| 0 |
255
|
mathoverflow
|
Is "Escherian metamorphosis" always possible?
|
$\DeclareMathOperator\int{int}\DeclareMathOperator\diam{diam}\DeclareMathOperator\area{area}\DeclareMathOperator\cl{cl}\DeclareMathOperator\ran{ran}\DeclareMathOperator\dom{dom}$_This is a tweaked version of a question which was asked and bountied[at MSE](https://math.stackexchange.com/q/4555946/28111); that question still hasn't quite been answered, but suggestions in the comments convinced me that this was the "right" version to ask instead:_
This question is motivated by Escher's series of **Metamorphosis** woodcuts (see e.g. [here](https://www.escherinhetpaleis.nl/story-of-escher/metamorphosis-i-ii-iii/?lang=en)), where one tesselating tile is gradually transformed into another. Basically, this is a precise way of asking the following: can we always "metamorphose" between any pair of tesselating shapes?
First, some definitions (everything is in $\mathbb{R}^2$ with the usual metric):
* A _shape_ is a compact connected set $X$ with $X=\cl(\int(X))$.
* A _tiling_ is a pair $(\mathscr{S},F)$ where $\mathscr{S}$ is a finite set of shapes and $F$ is a set of functions such that
* each $f\in F$ is an isometric embedding of some $S\in\mathscr{S}$ into $\mathbb{R}^2$,
* $\bigcup_{f\in F}\ran(f)=\mathbb{R}^2$, and
* if $f,g\in F$ are distinct with domains $S,T$ respectively and $x\in \int(S)$, then $f(x)\not\in \ran(g)$ ("shapes only meet at their boundaries").
* A _tile_ is a shape $X$ such that $(\\{X\\},F)$ is a tiling for some $F$.
* $d_H$ is [Hausdorff distance](https://en.wikipedia.org/wiki/Hausdorff_distance). _(Note that we could replace $d_H$ with the modified version $d_{H*}(U,V)$ = the infimum over planar isometries $p$ of $d_H(p(U), V)$ without changing the question substantively; all that would change is that the number of tiles needed would shrink, but I'm not looking at that here.)_
Now with apologies to Escher, given tiles $A,B$ and $\epsilon>0$ let a **strong $\epsilon$-metamorphism** from $A$ to $B$ be a tiling $(\mathscr{S},F)$ such that
* if $f\in F$ and $(x,y)\in \ran(f)$ with $x<0$ then $\dom(f)=A$,
* there is some $N$ such that if $g\in F$ and $(x,y)\in \ran(g)$ with $x>N$ then $\dom(g)=B$ (call the least such $N$ the _length_ of the $\epsilon$-metamorphism), and
* if $f,g\in F$ and $\ran(f)\cap \ran(g)\not=0$ then $$d_H(\dom(f), \dom(g))<\epsilon\cdot {\area_f+\area_g\over 2},$$ where $\area_j$ is the area of the domain of $j$ for $j\in F$. (This "scaling" of the Hausdorff distance means that we can't trivialize things by shrinking. This should have been included in the original problem, but I forgot it.)
Finally, say that an **$\epsilon$ -metamorphism** is as above but with the final inequality having right hand side merely $\epsilon$.
> **Question 1** : Is there always a strong $\epsilon$-metamorphism between $A$ and $B$ for any tiles $A,B$ and any positive $\epsilon$?
I recall seeing a theorem that the answer is **yes** , and that moreover we can always find metamorphisms with length "close to" the naive guess $${\max\\{\diam(A),\diam(B)\\}\cdot d_H(A,B)\over \epsilon},$$ but I haven't been able to track it down or prove it myself. I'm also interested in whether the situation changes as we appropriately tweak things to work in $\mathbb{R}^n$ for $n>2$.
> **Question 2** : Assuming for the moment that $\epsilon$-metamorphisms always exist (which seems plausible based on the comments at the MSE version of this question), what can we say about the length function $$\mathscr{l}_{A,B}: \epsilon\mapsto\inf\\{N: \mbox{an $\epsilon$-metamorphism from $A$ to $B$ with length $N$ exists}\\}$$ for a pair of tiles $A,B$? In particular, what is $\mathscr{l}_{\mbox{unitsquare, unithexagon}}$?
|
https://mathoverflow.net/questions/433479/is-escherian-metamorphosis-always-possible
|
[
"mg.metric-geometry",
"euclidean-geometry",
"plane-geometry",
"tiling"
] | 16 | 2022-10-29T10:38:49 |
[
"@MattF. Metamorphosis II has a portion that goes squares-lizards-hexagons, I believe.",
"Which of the Escher images do you find most suggestive for an answer about unit squares and unit hexagons?",
"Interesting question!"
] | 3 |
Science
| 0 |
256
|
math
|
A proof of $\dim(R[T])=\dim(R)+1$ without prime ideals?
|
**Background.** If $R$ is a commutative ring, it is easy to prove $\dim(R[T]) \geq \dim(R)+1$, where $\dim$ denotes the Krull dimension. If $R$ is Noetherian, we have equality. Every proof of this fact I'm aware of uses quite a bit of commutative algebra and non-trivial theorems such as Krull's intersection theorem. It is worth mentioning that Gelfand-Kirillov dimension satisfies $\mathrm{GK}\dim(R[T])=\mathrm{GK}\dim(R)+1$ for every $K$-algebra $R$.
T. Coquand and H. Lombardi have found a surprisingly elementary, first-order characterization of the Krull dimension that does not use prime ideals at all.
T. Coquand, H. Lombardi, A Short Proof for the Krull Dimension of a Polynomial Ring, The American Mathematical Monthly, Vol. 112, No. 9 (Nov., 2005), pp. 826-829 (4 pages)
You can read the article here. Here is a summary.
For $x \in R$ let $R_{\{x\}}$ denote the localization of $R$ at the multiplicative subset $x^{\mathbb{N}} (1+xR) \subseteq R$. Then we have
$$\qquad \dim(R) = \sup_{x \in R} \left(\dim(R_{\{x\}})+1\right)\!. \label{1}\tag{$\ast$}$$
It follows that for $k \in \mathbb{N}$ we have $\dim(R) \leq k$ if and only if for all $x_0,\dotsc,x_k \in R$ there are $a_0,\dotsc,a_k \in R$ and $m_0,\ldots,m_k \in \mathbb{N}$ such that
$$x_0^{m_0} (\cdots ( x_k^{m_k} (1+a_k x_k)+\cdots)+a_0 x_0)=0.$$
You can use this to define the Krull dimension.
This new characterization of $\dim(R) \leq k$ can be seen as a statement in first-order logic (whereas the usual definition with prime ideals uses second-order logic): Use two sorts $N,R$, the usual ring operations for $R$, but also the "mixed" operation $N \times R \to R$, $(n,x) \mapsto x^n$. Every ring becomes a model of this language.
A consequence of this is a new short proof of $\dim(K[x_1,\dotsc,x_n])=n$, where $K$ is a field. Using Noether normalization and the fact that integral extensions don't change the dimension, it follows that $\dim(R\otimes_K S)=\dim(R)+\dim(S)$ if $R,S$ are finitely generated commutative $K$-algebras. In particular $\dim(R[T])=\dim(R)+1$. This could be useful for introductory courses on algebraic geometry which don't want to waste too much time with dimension theory.
Question. Can we use the characterization \eqref{1} of the Krull dimension by Coquand-Lombardi to prove the formula $$\dim(R[T])=\dim(R)+1$$ for Noetherian commutative rings $R$?
Such a proof should not use the prime ideal characterization/definition of the Krull dimension. Notice that the claim is equivalent to $\dim(R[T]_{\{f\}}) \leq \dim(R)$ for all $f \in R[T]$. I suspect that this can only work if we find a first-order property of commutative rings (with powers) which is satisfied in particular by Noetherian rings and prove the formula for these rings.
Please read this first before answering. This question is only concerned with a proof of the dimension formula using the Coquand-Lombardi characterization. If you post an answer that doesn't mention the characterization, then it's not an answer to my question and therefore offtopic. As of writing this, 20 answers have been posted, all of which have been deleted.
|
https://math.stackexchange.com/questions/358423
|
[
"ring-theory",
"commutative-algebra",
"noetherian",
"krull-dimension",
"dimension-theory-algebra"
] | 873 | 2013-04-11T08:19:32 |
[
"@Alexey Back then I removed the remark about first-order because of your comment. But I have just added this back, because we can just change the language to make it first-order. I have added an explanation for this.",
"Why this question has no answer or answers were deleted?",
"Currently 18 answers, all deleted.",
"This may not be a direct answer to your question, but it is relevant to the problem and too long to share in the comments. You can use valuation overrings and valuative dimension instead of prime ideals and Krull dimension, respectively. Polynomial rings are well-behaved with respect to valuative dimension, and Jaffard rings are the natural context in which to study the condition you seek. All Noetherian rings and all Prufer rings are Jaffard, for example, and the class of Jaffard rings is much larger than that of the Noetherian rings. See, for example, reader.elsevier.com/reader/sd/pi",
"@WernerGermánBusch Of course $T$ is just one variable.",
"@moderators: Please make this question CW.",
"This question has been referred to at math.meta.stackexchange.com/questions/33453/…",
"@Malkoun en.wikipedia.org/wiki/First-order_logic",
"@Foobanana I know them. There is no such proof.",
"This might not be too helpful, but have you tried searching through Gathmann's old notes on algebraic geometry? We're going through his notes in my algebraic geometry this year and he gives a lot of alternative proofs to commutative algebra facts :D",
"Might I ask why the dimension definition using prime ideals isn't elementary enough? The proof isn't short, but it isn't terrible to see why the dimension increases by 1. Is there a reason this other construction is the one you'd prefer to use?",
"For my own education, what do you mean when you write \"first order property\"? What is the definition of a first order property?",
"I guess T is a set of polynomial variables, like $T= \\{x\\}$ gets you $R[x]$ polynomials in $x$ with coefficients in the ring $R$, $T= \\{x_1, x_2, ... , x_n\\}$ gets you the polynomials in $n$ variables, and so on. I'm not sure if OP specificies T to be numerable or even finite. +1 means just that, +1, the dimension is an integer number, so it makes sense to add 1 to it.",
"If you can explain in a few words, what is T and what does it mean +1 in your equation?",
"bourbaki's proof of this theorem uses only general results, not more partiuclar results such as krull IT",
"MO copy of the question: mathoverflow.net/questions/172350/a-short-proof-for-dimrt-dimr1",
"Shall I ask this on mathoverflow?",
"Indeed. By the compactness theorem, it can be seen that ${\\rm dim}R\\le l$ is not a first order statement for any finite $l$.",
"This characterization of $\\dim R\\le l$ does not look like first-order to me, at least in the language of rings: it uses quantifiers over $\\mathbb N$ and exponentiation with the exponent as an argument, which is not a ring operation.",
"Yes. But maybe there is a first-order characterization for noetherian rings, like the one for Krull dimension? Or maybe some weaker property already suffices? I don't know.",
"At first sight I can't see many hopes to do this as long as the characterization of the Krull dimension in that paper deals with elements instead of ideals, while the property of being noetherian relies on some properties of ideals. But who knows?"
] | 0 |
Science
| 1 |
257
|
math
|
Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
|
Let $(X,\tau), (Y,\sigma)$ be two topological spaces. We say that a map $f: \mathcal{P}(X)\to \mathcal{P}(Y)$ between their power sets is connected if for every $S\subset X$ connected, $f(S)\subset Y$ is connected.
Question: Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?
Full disclosure: I've now crossposted the question on MO.
Various remarks
If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = \sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $\mathbb{R}$: a bijection that preserves connectedness on $\mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:
Assume $f:\mathbb{R}^n\to\mathbb{R}^n$ is a bijection, where $\mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $\mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $\mathbb{R}^n$ with the standard topology to itself, self-maps of some arbitrary topological space, it is easy to make the answer go either way.
For example, if the topological space $(X,\tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by the cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = \mathbb{Z}$ be equipped with the topology generated by:
$$ \{\mathbb{N}\} \cup \{ \{k\} \mid k \in \mathbb{Z} \setminus \mathbb{N} \} $$
Then the map $k \mapsto k+\ell$ for any $\ell > 0$ maps connected sets to connected sets, but its inverse $k\mapsto k-\ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.
|
https://math.stackexchange.com/questions/952466
|
[
"general-topology",
"metric-spaces",
"examples-counterexamples",
"connectedness"
] | 632 | 2014-09-30T05:18:04 |
[
"@HelloWorld: thank you for the reminder. The cash prize is now posted on X.",
"Will there be a cash price in twelve days? ;)",
"Two dozen answers, all deleted.",
"@Carlyle: nothing immediate comes to mind. You can have highly discontinuous functions whose graph still remain connected.",
"How does this definition of a connected function relate, if at all, to the property of the graph of a function (in the product Topology) being connected?",
"@FurdzikZbignew: the question is asking about mappings from $\\mathbb{R}^n$ to itself, not from $\\mathbb{R}^n\\to\\mathbb{R}^m$; I am not sure how your observation is relevant?",
"A counterexample could be a space-filling Peano curve, which is a bijection from the unit interval to the unit square but does not preserve dimensions/connectedness. Therefore, the answer to the question is \"no\" for dimensions $n>1$. The total order on $\\mathbb{R}$ was crucial in the proof for $n=1$, but does not generalize to higher dimensions. Additional regularity like continuity would be needed.",
"@user65023: the function $f$ maps $(s,t) \\mapsto (s, t + \\sin(\\pi/s))$. And $J$ maps $(s,t) \\mapsto (-s,t)$. You want to show $J\\circ f \\circ J \\circ f$ is the identity, so just write it out $$ (s,t) \\mapsto (s,t + \\sin(\\pi/s)) \\mapsto (-s, t + \\sin(\\pi/s)) \\mapsto (-s, t + \\sin(\\pi/s) + \\sin(\\pi / (-s))) = (-s,t) \\mapsto (s,t) $$",
"@AlanU.Kennington: It seems like the example you gave above has the following properties: (1) $f$ is connected, (2) $f$ is discontinuous, (3) $f^{-1}$ is not connected, (4) $f$ is a bijection. $f(s,t) = (s,t+\\sin{({\\pi}/{s})})$ for $s\\neq 0$ and $f(0,t)=(0,t)$.",
"@PietroMajer: Do you have a proof of why AlanU.Kennington's example satisfies $f^{-1}=J\\circ f \\circ J$? I'm not seeing it. Also, it looks like for AlanU.Kennington's example $f^{-1}$ is not connected. I think $f^{-1}$ maps the connected set: $\\{(s,sin(\\frac{\\pi}{s})): s>0\\} \\cup \\{(0,t): 0 < t < 1\\}$ to the unconnected set: $\\{ (s,0): s>0\\} \\cup \\{ (0,t): 0 < t < 1\\}$. To see this set is unconnected, it is covered respectively by two open half planes: $\\{ (s,t): t < s\\}$ and $\\{ (s,t): t > s\\}$.",
"@WillieWong: Why do you think this would be embarrassing? Almost all of important branches of mathematics have been born just by such questions. For example think about Graph theory. Or think about one of my favorites unsolved problem in math i.e. square peg problem! a simple conjecture but so beautiful one. :). My suggestion is to offer this challenge to undergraduate and graduate students first. then history will find its way.",
"@C.F.G It would be embarrassing if, 30 years later, this question becomes my one legacy in mathematics. :-) That said, I'll perhaps save that (announcement of a cash prize) for the tenth anniversary of this post (which is only 1.5 years away).",
"@WillieWong: You Should (as the author of this problem :) ) offer a prize for any progress on this problem.",
"To be thorough: the statement is also true for $n = 0$. :-)",
"@AlexHeuman your $f$ is not a bijection. $f(x_1, x_2, \\ldots) = f(x_1, x_2 + 2\\pi, x_3,x_4 ,\\ldots)$.",
"@GerryMyerson: Now 17 answers, all deleted.",
"@WillieWong might it be possible to send borromian rings in $\\Bbb R^3$ to trefoil knots? I can see the \"rope\" would need to be spliced into possibly three strands, with twists in such a way that it continuously morphs between knots and rings. I've no idea if this can hang together rigorously.",
"Ok I think I see the issue. The piecewise assignment is a priori and causes us to voluntarily disconnect the domain. So the pre-image and the image are both disconnected. But the problem asks for a bijection with a connected pre-image but disconnected image, for example.",
"For n=1 with f(x) = x, if x is rational; and x+1, if x is irrational, f is a bijection that is nowhere continuous. The usual topology on R begins its life connected. What is the definition of connected that lets us discard this f and conclude n > 1 for R^n?",
"@GerryMyerson: though 6 of them are from the same user....",
"Currently, 12 answers have been posted – all deleted!",
"@MiloBrandt: for $S^2$ your suspicion is sound. See this answer on the MathOverflow version of this question. Thanks!",
"I suspect that there is no such map in the analogous problem for $S^2$ instead of $\\mathbb R^2$; the reason being that if we have a separating set $X$ (i.e. one such that $S^2\\setminus X$ is disconnected), then its preimage has to be a separating set as well. Separating sets are kinda circle-like and I suspect have properties like \"there are connected subsets $A,B,C\\subseteq X$, no pair disjoint, such that $A\\cap B\\cap C=\\emptyset$\". Given this lemma or a similar ones, one can show the non-existence of such a forward-but-not-backwards connected map $S^2\\rightarrow S^2$.",
"@WillieWong I think I can prove this if I succeed to prove that f maps closed connected sets onto closed connected sets.",
"@RobertFrost: this particular question I am asking about is turning out to be very subtle. This particular answer of the cross post to MathOverflow shows how to answer the question if we change some small number of conditions. But the original question is still, as far as I know, open.",
"@WillieWong I'm not competent with the subtleties of the concept of continuity but this reminds me of a function I'm doing some work with. The branch cuts of its inverse are all superimposed on top of each other so for any $x+\\epsilon$ there are $f^{-1}(x+\\epsilon)$ not necessarily nearby to $f^{-1}(x)$ but I'm not totally clear how the concept of continuity treats this and whether it would fit your conditions.",
"@RobertFrost: mostly for my curiosity. But I note that the answer to a related question is known, and that question is what inspired me to ask this one.",
"@RobertFrost: you are half right. See this previous comment",
"@AndréLevy: Can you justify that the bijection is connected from the 2 spheres to the 1 sphere? Each of the paradoxical pieces is dense in the sphere, this suggests that the corresponding bijection cannot be connected. (Let $P,Q\\subset \\mathbb{S}^2$ be two disjoint dense sets. And let $\\rho$ be a non-trivial rotation. Then for any $\\epsilon > 0$ there exists $p\\in P$ and $q\\in Q$ such that $d(p,q) < \\epsilon d(p,\\rho(q))$.)",
"Well, @WillieWong, isn't the disassembling and reassembling of the pieces in B-T a bijection, isn't this bijection connected from the 2 spheres to the 1 sphere, but not the reverse?"
] | 0 |
Science
| 1 |
258
|
math
|
Does every ring of integers sit inside a ring of integers that has a power basis?
|
Given a finite extension of the rationals, $K$, we know that $K=\mathbb{Q}[\alpha]$ by the primitive element theorem, so every $x \in K$ has the form
$$x = a_0 + a_1 \alpha + \cdots + a_n \alpha^n,$$
with $a_i \in \mathbb{Q}$.
However, the ring of integers, $\mathcal{O}_K$, of $K$ need not have a basis over $\mathbb{Z}$ which consists of $1$ and powers of a single element (a power basis). In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.
Question: Can every ring of integers $\mathcal{O}_K$ that does not have a power basis be extended to a ring of integers $\mathcal{O}_L$ which does have a power basis, for some finite $L/K$?
|
https://math.stackexchange.com/questions/1754860
|
[
"abstract-algebra",
"number-theory",
"ring-theory",
"algebraic-number-theory"
] | 185 | 2016-04-22T16:54:33 |
[
"The difficulty in finding a general approach to construct such an extension arises from the fact that the structure of the ring of integers can be quite complicated, and power bases may not always exist for every number field.",
"@Jacob Wakem, if you take a field extension, let's say $\\mathbb{Q}(\\sqrt{2})/\\mathbb{Q}$, then the ring of integers (of $\\mathbb{Q}(\\sqrt{2})$) refers in some sense to the 'analogon' of what $\\mathbb{Z}$ is to $\\mathbb{Q}$. In this case the ring of integers would be $\\mathbb{Z}[\\sqrt{2}]$. But I hope my comment is not misleading, because if $\\mathbb{Q}(\\alpha)/\\mathbb{Q}$ is a field extension, then the ring of integers of $\\mathbb{Q}(\\alpha)$ is not necessarily(!) $\\mathbb{Z}[\\alpha]$.",
"What I should say is this: The number of generators of a ring of integers, as an algebra, can be 1 (monogenic) or arbitrarily large.",
"I honestly don't understand your remark \"In fact, there exist number fields which require an arbitrarily large number of elements to form such a basis.\": rings of integers are $\\mathbb Z$-modules, so all their $\\mathbb Z$-bases have the same number of elements (namely the degree of the corresponding field extension).",
"What is a ring of integers? Merely a ring containing only integers?",
"Note a similar result holds for the ordinals. See: en.wikipedia.org/wiki/Ordinal_arithmetic#Cantor_normal_form",
"@Gro-Tsen and sadly it's not had much luck there either!",
"This has now been asked on MathOverflow",
"I think this question should really be reposted on MathOverflow. Also, before tackling the number field case, it's probably worth thinking about curves over a field: given a base curve $C$ and an affine open set $U$, say that $C_1\\to C$ (ramified covering) is \"monogenic\" for these data when there is $f$ regular on $U_1:=U\\times_C C_1$ such that $\\mathcal{O}(U_1)=\\mathcal{O}(U)[f]$. Is it true that there is always $C_2\\to C_1$ that is monogenic? (Maybe this is stupidly true or false, I'm not sure. But it could be easier to think about.)",
"Just a remark on terminology: If $\\mathcal O_K=\\mathbb Z[\\alpha]$, then $\\mathcal O_K$ is sometimes called monogenic.",
"Every $abelian$ number field $K$ sits inside $\\mathbb Q(\\zeta)$ for some root of unity $\\zeta$ and has a ring of integers $\\mathbb Z[\\zeta]$. So your question is answered in the positive for abelian $K/\\mathbb Q$."
] | 0 |
Science
| 1 |
259
|
math
|
If polynomials are almost surjective over a field, is the field algebraically closed?
|
Let $K$ be a field. Say that polynomials are almost surjective over $K$ if for any nonconstant polynomial $f(x)\in K[x]$, the image of the map $f:K\to K$ contains all but finitely many points of $K$. That is, for all but finitely many $a\in K$, $f(x)-a$ has a root.
Clearly polynomials are almost surjective over any finite field, or over any algebraically closed field. My question is whether the converse holds. That is:
If $K$ is an infinite field and polynomials are almost surjective over $K$, must $K$ be algebraically closed?
(This answer to a similar question gave a simple proof that $\mathbb{C}$ is algebraically closed from the fact that polynomials are almost surjective over $\mathbb{C}$. However, this proof made heavy use of special properties of $\mathbb{C}$ such as its topology, so it does not generalize to arbitrary fields.)
|
https://math.stackexchange.com/questions/1792464
|
[
"abstract-algebra",
"polynomials",
"field-theory"
] | 142 | 2016-05-19T19:06:55 |
[
"I guess the problem with the original reasoning is that it relies on the fact that the image of $\\mathbb{C}$ has to be simply connected, but this is hard to use for abitrary fields, the only topology you could use would be zariski but knowing zariski would give you info about the algebraic closeness. Maybe studying homology would give some results? The only nice thing is that being infinite gives you irreducibility",
"I do not know whether OP knew this but this was conjectured by Reinke in their 1975 paper \"Minimale Gruppen\", and as far as I know is still open. A similar conjecture in model theory is the so-called \"minimal fields conjecture\" of Podewski, asserting that infinite \"minimal\" fields are algebraically closed (where \"minimal\" means that every field formula with one free variable either defines a finite subset of the field or a cofinite one). Wagner solves the latter in his paper \"Minimal fields\" in positive characteristic, where he also solves a weaker variant of OP's question.",
"@student91: That's why I restrict $K$ to be infinite in the precise statement of the question.",
"I am slightly confused. If $K$ is finite, then any polynomial is \"almost surjective\" in this sense, because there are only finitely many points that are not in the image of $f$ (even if the image of $f$ consists of only one point). Furthermore, finite fields are not algebraically closed. Do you assume your characteristic to be 0?",
"Ooh you are right. I thought it was strange to have such a regular topology for arbitrary fields without being known.",
"@AndreaMarino: I think it is extremely unlikely that you can get such topology at the end of an infinite chain. Local compactness is also not enough (then you could have the discrete topology!), and you need $T_2$, not just $T_0$, since you need to know the limit of $y_n$ is unique to conclude $p(x)=s$. Note then that no countably infinite compact $T_2$ space is homogeneous, so you can never have such a topology on a countably infinite field.",
"Sorry for double comments but I can't edit. It is enough for the topology to be locally compact, so that we recover the cases of $\\mathbb{Q}, \\mathbb{C}$.",
"How I would build such a topology: start from $\\mathbb{Q}$ and $\\mathbb{F}_p$, then show that the topology extends by transcendent, algebraic extensions and by infinite chains. By Zorn, this gives the topology on whatever field. Don't know if this route map can work!!",
"Here it is my 1 cent also. Suppose there exist a $T_0$, compact and first countable topology on K such that polynomials are continuous. There are no isolated points x, otherwise $x+k$ would be also isolated because polynomials are continuous, thus the topology is discrete. But being compact it should be finite. Now take a polynomial p with non taken values $s_1,\\ldots, s_k$. Take $y_n$ in the $U_n $ (given by first countability) of $s_1$ , WLOG $x_n$ different by $s_i$. Take $x_n$ st $p(x_n)=y_n$. The space is sequentially compact, so WLOG $x_n\\to x$. Then $p(x)=s$ and p is surjective.",
"Or wait, can your question be rephrased as 'If $K$ is an infinite field and polynomials are almost surjective over $K$, must polynomials be fully surjective over $K$?' ?",
"I find the concept of almost surjective polynomials a bit confusing. Can you give an example of an almost surjective polynomial over $\\mathbb{C}$ that is not surjective?",
"Here's my 5 cents. $K$ must contain an $n^{th}$ root for each of its elements (take $f:=X^{n}$; for $\\alpha\\in K^{*}$, $f$ cannot miss the infinitely many elements $\\alpha\\cdot\\beta^{n}$ with $\\beta\\in K^{*}$). At least when $char(K)=0$, the roots of unity are expressible as radicals. This implies that $K$ does not admit any finite Galois extensions with solvable Galois group. I.e., any minimal non-trivial Galois extension of $K$ must have a simple non-abelian Galois group.",
"I believe this is open. At least, it was posted on MO with no definitive answer: mathoverflow.net/questions/6820/…"
] | 0 |
Science
| 1 |
260
|
math
|
Probability for an $n\times n$ matrix to have only real eigenvalues
|
Let $A$ be an $n\times n$ random matrix where every entry is i.i.d. and uniformly distributed on $[0,1]$. What is the probability that $A$ has only real eigenvalues?
The answer cannot be $0$ or $1$, since the set of matrices with distinct real eigenvalues is open, and also the set with distinct, but not all real, eigenvalues is open (the matrices with repeated eigenvalues have measure zero).
I don't see any easy transformation that links the two sets, and working on the characteristic polynomial seems quite impractical. Also, I have the feeling that $[0,1]^{n^2}$ is not a good space to work in, due to its lack of rotational invariance.
|
https://math.stackexchange.com/questions/3770846
|
[
"linear-algebra",
"probability",
"matrices",
"eigenvalues-eigenvectors",
"random-matrices"
] | 135 | 2020-07-27T03:59:33 |
[
"@EXodd Ok. Then we can try this way for $n\\geq 4$ case. Let $f(x)=det(xI−A)$ be the characteristic polynomial. Let $a_1 >⋯>a_{n−1}$ be roots of $f′(x)$. Then the roots of f are all real if $f(a_1)<0,f(a_3)<0,…$ and $f(a2)>0,f(a4)>0,…$. For $n=4$, this seem to give the explicit algebraic relation. For $n>4$, I am not sure.",
"@J1U already for degree 4, \"If the coefficients are real numbers and the discriminant is positive, then the roots are either all real or all non-real\" so it cannot predict if the roots are all real or not",
"@Exodd See this: en.m.wikipedia.org/wiki/Discriminant For degree 3 and 4, the discriminant is explicitly computed. For even higher degrees, the discriminant is still expressed as the algebraic relation of the coefficients of the polynomial, although it would be messier.",
"@J1U for the 2x2 it is easy indeed (actually the proobability is 1) but for larger degrees how do you ensure the polynomial has only real solutions?",
"Why don’t you just use the characteristic polynomial? All the eigenvalues are real if $det(xI-A)$ only has real solution. For example, for 2-by-2, $(a_{11}-a_{22})^2+4a_{12}a_{21}>0$ if and only if all the eigenvalues are real. Thus the probability is the volume of the set of inequality intersection $[0,1]^4$.",
"The probability the matrix is symmetric is zero, but maybe one can argue that a certain class of “almost symmetric” matrices (via a perturbation) have real eigenvalues and with positive probability, to get a lower bound on the desired probability in question.",
"@AdamCataldo yes, if you only consider real matrices then the eigenvalues are very biased towards the real line. Maybe finding a bigger space where to work may not be a bad idea, but the choice of the biger space has to reflect/expand somehow consistently the properties of the spectra of real matrices",
"I think what I'm stuck on is measure space you're trying to work in, since you're suggesting in the problem it shouldn't be $[0, 1]^{n^2}$ (presumably with Lebesgue measure). It sounds like it's also not the eigenspace (the subset of $\\mathbb{C}^n$ that is projected on by the set of matrices in $[0, 1]^{n \\times n}$). Or do you think the projection into the eigenspace is such that the eigenvalues on the real line have non-zero measure?",
"@AdamCataldo if the eigenvalues of the matrices were \"continuously\" distributed in the unit circle, then you would be correct. But if you take real matrices, the distribution of their eigenvalues is biased on certain areas of the unit circle, and in particular they greatly concentrate on the real line, so much that even if $[-1,1]$ is negligible in the unit circle, the probability of having real eigenvalues is not zero",
"Not sure I follow this, because what I'm saying is that in the unit circle in the complex plane, the real line, that is the real values between -1 and 1, has measure zero. It does in fact have an empty interior, because if you draw a ball of any radius around any of the points on the real line, the ball will include complex numbers, not just real values.",
"@AdamCataldo the probability Cannot be zero, since the set of all-real eigenvalues matrices has non-empty interior part. Your reasoning holds if you consider complex-entries matrices, while here we have only real entries",
"My intuition is that the probability is zero, and that intuition is based on a slightly simpler problem: Consider the set of n x n matrices where all eigenvalues satisfy $|\\lambda| < 1$. The subset of these matrices where all eigenvalues is real is a set of subset of measure zero relative to the entire set in eigenspace. I would expect something similar if you change the constraint from the one I suggested to the original problem, but I'm not sure what the mapping from n x n matrices with entries all in [0, 1] \"looks like\" in eigenspace, so that's where my intuition might fall apart.",
"Since this question is getting a lot of attention, I ask you to reformulate it. Every square matrix has at least one complex eigenvalue, since every polynomial of degree $\\geq 1$ has at least one complex root (note that real numbers are also complex numbers, by definition). What you ask is the probability for all eigenvalues to be real. Right?",
"just a quick thought. Assume $n=2$. If $A$ eigenvalue $\\lambda_2$ is complex, then $A-\\lambda_2 I $ cannot be drawn as a parallelogram in, say $XY$ plane i.e. its area (in $XY$ plane) does not exist. Since the number of eigenvalues increases as $n\\rightarrow \\infty$, it feels for sure that at least for one of them, say $\\lambda_n$, the $n$-dimenional area $det(A-\\lambda_n I)$ does not exist. Thus the sought probability is $0$.",
"@leonbloy (cont.) Then again, the \"large amount of independence\" from originally $n^2$ random values has to be \"squeezed\" into $(n-1)^2$ random values. So they are certainly independent indeed. Admittedly, this is still in a typical what could possible go wrong? scenario.",
"@leonbloy That is quite feasible: If we pick one (probably existing) real eigenvector and take it to $e_n$ via an orthonormal base change, the other entries in the resulting matrix are essentially linear combinations of a large number of iid uniform random variables, so by the law of big numbers are gaussian. Well, the independence is a bit hand-wavy as the original eigenvector depends on all entries and vice versa all entries in the new matrix are influenced by the eigenvector. (cont.)",
"This paper: shub.ccny.cuny.edu/articles/… says:Mehta [24, Conjectures 1.2.1 and 1.2.21 conjectures from extensive numerical experience that the statistical properties of matrices with independent identically distributed entries behave as if they were normally distributed as n -+ rn . Mehta focuses on the symmetric or Hermitian cases, but surely the idea is quite general.",
"with 10kk the results do not change much...",
"about one million (trusting Octave's random generator)",
"@leonbloy how many examples have you generated for the empirical evaluation?",
"It should be a coincidence, but the desired probability seems to be quite near the probability of an $n-1 \\times n-1$ random gaussian matrix having all real eigenvalues $$\\begin{array} n & & \\\\ 2 &1 & 1 \\\\ 3 &0.708 & 0.70711\\\\ 4 &0.346 & 0.35355\\\\ 5 &0.117 & 0.125\\\\ 6 & 0.028 & 0.03132\\\\ \\end{array} $$ Left column: empirical. Right column: from here: core.ac.uk/download/pdf/82140233.pdf",
"Empirically : $3$: $0.708$, $4$: $0.346$, $5$: $0.117$",
"@AdinaGoldberg We are on the set of real matrices, so the eigenvalues are either real or come in couples of conjugated complex numbers. If you have distinct real eigenvalues, then you can always find a small enough neighbourhood where the eigenvalues are not complex due to the continuity of the eigenvalues",
"I guess to clarify, I certainly agree that within the set of matrices with real eigenvalues, those with distinct eigenvalues is an open set. But that is relatively open, in the sense that $(0,1)$ is open in $\\mathbb{R}$ but not in $\\mathbb{R}^2$.",
"@Exodd Yes, indeed! But $\\mathbb{R}$ is not open as a subset of $\\mathbb{C}$. How can you complete your argument?",
"@AdinaGoldberg the eigenvalues of a matrix are the roots of the characteristic polynomial, and the roots are continuous functions of the coefficients",
"I'm curious why the set of matrices with distinct real eigenvalues is open (I'm assuming as a subset of the $n \\times n$ real matrices.)",
"Perhaps you could try to say something about the distribution of the discriminant of the characteristic polynomial",
"I'm not sure it will work, no. On a second look it seems like a key component of the argument for higher moments is that $\\sum_{j=1}^N(A^2)_{jj}=\\sum_{j,k=1}^N(A_{jk})^2$ which requires symmetric matrices.",
"@SamJaques they are dealing with symmetric matrices, and they work for big $n$.. Are you sure?"
] | 0 |
Science
| 1 |
261
|
math
|
A question about divisibility of sum of two consecutive primes
|
I was curious about the sum of two consecutive primes and after proving that the sum for the odd primes always has at least 3 prime divisors, I came up with this question:
Find the least natural number $k$ so that there will be only a finite
number of $2$ consecutive primes whose sum is divisible by $k$.
Although I couldn't go anywhere on finding $k$, I could prove the number isn't $1, 2, 3, 4$ or $6$, just with proving there are infinitely many primes $P_n$ so that $k\mid(P_n+P_{n+1})$ and $k$ is one of $1, 2, 3, 4, 6$:
The cases of $k=1$ and $k=2$ are trivial. The case of $k=3$, I prove as follows:
Suppose there are only a finite number of primes $P_k$ so that $3\mid(P_k+P_{k+1})$. We can conclude there exists the largest prime number $P_m$ so that $3\mid(P_m+P_{m-1})$ and thus, for every prime number $P_n$ where $n>m$, we know that $P_n+P_{n+1}$ is not divisible by $3$. We also know that for every prime number $p$ larger than $3$ we have: $p \equiv 1 \pmod 3$ or $p \equiv 2 \pmod 3$. According to this we can say for every natural number $n>m$, we have either $P_n \equiv P_{n+1} \equiv 1 \pmod 3$ or $P_n \equiv P_{n+1} \equiv 2 \pmod 3$, because otherwise, $3\mid(P_n+P_{n+1})$ which is not true. Now according to Dirichlet's Theorem, we do have infinitely many primes congruent to $2$ or $1$, mod $3$. So our case can't be true because of it.
We can prove the case of $k=4$ and $k=6$ with the exact same method, but I couldn't find any other method for proving the result for other $k$.
|
https://math.stackexchange.com/questions/527495
|
[
"number-theory",
"prime-numbers",
"prime-gaps"
] | 122 | 2013-10-15T11:52:59 |
[
"This is equivalent to: Find the least natural number $k$ so that there is no pair of $2$ consecutive primes whose sum is divisible by $k$. Let say $k$ is the least natural number such there is only a finite number of pair of $2$ consecutive primes whose sum is divisible by $k$, and that $p+q=kn$ is the largest of those sums. Then no sum of $2$ consecutive primes is divisible by $k\\left(n+1\\right)$ ...",
"Easily to check that each of the described sums for $n\\in[2,5000]\\;$ has one or more dividers among $\\pi(k),\\; k\\in[2, 999].$ That's why the required k should be too big.",
"OP, your first link goes to some weird Yahoo page. Please fix. Ty",
"@KeithBackman late response, but surely that can't be true: Schinzel's hypothesis H implies that for all $n$, there are infinitely many pairs of twin primes of the form $kn \\pm 1$. For example, $11$ divides $197+199$, $13$ divides $311+313$, $19$ divides $227+229$, etc.",
"@CODE: I'm just curious, isn't $k=2$ the answer?, of course this is except for $2 + 3$.",
"Anecdotally, there are a plethora of numbers that never divide the sum of twin primes, which are consecutive primes: $11,13,19,31,43, \\dots$. However, these numbers may well divide the sums of other consecutive primes.",
"The link to Yahoo Answers shut down today. Can you link elsewhere?",
"@RomainS It has, but no answer.",
"@CODE has this been crossposted to Math Overflow?",
"The question amounts to : does there exist a k, such that, only finitely many pairs of consecutive primes are additive inverses mod k.",
"My apologies. You are right.",
"@user254665 my reasoning for k=3 says that if the hypothesis is false, from some point on we would have all prime numbers are either congruent to 2 mod 3 or 1 mod 3, THAT can't be true according to Dirichlet's Theorem.",
"3+ years and no A. Maybe put it on MathOverflow and see what the pros say.",
"A particular case of Dirichlet's Theorem is that there are infinitely many primes congruent to 1 mod 3 and infintely many congruent to 2 mod 3. But Dirichlet's Theoerem does NOT say anything about CONSECUTIVE primes. Your reasoning for k=3 is invalid. ... However it is a good Q, and, I suspect, with no known A.",
"@ghosts not true, $12$ does not divide $103+107$, for instance.",
"You can make a kind of a prime counting function that counts the pairs of consecutive primes divisible by $k$ below $n$. Maybe call it $\\pi_k(n)$. After doing some computations it seems that $\\pi_k(n) \\sim c Li(n)$ for some constant $c$ which for a lot of $k$'s looks to be close to $1/\\phi(k)$",
"@CODE: Indeed, Schinzel's Hypothesis H implies it. In fact you don't even need that much: Dickson's conjecture suffices, with some cleverness.",
"I also doubt that this is going to be answered.why don't you try posting this to mathoverflow.net?",
"I think Schinzel's Hypothesis says it's true though.",
"Thanks, I myself think such a $k$ doesn't exist too, but proving it is another matter :P",
"this is a very nice question and it will seem weird if there exists such a $k$"
] | 0 |
Science
| 1 |
262
|
math
|
What is the largest volume of a polyhedron whose skeleton has total length 1? Is it the regular triangular prism?
|
Say that the perimeter of a polyhedron is the sum of its edge lengths. What is the maximum volume of a polyhedron with a unit perimeter?
A reasonable first guess would be the regular tetrahedron of side length $1/6$, with volume $\left(\frac16\right)^3\cdot\frac1{6\sqrt{2}}=\frac{\sqrt{2}}{2592}\approx 0.0005456$. However, the cube fares slightly better, at $\frac{1}{1728}\approx 0.0005787$.
After some experimentation, it seems that the triangular prism with all edges of length $1/9$ is optimal, at a volume of $\frac{\sqrt{3}}{2916}\approx0.00059398$. I can prove that this is optimal among all prisms (the Cartesian product of any polygon with an interval) and that there is no way to cut off a small corner from the shape and improve it.
Is the triangular prism the largest polyhedron with a fixed perimeter?
I can prove a weak upper bound of $\frac1{12\pi^2\sqrt{2}}\approx 0.00597$ on the volume of such a polyhedron, by combining the isoperimetric inequalities in both $2$ and $3$ dimensions (i.e., the fact that polygons cannot enclose more surface area than a circle and that a polyhedron cannot enclose more volume than a sphere of the same surface area) along with the observation that a single face of a polyhedron cannot take up the majority of its surface area. Note the number of leading zeros - this upper bound is a bit over $10$ times my lower bound!
Edit: A friend of mine has confirmed with Mathematica that no polyhedron with $5$ or fewer vertices, or anything combinatorially equivalent to the triangular prism, improves on this bound. (With some work, this approach might be extended to all polyhedra on at most $k$ vertices, for $k$ on the order of $7$ to $10$.)
|
https://math.stackexchange.com/questions/4044670
|
[
"geometry",
"optimization",
"volume",
"calculus-of-variations",
"polyhedra"
] | 112 | 2021-03-01T09:12:52 |
[
"The term \"skeleton\" in the title should be avoided because this term is used in Mathematical Morphology for something else, rather different from the meaning given here.",
"A good starting point would be to consider Gauss's Shoelace formula for volume en.wikipedia.org/wiki/Shoelace_formula#Generalization . Every polyhedron has a volume formulation which is the sum of tetrahedra radiating from a common centre (after triangulation of surface mesh).",
"This paper [Donald W. Grace, Search For Largest Polyhedra, MCOM, 1963] talks about the 8-vertex polyhedron formed by points on a unit sphere. From the article: \"The dual of Goldberg's conjecture can be loosely stated as follows: In the polyhedron of maximum volume formed by $n$ points on a sphere, the faces are triangular and the number of edges leading to any vertex is as nearly as possible equal to the average of such numbers.\" I thought this might be relevant. ams.org/journals/mcom/1963-17-082/S0025-5718-63-99183-X/…",
"Digging a bit on the Melzak conjecture literature, I found this paper: londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s2-6.2.382 (unfortunately not freely available), which gives the upper bound $V<\\frac{1}{432\\pi}\\approx 0.00073682$, which is pretty close to $0.00059$",
"This is essentially Melzak's Conjecture. ijgeometry.com/wp-content/uploads/2021/09/4.-63-73.pdf and scholar.rose-hulman.edu/cgi/…",
"@RavenclawPrefect Just like with cotangent area formula for regular polygons based on the amounts of sides, for now ignoring side lengths, there may a formula like this from n “pyramids”.",
"@TymaGaidash: What do you mean by a \"volume formula\"? The volume of a polyhedron is not uniquely determined by the multiset of its edge lengths, even given the associated edge graph (consider the family of parallelepipeds with unit edges), so there can be no deterministic mapping from a polyhedron's edge lengths to its volume.",
"If we could somehow find a volume formula and then optimize it, it should give the answer.",
"You can get a better upper bound by using the theorem 2 from this paper. From that together with polyhedron cannot enclose more volume than a sphere, you get the upper bound of $\\frac{1}{36\\sqrt{6}\\pi^2}\\approx 0.00115$",
"@StevenStadnicki: Just an update that I did investigate something like this for some small polyhedra and didn't find any improvements (see the edit).",
"I wonder whether you could approach this by specific combinatorial counting; the number of polyhedra with low edge counts is relatively small. An orange-slice shape suggests that it might be hard to lower-bound the number of 'long' edges, though. This is a great question."
] | 0 |
Science
| 1 |
263
|
math
|
Can Erdős-Turán $\frac{5}{8}$ theorem be generalised that way?
|
Suppose for an arbitrary group word $w$ ower the alphabet of $n$ symbols $\mathfrak{U_w}$ is a variety of all groups $G$, that satisfy an identity $\forall a_1, … , a_n \in G$ $w(a_1, … , a_n) = e$. Is it true, that for any group word $w$ there exists a positive real number $\epsilon (w) > 0$, such that any finite group $G$ is in $\mathfrak{U_w}$ iff $$\frac{\lvert\{(a_1, … , a_n) \in G^n : w(a_1, … , a_n) = e\}\rvert}{{|G|}^n} > 1 - \epsilon(w)?$$
How did this question arise? There is a widely known theorem proved by P. Erdős and P. Turán that states:
A finite group $G$ is abelian iff $$\frac{|\{(a, b) \in G^2 : [a, b] = e\}|}{{|G|}^2} > \frac{5}{8}.$$
This theorem can be rephrased using aforementioned terminology as $\epsilon([a, b]) = \frac{3}{8}$.
There also is a generalisation of this theorem, stating that a finite group $G$ is nilpotent of class $n$ iff $$\frac{|\{(a_0, a_1, … , a_n) \in G^{n + 1} : [ … [[a_0, a_1], a_2]… a_n] = e\}|}{{|G|}^{n + 1}} > 1 - \frac{3}{2^{n + 2}},$$ thus making $\epsilon([ … [[a_0, a_1], a_2]… a_n]) = \frac{3}{2^{n + 2}}$.
However, I have never seen similar statements about other one-word varieties being proved or disproved, despite such question seeming quite natural . . .
Actually, I doubt that the conjecture in the main part of question is true. However, I failed to find any counterexamples myself.
|
https://math.stackexchange.com/questions/3070332
|
[
"combinatorics",
"group-theory",
"finite-groups",
"conjectures",
"universal-algebra"
] | 94 | 2019-01-11T12:51:53 |
[
"Which group words w admit an ε(w) > 0 such that: A finite group G satisfies w = e globally ⟺ P_w(G) > 1 - ε(w)?",
"@Z. A. K. (Well, technically I proved them during my PhD, but I wrote them up earlier this year)",
"I proved analogous results about the equations $xy^2=y^2x$, $xy^3=y^3x$ and $xy=yx^{-1}$ (and infinitely many others derived from these by adding further variables) earlier this year: preprint.",
"@YaniorWeg This conjecture is false for $p=5$; see my answer at mathoverflow.net/a/337483/297 .",
"For some recent progress, see the paper Delizia et al., Gaps in probabilities of satisfying some commutator-like identities (2019), where the conjecture is proved for the metabelian and $2$-Engel word.",
"@MeesdeVries, there is also a conjecture, that $\\epsilon(x^p) = \\frac{p-1}{p^2}$ for prime $p$, however it remains unproven.",
"@MeesdeVries, for the case $n = 1$ only three results are currently known: $\\epsilon(x) = \\frac{1}{2}$, $\\epsilon(x^2) = \\frac{1}{4}$ and $\\epsilon(x^3) = \\frac{2}{9}$.",
"Is the $n=1$ case obviously true? Or is even that case difficult?",
"Not mentioned in a 2015 survey Farrokhi, D. G. (2015). ON THE PROBABILITY THAT A GROUP SATISFIES A LAW: A SURVEY (Research on finite groups and their representations, vertex operator algebras, and algebraic combinatorics), muroran-it.ac.jp/mathsci/danwakai/past/articles/201404-201503/…. Mentioned as open in a note by John D. Dixon, \"Probabilistic Group Theory\", people.math.carleton.ca/~jdixon/Prgrpth.pdf",
"I one wrote out an answer which is about generalising the Erdos-Turan result to infinite groups: math.stackexchange.com/a/2809964/10513 You might find it interesting/relevant."
] | 0 |
Science
| 1 |
264
|
math
|
Complete, Finitely Axiomatizable, Theory with 3 Countable Models
|
Does there exist a complete, finitely axiomatizable, first-order theory $T$ with exactly 3 countable non-isomorphic models?
A few relevant comments:
There is a classical example of a complete theory with exacly $3$ models. This theory is not finitely axiomatizable (For the trivial reason that the language is infinite).
In this post, Javier Moreno explains how to rephrase this example in a finite language. Still, the theory is not finitely axiomatizable.
Some less relevant comments:
I would like to know if finite axiomability has ever be asked in this context.
There have been research in connection with stability. Lachlan has proved that a superstable theory with finitely many countable models is $\omega$-categorical. And it is still open if this can be extended to all stable theories.
|
https://math.stackexchange.com/questions/913049
|
[
"logic",
"model-theory"
] | 92 | 2014-08-29T06:01:33 |
[
"and by Scott's theorem a complete $\\mathcal{L}_{\\omega_1,\\omega}$-theory has at most one countable model up to isomorphism (as long as the language is countable, anyways); actually, Scott showed that for every countable structure $\\mathcal{A}$ (in a countable language), there is a single $\\mathcal{L}_{\\omega_1,\\omega}$-sentence $\\varphi$ such that for all countable $\\mathcal{B}$ we have $\\mathcal{B}\\models\\varphi\\iff\\mathcal{B}\\cong\\mathcal{A}$ (google \"Scott sentence\"). So none of this is relevant.",
"@mohottnad It is not true that if $T$ is a conservative extension (even in the \"strong\" sense) of $S$ then $T$ has the same number of countable models (up to isomorphism) as $S$; the \"additional structure\" of $T$ may introduce additional models. All we can say is that there will be the same number of reducts to the language of $S$ of countable models of $T$, up to isomorphism, as there are countable models of $S$ up to isomorphism. But that's much weaker. As to changing to $\\mathcal{L}_{\\omega_1,\\omega}$, note that changing the logic also changes the meaning of \"complete,\" (cont'd)",
"@NoahSchweber thanks for your critique and I agree my above sentence is not well formed. However, similarly here's some post suggests \"every recursively axiomatizable theory in first-order logic with identity that has only infinite models, has a finitely axiomatized conservative extension... simply endow the given theory with a new sort..\" which seems applicable here for above theory with 3 countable models. Another way for a single sentence using $L_{w1,w}$ logic?",
"@mohottnad That doesn't help at all. \"$\\forall k.c_k<c_{k+1}$\" is not a first-order sentence: first-order logic doesn't let you quantify over symbol indices like that. And re: \"If you want it to be finitely axiomatized,\" finite axiomatizability is the whole point of the question (see e.g. the passage \"There is a classical example of a complete theory with exacly 3 3 models. This theory is not finitely axiomatizable (For the trivial reason that the language is infinite).\")",
"Let ∆ be the theory of dense linear orders without endpoints (finitely axiomatizable), then the classic Ehrenfeucht example theory T is axiomatized by $∆∪\\{c_k<c_{k+1} | k ∈ ω\\}$ which has 3 countable models up to isomorphism. If you want it to be finitely axiomatized, we can simply syntactically change above set of countably infinite number of axioms to a single $\\Pi_1$ sentence: $\\forall k. c_k<c_{k+1}$ where $k$ ranges over another sort of natural number type. This theory's alphabet is infinite, but it by no means we cannot have a finitely axiomatized theory using this language.",
"@DanielV There are 3 models up to isomorphism. \"Up to isomorphism\" is usually left implicit. If you do not count \"up to isomorphism\" there is always a proper class of models.",
"Do you mean that there are $n\\ge 3$ models, but only $3$ isomorphically distinct models, or that there are $3$ models, and they are not isomorphic?",
"This post might give you a lead, but I don't have David Marker's model theory book -- mathoverflow.net/questions/143394/…",
"Note though that the usual Ehrenfeucht example of a 3-model theory is unstable.",
"Wilfred Hodges' Model Theory theorem 12.2.18 states that a totally categorical theory $T$ (a) is not finitely axiomatisable and (b) is quasi-finitely axiomatisable; where the latter property means that it is definitionally equivalent to a single sentence together with all the sentences \"there are at least $n$ elements\" for each $n$. And further that an $\\omega$-stable and $\\omega$-categorical $T$ is not finitely axiomatisable either. (So your second question--has anybody thought about these questions before?--has answer \"yes\". Hodges says the theorem answers a conjecture of Vaught.)",
"Thanks for the comments, which make the question much more accessible to someone who hasn't thought about such things before.",
"@James pointed out my misreading of the Question that these 3 countable nonisomorphic models are the only models, interpreting instead that among its countable models, there are exactly three isomorphism classes."
] | 0 |
Science
| 1 |
265
|
math
|
Regular way to fill a $1\times1$ square with $\frac{1}{n}\times\frac{1}{n+1}$ rectangles?
|
The series $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=1$$ suggests it might be possible to tile a $1\times1$ square with nonrepeated rectangles of the form $\frac{1}{n}\times\frac{1}{n+1}$. Is there a known regular way to do this? Just playing and not having any specific algorithm, I got as far as the picture below, which serves more to get a feel for what I am looking for.
I think some theory about Egyptian fractions would help. It's nice for instance in the center where $\frac13+\frac14+\frac16+\frac14=1$. And on the right edge where $\frac12+\frac13+\frac16=1$.
Side note: The series is $\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots$. The similar looking $\left(\frac11-\frac12\right)+\left(\frac13-\frac14\right)+\left(\frac15-\frac16\right)+\cdots$ sums to $\ln(2)$, and there is a nice picture for that, if you interpret $\ln(2)$ as an area under $y=
\frac{1}{x}$:
|
https://math.stackexchange.com/questions/1164035
|
[
"sequences-and-series",
"visualization",
"egyptian-fractions"
] | 85 | 2015-02-24T15:08:28 |
[
"This question is basically identical to one posted on MathOverflow. I am leaving the question here as a reference for interested readers in the future.",
"And that's popularity....",
"I ❤ this question !!!!!!!!!",
"Nice problem. But there seems to be no method of arrangeing it analytically.",
"@alex.jordan Oh, ok, now I see.",
"@VividD He means that the rectangles from the $\\ln(2)$ example are not literally $\\frac{1}{2n-1}\\times\\frac{1}{2n}$. They have that same total area, but different dimensions.",
"@GregMartin What is \"not with given dimensions\"?",
"@alex.jordan : Thanks for the explanation; it's a cool fact. I note that the $\\ln2$ example tiles the curved area with rectangles of given area, but not with given dimensions in any reasonable way. The analogous problem would be to tile the unit square with rectangles of area $1/n(n+1)$ but not with given dimensions; and that's trivial. So I think we're still looking for an example of a natural known tiling with a sequence of rectangles of given dimensions.",
"@martycohen : look at either of the links in the above comments. Paulhus has fit them into a square of side length something like $1+10^{-9}$.",
"Since no one seems to know how to fit the rectangles into a square of area equal to the sum of their areas, how about this question: What is the smallest square that contains all the $\\frac1{n}\\times \\frac1{n+1}$ rectangles with no overlaps?",
"@GregMartin I can't take credit for that; I saw it somewhere at some point. After the big rectangle for $1-\\frac12$, the real pattern sets in. First, using $x=1+\\frac12$, we get the rectangle representing $\\frac13-\\frac14$. Then at $x=1+\\frac14$ and $x=1+\\frac34$, you get the rectangles for $\\frac15-\\frac16$ and $\\frac17-\\frac18$. Then at $x=1+\\frac18$, $x=1+\\frac38$, $x=1+\\frac58$, and $x=1+\\frac78$, you get rectangles for $\\frac19-\\frac1{10}$, $\\frac1{11}-\\frac1{12}$, $\\frac1{13}-\\frac1{14}$, and $\\frac1{15}-\\frac1{16}$. Do you see the pattern? Look for powers of $2$ in denominators.",
"What is your algorithm for placing the rectangle in the picture where the area sums to $\\ln 2$?",
"@GregMartin: Do you happen to know of a non-trivial tiling (not geometric like golden ratio or similar), with no two equal tiles, which does have some kind of enumeration for each different tiling?",
"@alex.jordan Generally, a question should not be deleted unless it is unsalvageably awful (e.g. do my homework for me), inappropriate, or spam. As it stands your question is none of these and does not deserve deletion at all.",
"What's protocol for when a question turns out to be research level? Should I delete the question?",
"I wrote a paper related to this (math.ubc.ca/~gerg/index.shtml?abstract=CTGP). Therein you can find a reference to this problem statement, as well as my reason for believing that such a packing is possible.",
"This was asked at MO: mathoverflow.net/questions/34145/… and it appears to still be an open problem.",
"This is a research level problem in \"Concrete Mathematics\" (2nd edition) by Grapham, Knuth, Patashnik (ISBN-10: 0201558025): see page 66, exercise 37. In the section with hints you will find that every one of the authors has a different opinion.",
"Wow! This is the most interesting question I've seen on this site in a while :)"
] | 0 |
Science
| 0 |
266
|
math
|
Does the $32$-inator exist?
|
Background
It is common popular-math knowledge that as we extend the real numbers to complex numbers, quaternions, octonions, sedenions, $32$-nions, etc. using the Cayley-Dickson construction, we lose algebraic properties at each step such as commutativity, associativity, alternativity, etc.
We can express this phenomenon in a more illustrative way. Consider the following multilinear maps $F_k$ of degree $k$:
Number two: $F_0: [\:] = 1 - (-1) = 2$.
Imaginary part: $F_1: [x] = x - x^*$.
Commutator: $F_2: [x, y] = xy - yx$.
Associator: $F_3: [x, y, z] = (xy)z - x(yz)$.
$16$-inator: $F_4: [x, y, z, w] = (x(yz))w+(w(yz))x-(xy)(zw)-(wy)(zx)$.
They measure the failure of the algebra to be of characteristic two, Hermitian, commutative, associative and Moufang, respectively. All of them are well-known except for the last one, which I adapted from the quadrilinear map described in section 5 of this paper and essentially comes from a linearized Moufang identity. (Note that depending on the convention, imaginary part, commutator and associator are often defined with an extra factor of $1/2$, and in the context of complex numbers, imaginary part almost always refers to $F_1$ divided by $2i$).
Since the maps are multilinear, they are determined by their values at the basis elements $1=e_0, e_1, e_2, e_3, \ldots, e_{2^n-1}$ (under the standard labeling) in the $n$th Cayley-Dickson algebra over the reals. All the maps above with degree $k \le n$ can then be checked to satisfy the following properties:
(A1) $[e_a, e_b, e_c, \ldots] = m e_i$ for some $i$, where $m\in\{-2,0,2\}$.
(A2) $[e_a, e_b, e_c, \ldots] = 0$ whenever $a, b, c, \ldots <2^{k-1}$.
(A3) $[e_1, e_2, e_4, \ldots, e_{2^{k-1}}] = 2 e_{2^{k}-1}$.
(A4) If $[x, y, \ldots] = z$, then $[\sigma(x), \sigma(y), \ldots] = \sigma(z)$ for any algebra automorphism $\sigma$.
Properties (A2) and (A3) together imply (rather trivially) that given a Cayley-Dickson algebra $\mathbb{A}$, $F_k$ vanishes in the next algebra $CD(\mathbb{A})$ if and only if both $F_k$ and $F_{k-1}$ vanish in $\mathbb{A}$; for $k < 4$ this fact holds in more generality for any $*$-algebra (see e.g. here) and provides a more uniform way to explain why commutativity, associativity, etc. break in sequence as we iterate the Cayley-Dickson process.
But rather than talk about algebraic properties, for the purposes of this question I would like to treat the maps themselves as the objects of interest. So let these properties (and multilinearity) be the axioms that define a $2^n$-inator in a real Cayley-Dickson algebra.
The first four maps $F_0, \ldots, F_3$ are in fact uniquely determined by these axioms when $k=n$. For example, (A3) implies $[\:]=2e_0=2$; (A2) and (A3) together with linearity imply $[a e_0+b e_1] = a[e_0]+b[e_1] = a\cdot 0 + b\cdot 2e_1 = 2b e_1$; the map $F_2$ in the quaternions is completely fixed by (A2), (A3) and (A4) applied twice, with an automorphism that cyclically permutes the imaginary basis elements and with the involution $e_1 \mapsto e_2, e_2 \mapsto e_1, e_3 \mapsto -e_3$; and similarly for $F_3$ in the octonions. I haven't checked whether the $16$-inator in the sedenions is uniquely recovered from the axioms, but I suspect it's probably nonunique, since the automorphism groups of the sedenions and above are "small" relative to the size of the algebras (the groups are all isomorphic to the $14$-dimensional octonionic automorphism group ($G_2$) times a finite group, while the algebras themselves double their dimension at each step). Nevertheless, it is natural to ask if one can continue the sequence by finding nontrivial examples of $F_k$ for $k>4$; my main question deals with the smallest case $k=n=5$.
This paper implies that if a $F_5$ exists, it is necessarily not expressible in terms of multiplication and real constants alone, unlike $F_0, F_2, F_3$ and $F_4$. There could still be a closed-form expression if we allow the use of conjugation as with $F_1$, but it looks unlikely to me. The possibility of using non-real constants seems more promising, because the automorphism groups of sedenions and above do not act transitively on imaginary elements of norm $1$, so some basis elements are "more special" than others. In fact, one can define new "conjugations" $x^{(8)} = e_{8} x e_{8}$, $x^{(16)} = e_{16} x e_{16}$, etc. which can be shown to be invariant under all automorphisms of the algebra; this follows from Lemma 2.1 here. An obvious followup to my main question would be whether the $2^n$-inators can be expressed in terms of multiplication, real constants, and the conjugations $x^*, x^{(8)}, x^{(16)}, x^{(32)}, \ldots, x^{(2^n)}$. But in principle it's not guaranteed that $F_n$ will be algebraically expressible at all, if it exists.
Question
My question is:
Is there a $32$-inator in the $32$-nions, i.e. a multilinear map $F_5: [x,y,z,w,v]$ satisfying axioms (A1)-(A4)? Followup: can it be expressed in terms of multiplication, real constants and conjugations?
Such a map would necessarily be nontrivial by (A3), and would be identically zero in any sedenion subalgebra by (A2) and (A4). Some remarks:
In page 12 of the paper I linked above, it is suggested that the possible existence of higher-order maps might be related to projective geometry over the field $\mathbb{F}_2$.
The first problem looks simple enough that it could be solved with a computer search, but the search space seems to be quite large at first glance. The obvious upper bound on possible cases to check, taking into account only multilinearity and (A1), would be $(2\cdot 32 + 1)^{32^5} = 65^{2^{25}} \lesssim 2^{2^{28}}$. Perhaps some clever argument could reduce this bound to a more manageable number.
|
https://math.stackexchange.com/questions/4498328
|
[
"abstract-algebra",
"recreational-mathematics",
"multilinear-algebra",
"octonions",
"sedenions"
] | 84 | 2022-07-22T14:58:59 |
[
"In regards to conjugation. One way to view it is as an invertible operator on a vector of reals, that can only change the signs and the order of elements but not the magnitudes of the elements themselves. For an $n$ vector there are $2^n n!$ such conjugation operators. That gives a lot of options to work with.",
"I see how $[w,x,y,z]$ is a polarization of the Moufang identity $x(yz)x=(xy)(zx)$, but I don't see how it's related to the other two Moufang identities. Are the three identities equivalent in any alternative algebra?"
] | 0 |
Science
| 0 |
267
|
math
|
Is there a "ping-pong lemma proof" that $\langle x \mapsto x+1,x \mapsto x^3 \rangle$ is a free group of rank 2?
|
Let $f,g\colon \mathbb R \to \mathbb R$ be the permutations defined by $f\colon x \mapsto x+1$ and $g\colon x \mapsto x^3$, or maybe even have $g\colon x \mapsto x^p$, $p$ an odd prime.
In the book, by Pierre de la Harpe, Topics in Geometric Group Theory section $\textrm{II.B.40}$, as a research problem, it asks to find an appropriate "ping-pong" action to show that the group, under function composition, $G=\langle f,g \rangle$ is a free group of rank two.
Is there such a proof? That is, is there a proof where the key insight is having that group act in such a way to apply the ping-pong lemma(table-tennis lemma)? I have not been able to find such a proof either by working on it, or in the literature.
Maybe we don't have such a proof but do we have a proof that $G$ contains a free subgroup of rank two, akin to proofs for torsion-free hyperbolic group, or the Tits alternative. I am not sure how obvious it is that $G$ is hyperbolic, or linear. I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly.
Note that there are proofs of this theorem, but as far as I know, they do not use the ping-pong lemma.
The only proofs of the result(and more general things) I know of are in :
Free groups from fields by Stephen D. Cohen and A.M.W. Glass
The group generated by $x \mapsto x+1$ and $x \mapsto x^p$ is free. by Samuel White
Arithmetic permutations by S.A. Adeleke and A.M.W. Glass
|
https://math.stackexchange.com/questions/1295967
|
[
"group-theory",
"reference-request",
"group-actions",
"geometric-group-theory"
] | 80 | 2015-05-23T15:06:23 |
[
"I don't expect that linearity would be easier than free. For instance, the subgroup $H$ generated by $f,g$ and in addition $h:x\\mapsto 2x$ is not linear because of the relations $hfh^{-1}=f^2$, $ghg^{-1}=h^3$. Indeed the last implies that $g^nhg^{-n}f^{\\pm 1}g^nh^{-1}g^{-n}=f^{\\pm 2^{(3^n)}}$. If these are complex matrices, the norm of the left hand is bounded exponentially, and hence the norm of $f^m$ partially growth logarithmically (both for $m\\ge 0$ and $m\\le 0$). This is possible only if $f$ has finite order, so every representation of $H$ maps $f$ to an element of finite order.",
"\"I am guessing it is not obvious that it is linear since I would suspect a ping-pong proof would come out of that pretty quickly\": this is not quite true. To find free groups in linear groups is now easy, but to show that two given elements generate a free group is usually delicate. For instance I think the following is open (or known but hard): let $G$ be a Zariski-dense subgroup of $SL(n,\\mathbf{C})$ be non-virtually-solvable, and let $x\\in G$ be of infinite order: does there exist $y\\in G$ such that $\\langle x,y\\rangle$ is free? Typically if $x$ is unipotent, it's hard.",
"Given that I found the idea of such a proof quite compelling, I figured others had the same though, and maybe someone had found one since the book was written.",
"@MartinBrandenburg I agree, the Andrews-Curtis conjecture, 3d Poincare conjecture, \"is every hyperbolic group residually finite?\", are only some of the research problems he has in the book. It has been 15 years or so since the book was published, so I am wondering is such a proof has been done, and I don't think at the time of the book being published there was such a proof. It is also really difficult to find papers on this, the only reason I found the ones sited above was because he sited them! (It is sometimes a little ambiguous as to \"how solved\" the research problems are)",
"The book contains quite a lot of these \"research problems\". Could it be that the author hasn't worked them out and instead offers them as, say, research proposals for bachelor or master theses? Perhaps they are open problems, including this one here? Notice that the existing proofs are quite complicated."
] | 0 |
Science
| 0 |
268
|
math
|
Dedekind Sum Congruences
|
For $a,b,c \in \mathbb{N}$, let $a^{\prime} = \gcd(b,c)$, $b^{\prime} = \gcd(a,c)$, $c^{\prime} = \gcd(a,b)$ and $d = a^{\prime} b^{\prime} c^{\prime}$. Define $\mathfrak{S}(a,b,c) = a^{\prime} \mathfrak{s}( \tfrac{bc}{d}, \tfrac{a}{b^{\prime} c^{\prime}}) + b^{\prime} \mathfrak{s}( \tfrac{ac}{d}, \tfrac{b}{a^{\prime} c^{\prime}}) + c^{\prime} \mathfrak{s}( \tfrac{ab}{d}, \tfrac{c }{a^{\prime} b^{\prime}} )$, where $\mathfrak{s}$ is the Dedekind sum. I can prove the following: For $a, b, c \in \mathbb{N}$ with $\gcd(a,b,c) = 1$,
\begin{align}
12 a b c \ \mathfrak{S}(a,b,c) \equiv (ab)^{2} + (bc)^{2} + (ca)^{2} + d^{2} \ \pmod{a b c}.
\end{align}
Is this congruence in the literature? I'm aware of the 3-term generalization of Rademacher and Pommersheim and of the work of Beck on the Carlitz-Dedekind Sums, but these don't seem to include this particular congruence as a special case.
I generalize. For $a_{1}, \dots, a_{n} \in \mathbb{N}$ such that $\gcd(a_1, \dots, a_{n}) = 1$, let $a_{i}^{\prime} = \gcd(a_{1}, \dots, \hat{a}_{i}, \dots, a_{n})$. Define the symmetric summation
$$
\mathfrak{S}(a_{1},\dots, a_{n}) = \sum_{i = 1}^{n} a_{i}^{\prime} \, \mathfrak{s}\left( \tfrac{a_{1} \cdots \hat{a}_{i} \cdots a_{n}}{a_{n+1}}, \tfrac{a_{i}}{a_{1}^{\prime} \cdots \hat{a}_{i}^{\prime} \cdots a_{n}^{\prime}} \right).
$$
where $\hat{}$ denotes omission. For pairwise coprime $a_{1}, \dots, a_{n} \in \mathbb{N}$ (all primed variables equal $1$), I conjecture the following congruence holds for $n = 4$ (and I have a hunch that it continues to hold for $n > 4$):
$$
12 a_{1} \cdots a_{n} \, \mathfrak{S}(a_{1}, \dots, a_{n}) \equiv \left( \sum_{i = 1}^{n} (a_{1} \cdots \hat{a}_{i} \cdots a_{n})^{2} \right) + 1 \ \pmod{a_{1} \cdots a_{n}}.
$$
Is this more general congruence known?
|
https://math.stackexchange.com/questions/61026
|
[
"number-theory",
"reference-request"
] | 73 | 2011-08-31T13:39:55 |
[] | 0 |
Science
| 0 |
269
|
math
|
Determinant of a matrix that contains the first $n^2$ primes.
|
Let $n$ be an integer and $p_1,\ldots,p_{n^2}$ be the first prime numbers. Writing them down in a matrix
$$
\left(\begin{matrix}
p_1 & p_2 & \cdots & p_n \\
p_{n+1} & p_{n+2} & \cdots & p_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
\cdots & \cdots & \cdots & p_{n^2}
\end{matrix}
\right)
$$
we can take the determinant. How to prove that determinant is not zero for every $n$?
|
https://math.stackexchange.com/questions/2355787
|
[
"linear-algebra",
"number-theory",
"prime-numbers",
"determinant"
] | 69 | 2017-07-11T23:55:08 |
[
"Is it possible to prove this by showing that inverse of given matrix always exists as there always exists trivial solutions to Ax = 0 equation?. Trivial solutions are only possible as the rows of the matrix will always be independent, as each element of matix is a distinct prime number.",
"Has anyone thought of posting this on math overflow?",
"@Gerald Did you misread \"not zero\" as \"nonnegative\"?",
"@Boris: Although the absolute value increases most of the time, there are exceptions. For example with $n=57,58$ the determinants are about $2.14 \\times 10^{89}$ and $1.25 \\times 10^{89}$, respectively. There are more exceptions later.",
"@Elborito This is because the Perron-Frobenius Eigenvalue grows faster than linearly and is much larger than the other eigenvalues (which are rarely smaller than one). Here is a list of the eigenvalues for $n=14$: $7843.09, -168.956, -11.4222 + 2.69341 i, -11.4222 - 2.69341 i, -5.68197 + 5.79415 i, -5.68197 - 5.79415 i, 0.121105 + 6.20539 i, 0.121105 - 6.20539 i, 4.28891 + 4.0145 i, 4.28891 - 4.0145 i, 11.132, 7.30824, -4.67599, -1.51002$. I've also made a plot of the lowest singular value for $n=10,\\ldots,100$. But I can't see any pattern. E.g. , at $n=28$ the lowest sing. value is $0.006$.",
"The absolute value of the determinant of the matrix is strictly crescent. maybe by recursivity",
"$n=951$ and there isn't counterexample",
"But I want to tell those interested that the bound that arrives is $n=845$ with $714025$ primes in the maximum matrix.",
"No answer: You might look at \"Integral Matrices\" by Morris Newman. In particular the \"Smith Normal Form\". It discusses divisibility and such. As I recall some of the results were surprising. The class is just a distance memory (but I have the book) so it might lead nowhere.",
"@RoflUkulus We can also to check whether starting series from $1$ gives similar results. So far it was checked up to $n=200$.",
"From my point of there are two things to study: 1.) Are there any permutations $\\sigma \\in S_{n^2}$ that permutate the entries such that the determinant goes to zero? If so, what do they look like? 2.) What are the least gaps (least is not really defined properly), that primes can have, to produce determinant zero, like the matrices Robert posted, for example maybe is there any $k$ such that $p_1,p_{1+k},p_{2+k},\\ldots$ determinant gets zero.",
"@RobertIsrael The number of digits for determinant seems to increase (or at least not to decrease - almost linear function) . Could it be possible to compute with your software also series with this number of digits?",
"@RobertIsrael Very inspiring, I hope some day a limit $n= 1000$ will be achieved.",
"At $n= 460$ the determinant has $1001$ digits. I was making a b-file for sequence A067276, and the OEIS doesn't like numbers with more than $999$ digits. I could go further, but computations start to slow down...",
"@RobertIsrael so now 460 is the limit. I wonder why exactly 460 in this case?",
"@Robert Israel thank you for these examples. Indeed, this makes it harder to prove and there might be some evil matrix around which determinant goes to zero :-) I looked up the prime factors of the determinants but did not find any pattern. There are large powers of 2 appearing in the factorization, but there not even increasing monotonely.",
"Also with determinant $0$: $$ \\pmatrix{2 & 3 & 5 & 7\\cr 11 & 13 & 17 & 19\\cr 23 & 29 & 31 & 37\\cr 41 & 47 & 67 & 73\\cr }$$",
"I think this is going to be an intractable problem. Note that there are square matrices with determinant $0$ made up of distinct primes, e.g. $$\\pmatrix{2 & 3 & 5\\cr 7 & 11 & 13\\cr 19 & 23 & 97\\cr}$$ Thus you somehow have to depend on the fact that you're using the consecutive primes. And those just don't have enough regularity.",
"No zeros up to $n=460$.",
"@RoflUkulus I think that Peter is the best expert on prime numbers ( much better than me) , ask him when he will be available. BTW he has asked many other interesting questions about primes..",
"@Widawensen Oh, you already asked the questions. Did you try the flintc library for checking n > 201. Not sure if it can handle those big matrices. I strongly believe that the proof will be difficult, too. Maybe it would be interesting to study the prime factorization of those determinants.",
"Read comments from math.stackexchange.com/questions/2047879/…, especially Peter's comments are valuable.",
"The sequence of determinants is OEIS sequence A067276. Not that this helps..."
] | 0 |
Science
| 0 |
270
|
math
|
Does the average primeness of natural numbers tend to zero?
|
Note 1: This questions requires some new definitions, namely "continuous primeness" which I have made. Everyone is welcome to improve the definition without altering the spirit of the question. Click here for a somewhat related question.
A number is either prime or composite, hence primality is a binary concept. Instead I wanted to put a value of primality to every number using some function $f$ such that $f(n) = 1$ iff $n$ is a prime otherwise, $0 < f(n) < 1$ and as the number divisors of $n$ increases, $f(n)$ decreases on average. Thus $f(n)$ is a measure of the degree of primeness of $n$ where 1 is a perfect prime and 0 is a hypothetical perfect composite. Hence $\frac{1}{N}\sum_{r \le N} f(r)$ can be interpreted as a measure of average primeness of the first $N$ integers.
After trying several definitions and going through the ones in literature, I came up with:
Define $f(n) = \dfrac{2s_n}{n-1}$ for $n \ge 2$, where $s_n$ is the
standard deviation of the divisors of $n$.
One advantage of using standard deviation is that even if two numbers have the same number of divisor their value of $f$ appears to be different hence their measure of primeness will be different.
Question 1: Does the average primeness tend to zero? i.e. does the following hold?
$$
\lim_{N \to \infty} \frac{1}{N}\sum_{r = 2}^N f(r) = 0
$$
Question 2: Is $f(n)$ injective over composites? i.e., do there exist composites $3 < m < n$ such that $f(m) = f(n)$?
My progress
$f(4.35\times 10^8) \approx 0.5919$ and decreasing so the limit if it exists must be between 0 and 0.5919.
For $2 \le i \le n$, the minimum value of $f(i)$ occurs at the largest highly composite number $\le n$.
Note 2: Here standard deviation of $x_1, x_2, \ldots , x_n$ is defined as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n}$. Also notice that even if we define standard deviation as $\sqrt \frac{\sum_{i=1}^{n} (x-x_i)^2}{n-1}$ our questions remain unaffected because in this case in the definition of $f$, we will be multiplying with $\sqrt 2$ instead of $2$ to normalize $f$ in the interval $(0,1)$.
Note 3: Posted this question in MO and got answer for question 1. Indeed the limit tends to zero. Question 2 is still open.
|
https://math.stackexchange.com/questions/3176228
|
[
"number-theory",
"limits",
"statistics",
"prime-numbers",
"natural-numbers"
] | 62 | 2019-04-05T10:59:06 |
[
"If $f(n)=1$ iff $n$ is a prime then I would have thought $f(n)$ cannot uniquely identify $n$ (i.e. $f(n)$ is not $1-1$ injective), since it cannot when $n$ is a prime.",
"@daniel One of the reasons why I invented this definition was because even if two numbers have the same number of divisors or the same number distinct prime divisors, their value of $f(n)$ was found to be unique. In that way, $f(n)$ can uniquely identify $n$ but $\\omega(n)$ or $d(n)$ cannot uniquely identify $n$.",
"@daniel After I posted in MO, it was answered. But I am not sure if a similar question was asked in MSE and answered",
"@daniel I do not see how Erod Kac theorem directly implies that the avove limit tends to zero?",
"$0.59$ is the average of all the values for $n \\le 4.35 \\times 10^8$",
"Yes, that is what I got. I was confused when you reported getting 0.59 above.",
"@Scott You must have done an error. Try this Sagemath code (N(variance(divisors(320700000), bias = True), digits = 10))^0.5 / (320700000-1) you will get 0.1009653420",
"In that case, for 320700000, the divisors are: $[1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 32 ......... 13362500, 16035000, 20043750, 21380000, 26725000, 32070000, 40087500, 53450000, 64140000, 80175000, 106900000, 160350000, 320700000]$ with a variance of $1048437529950912.5$. However, this gives $f(320700000) \\approx 0.201930684009565$ = 2 times the square root of the variance divided by 320700000 - 1. What am I missing here?",
"@Scott E.g. For $n = 19$, the divisors are $1,19$ and the variance of these two divisors is $82$. Similarly for $n = 20$, the divisors are $1,2,4,5,10,20$ and their variance is $42$",
"I can't seem to recreate your test cases. Do you mean proper divisors, etc? Could you post a couple of low-valued test cases for this so that we can make sure we're evaluating the same thing?",
"@NilotpalKantiSinha I'm not sure I follow why Question 2 is a question. Isn't $f(p) = f(q) = 1$ for any primes $p,q$ by construction?",
"@JavaMan Yes, standard deviation gives more information than just the number of divisors. But to be honest, I don't exactly know how to interpret standard deviation completely in the context of primeness. One of the reason for using standard deviation was because some progress was already made on the standard deviation of the divisors of the first $n$ natural numbers. Check this post. math.stackexchange.com/questions/2773289/…",
"Why do you care about the standard deviation? I understand that $s_n$ may be different even for a number with the same amount of divisors, but if all you want is to measure “primeness,” then why complicate this with standard deviation? It seems like this question now accounts for more than just primeness. It now considers smoothness of an integer (or something like it).",
"@TheSimpliFire Sure joining",
"We can continue this discussion in chat",
"@TheSimpliFire My graph is identical to yours but instead of converging to about $0.4$ as in your case, it is converging to about $0.4 \\times \\sqrt 2 \\approx 0.59$. This is because of bais = True or False in your formula for calculating standard deviation. You have used bias = False and got the limit 0.4. I have used bias = True and got $f(320700000) = 0.594127573872085$",
"I meant your second sentence below the Note at the start of your post. Here is a plot of the limit for $N\\le 10000$.",
"@TheSimpliFire The difference is because of the the parameter 'bias = ' in the formula for calculating standard deviation in Sagemath. Check my source code above. For a prime $p$ we have 2*std(divisors(p), bias = True).n()/(p-1) = 1 and 2*std(divisors(p), bias = True).n()/(p-1) = $\\sqrt 2$. I have used bias = True because the range $(0,1)$ looks more elegant than $(0, \\sqrt 2)$. Also $r = 2$ is correct. However, regardless of bias = True or False, the the question remain unchanged as $f$ is only gets multiplied by a scaling factor so we are good.",
"I don't get why you say that $f(n)=1$ if $n$ is prime. For a prime $p$, $$f(p)=\\frac{2\\sqrt{\\frac{\\left(\\frac{p-1}2\\right)^2+\\left(\\frac{1-p}2\\right)^2}{2-1}}}{p-1}=\\sqrt2$$ so shouldn't it be $f(n)=\\frac{\\sqrt2 s_n}{n-1}$? Furthermore the sum should be from $r=2$ to $N$ as $1/(r-1)$ is undefined when $r=1$.",
"@YiFan Sure done.",
"You might want to rephrase \"is $f(n)$ unique\" to \"is $f(n)$ injective\". At first glance I thought you didn't know whether $f(n)$ was well-defined.",
"@TheSimpliFire I have added my Sagemath code above",
"I have written some code for this in R: stanfun <- function(n){sd(divisors(n))/(n-1)};funstan <- function(m){sum(sapply(2:m, function(i){stanfun(i)}))/m}, so $\\frac1{10000}\\sum\\limits_{r=1}^{10000}f(r)$ is given by funstan(10000) which outputs $0.404801$.",
"@GerryMyerson: Here is a more technical answer why standard deviation. If two numbers have the same number of divisors then the value $2/d(n)$ is same for both but the values of $f(n)$ is different. So under my definition, I will consider the number with smaller value of $f(n)$ to have a greater primness because we are not just measuring how many divisors a number has but also how scattered these divisors are. At the moment, I don't know if $f(n)$ is unique. I will add this to the question.",
"Why involve the standard deviation? Why not something simpler, like $2/d(n)$, where $d(n)$ is the number of divisors of $n$?",
"Is your second statement about the minimum of $f$ obvious? I'm unable to come up with an argument."
] | 0 |
Science
| 0 |
271
|
math
|
On the equivalence relation $(a,b) \sim (c,d)\iff a+b=c+d$
|
Setup
Let $A=\{a_1<a_2<\cdots<a_p\}$ and $B=\{b_1<b_2<\cdots<b_q\}$ be two finite sets of real numbers. Define an equivalence relation $\sim_{A,B}$ on $R_{p,q}=\{1,\dots,p\}\times\{1,\dots,q\}$ by setting $$(i,j)\sim_{A,B}(k,l)\iff a_i+b_j=a_k+b_l$$ and further define a total order on the set of equivalence classes $R_{p,q}/\sim_{A,B}$ by setting, for any equivalence classes $c,d$ and representatives $(i,j)\in c$ and $(k,l)\in d$ $$c<_{A,B}d\iff a_i+b_j<a_k+b_l$$ This question is about characterizing such equivalence relations $\sim_{A,B}$ and total orders $<_{A,B}$. An equivalence relation $\sim$ on $R_{p,q}$ for which there exist $A,B$ with $\sim\;=\;\sim_{A,B}$ is called realizable. If $\sim$ is realizable, a total order $<$ on $R_{p,q}/\sim$ is called realizable if we can find $A,B$ with $\sim\;=\;\sim_{A,B}$ and $<\;=\;<_{A,B}$. In both cases we say that the pair $(A,B)$ realizes $\sim$ (resp. $(\sim,<)$.)
Problem 1. Characterize realizable equivalence relations on $R_{p,q}$.
Problem 2. Suppose $\sim$ is realizable, can one find $\mathcal{A},\mathcal{B}\subset\Bbb{N}$ with $\sim\;=\;\sim_{\mathcal{A}, \mathcal{B}}$?
Problem 3. Characterize realizable orders on $R_{p,q}/\sim$ (where $\sim$ is realizable.)
Have these questions been considered / solved somewhere?
Motivation
As suggested in the comments I should give some context. The question arose while solving Problem 3.5 in David Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry. In this problem one considers a graded ring $R=\bigoplus_{\gamma\in\Gamma} R_\gamma$ and a graded module $M=\bigoplus_{\lambda\in\Lambda}M_\lambda$ where $(\Gamma,+,0,<)$ is a totally ordered abelian monoid acting freely and compatibly on the totally ordered set $(\Lambda,<)$. When working out the solution to this exercise I found myself drawing rectangles $R_{p,q}=\{1,\dots,p\}\times\{1,\dots,q\}$ and writing $\gamma_i+\lambda_j$ at position $(i,j)$, where I had picked some subsets $\{\gamma_1<\gamma_2<\cdots<\gamma_p\}\subset\Gamma$ and $\{\lambda_1<\lambda_2<\cdots<\lambda_q\}\subset\Lambda$. I was trying to figure out when $R_{\gamma_i}\cdot M_{\lambda_j}$ and $R_{\gamma_k}\cdot M_{\lambda_l}$ lie in the same $M_{\lambda}$ i.e. what and one say about the equivalence relation on $R_{p,q}$ defined by $(i,j)\sim(k,l)\iff\gamma_i+\lambda_j=\gamma_k+\lambda_l$? For concreteness I considered $\Gamma=\Lambda=\Bbb{N}$ or $\Gamma=\Lambda=\Bbb{R}_+$ but the problem makes sense in the broader context described by Eisenbud. I realized that there were some nice pictures to be drawn and that there was a not so obvious combinatorial problem of independent interest which is Problem 1. Problem 2 arose when comparing the cases $\Bbb{N}$ and $\Bbb{R}_+$: do we recover the same equivalence relations? Or would some not be realizable using integers? Problem 3 arose when drawing not only the blobs representing equivalence classes but the total order between them. What constraints must such orders satisfy?
Necessary conditions and proposed characterizations
While thinking about these problems I made some progress and was able to identify some necessary conditions for realizability of $\sim$ and $<$. I wouldn't call any of these "conjectures" per se ... Is there an acceptable phrase for "half-assed guesses"? Using terminology introduced below I wonder
Observation and Guess 1. Any realizable equivalence relation $\sim$ on $R_{p,q}$ satisfies Inclination, Non Crossing and both Antidiagonal Propagation conditions. Conversely, are these conditions enough to characterize realizability of $\sim$ ?
I'm not sure these conditions alone will do the job ... but they may. It seems in that for $p=2$ these conditions are sufficient (with an algorithm for finding explicit sets $A,B$ of rational numbers realizing $\sim$.)
Using terminology introduced below, there are some simple necessary Visibility conditions satisfies by realizable orders. There are also some less obvious but still simple Horizontal and Vertical Coherence conditions that such orders must satisfy.
Observation and Guess 2. Any realizable order satisfies Visibility and both Horizontal and Vertical Coherence conditions. Conversely, are these enough to characterize realizability of $<$ ?
Examples
Here are some simple examples representing equivalence classes of $\sim_{A,B}$ and the order relation $<_{A,B}$. In the pictures below we represent equivalence classes as "blobs" and signify $c<_{A,B}d$ by drawing an arrow from one equivalence class to another.
Example 1. Let $A=\{0<1<2<\cdots<p-1\}$ and $B=\{0<p<2p<\cdots<(q-1)p\}$. The equivalence classes of $\sim_{A,B}$ are singletons (this is uniqueness of euclidean division).
Example 2. Let $A=\{0<1<2<\cdots<p-1\}$ and $B=\{0<1<2<\cdots<q-1\}$. The equivalence classes of $\sim_{A,B}$ are antidiagonals.
The answer to Problem 2 may be positive. Something along the lines of "the condition is open and as long as one moves in tandem the things that are correlated (i.e. related by $\sim_{A,B}$) one can change coordinates to rational ones" ought to "prove" the assertion. Maybe results from real algebraic geometry and semi-rational sets of degree 1 could also answer the question, but I know nothing about this material. Once one has rational $A$ and $B$ one can multiply everything by some integer and get the desired integer $\mathcal{A}$ and $\mathcal{B}$.
There may be a connection with configuration spaces. There certainly is a connection with geometry (existence of rational points on a rational algebraic set) and hyperplane arrangements but there, too, I'm not quite sure how to make it precise.
Notation
Given an equivalence class $c$ for $\sim_{A,B}$ we define its sum to be $a_i+b_j$ for any $(i,j)\in c$. Given $(i,j)\in R_{p,q}$ we define the vertical set $V_i=\{(i,a)\mid a=1,\dots,q\}$ and the horizontal set $H_j=\{(a,j)\mid a=1,\dots,p\}$. Given an equivalence class $c$ we define two subsets of the rectangle: $c^+$ (the green shaded region in the picture below) is the set of all $(i,j)$ such that there exists some $(a,b)\in c$ with $(a,b)\neq(i,j)$ and $a\leq i$ and $b\leq j$ and $c^-$ (the orange shaded region in the picture below) is the set of all $(i,j)$ such that there exists some $(a,b)\in c$ with $(a,b)\neq(i,j)$ and $a\geq i$ and $b\geq j$.
In the picture below the blue dots represent an equivalence class, the green region is $c^+$ and the orange region is $c^-$.
Inclination and Non Crossing (necessary conditions for realizability of $\sim$)
The following are simple necessary conditions on an equivalence relation $\sim$ on $R_{p,q}$ for it to be of the form $\sim_{A,B}$:
Inclination. Two equivalent elements may not have the same first or second coordinate, i.e. any intersection $c\cap V_i$ and $c\cap H_j$ contains at most one element.
Non Crossing. Given distinct equivalence classes $c,c'$, the equivalence class $c'$ may not intersect both $c^+$ and $c^-$.
The necessity of both conditions is clear (the sum of an equivalence class meeting $c^+$ (resp. $c^-$) is greater (resp. smaller) than that of $c$). Actually the first condition is redundant since it is contained in the second. These conditions alone, however, don't characterize equivalence classes of the form $\sim_{A,B}$, see the counter example below.
Constraints and counter example
Some configurations of equivalence classes have a constraining effect on the rest of the equivalence classes. Consider for example the case where we have an antidiagonal and subantidiagonal equivalence class (say with $p=q=n+1\geq 3$) i.e. $(1,n)\sim_{A,B}(2,n-1)\sim_{A,B}\cdots\sim_{A,B}(n,1)$ and $(1,n-1)\sim_{A,B}(2,n-2)\sim_{A,B}\cdots\sim_{A,B}(n-1,1)$. Then $(i,j)\sim_{A,B}(k,l)\iff i+j=k+l$.
This implies that some equivalence relations are not of the form $\sim_{A,B}$ even though they satisfy the Inclination and Non Crossing conditions. For example the one depicted below:
These can't represent the equivalence classes of a relation $\sim_{A,B}$: the presence of the $3$ element equivalence classes would impose the relations $(1,3)\sim_{A,B}(2,2)$ and $(2,7)\sim_{A,B}(3,6)$ (both represented in blue on the right).
We thus get a third necessary condition:
Suppose $A'=\{a_1',\dots,a_r'\}\subset A$ and $B'=\{b_1',\dots,b_r'\}\subset B$ for $r\geq 3$ satisfy either that $(1,r-1)\sim_{A',B'}(2,r-2)\sim_{A',B'}\cdots\sim_{A',B'}(r-1,1)$ and $(1,r-2)\sim_{A',B'}(2,r-3)\sim_{A',B'}\cdots\sim_{A',B'}(r-2,1)$ or the analoguous condition with the antidiagonal above the principal antidiaongal. Then for all $1\leq i,j,k,l\leq r$, $(i,j)\sim_{A',B'}(k,l)\iff i+j=k+l$.
It turns out we can improve on this condition.
Antidiagonal Propagation (necessary condition for realizability of $\sim$)
There are two versions of antidiagonal propagation: the one described below where we consider a main antidiagonal and one below it, but one should add the other possibility too, where the smaller antidiagonal is above the larger one.
Antidiagonal Propagation. Suppose $A=\{a_1<a_2<\cdots<a_n\}$, $B=\{b_1<b_2<\cdots<b_n\}$ with $n=mT+r$ with $m+1\geq 3$, $T\geq 1$, $\newcommand{\T}{[\![1,T]\!]}r\in\T$. Suppose
$$\newcommand{\n}{[\![1,n]\!]}
\forall i,j\in\n,
\quad
\begin{cases}
i+j=n+1:& a_i+b_j=C\\
i+j=n+1-T:& a_i+b_j=D
\end{cases}$$
Then
$$\newcommand{\n}{[\![1,n]\!]}
\forall i,j,k,l\in\n,
\quad
\left\{
\begin{array}{l}
i\equiv k\mod T,\\
j\equiv l\,\mod T,\\
\text{and }i+j=k+l
\end{array}
\right\}
\implies (i,j)\sim_{A,B}(k,l)$$
In other words if $(i,j)$ and $(k,l)$ lie on the same antidiagonal and are a multiple of $T$ positions apart then they are $\sim_{A,B}$-equivalent.
Proof. The condition implies that, setting $\Delta=C-D$,
$$
\left\{
\begin{array}{lll}
a_{1+kT}=a_1+k\Delta, & b_{1+kT}=b_1+k\Delta & ~\text{for }k=0,1,\dots,m-1,m\\
a_{2+kT}=a_2+k\Delta, & b_{2+kT}=b_2+k\Delta & ~\text{for }k=0,1,\dots,m-1,m\\
\qquad\vdots&\qquad\vdots&\qquad\vdots\\
a_{r+kT}=a_r+k\Delta, & b_{r+kT}=b_r+k\Delta & ~\text{for }k=0,1,\dots,m-1,m\\[2mm]\hline
a_{r+1+kT}=a_{r+1}+k\Delta, & b_{r+1+kT}=b_{r+1}+k\Delta & ~\text{for }k=0,1,\dots,m-1\\
\qquad\vdots&\qquad\vdots&\qquad\vdots\\
a_{T+kT}=a_T+k\Delta, & b_{T+kT}=b_T+k\Delta & ~\text{for }k=0,1,\dots,m-1
\end{array}
\right.
$$
and thus if $i\equiv k\mod T$, say $i=IT+\tau$, $k=KT+\tau$, $j\equiv l\mod T$, say $j=JT+\sigma$, $l=LT+\sigma$, and $i+j=k+l$ i.e. $I+J=K+L$, then
$$
a_i+b_j
=
a_\tau+b_\sigma+(I+J)\Delta
=
a_\tau+b_\sigma+(K+L)\Delta
=
a_k+b_l
$$
Here is an illustration. You have two large equivalence classes:
You focus on the lines and columns that contain these equivalence classes and chuck the others out for now. Then there are some automatic equivalences depicted below. Equivalence classes are always subsets of antidiagonals, i.e. sets of the form $\{(i,j)\mid i+j=\mathrm{cst}\}$ and are color coded there. For extra clarity I've also added some markings to some of the equivalence classes. Funnily these look like designs straight out of the 1960ies and 1970ies.
Visibility (necessary condition for realizability of $<$)
There is an obvious necessary condition for a total order on $R_{p,q}/\sim$, where $\sim$ is realizable, to be realizable. Suppose $c,d$ are equivalence classes. We say that $d$ is visibly greater than $c$ if there exists some $i$ with $V_i$ intersecting both $c$ and $d$ and such that if $(i,a)\in c$ and $(i,b)\in d$ then $a<b$, or if there is some $j$ with $H_j$ intersecting both $c$ and $d$ and if $(a,j)\in c$ and $(b,j)\in d$ then $a<b$. We allow ourselves to say that $c$ is visibly greater than $d$ if there is a chain $c=c_0,c_1,\dots,c_n=d$ with $c_i$ visibly greater than $c_{i-1}$ (i.e. we consider the transitive closure of the previously defined relation).
Visibility. If $c$ is visibly greater than $d$ then $c<d$.
Visibility is a necessary property satisfied by realizable $<$ but it is not enough.
Counter Example for realizability of $<$
There's more to this, yet. Consider the following discrete equivalence relation (i.e. equivalence classes are singletons) and total order $<$. This total order satisfies all previously proposed conditions but can't arise as a $\sim_{A,B}$ and $<_{A,B}$.
Here's a simpler counter example showing that the individual $3\times 3$ orders are realized.
Horizontal and Vertical Coherence (necessary conditions for realizability of $<$)
The previous example shows that Visibility is not enough to characterize realizable orders. A problem arose where traversing too many horizontal gaps when following $<$ along equivalence classes led to inconsistencies in the values of those gaps. There are some horizontal and vertical consistency conditions that a realizable $<$ must satisfy. Let's start with Horizontal Consistency conditions. Take $(i,j)$ with $1<i$. The sum of the classes containing $(i-1,j)$ and $(i,j)$ differ by $(\Delta a)_i=a_i-a_{i-1}$. If we follow the $<$-path from the $\sim$-class containing $(i-1,j)$ to that containing $(i,j)$ we get
$$(\Delta a)_i=\text{sum of positive terms, some $(\Delta a)_{k}$, some $(\Delta b)_l$}$$
from which we extract a condition
$$H_{ij}:(\Delta a)_i>\text{sum of some $(\Delta a)_k$ and some $(\Delta b)_l$}$$
where the $(\Delta a)_k$ and $(\Delta b)_l$ to take into account are found using visibility relations. Similarly there are Vertical Consistency conditions of the same ilk
$$V_{ij}:(\Delta b)_j>\text{sum of some $(\Delta a)_k$ and some $(\Delta b)_l$}$$
where again, the $(\Delta a)_k$ and $(\Delta b)_l$ that appear on the right hand side are deduced from visibility relations along the $<$-path of equivalence classes linking the $\sim$ equivalence class containing $(i,j-1)$ to the $\sim$ equivalence class containing $(i,j)$
Beginnings of a positive solution for $p=2$
I believe the answer is positive to both questions at least when $p=2$. WLOG we can take $a_1=0$, $a_2=1$ and $b_0=0$. There seems to exist an explicit algorithm to find a rational solution to the problem. I've illustrated it below. The necessary and sufficient condition on $\sim$ in the case $p=2$ seems to be that the lines associated to $2$-element equivalence classes don't intersect. The algorithm is simple but I haven't conceptualized it really:
You start with $b_1=0$, the sum in the the lower left corner $(1,1)$ is thus $0$ since $a_1=0$
You put $1$ in the lower right corner $(2,1)$, the sum is $1$ since $a_2=1$
If the lower right corner is part of a $2$ element equivalence class you put are forced to attribute $1$ to the other member $(1,r)$.
if $r>2$ you set $b_r=1$. This is the case in the example below so you set $b_3=1$. Then you are forced to put a $2$ in position $(2,r)$ since $a_2=1$. You keep on going until you don't fall into a 2 element equivalence class.
Then you have to insert values for $b_2,\dots,b_{r-1}$. You pick them uniformly apart between $b_1=0$ and $b_r=1$, that is $b_k=\frac{k-1}r$.
Then you are forced to inscribe $b_k+1$ on the right hand side. If the right hand side is part of a $2$ element equivalence class you are forced to set the same value for some new $b_s$ etc ...
I hope the picture below (and the color coded rounds) are more explicit than the half-baked algorithm above.
I don't know if this will work for $p\geq 3$.
A connection with topology
$\newcommand{\O}{\mathcal{O}}\newcommand{\R}{\Bbb{R}}$
Fix $\sim$ an equivalence relation on $R_{p,q}$ and $<$ a total order on $R_{p,q}/\sim$. Define
$$\O_{p,q}[\sim,<]=\{(A,B)\in\O_p\times\O_q\text{ inducing $\sim$ and }<\}$$
where $\O_n=\{(x_1,\dots,x_n)\in\R^n\mid x_1<x_2<\cdots<x_n\}$. Then $\O_{p,q}[\sim,<]$ is nonempty iff $(\sim,<)$ is realizable. When it is nonempty it is a convex cell. The $\O_{p,q}[\sim,<]$ partition $\O_p\times\O_q\simeq \R^{p+q}$. It would be interesting to understand the poset structure associated to containment of closures of these. It seems natural to expect:
the poset structure is closely related to refinements of $(\sim,<)$;
maximal elements correspond to the discrete equivalence relation with the compatible total orders;
$(\sim_{A,B},<_{A,B})$ for $A=\{1<2<\cdots<p\}$ and $B=\{1<2<\cdots<q\}$ is minimal but there may be other minimal elements.
Let us put $\newcommand{\real}{\mathfrak{R}}\real_{p,q}$ the set of realizable pairs $(\sim,<)$. We define an order relation on $\real_{p,q}$ by setting
$$(\sim,<)\prec(\sim',<')\iff\text{there is an onto map of posets }
(R_{p,q}/\sim',<')\to(R_{p,q}/\sim,<)$$
Lemma. Let $(\sim,<),(\sim',<')\in\real_{p,q}$ be realizable. The following are equivalent:
$\overline{\O_{p,q}[\sim,<]}\subset\overline{\O_{p,q}[\sim',<']}$
$\O_{p,q}[\sim,<]\subset\overline{\O_{p,q}[\sim',<']}$
$\O_{p,q}[\sim,<]\cap\overline{\O_{p,q}[\sim',<']}\neq\emptyset$
$(\sim,<)\prec(\sim',<')$
Sketch of Proof. $1$ and $2$ are always equivalent and clearly imply $3$. If we let $(A_n',B_n')\in\O_{p,q}[\sim',<']$ tend to $(A,B)\in\O_{p,q}[\sim,<]$ then what can happen is fusion of $<'$-intervals of $\sim'$-equivalence relations and $<$ and $<'$ have to be compatible which amounts to the existence of an onto homomorphism. This will prove $3\implies 4$. To finish the proof we show that $4\implies 2$. It is enough to notice that if $(\sim,<)\prec(\sim',<')$ and $(A,B)\in\O_{p,q}[\sim,<]$ and $(A',B')\in\O_{p,q}[\sim',<']$ then for all $t\in[0,1)$, $$\qquad\qquad\qquad(tA+(1-t)A', tB+(1-t)B')\in\O_{p,q}[\sim',<'].\qquad\qquad\qquad\square$$
This has desirable homotopical consequences: one can use these cells to compute homotopy colimits of subcategories of the poset $(\real_{p,q,\prec})$ if one ever wanted to do so.
|
https://math.stackexchange.com/questions/4103564
|
[
"combinatorics",
"discrete-mathematics",
"equivalence-relations",
"additive-combinatorics",
"configuration-space"
] | 56 | 2021-04-15T10:36:32 |
[
"Sets where you have at most one pair from each equivalence class are called Sidon sets and there's huge theory behind them. Might be worth a shot to look into that",
"@SamuelMuldoon Thanks! I'm using Goodnotes on ipad",
"@OlivierBégassat The diagrams having a graph-paper/grid background are very pretty. What computer program did you use to draw those diagrams? I like how nice and straight the colored arrows are.",
"By the way, the slope condition is indeed sufficient in the p=2 case. I didn't write it into the question yet, but there is indeed an explicit algorithm in this case.",
"On the topic of negative slopes, I was thinking along similar lines but maybe not taking it as far as you suggest: I wanted to rename the Inclination property to Negative slope property which is the fact that the previously named sets are disjoint. I'll have to draw the map $F$ tomorrow to get a clearer picture of what you're suggesting.",
"@BenBlum-Smith I just saw your comments and only started reading them. On the unnaturality of horizontal / vertical sets: I agree, there is a better characterization purely in terms of $c^+$, $c$ and $c^-$ which is that these sets be disjoint. Let me read the rest of your comments : )",
"constraint on the slopes? Actually, I do believe so, i.e., any choice of a negative slope for each $s_{i,j}$ satisfying this set of \"cross-product\" constraints is realizable by a choice of $a_i$'s and $b_j$'s. So this constraint must give rise to the \"antidiagonal propagation\" property you picked up. Anyway, I'm optimistic that thinking this way can yield a cleaner characterization of permissible level curves, and these then characterize both the possible partitions and the orders they induce.",
"crosses these curves. The level curves are determined completely by the slope in each square $s_{i,j} := [i,i+1]\\times [j,j+1]$. Besides always being negative, the slopes of the curves in the different squares are constrained by the fact that... ok for notational convenience define $d_i = a_{i+1}-a_i$, $\\delta_j = b_{j+1}-b_j$, so the slope in $s_{i,j}$ is $-d_i/\\delta_j$. Then because $(d_i/\\delta_j)(d_k/\\delta_\\ell) = (d_i/\\delta_\\ell)(d_k/\\delta_j)$, the slopes in $s_{i,j}$ and $s_{k,\\ell}$ have to have the same product as the slopes in $s_{i,\\ell}$ and $s_{k,j}$. Perhaps this is the only",
"rectangle $[1,p]\\times [1,q]$, you can see their shape: they're all defined by piecewise-linear functions with each linear segment having negative slope: a level set's intersection with the square $[i,i+1]\\times[j,j+1]$ will have slope $-(a_{i+1}-a_i)/(b_{j+1}-b_j)$ if I'm thinking about this right. Level sets can change slope only when they cross from one of these squares to another. A partition is realizable iff it can be obtained by drawing the appropriate kinds of level curves across the diagram. The order such a partition realizes is the order in which any path from $(1,1)$ to $(p,q)$",
"be the PL function that linearly interpolates $j\\mapsto b_j$. Then the mapping $F$ defined by $(x,y)\\mapsto (f(x),g(y))$ is a PL function from the rectangle $[1,p]\\times[1,q]$ to the rectangle $[a_1,a_p]\\times [b_1,b_q]$. The set $R_{p,q}$ is the set of integer lattice points in the domain of $F$. Let $S:\\mathbb{R}^2\\rightarrow\\mathbb{R}$ be the linear function $(x,y)\\mapsto x+y$. Then the equivalence classes of $R_{p,q}$ given by these choices of $a_i$ and $b_j$ are the level sets of $S\\circ F$ restricted to $R_{p,q}$. If you don't restrict to $R_{p,q}$ but consider level sets on the whole",
"@OlivierBégassat - Okay, I just spent far longer thinking about this than I probably should've lol. I don't have a complete answer but my guess is that if your conditions aren't sufficient, they're pretty close; however, I think your focus on the horizontal and vertical sets is unnatural, and there will be a cleaner (complete) characterization that doesn't look at them. I think the following way of thinking is useful, though I haven't played it all the way out: Let $f:[1,p]\\rightarrow[a_1,a_p]$ be the PL function that linearly interpolates $i\\mapsto a_i$, and let $g:[1,q]\\rightarrow[b_1,b_q]$",
"I haven't given this more than cursory thought, but I think the answer to (2) has to be affirmative. A given realizable equivalence relation is defined by a finite system of linear equalities and inequalities with integer coefficients. If there is any solution, there is a rational solution; then multiply through by a common denominator to get integers, and add a big integer to everything to get nonnegative integers.",
"@belwarDissengulp Goodnotes on ipad",
"@OlivierBégassat Which program did you use to draw the diagrams?",
"To give even further context, the question arose when thinking about Exercise 3.5 in Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry ... I have been learning commutative algebra and algebraic geometry in my free time in the hopes of understanding elliptic curves. This is an idle question that came up in that effort which really caught my attention for the last 2-3 days.",
"@Christoph I debated including my motivation for the question but decided against it. It seemed like a problem of independent interest more related to the topology of the real line than to graded modules and rings. I can always include my motivation into the post if you see value in it.",
"@Christoph It came up when thinking about graded modules $M=\\bigoplus_{\\lambda\\in\\Lambda} M_\\lambda$ over a graded ring $R=\\bigoplus_{\\gamma\\in\\Gamma}$ and thinking about the possible ways in which $R_\\gamma \\cdot M_\\lambda \\subset M_{\\mu}$ can happen. I drew pictures and was led to drawing blob diagrams of the sort pictured above. I realized that there was a combinatorial question of interest and tried answering that. This led to the question.",
"This question is certainly elaborated and it took time and work to write it out, which is why I'm hesitant to vote to close it. Yet, I agree with @Nij that it is not the type of question suited for MSE. It looks more like a full (small) research project. Indeed, despite being extraordinarily long it misses important context: where did you come across this problem and why is it relevant to you?",
"@Nij There are really three questions: (1) characterize realizable $\\sim$ (2) decide if realizable $\\sim$ can be realized with sets of integers and (3) characterize realizable $<$. These questions don't require a build up beyond the definitions given in the first paragraph and are related enough that batching them up in a single post makes sense IMO. The other points are proposals for a characterization and these do require definitions, which is what I provide along with some illustrations. Everything beyond the three questions above represents my work and partial progress.",
"How is this remotely a question for Maths SE? There are at least four separate questions, each of which has more buildup than is expected of a single whole question here. At the very least break it up into the four parts. No question short of a thesis would sufficiently answer all four, and no answer that fully addresses just one could be compared to that which answers one other.",
"I guess what I'm getting at is that there is a stratification on the set $\\mathcal{O}^2$ where $\\mathcal{O}=\\{(x_1<\\cdots<x_n)\\in\\Bbb{R}^n\\}$ where strata are defined by the requirement that $(a,b)\\in\\mathcal{O}^2$ define the same equivalence relation. And then there should be some relation between closures of strata and refinement of the associated equivalence relations.",
"@ErickWong Additive combinatorics makes sense. The connection with configuration spaces is indeed debatable. I included it because (a) it somehow reminded me of Fox Neuwirth cells and (b) I imagine that if one were to consider refinement of these equivalence relations then refinement would be related to closure of sets of pairs of n-tuples defining the same equivalence relation.",
"I tagged additive-combinatorics since this seems closely related to additive energy. I’m afraid I don’t see the connection to configuration spaces."
] | 0 |
Science
| 0 |
272
|
math
|
Does there exist a polynomial $P(x,y)$ which detects all non-squares?
|
Problem. Does there exist a two-variable polynomial $P(x, y)$ with integer coefficients such that a positive integer $n$ is not a perfect square if and only if there is a pair $(x, y)$ of positive integers such that $P(x, y)=n$?
Context. The answer is positive for polynomials in 3 variables! This appeared as a problem in USA Team Selection Test in 2013. It turns out that the polynomial $P(x, y, z)=z^2\cdot (x^2-zy^2-1)^2+z$ enjoys the following property: a positive integer $n$ is a not a perfect square if and only if $P(x, y, z)=n$ has a solution in positive integers $(x, y, z)\in \mathbb{N}^{3}$. This construction works nicely due to Pell's equation. If $n$ is not a perfect square, then Pell's equation $x^2-ny^2=1$ has a solution in positive integers $(x_0, y_0)$, and so we get $P(x_0, y_0, n) = n$. Conversely, if $P(x, y, z)=n$, then one can show that $n$ cannot be a perfect square because $n=z^2(x^2-zy^2-1)^2+z$ can be squeezed between two consecutive perfect squares:
$$
(z(x^2-zy^2-1))^2 < n < (z(|x^2-zy^2-1|+1)^2
$$
Remark. It is clear that there is no single-variable polynomial $P(x)$ which could achieve the desired property. Indeed, there are arbitrary number of consecutive non-squares, and a polynomial $P(x)$ of degree $n>1$ cannot output a consecutive list of $n+1$ numbers. This last claim itself is a nice problem; for a solution, see Example 2.24 in page 11 of Number Theory: Concepts and Problems by Andreescu, Dospinescu and Mushkarov.
|
https://math.stackexchange.com/questions/4438559
|
[
"number-theory",
"polynomials",
"contest-math"
] | 49 | 2022-04-28T11:58:00 |
[
"@Sil Great suggestion! Just posted the problem on MathOverFlow at this link.",
"Interesting problem, you might want to consider asking on MathOverflow as it was not resolved here even after bounty.",
"In fact, the polynomial $P=p*q=[(p-1)/2 + (p+1)/2]*[(q-1)/2 + (q+1)/2]$ can represent non-squares and non-primes. Example: $91=7*13=(3+4)*(6+7)=3*6 +3*7 +4*6 + 4*7$ with $(91-1)/2=3*7 + 4*6=45$ and $(91+1)/2=3*6 + 4*7=46$.",
"I cannot answer the above question but I can say that every odd square can be represented by $P(x,y)=(x+y)^2=x^2 + 2xy +y^2$ with $xy=2T_n$ with $T_n$ twice a triangular number and also $y=x+1$. Ex: $81=9^2=(4+5)^2=4^2 + 2*4*5 + 5^2$. Example of P(x,y) for a non-square: $73=(72/2 + 74/2)=1+2*18+6^2$. Note that $18$ is not twice a triangular number. This means than non-squares can be represented by $P(x,y)$ but $x$ and $y$ will necessarily differ by more than $1$. If $(N-1)/2$ is odd, the number can never be a square. Ex $71$ with $(71-1)/2=35$ or $91$ with $(91-1)/2=45$.",
"A possibly helpful note: This can be done with two very similar polynomials $P_1(x,y)=(x+y-1)^2+y$ and $P_2(x,y)=(x+y-1)^2+(x+y-1)+y$. These two polynomials have disjoint images, with $\\operatorname{Im}(P_1)\\cup \\operatorname{Im}(P_2)$ representing all non-squares: the image of $P_1$ is the union of the intervals $[n^2+1,n^2+n]$ for each $n\\geq 1$, while the image of $P_2$ is the union of the intervals $[n^2+n+1,n^2+2n]$ for each $n\\geq 1$. (This can be used to construct a $3$-variable polynomial $$P(x,y,z)=\\big(1-(z-1)(z-2)\\big)\\big((x+y-1)^2+(z-1)(x+y-1)+y\\big)$$ with the desired property."
] | 0 |
Science
| 0 |
273
|
math
|
The sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms
|
PROBLEM. Show that the sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms.
It suffices to show that the terms of the sequence
$$\,b_n=\mathrm{e}^n\,\mathrm{mod}\, 2,\,\,\,n\in\mathbb N,$$
are dense in $[0,2]$.
Unfortunately, Weyl's Theorem does not look helpful in this case.
EDIT. As Chris Culter said, the claim that the terms of the sequence
$$\,b_n=\mathrm{e}^n\,\mathrm{mod}\, 2,\,\,\,n\in\mathbb N,$$
are dense in $[0,2]$ is (or might be) an open problem. Nevertheless, this does not imply that the claim that the sequence $\,a_n=\lfloor \mathrm{e}^n\rfloor$ contains infinitely many odd and infinitely many even terms is necessarily an open problem as well. It is also noteworthy that it is relatively easy to construct an irrational $\alpha$ with the property that the sequence $\,\alpha^n\,\mathrm{mod}\, 2,\,\,n\in\mathbb N,$ is NOT dense in $[0,2]$.
|
https://math.stackexchange.com/questions/2621776
|
[
"real-analysis",
"sequences-and-series",
"uniform-distribution"
] | 46 | 2018-01-25T23:17:44 |
[
"For any $a\\in\\mathbb N$, $\\lfloor a^n\\rfloor$ will have the same parity as $a$. For $0 < a < 1$ we always have $\\lfloor a^n\\rfloor = 0$. But for any other values of $a$ I don't see obvious way to prove or disprove the property from the problem. Maybe it has nothing to do specifically with $e$ and a proof would cover a class of reals (potentially all non-integers $>1$ ?).",
"Exercises for university students? Seems hard. The sequence is tabulated at oeis.org/A000149 but I don't think anything at that page helps to resolve the question.",
"I had found it in a Greek site with Mathematical problems for university students.",
"What's the source of this question, please?",
"@MathematicsStudent1122 Does \"look\" helpful, but not really helpful.",
"If $e^n$ is dense modulo $2$, then it is also dense modulo $1$, yet the latter is an open problem, at least circa 2003 per this paper."
] | 0 |
Science
| 0 |
274
|
math
|
Is every finite list of integers coprime to $n$ congruent $\pmod n$ to a list of consecutive primes?
|
For example the list $(2, 1, 2, 1)$ is congruent $\pmod 3$ to the consecutive primes $(5, 7, 11, 13)$. But how about the list $(1,1,1,1,1,1,1,1,2,3,4,3,2,3,1) \mod 5$?
More generally, we are given some integer $n \geq 2$ and a finite list of integers that are coprime to and less than $n$. Is it always possible to produce the same list by consecutive primes $\pmod n$?
Formally: given $n \geq 2$ and $(a_0,a_1,\cdots \,a_k)$ such that for all $i$, $GCD(a_i, n) = 1$, is there a list of consecutive primes such that each $p_i \equiv a_i \pmod n$?
|
https://math.stackexchange.com/questions/2194973
|
[
"number-theory",
"elementary-number-theory",
"prime-numbers",
"prime-gaps"
] | 40 | 2017-03-20T05:52:33 |
[
"For those seeking Shiu's paper, here is the citation: Shiu, D.K.L. (2000), Strings of Congruent Primes. Journal of the London Mathematical Society, 61: 359-373. doi.org/10.1112/S0024610799007863",
"Very interesting question. Just as an observation, if we took $n$ to be a prime and a finite list of values less than $n$ as stated in the problem and if the list of values is congruent to successive primes mod $n$, then we could encode that entire list using just the initial prime's index and deterministically recover the elements of the list with no additional information (except the size of the list). Perhaps, this is intricately linked to Shannon Entropy.",
"It's not, and I never said it was...",
"@RushabhMehta How is that Greenwood-Tao result an advance on Dirichlet's theorem that every AP of integers $a, a+d,\\dots$ with $d>0$ coprime to $a$ contains arbitrarily many primes?",
"I Think the question is more interesting if we consider the case where $n$ is a prime itself.",
"What about using a generalized Dirichlet series but with multiple entries, i.e. characters $\\chi: (\\mathbb{Z}/n\\mathbb{Z} )^* \\times \\ldots \\times (\\mathbb{Z}/n\\mathbb{Z} )^* \\to \\mathbb{C}^*$ ? I just don't know how to implement the \"consecutive primes\" constraint..",
"Well lets first note, that even residues modulo odd $n$ aren't possible to be prime unless the multiplier on $n$ is odd. Likewise, odd residues aren't representing primes unless the multiplier on $n$ is even in this case. This at least forces gap size constraints (1,4,2,3) mod 5 won't happen below gap sizes 8,8,6 with gaps 3 mod 5, 3 mod 5, and 1 mod 5 Your long one needs gaps of 10,10,10,10,10,10,10,6,6,6,4,4,6,8",
"@cr001 Can't find a wiki link. TLDR, it was proven by Daniel Shiu in the 90s, and the statement is as follows: for any coprime $a,b\\in\\mathbb N$, there exists arbitrarily long sequences of consecutive primes, all congruent to $a\\bmod b$. Similar but different than Greenwood-Tao, which doesn't require consecutiveness of the primes.",
"nobody intrested in my bounty ??",
"Is there a Wikipedia link to the Shiu Theorem? I wan't able to find any statement of the Theorem via google search.",
"For the record, you can find $\\{1,1,1,1,1,1,1,1,2,3,4\\} \\pmod {5}$ starting at $1313286451$, which looks about in line with what I'd expect from the random model. So yeah, it certainly seems likely to me that arbitrarily long sequences of this sort exist.",
"This is an open problem except in the case where all $a_i$ are equal, which is a famous theorem of Shiu. There are a few other cases that are consequently true for simple combinatorial reasons, such as $(1,1,1,1,1,1,1,1,2)$ mod $3$, but essentially Shiu's theorem is the best we have, AFAIK.",
"Very interesting question. Altough I would be surprised if this wasn't true, I would be even more surprised if there was an easy proof.",
"@David Schneider-Joseph thanks.",
"No, for example $(2, 2) \\pmod 4$ would require two even primes. The question might be more interesting if you require that entries in the list are coprime with $n$ (a generalization of disallowing 0)."
] | 0 |
Science
| 0 |
275
|
math
|
Are these generalizations known in the literature?
|
By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(2k)}{(a-2k)!}\ln^{a-2k}(z)\tag{b}$$
I managed to find:
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{\sqrt{y}}\sum_{k=0}^n\binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}\ln^{2k}(y)\tag{c}$$
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2\sqrt{y}}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}\ln^{2k+1}(y)\tag{d}$$
$$\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$
$$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{e}$$
$$\int_0^\infty\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!}*$$
$$\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{f}$$
$$\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\frac12|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$
$$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{2(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$
$$+2\sum_{k=0}^a\frac{(2k+2n+1)!}{(2k+1)!}4^k \eta(2a-2k)\beta(2k+2n+2)\tag{g}$$
$$\int_0^1\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=$$
$$\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{4(2a-1)!}\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$
$$+\sum_{k=0}^a\frac{(2k+2n-1)!}{(2k)!}4^k \eta(2a-2k)\beta(2k+2n)\tag{h}$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\frac1{2(2n)!}|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$
$$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{2(2n)!(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$
$$+\frac{2}{(2n)!}\sum_{k=0}^a\frac{(2k+2n+1)!}{(2k+1)!}4^k \eta(2a-2k)\beta(2k+2n+2)\tag{i}$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\frac{\left(\frac{\pi}{2}\right)^{2n-1}}{4(2n-1)!(2a-1)!}*$$
$$\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$
$$-\frac{1}{(2n-1)!}\sum_{k=0}^a\frac{(2k+2n-1)!}{(2k)!}4^k \eta(2a-2k)\beta(2k+2n)\tag{j}$$
Question: Are the results of $(c)$ to $(j)$ known in the literature?
If the reader is curious about the correctness of the results above and wants to verify them on Mathematica, the Mathematica command of $|E_r|$ is Abs[EulerE[r]]
Thanks,
Proof of (a): By using Euler's reflection formula
$$\Gamma(m)\Gamma(1-m)=\pi \csc(m\pi),\quad m\notin\mathbb{Z}$$
and beta function
$$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}dx$$
with $a=m$ and $b=1-m$, we have
$$\pi\csc(m\pi)=\int_0^\infty\frac{x^{m-1}}{1+x}dx$$
differentiate both sides $2n$ times with respect to $m$ then let $m$ approach $1/2$
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=\frac{\pi}{2^{2n+1}}\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)$$
the proof completes on using
$$\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=|E_{2n}|\pi^{2n}$$
which is explained here.
Proof of (b): Divide both sides of the common dilogarithm identity
$$\text{Li}_2(-z)+\text{Li}_2(-1/z)=-\frac{\ln^2(z)}{2}-2\eta(2)$$
by $z$ then integrate repeatedly.
Proof of (c): Let $yx^2=t^2$,
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac{1}{\sqrt{y}}\int_0^\infty\left(\ln(\sqrt{y})-\ln(t)\right)^{2n}\frac{dt}{1+t^2}$$
$$=\frac{1}{\sqrt{y}}\int_0^\infty\left(\sum_{k=0}^{2n} \binom{2n}{k}(-\ln(t))^{2n-k}\ln^k(\sqrt{y})\right)\frac{dt}{1+t^2}$$
$$=\frac1{\sqrt{y}}\sum_{k=0}^{2n}\binom{2n}{k}\ln^k(\sqrt{y})\int_0^\infty\frac{(-\ln(t))^{2n-k}}{1+t^2}dt$$
Since we care for only the even powers of $\ln(t)$ as the odd powers make the integral zero, we have
$$\sum_{k=0}^{2n}f(k)=\sum_{k=0}^n f(2k)$$
and so
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac1{\sqrt{y}}\sum_{k=0}^{n}\binom{2n}{2k}\ln^{2k}(\sqrt{y})\int_0^\infty\frac{\ln^{2n-2k}(t)}{1+t^2}dt$$
The proof completes on using the result in $(a)$.
Proof of (d): We follow the same steps in proof $(c)$:
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-1}{\sqrt{y}}\int_0^\infty\left(\ln(\sqrt{y})-\ln(t)\right)^{2n-1}\frac{dt}{1+t^2}$$
$$=\frac{-1}{\sqrt{y}}\int_0^\infty\left(\sum_{k=0}^{2n-1} \binom{2n-1}{k}(-\ln(t))^{2n-k-1}\ln^k(\sqrt{y})\right)\frac{dt}{1+t^2}$$
$$=\frac{-1}{\sqrt{y}}\sum_{k=0}^{2n-1}\binom{2n-1}{k}\ln^k(\sqrt{y})\int_0^\infty\frac{(-\ln(t))^{2n-k-1}}{1+t^2}dt$$
Since we care for only the even powers of $\ln(t)$, we have
$$\sum_{k=0}^{2n-1}f(k)=\sum_{k=0}^{n-1} f(2k+1)$$
and so
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-1}{\sqrt{y}}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}\ln^{2k+1}(\sqrt{y})\int_0^\infty\frac{\ln^{2n-2k-2}(t)}{1+t^2}dt$$
The proof completes on using the result in $(c)$.
Proof of (e): Using the integral representation of the polylogarithm function:
$$\text{Li}_{a}(z)=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\frac{z\ln^{a-1}(t)}{1-zt}dt$$
We have
$$\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}\left(\frac{1}{(2a)!}\int_0^1\frac{-x^2\ln^{2a}(y)}{1+yx^2}dy\right)dx$$
$$=\frac{1}{(2a)!}\int_0^1\ln^{2a}(y)\left(\int_0^\infty\frac{-x^2\ln^{2n}(x)}{(1+x^2)(1+yx^2)}dx\right)dy$$
$$=\frac{1}{(2a)!}\int_0^1\frac{\ln^{2a}(y)}{1-y}\left(\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx-\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx\right)dy$$
use $(a)$ and $(c)$ for the inner integrals
$$=\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2a)!}\left(|E_{2n}|\int_0^1\frac{\ln^{2a}(y)}{1-y}dy-\sum_{k=0}^n\binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}\int_0^1\frac{\ln^{2k+2a}(y)}{\sqrt{y}(1-y)}dy\right)$$
The proof completes on using:
$$\int_0^1\frac{\ln^a(x)}{1-x}dx=(-1)^aa!\zeta(a+1)$$
$$\int_0^1\frac{\ln^a(x)}{\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\zeta(a+1)$$
Proof of (f): Following the same steps in proof $(e)$:
$$\int_0^\infty\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n-1}(x)}{1+x^2}\left(\frac{1}{(2a-1)!}\int_0^1\frac{x^2\ln^{2a-1}(y)}{1+yx^2}dy\right)dx$$
$$=\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\int_0^\infty\frac{x^2\ln^{2n-1}(x)}{(1+x^2)(1+yx^2)}dx\right)dy$$
$$=\frac{1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1-y}\left(\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx-\int_0^\infty\frac{\ln^{2n-1}(x)}{1+x^2}dx\right)dy$$
substitute $(d)$ and notice that the second inner integral is zero
$$=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}\int_0^1\frac{\ln^{2k+2a}(y)}{\sqrt{y}(1-y)}dy$$
The proof completes on using
$$\int_0^1\frac{\ln^a(x)}{\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\zeta(a+1)$$
Proof of (g):
$$\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx}_{x\to1/x}$$
$$=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-1/x^2)}{1+x^2}dx$$
add the integral to both sides
$$2\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$
$$+\int_0^1\frac{\ln^{2n}(x)[\text{Li}_{2a+1}(-x^2)-\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx$$
the first integral is given in $(e)$. For the second integral, replace $a$ by $2a+1$ in $(b)$ then use $\sum_{k=m}^na_k=\sum_{k=m}^{n}a_{n-k+m}$
$$\text{Li}_{2a+1}(-z)-\text{Li}_{2a+1}(-1/z)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k+1)!}\ln^{2k+1}(z)$$
and so
$$\int_0^1\frac{\ln^{2n}(x)[\text{Li}_{2a+1}(-x^2)-\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx=-4\sum_{k=0}^a\frac{\eta(2a-2k)}{(2k+1)!}4^k\int_0^1\frac{\ln^{2k+2n+1}(x)}{1+x^2}dx$$
and the proof completes on using
$$\int_0^1\frac{\ln^a(x)}{1+x^2}dx=(-1)^a a! \beta(a+1)$$
Proof of (h): We follow exactly the same steps in proof $(g)$ but here we use
$$\text{Li}_{2a}(-z)+\text{Li}_{2a}(-1/z)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(z)$$
Proof of (i) and (j): Using the generating function
$$\sum_{k=1}^\infty H_k^{(a)} x^k=\frac{\text{Li}_a(x)}{1-x}$$
and the fact that
$$\int_0^1 x^{k}\ln^n(x)dx=\frac{(-1)^nn!}{(k+1)^{n+1}}$$
we have:
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\frac1{(2n)!}\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\frac{-1}{(2n-1)!}\int_0^1\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx$$
These two integrals are given in $(g)$ and $(h)$.
Edit 3/26/2024
More generalized polylogarithmic integrals can be found in this preprint.
|
https://math.stackexchange.com/questions/4397045
|
[
"integration",
"sequences-and-series",
"reference-request",
"harmonic-numbers",
"polylogarithm"
] | 40 | 2022-03-05T22:37:13 |
[
"Very impressive and nice work !",
"Marvelous job. A lot of work.",
"Thanks for great effort in exploring to such an amazing depth!",
"@AliShadhar Thank you so much for your comment. I meant significance in a broad sense... Have similar identities been applied in other areas of math (to give better approximation or better bounds, etc)?",
"I'm sorry for my limited understanding, but can somebody elaborate on the significance of these identities (besides their novelty)?",
"It might be interesting to know that these results have been submitted virtually literally by a group of authors to a scientific journal 25 days after the publication here and have been published without mentioning the source. I wrote an e-mail to the responsible author pointing out this violation of the scientific code of conduct. He replied finally, quote, \"I have e-mailed the editor in chief of the journal asking him to retract the paper.\"",
"Oh yes, now these generalizations are known in the (online) literature !!",
"(+1) Great work. Nice proofs, and nicely laid out, as you always do. I'll put a printout into my copy of your book.",
"Nice results . We can find a recursive relationship using complex analysis",
"Very impressive.",
"(+1) Great pack of results to have in hand and refer to! :-)"
] | 0 |
Science
| 0 |
276
|
math
|
If $K\cong K(X)$ then must $K$ be a field of rational functions in infinitely many variables?
|
If $k$ is any field, then the field $K=k(X_0,X_1,\dots)$ of rational functions in infinitely many variables satisfies $K(X)\cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_{n+1}$). My question is, does the converse hold? That is:
Suppose $K$ is a field such that $K\cong K(X)$. Must there exist a field $k$ such that $K\cong k(X_0,X_1,\dots)$?
(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $R\cong R[X]$ but $R\not\cong S[X_0,X_1,\dots]$ for any ring $S$. A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one. However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group. (In particular, an answer to the former question would give a field $F$ such that $F\cong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)
|
https://math.stackexchange.com/questions/2856060
|
[
"abstract-algebra",
"field-theory"
] | 37 | 2018-07-18T15:05:43 |
[
"@JeremyRickard Nevermind, those are the same thing. And I see now why my idéa doesn't work.",
"@JeremyRickard I thought that $K(X)$ denoted $\\text{Frac} (K[X])$, not field extension.",
"@Kilian I don't see why $K$ would be isomorphic to $K(X)$. $\\mathbb{R}((X_0,X_1,\\dots))$ is not generated as a field extension of $\\mathbb{R}((X_1,X_2,\\dots))$ by $X_0$.",
"Just an idea, haven't fully thought it through, but if we let $K=\\mathbb{R}((X_0,X_1...))$, i.e. the field of formal Laurent series in infinitely many variables over $\\mathbb{R}$, it seems like we would have $K\\cong K(X)$ but I don't see how it would be isomorphic to $k(X_0,X_1,...)$ for any field $k$. Possibly a counterexample?",
"It’s an example of something. Either it’s a counterexample for your question, or the rational group algebras of the nonisomorphic groups $\\mathbb{Z}^\\omega$ and $\\mathbb{Z}^\\omega\\oplus\\mathbb{Z}^{(\\omega)}$ have isomorphic fields of fractions, answering my question.",
"@JeremyRickard: That's a neat idea. I don't actually recall thinking through it before (I think the constructions I was trying were all countable).",
"Do you know that the field of fractions of the rational group algebra of the Baer-Specker group is not a counterexample? Given the links in the question, I’m guessing you’ve thought about it.",
"Sure. Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though. Feel free to crosspost it if you want.",
"Might it be worth asking this interesting question on MathOverflow?"
] | 0 |
Science
| 0 |
277
|
math
|
Constructing an infinite chain of subfields of 'hyper' algebraic numbers?
|
This has now been cross posted to MO.
Let $F$ be a subset of $\mathbb{R}$ and let $S_F$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $F$.
That is, we let $S_F$ denote
$$\bigg \{x \in \mathbb{R}: 0=\sum_{i=1}^n{a_i x^{e_i}}: e_i \in F \text{ distinct}, a_i\in F \text{ non-zero}, n\in \mathbb{N} \bigg \}$$
Then $S_{\mathbb{\mathbb{Q}}}$ is the set of algebraic real numbers and we start to see the beginnings of a chain:
$
\mathbb{Q}
\subsetneq S_\mathbb{Q} \subsetneq
S_{S_\mathbb{Q}} $
Main Question
Does this chain continue forever? That is, we let $A_0= \mathbb{Q}$ and let $A_{n+1}=S_{A_{n}}$. Is it the case that $A_n \subsetneq A_{n+1}$ for all $n\in\mathbb{N}$?
Other curiosities:
Is $A_i$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $F \subset \mathbb{R}$, a field implies that $S_F$ is a field?
Is it possible to see that $e\notin \cup A_i$? Perhaps this is just a tweaking of LW Theorem.
|
https://math.stackexchange.com/questions/3014759
|
[
"real-analysis",
"field-theory",
"transcendental-numbers"
] | 36 | 2018-11-26T10:55:38 |
[
"@JackM, $x=0$ satisfies this and is algebraic. I am not sure I understand your meaning... But if you take $x=y^6$ then we can replace your equation with $y^3+y^2=0$ and we can always manage to do this type of change when the exponents are rational.",
"Why is $S_\\mathbb Q$ equal to the set of algebraic real numbers? For instance, is any solution to $\\sqrt x+\\sqrt[3]x=0$ algebraic?",
"I am not so sure but maybe this piece by B Zilbner speaks to the question.",
"Actually... $S_F$ is clearly closed under multiplicative inverses because $-e_i$ is in a field $F$ whenever $e_i$ is in $F$",
"It seems like we might be able to achieve multiplicative inverses using a technique analogous to the one presented here. I haven't worked out all the details but it looks like it could be a good lead",
"How do we prove that $4^\\sqrt3+1\\in\\cup A_n$? This seems like an argument for defining $S_F$ to include closure under field operations.",
"I don’t think we can rule out $e\\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\\sqrt6-4^\\sqrt3-1$.",
"@JyrkiLahtonen, for this purpose I think we can define $(-|x|)^\\alpha$ as $-(|x|^\\alpha)$.",
"Just a question: is $n$ (defined inside the definition of $S_F$ as a natural number) also arbitrary for any member of $S_F$?",
"(continued) We should answer $e^{i \\pi \\sqrt{2}}$ but all of sudden it's start look like the question is better raised in the complex numbers. But maybe there is a different polynomial that satisfies $-x$?",
"@JyrkiLahtonen. Just to be clear about the concern raised: $ x^{\\sqrt2}-2=0 \\implies x= \\sqrt{2}^{1/\\sqrt{2}}$ What (generalized) polynomial is satisfied by $-x$? It could be that $ (-x)^{\\sqrt2}-2=0$ is not a satisfactory answer. No. I would think this is indeed a satisfactory answer. The first polynomial has a non negative domain and the second has a non positive domain. What trouble do we get into if we take this naive approach? This route at least wins us closure under negation. Oh. The trouble we get into is when someone asks what the coefficient of the $x^{\\sqrt 2}$ term is...",
"Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?",
"I'm still somewhat concerned about the definition of $x^\\alpha$ when $x<0$ and $\\alpha$ is not an integer. But, there is a cool question in here (possibly a very difficult one).",
"$2^{\\sqrt2}$ is transcendental, but it satisfies $x^{\\sqrt2}-4=0$, so it's in $A_2$. I'd conjecture that $2^{2^{\\sqrt2}}$ is in $A_3$ and not $A_2$, that $2^{2^{2^{\\sqrt2}}}$ is in $A_4$ and not $A_3$, etcetera.",
"How do you propose to define $x^\\alpha$ if $\\alpha \\notin \\mathbb{Q}$?",
"Every set in the chain is countable, so you certainly don't hit $\\mathbb{R}$ at any point."
] | 0 |
Science
| 0 |
278
|
math
|
An iterative logarithmic transformation of a power series
|
Consider the following iterative process. We start with the function having all $1$'s in its Taylor series expansion:
$$f_0(x)=\frac1{1-x}=1+x+x^2+x^3+x^4+O\left(x^5\right).\tag1$$
Then, at each step we apply the following transformation:
$$f_{n+1}(x)=x^{-1}\log\left(\frac{f_n(x)}{f_n(0)}\right).\tag2$$
A few initial iterations give us:
$$
\begin{array}{l}
f_1(x)=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\frac{x^4}{5}+O\left(x^5\right), \\
f_2(x)=\frac{1}{2}+\frac{5 x}{24}+\frac{x^2}{8}+\frac{251 x^3}{2880}+\frac{19 x^4}{288}+O\left(x^5\right), \\
f_3(x)=\frac{5}{12}+\frac{47 x}{288}+\frac{2443 x^2}{25920}+\frac{5303 x^3}{82944}+\frac{412589x^4}{8709120}+O\left(x^5\right),
\end{array}\tag3
$$
and their initial terms form the following sequence:
$$1,\,\frac{1}{2},\,\frac{5}{12},\,\frac{47}{120},\,\frac{12917}{33840},\,\frac{329458703}{874222560},\,\dots,\tag4$$
whose denominators grow pretty quickly, but which appear to slowly converge to a value $$\lambda\stackrel{\color{#aaaaaa}?}\approx0.3678\dots\tag5$$
If we assume that the process with the iterative step $(2)$ converges to a fixed point, we can see that it must have a form:
$$f_\omega(x)=-x^{-1}\,W(-c\,x)=c+c^2 x+\frac{3\, c^3\, x^2}{2}+\frac{8\, c^4\, x^3}{3}+\frac{125\, c^5\, x^4}{24}+O\left(x^5\right),\tag6$$
where $W(\cdot)$ is the Lambert 𝑊-function, and $c$ is a coefficient that is not uniquely determined but depends on the choice of the initial function $f_0(x)$. In our case, $c=\lambda$.
Questions: Does this process actually converge to a fixed point? If yes, then what is a closed-form expression (or another useful description) for $\lambda$?
Update: An explicit recurrent formula for the coefficients (the parenthesized superscript $m$ in $a_n^{\small(m)}$ is just the second index of the coefficient; the sum $\sum_{\ell=1}^m$ is assumed to be $0$ when $m=0$):
$$f_n(x)=\sum_{m=0}^\infty a_{n\vphantom{+0}}^{\small(m)}x^m,$$
where
$$a_{0\vphantom{+0}}^{\small(m)}=1,\quad a_{n\vphantom{+0}}^{\small(m)}=\frac1{\,a_{n-1}^{\small(0)}\,}\left(a_{n-1}^{\small(m+1)}-\frac1{m+1} \sum_{\ell=1}^m\ell\;a_{n\vphantom{+0}}^{\small(\ell-1)}\,a_{n-1}^{\small(m-\ell+1)}\right).$$
The sequence of coefficients $(4)$ is $\big\{a_{n\vphantom{+0}}^{\small(0)}\big\}$.
|
https://math.stackexchange.com/questions/4338032
|
[
"limits",
"logarithms",
"power-series",
"closed-form",
"lambert-w"
] | 36 | 2021-12-19T19:33:11 |
[
"He also gives the first terms of the next power series in the sequence. Then, he writes the first ever recorded insights on the formula for iteration of power series $f \\circ \\ldots \\circ f(x)$, which then insipired Schröder in his 1872 paper, in which he first considered, what now call, the \"Schröder equation\". This is one of the first paper on iteration theory, which unfortunately doesn't seem to apply to your case.",
"Interestingly enough, Caylay briefly considered this problem in a note called \"On Some Numerical Expansions\" from the Quarterly Journal of Pure and Applied Mathematics, vol. III. (1860), pp. 366-369, but he didn't go much farther than you. What stroke me is that he got the same first few terms as you, but he was probably wrong for the last term \"$\\log\\left(-\\frac2x\\log\\left(\\frac1x\\log(1+x)\\right)\\right)=-\\frac5{12}x+\\frac{47}{288}x^2+\\frac{2443}{25920}x^3-\\frac{5303}{82944}x^4+\\frac{19631}{580608}x^5+\\&c$\".",
"@VladimirReshetnikov It seems that, if $f_n(0)=a/b$ (in lowest terms), then the denominator of $f_{n+1}(0)$ is a small multiple ($2$ or $6$) of $ab$. Let $b_n$ be the denominator of $f_n(0)$. If $f_n(0)$ converges to a limit $\\lambda$, then this would imply that $b_{n+1}$ is multiplicatively not far off from $\\lambda b_n^2$, which gives double-exponential growth.",
"Another very interesting question is how fast do denominators grow (or would grow if we ignored occasional cancellations of some of their factors with numerators). Empirically, they appear to grow (very roughly) doubly exponentially: $2^{2^{n-0.97942\\dots}}$.",
"Originally I found this sequence when I tried to construct the continued exponent expansion of the Fibonacci numbers: pbs.twimg.com/media/FHAQfurVQAANsLx?format=png&name=large, but later realized I can get it simply from $1/(1-x)$.",
"@mathworker21 Usually, OEIS keeps numerators and denominators as distinct sequences. This entry mixes them into one interleaved. Probably this is why I was not able to find the coefficients in OEIS after I computed a few of them.",
"@mathworker21 Thanks! It would be nice to see a rigorous proof. I haven't been able to prove it myself so far.",
"it says the limit is $1/e$ here.",
"I concur with Claude and will go further saying this is a very nice problem in every respect.",
"I think $\\lambda =1/e=0.367879441\\ldots$. A rough argument is as follows. It seems that each $f_n(x)$ is singular at $x=1$. Since the limit function is $-x^{-1}W(-\\lambda x)$ and $W(z)$ is singular at $z=-1/e$, $\\lambda$ should be $1/e$ in order for $f_\\omega(x)$ to be singular at $x=1$.",
"The formula is correct. We assume that $\\sum_{\\ell=1}^0(\\dots)=0$, because there are no indices satisfying $1\\le\\ell\\le0$, so the sum is empty .",
"is there a typo in your explicit formula? We are after $a_n^{(0)}$ and when I tried plug $m = 0,$ the sum starts at $l=1$ and ends at $l = m.$",
"Whatever the answer could be, this is a nice problem.",
"By the mean value theorem, taking $x\\to 0,$ you get $$f_{n+1}(0)=\\frac{f_n’(0)}{f_n(0)}.$$ Not sure if that helps in any way."
] | 0 |
Science
| 0 |
279
|
math
|
Smallest Subset of $\mathbb{R}_{>0}$ Closed under Typical Operations
|
Let $S$ denote the smallest subset of $\mathbb{R}_{>0}$ which includes $1$, and is closed under addition, multiplication, reciprocation, and the function $x,y \in \mathbb{R}_{>0} \mapsto x^y.$
Questions:
1. Do either $S$ or its elements have an accepted name?
2. Where can I learn more about the set $S$ and it elements? (Reference Request)
3. Do there exist positive, real algebraic numbers which are not in $S$?
4. Are either $e$ or $\pi$ elements of $S$?
What I Already Know:
By the Gelfond-Schneider Theorem, $S$ includes some transcendental numbers, like $2^{\sqrt{2}}$.
|
https://math.stackexchange.com/questions/1081468
|
[
"number-theory",
"reference-request",
"algebraic-number-theory",
"real-numbers",
"transcendental-numbers"
] | 36 | 2014-12-26T06:48:47 |
[
"Could we build this set by transfinite recursion and then study its properties (like for the Borel Hierarchy)?",
"@EricWofsey, if S is closed under subtraction it would include $2^\\sqrt{2}-1$, and that seems unlikely.",
"S is a subset of the positive \"elementary-logarithmic\" or \"closed-form\" numbers, as defined by Chow. (timothychow.net/closedform.pdf) So his paper proves, assuming Schanuel's conjecture, that the positive roots of $2x^5-10x+5$ are not in S.",
"@YoTengoUnLCD, in any partially ordered set $P$, every subset $A$ of $P$ has at most one minimum element (but may have many \"minimal\" elements). So let $P$ denote the poset $(\\mathcal{P}(\\mathbb{R}_{>0}),\\subseteq).$ Let $A$ denote those elements of $P$ that are closed under the relevant operations. Then $A$ has at most one least element. Regarding existence of a minimum element, it can be constructed as $\\bigcap A$.",
"Just a question: what tells you that this characterization determines $S$? As inclusion is not a total order...",
"I might add another question: is $S$ closed under subtraction (when the difference is positive)?",
"@Jihad, I suppose that any incomputable would not belong to $S,$ and there exist incomputable numbers that can be defined explicitly. Although my knowledge of computability theory is sufficiently poor that I am not 100% certain of this statement. More interestingly perhaps, I have hunch that $e$ and $\\pi$ do not belong to $S$, and that there might exist a fifth-degree polynomial (or higher) over $\\mathbb{R}$ with a solution that does not belong to $S$.",
"Do you know any number that does not belong to this set?",
"@Jihad, it certainly is. That doesn't tell us much, though, since the set of algebraic numbers is also countable.",
"Seems like this set should be countable."
] | 0 |
Science
| 0 |
280
|
math
|
Determining the kernel of a Vandermonde-like matrix
|
The kernel of a Vandermonde matrix can be determined using this formula.
The following type of matrix has a similar structure, and should also have a one-dimensional kernel.
$$V=
\begin{bmatrix}
1 & 1 & 1 & \ldots & 1 \\
x_1 & x_2 & x_3 & \ldots & x_n \\
x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \ldots & x_n^{m-1}\\
y_1 & y_2 & y_3 & \ldots & y_n \\
y_1x_1 & y_2x_2 & y_3x_3 & \ldots & y_nx_n \\
y_1x_1^2 & y_2x_2^2 & y_3x_3^2 & \ldots & y_nx_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y_1x_1^{m-1} & y_2x_2^{m-1} & y_3x_3^{m-1} & \ldots & y_nx_n^{m-1}\\
y_1^2x_1 & y_2^2x_2 & y_3^2x_3 & \ldots & y_n^2x_n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y_1^{m-1}x_1^{m-1} & y_2^{m-1}x_2^{m-1} & y_3^{m-1}x_3^{m-1} & \ldots & y_n^{m-1}x_n^{m-1}\\
\end{bmatrix} \in \mathbb{R}^{(n-1)\times n}$$
where $n = m^2+1$ and $(x_i, y_i) \neq (x_j, y_j)$ for $i \neq j$; i.e. there are $m$ groups of $m$ rows with all possible combinations of powers $y^ax^b$ and one more column than rows.
Does a similar analytical form exist for it? Or, would additional constraints be required, like $x_i^ay_i^b \neq x_i^cy_i^d$ for $i \neq j$?
(Crossposted to MO)
|
https://math.stackexchange.com/questions/407484
|
[
"linear-algebra",
"matrices",
"determinant"
] | 35 | 2013-05-31T00:32:53 |
[
"Wouldn't there be an additional row of $y_1^2 \\cdots y_n^2$ before the row $y_1^2x_1 \\cdots y_n^2x_n$?",
"Assuming you're in the complex field, what you're asking for is exactly which are the two-variable polynomials in $\\mathbb{C}[x,y]$ with degree $m-1$ in both $x$ and $y$, such that they are zero on a given set of points $\\{ (x_1,y_1), (x_2,y_2) , ... (x_n, y_n)\\}$. This is just Lagrange interpolation.",
"Questions: have you tried using Lagrange interpolation, as in the regular proof of the kernel of the van der monde matrix? Also, have you tried getting people on mathematica.stackexchange to experiment with different $m$?",
"Edited the question to make it clearer.",
"Ah, I see now. Nevermind.",
"There are $m$ groups of $m$ rows (All possible combinations of powers $x^iy^j$); the matrix should have one more column than rows to give a 1-dimensional kernel, therefore $n=m^2+1$.",
"Shouldn't $n=2m+1$ rather than $n=m^2+1$?"
] | 0 |
Science
| 0 |
281
|
math
|
Is the power of $2$ in the Euclidean norm related to the fact that the algebraic closure of the reals is $2$-dimensional?
|
Consider any local field $K$, endowed with its topological field structure. We define the function $| \cdot | : K \to \mathbb{R}_{\ge 0}$ as
$$|x| = \frac{\mu(xS)}{\mu(S)},$$
where $\mu$ is any Haar measure (which exists because a local field is additively a locally compact group), and $S$ is any measurable set of nonzero measure (see e.g. Definition 12.28 here). This definition can be shown to be independent of $\mu$ and $S$.
Now consider a vector space $V \simeq K^n$ over $K$, and equip it with the $p$-length function $$\|(x_1,x_2,\ldots,x_n)\|_p = (|x_1|^p+|x_2|^p+\ldots+|x_n|^p)^{1/p}$$ for some $p \in [1,\infty]$ (where as usual we define $\| \cdot \|_\infty$ as the limit of $\| \cdot \|_p$ as $p\to \infty$). There is a unique critical value $p=p^*(K)$ for which this length function becomes especially symmetric, in that its associated group of isometries (i.e., the subgroup of $GL(V)$ that preserves $\| \cdot \|_{p^*(K)}$) acts transitively on the corresponding unit sphere. The amazing fact is that this critical value $p^*(K)$ always coincides with the degree $[\bar{K}:K]$ of the algebraic closure of $K$!
Here is a proof sketch:
From the classification of local fields, it is known${}^{(\dagger)}$ that $K$ is either $\mathbb{R}$, $\mathbb{C}$ or a non-Archimedean local field (that is, a finite extension of either $\mathbb{Q}_P$ or $\mathbb{F}_P((t))$ for some prime $P$).
If $K=\mathbb{R}$, $|\cdot|$ is the usual absolute value, and the critical length function turns out to be $\| \cdot \|_2$ (the usual Euclidean length), invariant under $O(n)$. On the other hand, $[\bar{\mathbb{R}}:\mathbb{R}]=[\mathbb{C}:\mathbb{R}]=2$.
If $K=\mathbb{C}$, we have $|x|=\bar{x} x$, and the corresponding critical length function is $\| \cdot \|_1$, invariant under $U(n)$.${}^{(\ddagger)}$ On the other hand, $[\bar{\mathbb{C}}:\mathbb{C}]=[\mathbb{C}:\mathbb{C}]=1$. Note that neither $|\cdot|$ nor $\|\cdot\|$ satisfy the triangle inequality in this particular case (this is why I avoided using the terms absolute value and norm throughout the question).
If $K$ is non-Archimedean, $|\cdot|$ is a suitably normalized $\chi$-adic absolute value for some $\chi$, and the corresponding critical length function is the supremum norm $\| \cdot \|_\infty$, which can be shown to be invariant under $GL(\mathcal{O}_K^n)$, where $\mathcal{O}_K$ is the ring of integers of $K$. On the other hand, it is known that the algebraic closure of $K$ has infinite degree, i.e. $[\bar{K}:K]=\infty$.
Thus in all cases we have $p^*(K) = [\bar{K}:K]$.
A problem with this proof is that it gives no indication as to why the relationship holds: after all, the values could just happen to be the same by complete coincidence. For that reason, I am wondering if there exists an alternative proof of this fact that is "$K$-agnostic", i.e. a proof that does not use at any point the Artin-Schreier theorem, nor the classification of local fields, nor any properties from a specific such field (such as the existence of a symmetric/Hermitian inner product) other than the degree of an algebraic closure and properties derived from it.
Such a proof would allow one to meaningfully speak of a relationship between those two quantities, and to state things like "in a field with a $3$-dimensional algebraic closure, the natural analogue of the Pythagorean theorem would be $|a|^3+|b|^3=|c|^3$" without appealing to the principle of explosion. I know it is hard in general to formally determine whether a proof does or does not use a fact, hence why I'm labeling this with the soft-question tag, but hopefully the kind of proof I want is clear, at least informally.
In summary, my question is:
Given a local field $K$, with $| \cdot |$ defined as above, is there any $K$-agnostic proof that the group of isometries of $\|\cdot\|_p$ (defined in terms of $| \cdot |$ as above) acts transitively on unit spheres if and only if $p = [\bar{K}:K]$?
It seems reasonable to try proving first that $p^*(K) = 1$ iff $K$ is algebraically closed. That is where I am currently stuck. The thing that makes linear algebra over an algebraically closed field "special" is that every linear transformation has an eigenvalue, but I am not sure how to apply that fact to show that the critical length function is $\|\cdot\|_1$.
UPDATE: As suggested by Torsten Schoeneberg in the comments, perhaps it will be easier to prove the relationship $p^*(K) = [L:K]\: p^*(L)$ for a finite field extension $L/K$. Consider such an extension with $[L:K]=n$, and suppose we know $L$ has a critical $p$-length function for some $p$. Now, $L \simeq K^n$ is additively a vector space over $K$, which is equipped with the product topology. This means that, if we identify $k\in K $ with its image under the inclusion $K \subseteq L$, we have
$$|k|_L = \frac{\mu_L(k S^n)}{\mu_L(S^n)} = \left(\frac{\mu_K(k S)}{\mu_K(S)}\right)^n = |k|_K^n.$$
On the other hand, note that the subgroup of $GL_1(L) = L^\times$ preserving $|\cdot|_L$, i.e. the group of invertible elements of norm $1$, acts transitively on the associated unit sphere (which coincides with the group itself) in a tautological way, since any element $x$ of norm $1$ can be sent to any other element $y$ of norm $1$ through multiplication by $x^{-1}y$. Since $GL_1(L) \subseteq GL_n(K)$, this proves that the length function on $K^n$ defined through the identification $L \simeq K^n$ as $\|\cdot\| = |\cdot|_L^{1/n}$ is a critical length function, that moreover satisfies the homogeneity property $\|kv\|=|k|_L \|v\|$ (which follows by the above relationship between the Haar measures and multiplicativity of $|\cdot|$). In a similar way, from the inclusion $GL_m(L) \subseteq GL_{mn}(K)$ we can prove that any critical length function $||\cdot||$ on $L^m$ induces a corresponding critical length function on $K^{mn}$.
The only ingredient left would be to prove is that there always exists some basis $\{1=a_0, a_1, \ldots, a_{n-1}\}$ of $L$ such that the above defined length functions are of the required form, that is, $|\sum_{i=0}^{n-1} k_i a_i|_L = \sum_{i=0}^{n-1} |k_i|_K^n$. Once we have that, we can show using isotropy subgroups that the length function induced from $L$ is critical in any dimension, not just multiples of $n$, so that $p^*(K) = [L:K]\: p^*(L)$. However, I don't know how to prove the existence of such a basis in a $K$-agnostic way.
EDIT: A possible way to restate the problem is by noticing that for any $(v,w)$ in the direct sum $V \oplus W$, we have $\|(v,w)\|^p_p = \|v\|^p_p + \|w\|^p_p$, i.e. the $p$th power of $\|\cdot\|_p$ acts additively on "independent" vectors (for finite $p$, of course).
Given a length function on vector spaces $V$, we can define the associated Gaussian function as an integrable function $g : V \to \mathbb{R}_{\ge 0}$ depending only on length (i.e. factorizing as the composition $f \circ \|\cdot\|_p : V \to \mathbb{R}_{\ge 0} \to \mathbb{R}_{\ge 0}$ for some $f$), and satisfying the independence property
$$g( (v,w) ) = g(v) g(w).$$
With the above choice of maximally symmetric length functions, using the aforementioned additivity property immediately leads to the usual Gaussian function $\exp(-k \|x\|_2^2)$ for $\mathbb{R}$ and to $\exp(-k' \|x\|_1)$ for $\mathbb{C}$ (where the constants $k, k' >0$ depend on the normalization convention). For $p=\infty$, this reasoning can't be used directly, but if we interpret it as a limit like before, we do recover the standard Gaussian function for non-Archimedean fields
$$\lim_{p\to \infty} \exp(-k'' \|x\|_p^p) = \begin{cases} 1 & \|x\|_p \le 1 \\ 0 & \text{otherwise} \end{cases} = \mathbf{1}_\mathcal{O_K^n}(x),$$
which is the indicator function of $\mathcal{O}_K^n$ in $V=K^n$. Focusing on the $1$-dimensional case $V=K$, these Gaussian functions coincide with the standard ones as used e.g. in Tate's thesis.
So if I'm not mistaken, the problem can be restated as proving that $[\bar{K} : K] = p$ if and only if the standard Gaussian function associated to $K$ is $\exp(-k \|x\|_p^p)$ (with a limit intended if $p=\infty$). A possible advantage of this restatement is that standard Gaussian functions have different characterizations that do not previously assume any length function (for example, they are their own Fourier transform, or satisfy analogues of the central limit theorem, see e.g. here).
$(\dagger)$ Note that by the Artin-Schreier theorem, the only degrees $[\bar{K}:K]$ that could possibly occur for any field $K$ are $1, 2, \infty$ (this, in turn, seems to be related to the fact that a nonzero integer can only have multiplicative order $1, 2$ or $\infty$, but that is a topic for another question).
$(\ddagger)$ As a fun fact, this can be shown to imply that the hypervolume of the $n$-dimensional complex unit ball $|x_1|+|x_2|+\ldots+|x_n| = 1$ , i.e. the ordinary $2n$-dimensional unit ball, equals $V(B_{2n})=V(B_{2})^n /n!=\pi^n /n!$; compare with the hypervolume $V(X_{n})=V(X_{1})^n /n!=2^n /n!$ of the $n$-dimensional cross-polytope $|x_1|+|x_2|+\ldots+|x_n| = 1$, where $| \cdot |$ now denotes the real absolute value.
|
https://math.stackexchange.com/questions/4353238
|
[
"linear-algebra",
"number-theory",
"measure-theory",
"soft-question",
"local-field"
] | 34 | 2022-01-10T04:38:48 |
[
"The question has been posted on MathOverflow.",
"I also suggest asking on MO, as Nolord suggested. You can repost it there and link the questions to each other (for example, with a brief heading saying \"Cross-posted to/from MO/MSE.\").",
"@user76284 Edited, thanks for the suggestion!",
"Minor suggestion: You can typeset double vertical bars as \\|. This shows up as $\\|x\\|$ rather than $||x||$ (notice the lack of excess space between the bars).",
"I guess in the last paragraph should be $|x_1|+|x_2|+\\ldots+|x_n| {\\color{red}\\le}1 $ instead of $|x_1|+|x_2|+\\ldots+|x_n| = 1$.",
"This should go on MathOverflow!",
"@TorstenSchoeneberg That seems like a good idea, thank you! I've updated the question with some of my thoughts about it.",
"Well, maybe there is a \"field-agnostic\"/\"Artin-Schreier-less\" proof that if $L \\vert K$ is a finite dimensional field extension, $p^*(K) =[L:K] \\cdot p^*(L)$. (Or show it at least for some, e.g. cyclotomic, extensions. But as you say it actually holds even for skew field extensions.) Then we would be done in the non-archimedean case once we show further that the possible $p^*$'s are bounded away from zero.",
"@TorstenSchoeneberg As I mention in a footnote, the fact that the multiplicative order of $n\\in\\mathbb{Z}$ (and thus the order of this generator) can only be $1, 2, \\infty$ seems to be the \"source\" of many (if not all?) appearances of the trichotomy in the theory.",
"@TorstenSchoeneberg Yes, that trichotomy is realized as $\\mathrm{trivial}, \\mathbb{Z}_2, \\mathbb{Z}$ in the Brauer groups of local fields. There are many things to say about this which I want to save for my next question, but here is something I can't resist mentioning: for all local $K$, the topological generator of the Galois group of $K^{ur}$ acts as $\\omega \\mapsto \\omega^n$ on roots of unity for some $n$. In the nonarchimedean case this is Frobenius, and $n$ is a prime power; for $\\mathbb{R}$ it is complex conjugation with $n=-1$, and for $\\mathbb{C}$ it is the identity ($n=1$).",
"@TorstenSchoeneberg Good questions! For the quaternions one can mimic the construction of $|\\cdot |$, with the slightly weird result that the critical length function invariant under $USp(n)$ looks like $||(x_1,\\ldots,x_n)||_{1/2} = (\\sqrt{|x_1|}+\\ldots+\\sqrt{|x_n|})^2$. I haven't checked it but I think the same should happen for the octonions with $||\\cdot ||_{1/4}$ invariant under several exceptional groups, with the usual caveat that octonionic vector spaces are only well-behaved for $n\\le 3$.",
"By the way, what happens over the quaternions $\\mathbb H$? --- Superficially related, the trichotomy $1,2,\\infty$ also matches the cardinality of the Brauer group $Br(k)$ for the respective local fields $k$, right?",
"I have a feeling that at least in the two archimedean cases, this is actually related to the fact that what we have decided on as codomain for measures, $\\mathbb R_{\\ge 0}$, is made up of the squares in $(\\mathbb R, \\cdot)$, or the norms from $\\mathbb C \\vert \\mathbb R$. --- Anyway, maybe one can get insight by checking what happens in a non-local case like some cubic or higher extension of $\\mathbb Q$, or $\\mathbb Q(i)$, with restricting the possible Haar measures (they still exist, just not unique) to those."
] | 0 |
Science
| 0 |
282
|
math
|
Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?
|
It seems preposterous at first glance. I just want to be sure. Even in 3D the behaviour of rotating objects can be surprising (see the Dzhanibekov effect); in 4D it could be more surprising.
A 2D or 3D spacecraft (with no reaction wheels or gimbaling etc.) needs at least two thrusters to control its spin. See my answer on space.SE.
For any two $4\times4$ matrices $X$ and $Y$, define the commutator $[X,Y]=XY-YX$, the anticommutator $\{X,Y\}=XY+YX$, and the Frobenius inner product $\langle X,Y\rangle=\operatorname{tr}(X^\top Y)$, where $\operatorname{tr}$ is the trace and $\top$ is the transpose.
Let $M$ be a symmetric positive-definite $4\times4$ matrix, and $T=\mathbf f\,\mathbf r^\top-\mathbf r\,\mathbf f^\top$ an antisymmetric $4\times4$ matrix. ($M$ describes the distribution of mass in the spacecraft, $\mathbf r$ is a vector locating the thruster relative to the centre of mass, $\mathbf f$ is the force produced by the thruster, and $T$ is the torque produced by the thruster. More details here. Everything is described in the rotating reference frame.)
The angular velocity $\Omega(t)$, an antisymmetric $4\times4$ matrix, changes with time $t$ according to Euler's equation
$$\{M,\Omega'(t)\}+[\Omega(t),\{M,\Omega(t)\}]=f(t)\,T$$
where $f(t)\geq0$ is a function (continuous, piecewise-constant, or just integrable) describing when and how strongly the thruster is used.
Question: Can $M$ and $T$ be chosen such that, for any two antisymmetric $4\times4$ matrices $\Omega_0$ and $\Omega_1$, there exist $t_1>0$ and $f$ such that the solution $\Omega$ to Euler's equation with initial value $\Omega(0)=\Omega_0$ has final value $\Omega(t_1)=\Omega_1$?
To simplify things, we may take $f$ to be a combination of Dirac deltas instead of an ordinary function. Then the angular velocity satisfies $\{M,\Omega'(t)\}+[\Omega(t),\{M,\Omega(t)\}]=0$ except at a finite set of times when the angular momentum $L(t)=\{M,\Omega(t)\}$ changes by a positive multiple of $T$, or equivalently when $\Omega(t)$ changes by a positive multiple of $A=\{M,\}^{-1}(T)$, where $A$ is the angular acceleration produced by the thruster.
Here is Euler's equation in terms of the components $\omega_{ij}$ of the angular velocity. Assume that $M$ is diagonal, with components $m_i>0$.
$$(m_1+m_2)\,\omega_{12}'(t)+(m_2-m_1)\big(\omega_{13}(t)\,\omega_{32}(t)+\omega_{14}(t)\,\omega_{42}(t)\big)=f(t)\,\tau_{12}$$
(And permute the indices to get 6 equations like this.) Equivalently:
$$\omega_{12}'(t)=\frac{m_1-m_2}{m_1+m_2}\big(\omega_{13}(t)\,\omega_{32}(t)+\omega_{14}(t)\,\omega_{42}(t)\big)+f(t)\,\alpha_{12}$$
If $m_1=m_2$, then $\omega_{12}'$ has constant sign; $\omega_{12}$ is either always non-increasing, or always non-decreasing. So, if the difference between initial and final values of $\omega_{12}$ has the wrong sign compared to $\tau_{12}$, then there is no solution. Thus, we must take $m_1\neq m_2$, and similarly $m_i\neq m_j$ for $i\neq j$.
The Frobenius inner product of two antisymmetric matrices has every term duplicated: $\sum_{i,j}x_{ij}y_{ij}=2\sum_{i<j}x_{ij}y_{ij}$ (since $x_{ji}=-x_{ij}$ and $y_{ji}=-y_{ij}$). So it's natural to take half of this.
The angular momentum's squared magnitude is $\tfrac12\langle L(t),L(t)\rangle$; the derivative of this is $\langle L(t),f(t)T\rangle$. So the angular momentum has constant magnitude whenever $f(t)=0$. (The angular momentum itself would be constant in an inertial reference frame, but here we're using a rotating reference frame.)
The rotational energy is $\tfrac14\langle\Omega(t),L(t)\rangle$; the derivative of this is $\tfrac12\langle\Omega(t),f(t)T\rangle$. So the energy is constant whenever $f(t)=0$.
All of that applies in any dimension. But in 4D there's another constant, the angular momentum bivector's exterior square, or equivalently its inner product with its Hodge dual.
Here are those constants in component form. Squared magnitude of angular momentum:
$$(m_1+m_2)^2\omega_{12}^2+(m_1+m_3)^2\omega_{13}^2+(m_2+m_3)^2\omega_{23}^2\\+(m_1+m_4)^2\omega_{14}^2+(m_2+m_4)^2\omega_{24}^2+(m_3+m_4)^2\omega_{34}^2$$
Doubled energy:
$$(m_1+m_2)\omega_{12}^2+(m_1+m_3)\omega_{13}^2+(m_2+m_3)\omega_{23}^2\\+(m_1+m_4)\omega_{14}^2+(m_2+m_4)\omega_{24}^2+(m_3+m_4)\omega_{34}^2$$
Halved exterior square of angular momentum:
$$(m_1+m_2)(m_3+m_4)\omega_{12}\omega_{34}-(m_1+m_3)(m_2+m_4)\omega_{13}\omega_{24}+(m_2+m_3)(m_1+m_4)\omega_{23}\omega_{14}$$
(Its derivative is the exterior product of angular momentum and torque: $(m_1+m_2)\omega_{12}\,f\,\tau_{34}+\cdots$ .)
These three constants define two ellipsoids and another quadratic surface (with signature $+^3-^3$) in the 6D space of antisymmetric matrices. When $f=0$, $\Omega$ must remain on the intersection of these three quadratic surfaces.
In fact there are more than three, as I recently found! And these aren't specific to 4D. The general form is $Q_i=\sum_{j\neq i}\frac{m_i+m_j}{m_i-m_j}\omega_{ij}^2$; check that $Q_i'=0$, and also note $\sum_iQ_i=0$. Then define $P_i=\sum_{j\leq i}Q_j$; these are positive-semidefinite, on the assumption (WLOG) that $m_i$ are sorted, $m_1>m_2>m_3>m_4>0$. The doubled energy is $\sum_i(m_i-m_{i+1})P_i$, and the momentum's squared magnitude is $\sum_i(m_i^2-m_{i+1}^2)P_i$. But what we should focus on are the quadratic forms $N_i=P_{i-1}-P_i+P_{i+1}$ which are not positive-semidefinite but have a single negative term:
$$N_1=\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2+\frac{m_2+m_4}{m_2-m_4}\omega_{24}^2-\frac{m_1+m_2}{m_1-m_2}\omega_{12}^2$$
$$N_2=\frac{m_1+m_2}{m_1-m_2}\omega_{12}^2+\frac{m_1+m_4}{m_1-m_4}\omega_{14}^2+\frac{m_3+m_4}{m_3-m_4}\omega_{34}^2-\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2$$
$$N_3=\frac{m_1+m_3}{m_1-m_3}\omega_{13}^2+\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2-\frac{m_3+m_4}{m_3-m_4}\omega_{34}^2$$
These are all constant. The first expression (if it's equated to $0$) defines a cone, or a pair of opposite cones, around the $\omega_{12}$ axis in 6D. The second expression defines a pair of opposite cones around the $\omega_{23}$ axis; and the third, around the $\omega_{34}$ axis. If the applied acceleration $A$ is inside one of these cones, and the initial angular velocity $\Omega_0$ is in the same cone, then $\Omega$ stays in that cone, and there's no solution with the final angular velocity $\Omega_1$ outside of the cone. Thus, we must take $A$ to be outside of all six of these cones. That is, all three expressions (with $\alpha_{ij}$ in place of $\omega_{ij}$) must be positive.
Notice that Euler's equation has a time symmetry: Given a solution $\Omega(t)$, we can construct another solution $\tilde\Omega(t)=-\Omega(t_1-t)$, so that $\{M,\tilde\Omega'(t)\}+[\tilde\Omega(t),\{M,\tilde\Omega(t)\}]=f(t_1-t)\,T$, and the initial and final values are swapped (and negated): $\tilde\Omega(0)=-\Omega_1$ and $\tilde\Omega(t_1)=-\Omega_0$. Therefore, any angular velocity can be reached from any other, if and only if any angular velocity can be reached from $0$. (We can reverse a path from $0$ to $-\Omega_0$ to get a path from $\Omega_0$ to $0$, and concatenate that with a path from $0$ to $\Omega_1$, to get a path from $\Omega_0$ to $\Omega_1$.)
Also, Euler's equation has a scale symmetry: Given a solution $\Omega(t)$, for any $k>0$ we can construct another solution $\tilde\Omega(t)=k\,\Omega(kt)$, so that $\{M,\tilde\Omega'(t)\}+[\tilde\Omega(t),\{M,\tilde\Omega(t)\}]=k^2f(kt)\,T$, and the final value is $\tilde\Omega(t_1/k)=k\,\Omega(t_1)=k\,\Omega_1$. Therefore, any angular velocity can be reached from $0$, if and only if any angular velocity in a small neighbourhood of $0$ can be reached from $0$.
(It looks like these two symmetries are related by $k=-1$, but we should keep a distinction between positive time and negative time.)
This has the form of a quadratic differential equation, $\mathbf x'(t)=\mathbf x(t)\odot\mathbf x(t)$ where $\odot$ is some bilinear function. (Specifically, for antisymmetric matrices $X$ and $Y$, define $X\odot Y=-\{M,\}^{-1}([X,\{M,Y\}])$, so Euler's equation is $\Omega'=\Omega\odot\Omega$, as long as $f=0$. Alternatively, define $X\odot Y=-[\{M,\}^{-1}(X),Y]$, so Euler's equation is $L'=L\odot L$.)
(I ought to move this power series stuff to a different page, since it's applicable far beyond this particular Question, and anyway it's bloating this page.)
Fix a norm on the space, and find some constant $\lVert\odot\rVert>0$ such that $\lVert\mathbf x\odot\mathbf y\rVert\leq\lVert\odot\rVert\lVert\mathbf x\rVert\lVert\mathbf y\rVert$ for all $\mathbf x,\mathbf y$.
Given initial value $\mathbf x(0)=\mathbf a$, the equation $\mathbf x'=\mathbf x\odot\mathbf x$ has the power series solution
$$\mathbf x(t)=\sum_{n=0}^\infty t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}$$
where the coefficient of $t^n$ is the average of all possible ways of evaluating $\mathbf a^{n+1}$ using the product $\odot$. For example, the coefficient of $t^3$ is $\tfrac16$ of
$$\sum^6\mathbf a^4=((\mathbf a\mathbf a)\mathbf a)\mathbf a+(\mathbf a(\mathbf a\mathbf a))\mathbf a+2(\mathbf a\mathbf a)(\mathbf a\mathbf a)+\mathbf a((\mathbf a\mathbf a)\mathbf a)+\mathbf a(\mathbf a(\mathbf a\mathbf a))$$
$$\newcommand{\aaaa}[3]{\mathbf a\underset{#1}\odot\mathbf a\underset{#2}\odot\mathbf a\underset{#3}\odot\mathbf a} =\aaaa{1}{2}{3}+\aaaa{2}{1}{3}\begin{matrix}{}+\aaaa{1}{3}{2} \\ {}+\aaaa{2}{3}{1}\end{matrix}+\aaaa{3}{1}{2}+\aaaa{3}{2}{1}.$$
The terms in the series can be obtained recursively:
$$\mathbf u_n=\frac{t^n}{n!}\sum^{n!}\mathbf a^{n+1}$$
$$=\frac tn\sum_{k=0}^{n-1}\mathbf u_{n-1-k}\odot\mathbf u_k.$$
We also have $\lVert\mathbf u_n\rVert\leq|t|^n\lVert\odot\rVert^n\lVert\mathbf a\rVert^{n+1}$, which ensures absolute convergence, for small $t$:
$$\sum_{n=0}^\infty\left\lVert t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}\right\rVert\leq\sum_{n=0}^\infty|t|^n\lVert\odot\rVert^n\lVert\mathbf a\rVert^{n+1}=\frac{\lVert\mathbf a\rVert}{1-|t|\lVert\odot\rVert\lVert\mathbf a\rVert},$$
$$|t|<\frac{1}{\lVert\odot\rVert\lVert\mathbf a\rVert}.$$
And at certain times (when the thruster is used) an impulse may be applied to $\mathbf x$, with a fixed direction $\mathbf b$ but an arbitrary magnitude $c>0$, thus: $\mathbf x(t+)=\mathbf x(t-)+c\mathbf b$. These impulses, and the time intervals between them, are many variables that we can control. If the number of variables is at least $6$ (or the dimension of the space $\mathbf x$ is in), then there is hope of surrounding $0$ in an open set, and thus reaching everywhere in the space (according to the previous section).
$$\mathbf x(0+)=0+c_0\mathbf b$$
$$\mathbf x(t_1-)=\sum_{n=0}^\infty\frac{t_1^n}{n!}\sum^{n!}\mathbf x(0+)^{n+1}$$
$$\mathbf x(t_1+)=\mathbf x(t_1-)+c_1\mathbf b$$
$$\mathbf x(t_1+t_2-)=\sum_{n=0}^\infty\frac{t_2^n}{n!}\sum^{n!}\mathbf x(t_1+)^{n+1}$$
$$\mathbf x(t_1+t_2+)=\mathbf x(t_1+t_2-)+c_2\mathbf b$$
$$\mathbf x(t_1+t_2+t_3-)=\sum_{n=0}^\infty\frac{t_3^n}{n!}\sum^{n!}\mathbf x(t_1+t_2+)^{n+1}$$
$$c_0,t_1,c_1,t_2,c_2,t_3\geq0$$
$$\mathbf x(t_1+t_2+t_3-)\approx0\quad?$$
|
https://math.stackexchange.com/questions/4472850
|
[
"ordinary-differential-equations",
"analysis",
"dynamical-systems",
"physics"
] | 34 | 2022-06-14T18:13:22 |
[
"Cross-posted to MO: mathoverflow.net/questions/448773/…",
"Oh, I didn't explain what $M$ is (though the linked physics.SE post did explain it). It's the matrix of second moments of mass: $M=\\int\\mathbf r\\,\\mathbf r^T\\,\\rho\\,dV$, where $\\mathbf r$ is position relative to the centre of mass, $\\rho$ is density, and $V$ is volume. But that's not necessary for understanding this Question.",
"Given that $M$ is diagonal (which is no loss of generality, since any symmetric matrix has an orthogonal eigenbasis), the inertia tensor is also diagonal, with components $m_1+m_2$, $m_1+m_3$, $m_2+m_3$, $m_1+m_4$, $m_2+m_4$, $m_3+m_4$.",
"And of course I'm considering bivectors as antisymmetric matrices.",
"The inertia tensor is the function $\\Omega\\mapsto L=\\{M,\\Omega\\}=M\\Omega+\\Omega M$.",
"What's the relationship of $M$ with the inertia tensor? My understanding was that the inertia tensor is a linear transformation on bivectors, which in this case are 6-dimensional, so it would be a symmetric 6x6 matrix. Is that not correct?"
] | 0 |
Science
| 0 |
283
|
math
|
Semirings induced by symmetric monoidal categories with finite coproducts
|
A symmetric monoidal category with finite coproducts is by definition a symmetric monoidal category $(\mathcal{C},\otimes,1,\dotsc)$ such that the underlying category $\mathcal{C}$ has finite coproducts (this includes an initial object $0$) with the property that $\otimes$ preserves finite coproducts in each variable. In other words, we have a distributive law
$$X \otimes (\bigoplus_{i \in I} Y_i) \cong \bigoplus_{i \in I} (X \otimes Y_i).$$
This is also known as a finitary distributive symmetric monoidal category. Notice that to each symmetric monoidal category with finite coproducts we may associate a commutative semiring: The elements are the isomorphism classes of objects in $\mathcal{C}$. The addition is $[X] + [Y] := [X \oplus Y]$, the multiplication is $[X] \cdot [Y] := [X \otimes Y]$. The zero element is $[0]$, the multiplicative unit is $[1]$. To avoid set-theoretic issues, let us assume that $\mathcal{C}$ is essentially small. As a consequence, we may think of symmetric monoidal categories with finite coproducts as a possible categorification of semirings.
A familar example is the symmetric monoidal category of vector bundles on a space $X$, which produces the semiring of isomorphism classes of vector bundles on $X$. If $X$ is a point, we see that the semiring associated to $\mathsf{FinSet}$ is $\mathbb{N}$.
Note, however, that not every commutative semiring arises from the construction above. Basically this is because we have agreed that the addition is not induced by some symmetric monoidal structure which distributes over $\otimes$, but rather from the coproduct. The coproduct satisfies $X \oplus Y \cong 0 \Rightarrow X \cong Y \cong 0$. It follows that in the associated semiring only $0$ has an additive inverse.
Question. How can we classify those commutative semirings which are induced by essentially small symmetric monoidal categories with finite coproducts?
Every commutative monoid with the property $a+b=0 \Rightarrow a=b=0$ arises from an essentially small category with finite coproducts, see SE/834869. But it is not clear if we can use this here.
|
https://math.stackexchange.com/questions/813676
|
[
"ring-theory",
"category-theory",
"examples-counterexamples",
"monoidal-categories"
] | 33 | 2014-05-29T05:39:16 |
[
"@Arducode This does not define a functor for $\\otimes$. Or how do you define the tensor product of morphisms? Also notice that it is required that the tensor product is distributive - this is not guaranteed by arbitrary isomorphisms, they have to be the canonical ones. Therefore I do not think that the skeleton will help us.",
"If you have a commutative semiring $(R, +, \\cdot)$ that satisfies $a+b = 0 \\Rightarrow a=b=0$, can't you just take an essentially small category $C$ realizing $(R,+)$, take a skeleton $\\mathrm{Sk}(C)$ of it and define on $\\mathrm{Sk}(C)$ the symmetric monoidal structure induced by $(R,\\cdot)$? That is, $[x] \\otimes [y] \\colon= [x \\cdot y]$ (where $[x]$ denotes the object in $\\mathrm{Sk}(C)$ that is associated to the object $x \\in R$)"
] | 0 |
Science
| 0 |
284
|
math
|
Can you obtain $\pi$ using elements of $\mathbb{N}$, and finite number of basic arithmetic operations + exponentiation?
|
Is it possible to obtain $\pi$ from finite amount of operations
$\{+,-,\cdot,\div,\wedge\}$ on $\mathbb{N}$ (or $\mathbb{Q}$, the answer will still be the same)? Note that the set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental).
Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $\mathbb{N}$) ?
|
https://math.stackexchange.com/questions/2865253
|
[
"number-theory",
"pi",
"transcendental-numbers"
] | 33 | 2018-07-28T06:29:14 |
[
"As mentioned above, it is probably false, but a disproof is difficult. A hope could be to describe a subfield of $\\Bbb{R}$ containing all elements generated by the method you describe, but not containing $\\pi$. How such a field would look is difficult to imagine though.",
"The bonus question is the same question except allowing the use of $\\log$, since all the operations are invertible, and their inverse is included in the allowed operations - except for $^$, which needs $\\log$.",
"If you only use +,-,x,/, you can't get $\\pi$ because it will only yield a rational number, so you have to use the exponential operator sometime.",
"It can be obtained using calculus, check Wikipedia.",
"Could Machin-like formulas be of any use?",
"Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.",
"Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be.",
"(this one is more general. If the answer to this one is \"no\" then the answer to the other one is \"no\" as well)",
"Related: math.stackexchange.com/questions/2611084/…"
] | 0 |
Science
| 0 |
285
|
math
|
A holomorphic function sending integers (and only integers) to $\{0,1,2,3\}$
|
Does there exist a function $f$, holomorphic on the whole complex plane $\mathbb{C}$, such that $f\left(\mathbb{Z}\right)=\{0,1,2,3\}$ and
$\forall z\in\mathbb{C}\ (f(z)\in\{0,1,2,3\}\Rightarrow z\in\mathbb{Z})$?
If yes, is it possible to have an explicit construction?
Note that, for example, $h(z)=\frac{1}{6} \left(9-8 \cos \left(\frac{\pi z}{3}\right)-\cos (\pi z)\right)$ is not a valid solution, since, in particular, certain roots of the equation $h(z)=0$ are not integers, but complex numbers.
Also note that this question is answered in positive regarding the function $g$ such that $g\left(\mathbb{Z}\right)=\{0,1,2\}$ and $\forall z\in\mathbb{C}\ (g(z)\in\{0,1,2\}\Rightarrow z\in\mathbb{Z})$. In this case, an example of such function is $g(z)=1-\cos \left(\frac{\pi z}{2}\right)$.
|
https://math.stackexchange.com/questions/4624439
|
[
"complex-analysis",
"galois-theory",
"riemann-surfaces",
"elliptic-functions"
] | 32 | 2023-01-23T11:39:57 |
[
"It is already solved on MO",
"Crossposted to MathOverflow: A holomorphic function sending integers (and only integers) to $\\{0,1,2,3\\}$",
"I suggest you post this to MathOverflow, adding a link to this MSE post to let readers know you've crossposted.",
"I'd ask for a Laurent polynomial solution in a separate post (ie. $g\\in \\Bbb{C}[q,q^{-1}]$ such that $g^{-1}(\\{0,1,2,3\\}) = \\mu_n$ for some $n$)",
"@user1950 - one has to be careful as the cited theorem applies only for multiplicity one values, so in other words if $f^{-1}(\\{0,1,2,3\\})=\\mathbb Z$ and $f'(n) \\ne 0$ for any $n \\in \\mathbb Z$; probably not hard to generalize if the multiplicities are uniformly bounded but unclear what happens if multiplicities at integers can be unbounded",
"So from @reuns remark it follows that the sought function, if it exists, must be a Laurent polynomial of $e^{\\frac{2\\pi i z}{n}}$. Looks like the problem is within reach of an expert in Galois theory.",
"It follows from Theorem 1 of G.G. Gundersen and C.C. Yang's \"On the Preimage Sets of Entire Functions\" that the sought function must be periodic. (projecteuclid.org/journalArticle/…)",
"one way to go is to interpolate the values $0,1,2,3$ evenly at integers (first for example the values may be symmetrized and the function made odd say on the integers as that gives better interpolation convergence) and get a function $f(z)=\\frac{\\sin \\pi z}{\\pi}(f(0)/z +\\sum_{n \\ne 0} (-1)^n f(n)/(z-n)+c)$ of order $1$ where $f(n)$ are precisely the choices you make and see if some clever choice of the interpolation values, the function obtained can be show to satisfy the property that only integers are sent there",
"why does it need to be periodic? the above applies only to periodic functions",
"@Conrad: So any valid solution to the main question must be of the form $\\sum_{n=a}^{b} a_n q^n$, where $-\\infty<a,b<+\\infty$ and $q=e^{\\frac{2\\pi i z}{k}}$?",
"the problem with @joricki function is that is of infinite order so any $f(z)=a$ should have tons of solutions (many more than integers); for example $f(z)=0$ means $\\frac{\\pi z}3+\\alpha\\sin \\frac{2\\pi z}3=(2k+1)\\pi$ and for a fixed large $k$ that has only a few integral solutions since $z/3$ needs to be around $2k+1$ as the other term is bounded on the integers, but (by Picard) it must have infinitely many solutions for all but at most one $k$; a function with the given property kind of have to be of finite order $1$ so the roots of $f(z)=0,1,2,3$ are the right number in any disc of radius $R$",
"@joriki $f$ is entire $n$-periodic iff $g(q) = f(\\frac{n\\log(q)}{2i\\pi })$ is analytic on $\\Bbb{C}^*$. Due to great Picard's theorem $g(q)$ must be in the Laurent polynomials ring $\\Bbb{C}[q,q^{-1}]$.",
"@reuns: I'm afraid I don't quite follow. What are $q$ and $\\mathbb C[q,q^{-1}]$? (And in case this is an argument that my proposed function takes on the desired values at non-integers, why doesn't it apply to the $4$-periodic function in the question?)",
"@joriki Your function is $6$-periodic. If there is a $n$-periodic solution $f(z)=g(e^{2i\\pi z/n})$ then $g$ is analytic on $\\Bbb{C}^*$ so it must be in $\\Bbb{C}[q,q^{-1}]$ as otherwise $g(q)$ or $g(q^{-1})$ has an essential singularity at $0$ and we can apply great Picard's theorem",
"@user1950: I'm sorry, I was missing a factor of $2$, it should be $\\frac32\\left(1+\\cos\\left(\\frac{\\pi z}3+\\alpha\\sin \\frac{2\\pi z}3\\right)\\right)$ (with the same $\\alpha$).",
"@joriki - can you recheck the formula? With $z=2$ I get $f(2)=3/2 (1 + \\sin[π/6 - \\arccos[1/3]])$, which is not an integer.",
"A candidate would be $\\frac32\\left(1+\\cos\\left(\\frac{\\pi z}3+\\alpha\\sin \\frac{\\pi z}3\\right)\\right)$ with $\\alpha=\\left(\\arccos\\left(\\frac13\\right)-\\frac\\pi3\\right)/\\sin\\left(\\frac\\pi3\\right)\\approx0.21219$, which takes the right values on the integers, but I suspect that it also takes them elsewhere, and I don't know how to easily check for that. Wolfram|Alpha doesn't return any other solutions.",
"@Conrad - well, first I've noticed the fact that this is easily solvable if the discrete set is composed of all the $n$-th roots of unity. And then started to check if generalizations are possible.",
"@Dmitry - the function is entire so is analytic on the plane; the condition just talks about specific values of the function at specific points - it is very easy to construct a function like that st it sends the integers surjectively to the given set, the problem is to ensure that no other numbers go there and that is harder; for the smaller set $0,1,2$ a simple solution is actually presented",
"No, such a function does not exist. A holomorphic function on the complex plane is infinitely differentiable, and thus must have derivatives of all orders. However, since the set of integers is countable, a function that is only defined on the integers cannot be infinitely differentiable. Therefore it cannot be holomorphic on the whole complex plane.",
"where does this problem come from?"
] | 0 |
Science
| 0 |
286
|
math
|
A conjecture about Catalan sequence
|
For $n=2$, consider the free abelian group generated by polynomials corresponding to $\frac{(2 n)!}{2^n n!}=3$ partitions of $2n=4$ vertices into pairs (chords):
$$
(x_1-x_3)(x_2-x_4),(x_1-x_2)(x_3-x_4),(x_1-x_4)(x_2-x_3)
$$
Euler's Identity: $
(x_1-x_3)(x_2-x_4)=(x_1-x_2)(x_3-x_4)+(x_1-x_4)(x_2-x_3)$.
So$$\{(x_1-x_2)(x_3-x_4),(x_1-x_4)(x_2-x_3)\}\tag1\label2$$
is a basis.
So the rank of the free abelian group is $2$.
For $n=3$, consider the free abelian group generated by polynomials corresponding to $\frac{(2 n)!}{2^n n!}=15$ partitions of $2n=6$ vertices into pairs (chords):
\begin{array}r
\{(x_1-x_2) (x_3-x_4) (x_5-x_6),\\(x_1-x_2) (x_3-x_5) (x_4-x_6),\\(x_1-x_2) (x_4-x_5) (x_3-x_6),\\(x_1-x_3) (x_2-x_4) (x_5-x_6),\\(x_1-x_3) (x_2-x_5) (x_4-x_6),\\(x_1-x_3) (x_4-x_5) (x_2-x_6),\\(x_2-x_3) (x_1-x_4) (x_5-x_6),\\(x_1-x_4) (x_2-x_5) (x_3-x_6),\\(x_1-x_4) (x_3-x_5) (x_2-x_6),\\(x_2-x_3) (x_1-x_5) (x_4-x_6),\\(x_2-x_4) (x_1-x_5) (x_3-x_6),\\(x_3-x_4) (x_1-x_5) (x_2-x_6),\\(x_2-x_3) (x_4-x_5) (x_1-x_6),\\(x_2-x_4) (x_3-x_5) (x_1-x_6),\\(x_3-x_4) (x_2-x_5) (x_1-x_6)\}
\end{array}
Using Euler's Identity we can decompose the product of intersecting chords into the sum of products of the non-intersecting chords. For example $(x_1-x_4)(x_2-x_5)(x_3-x_6)$ as in Fuhrmann's Theorem,
There are 5 partitions of 6 vertices into non-intersecting chords
$$\tag2\label1\begin{array}l\{(x_1-x_6)(x_2-x_5)(x_3-x_4),\\ (x_1-x_6)(x_2-x_3)(x_4-x_5),\\ (x_1-x_4)(x_2-x_3)(x_5-x_6),\\(x_1-x_2)(x_3-x_6)(x_4-x_5),\\(x_1-x_2)(x_3-x_4)(x_5-x_6)\}\end{array}$$
Under lexicographic order, the leading terms of \eqref{1} are distinct: $\{x_1x_2x_3,x_1x_2x_4,x_1x_2x_5,x_1x_3x_4,x_1x_3x_5\}$. Suppose $c_1(x_1-x_6)(x_2-x_5)(x_3-x_4)+\dots+c_5(x_1-x_2)(x_3-x_4)(x_5-x_6)=0$ for some $c_1,\dots,c_5\in\mathbb{Z}$.
Comparing coefficients of term $x_1x_2x_3$ to get $c_1=0$, comparing coefficients of term $x_1x_2x_4$ to get $c_2=0$, comparing coefficients of term $x_1x_2x_5$ to get $c_3=0$, comparing coefficients of term $x_1x_3x_4$ to get $c_4=0$, comparing coefficients of term $x_1x_3x_5$ to get $c_5=0$, so $c_1=\dots=c_5=0$, \eqref{1} is linearly independent.
So the rank of the free abelian group is 5.
Here is a proof that the number of non-intersecting chords is Catalan number $\frac1{n+1}\binom{2n}n$.
Each polynomial in \eqref{1} can be written as a $2×n$ "non-intersecting tableaux"
$$
\Biggl\{\begin{array}{|c|c|}\hline1&2&3\\\hline6&5&4\\\hline\end{array},
\begin{array}{|c|c|}\hline1&2&4\\\hline6&3&5\\\hline\end{array},
\begin{array}{|c|c|}\hline1&2&5\\\hline4&3&6\\\hline\end{array},
\begin{array}{|c|c|}\hline1&3&4\\\hline2&6&5\\\hline\end{array},
\begin{array}{|c|c|}\hline1&3&5\\\hline2&4&6\\\hline\end{array}\Biggr\}
$$
To show that a "non-intersecting tableux" is determined by its first row, we give a procedure to fill in the second row.
For example $\begin{array}{|c|c|}\hline1&2&5&7\\\hline?&?&?&?\\\hline\end{array}$ to fill $3,4,6,8$ in the second row such that there are no intersecting chords, we must put $3$ under the largest number less than $3:\begin{array}{|c|c|}\hline1&2&5&7\\\hline?&3&?&?\\\hline\end{array}$ then we must put $4$ under the largest remaining number less than $4:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&?&?\\\hline\end{array}$, then we must put $6$ under the largest remaining number less than $6:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&6&?\\\hline\end{array}$, then we must put $8$ under the largest remaining number less than $8:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&6&8\\\hline\end{array}$
For all $n$ we have shown the first rows of $\frac1{n+1}\binom{2n}n$ number of "non-intersecting tableaux" are distinct.
Under lexicographic order, the leading terms of \eqref{1} correspond to the first row of tableaux, so the leading terms are distinct, so the set of polynomials corresponding to "non-intersecting tableux" are linearly independent for all $n$.
So we have proved:
For all $n$, the rank of the above free abelian group in $2n$ variables is $\frac1{n+1}\binom{2n}n$.
For $n=4$, similarly decompose the product of intersecting chords into the sum of products of the non-intersecting chords. For example $(x_1-x_5)(x_2-x_6)(x_3-x_7)(x_4-x_8)$
As you can see, there will be two identical pairs in the end, so after collecting terms there will be two terms with coefficient $2$
\begin{align} (x_1 - x_5)(x_2 - x_6)(x_3 - x_7)(x_4 - x_8)\\
= (x_1 - x_4)(x_2 - x_3)(x_5 - x_8)(x_6 - x_7)\\
+ \color{red}{2}(x_1 - x_8)(x_2 - x_3)(x_4 - x_5)(x_6 - x_7)\\
+ (x_1 - x_2)(x_3 - x_8)(x_4 - x_7)(x_5 - x_6)\\
+ (x_1 - x_2)(x_3 - x_8)(x_4 - x_5)(x_6 - x_7)\\
+ \color{red}{2}(x_1 - x_2)(x_3 - x_4)(x_5 - x_6)(x_7 - x_8)\\
+ (x_1 - x_2)(x_3 - x_4)(x_5 - x_8)(x_6 - x_7)\\
+ (x_1 - x_4)(x_2 - x_3)(x_5 - x_6)(x_7 - x_8)\\
+ (x_1 - x_8)(x_2 - x_3)(x_4 - x_7)(x_5 - x_6)\\
+ (x_1 - x_2)(x_3 - x_6)(x_4 - x_5)(x_7 - x_8)\\
+ (x_1 - x_8)(x_2 - x_7)(x_3 - x_4)(x_5 - x_6)\\
+ (x_1 - x_8)(x_2 - x_7)(x_3 - x_6)(x_4 - x_5)\\
+ (x_1 - x_8)(x_2 - x_5)(x_3 - x_4)(x_6 - x_7)\\
+ (x_1 - x_6)(x_2 - x_3)(x_4 - x_5)(x_7 - x_8)\\
+ (x_1 - x_6)(x_2 - x_5)(x_3 - x_4)(x_7 - x_8)\end{align}
Conjecture. Express $(x_1-x_{n+1})(x_2-x_{n+2})\dots(x_n-x_{2n})$ as a linear combination of polynomials corresponding to the non-intersecting chords, then the term $(x_1-x_2)(x_3-x_4)\cdots(x_{2n-1}-x_{2n})$ and the term $(x_{2n}-x_1)(x_2-x_3)\cdots(x_{2n-2}-x_{2n-1})$ have largest coefficient in absolute value.
Mathematica code to compute the largest coefficients in absolute value
Needs["Combinatorica`"];
coeff[n_]:=Max[Abs[PolynomialReduce[Product[Indexed[x,i]-Indexed[x,n+i],{i,n}],Table[Module[{t1=t[[1]],t2=t[[2]],poly=1,i=1},Do[i=1;While[Length[t1]>i&&t1[[i+1]]<t2[[1]],i++];poly*=Indexed[x,t1[[i]]]-Indexed[x,t2[[1]]];t1=Delete[t1,i];t2=Rest[t2],n];poly],{t,Tableaux[Table[n,2]]}],Table[Indexed[x,i],{i,2n}]][[1]]]]
the terms $n=2,\dots,9$ in the sequence agree with A099960:
\begin{array}{c|c}
n&2&3&4&5&6&7&8&9\\\hline
\text{largest abs coefficient}&1&1&2&3&8&17&56&155
\end{array}
|
https://math.stackexchange.com/questions/3792709
|
[
"polynomials",
"conjectures",
"catalan-numbers",
"free-abelian-group"
] | 31 | 2020-08-16T03:38:47 |
[
"@DietrichBurde The first example you gave only starts failing at the $x^{10}$ term. $132$ and $429$, as well as the two terms that follow $1430$ and $4862$, are Catalan numbers. The second example evaluates to $3$ when $k=3$. Though I do get your point that it's best not to be too confident about integer sequences when you only know the first few terms.",
"The expansion of $(1-8x+21x^2-20x^3+5x^4)/(1-9x+28x^2-35x^3+15x^4-x^5)$ also produces $1,1,2,5,14,42$ as Catalan, but then $132,429$, which is not Catalan. Also $binomial(fibonacci(k) + 1, 2)$ produces $1,1,2,5,14,42,132$ and so on."
] | 0 |
Science
| 0 |
287
|
math
|
How do I find the common invariant subspaces of a span of matrices?
|
Let $G_1, \ldots, G_n$ be a set of $m\times m$ linearly-independent complex matrices.
Let $\mathcal{G} = \operatorname{span}\left\{ G_1, \ldots , G_n\right\}$ be the vector space that spans the set of $G$ matrices, and let $\mathcal{S}$ be any linear subspace of $\mathcal{G}$.
Finally, let $\mathcal{I}$ be a linear subspace of $\mathbb{C}^m$.
Is there a way to determine all pairs $\left( \mathcal{S}, \mathcal{I}\right)$ such that for all $S \in \mathcal{S}$, the subspace $\mathcal{I}$ is an invariant subspace of $S$? In other words, is there a way to determine the linear subspaces of $\mathcal{G}$ whose elements share a (non-trivial) invariant subspace?
This problem is vaguely related to this one posted in 2013.
I know that if we arbitrarily pick an $\mathcal{I}$, we can determine the largest subspace $\mathcal{S}$ for which $\mathcal{I}$ is invariant. The problem is that there are infinitely many possible $\mathcal{I}$ to choose from. Likewise, if we arbitrarily choose an $\mathcal{S}$, we can deduce $\mathcal{I}$, but there are infinitely many $\mathcal{S}$ to choose from.
I haven't made any progress trying to determine both subspaces simultaneously. Any assistance would be greatly appreciated. Even some suggestions on where I might look to find a solution would be much appreciated.
|
https://math.stackexchange.com/questions/1122655
|
[
"linear-algebra",
"matrices",
"vector-spaces",
"eigenvalues-eigenvectors"
] | 31 | 2015-01-27T16:06:22 |
[
"How about computing products $G_1G_2\\cdots G_k$, $\\forall k \\in \\{1,N\\}$ and multiply a random vector with that. Then put all vectors with some large enough norm into a new matrix and then iterate by the new matrix with the sequence of matrices. Then iterate the procedure. This (I hope) will give something similar to the \"power method\" for finding largest eigenvalue, eigenvector pair. Oh yes, probably some re-normalization should be done also.",
"Does this help?",
"@DavidSpeyer - Many thanks for the heads up. I'll look into it.",
"More at math.stackexchange.com/questions/380730/… and math.stackexchange.com/questions/184960/…",
"See mathematica.stackexchange.com/questions/6519/…"
] | 0 |
Science
| 0 |
288
|
math
|
Minimal time gossip problem
|
The gossip problem (telephone problem) is an information dissemination problem
where each of $n$ nodes of a communication network has a unique piece of information
that must be transmitted to all the other nodes using two-way communications
(telephone calls) between the pairs of nodes. Upon a call between the given two
nodes, they exchange the whole information known to them at that moment.
The minimum number of calls needed to guarantee the spread of whole information is $2n-4$.
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/gossips.pdf
Assuming that the calls between non-overlapping pairs of nodes can take place simultaneously, the minimum amount of time $T(n)$ required to complete gossiping is
$\lceil \log_2 n \rceil$ for even $n$ and $\lceil \log_2 n \rceil + 1$ for odd $n$.
http://onlinelibrary.wiley.com/doi/10.1002/net.3230180406/abstract
As it is written in the paper by R. Labahn, "Kernels of minimum size gossip schemes",
Discrete Mathematics 143 (1995) 99-139:
Theorem 5.3 Any gossip scheme on $n \geq 4$ vertices with $2n-4$ calls has at least $2\lceil \log_2 n \rceil - 3$ rounds.
My question is following. Is there a result on the problem to find the minimum number of calls of a gossip scheme with time (rounds) $T(n)$.
|
https://math.stackexchange.com/questions/576326
|
[
"combinatorics",
"graph-theory",
"discrete-mathematics"
] | 31 | 2013-11-21T11:25:40 |
[
"Yes, this paper is about fault-tolerance of the gossip schemes. I have some ideas, and I want simply find any results about minimum number of calls of a gossip scheme with minimal time $T(n)$ to develop the theory. Presumably it is an open question.",
"It was this one:arxiv.org/abs/1304.5633 but I didn't read the second half of the title. Are you looking for a reference to add to your paper?",
"@BrianRushton, thank you for the comment. Which paper you mean ?",
"It looks like your own paper has such results."
] | 0 |
Science
| 0 |
289
|
math
|
The limit of $(\sin(n!)+1)^{1/n}$ as n approaches infinity
|
Calculate the limit
$$
\lim_{n\rightarrow\infty}(\sin(n!)+1)^{1/n}
$$
or prove that the limit does not exist.
This appeared as a problem in my mathematical analysis test, and the answer was that the limit exists and it was $1$. But later the teacher found a mistake in his proof and eventually removed the problem from the test. But I'm just curious. Does this problem have a certain answer?
The biggest question for me, is that I can't show that there does not exist any $n_0$ so that $\sin(n_0!)$ is close to $-1$ enough so that the original term might not converge. Any help would be appreciated!
|
https://math.stackexchange.com/questions/4577234
|
[
"sequences-and-series",
"limits",
"analysis",
"trigonometry",
"factorial"
] | 30 | 2022-11-15T07:22:14 |
[
"One way to show that this sequence diverges would be to construct a subsequence which diverges.",
"AFAIK, $1$ will be an accumulation point (limit of a subsequence). But it is much harder to say that $0$ is not another one...",
"Factorial is not defined on not-integers. If the gamma function is used to extend the factorial function, the function oscillates between 0 and 1.",
"Related math.stackexchange.com/questions/677719/… and mathoverflow.net/questions/45665/…",
"I was being sloppy with my language, I meant one thing and wrote another @AdamRubinson . I was thinking: \"... if $\\sin(n!)$ is bounded away from $-1$, then...\" but although: equidistributed --> not bounded away from $-1$, the converse fails! You are quite right, it's unjustified at present",
"@FShrike Please can you justify your claim? It seems wrong, or at best, unjustified to me. For example, it's not clear to me why we cannot have: $\\ (\\sin(n!))_n\\ $ not equi-distributed and $\\liminf_{n\\rightarrow\\infty}(\\sin(n!)+1)^{1/n}<1.$",
"@FShrike: The problem of the density or lacunarity of the limit points of the sequence $\\sin n!$ is probably as hard as that of $\\cos(n!)$.",
"@MartinR we know $(\\sin(n))_n$ is equidistributed, but do we know $(\\sin(n!))_n$ is equidistributed? If it isn't, then we know for sure that the limit is $1$",
"Here is a similar question about $\\cos(n!)$. The answer (as I understand it) is “we don't know.”"
] | 0 |
Science
| 0 |
290
|
math
|
Wave equation: predicting geometric dispersion with group theory
|
Context
The wave equation
$$
\partial_{tt}\psi=v^2\nabla^2 \psi
$$
describes waves that travel with frequency-independent speed $v$, ie. the waves are dispersionless. The character of solutions is different in odd vs even number of spatial dimensions, $n$. A point source in odd-$n$ creates a disturbance that propagates on the light cone and vanishes elsewhere: if the point source is a flash of light, an observer sees darkness, then a flash, then darkness. When $n$ is even, a disturbed media never returns to rest: the observer sees darkness, then brightness that lingers for all $t$. This phenomena is known as geometric dispersion.
Question
Is it possible to show that geometric dispersion is predicted by the wave equation, using group theory? For a point source at the origin, we would be searching for spherically symmetric solutions, and the rotation group $SO(n)$ has a different structure depending on whether $n$ is odd or even. In particular, I am interested in doing this without actually solving the wave equation. Unfortunately, I don't know enough group theory to know if this is even possible.
What I know
I can 'show' geometric dispersion by solving the wave equation with an initial condition, or computing the Green's function for the wave equation and noting that it is either supported only on the light cone (odd $n$), or everywhere within the light cone (even $n$).
I know some group theory 'for physicists'.
Related
This unanswered question is similar. I think my question is more specific: I'm asking about a way to predict (rather than explain) geometric dispersion using group theory.
Update: (thanks to comments of Alp Uzman and GiuseppeNegro)
It appears to be possible using group theoretic machinery, described in the book Nonabelian harmonic analysis by Howe and Tan. The relevant section is 4.3.1. So an equivalent question becomes: can someone explain the result from Howe and Tan in a more accessible way? The book is beyond my level of group theory at the moment.
|
https://math.stackexchange.com/questions/4278597
|
[
"group-theory",
"partial-differential-equations",
"mathematical-physics",
"wave-equation",
"dispersive-pde"
] | 29 | 2021-10-16T12:35:52 |
[
"@GiuseppeNegro I had tried studying the Howe-Tan account of the Huygens' principle about five years ago and at the time I too could not handle the representation theory involved (I'm afraid I still can't). I don't really have any useful insights, as such I can't claim any credits, though thank you for acknowledging my comment elsewhere.",
"I wonder if we can @AlpUzman to notify them. It's them, not me, who gave that useful reference",
"@GiuseppeNegro Thank you, that reference seems to answer precisely my question- I do not understand it yet either",
"Following the comments in the linked question, I arrived at Theorem 4.3.1 in the book \"Nonabelian harmonic analysis\" of Howe and Tan (thank you Alp Uzman). That theorem contains exactly the answer to this question. Unfortunately, it uses heavy machinery from representation theory and I cannot fully understand it.",
"I would love to see an answer to this. Group theory is one of my weakest areas though, so I won't be able to produce one...."
] | 0 |
Science
| 0 |
291
|
math
|
Binomial coefficients modulo a prime
|
Consider an odd prime $p\equiv1 \pmod {16}$ and set $M=\frac{p-1}{2}$ for notational convenience. Then is there even a single prime $p$ of the above form for which the following congruence holds? $$\binom{M}{M/2}\binom{M}{M/4}\equiv \pm \binom{M}{3M/8}\binom{M}{7M/8}\pmod p ?$$
Here the symbol $\binom{n}{r}$ is a binomial coefficient and is defined as $\frac{n!}{r!(n-r)!}$
I am trying to prove something significantly more abstract but have managed to reduce it down to proving that the above statement can never hold. I have written some code to check this for primes of the above form less than $1000$ and found no counter examples. My attempts to find results on binomial coefficients $\pmod p$ have mostly involved Lucas' theorem which does not seem very applicable here. I would appreciate a solution, reference or possible strategies I might try.
Added later: The more general problem.
Let $$f_a(x)=x(x^4-1)(x^4+ax^2+1)$$ be a family of polynomials in $
\mathbb F_p[t]$ in the parameter $a\in \overline {\mathbb F}_p$.
Let $$\sum c_{\alpha}x^\alpha := f_a(x)^{\frac{p-1}{2}} $$ and note that the $c_\alpha$ are polynomials in $a$. Now make the $4
\times 4$ matrix whose $(i,j)$th entry is $c_{pi-j}$ for $1\leq i,j \leq 4$. Then I want to show that this matrix is non-invertible for values of $a$ other than $a=2,-2$. In other words, I want to show that the determinant $d(a)$ has irreducible factors other than $(a-2)$ and $(a+2)$.
What I have managed to prove is that $d(a)$ is a polynomial of degree $\frac{3(p-1)}{2}$ and also that $(a-2)$ and $(a+2)$ occur with the same multiplicity in the factorization of $d(a)$ in $\mathbb F_p[a]$. So for a contradiction, suppose $d(a)=\alpha(a^2-4)^{\frac{3(p-1)}{4}}$ for some $\alpha\in \mathbb F_p$. Then, the ratio of the leading term and the constant term must be 1. The leading term is given by the square of the right hand side in the top most equation and the constant term is given by the square of the left hand side. The other coefficients are very difficult to extract and it is lucky that I even managed to get the leading and constant terms. So I am hoping that if by some combinatorial argument I can rule out this congruence, then I will have proved that the determinant has to have an irreducible factor other than $(a-2)$ and $(a+2)$.
|
https://math.stackexchange.com/questions/3411958
|
[
"modular-arithmetic",
"binomial-coefficients",
"finite-fields"
] | 29 | 2019-10-27T21:43:24 |
[
"@JyrkiLahtonen it's been a while but yes the presence of that last factor is critical. For some more information, this is about superspecial hyperelliptic curves in the family $y^2 =x(x^4-1)(x^4+ax^2+1)$. I was trying to prove that for all primes $p = 1,7 \\mod 8$, and $p>7$, there is a superspecial curve in this family.",
"You can find a bunch of congruences of similar shape in Binomial coefficients and Jacobi sums, by Hudson and Williams.",
"I still wonder what the original problem is? The exponent in $f_a(x)^{(p-1)/2}$ somehow suggests a character sum involving a quadratic character, but I may be off :-)",
"Just to check that I have the right determinant. When $p=17$, the determinant of that 4x4 matrix is $$8(a+2)^8(a-2)^8(a^2-5)^4,$$ and the presence of that last factor is crucial for your purpose?",
"@mathworker21 Per your request, I have added some additional information.",
"@RKD can you say the significantly more abstract thing you want to prove? that might actually help to prove the thing you asked as your question",
"(Added for completeness of comment: I wrote some code to check up to $10^7$ - didn't find anything)",
"(cont.) My point being that strong computational evidence would require that one examine way more than the primes up to $10000$; I got Mathematica to go up to $244737$ and found nothing, but computing the binomial coefficients seems to be rather expensive (...though maybe less so if I could tell Mathematica to do this computation mod $p$, though I still don't know of any methods better than the usual formula, which uses $O(p)$ multiplication mod $p$ - which is really bad if you want a long search)",
"A heuristic note: if we let $P'_n$ be the $n^{th}$ prime congruent to $1$ mod $16$ and if we chose two numbers $(a_n,b_n)$ at random mod each $P'_n$, the probability that no solution to $a_n=\\pm b_n$ exists in the first $N$ terms is bounded below by $$\\prod_{n=1}^{N}(1-2/P_n).$$ This product is zero in the limit, suggesting a counterexample exists unless a pattern does - but of the primes up to $10^8$, this product is still $0.619$ - so even if we checked that far on a random sequence, we would still be less likely to find a $a_n=\\pm b_n$ pair than not to.",
"I have no clue. They seem to be completely at random.",
"Do you have any clue what LHS/RHS (modulo $p$) might be equal to?"
] | 0 |
Science
| 0 |
292
|
math
|
$f\colon I\rightarrow G$ and Gromov $\delta$-hyperbolicity
|
Please recall that $\left|\int_0^1 f(t)\,dt -w\right|\leq \int_0^1|f(t)-w|\,dt$. In general, let $(X,d)$ be a metric space. Given a function $f:I\to X$ let $m_f\in X$ be such that $d(m_f,w)\leq \int_0^1 d(f(t),w)\,dt$ for every $w\in X$. If such $m_f\in X$ exists then we call $f$ as $D$-integrable with $D$-integral $m_f$. ($I=[0,1]$.)
Question: Given a finitely generated group $G$ (with a usual word distance), does there exist a condition on the set of $D$-integrable functions $f: I\to G$ (which uses $D$-integrability) for the Gromov $\delta$-hyperbolicity of $G$? I.e, what is the characterization of hyperbolicity in terms of $D$-integrability?
(It seems to me that the real question here should be: how can one extract group theoretic intel from $D$-integrability?)
|
https://math.stackexchange.com/questions/49208
|
[
"calculus",
"group-theory",
"metric-spaces"
] | 28 | 2011-07-03T07:06:02 |
[] | 0 |
Science
| 0 |
293
|
math
|
Lowest dimensional faithful representation of a finite group
|
How does one compute the lowest dimensional faithful representation of a finite group?
This question originated in the context of given a finite group $G$: trying to find the lowest dimensional shape whose rotational/reflection symmetries form $G$. (Formally stated as finding the lowest dimensional faithful representation of $G$ into the orthogonal group $O(n)$.)
Now just listing out examples is hard, because even in just $\mathbb{R}^2$ outside of the symmetries of simple polygons we see things like $\mathbb{Z}_2 \times \mathbb{Z}_2 $ crop up, yet it appears that $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ doesn't have any obvious two dimensional representations. Grouppropswiki doesn't even have a single representation for it at all.
So I thought it was wise before I tackle the orthogonal group question I should know how to just, in general, find a low dimensional faithful representation. Some trivialities are that the dimension for a group $G$ will be less than or equal to the smallest $j$ such that $G \subset S_j$ since each symmetric group can be realized as the symmetries of a $j$-dimensional simplex space due to Burnside.
But this doesn't say much because even something like $\mathbb{D}_{\text{Graham's Number}}$ can be realized in $\mathbb{R}^2$ (and we still are dealing with isometries here, what about represenatations that transcend that!?)
|
https://math.stackexchange.com/questions/1688173
|
[
"linear-algebra",
"group-theory",
"finite-groups",
"representation-theory",
"linear-transformations"
] | 28 | 2016-03-08T00:31:03 |
[
"For future visitors: some research developments since this question was asked mathoverflow.net/questions/351938/… as well it is interesting to ask even whether there are small irreducible representations, as discussed here: mathoverflow.net/questions/400864/… Finally, note that $j$ in the question statement is called the minimal degree in group theory; computing is still open in many cases.",
"Is there a list of these smallest dimensions somewhere for small groups? @DerekHolt",
"It's a difficult problem in general over any field. For an abelian group with invariant factor decomposition $C_{k_1} \\times \\cdots C_{k_r} \\times C_2^s$ with each $k_i > 2$, I think the smallest dimension of a real faithful representation is $2r+s$. It is certainly $s$ for $C_2^s$.",
"I'm focusing on real representations. Does that make matters worse?",
"Are you talking about complex representations, or representations over an arbitrary field? For complex representations, the smallest dimension of a faithful representation of a finite abelian group with minimal generating set of size $d$ is $d$. But $C_2 \\times C_2 \\times C_2$ has a faithful $2$-dimensional representation over the finite field of order $8$. In general, this is a difficult problem - even finding the smallest degree faithful permutation representation can be difficult."
] | 0 |
Science
| 0 |
294
|
math
|
Dividing the whole into a minimal amount of parts to equally distribute it between different groups.
|
Suppose we have a finite amount of numbers $x_1, x_2, ..., x_n$ ($x_i\in\mathbb{N}$) and an object that should be divided into parts in such a way that it can be without further dividing distributed in $x_i$ equal piles for any $1\leq i\leq n$.
The question is what is the minimal amount of parts for some $x_1, x_2, ..., x_n$ that would allow us to do that?
Example: we have a cake and expect some friends, but we don't know for sure how many of them are coming. But somehow we know, that the total amount of people will be either 5 or 6 $(x_1 = 5, x_2 = 6)$. So we divide a cake beforehand into $5\times 6 = 30$ parts, and so in each scenario we can just give the guest his share of the cake.
Optimization. Of course, 30 parts is too many. In the example above we could just divide the cake into 5 parts, then ignore it and divide the cake (like it was whole) into 6 parts. First we do 4 cuts, second we do 5 cuts; they don't coincide because 5 and 6 are coprimes. Total 9 cuts gives us 10 pieces. So instead of cutting the cake in 30 pieces, we can only cut it in 10. And that is the minimum here.
Simple case. Let $x_1$ and $x_2$ be two coprime numbers. Then, using the same argumentation as above with the case $x_1=5, x_2=6$, we can note that the minimal amount of parts is $(x_1-1)+(x_2-1)+1$ which is equal to $x_1+x_2-1$.
Generalization. First I thought that for $n$ coprime numbers the solution is simple and beautiful:
$$
\sum_{i=1}^n (x_i-1) + 1
$$
But this formula is wrong for $n>2$ as shows the following counterexample.
Counterexample. For $x_1=3, x_2 = 4, x_3 = 5$ the formula gives the answer 10. But that's not the minimum, as we can divide at least in nine portions:
$$
\frac{1}{60} + \frac{2}{60} + \frac{4}{60} + \frac{5}{60} + \frac{7}{60} + \frac{8}{60} + \frac{10}{60} + \frac{11}{60} + \frac{12}{60} = 1
$$
so we can split it equally in 3 pieces:
$$
\left(\frac{1}{60} + \frac{2}{60} + \frac{7}{60} + \frac{10}{60}\right) + \left(\frac{4}{60} + \frac{5}{60} +\frac{11}{60}\right) + \left(\frac{8}{60} + \frac{12}{60}\right) = 1
$$
in 4 pieces:
$$
\left(\frac{1}{60} + \frac{2}{60} + \frac{12}{60}
\right) + \left(\frac{4}{60} +\frac{11}{60}\right) + \left(\frac{5}{60} + \frac{10}{60}\right) + \left(\frac{7}{60} + \frac{8}{60}\right) = 1
$$
in 5 pieces:
$$
\frac{12}{60}+\left(\frac{1}{60} + \frac{11}{60}
\right) + \left(\frac{2}{60} +\frac{10}{60}\right) + \left(\frac{4}{60} + \frac{8}{60}\right) + \left(\frac{5}{60} + \frac{7}{60}\right) = 1
$$
To sum up, the posed question has an evident answer in the case $n=1,2$, but in the case $n=3$ (and higher) I didn't discover any pattern.
[Edit] Related Topics:
Minimum Cake Cutting for a Party
https://puzzling.stackexchange.com/questions/19870/nine-gangsters-and-a-gold-bar
https://mathoverflow.net/questions/214477/minimal-possible-cardinality-of-a-a-1-a-k-distributable-multiset
|
https://math.stackexchange.com/questions/1381042
|
[
"number-theory",
"elementary-number-theory",
"divisibility"
] | 28 | 2015-08-01T03:46:39 |
[
"Note that the same question has been asked at math.stackexchange.com/questions/1383406/…",
"OK, so for $n$ coprime numbers, we can always do it with $1+\\sum(x_i-1)$ pieces, and the question is, under what circumstances can we do it with fewer, and how many fewer. So I wonder whether the $3,4,5$ construction generalizes to, say, $3,3k+1,3k+2$, or to $2k-1,2k,2k+1$, or some other infinite family of examples.",
"@rywit, but if we shift a little the second series of cuts like this (dividing with shift), we will use one additional cut but at the same time we will be able to assemble one-third parts and no more cuts will be needed.",
"@rywit, about the counterexample. Dividing into 3, 4 and 5 part using the formula in the question (let's call it the naive formula), can be graphically presented like this: naive dividing. When we divide the object into 5 parts and then into 6, it can be pictured like this (numbers are sizes of pieses in one-sixtieth portions): first two series of cuts. We cannot assemble any one-third parts using this pieses.",
"@GerryMyerson, at least that small (you can make it as big as you want by dividing into $kx_1x_2$ parts with any positive integer $k$). In order to divide into $x_1$ parts we do $x_1 - 1$ cuts. To divide into $x_2$ parts we do $x_2-1$ cuts. Number of pieces is the number of cuts plus 1, so we have $(x_1 - 1)+(x_2-1)+1 = x_1 +x_2-1$. Geometrical point of view shows that this is minimum, but that is more intuitive approach than a strict proof; that's why I wrote \"seems to be\".",
"@GerryMyerson Ah, you're right. Thanks!",
"@rywit, I think we're trying to minimize the number of pieces, rather than the number of physical cuts.",
"When you write, in the \"simple case\", that the answer \"seems to be\" $x_1+x_2-1$, do you mean you can prove it's at least that big? or do you mean you can prove it's at least that small? or what?",
"Also, in your example with 5 and 6 people you say you make 4 cuts and then separately 5 cuts for a total of 9. But you could align the first cut in each set. So you could make 4 cuts (for 5 people) and then 4 more cuts (reusing one of the existing cuts) for 6 people. Thus a total of only 8 cuts is required in this case. Am I right?",
"How did you discover this counterexample?"
] | 0 |
Science
| 0 |
295
|
math
|
Circles of radius $1, 2, 3, ..., n$ all touch a middle circle. How to make the middle circle as small as possible?
|
Non-overlapping circles of radius $1, 2, 3, ..., n$ are all externally tangent to a middle circle.
How should we arrange the surrounding circles, in order to minimize the middle circle's radius $R$?
Take $n=10$ for example.
On the left, going anticlockwise the radii are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, and $R\approx 10.77$.
On the right, going anticlockwise the radii are $10, 2, 9, 4, 7, 6, 5, 1, 8, 3$ and $R\approx 9.98$.
How should the circles be arranged to minimize $R$?
Here is a desmos graph where you can try different arrangements for the case $n=10$.
My attempt
For general $n$, call the radii of the surrounding circles going anti-clockwise, $r_1, r_2, r_3, ..., r_n$.
Draw line segments from the centre of the middle circle to the centre of each surrounding circle. So we have $n$ angles at the centre of the middle circle. Each of these angles can be expressed in terms of $R, r_k, r_{k+1}$, using the law of cosines. The sum of the $n$ angles is $2π$. So we have:
$$\sum\limits_{k=1}^n \arccos{\left(\frac{(R+r_k)^2+(R+r_{k+1})^2-(r_k+r_{k+1})^2}{2(R+r_k)(R+r_{k+1})}\right)}=2\pi \text{ (where }r_{n+1}=r_1)$$
This simplifies to:
$$\sum\limits_{k=1}^n \arccos{\left(1-\frac{2r_k r_{k+1}}{(R+r_k)(R+r_{k+1})}\right)}=2\pi$$
We want to assign each $r$ a unique value among $1, 2, 3, ..., n$ so that $R$ is minimized. But how? Is there a general pattern?
Conjectured answer
I think the following general procedure will make the middle circle as small as possible. Take $n=10$ for example. We ignore the $1$ at first; it will be placed last. First put down the numbers $10, 9, 8, 7, 6$ in pyramid-fashion, from top to down and from left to right, like the red numbers below. Then put down the numbers $2, 3, 4, 5$ in pyramid-fashion also, but between the previous rows, like the blue numbers below.
This gives the order of radii going around the circle: $7, 4, 9, 2, 10, 3, 8, 5, 6$. Then put the circle with radius $1$ anywhere you like, as long as it fits without disturbing other circles.
For larger values of $n$, there may be multiple small circles that can fit between other circles without disturbing them. Put those last.
Intuitively, this procedure minimizes the tendency of the large surrounding circles to take up space around the middle circle. You can see this in the diagrams above with $n=10$, looking specifically at the two largest surrounding circles in each case.
Anyway, it's just a conjecture for now. Notice that this conjecture implies that, for $n=10$, the arrangement I showed above on the right, yields the minimum radius of the middle circle. I don't think the middle circle can be any smaller.
|
https://math.stackexchange.com/questions/4619480
|
[
"geometry",
"optimization",
"circles",
"discrete-geometry"
] | 27 | 2023-01-16T03:46:59 |
[
"I wrote a program that loops over all permutations of 2...10 (excluding the circle of radius 1), and finds an $R$ that solves the equation that you wrote (the sum of the arc-cosines). It confirmed that the optimal permutation is the one that you conjectured: 2, 9, 4, 7, 6, 5, 8, 3, 10, where the radius of the inner circle is ~9.979907. I had to exclude the circle of radius $1$ from the sum of course.",
"It seems the right drawing the small circle may be removed without impacting the result.",
"@VarunVejalla Good point. I took this point into account in my desmos graph, but forgot to mention this in my question. We should ignore the smallest few circles, because they will each fit between some two other circles that touch each other. How many circles should be ignored, depends on $n$, with an upper limit of $n/4$ (based on Descartes' Circle Theorem).",
"The problem with the current sum of $\\arccos$ is that you can have gaps between circles. For example, in the arrangement on the right, there would have to be some wiggle room for the circle of radius $1$ between the circles of radius $5$ and $8$ (i.e. it can't be tangent to both of those and the middle circle), and numerically solving for $R$ using the sum of $\\arccos$ yields $\\approx 9.90$, different from the $\\approx 9.98$ that it would actually be.",
"@Dan Yes, absolutely. Actually, I think there's no point trying that. For small numbers (say, $n=7,8,9,10$) one might solve the equation numerically for all the possible combinations of $r_k$. Perhaps that gives a clue on how the $r_k$ must be arranged for general $n$.",
"Yeah, I thought the same",
"@Dog_69 That approach is very difficult due to the $\\sum$, I think.",
"First thing tthat comes to my mind is to take the derivative with respect to $R$ and see the condition for it to vanish."
] | 0 |
Science
| 0 |
296
|
math
|
A curious identity on $q$-binomial coefficients
|
Let's first recall some notations:
The $q$-Pochhammer symbol is defined as
$$(x)_n = (x;q)_n := \prod_{0\leq l\leq n-1}(1-q^l x).$$
The $q$-binomial coefficient (also known as the Gaussian binomial coefficient) is defined as $$\binom{n}{k}_q := \frac{(q)_{n}}{(q)_{n-k}(q)_{k}}.$$
I found the following curious identity on $q$-binomial coefficients, and I would like to ask for some ideas on how to prove it.
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} (q)_n$$
I am familiar with the $q$-binomial theorem and $q$-Vandermonde identity, which I think might be useful. If there are some other well-known identities on $q$-binomial coefficients that might be helpful in figuring this out, I would like to know them too. Thank you!
Some thoughts:
The classical version of this identity is simply $$\sum_{0\leq j\leq 2n}(-1)^{j}(2n+1-j)\binom{2n}{j} = \sum_{0\leq j\leq 2n}(-1)^{j}(j+1)\binom{2n}{j} = \delta_{n,0},$$
which can be proved simply by $$\sum_{0\leq j\leq 2n}(-1)^{j}\binom{2n}{j} + \sum_{0\leq j\leq 2n}(-1)^{j}j\binom{2n}{j} = (1-x)^{2n}+\frac{d}{dx}(1-x)^{2n} \bigg\vert_{x=1} = \delta_{n,0}.$$
There is a natural $q$-analogue of the above identity, obtained by replacing $(1-x)^{2n}$ by $(x)_{2n}$ and the derivative by the $q$-derivative, but unfortunately this is not the desired identity.
There seems to be other closely related identities:
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n-1)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} \delta_{n,0},$$
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n+1)+\frac{j(j-1)}{2}}\binom{2n}{j}_q \overset{?}{=} q^{n}(q)_{2n}.$$
Proving these other identities (or a family of similar identities) might be helpful if we were to try to use induction on $n$.
|
https://math.stackexchange.com/questions/4434174
|
[
"binomial-coefficients",
"q-analogs",
"q-series"
] | 27 | 2022-04-22T23:52:21 |
[
"So we begin with a combinatorial interpretation : Consider all $n$-tilings using $k$ green tiles and $n-k$ red tiles, such that every green tile's weight is $q^r$ where $r$ is the number of red tiles preceding this green tile. The weight of a tiling is the product of weights of all green tiles, and the sum over all such tilings of these weights equals $\\binom{n}{k}_q$. We get that the LHS is a sum over all tilings of length $2n$ and $k \\geq $ the number of green tiles, of sth. involving $k$ and the green tiles. Using $(-1)^j$ and some work, cancellations can be figured out that eliminate $k$.",
"I managed to obtain a partial explanation using combinatorics. I don't have a full answer to fall back upon ,but I eliminated the role of $k$ and figured out some further cancellations using the ideas. I don't want to post a partial answer (I'd rather it be complete and worthy of evaluation), but if someone's interested I'll put the details up in the next comment(s).",
"It might be useful to post the same question on Mathoverflow. It is definitely worth it.",
"@Henry That is nice. I will have a look today at it. The quadratic form on the exponent is unusual. Nice identity tho!",
"@Phicar Okay, I have turned the desired identity into a single-sum identity and was able to verify it using qZeil. This is nice, and thanks for the suggestion. Still, it would be nice to have a non-algorithmic proof of the identity.",
"@SarveshRavichandranIyer Unfortunately I am not aware of any useful reference either.",
"@Phicar Can qZeil handle double summations? I thought it was mostly for single summations.",
"Can you provide some source for these identities? I had a general look and unfortunately there's very little on double-sum identities that I found lying around for the $q$-binomial.",
"Perhaps $q-$Zeilbergs gets you a nice recursion? risc.jku.at/sw/qzeil"
] | 0 |
Science
| 0 |
297
|
math
|
What is $\int_0^1 \left(\tfrac{\pi}2\,_2F_1\big(\tfrac13,\tfrac23,1,\,k^2\big)\right)^3 dk$?
|
As in this post, define the ff:
$$K_2(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)}$$
$$K_3(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac13,\tfrac23,1,\,k^2\right)}$$
$$K_4(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac14,\tfrac34,1,\,k^2\right)}$$
$$K_6(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac16,\tfrac56,1,\,k^2\right)}$$
We find that,
$$\int_0^1 K_2(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;1,\tfrac32;1\right)}=2G$$
$$\int_0^1 K_3(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac13,\tfrac23;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}2\, \ln2$$
$$\int_0^1 K_4(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac14,\tfrac34;1,\tfrac32;1\right)}=2\ln(1+\sqrt2)$$
$$\int_0^1 K_6(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac16,\tfrac56;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}4\, \ln(2+\sqrt{3})$$
where $G$ is Catalan's constant. However, this post gives a third power,
$$\int_0^1\big(K_2(k)\big)^3 dk=\frac35 \bigg(\tfrac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12,1,\tfrac12\big)\bigg)^4=\frac35 \big(K(k_1)\big)^4 = \frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2} \approx 7.0902$$
where $K(k_d)$ is an elliptic integral singular value while YuriyS in his comment below gives,
$$\int_0^1\big(K_3(k)\big)^3 dk \approx 6.53686311168760876289835638374 $$
Questions:
What are the closed-forms of:
$$\int_0^1\big(K_n(k)\big)^m dk=\,?$$
for power $m=2$ or $m=3$?
Or at least their numerical evaluation up to 20 digits?
P.S. My old version of Mathematica can't evaluate it with sufficient precision, nor does WolframAlpha. (Enough digits may make it amenable to the Inverse Symbolic Calculator.)
|
https://math.stackexchange.com/questions/3103805
|
[
"integration",
"definite-integrals",
"closed-form",
"hypergeometric-function",
"elliptic-functions"
] | 27 | 2019-02-07T05:32:18 |
[
"At one point, I convinced myself that dlmf.nist.gov/15.5 (at least a couple of them) worked for fractional derivatives. If so, then the order might be reduced To 1F1(a,b;k^2), which might be easier to evaluate (and invert). If you're still interested, I might take a stab at that. No warranties implied :)",
"A related case with a closed form in this question: math.stackexchange.com/q/3325678/269624",
"@YuriyS: I revised the post to give a clearer version of the problem.",
"Mathematica gives $I_3=6.53686311168760876289835638374$ with WorkingPrecision->30 $$ $$ It has no problem evaluating the integral, only takes a few seconds",
"@YuriyS: The closed-form is not the titular integral $I_3$. Note the closed-form involves $\\frac12,\\tfrac12$, while the titular integral $I_3$ involves $\\frac13,\\frac23$. P.S. If you can provide a numerical evaluation of $I_3$ for 20 digits or more, that would be nice.",
"Could you clarify: does your 2nd question ask for numerical evaluation of the titular integral? Or the general integral from the 1st question? Above you have already provided the closed form for the titular integral, so I'm confused",
"Please use \\left and \\right rather than \\bigg, especially in titles."
] | 0 |
Science
| 0 |
298
|
math
|
Is there a group $G$ for which $\mathrm{Aut}(G) \simeq (\mathbb{R},+)$?
|
I know the classic theorem that $(\mathbb{Q},+)$ cannot be expressed as an automorphism group, i.e. there is no group $G$ such that $\mathrm{Aut}(G)\simeq (\mathbb{Q},+)$.
Theorem A. If $L$ is a locally cyclic group with no element of order $2$, then $L$ cannot be expressed as an automorphism group.
But how about $(\mathbb{R},+)$?
I think the answer might be no, and the proof proceeds by showing that if $\mathrm{Aut}(G) \simeq \mathbb{R}$, then some subgroup or quotient $H$ of $G$ will satisfy $\mathrm{Aut}(G) \simeq \mathbb{Q}$, which contradicts Theorem A.
So how can we construct this $H$? Assuming the axiom of choice, we can say $\mathbb{R} = \mathbb{Q}\oplus B$ for some additive subgroup $B$ of $\mathbb{R}$ (just by picking a $\mathbb{Q}$-basis for $\mathbb{R}$). Now I'm tempted to do some Galois-type thing, where you use a "fixed" subgroup $$H=\mathrm{Fix}(B)=\{g\in G : b(g) = g \text{ for all } b\in B \},$$ and then try to say something about $\mathrm{Aut}(H)$ or $\mathrm{Aut}(G/H)$ (where, for the latter, maybe we take the normal closure of $H$). But I can't complete this line of reasoning. It seems key that $\mathrm{Aut}(G)$ splits as a direct sum --- that seems special.
Am I just totally off-base here? Is there some obvious group $G$ whose automorphism group is $\mathbb{R}$?
Related questions
If I can prove that $\mathbb{R}$ is not an automorphism group, then the same would hold for the isomorphic group $\mathbb{R}_{>0}$ of positive reals under multiplication. But how about $\mathbb{R}^\times \simeq \mathbb{R}_{>0}\times C_2\simeq \mathbb{R}\times C_2$ --- can this be achieved as an automorphism group? If $\mathrm{Aut}(G)$ is a direct product of abelian groups, what can be said about $G$?
A related observation is that if $\mathrm{Aut}(G)$ is abelian, then $G$ is nilpotent of rank $\leq 2$. So there is a sense in which $G$ is "almost" abelian.
I know $\mathbb{Q}^\times$ is the automorphism group of $(\mathbb{Q},+)$, and that $\mathbb{R}^\times$ constitutes the continuous automorphisms of $(\mathbb{R},+)$ ... Then there's multiplicative/additive groups of other rings/fields ...
I can't even resolve this question in the case of $\mathrm{Aut}(G) \simeq \mathbb{Z}\times \mathbb{Z}$. For this one I can show that $G'$ would necessarily by cyclic, but that's about it.
I'm very curious about which types of groups can be achieved as automorphism groups.
A quick observation
If $\mathrm{Aut}(G) \simeq \mathbb{R}$, then $G$ must be infinitely generated (or else its automorphism group is countable) and nilpotent of rank $\leq 2$ (this follows for any group whose automorphism group is abelian).
|
https://math.stackexchange.com/questions/3925067
|
[
"group-theory",
"automorphism-group"
] | 27 | 2020-11-27T07:48:46 |
[
"Thanks, I did already know that! Same idea to prove that $\\mathbb{Q}$ is not an automorphism group.",
"Partial answer: $G$ cannot be abelian. If $G$ is abelian, then $x\\mapsto x^{-1}$ is an automorphism of $G$ of order $\\leq 2$. $\\mathbb{R}$ has no nonidentity elements of finite order, so $x=x^{-1}$ for all $x\\in G$. This implies that $G$ is the underlying abelian group of an $\\mathbb{F}^2$-vector space $V$, and we have $\\operatorname{Aut}(G)=\\operatorname{Aut}_{\\mathbb{F}_2}(V)$. We must have $\\dim V\\geq 2$, but this means $\\operatorname{Aut}(G)$ contains $\\operatorname{GL}_2(\\mathbb{F}_2)$ as a subgroup. This would make $\\operatorname{Aut}(G)$ nonabelian, contradiction!",
">it's consistent with ZF that every endomorphism of R is measurable --- Wow.",
"@Ehsaan: re: your related questions, it's consistent with ZF that every endomorphism of $\\mathbb{R}$ is measurable, hence continuous; if this is true then $\\text{Aut}(\\mathbb{R}) \\cong \\mathbb{R}^{\\times} \\cong \\mathbb{R} \\times C_2$. (With the axiom of choice $\\text{Aut}(\\mathbb{R})$ is much larger and in particular nonabelian.)",
"@DietrichBurde thank you --- that's related for sure. I wonder which if there are groups $A$ for which it is true, but difficult, to prove that $A$ cannot be achieved as an automorphism group.",
"See also this post.",
"Yes, it would have to be some special property of $\\mathbb{R}$ that comes up here. Subgroup or quotient would be OK here --- just as long as we can use the fact that $\\mathbb{Q}$ is a direct summand of $\\mathbb{R}$ to reduce to the case of $\\mathbb{Q}$ (which is a known exercise).",
"It's certainly not the case 'generically' that if $H\\subset Aut(G)$ then there's some subgroup $G'$ of G with $Aut(G')\\equiv H$, even if $Aut(G)=H\\times K$; consider the automorphism group of $C_p$ for $p$ prime. I don't see anything in your reasoning offhand that uses any properties of $\\mathbb{R}$ that don't mirror to this case."
] | 0 |
Science
| 0 |
299
|
math
|
If $r\in\mathbb{Q}\setminus\mathbb{Z}$ is it possible that $r^{r^{r^r}}\in \mathbb{Q}$?
|
It's straightforward to prove that $r^r\notin\mathbb{Q}$, which furthermore allows us to use the Gelfond-Schneider theorem to prove that $r^{r^r}\notin\mathbb{Q}$.
Is it true that $r^{r^{r^r}}\notin\mathbb{Q}$?
It seems like it ought to be true, though my guess would be that this is an open problem. If not, does anyone know a proof? And if so, do there exist any discussions of it in the literature?
|
https://math.stackexchange.com/questions/3655352
|
[
"reference-request",
"exponentiation",
"irrational-numbers",
"tetration"
] | 27 | 2020-05-02T09:25:14 |
[
"en.wikipedia.org/wiki/… - the techniques for the proof by infinite descent here can be generalized in a straightforward way",
"@DarkMalthorp how do u prove $r^r \\not \\in \\mathbb{Q}$?",
"Yes, indeed it is a stronger result :) Of course, I fully expect that $r^{r^{r^r}}$ is also transcendental.",
"@darkmalthorp For the record, Gelfond-Schneider says that $r^{r^r}\\notin\\overline{\\Bbb Q}$, not just $\\notin\\Bbb Q$ (though I presume you meant that).",
"I would also guess that this is an open problem. It is even unknown whether the power-tower is an integer for $\\ r=\\pi\\ $ (which would however be a miracle)"
] | 0 |
Science
| 0 |
300
|
math
|
What functions from $\Bbb N\to\Bbb N$ can be made from $+$, $-$, $\times$, $\div$, exponentiation, and $\lfloor\cdot\rfloor$?
|
What functions from $\Bbb N\to\Bbb N$ can be made from $+$, $-$, $\times$, $\div$, exponentiation, and $\lfloor\cdot\rfloor$? Call this class of functions $\mathcal Flex$ (for "floor and exponentiation").
The $\rm mod$ function $a\operatorname{mod}b$ can be defined by $a-b\lfloor a/b\rfloor$. In addition,
$$\binom nk=\left\lfloor\frac{(u+1)^n}{u^k}\right\rfloor\operatorname{mod}u$$
for $u>2^n$, and
$$n!=\left\lfloor\frac{r^n}{\binom rn}\right\rfloor$$
for $r>(2n)^{n+1}$. Thus the factorial function is in $\mathcal Flex$.
The function $\left\lfloor\lfloor\frac ab\rfloor-\frac ab\right\rfloor+1$ equals $1$ if $a$ divides $b$ and $0$ otherwise. According to Wilson's theorem, $n$ is prime iff $n$ divides $(n-1)!+1$; therefore, the function that is $1$ for primes and $0$ otherwise is in $\mathcal Flex$.
The function $\delta_{0,n}$ is in $\mathcal Flex$ since it equals $\left\lfloor\frac1{1+n}\right\rfloor$. I feel like there should be some way to use the MDRP theorem to show that all primitive recursive functions with images in $\{0,1\}$ are in $\mathcal Flex$, but I'm not quite sure how.
In fact, my big conjecture so far is that $\mathcal Flex$ actually equals the set of all primitive-recursive functions. (And I have a suspicion that exponentiation might be unnecessary after all...)
|
https://math.stackexchange.com/questions/4605213
|
[
"exponential-function",
"diophantine-equations",
"recursion",
"computability",
"ceiling-and-floor-functions"
] | 26 | 2022-12-24T20:59:34 |
[
"Given that we have binomials and modular arithmetic, the function $\\Big\\lfloor 10^n\\frac{16^{f(n)}}{f(n){2f(n)\\choose f(n)}^2} \\pmod{10}\\Big\\rfloor$ probably* gives the $n^\\textrm{th}$ digit of $\\pi$ for sufficiently fast-growing $f(n)$ like $n^{n^{n^n}}$. *Unless pi has an incredibly long string of zeros somewhere in its decimal expansion.",
"Thanks for your explanation. This got into my review queue as a test, which got me very confused.",
"@youthdoo Well, if someone provided a nice criterion for something to fall into this class, it would be a good solution. As it stands, I think this coincides with the ELEMENTARY class I found a month ago.",
"Very possibly I didn't understand this well, but isn't this question too broad?",
"I vote to call these \"flexible functions\"",
"Maybe it's this?: en.wikipedia.org/wiki/ELEMENTARY I have some reading to do.",
"Incidentally, I don't think it really matters whether we look at $\\Bbb N\\to\\Bbb N$ or $\\Bbb Z\\to\\Bbb Z$. In the latter case, the function $\\lfloor\\frac{3n}{3n-1}\\rfloor$ seems to give $1$ if $n$ is positive and $0$ otherwise, which seems useful.",
"It is also possible to show that if $h(a,b,y):=\\prod_{k=1}^y(a+bk)$ then $h\\in\\mathcal Flex$. (I believe this is $b^yy!\\binom{q+y}y\\operatorname{mod}(bq-1)$ where $q>(a+by)^y$.) In the context of proving the MDRP theorem, this is function is used in conjunction with the Sunzi (Chinese) Remainder Theorem to encode bounded quantifiers.",
"Most of these equations come from Martin Davis's exposition of the proof of the MDRP theorem: math.umd.edu/~laskow/Pubs/713/Diophantine.pdf"
] | 0 |
Science
| 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.