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101
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mathoverflow
|
Supercompact and Reinhardt cardinals without choice
|
A friend of mine and I ran into the following question while reading about proper forcing, and have been unable to resolve it:
_Definition._ A cardinal $\kappa$ is _supercompact_ if for all ordinals $\lambda$, there exists a transitive inner model $M_\lambda$ of $V$ and an elementary embedding $j_\lambda: V\rightarrow M_\lambda$ such that $crit(j_\lambda):=\min\lbrace\alpha\in ON: j(\alpha)\not=\alpha\rbrace=\kappa$ and $M_\lambda$ is closed under $\lambda$-sequences (that is, $^\lambda M_\lambda\subseteq M_\lambda$).
Now in the definition of supercompactness, the inner models $M_\lambda$ are allowed to vary with $\lambda$. We could define a large cardinal notion, _uniform supercompactness_ , by removing this possibility: $\kappa$ is uniform supercompact if there is a single $M$ and $j$ such that $M$ is closed under all sequences of ordinal length and $j: V\rightarrow M$ is an elementary embedding with critical point $\kappa$. But in ZFC, this is not an interesting notion. First, assuming choice, $M$ is closed under sequences of ordinal length if and only if $M=V$ (since two models with the same ordinals and sets of ordinals are equal), so $ZFC\models$ "$\kappa$ is uniformly supercompact $\iff$ $\kappa$ is Reinhardt;" second, Kunen showed that Reinhardt cardinals are inconsistent with $ZFC$. (EDIT: as Joel points out, this statement doesn't actually make sense, as Reinhardt-ness is not a first-order property. One needs to pass to some extension of ZFC which can talk about proper classes directly, like Morse-Kelly set theory, which is the setting Kunen used for his proof.)
However, assuming $\neg$Choice, both of these points seem suspect. It is still open whether Reinhardt cardinals are consistent with $ZF$, and it is unclear to me that $ZF$ alone proves that uniformly supercompact cardinals are Reinhardt. So my questions are the following:
Question 1: Does $ZF\models$ every uniformly supercompact cardinal is Reinhardt?
Question 2: What is the consistency strength of $ZF+$ there exists a uniformly supercompact cardinal?
Obviously $ZF$ proves that every Reinhardt cardinal is uniformly supercompact; also obviously, Question 2 is really only interesting in lieu of a positive answer to Question 1. The most basic obstruction to a positive answer to Question 1 is the fact that given an elementary embedding $j: V\rightarrow M$ with $M$ closed under ordinal-length sequences and $crit(j)=\kappa$, the restriction of $j$ to $M$, $j\upharpoonright M: M\rightarrow M$ is not obviously (to me at least) an elementary embedding.
|
https://mathoverflow.net/questions/94900/supercompact-and-reinhardt-cardinals-without-choice
|
[
"lo.logic",
"large-cardinals",
"set-theory",
"axiom-of-choice"
] | 27 | 2012-04-22T18:53:23 |
[
"The point here is that we require $j(\\kappa)$ to be the same in all of these embeddings. Clearly if there is a class embedding its restrictions satisfy this property. And now we can quantify over all $\\lambda$ if need be.",
"I know that this is quite old by now, and I'm sure that Noah and several others here would have no problem tackling @Joel's last comment. Here is my \"starting suggestion\" to modify Woodin's definition: $\\kappa$ is $\\lambda$-supercompact* if there is $\\lambda'\\geq\\lambda$ s.t. for every large enough $\\alpha$ there is an elementary $j\\colon V_\\alpha\\to N$, where $N$ is transitive, the critical point is $\\kappa$, $j(\\kappa)=\\lambda'$, $V_\\lambda\\subseteq N$, and $N^\\lambda\\subseteq N$. (Some indices might need a small tweak, e.g. $V_{\\lambda+1}$ or so to be \"faithful to the idea\".)",
"Hm, good point. Is there a \"uniformized\" version of the existence of normal fine measures that might be to the existence of normal fine measures what the second-order definition of uniform supercompact is to the second-order definition of supercompact? (If that makes any sense.) Looking at it now, I don't see one . . .",
"I noticed that the definition you give of supercompact is explicitly second-order. With AC, this is equivalent to the existence of normal fine measures, but without AC, one no longer has the Los theorem and so it isn't clear that your second-order definition has a first-order equivalent. ",
"Thanks for your thoughtful response, Noah. Elliott Mendelson dispenses with global AC by positing, in place of Limitation of Size, AC for sets plus a form of Replacement. In that case, ZFC only holds for sets, not for proper classes (\"pc's\") such as V and M (for M an inner transitive model of V). Hence, ZFC doen't rule out the existence of a suitable pc M, i.e. a pc that witnesses a negative answer to Question 1. Also, a negative answer to 1 is compatible with the existence of Reinhardts; it simply says that the class of Reinhardts doesn't include all of the uniform supercompacts. ",
"will yield a theory which still proves ZFC; in this case, there will be no such suitable model. Specifically, I believe that given any (I suppose wellfounded) model $W$ of this (admittedly vague) version of NBG, it will still be the case that any submodel of $W$ with the same ordinals and sets of ordinals has the same sets as $W$ itself. If, on the other hand, we weaken Limitation of Size so much that even normal choice is not provable, then we can have such an $M$, but we no longer no that no Reinhardts exist; so we still can't answer 1 negatively.",
"Oh, I see what you mean. Sorry, I missed the \"global\" in your comment and didn't understand what you were saying. Actually addressing the point you made: The way I'm familiar with it, NBG doesn't actually have global choice as an axiom; instead, it follows from Limitation of Size. So there are two ways I could understand \"dropping global choice\" from NBG: weakening Limitation of Size so that only normal AC is provable, or weakening Limitation of Size so that not even normal AC is provable. Both are vague, but some things can be said about them. Any suitable way of doing the former (cont'd)",
"I agree that the mere existence of an M with certain properties doesn't mean we have an elementary embedding. Question 1, however, doesn't ask for an e.e.; it asks if all uniform supercompact cardinals are Reinhardt (given ZF). I.e., it assumes that uniform supercompacts and their e.e.'s exist (including a suitable M), and asks if this entails an e.e. of V into V. In NBG without global AC, there exists (I claim) a suitable M distinct from V, thus answering Question 1 negatively. I may be wrong, of course; but even if I am, I don't think it's due to not showing that a certain e.e. exists.",
"@wmitt, I don't think that works - uniform supercompactness (and, for that matter, Reinhardtness) doesn't just ask for an inner model with some properties, it asks for an elementary embedding. So even given an $M\\subsetneq V$ closed under sequences of ordinal length, that doesn't mean that we have an elementary embedding $V\\preccurlyeq M$. (I think what I've said is right; somebody who knows more, please correct me if I'm wrong!)",
"Thanks. If we do drop global AC from NBG, doesn't the resulting lack of a bijection between ON and V entail the existence of an inner model M that's (i) closed under sequences of ordinal length, and (ii) not equal to V? If so, we would seem to have a negative answer to Question 1, making the idea of uniform supercompactness interesting. ",
"wmitt, Yes, it does not require any AC or global AC to formulate the Reinhardt cardinal concept in second-order Godel-Bernays set theory (without AC). So you are correct, wmitt, that we should speak of formalizing it in GB rather than in GBC. The Kunen inconsistency, which rules out Reinhardt cardinals, is a theorem of GBC, with choice. ",
"Is there a way to formalize Reinhardt cardinals and uniform supercompactness without using Choice? The above examples of using NBG or KM to formalize these notions both involve the use of Choice. But we need to work with a Choiceless universe in order to answer Questions 1 and 2.",
"We are jumping over each other's comments! Yes, by \"second-order GBC\", I just meant that GBC was a second-order theory. Sorry to confuse...",
"Yes, NBG and GBC are the same; just an alternative acronym. N = von Neumann. It is conservative over ZFC, because you can force to add a global choice class and then just take the definable classes. So every model of ZFC is the first-order part of a model of GBC. This is not true for KM, since KM proves the existence of satisfaction classes for first-order truth, and hence proves Con(ZFC). ",
"(Not to keep spamming comments, but:) Your most recent comment confuses me somewhat. Is GBC a theory in second-order logic, or by \"second-order GBC\" did you just mean something similar to \"second-order arithmetic?\" ",
"The super compactness of $\\kappa$ is expressible in first-order as the assertion: for every ordinal $\\lambda$, there is a $\\kappa$-complete normal fine ultrafilter on $P_\\kappa\\lambda$. ",
"Also, you say that it is also possible to formalize Kunen's inconsistency argument in GBC. I'm not familiar with GBC; is it the same as NBG, and if not, is it a conservative extension of ZFC?",
"Noah, while we can prove that Reinhardt cardinals are not first order expressible (otherwise let $\\kappa$ be the least one, and then consider $j(\\kappa)$...), it seems that we should similarly expect the same for uniform supercompactness. But it can be formalized in second-order GBC, as an assertion about a class $j$ and a class $M$. One replaces the assertion that $j$ is fully elementary with the assertion that $j$ is $\\Sigma_1$-elementary and cofinal, which implies $\\Sigma_n$-elementarity by a meta-theoretic induction. ",
"@Joel: good points. I've corrected the question. Also, building off of your first comment, is it obvious that uniform supercompactness is first-order expressible? For that matter, how is supercompactness itself first-order expressible?",
"I believe (from your remark about ordinal lengths) that you intend that the assertion \"for all $\\lambda$...\" in the definition is meant to be interpreted as, \"for all ordinals $\\lambda$...\". The difference matters, as you likely know, because if you have $j:V\\to M$ and $j\"A\\in M$ for all sets $A$, then it follows easily that $M=V$ even in ZF, since if $A$ is transitive, then $A$ is the Mostowski collapse of $j\"A$.",
"Very nice question. One quibble: it isn't really correct to say that ZFC refutes Reinhardt cardinals, since there is no first-order characterization of Reinhardt cardinals; that is, you can't say \"$\\kappa$ is a Reinhardt cardinal\" in a first-order way. Kunen proved his inconsistency in the second-order Kelly-Morse set theory, but it is also possible to formalize the result in Goedel-Bernays set theory GBC. See jdh.hamkins.org/generalizationsofkuneninconsistency for a further discussion of this point. ",
"This is an interesting question. The contradiction in ZFC comes from the fact that models with the same sets of ordinals are isomorphic. In ZF it is not true. Monro proved that there is a chain of $\\omega$ models, where $V_n\\subseteq V_{n+1}$ and they both have the same sets-of-sets-..$n$-times of ordinals, but they are not isomorphic (Jech The Axiom of Choice, Ch. 5, Problem 17). This, however is \"going up\" while inner models are \"going down\" so there might be some difference there.",
"Like you say in the last paragraph, it is not clear that the restriction of j to M is going to be an elementary embedding into M in general. Instead you could consider the restriction of j to smaller models, like the class N consisting of all sets constructible from sequences of ordinals. Then j does restrict to an embedding N -> N but you still have might have the problem that the embedding is not amenable to the model. This seems like the bigger issue to me."
] | 23 |
Science
| 0 |
102
|
mathoverflow
|
Orders in number fields
|
Let $K$ be a degree $n$ extension of ${\mathbb Q}$ with ring of integers $R$. An order in $K$ is a subring with identity of $R$ which is a ${\mathbb Z}$-module of rank $n$.
Question: Let $p$ be an unramified prime in $K$. Is it true that the number of orders in $R$ of index equal to $p^r$, for some natural number $r$, is less than or equal to the number of subrings with identity of ${\mathbb Z}^n$ of index equal to $p^r$?
Nathan Kaplan and I need this fact for $n=5$ in a project where we are trying to find asymptotic formulae for the number of orders of bounded index in a given quintic field.
We've been staring at Jos Brakenhoff's [thesis](https://openaccess.leidenuniv.nl/bitstream/handle/1887/14539/proefschrift-brakenhoff.pdf?sequence=2) for a while, but I haven't gotten anywhere. Any advice will be greatly appreciated. Thanks.
Added in edit: Here is an elementary reformulation of this problem. Let $r(x)$ be a polynomial with integer coefficients. Then show that for any natural number $a$ in order to maximize the number of subrings of $({\mathbb Z}/p^a {\mathbb Z})[x]/(r(x))$ of a given index, the polynomial $r(x)$ has to be a product of linear factors modulo $p$.
Update: In a soon-to-be-posted preprint Nathan Kaplan, Jake Marcinek, and I have proved an asymptotic formula for the number of orders in a given quintic field of bounded discriminant, using an argument independent of the above question. We can also give non-trivial upper bounds for a general number field. I'm now convinced the answer to the question is yes, but I still don't know how to prove it.
|
https://mathoverflow.net/questions/130089/orders-in-number-fields
|
[
"number-fields",
"nt.number-theory",
"algebraic-number-theory"
] | 26 | 2013-05-08T08:50:58 |
[
"Thanks everyone. You can actually count the number of subrings of index $p$. Namely, if $r(x) \\mod p$ has $u$ factors of degree $1$ and $w$ factors of degree $2$, and the rest of higher degrees, then the number of subrings of ${\\mathbb Z}/p{\\mathbb Z}[x]/(r(x))$ of index $p$ is ${u \\choose 2} + w$. ",
"The $r=1$ case can be handled as follows. Each order in $R$ of index $p$ is uniquely determined by its image in $R/pR$. Now $R/pR$ is an \\'etale $k := \\mathbf{F_p}$-algebra of rank $n$ as $p$ is unramified. Now the set of such rings injects into the set of index $p$ $k$-subalgebras of $R/pR \\otimes_k \\overline{k} \\simeq \\overline{k}^n$. The latter set is a combinatorial object (it's the set of surjective maps from an $n$-element set to an $(n-1)$-element set), and can be bounded above by the set of index $p$ subrings of $\\mathbf{Z}^n$. ",
"I found a paper by Melanie Wood which might be relevant, although I did not check it carefully: arxiv.org/pdf/1007.5508v1.pdf",
"Have you tried (probably you have) writing down a binary $n$-ic form and playing the same game as in Theorem 9 to Lemma 13 of Bhargava, Shankar, and Tsimerman's paper arxiv.org/pdf/1005.0672.pdf ? This doesn't \"just work\", the details are much messier, but since you need something weaker I wonder if it is possible to salvage the result you are looking for?",
"In our work we only need this for unramified primes. ",
"It is clear that this question can be studied locally at $p$. In other words, this depends only on $R \\otimes \\mathbb Z_p$, so it depends only on $K \\otimes \\mathbb Q_p$, which is just a product of $p$-adic fields.",
"Is it obvious that the elementary reformulation is the same, given that one is restricted to $p$ unramified and the other is not?",
"Who voted to close? And why?"
] | 8 |
Science
| 0 |
103
|
mathoverflow
|
Where do uncountable models collapse to?
|
Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\mathcal{M}\models T$ of cardinality $\kappa$, we can ask whether $\mathcal{M}$ can be made isomorphic to any countable model $\mathcal{N}\models T$ in the ground model; that is, whether there is some $\mathcal{N}\models T$, which is countable, such that $V[G]\models \mathcal{M}\cong\mathcal{N}$ for $G$ $Col(\kappa,\omega)$-generic over $V$. **For simplicity, we'll say "$\mathcal{M}$ collapses to $\mathcal{N}$".**
In case $T$ has _countably many_ countable models, the answer is yes: the set of models of $T$ is a countable Borel set, and so is absolute. The uncountable models of such a $T$ are thus partitioned into $\omega$-many classes. My question is about what the nonempty classes are. For example, suppose $T$ is $\aleph_1$-categorical. Then the countable models of $T$ form an elementary chain, with a "top" model $\mathcal{N}_\omega$, and it is easy to see that this is the only countable model which uncountable models can collapse to.
However, there are plenty of non-$\aleph_1$-categorical theories with only countably many countable models, and beyond the $\aleph_1$-categorical setting things are much less clear. Some examples:
* The language has predicates $P_i$ ($i\in\omega$); $T$ asserts that the $P_i$ name infinite disjoint sets. Then the countable models of $T$ are classified by how many elements they have not in $\bigcup_{i\in\omega} P_i$, and so there are countably many countable models. For each countable model of $T$, there is an uncountable model of $T$ which collapses to it.
* The language has a single binary relation $E$; $T$ asserts that $E$ is an equivalence relation and that there is exactly one class of cardinality $n$, for each $n$. Countable models of $T$ are determined by how many infinite classes they have; and every countable model of $T$ **except the prime** is collapsed to by some uncountable model of $T$.
I'm curious what we can say in general about the set of models which can be collapsed to. There's a lot of questions around here that one can ask; let me focus on:
> Is there a theory $T$ with countably many countable models, including $\mathcal{M}_0\prec\mathcal{M}_1$, such that $\mathcal{M}_0$ can be collapsed to by some uncountable structure but $\mathcal{M}_1$ cannot?
That is, can there be "gaps" in the range of the collapse?
* * *
EDIT: As Paul Larson pointed out in the comments below, a countable model is collapsed to by some uncountable model iff it is **extendible** (see Definition 2.6 in <http://shelah.logic.at/files/1003.pdf>) - that is, iff it has an uncountable $L_{\omega_1\omega}$-elementary extension. One direction is trivial. To show that every extendible model is collapsed to, let $\mathcal{M}$ be extendible with $\mathcal{M}\prec_{\omega_1\omega}\mathcal{N}$ for $\mathcal{N}$ uncountable. Let $\varphi$ be the Scott sentence of $\mathcal{M}$, and let $V[G]$ be a forcing extension in which $\mathcal{N}$ is made countable. Then $\mathcal{N}\models\varphi$ in both $V$ and $V[G]$; moreover, the statement "$\varphi$ is the Scott sentence of $\mathcal{M}$" is $\Pi^1_2$ (actually, better, but this is enough) and so absolute between $V$ and $V[G]$. So $V[G]\models\mathcal{M}\cong\mathcal{N}$.
It's worth keeping in mind the broader result, due to Barwise and Karp if I recall correctly, that $\mathcal{M}\equiv_{\infty\omega}\mathcal{N}$ iff $\mathcal{M}\cong\mathcal{N}$ in some forcing extension.
|
https://mathoverflow.net/questions/220850/where-do-uncountable-models-collapse-to
|
[
"lo.logic",
"set-theory",
"model-theory",
"forcing"
] | 26 | 2015-10-13T21:45:52 |
[
"I asked Saharon my version of your question, and it turns out to have a relatively easy positive answer. Here is a write-up, which could probably be improved : users.miamioh.edu/larsonpb/extendible.pdf",
"I've been wondering about this too. I don't see that it matters. By the way, Su Gao's \"On automorphism groups of countable structures\" shows that a countable structure is extendible if and only if its automorphism group is not closed in the group of permutations of its domain. So one could rephrase the problem using this. Here's a variation I've wondered about : can you have countable $M \\prec N$ such that $M$ is extendible and $N$ is rigid?",
"@PaulLarson Coming back to this a year later, it occurs to me that we can ask the same question of theories with uncountably many countable models. That is, are there $\\mathcal{M}_0\\prec \\mathcal{M}_1$ with $\\mathcal{M}_0$ extendible but $\\mathcal{M}_1$ not? It's not obvious to me that this makes the question any easier - I don't see how to get such a pair, even if I allow lots of countable models - but maybe you know an easy answer in this case?",
"For instance, it seems that the model-theoretic forcing approach gives : if $T$ is a complete first order theory having an uncountable model and only countably many countable models, then there is a countable extendible model of $T$ having a copy of each countable model of $T$ as an elementary submodel. Again, I don't know how much insight this gives into your question.",
"@PaulLarson That looks really cool, thanks!",
"You might also get some mileage out of Shelah's theory of model-theoretic forcing, from his publication #88r, depending on what you're interested in.",
"John Baldwin used \"extendible\" in our paper with Saharon (shelah.logic.at/files/1003.pdf). I assume that it was used previously, but I don't really know.",
"@PaulLarson I didn't know about extendible models - yes, it looks like a model is extendible iff it is collapsed to. But googling around I can't find much about them - can you point me to a good source?",
"It seems that you could naturally rephrase this in terms of extendible models (i.e., countable models which have the same Scott sentence as some uncountable model) and then drop the mention of forcing."
] | 9 |
Science
| 0 |
104
|
mathoverflow
|
Which sets of roots of unity give a polynomial with nonnegative coefficients?
|
> **The question in brief:** When does a subset $S$ of the complex $n$th roots of unity have the property that $$\prod_{\alpha\, \in \,S} (z-\alpha)$$ gives a polynomial in $\mathbb R[z]$ with nonnegative coefficients?
Some trivial necessary conditions include $1\not\in S$ and also that $S$ is self-conjugate so that the coefficients will at least be real.
A sufficient condition that is not obvious (to me) is that the roots of unity in $S$ are precisely those lying outside of some wedge-shaped region of the complex plane, i.e., those with $\left|\arg(z)\right| > t$ for some fixed $t > 0$. This follows from the main theorem of
> Barnard, R. W., W. Dayawansa, K. Pearce, and D. Weinberg. "Polynomials with nonnegative coefficients." _Proceedings of the American Mathematical Society_ 113, no. 1 (1991): 77-85. [Journal link](http://www.ams.org/journals/proc/1991-113-01/S0002-9939-1991-1072329-2/S0002-9939-1991-1072329-2.pdf)
which says that, given any polynomial with nonnegative coefficients, dividing it by the linear factors corresponding to exactly those of its roots lying in such a region produces a polynomial with only positive coefficients. Applying this to $z^{n-1}+\cdots+z+1$ gives the desired result.
But that theorem is not number-theoretic; it doesn't care at all that we are dealing with roots of unity. So maybe there is a nicer proof in this special case?
More generally, what other necessary and/or sufficient conditions exist? Is there some reasonable number-theoretic or combinatorial characterization of such sets? What tools seem likely to shed some light on this question?
**ADDED LATER:** A couple of very interesting suggestions have been made in the comments so far, and I'm eager to see where they lead. However, I remain quite curious as to what can be said from a number-theoretic perspective. For example, if $n=p^r > 1$ for some prime $p$, then the cyclotomic polynomial $\Phi_n(x)=\Phi_p(x^{n/p})$ has only nonnegative coefficients. So a sufficient condition that I did not mention above is that $S$ is the set of primitive $n$th roots of unity for some prime power $n$. I believe this is the only way to obtain such a polynomial that is a _cyclotomic_ polynomial, but that doesn't imply that nothing else can be gained from a number-theoretic perspective on this question – however I myself lack the expertise necessary to make the most of such a perspective.
I should also mention that (ironically, given my desire for a number-theoretic perspective on this) the cyclotomic result actually follows from a more general, not number-theoretic result of
> Evans, Ronald, and John Greene. "Polynomials with nonnegative coefficients whose zeros have modulus one." _SIAM Journal on Mathematical Analysis_ 22, no. 4 (1991): 1173-1182. [Author's link](http://math.ucsd.edu/~revans/PolynomialsGreene.pdf)
which gives that for any proper divisor $d$ of $n$, $(x^n-1)/(x^d-1)$ has only nonnegative coefficients. So that generalizes the cyclotomic case, but not via number theory!
|
https://mathoverflow.net/questions/214784/which-sets-of-roots-of-unity-give-a-polynomial-with-nonnegative-coefficients
|
[
"nt.number-theory",
"co.combinatorics",
"polynomials",
"additive-combinatorics",
"roots-of-unity"
] | 26 | 2015-08-14T07:23:56 |
[
"\"I believe this is the only way to obtain such a polynomial that is a cyclotomic polynomial\": yes, you're correct. The easiest way to see this is plugging in x=1: it's easy to show that $\\Phi_m(1) = \\begin{cases}0 & m=1 \\\\ p & m=p^k \\\\ 1 & \\text{otherwise}\\end{cases}.$ This essentially says there must be cancellation outside of the prime-power case.",
"@Josep: The reference you gave is extremely interesting. I was totally unaware of Suffridge's result.",
"In fact the Suffridge polynomials cover the more general case when $1\\not\\in S,$ S self-conjugate and the arguments of the roots in S are separated by the same angle, except for a pair.",
"In the wedge case you can compute explicitly the coefficients and see they are positive. It's a particular case of Suffridge's extremal polynomials- you have the expression deduced in Sheil-Small, Complex polynomials, p. 251-252.",
"Maybe, the following related result of Kellog is helpful: Let $A$ be a complex $n\\times n$ matrix. If all its elementary symmetric functions are positive (so that the characteristic polynomial has alternating signs), then the spectrum of $A$ lies in the set $\\{z : |\\text{arg}z| \\le \\pi - \\pi/n\\}$...."
] | 5 |
Science
| 0 |
105
|
mathoverflow
|
Chromatic Spectra and Cobordism
|
I apologize in advance, if some of the things I've written are incorrect.
The cobordism hypothesis states that $\mathbf{Bord}^\mathrm{fr}_n$ is the free symmetric monoidal $(\infty,n)$-category with duals generated by a point. There is a geometric realization functor $|-|:\text{Cat}_{(\infty,n)}\rightarrow \text{Gpd}_\infty$, where $\text{Cat}_{(\infty,n)}$ is the $\infty$-category (I mean a quasicategory, but you can think of it as a model category, or whatever you like) of (small) $(\infty,n)$-categories, and $\text{Gpd}_\infty$ is $\infty$-category of $\infty$-groupoids, otherwise known as spaces. This functor is a left adjoint (in the homotopical sense) to the inclusion functor $\text{Gpd}_\infty\hookrightarrow \text{Cat}_{(\infty,n)}$. Therefore, $|-|$ preserves (homotopy) colimits. If we "restrict" ourselves to the symmetric monodical $(\infty,n)$-categories, we would get a functor $|-|:\text{SymMon}_{(\infty,n)}\rightarrow \text{Alg}_{\mathbb{E}_\infty}$, where $\text{SymMon}_{(\infty,n)}$ is the $\infty$-category of symmetric monoidal $(\infty,n)$-categories and $\text{Alg}_{\mathbb{E}_\infty}$ is the $\infty$-category of infinite loop spaces. This should also preserve colimits. It manifests itself in the fact that $|\mathbf{Bord}^\text{fr}_n|$ is the infinite loop space of the sphere spectrum.
There is a universal characterization of the bordism $(\infty,n)$-category of manifolds with singularities [Theorem 4.3.11 in [Lurie's paper](http://arxiv.org/abs/0905.0465)]. The type of singularities considered are those of Baas-Sullivan theory, which produce a wide array of chromatic spectra, such as the Brown-Peterson spectrum $\text{BP}$, the Johnson-Wilson spectra $\text{E}(n)$, and, of course, the Morava K-theory spectra $\text{K}(n)$. I am wondering:
> Is it possible to give these spectra a universal characterization using the fact that geometric realization preserves colimits?
**EDIT:** When I think about it, I am not sure whether there is a universal description of $|\mathbf{Bord}^{(X,\zeta)}_n|$, where $X$ is topological space, $\zeta$ a vector bundle on it of dimension $n$, other than the case when $X$ is point. We _can_ write it down. It is simply $\Omega^{\infty-n}M\zeta$, where $M\zeta$ is the Thom spectrum of $\zeta$.
> Do we get a different characterization of these infinite loopspaces from the cobordism hypothesis?
|
https://mathoverflow.net/questions/184261/chromatic-spectra-and-cobordism
|
[
"homotopy-theory",
"stable-homotopy",
"infinity-categories",
"cobordism",
"chromatic-homotopy"
] | 26 | 2014-10-12T17:49:07 |
[] | 0 |
Science
| 0 |
106
|
mathoverflow
|
Planar minor graphs
|
The theorem of Robertson-Seymour about graph minors says that there exists no infinite family of graphs such that none of them is a minor of another one.
Apparently, it came as a generalization of the Kruskal's theorem that states that there exists no infinite family of rooted ordered trees such that none is a minor of another one. Here, _rooted ordered_ means that the tree has a root, and that the edges escaping from a vertex are ordered. In other words, the trees are assumed to be embedded in the plane, and the minor operation has to respect this embedding.
Here comes the question: is Robertson-Seymour theorem true for planar graphs, when we add the condition that the minor operation respects the embedding? (_i.e._ we not only ask $G_i$ to be a minor of $G_j$ as an abstract graph, but also as an embedded graph.)
It is not clear to me that this should be a direct corollary of the original theorem, because of the amount of possible embeddings into the plane for a given planar graph.
|
https://mathoverflow.net/questions/92421/planar-minor-graphs
|
[
"graph-minors",
"graph-theory"
] | 26 | 2012-03-27T16:28:06 |
[
"@PierreDehornoy: I think it would improve the question if \"true for planar graphs\" was replaced with 'true for plane graphs', because what you write immediately afterwards suggests that you are thinking about planar graphs together with a specified embedding into the plane, and that is called a plane graph.",
"@Pierre Dehornoy: Pierre, sorry, you a right, this argument does not answer your question. ",
"@ Misha: This looks almost convincing, but I do not understand the end of the argument. One of the embedding of $G_m$ appears infinitely offen, but how does it says that this particular embedding is the one that was given at the beginning, i.e. that corresponds to the given embedding of $G_m$?",
"I guess, my understanding of R-S theorem is correct: If a sequence of graphs were to contain infinitely many minimal elements $G_m$ then the new sequence of graphs $(G_m)$ would violate R-S theorem. ",
"@Pierre Dehornoy: I thought that the conclusion of Robertson-Seymour theorem could be strengthened to \"A certain graph $G_m$ occurs as a minor infinitely many times in a given infinite sequence\" ($G_m$ would be one of the finitely many minimal elements of the sequence, with respect to the \"minor\" order). If my understanding is correct, then the answer to your question is positive since $G_m$ has only finitely many planar embeddings, up to isotopy, so one of them would be repeated (infinitely often). "
] | 5 |
Science
| 0 |
107
|
mathoverflow
|
Is every $p$-group the $\mathbb{F}_p$-points of a unipotent group
|
Let $\Gamma$ be a finite group of order $p^n$. Is there necessarily a unipotent algebraic group $G$ of dimension $n$, defined over $\mathbb{F}_p$, with $\Gamma \cong G(\mathbb{F}_p)$?
I have no real motivation for this question, it just came up in conversation and no one knew the answer. There does not appear to be any sort of uniqueness to $G$; both the groups $\mathbb{Z}/p^2 \mathbb{Z}$ and $(\mathbb{Z}/p \mathbb{Z})^2$ have infinitely many lifts to unipotent groups.
* * *
I am bumping this question to the top because I now suspect I have a counterexample, namely the dihedral group of order $16$, also known as $C_2 \ltimes C_8$, where $C_n$ is cyclic of order $n$. The "obvious" lift of $C_8$ is the additive group of the $3$-truncated Witt vectors, $\mathcal{W}_3$. So the obvious thing to try is a semidirect product $\mathbb{G}_a \ltimes \mathcal{W}_3$. But $\mathbb{G}_a$ has exponent $2$, so an action of $\mathbb{G}_a$ on $\mathcal{W}_3$ must live in the $2$-torsion elements of $\mathrm{Aut}(\mathcal{W}_3)$. I compute that the space of $2$-torsion elements in $\mathrm{Aut}(\mathcal{W}_3)$ has two connnected components, one of which maps $1$ to things that are $1 \bmod 4$ and one of which maps $1$ to things which are $-1 \bmod 4$. Since $\mathbb{G}_a$ is connected, it must land in one component, but the two elements of $C_2$ live in different components.
I've made some partial progress thinking about the structure of connected unipotent groups whose $\mathbb{F}_2$ points are $C_2 \ltimes C_8$, but I've gotten stuck, so I am putting up a bounty for progress either on this case or the questions as a whole.
|
https://mathoverflow.net/questions/262745/is-every-p-group-the-mathbbf-p-points-of-a-unipotent-group
|
[
"ag.algebraic-geometry",
"gr.group-theory",
"finite-groups",
"algebraic-groups",
"nilpotent-groups"
] | 26 | 2017-02-20T19:26:35 |
[
"Do you require $G$ to be connected?",
"@DrorSpeiser the Zariski closure of a finite set is itself (so not unipotent). I don't see any reason why we should expect any given embedding of D_8 in a unipotent group to come from the F_2-points of a connected subgroup.",
"Take $a=1+E_{12}+E_{34}+E_{56}$, $b=1+E_{23}+E_{45}+E_{67}\\in Mat_{7\\times7}(\\mathbb{F}_2)$, where $E_{ij}$ has a unique nonzero element at row $i$ and column $j$ equal to 1. Set $c=ab$. Let $V$ be $\\overline{\\mathbb{F}}_2^7$, and let $D_{16}\\cong<x, y|x^2=y^8=xyxy=1>$ act on $V$ by sending $x$ to $a$ and $y$ to $c$. With this action $V$ is an indecomposable module for $D_{16}$. Let $\\overline{D_{16}}$ be the Zariski closure of $<a, c>$ in $GL(V)$, which is unipotent. Does it satisfy $\\overline{D_{16}}(\\mathbb{F}_2)=<a,c>\\cong D_{16}$?",
"@DrorSpeiser I decided to put the bounty on my own version, rather than the duplicate because I'd rather edit my own question and because it has a bunch of useful comments. If I had (two years ago) seen Gjergji's link to the duplicate before a bunch of those comments showed up, I probably would have self-closed, but it doesn't seem like a big deal either way. Of course, if someone posts a good answer, I'll link it on that question as well to make sure searchers find it.",
"Given that this is a duplicate of a question six years earlier, how is this not closed?",
"Okay, I'll take a look at Abhyankar's conjecture and see what that does for me. In the mean time, I'll record that the formula was meant to be $(x_1, y_1, z_1) \\ast (x_2, y_2, z_2) = (x_1+x_2, y_1+y_2, z_1+z_2 +(x_1 y_2)^p-x_1 y_2)$, sorry for the incorrect subscripts in the previous one.",
"I am specifically suggesting to use the faithful representations of $p$-groups arising from Abhyankar's conjecture. If we realize $\\Gamma$ as the Galois group of a finite etale extension $\\mathbb{F}_q[t] \\subset R$, then for every subgroup $\\Lambda \\subset \\Gamma$, then $R$ is a finite etale extension of the $\\Lambda$-invariant subring $S$, with Galois group $\\Lambda$. This is a \"complete reducibility\" result that is missing for arbitrary representations. I suggest we projectively complete these rings at $\\infty$, and quotient by the maximal ideal of $\\infty$ in $\\mathbb{F}_q[t^{-1}]$.",
"Suppose there were such a $G$, and let $Z$ be the center $\\{ (0,0,z) \\}$ of $U$. Case 1: $G \\supset Z$. Then $G(\\mathbb{F}_p) \\supset \\{ (0,0,z) : z \\in \\mathbb{F}_p \\}$, which is not in $\\Gamma$. Case 2: $G \\cap Z$ is finite. In that case, the projection $G \\to U/Z$ is surjective. But any subvariety of $U$ which surjects onto $U/Z$ generates $U$, contradicting that $G$ is two dimensional.",
"I think there is a fatal flaw in @JasonStarr's approach. It seems that Jason's method would show that, if $U$ is a unipotent group, and $\\Gamma$ is a subgroup of $U(\\mathbb{F}_p)$, then there is a unipotent subgroup $G$ of $U$ with $G(\\mathbb{F}_p) = \\Gamma$. This isn't true. Let $U$ be the group structure on $\\mathbb{A}^3$ given by $(x_1,y_1,z_1) \\ast (x_2,y_2,z_2) = (x_1+x_2, y_1+y_2, z_1+z_2+x_1^p y_1^p - x_1 y_1)$ and let $\\Gamma = \\{ (x,y,0) : x,y \\in \\mathbb{F}_p \\} \\subset U(\\mathbb{F}_p)$. Then $\\Gamma$ is not $G(\\mathbb{F}_p)$ for any $2$-dimensional subgroup of $U$. (continued)",
"@DavidSpeyer. I should not have called this the unique smallest subgroup. I am saying that for a vector space $V$ and a finite subgroup $Z$ of the group of $\\mathbb{F}_p$-points of the scheme $\\text{Spec}\\ \\text{Sym}_{\\mathbb{F}_p}(V)$, the kernel of the restriction map $V\\to \\prod_{z\\in X} \\kappa(z)$ is an $\\mathbb{F}_p$-linear subspace of $V$ generating an ideal whose associated closed subscheme is a closed subgroup scheme whose associated group of $\\mathbb{F}_p$-points equal to $Z$.",
"@JasonStarr It seems like you are saying that, given an connected abelian group $H$ (in your setting, $Z(U)$), and a discrete subgroup $B$, there is a smallest connected group $A$ containing $B$. But that isn't true: consider $H = \\mathbb{G}_a^2$ and $B = \\{ (x,0) : x \\in \\mathbb{F}_p \\}$. Then $B$ is contained in $\\mathbb{G}_a$ embedded in $H$ by either $t \\mapsto (t,0)$ or $t \\mapsto (t,t^p-t)$. What am I missing?",
"@JasonStarr: I was wondering about that connectedness issue too but thought I might be overlooking something.",
"@nfdc23. You are correct. I realize now the problem with the approach I suggest: why should the centralizer of $Z(\\Gamma)$ in $U$, much less the center of the centralizer, be a connected group?",
"@JasonStarr: In view of your inductive suggestion it seems that what you call \"$n$\" (from SL$_n$) is not \"$n$\" as in the statement of the question, just some unknown integer, in which case such $\\rho$ is provided by the regular representation of $\\Gamma$ on itself over $\\mathbf{F}_p$ (that gives $\\rho$ into ${\\rm{GL}}_N(\\mathbf{F}_p)$, necessarily landing inside SL$_N({\\mathbf{F}}_p)$ and even conjugating into $U_N(\\mathbf{F}_p)$ since the latter is a $p$-Sylow subgroup).",
". . . So there is a unique minimal, connected subgroup $A\\subset Z(U)$ such that $A(\\mathbb{F}_p)$ equals $Z(\\Gamma)$. Now replace $U$ by $U/A$, and use induction on the rank.",
"Here is how I would try to prove this. There exists a faithful homomorphism, $\\rho:\\Gamma\\to \\textbf{SL}_n(\\mathbb{F}_p)$. This follows, for instance, from the solution of the Abhyankar conjecture by Raynaud and Harbater. By counting $\\mathbb{F}_p$-points of the flag variety of $\\textbf{SL}_n$, there exists a Borel subgroup $B$ that contains the image of $\\rho$. Thus, there exists a faithful homomorphism from $\\Gamma$ to a unipotent group. Replace that unipotent group by the centralizer $U$ of $Z(\\Gamma)$. Thus $Z(U)(\\mathbb{F}_p)$ is an additive group that contains $Z(\\Gamma)$ . . .",
"A step towards an inductive proof: if $G$ is a unipotent $k$-group, and $c \\in Z^2(G(k), Z / p Z)$ is a 2-cocycle (for trivial action), does there exist a polynomial 2-cocycle $\\gamma \\in Z^2(G, G_a)$ such that $\\gamma$ gives $c$ upon taking $k$-points? I want to attack this one inductively with inflation-restriction with respect to $G$ and $[G,G]$. But no time for more than a sketchy comment.",
"This was asked a while ago, as well, but there was no answer. mathoverflow.net/questions/69397/…",
"If $U$ is smooth connected unipotent of dimension $n$ over $k=\\mathbf{F}_p$ then $\\#U(k)=p^n$ (use composition series with successive quotients $\\mathbf{G}_a$). Thus, if $\\Gamma=G(k)$ for a unipotent $k$-group $G$ of dimension $n$ then by passing to $G_{\\rm{red}}$ so that $G$ is smooth we see that if $G$ is not connected then $G^0(k)$ of size $p^n$ must exhaust $G(k)$. In other words, we lose nothing by assuming $G$ is smooth and connected. So it is the same as asking for smooth connected unipotent $k$-groups $U$ if there is any constraint on $U(k)$ beyond its order being $p^n$. Good puzzler."
] | 19 |
Science
| 0 |
108
|
mathoverflow
|
Does the Tate construction (defined with direct sums) have a derived interpretation?
|
Any abelian group M with an action of a finite group $G$ has a Tate cohomology object $\hat H(G;M)$ in the derived category of chain complexes. There are several ways to define this. One is as the tensor product $W \otimes_G M$ with a particular (typically unbounded) complex of projective $\Bbb Z[G]$-modules which happens to have trivial homology. One is to instead use $Hom^G(W',M)$ for $W'$ a similar complex. One is as the mapping cone of a norm map $$ M_{hG} \to M^{hG} $$ from derived coinvariants to derived invariants, where we view $M$ as a complex concentrated in degree zero.
We can attempt to generalize this construction to the derived category of $\Bbb Z[G]$-modules, or of $R[G]$-modules for $R$ a commutative ring. If $M_*$ is a chain complex of $\Bbb Z[G]$-modules, these give us three definitions of the Tate construction on $M_*$: $$ \begin{align*} Tate^\oplus(M)_n &= \bigoplus_{p+q=n} W_p \otimes_G M_q\\\ Tate^\prod(M)_n &= \prod_{p+q=n} W_p \otimes_G M_q\\\ Tate(M)_n &= \bigcup_N \prod_{p+q = n, p \leq N} W_p \otimes_G M_q \end{align*} $$ (There are natural maps $Tate^\oplus \to Tate \to Tate^{\prod}$ which are isomorphisms on bounded complexes.)
The first construction is nice because it commutes with filtered colimits, and the second with filtered limits. The third is neither, but it does have a norm-cofiber sequence and a Tate cohomology spectral sequence $$ \hat H^p(G; H^q(M)) \Rightarrow H^{p+q}(Tate(M)). $$ The third construction is also the only one that preserves quasi-isomorphisms, and thus the only one that (naturally) extends to the _unbounded_ derived category.
Are there derived-category interpretations for the constructions $Tate^\oplus$ or $Tate^\prod$?
Such an interpretation probably needs to take in more information than an object of the derived category of $\Bbb Z[G]$-modules. The main motivation for asking is that I'd like to see what is necessary in more general contexts (e.g. modules over a differential graded algebra).
|
https://mathoverflow.net/questions/234577/does-the-tate-construction-defined-with-direct-sums-have-a-derived-interpretat
|
[
"at.algebraic-topology",
"homological-algebra",
"group-cohomology"
] | 25 | 2016-03-26T13:01:27 |
[
"Another characterisation of Tate cohomology $\\hat H^n(G, M)$ is as $Hom(\\mathbf Z, M[n])$, where the Hom-set is in the category of singularities, defined as the quotient of the usual derived category $D^b(Mod_{\\mathbf Z G})$ modulo its subcategory of perfect complexes. A natural extension thus seems to be (I remember this is also mentioned somewhere in Gaitsgory's work, but don't seem to find it right now) to consider the Hom-set or -space in the quotient of Ind-Coherent sheaves modulo Quasi-coherent sheaves (in the $\\infty$-sense).",
"@Jesper Both of those are good suggestions. I haven't asked either yet, and it looks like I'll probably be doing so soon.",
"Dave Benson or John Greenlees would be my go-to people for this. Maybe you've already asked them..."
] | 3 |
Science
| 0 |
109
|
mathoverflow
|
Are amenable groups topologizable?
|
I've learned about the notion of topologizability from "On topologizable and non-topologizable groups" by Klyachko, Olshanskii and Osin (<http://arxiv.org/abs/1210.7895>) - a discrete group $G$ is topologizable iff there exists a topology on $G$ which makes it into a Hausdorff non-discrete topological group.
> **Main question:** Is every infinite amenable group topologizable?
* * *
Main motivation for this question is that perhaps naively non-topologizability seemed to me such a strange property that I hoped it could be used to show existence of non-sofic groups.
> **Question:** Is there an infinite non-topologizable sofic group?
After failing to prove that sofic groups are topologizable I thought it would be still interesting to prove that infinite "elementary sofic" groups are topologizable. Elementary sofic groups are for the purpose of this discussion the class of "groups which are provably sofic by current methods", i.e. it contains all amenable groups, is closed under taking free products amalgamated over amenable groups, extensions with sofic kernel and amenable quotient, /any other results which are in the literature/, and with the property that if G is residually elementary sofic then G is elementary sofic.
> **Question:** Is there an infinite non-topologizable elementary sofic group?
Unfortunately my plan to answer the above question negatively failed at step 1, and hence the Main question.
|
https://mathoverflow.net/questions/114688/are-amenable-groups-topologizable
|
[
"gr.group-theory",
"gn.general-topology"
] | 25 | 2012-11-27T10:42:34 |
[
"I essentially stopped thinking about this, because I don't know how to proceed, but I thought I'd share one idea which at one point I thought was hopeful: G acts on a certain metric space X by isometries - namely fix a mean m on G and define X to be the set of subsets of G up to sets of mean 0. The metric on X is d(A,B) = m(A-B \\cup B-A). There are various topologies on Isom(X) but I failed to prove any of them gives a Hausdorff non-discrete topology on G. ",
"@Simone: AFAI understand, Bohr compactif'n B(G) is in particular compact, so by Peter-Weyl thm if G embeds in B(G) then G is res. linear, and so by Malcev thm if it is fin. generated then it is res. finite. So to produce example of an amenable group G which doesn't embed into B(G) take a fin. generated simple amenable group, or easier take a fin.gen. solvable non-res. finite group (e.g. BS(1,n)). Taking a simple group is more convincing though, because the image in B(G) is trivial, whereas if the image in B(G) is infinite one could still hope for inducing a (Hausdorff) topology on G somehow...",
"probably Corollary 3 on page 213 in mscand.dk/article.php?id=2006 will help...",
"unfortunately it seems that the bohr topology is T2 only for MAP groups... is there any known example of amenable non-MAP group (with the discrete topology)?",
"Any infinite abelian group is topologizable (e.g. by its Bohr topology).",
"Just to clarify, Łukasz's first sentence actually characterizes the pro-p topology for fixed p; the profinite topology on $\\mathbb{Z}$ is generated by all the pro-p topologies.",
"@Ben: you can use profinite topology, e.g. fix a prime number $p$; then integer $n$ is \"near\" to $0$ if large power of $p$ divides $n$. Other way to topologize integers is - take the action of integers $Z$ on the circle such that the generator $t$ of $Z$ acts by irrational rotation, and define $t^k$ to be \"close\" to the identity element iff $t^k$ is a rotation by a \"small\" angle. This is special case of \"find an infinite cyclic subggroup of a compact group and induce the topology\"",
"Wouldn't the evenly spaced integer topology work?",
"Are the integers topologizable? What does that topology look like?"
] | 9 |
Science
| 0 |
110
|
mathoverflow
|
Caramello's theory: applications
|
In [this](https://www.laurentlafforgue.org/math/TheorieCaramello.pdf#page=9) text, the author says (well, he says it in French, but I am too lazy to fix all the accents, so here is a Google translation):
_In any case, contemporary mathematics provides an example of extraordinarily deep and highly studied equivalence which, without having so far been formulated as an equivalence of Morita, still seems very close to the general framework of Caramello's theory: it is the correspondence from Langlands._
I am trying to understand this remark. The entire mathematical world has produced few, if any, correspondences which unify lots of non-trivial mathematics as elegantly as the Langlands program does (other than the Langlands, that is). Several genuises (including L. and V. Lafforgue, A. Wiles, P. Scholze) have contributed to its study and our understanding of it is still far from complete.
> I wonder if there were any papers confirming the overall sentiment of the author, i.e. papers where Caramello's theory is applied to Langlands. Were there at least some long-standing problems that historically were not believed to have any serious connection with logic or topoi, and were solved using Caramello's techniques?
There were some similar questions, but [this](https://mathoverflow.net/q/330087/140765) one seemed to be largely about understanding an incorrect translation of a quote of another distinguished mathematician (hopefully, our translation is correct!), and [this](https://mathoverflow.net/q/29232/140765) one is pretty similar but not the same (it asks about producing theorems without any creative effort, while we allow additional creative input but just want to see theorems that rely in an essential way on ideas of Caramello, among other things). The second question did receive a response that might have qualified as an answer to our question, but that response also received [criticism](https://mathoverflow.net/questions/29232/the-unification-of-mathematics-via-topos-theory?noredirect=1&lq=1#comment70456_29826) from BCnrd (which was not addressed AFAICT).
P.S. The kind of response we are not looking for in this question:
* some philosophical argument why topoi are great that does not give a specific resut. While such arguments can be of interest in some situations, not right now.
* something along the lines of "One can produce thousands of deep theorems easily, so easily in fact that I won't even bother producing a single example".
The kind of response we are looking for:
* a link to a paper proving a result outside of logic or set theory and whose statement does not involve topoi, and an explanation of how Caramello's techniques enter the proof as an important ingredient.
|
https://mathoverflow.net/questions/332279/caramellos-theory-applications
|
[
"ag.algebraic-geometry",
"ct.category-theory",
"topos-theory",
"langlands-conjectures",
"applications"
] | 25 | 2019-05-23T03:04:44 |
[
"@TimCampion OK, maybe somebody who actually read those papers will comment on them. But who knows, maybe logic will give us the motivic t-structure!",
"The first two publications listed under \"Publications\" on her website are both papers in English written jointly with Lafforgue. In particular, the second appears to do something reasonably concrete in algebraic geometry / number theory. The first appears to be more category-theoretic, but perhaps closer inspection would reveal some algebraic geometry / number theory.",
"@TimCampion I browsed through some of them and did not manage to find any results of geometric or topological interest (though those that I saw were in French, so I did not try super hard).",
"Caramello's website lists 3 papers joint with Lafforgue by my count. Presumably they contain some kind of working-out of her theoretical framework, if not directly in the area of the Langlands program, then at least somewhere in that area."
] | 4 |
Science
| 0 |
111
|
mathoverflow
|
$\infty$-topos and localic $\infty$-groupoids?
|
It's known that every classical (Grothendieck) topos is equivalent to the topos of sheaves on a localic groupoid (a groupoid in the category of locales).
For the record, this is proved by, starting form a topos $T$, constructing a locale $L$ and a surjection $L \rightarrow T$ 'nice enough' (like a proper surjection, or an open surjection depending on the proof). Then $(L, L \times_T L, L \times_T L \times_T L)$ is a truncated simplicial locale, which can be seen as a localic groupoid. There is a canonical geometric morphism from the topos of sheaves on this groupoid to $T$, and if the surjection $L \rightarrow T$ was nice enough it's an isomorphism.
My question is : Can we hope for a similar result for $\infty$-toposes ? for example by replacing localic groupoids by a localic $\infty$-groupoids, defined as a simplicial locale satisfying some form of Kan complex condition and defining the associated $\infty$-topos as the colimits of the corresponding simplicial diagram of localic $\infty$-toposes.
|
https://mathoverflow.net/questions/93517/infty-topos-and-localic-infty-groupoids
|
[
"ct.category-theory",
"topos-theory",
"infinity-topos-theory",
"locales",
"infinity-categories"
] | 25 | 2012-04-08T13:23:10 |
[
"@SimonHenry In classical topos theory, the Diaconescu localic cover theorem gives a connected locally connected geometric morphism. Often people only stress on the fact that it is open.",
"@IvanDiLiberti the stability under pullback of localic morphisms might be useful at some point yes. But the crucial point would be some form of Diaconescu Localic open cover theorem for $\\infty$-toposes (though \"open\" should probably be replaced by locally $\\infty$-connected or something like that...).",
"@SimonHenry I think that would be a very good starting point to prove the result you want.",
"@IvanDiLiberti Probably - but as far as I remember it hasn't been written (from what I remember Lurie in HTT talks about n-localic toposes but does not study the relative version)",
"Do we have a hyperconnected-localic fact system?",
"So, you're asking if every classical $\\infty$-topos is equivalent to the category of sheaves on a localic $\\infty$-topos, correct?",
"If I found any things about this, I will. But for now I have a lot of thing to learn before...",
"This is something I have thought about in the back of my head for several years. If you come up with something, or would just like to brainstorm, let me know."
] | 8 |
Science
| 0 |
112
|
mathoverflow
|
Status of the Euler characteristic in characteristic p
|
In the introduction to the Asterisque 82-83 volume on `Caractérisque d'Euler-Poincaré, Verdier writes:
> Enfin signalons que la situation en caractéristique positive est loin d'être aussi satisfaisante que dans le cas transcendant. On aimerait pouvoir disposer d'un groupe de Grothendieck `sauvage' des faisceaux constructibles tels que deux faisceaux ayant même rang et même comportement sauvage en chaque point aient même classe dans ce groupe de Grothendieck et tel que ce groupe soit foncteur covariant. Des resultats récents de Laumon donnent des indications dans ce sens. On aimerait aussi disposer d'une théorie des classes d'homologie de Chern qui sereait une transformation naturelle de ce groupe de Grothendieck sauvage dans les groupes d'homologie étale.
My question is:
> > Has the situation improved in the intervening 32 years?
I recall in detail below some background and ask some more precise questions.
For a map of complex algebraic varieties $f: X \to Y$, one can define the pushforward of a constructible function $\xi$:
$$(f_* \xi)(p) = \sum_n n \chi(\xi^{-1}(n) \cap f^{-1}(p) )$$
This is a relative notion of Euler characteristic: $f_* (1_X)$ is the function on $Y$ whose value at $y$ is the Euler characteristic $\chi(X_y)$.
This notion is functorial: $(gf)_* = g_* f_*$.
The (naively) analogous notion in characteristic $p$ is not functorial. The standard counterexample is the Artin-Schreier map
$g:\mathbb{A}^1 \to \mathbb{A}^1$
$z \mapsto z^p - z$
Note $g_* 1 = p$. On the other hand, taking $f: \mathbb{A}^1 \to \mathrm{Spec}\,k$ the structure map, one has $f_* (g_* 1)) = p \ne 1 = (fg)_* 1$. The problem seems to be that the map is wildly ramified at infinity. The beginning of the above paragraph of Verdier is a request that this situation be remedied by introducing some intermediary between constructible sheaves and their stalkwise Euler characteristics which would record the possibility of wild ramification and thus be functorial under pushforward.
> Is there now some such object?
The second part of Verdier's paragraph presumes that the answer to the question above is yes, and requests an analogue of Macpherson's Chern class transformation. Over $\mathbb{C}$, this is a natural transformation $c_{SM}: Con(\cdot) \to H_*(\cdot)$ between the functors (covariant with respect to proper pushforward) which respectively assign to a variety its constructible functions and homology; it has the property that if $X$ is proper and smooth then $c_{SM}(X) = c(TX) \cap [X]$.
> Is there an analogue of the Macpherson Chern class transformation in characteristic $p$?
The Macpherson Chern class is defined in terms of something called the 'local Euler obstruction'. This is a constructible function $Eu_V$ attached to a variety $V$ which, roughly speaking, at a point $v\in V$ records the virtual number of zeroes of any extension of the radial vector field from the boundary of ball around $v$ to its interior. (One passes to the Nash blowup in order to make sense of this vector field.)
The Euler obstruction enjoys the following hyperplane formula: locally embedding $(V, v)$ in a smooth variety $Y$, taking some sufficiently small $\epsilon$ ball $B_\epsilon(v)$, and a generic projection $l: B_\epsilon(v) \to \mathbb{C}$, one finds that $l_* Eu_V$ is locally constant near $l(v)$. In other words, the Euler obstruction is annihilated by taking vanishing cycles with respect to a general linear function.
> Is there an analogue of the local Euler obstruction in characteristic $p$, and is it annihilated by taking vanishing cycles with respect to a general linear function?
|
https://mathoverflow.net/questions/135983/status-of-the-euler-characteristic-in-characteristic-p
|
[
"ag.algebraic-geometry",
"characteristic-p",
"characteristic-classes",
"euler-characteristics"
] | 25 | 2013-07-07T00:32:03 |
[
"You've probably seen this already, but I'm adding it here anyway for others: arxiv.org/abs/1510.03018 by Takeshi Saito is relevant."
] | 1 |
Science
| 0 |
113
|
mathoverflow
|
0's in 815915283247897734345611269596115894272000000000
|
> Is 40 the largest number for which all the 0 digits in the decimal form of $n!$ come at the end?
Motivation: My son considered learning all digits of 40! for my birthday. I told him that the best way to remember them would be to come up with a mnemonic. He asked me what word he should come up with if the digit is 0. The rest is history.
|
https://mathoverflow.net/questions/393993/0s-in-815915283247897734345611269596115894272000000000
|
[
"nt.number-theory",
"open-problems",
"digits"
] | 25 | 2021-05-28T09:10:15 |
[
"Maybe it will easy to solve in binary first. Where satisfied for 0!, 1!, 2!, 3! and 4!.",
"The point is that the question is not about the zeros in this number (which is a trivial question) but about the zeros in infinitely many other numbers.",
"I prefer to keep an air of mystery about 815915283247897734345611269596115894272000000000.",
"I changed the title to \"$0$'s in 40!=...\", that is, added \"40!\". This makes the title less cryptic within the list of questions. I regret you reverted it.",
"The title would be improved (and less puzzling) if it were replaced by the sentence that is the actual question.",
"@Malkoun I did a similar check before seeing your post, and ran it up to 40,000. OEIS A182049 mentioned by Jeppe Stig Nielsen implies it's been checked up to 100,000. I guess I could check somewhat further by letting a computer run through the weekend, but I doubt there's much point.",
"$41!$ seems to be the last factorial number with no digit 9 in it. It looks like for $n\\ge 42$, the decimal representation of $n!$ will contain all ten digits, even after trailing zeros have been removed. Edit: Found OEIS entry.",
"Ok, it seems you're Hungarian. Well, finding such words should not be too much of an issue :-)",
"As for the mnemonic, he can try to use 10 letters words, if they're not too uncommon in your native language.",
"Interesting... I could not find another example for $41 \\leq n \\leq 16000$. I wrote a small Python program for that purpose, which checked the $n$s in the previous range one by one.",
"Then I suppose this answers my question as 'known open problem'.",
"One could call it the Zumkeller conjecture.",
"@domotorp: oh, that's what I feared... :D",
"I would learn a cyclic number instead, that he can use for magic tricks en.wikipedia.org/wiki/Cyclic_number",
"@Loïc Thanks, though a bit early, and it is not the 40th, but the 815915283247897734345611269596115894272000000000th ;)",
"Oh, I understood it the wrong way! My bad.",
"@JukkaKohonen: I mean that 40 probably is the largest number $n$ for which all the 0's in $n!$ appear at the end.",
"@Will, did your back-of-envelope calculation provide any estimate on when it would happen?",
"Using the heuristic that the digits of $n!$ are just random, followed by some zeroes at the end, you can calculate without too much trouble that the answer to your question is: probably yes. My (very crude) estimate puts it at a better than 99% chance. I don't know whether this heuristic is any good, and I doubt this way of thinking will lead to a real answer . . . but there you have it.",
"What follows doesn't answer your question, and probably won't contribute to an answer either, but a result by John Edward Maxfield -- A note on $N!$, Mathematics Magazine 43 #2 (March 1970), pp. 64-67 -- might be of interest: Given any positive integer $n,$ there exists a positive integer $N$ such that the decimal digits of $N!$ begin with all the decimal digits of $n,$ in their correct order. For more about this, see my answer to Short papers for undergraduate course on reading scholarly math.",
"And by the way, happy 40th birthday ;)",
"Is there an integer n which does not divide 10 such that for any k integer with 10 does not divide a = n * k and a has a zero in its decimal places ?",
"Interesting question, but very difficult. I am not sure we even know if there are an infinite number of 0s in the decimal expansion of pi. (But it's believed this is the case)"
] | 23 |
Science
| 0 |
114
|
mathoverflow
|
What's the smallest $\lambda$-calculus term not known to have a normal form?
|
For Turing Machines, the question of halting behavior of small TMs has been well studied in the context of the Busy Beaver function, which maps n to the longest output or running time of any halting n state TM. The smallest TM with unknown halting behavior has 5 states. It takes $5\cdot2\cdot\log_2(5\cdot4 + 4)$ or roughly 46 bits to describe, 50 bits if we assume a straightforward encoding.
In comparison, $\lambda$-calculus terms have a [simple binary encoding](http://tromp.github.io/cl/Binary_lambda_calculus.html): $00$ for lambda, $01$ for application, and $1^n0$ for variable with de Bruijn index $n$. It's natural to define a $\lambda$-calculus analog of the busy beaver function as the maximum normal form size of any size $n$ closed lambda term.
As the smallest closed lambda term is $\lambda\,1$, with encoding $0010$, we determine $$ BB_{\lambda}(4) = 4 $$ The next smallest ones, $\lambda\,\lambda\,1$ and $\lambda\,\lambda\,2$ are similarly already in normal form, and give $$ BB_{\lambda}(6) = 6,\qquad BB_{\lambda}(7) = 7 $$ $BB_{\lambda}(n)$ will have to remain undefined for $n < 4$ or $n = 5$.
The first enlarged normal form shows up at $\lambda\,(\lambda\,1\,1)\,(1\,(\lambda\,2))$ which gives $$ BB_{\lambda}(21) = 22 $$ $BB_{\lambda}$ starts to grow rapidly at $n \geq 30$, since tripling Church numeral two, $(\lambda\,1\,1\,1)\,(\lambda\,\lambda\,2\,(2\,1))$ with normal form Church numeral $2^{2^2}= 16$, gives $$ BB_{\lambda}(30) \geq 5 \cdot 16 + 6 = 86 $$ and quadrupling/quintupling give $$ BB_{\lambda}(34) \geq 5 \cdot 2^{16} + 6 $$ $$ BB_{\lambda}(38) \geq 5 \cdot 2^{2^{16}} + 6 $$ which exceed the TM Busy Beavers for 4 and 5 states.
An Ackermann-like function takes a mere [29 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/ackermann.lam). A twisted application to Church numeral $2$ yields a $BB_{\lambda}(51)$ exceeding 2↑↑↑↑5.
Graham's number is exceeded in at most [49 bits](https://github.com/tromp/AIT/blob/master/fast_growing_and_conjectures/melo.lam), giving $$ BB_{\lambda}(49) \geq 5 \cdot G + 6 $$ (compared with a 16 state TM that needs over 192 bits to describe).
What's the smallest n for which $BB_{\lambda}(n)$ is unknown in ZFC?
One upper bound is [213 bits](https://codegolf.stackexchange.com/questions/79620/laver-table-computations-and-an-algorithm-that-is-not-known-to-terminate-in-zfc).
Let's try to narrow it down some more.
Function $BB_{\lambda}$ has been added to the [Online Encyclopedia for Integer Sequences](https://oeis.org/A333479).
|
https://mathoverflow.net/questions/353514/whats-the-smallest-lambda-calculus-term-not-known-to-have-a-normal-form
|
[
"lo.logic",
"lambda-calculus"
] | 24 | 2020-02-25T03:18:56 |
[
"Yes, I meant unknown in ZFC. I'll make that explicit.",
"Does \"unknown\" mean \"unknown in ZFC\"? I infer that this is the case from the upper bound being a reference to a program which is known to halt under the assumption of a rank-into-rank cardinal.",
"I mean \"unknown halting behaviour\". We know that (\\ 1 1)(\\ 1 1) is the smallest nonhalting term, of size 18.",
"By “not known to halt”, do you mean “unknown halting behavior” or “known to not halt or unknown halting behavior”?",
"I also have an upper bound of 247 bits for finding an odd perfect number at github.com/tromp/AIT/blob/master/oddperfect.lam",
"Yes; Goldbach is 267 bits, as can be seen at github.com/tromp/AIT/blob/master/goldbach.gif which is compiled from goldbach.lam; larger than the 213 bit Laver statement above.",
"For an upper bound, do you know what a lambda calculus formulation of twin primes / Goldbach / similar number-theoretic statements might look like?",
"Curiously, triple Church_2 is not the best 30 bit beaver. A very minor tweak gives a 160 bit normal form!",
"oops; good catch! will fix right away.",
"It seems that application is also the other way round... quoting yourself :-)",
"When i changed from 0-based to 1-based variables, I forgot to update the encoding. It's fixed now.",
"Shouldn't $\\lambda 1$ have encoding 00110 (of length five), or am I missing something in the notation here?",
"Yes, they were. I replaced them by actual lambdas now.",
"Are all those slashes supposed to represent lambda?"
] | 14 |
Science
| 0 |
115
|
mathoverflow
|
conjectures regarding a new Renyi information quantity
|
In a recent paper <http://arxiv.org/abs/1403.6102>, we defined a quantity that we called the "Renyi conditional mutual information" and investigated several of its properties. We have some open conjectures, and I would like to pose them as open questions to the mathematics community on this forum. Perhaps there is a simple resolution, but we have considered known approaches and it appears that new tools are necessary for solving these problems (or maybe known tools could work?).
## Background
Let $\mathcal{H}$ denote a Hilbert space, and let $\mathcal{D}(\mathcal{H})$ denote the set of [density operators](https://en.wikipedia.org/wiki/Density_matrix#Formulation) acting on this Hilbert space (positive semi-definite operators with trace one). (It suffices for our purposes to consider finite-dimensional Hilbert spaces, but of course the infinite-dimensional case is interesting as well.) We are interested in "three-party" density operators $\rho_{ABC}$ acting on the tensor-product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$. Let $\rho_A$ denote the "marginal" density operator of $\rho_{ABC}$, obtained by taking a partial trace over the spaces $\mathcal{H}_B \otimes \mathcal{H}_C$: $$ \rho_A = \operatorname{Tr}_{BC} \\{ \rho_{ABC}\\} $$ In a similar way, we can define $\rho_{B}$, $\rho_{C}$, $\rho_{AB}$, $\rho_{BC}$, and $\rho_{AC}$.
We define the "Renyi conditional mutual information" of order $\alpha \geq 0 $ as follows: $$ I_{\alpha}(A;B|C)_{\rho} \equiv \frac{1}{\alpha - 1} \log \operatorname{Tr} \\{ \rho_{ABC}^{\alpha} \rho_{AC}^{(1-\alpha) / 2} \rho_C^{(\alpha-1)/2} \rho_{BC}^{1-\alpha} \rho_C^{(\alpha-1)/2} \rho_{AC}^{(1-\alpha) / 2} \\} $$ In the above, identity operators are implicit, so that, e.g., $\rho_C^{(\alpha-1)/2} = I_{AB} \otimes \rho_C^{(\alpha-1)/2}$, where $I_{AB}$ is the identity operator acting on $\mathcal{H}_A \otimes \mathcal{H}_B$. For more motivation for the above quantity, please consult our paper.
We can easily prove that the above quantity is "monotone under a local quantum operation acting on system $B$" for all $\alpha \in [0,2]$, i.e., that the following inequality holds $$ I_{\alpha}(A;B|C)_{\rho} \geq I_{\alpha}(A;B|C)_{\omega}, $$ where $\omega_{ABC} \equiv (\operatorname{id}_A \otimes \mathcal{M}_B \otimes \operatorname{id}_C)(\rho_{ABC})$, $\operatorname{id}$ denotes the identity map, and $\mathcal{M}_B$ is a completely positive trace preserving linear map acting on system $B$. A proof follows by applying a standard approach with the [Lieb concavity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Lieb.27s_concavity_theorem) and the [Ando convexity theorem](https://en.wikipedia.org/wiki/Trace_inequalities#Ando.27s_convexity_theorem) (proof is detailed in the paper).
## Conjectures/Questions
We have several conjectures which (to fit the MathOverflow format) could be converted into questions in obvious ways (e.g., are the statements true? please provide an argument or counterexample, etc.).
Conjecture 1:
> The quantity $I_\alpha(A;B|C)$ is "monotone under a local quantum operation acting on system $A$" for all $\alpha \in [0,2]$, i.e., the following inequality holds $$ I_{\alpha}(A;B|C)_{\rho} \geq I_{\alpha}(A;B|C)_{\tau}, $$ where $\tau_{ABC} \equiv (\mathcal{N}_A \otimes \operatorname{id}_B \otimes \operatorname{id}_C)(\rho_{ABC})$ and $\mathcal{N}_A$ is a completely positive trace preserving linear map acting on system $A$. (Numerical evidence indicates that this conjecture should be true.)
Conjecture 2:
> The Renyi conditional mutual information of order $\alpha$ should be monotone increasing in the Renyi parameter $\alpha$, i.e., if $0 \leq \alpha \leq \beta$, then $$ I_{\alpha}(A;B|C)_{\rho} \leq I_{\beta}(A;B|C)_{\rho}. $$ (Numerical evidence indicates that this conjecture should be true as well, and we furthermore have proofs that it is true in some special cases (for $\alpha$ in a neighborhood of one and when $\alpha + \beta = 2$).)
These are the basic conjectures, but please consult our paper for more general forms of them. Solving the first conjecture would be very interesting for us, as it would establish the Renyi conditional mutual as a true "Renyi generalization" of the "von Neumann style" quantum conditional mutual information. Solving the second conjecture would have widespread implications throughout quantum information theory and even for condensed matter physics ([topological order](http://dao.mit.edu/~wen/topartS3.pdf) \- pdf).
|
https://mathoverflow.net/questions/172684/conjectures-regarding-a-new-renyi-information-quantity
|
[
"pr.probability",
"linear-algebra",
"it.information-theory",
"matrix-analysis",
"quantum-mechanics"
] | 24 | 2014-06-25T20:21:44 |
[
"There has now been even more progress on Conjecture 2 that is useful enough for applications in quantum information theory. The idea was to make use of the Hadamard three-line theorem (in particular, Riesz-Thorin interpolation). This is detailed in the following paper: arxiv.org/abs/1505.04661 . See also arxiv.org/abs/1506.00981 for follow-up work.",
"This is just to say that there has been what I would consider siginificant progress on Conjecture 2 since it was posted. Fawzi and Renner have proved a variation of the case when $\\beta = 1$ and $\\alpha = 1/2$, in the paper arxiv.org/abs/1410.0664 . My coauthors and I have also generalized the notion of Renyi conditional mutual information Renyi relative entropy differences in the paper arxiv.org/abs/1410.1443 and we generalized the methods of Fawzi and Renner in the paper arxiv.org/abs/1412.4067 . Of course, the conjectures stated here still remain open.",
"In response to earlier comments (now deleted), I have reformatted the post in order to better highlight the apparent questions. With that, let's please focus on the mathematics.",
"I really like this question --- I hope I manage to get some time to think about it!"
] | 4 |
Science
| 0 |
116
|
mathoverflow
|
Is A276175 integer-only?
|
The terms of the sequence [A276123](https://oeis.org/A276123), defined by $a_0=a_1=a_2=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-3}}\;,$$ are all integers (it's easy to prove that for all $n\geq2$, $a_n=\frac{9-3(-1)^n}{2}a_{n-1}-a_{n-2}-1$).
But is it also true for the sequence [A276175](https://oeis.org/A276175) defined by $a_0=a_1=a_2=a_3=1$ and $$a_n=\dfrac{(a_{n-1}+1)(a_{n-2}+1)(a_{n-3}+1)}{a_{n-4}} \;\;?$$
**Remark** : This question has been [asked previously on math.SE](https://math.stackexchange.com/questions/1905063/is-a276175-integer-only) ; one participant gave an interesting answer, but partial.
|
https://mathoverflow.net/questions/248604/is-a276175-integer-only
|
[
"nt.number-theory",
"sequences-and-series",
"integer-sequences",
"cluster-algebras"
] | 24 | 2016-08-30T02:54:11 |
[
"@YCor the sequences are examples of Y-frieze patterns, which are naturally associated to Fock and Goncharov's $\\mathcal{X}$-cluster varieties.",
"More precisely, it seems that $a^{(5)}$ is still integral (checked for $n<35$), and $a^{(k)}$ is non-integral for $k>5$.",
"Initially I thought all of the recurrences $a^{(k)}_n = \\prod_{i=1}^{k-1} (a^{(k)}_{n-i}+1)/a^{(k)}_{n-k}$ (with initial values all $1$) might have all terms integral - where $a^{(3)}$ is oeis.org/A276123 and $a^{(4)}$ is oeis.org/A276175. But it seems that $a^{(6)}_{12}$ is non-integral.",
"@YCor The first sequence (the one with the explicit quasi-linear recurrence) is one of the period-1 Laurent phenomenon sequences from Alman, Cuenca and Huang's paper, thus presumably related to cluster algebras (not sure about this).",
"@CarloBeenakker The proof relies on a computer calculation that no one but mercio seems to have seen let alone checked. And from the sound of his comments, mercio doesn't seem 100% sure of it either.",
"@FedorPetrov what's the connection with cluster algebras?",
"@darij grinberg : perhaps in view of today's edit it would be helpful to indicate why the MSE proof is not complete (I note that also OEIS lists this as a proven statement).",
"@FedorPetrov they are actual polynomials in the first eight $a_{n-1}a_{n+1}/(a_n(a_n+1))$ but unfortunately for this particular sequence, those quantities are not all integers.",
"At least a priori it could be Laurent polynomial not literally in $a_0,\\dots,a_3$, but in something different, like, say, also $(a_1a_2+1)/(a_2+a_3)$ or whatever.",
"So it is unlikely 'cluster algebras'.",
"When using generic $a_0,a_1,a_2,a_3$, one does not get Laurent polynomials (starting with $a_8$).",
"I feel that 'cluster algebras' tag may be appropriate."
] | 12 |
Science
| 0 |
117
|
mathoverflow
|
Subfields of $\mathbb{C}$ isomorphic to $\mathbb{R}$ that have Baire property, without Choice
|
While sitting through my complex analysis class, beginning with a very low level introduction, the teacher mentioned the obvious subfield of $\mathbb{C}$ isomorphic to $\mathbb{R}$, and I then wondered whether this was unique. When I got back to my computer, I looked on google and came up with [this](http://camoo.freeshell.org/cfield.isor.html), which indicated that is not. However, the proof given there makes heavy use of Choice, and based on similar things, I am guessing that this cannot be avoided.
Define DC($\omega_1$) as:
For all [trees](http://en.wikipedia.org/wiki/Tree_%28set_theory%29) $T$, $\quad$ $T\:$ has a branch $\ $ or $\ $ $T\:$ has a chain of length $\omega_1$ $\quad$.
which, if I haven't messed up the simplification, ZF proves is equivalent to the definition given at shelah.logic.at/files/446.ps .
Does $\ ZF+DC(\omega_1)\ $ prove that there is a unique subfield of $\mathbb{C}$ which is both isomorphic to $\mathbb{R}$ and has Baire property?
|
https://mathoverflow.net/questions/51085/subfields-of-mathbbc-isomorphic-to-mathbbr-that-have-baire-property-w
|
[
"set-theory",
"gn.general-topology",
"axiom-of-choice",
"fields"
] | 24 | 2011-01-03T19:36:46 |
[
"Just a comment - The first link should be camoo.freeshell.org/cfield.isor.html instead of camoo.freeshell.org/cfield.isor to make it more readable.",
"@Ricky: Your translation of DC_omega1 to trees is not correct. The above statement is obviously false, for instance since clearly the one element tree has no chain of length omega1.",
"@Harry, What does there seem to be wrong with the formatting?",
"Ah, thanks for the explanation Tobias. I found the result very counterintuitive until you pointed out that there was no reason to have that intuition to begin with :)",
"There seems to be something wrong with the formatting?",
"@Zev: The fact that $\\mathbb C$ has uncountably many field automorphisms does not seem so surprising to me. After all the only obstruction that keeps $\\mathbb R$ from having many automorphisms is the fact that its order can be described algebraically, hence is preserved. Since there is no such obstruction on $\\mathbb C$, there should be plenty of automorphisms, and in fact there are.",
"+1 just for blowing my mind with the fact that there are subfields of $\\mathbb{C}$ isomorphic to $\\mathbb{R}$ other than $\\mathbb{R}$!"
] | 7 |
Science
| 0 |
118
|
mathoverflow
|
Example of a quasi-Bernoulli measure which is not Gibbs?
|
Let $X=\\{0,1\\}^{\mathbb{N}}$. For simplicity I consider measures on $X$ only.
A measure $\mu$ is **quasi-Bernoulli** if there is a constant $C\ge 1$ such that for any finite sequences $i,j$, $$ C^{-1} \mu[ij] \le \mu[i]\mu[j] \le C\mu[ij]. $$
(Here as usual $ij$ is the juxtaposition of $i$ and $j$ and $[k]$ is the cylinder of all infinite sequences starting with $k$.)
Let $f:X\to \mathbb{R}$ be continuous. The measure $\mu$ is a **Gibbs measure** with potential $f$ if there are $C>0$ and $P\in\mathbb{R}$, such that for every infinite sequence $i_1 i_2\ldots$ and all natural $n$, $$ C^{-1} \le \frac{\mu[i_1\ldots i_n]}{\exp(-nP+f(i)+f(\sigma i)+\cdots+f(\sigma^{n-1}i))} \le C, $$ where $\sigma$ is the left shift.
Of course, there are other definitions of Gibbs measure, but they all agree if the potential $f$ is Hölder. In this case, it follows readily that a Gibbs measure is quasi-Bernoulli.
If a measure is quasi-Bernoulli, there is an equivalent measure (mutually absolutely continuous with bounded densities) which is invariant and ergodic under the shift.
* * *
**Question** : Are all quasi-Bernoulli measures Gibbs? (for some continuous potential, not necessarily Hölder). If not, what is a counterexample?
**Motivation** : Gibbs measures (with Hölder potentials) enjoy many nice statistical properties. Sometimes I have a measure that is quasi-Bernoulli or satisfies some similar but weaker property. I would like to understand if and to what extent good statistical properties continue to hold in that setting.
|
https://mathoverflow.net/questions/53406/example-of-a-quasi-bernoulli-measure-which-is-not-gibbs
|
[
"ds.dynamical-systems",
"measure-theory"
] | 24 | 2011-01-26T13:19:29 |
[
"Dear Vaughn, dear John, could you give us references about both of your arguments? That would be very useful!",
"Let me point out that there is a standard procedure associating to each almost additive sequence of functions a measure on the algebra up to each level (say $n$ if you consider the element $\\varphi_n$ of the sequence of functions). Moreover, any weak sublimit of an almost additive sequence with bounded variation is a Gibbs measure. This implies that your question has a positive answer at each level $n$, that is, indeed any quasi-Bernoulli measure (in your sense) is Gibbs at the algebra of level $n$. Moreover, any of its weak sublimits is also Gibbs, now on the whole $\\sigma$-algebra.",
"Just a thought that doesn't necessarily go anywhere. If $\\mu$ is a Gibbs measure, then you can recover the corresponding potential $f$ (up to a constant) by $f = \\log(\\frac{d\\mu}{d(\\mu\\circ\\sigma)})$. So if we take an arbitrary quasi-Bernoulli measure and take $f$ to be its log-Jacobian, can we deduce anything about the regularity of $f$ and/or a Gibbs relationship based on the quasi-Bernoulli property?"
] | 3 |
Science
| 0 |
119
|
mathoverflow
|
Is the Poset of Graphs Automorphism-free?
|
For $n\geq 5$, let $\mathcal {P}_n$ be the set of all isomorphism classes of graphs with n vertices. Give this set the poset structure given by $G \le H$ if and only if $G$ is a subgraph of $H$.
> Is it true that $\mathcal {P}_n$ has no nontrivial automorphisms?
**Remarks:**
This follows if one can recognize a graph from the set of isomorphism classes of its edge-deleted subgraphs. However, since recognizing a graph from the set of isomorphism classes of its edge-deleted subgraphs is stronger than edge reconstruction, I'm wondering if there is an alternative way of proving this.
|
https://mathoverflow.net/questions/153208/is-the-poset-of-graphs-automorphism-free
|
[
"co.combinatorics",
"graph-theory",
"posets",
"graph-reconstruction"
] | 24 | 2013-12-30T20:23:03 |
[
"Probably has something to do with: math.stackexchange.com/questions/3052384/…",
"Of course there is in fact no need to check, as it anyhow follows from the set edge reconstruction conjecture but I could not find anything about for what values that has been checked.",
"$\\mathcal P_5$ can pretty easily be checked by hand. As for anything larger, I'm not sure.",
"Maybe it is useful to add that $\\cal P_4$ has a nontrivial automorphism. Did you ckeck for higher values, e.g. using the tables from maths.uq.edu.au/~pa/research/posets4to8.html?"
] | 4 |
Science
| 0 |
120
|
mathoverflow
|
An $n \times n$ matrix $A$ is similar to its transpose $A^{\top}$: elementary proof?
|
A famous result in linear algebra is the following.
> An $n \times n$ matrix $A$ over a field $\mathbb{F}$ is similar to its transpose $A^T$.
I know one proof using the Smith Normal Form (SNF). However, I want to find an elementary proof avoiding any concepts related to the SNF. My question is: is there an elementary way to prove this?
The requirements are:
* Do NOT use the structure theorem over PID.
* Do NOT use the Smith Normal form (nor Jordan canonical form).
* Do NOT use the concept of invariant factors.
* Provide an explicit invertible matrix $P$ such that $A=PA^T P^{-1}$.
|
https://mathoverflow.net/questions/122345/an-n-times-n-matrix-a-is-similar-to-its-transpose-a-top-elementary-pr
|
[
"linear-algebra",
"matrices",
"matrix-theory",
"similarity"
] | 24 | 2013-02-19T11:40:49 |
[
"After many years this question was first asked, I cannot find anything meeting all of my requirement. Actually, finding Smith Normal Form can be done through elementary row/column operations. So, it is already elementary enough, and I do not have any reason left for avoiding such a simple and powerful tool. Also, SNF provides not only the existence of $P$, but also an explicit way to write down a such $P$.",
"@user46896 The method is interesting, but not very far from Jordan canonical form (see questions V.B.1. and V.B.2.). One dreams about an algorithm providing $P$ from scratch ...",
"@i707107 So, I hope you can work with this : sujets-de-concours.net/sujets/centrale/2003/tsi/math2.pdf , look at Partie V, it's an elementary approach. The solution here : pomux.free.fr/corriges-2003/index.html",
"@Julien I know very little french, but there is google translate.",
"@i707107 Can you read some french papers ?",
"See projecteuclid.org/euclid.pjm/1103039127. Maybe math.stackexchange.com/questions/62497/… is related.",
"@BenjaminDickman The first link shows that the solution space of $XA=A^T X$, $X=X^T$ has at least dimension $n$. However, the paper does not discuss how to obtain a \"nonsingular\" matrix in the solution space. That was actually a motivation of my question.",
"Is the proof given here msp.org/pjm/1959/9-3/pjm-v9-n3-p25-p.pdf sufficiently \"elementary\"? Incidentally, note the remark on p.895 (pdf 4/7) comes from M. Newman; there was an earlier MO question answered using Smith's work (which you wish to avoid here...) that turned out to have a more elementary solution from M. Newman (mathoverflow.net/questions/151166/…). If the linked proof here does not suffice, perhaps Newman is the fellow to look to...",
"@MarkSapir : Yes, of course the fact that $A$ and $A^{T}$ have the same minimum polynomial is obvious: it was the use of rational canonical form ( which is really more general that Jordan Normal Form) that seemed to be outside the spirit of the question.",
"This is slightly related to my question <a href=\"http:mathoverflow.net/questions/42072</a>",
"@Tom: This link projecteuclid.org/…",
"@Geoff: the fact that $A$ and $A^T$ have the same minimal polynomial is obvious: if $p(A)=0$, then $p(A^T)=p(A)^T=0$. ",
"In the two by two case $P$ can always be chosen to be symmetric. Is this true in general? ",
"It's easy to prove using normal forms, but you can reframe it as the statement that if $A:V\\to V$ is linear then there is a nondegenerate bilinear form on $V$ such that $\\langle Av,w\\rangle=\\langle v, Aw\\rangle$ identically. Is there any chance of a proof of this that is quite different from the normal-form proofs? ",
"There is a proof using rational canonical form, which basically just needs the fact that $A$ and $A^{t}$ have the same minimum polynomial, but I don't think that will be considered explicit .enough",
"@Mariano: Yeah, it might not be possible to avoid those to prove it. The matrix P coming from Jordan canonical form proof is already complicated enough. Maybe requiring to provide explicit P is not possible without those. The generalized eigenvectors are involved and an inversion that transforms Jordan block to its transpose. ",
"This is a lot like looking for a proof which does not need the letter a in order to be written down :-) What exactly do you mean by elementary? The Jordan canonical form is basic linear algebra, really."
] | 17 |
Science
| 0 |
121
|
mathoverflow
|
What is the symmetric monoidal functor from Clifford algebras to invertible K-module spectra?
|
There ought to be a symmetric monoidal functor from the symmetric monoidal $2$-groupoid whose objects are Morita-invertible real superalgebras (precisely the Clifford algebras), morphisms are invertible superbimodules, and $2$-morphisms are invertible superbimodule homomorphisms to the symmetric monoidal ($\infty$-)groupoid whose objects are invertible $KO$-module spectra, morphisms are invertible morphisms of $KO$-module spectra, etc. Everything I'm about to say should generalize to the complex case and $KU$ but let me only address the real case.
On $\pi_0$ this functor ought to induce a map from the graded Brauer group of $\mathbb{R}$ to the Picard group of $KO$, both of which turn out to be $\mathbb{Z}_8$; this directly relates Bott periodicity for Clifford algebras and for K-theory.
Recently I tried to write down this functor "explicitly" (to prepare for a talk and to confirm that I wasn't lying when I wrote [this](https://mathoverflow.net/questions/87345/brauer-groups-and-k-theory/177321#177321)) and found that I couldn't quite get some details to work out the way I wanted them to. The idea should be, starting with a real (resp. complex) superalgebra $A$, to
1. Construct the topological groupoid of finitely generated projective $A$-supermodules (automorphism groups have their usual topologies as Lie groups),
2. Pass to the algebraic K-theory spectrum of this groupoid, regarded as a symmetric monoidal groupoid under direct sum,
3. At some point during this process, kill off "trivial objects."
The correct version of this process should, when given $A = \mathbb{R}$ itself, reproduce $KO$, and more generally, when given the Clifford algebra $\text{Cliff}(p, q)$, reproduce a shift of $KO$ (as a $KO$-module spectrum) by $p - q$ either up or down depending on some [sign conventions](https://mathoverflow.net/questions/185645/what-are-the-correct-conventions-for-defining-clifford-algebras).
I ran into two problems getting this picture to work out:
1. The construction I know of the algebraic K-theory spectrum of a symmetric monoidal groupoid produces a connective spectrum. For example, applied to $\text{Vect}_{\mathbb{R}}$ it reproduces $ko$ rather than $KO$.
2. It seems like there are two inequivalent ways to kill off trivial objects, and they don't produce the same answer.
In detail, to fix the sign conventions, let $\text{Cliff}(p, q)$ be the Clifford algebra generated by $p$ anticommuting square roots of $1$ and $q$ anticommuting square roots of $-1$, so that $\text{Cliff}(1, 0) \cong \mathbb{R} \times \mathbb{R}$ and $\text{Cliff}(0, 1) \cong \mathbb{C}$ (as ungraded algebras). Let $K(\text{Cliff}(p, q))$ be the algebraic K-theory of the category of finite-dimensional $\text{Cliff}(p, q)$-supermodules (so some connective spectrum). There are two natural forgetful functors inducing maps
$$K(\text{Cliff}(p + 1, q)) \to K(\text{Cliff}(p, q)), K(\text{Cliff}(p, q + 1)) \to K(\text{Cliff}(p, q))$$
of connective spectra; as maps on categories, they pick out the supermodules admitting an additional odd automorphism squaring to $1$ or an odd automorphism squaring to $-1$ respectively. For example, when $p = q = 0$, these are two ways of identifying the supervector spaces you want to regard as trivial for the purposes of defining $KO$ in terms of supervector spaces.
I computed that the homotopy cofibers of these maps (in connective spectra) are $\Omega^{p-q} ko$ and $B^{p-q} ko$ respectively; that is, we get the infinite loop spaces making up $ko$, but in two different orders.
> How can this construction be fixed so that it outputs $KO$-module spectra rather than $ko$-module spectra, and what's the "correct" way to kill trivial objects?
One desideratum for "correct" is that it should generalize cleanly to the case where we replace $\mathbb{R}$ with a discrete field $k$ and replace $KO$ with the algebraic K-theory spectrum $K(k)$ of $k$. The end result of the "correct" process is that it should naturally give a symmetric monoidal functor inducing a map from the graded Brauer group of $k$ to the Picard group of $K(k)$. But here there are not only two choices for what an odd automorphism could square to: instead the set of choices is parameterized by
$$k^{\times}/(k^{\times})^2 \cong H^1(\text{Spec } k, \mu_2).$$
So I don't know what to do. Maybe always pick squaring to $1$?
One thing that made me uncomfortable about the above is that it's conceptually important that the supermodule categories should be regarded as enriched and tensored over supervector spaces, but I ignored their enrichment over supervector spaces when I passed to algebraic K-theory spectra. The enrichment seems like it ought to be important for defining the "correct" way to kill of trivial objects.
|
https://mathoverflow.net/questions/186148/what-is-the-symmetric-monoidal-functor-from-clifford-algebras-to-invertible-k-mo
|
[
"at.algebraic-topology",
"kt.k-theory-and-homology",
"clifford-algebras"
] | 23 | 2014-11-03T18:03:26 |
[
"Update: awhile back I asked Lurie this question and his sense is that the functor I want isn't symmetric monoidal but at best lax / oplax.",
"Have you worked out the complex version (graded complex central simple algebras and invertible $KU$-module spectra) of this story? If so, perhaps you can go down to $KO$-module spectra by taking $\\mathbb{Z}/2$-fixed points."
] | 2 |
Science
| 0 |
122
|
mathoverflow
|
Base change for $\sqrt{2}.$
|
This is a direct follow-up to [Conjecture on irrational algebraic numbers](https://mathoverflow.net/questions/173414/conjecture-on-irrational-algebraic-numbers).
Take the decimal expansion for $\sqrt{2},$ but now think of it as the base $11$ expansion of some number $\theta_{11}.$ Is there an easy (or, failing that, hard) proof that $\theta_{11}$ is transcendental? Of course, same question stands for $\theta_k,$ for your favorite $k>10.$
|
https://mathoverflow.net/questions/173442/base-change-for-sqrt2
|
[
"nt.number-theory",
"ds.dynamical-systems"
] | 23 | 2014-07-06T08:36:56 |
[
"@NikitaSidorov I, for one, am completely agnostic on the subject, and have absolutely no clue why anyone would believe what you say everyone believes. I am, of course, just a caveman.",
"@NikitaSidorov And of course, we all speak for ourselves.",
"Of course, we all believe it is transcendental. And of course, we all know that this is way beyond our reach at present. Why ask, then?",
"@AnthonyQuas I cannot actually understand your comment...",
"Since the digits are the same but the base is larger, in a certain sense, $\\theta_{11}$ has a quicker rational approximations than $\\sqrt 2$. Can this fact affect the irrationality measure of $\\theta_{11}$?",
"@AnthonyQuas: Are there reasons to believe in this independency, besides mere numerical observations?",
"Algebraic irrationality \"is independent of\" base expansions; there are countably many algebraic irrationals; Therefore \"with probability 1\", all algebraic irrationals are normal to all bases.",
"@GeraldEdgar What is the justification for the well-known conjecture?",
"$\\theta_{11}$ is certainly irrational. There is a well-known (but still far from proved) conjecture that all algebraic irrationals are normal in all bases. So (of course) your number is transcendental. But your question (whether there is an easy proof of it) is not answered by this."
] | 9 |
Science
| 0 |
123
|
mathoverflow
|
Do all possible trees arise as orbit trees of some permutation groups?
|
## I.Motivation from descriptive set theory
(Contains some quotes from Maciej Malicki's paper.)
The classical theorem of Birkhoff-Kakutani implies that every metrizable topological group G admits a compatible left-invariant metric, that is, a metric d inducing the group's topology, and such that d(zx,zy)=d(x,y) for all x,y,z in G. However, even if G is completely metrizable, it is not always true that G admits a metric that is left-invariant and complete at the same time.
Polish groups admitting a complete left-invariant metric are called CLI groups. We only focus on CLI subgroups of $S_{\infty}$, the symmetry group of natural numbers, for now. Malicki proved that a permutation group G is CLI iff the _orbit tree_ of G is well-founded, i.e. every branch of the tree terminates at some finite level, where the concept _orbit tree_ is defined as following.
* * *
## II.The problem
Let $G\leq S_{\infty}$ be a permutation group acting on $\mathbb{N}$ (starts from 1 for convenience). Define $G_n$ to be the n-point stabilizer, $G_n=\\{g\in G|g(i)=i,1\leq i\leq n\\}$, and $G_0=G$. We represent an infinite orbit $O$ of $G_n$ by a node $N$ on level $n$ (finite orbits do not count).
If $O_1$ is an infinite orbit of $G_k$, $O_2$ an infinite orbit of $G_{k+1}$ and $O_2\subset O_1$ (possibly the same), we correspondingly draw $N_2$ as a one-step extension of $N_1$.
Assuming $G$ is transitive, this process defines a map from $G$ to a rooted tree $T_G \subset\omega^{<\omega}$. We say it is the _orbit tree_ of $G$ and it roughly describes how the orbits of $G$ split when we keep fixing more and more points. The natural inverse question is
> For every tree $T\subset\omega^{<\omega}$, is there a group $G$ such that the orbit tree $T_G$ is isomorphic to $T$? Can we construct it?
* * *
## III.Some attempts
Since I do not know too much about group theory, especially not about infinite permutation groups, I started with some concrete examples. The construction was hard even for very simple finite trees, though. For instance, $\mathbb{Z}$ acting on $\mathbb{Z}$ realizes the tree of a single node. $\mathbb{Z}_{wr}\mathbb{Z}_2$(wr for wreath product) on $\mathbb{Z}\times 2$ realizes a "chain" of any finite depth, etc. But these seem to be too _ad hoc_.
I also received a generous hint from an expert, saying that
> 1\. If $G$ is a permutation group on $\Omega$ and $\alpha\in\Omega$ and the orbits of $G_{\alpha}$ are $\Omega_{\lambda}(\lambda\in\Lambda)$ then $G_{\alpha}$ is a subdirect product of the restrictions $H_{\lambda}:=G_{\alpha}^{\Omega_{\lambda}}$. Specifically $G_{\alpha}$ can be embedded in the Cartesian product $\prod_{\lambda}H_{\lambda}$ such that the projections onto each factor are surjective.
>
> 2\. Now let $G$ be a free group of countably infinite rank. We want to find a corefree subgroup $H$ of $G$ of countable index such that $H$ is free of infinite rank and has a specified number of double cosets $HxH$ for which $|H:x^{-1}Hx|$ is infinite ($H$ is going to play the role of $G_{\alpha}$). We know that every subgroup of a free group is free and many (most of them?) have infinite rank when $G$ has infinite rank. I don't think it is that hard to satisfy the condition about the number of cosets. If this is true then we can go from level 0 to level 1.
>
> 3\. To go to level 2 we know that $H$ is a subdirect product of the $H_{\lambda}$. Suppose that all the infinite $H_{\lambda}$ are free of infinite rank and that the embedding of $H$ induces an isomorphism onto each of these. Now for each infinite $H_{\lambda}$ choose (corefree, infinite rank) $K_{\lambda}$ so that $H_{\lambda}$ and $K_{\lambda}$ play the roles of $G$ and $H$ in 2 to provide the correct number of infinite orbits at the next level. Can we choose the isomorphisms $H\rightarrow H_{\lambda}$ and a subgroup $K$ of $H$ such that $K$ maps onto $K_{\lambda}$ for each $\lambda$?
>
> 4.If the answers to these questions are yes, I think that you can then prove what you want.
It is not easy for me to comprehend his thought since I have no training in free groups. I have some vague doubts that
1. What is the action of $G$ on countably many things? $G$ naturally acts on itself by concatenation but this seems not correct.
2. Why must such an $H$ be a one-point stabilizer?
3. Moreover, the free group of countably infinite rank is countable itself. It is known that every countable group is CLI, hence its orbit tree must be well-founded. That is to say, $G$ may never realize ill-founded trees. I suspect we need to use other methods for non-CLI groups, taking inverse limit for example.
* * *
## IV.Possible modifications of the problem (not strict)
From the viewpoint of descriptive set theory, it is not necessary to realize every tree, "a subcollection of trees" which has the same Borel complexity as "all trees" will suffice. For example, we may properly "stretch" the tree so that there is at most one "event" (a node terminates or a node splits into more than one nodes) happens on each level. we may also require every splitting node to split infinitely, etc. As long as the transformation is Borel, we may choose the easiest class of trees to realize.
To realize the tree, we may also choose to fix finitely many points at one time if necessary.
If we only look at the well-founded trees, we may construct a big tree recursively. We know that the one-point stabilizer is isomorphic to a subdirect product of the restrictions on each orbit. But
> Given transitive permutation groups $H_1$, $H_2$... (finitely or countably many) acting on $\mathbb{N}$, is there a transitive group $G$ such that $G_{\alpha}$ is isomorphic to the direct product of these $H_{\lambda}$?
My greedy hope (possibly incorrect!) is, there is such a group. If this is true, noticing that factors of direct product are "independent" of each other, we may fix points on orbits "one by one" and let other orbits "wait". Say the orbit tree of $H_{\lambda}$ is $T_{\lambda}$, then the orbit tree of $G$ will be of the shape that each subtree is a "stretched" $T_{\lambda}$. It will work since we only need to realize all "stretched" trees.
Please do not laugh at my silliness if I am miserably wrong :-P Thank you so much for your time and effort.
|
https://mathoverflow.net/questions/61361/do-all-possible-trees-arise-as-orbit-trees-of-some-permutation-groups
|
[
"gr.group-theory",
"co.combinatorics",
"descriptive-set-theory",
"topological-groups",
"permutations"
] | 23 | 2011-04-11T21:02:55 |
[
"a rooted tree is a subset(well, my fault) in \\omega^{<\\omega} that is closed under taking initial segment. For example, the set {(0),(0,0),(0,1),(0,2)} is the tree that has one root and three \"branches\", each terminates at level 1. But we usually think of it intuitively(as drawn on paper). As we fix more and more points in N, the orbits become smaller and smaller as a set. We intuitively draw a node for each orbit with infinite size. If O2 is a subset of O1, then N2(on the level below) is linked to N1. Thus we may draw the orbit tree.",
"Thank you for your comment. G_1 could have infinitely many orbits. But your construction will not obtain the wanted direct product. This is because g(a1,a2...)g^{-1}=(g(a1),g(a2)...) and it will introduce a lot more permutations.",
" How are you relating a tree to an element in \\omega^{<\\omega}? Does this mean all finite sequences in natural numbers? If so why does the tree need to be finite? Probably I am misunderstanding the notation. Can you clarify?",
"Can't G1 have infinitely many orbits? Construct your group in the following fashion. Divide N into {1} and countably more disjoint infinite sets Sis. Let G be generated by transpositions between the 1 and the first element of Sis and the full permutation group on Si. I was not sure how G corresponded to the rooted tree"
] | 4 |
Science
| 0 |
124
|
mathoverflow
|
Boundaries of noncompact contractible manifolds
|
It is known that a manifold $B$ bounds a compact contractible topological manifold if and only if $B$ is a homology sphere. The "only if" direction follows by excising a small ball in the interior of the contractible manifold, and noting that its boundary sphere has the same homology as $B$ by Poincare duality. The "if" direction is due to Freedman in dimension $3$ and to Kervaire in dimensions $>3$.
**Question.** Is there a characterization of boundaries of _noncompact_ contractible manifolds?
Note that if $B$ is a boundary component of a contractible $n$-manifold $W$, then the following holds.
1. If $B$ is compact, then $W$ is compact and $\partial W=B$.
(If $V$ denotes the union of $B$ and the interior of $W$, then the long exact sequence of the pair gives $H_{n-1}(B)\cong H_{n}(V,B)$ and Poincare duality gives implies that $H_{n-1}(B)$ is nontrivial, and hence so is $H_{n}(V,B)\cong H^0_c(V)$ which is only possible if $V$ is compact.)
2. $B$ is stably parallelizable (because $W$ is parallelizable).
3. If $U$ is a proper open subset of $B$, then $U$ bounds a noncompact contractible manifold, namely, $U\cup\mathrm{Int}(W)$.
4. The product of $B$ with any open contractible manifold bounds a noncompact contractible manifold.
5. If $B'$ bounds a contractible manifold $W'$, then the connected sum $B\\# B'$ bounds a contractible manifold, namely, the boundary connected sum of $W$, $W'$.
The above question may be too hard, so I would be happy with any addition to 1--5.
|
https://mathoverflow.net/questions/89604/boundaries-of-noncompact-contractible-manifolds
|
[
"gt.geometric-topology",
"mg.metric-geometry"
] | 23 | 2012-02-26T12:43:35 |
[
"@User See theorem 11.3 in Massey \"Homology and cohomology theory\", or the section on Poincare duality in Bredon's \"Sheaf theory\". There two references involve (co)homology of general spaces and may be hard for a beginner. Alternatively, solve exercise 35 in section 33 in Hatcher's \"Algebraic topology\".",
"Could you please tell me a source of the proof of $H^k_\\textbf{c}(M)\\cong H_{n-k}(M,\\partial M)$, where $n=\\dim M$?",
"Igor, of course, one has to construct a Gromov-hyperbolic metric on $W$ \"relative\" to $B$, so that $B$ embeds. The difficulty with my approach, I think, is in two absolute cases: (1) constructing a Gromov-hyperbolic metric on $W$ so that the entire boundary is a homology sphere; (2) Given a compact contractible manifold $W$ with boundary $B$, construct a metric on $W$ so that $B$ is the ideal boundary. Part (2), I think, is quite doable and mildly interesting. Part (1), I think, is difficult, if not impossible. ",
"Misha, thank you for bringing geometry into the game. Yet I do not see how this gives an embedding of the boundary of $W$ into the ideal boundary of (a hyperbolic metric on) the interior of $W$. The ideal boundary is not necessarily the largest compactification. Say start with a compact contractible manifold whose boundary is a homology sphere $B$, remove a proper closed subset of $B$, and try to use your idea to embed what is left in a homology manifold; this looks hard.",
"A remark: The approach I suggested could work only for $n\\ge 4$, since every homology 2-sphere is homeomorphic to the usual sphere. But for $n\\ge 4$ I do not see any obvious obstructions. ",
"Recall that Bestvina proved in 1996 that a hyperbolic group $G$ is a $PD(n)$ group if and only if its ideal boundary is a homology manifold which is also a homology sphere (of dimension n-1 of course). One can also ask an absolute version of your question in the hyperbolic context: Suppose $W$ is contractible $n$-manifold. Does $W$ admit a bounded geometry hyperbolic metric so that the ideal boundary is a homology manifold/homology sphere (of dimension n-1)? No group is involved here, just pure geometry. ",
"I do not have an answer, but only a vague suggestion: Instead of trying to embed $B$ in a homology (n-1)-manifold, one can try to embed it in a homology sphere which is also a homology (n-1)-manifold. This might be easier. This situation could be similar to Mladen Bestvina's theorems on ideal boundaries of hyperbolic PD(n) groups. One approach would be to construct a suitable Gromov-hyperbolic metric on $W$ so that $B$ is a part of the ideal boundary. Alternatively, one can try to imitate Bestvina's arguments without having a metric (or, instead, using a weaker structure a la John Roe). ",
"Tom's comment on intersection product together with classification of noncompact 2-manifolds implies that if $B$ has dimension $2$, then it is homeomorphic to an open subset of $S^2$, and clearly any open subset of $S^2$ occurs as $B$. Indeed, if the surface is not planar, then it contains a handle, and hence there are two curves that only intersect once.",
"@Igor: I meant the $Q/Z$ valued linking form on the torsion homology classes. I was thinking a warm up problem is to determine which $B=$interior$(M)$, $M$ compact, are boundaries, and suggesting maybe all such embed as codim 0 submanifolds of a homology sphere. ",
"@Paul, linking numbers do not vanish, not even in the sphere bounding a disk. I am not sure whether \"open subsets of homology spheres\" is the answer. Certainly, $B\\times\\mathbb R^n$ is homeomorphic to an open subset of $\\mathbb R^{2n-1}$: just embed $B$ properly in $\\mathbb R^{2n-1}$ and note that the normal bundle is trivial by 2. But I cannot see why $B$ itself would be homeomorphic to an open subset of $\\mathbb R^{2n-1}$. Would not it be like showing that any open contractible manifold has a manifold compactification (which is surely false)? ",
"Maybe the answer is any open subset of a homology sphere? Linking numbers vanish also. I can't see how to find a compact $n$-manifold with boundary $M$ whose interior bounds a contractible manifold but which doesn't embed in a homology $n$-sphere. For $n=3$, it seems like you can add 2-handles to $M$ to get a homology ball, and perhaps in higher dimensions you could add handles to get a homology ball.",
"@Tom, oh, and the intersection number is trivial because the $0$-cycle is a boundary, so its points come in pairs with opposite signs, and cancel. Thank you for suggesting this!. I think this \"intersection product\" is Poincare dual to the cup product, as discussed in Dold's \"Algebraic Topology\", section VIII.13.",
"In particular I am referring to intersection numbers, not to the ring structure on homology.",
"Every cycle in $B$ is the boundary of a chain in $W$. So basically given a $p$-cycle and an $(n-p-1)$-cycle in $B$ their intersection is the boundary of the intersection of a $(p+1)$-chain and an $(n-p)$-chain in $W$, so the boundary of a $1$-chain, so a homologically trivial $0$-chain. To make this rigorous I guess you would need to say something about Poincare duality and compactly supported cohomology.",
"@Tom, I am struggling to make the notion of an intersection product precise. Is triviality of the intersection product based on the fact that $B$ has trivial normal bundle in $W$, or does the contractibility of $W$ get used?",
"By 3, $B$ can have the homotopy type of any finite complex, by taking a thickening of some embedding in Euclidian space. So for example any $k$ dim'l complex has the homotopy type of a $n=2k+1$ manifold which is the boundary of a contractible $n+1$ manifold.",
"6. Intersection products in $B$ are trivial."
] | 17 |
Science
| 0 |
125
|
mathoverflow
|
Riemannian manifolds etc. as locally ringed spaces?
|
There are, among others, three general ways of equipping a "space" (which for the purposes of this question could be a topological space or a differentiable manifold, according to the case) with further structure:
(1) "Specifying regular funcions", which leads to locally ringed spaces, e.g. real-analytic manifolds or holomorphic manifolds.
(2) "Choosing a section (perhaps with some nondegeneracy properties) of some (usually tensor) bundle", which leads for example to Riemannian manifolds, quasi-symplectic manifolds and quasi-holomorphic manifolds.
(2') A variant of the latter: "Choosing a nondegenerate section _with some integrability properties_ of some (usually tensor) bundle", which leads e.g. to symplectic geometry and holomorphic geometry.
(3) "Choosing a sub-bundle (with some properties) of some bundle" , and this leads for example to foliated manifolds and contact structures.
There are certainly some overlaps between the above approaches. For example, one can think of a complex manifold as a topological space equipped with a sheaf of local rings (the sheaf of holomorphic functions of the complex manifold), which is an instance of (1), or as an even dimensional smooth manifold equipped with an "integrable" field of endomorphisms of its tangent bundle which square to $-\mathrm{id}$, and that's an instance of (2), or rather (2'). Another example is that of a foliated manifold: it's usually seen as an instance of (3), but you can construct the sheaf of locally-constant-on-the-leaves smooth functions, and that reduces, in some sense, to the point of view (1).
Furthermore, smooth (in the sense of differentiable) maps between complex manifolds that preserve the tensor $J$ giving the complex structure correspond exactly to locally ringed space maps between the corresponding complex analytic spaces. And I would imagine that some foliation-preserving map in the case of foliated manifolds translate fully faithfully into maps for the locally ringed structure mentioned above.
In either case, if I'm not mistaken, you get a fully faithful embedding of some "geometric category" into the category of locally ringed spaces (such that, say, the forgetful functor to $\mathrm{Top}$ is respected). So it seems that the approach (1) is quite general. It's then spontaneous to ask:
> Is the category of symplectic manifolds (with symplectomorphisms as morphisms) fully faithfully embeddable (say, preserving the forgetful functor to $Top$) into the category of locally ringed spaces?
and
> Is there a good notion of morphisms between Riemannian manifolds (e.g. local isometries? Riemannian submersions? compositions thereof?) such that the resulting category has a fully faithful embedding into locally ringed spaces as above?
* * *
The first (perhaps useless?) construction that comes to my mind is the following. Given a smooth manifold $X$ with sheaf of differentiable functions $\mathcal{O}_X$, let $\mathcal{T} \\!en_X^{\bullet}:=\oplus_{i\geq 0} (T_X^{\; *})^{\otimes i}$ be the sheaf of covariant tensors on $X$. If $g$ is a Riemannian metric on $X$, one could consider the sheaf of (commutative) local rings generated by $\mathcal{O}_X$ and $g$:
$\mathcal{O}_X[g]\subseteq \mathrm{Sym}^{\bullet}(T_X^{\; *})\subseteq \mathcal{T} \\!en_X^{\bullet}$
and then consider the locally ringed space $(X,\mathcal{O}_X[g])$.
One can do an analogous thing with a symplectic manifold $(X,\omega)$ viewing $\omega$ as an anti-symmetric tensor:
$\mathcal{O}_X[\omega]\subseteq \Omega_X^{\mathrm{even}} \subseteq \mathcal{T} \\!en_X^{\bullet}$
to get a locally ringed space $(X,\mathcal{O}_X[\omega])$.
Can this construction be of any help to answer the above questions?
|
https://mathoverflow.net/questions/56833/riemannian-manifolds-etc-as-locally-ringed-spaces
|
[
"locally-ringed-spaces"
] | 23 | 2011-02-27T08:08:01 |
[
"From scheme theoretic perspective, how do wish to 'deal' with additional structures like metrics, symplectic forms, hermitian or kahler structures?",
"I'm quite interested in this question. Made any progress since Feb?",
"@JohanesEbert: recovering $g$ up to conformal equivalence would already be nice. And what about LRS morphisms: would they correspond to conformal mappings? ",
"@Zack: ya.. it looks like so. Perhaps my definition is just useless. But I don't know how morphisms of symplectic/Riemannian manifolds induce homomorphisms on that sheaves. ",
"Isn't $\\mathcal{O}[\\omega]$ just $\\mathcal{O}[x]/x^{n+1}$?",
"You cannot recover $g$ from $\\mathcal{O}[g]$, only up to conformal equivalence.",
"@André: In the theory of \"supermanifolds\" people already use sheaves of $\\mathbb{Z}/2\\mathbb{Z}$-graded rings. And there are some recent theories of \"derived manifolds\" (of which I don't know anything). But my question was perhaps more down to earth. Anyway, why do you suggets using differential graded rings? Is it to keep track that the \"generator\" $g$ in $\\mathcal{O}_X[g]$ is placed in (tensor) degree 2?",
"Maybe it could be useful to generalize the notion of locally ringed space, and allow \"structure sheaves\" with values in categories other than rings (e.g. differential graded rings)."
] | 8 |
Science
| 0 |
126
|
mathoverflow
|
When does a representation admit a spin structure?
|
Let $G$ be a finite group, and let $V$ be an $n$-dimensional _real_ representation of $G$. Think of $V$ as given by a homomorphism $$ \rho_V\colon G\to O(n).$$ Write $\chi_V$ for the character of $V$.
Here are two problems.
1. Using _only_ the character $\chi_V$ of $V$, determine whether $\rho_V(G)\subset SO(n)$.
2. Using _only_ the character $\chi_V$ of $V$, and assuming $\rho_V(G)\subset SO(n)$, determine whether $\rho_V$ admits a factorization through a homomorphism $\widetilde{\rho}_V:G\to Spin(n)$.
Here's one answer to 1: the identity of formal power series $$ \sum_{k\geq0} \chi_{\Lambda^kV}(g)\,T^k = \exp\Bigl( -\sum_{k\geq1} \frac{1}{k}\chi_V(g^k)\, (-T)^k \Bigr)$$ where $\Lambda^kV$ is the $k$-th exterior power representation of $V$, gives $\chi_{\Lambda^nV}(g)$ as a polynomial in $\chi_V(g),\dots,\chi_V(g^n)$, and $\rho_V(G)\subset SO(n)$ if and only if all $\chi_{\Lambda^nV}(g)>0$.
Is there a better answer for 1? Is there any answer in a similar spirit for 2?
|
https://mathoverflow.net/questions/52643/when-does-a-representation-admit-a-spin-structure
|
[
"rt.representation-theory"
] | 23 | 2011-01-20T08:10:12 |
[
"For 1, it depends what you mean by \"using the character ONLY\". You can't tell the number of elements of each order from the character alone. If you do know the order of each element, then 1 requires just that each element of $G$ has the eigenvalue $-1$ with even (possibly zero) multiplicity, and this is an easy calculation, given the character values ( in fact, you only need to check elements whose order is a power of $2$).",
"Charles, I can see at least a necessary condition on the character (that you probably know of). It is that for any involution $s\\in G$, $\\chi_V(s)\\equiv\\chi_V(1) \\mod 8$ (this comes from restriction to cyclic subgroups). Given the \"miracles\" of Brauer theory, might it be sufficient also ? A perhaps relevant reference is this Deligne's 1976 [paper][1], related to Marty's first remark about $\\epsilon$ factors. [1]: digizeitschriften.de/de/dms/img/?PPN=GDZPPN002092654",
"Marty, your observation about pairs of conjugacy classes seems convincing to me.",
"@BS: Ah yes.. that was dumb on my part. I had in mind the lifting of subgroups isomorphic to $Z$, not $Z / kZ$, when I wrote that. The remainder of my comment -- that one needs information about pairs, not just individual elements -- should hold (though it's vague). ",
"@marty : are you saying that any morphism $Z/kZ \\to SO(n)$ lifts $Spin(n)$ ? This seems not true if $k$ is even.",
"Here's another reason why such an answer shouldn't exist for (2). Let's say you know all the eigenvalues for $\\rho(g)$, $\\rho(g^2)$, etc., information given by the character. Well, that information tells you nothing about whether $\\rho$ factors through the 2-fold Spin cover, because every element of $SO(n)$ lifts (compatibly with its powers) to an element of $Spin(n)$. Instead, you need information about $\\rho(g_1), \\rho(g_2), \\rho(g_1 g_2)$ for all pairs $(g_1, g_2)$ -- information depending probably on the conjugacy class of the pair $(g_1, g_2)$ (not individual conjugacy classes).",
"Okay. But I don't see why, a priori, having a cohomological obstruction means you can't compute it from the character; my question 1 is really about the vanishing of $w_1(\\rho_V)$, after all. On the other hand, I could certainly believe that such a problem could be intractible.",
"(I suppose that the $SU(2)$ and $SO(3)$ case is just the case $n = 3$ of the question...)",
"I don't think that any such answer exists for (2). The obstruction to lifting is called the second Steifel-Whitney class of the orthogonal real representation $\\rho_V$, written $w_2(\\rho_V)$. It's difficult to compute and quite important for Galois representations (related to $\\epsilon$-factors). Such results lie beyond simple character calculuations, I think. Compare, for instance, the problem of deciding whether a rep. into $SO(3)$ arises as $Sym^2$ of a rep into $SU(2)$. Can you tell by the character? ",
"I guess there is a cohomological obstruction for 2 in $H^2(G,\\mathbb{Z}/2\\mathbb{Z})$. Although, I have no idea how to express this in character theory."
] | 10 |
Science
| 0 |
127
|
mathoverflow
|
Is the set of integers of the form $a/(b+c)+b/(a+c)+c/(a+b)$ computable?
|
The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying
$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$
are [absurdly high](https://web.archive.org/web/20170813162605/https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM), namely $$(154476802108746166441951315019919837485664325669565431700026634898253202035277999,$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579 ,$$ $$ 4373612677928697257861252602371390152816537558161613618621437993378423467772036) .$$ This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \\{n\in\mathbb{N}\setminus\\{0\\}: (\exists a,b,c \in\mathbb{N}\setminus\\{0\\}):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\\}$$ computable? (As user [Watson](https://mathoverflow.net/users/84923/watson) points out in the comment section below, $C$ contains no odd numbers. It would also be great to see an even number $\geq 6$ not contained in $C$.)
|
https://mathoverflow.net/questions/278747/is-the-set-of-integers-of-the-form-a-bcb-acc-ab-computable
|
[
"nt.number-theory",
"computability-theory",
"diophantine-equations",
"computer-science"
] | 22 | 2017-08-14T12:16:31 |
[
"The goo.gl link in a previous comment points to Quora.",
"mathoverflow.net/questions/264754/…",
"Also discussed at mathematica.stackexchange.com/questions/184956/…",
"Cross-posted to cstheory.stackexchange.com/questions/39383/…",
"(continued) Note that there are odd $n$ for which the associated elliptic curve has positive rank, but does not have rational points on the component that contains the positive points. (IIRC, $n = 19$ is an example.) This gives an additional twist to the question.",
"This is somewhat similar to the question which natural numbers $n$ are congruent numbers (i.e. the area of a right-angled triangle with rational sides), which comes down to asking whether the elliptic curve $E_n \\colon y^2 = x^3-n^2x$ has positive rank over $\\mathbb Q$. In this case, there is a conjectural answer (Tunnell's Theorem, conditional on the BSD conjecture), which relies on the fact that the $E_n$ are all quadratic twists of a fixed curve. The question asked here is likely to be harder, since the resulting curves are not twists, and there is the positivity condition. -->",
"So we're half done!",
"Here is explained why $C$ doesn't contain any odd number.",
"See also math.stackexchange.com/questions/402537/… and Bremner and Macleod, An unusual cubic representation problem, Annales Mathematicae et Informaticae 43 (2014) 29-41, ami.ektf.hu/uploads/papers/finalpdf/AMI_43_from29to41.pdf",
"I don't have the time to look into the details, but from a cursory glance at this quora answer it looks like the problem amounts to deciding whether a certain elliptic curve has solutions over $\\mathbb{Q}$. Now IIRC there is an algorithm for that (find solutions locally and try to globalize) provided Ш is finite, which is conjecturally always the case. So I think conjecturally your set is indeed computable.",
"A quick heuristic comment: for each $n$, the existence of such $a, b, c$ corresponds to the existence of a solution to a degree-$3$ Diophantine equation in $3$ variables; and this is, I believe, a bit beyond what is generally known to be decidable. So I suspect that there will be no general reason why this set is computable; rather, if it is computable (which I extremely strongly suspect it is), the proof is likely to be a corollary of a complete characterization of these $n$s, which will not use any computability theory."
] | 11 |
Science
| 0 |
128
|
mathoverflow
|
A mysterious paper of Stallings that was supposed to appear in the Annals
|
In Stallings's paper
* Stallings, John, _[Groups with infinite products](https://doi.org/10.1090/S0002-9904-1962-10817-9)_ , Bull. Amer. Math. Soc. **68** (1962), 388–389.
he briefly discusses how to prove "several generalizations" of Brown's theorem saying that monotone union of open $n$-cells is an open $n$-cell; however, he never formulates a precise statement of what he proves.
For the details, he refers to the paper
* J. Stallings, On a theorem of Brown about the union of open cones, Ann. of Math, (to appear).
But as far as I can tell this paper never appeared. Does anyone know what happened and what the theorem is?
|
https://mathoverflow.net/questions/405617/a-mysterious-paper-of-stallings-that-was-supposed-to-appear-in-the-annals
|
[
"reference-request",
"gt.geometric-topology",
"soft-question"
] | 22 | 2021-10-06T13:41:53 |
[
"I feel the pain. There's more than one of these in category theory from the 70s-80s, announced to appear in JPAA, but which never turned up.",
"With regards to the publication of this paper, it looks like the journal may have been... stalling",
"@CarloBeenakker: Thanks for noticing that reference! I think that the lack of a journal might just be the Annals's style guide -- there are several papers in that bibliography that are \"to appear\", and none of them specify a journal.",
"in a 1963 Ann. of Math. article by Stallings that paper is still cited as \"to appear\", however now without a journal being mentioned."
] | 4 |
Science
| 0 |
129
|
mathoverflow
|
The multiplication game on the free group
|
Fix $W\subseteq\mathbb F_2$ and consider the following two-person game: Player 1 and Player 2 **simultaneously** choose $x$ and $y$ in $\mathbb F_2$. The first player wins, say one dollar, iff $xy\in W$. Does this game admit Nash equilibria?
Of course, for very particular choices of $W$ it does, for instance if $W$ is either finite or cofinite. But I am not interested in these cases, I am interested in _solvability_ of the _multiplication game_ for all $W$.
One possible mathematical reformulation (game theorists would say that the mixed extension of this game is not uniquely defined and then there are _several_ different mathematical reformulations) is the following:
> **Question.** Do there exist, for all $W\subseteq\mathbb F_2$, finitely additive probability measures $\overline\mu,\overline\nu$ (of course depending on $W$) on the power set of $\mathbb F_2$ such that for all other finitely additive probability measures $\mu,\nu$ one has $$ \int\int\chi_W(xy)d\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\overline\nu(y)\leq\int\int\chi_W(xy)d\overline\mu(x)d\nu(y) $$ ?
**Motivation.** If we play the same game on a (countable) amenable group, one can show that a Nash equilibrium exists and is given by a right-invariant mean $\overline\mu$ and a bi-invariant mean $\overline\nu$. I am wondering what can happen for non amenable groups... I suspect (motivated by some possibly wrong speculative consideration) that the _multiplication game_ is actually solvable for all $W$ only on amenable groups, giving an hopefully nice new characterization of amenability. However, any approach to proving that solvability implies amenability so far got stuck.
Thanks in advance,
Valerio
|
https://mathoverflow.net/questions/98493/the-multiplication-game-on-the-free-group
|
[
"gr.group-theory",
"game-theory",
"amenability"
] | 22 | 2012-05-31T07:19:39 |
[
"Ah, that was my misunderstanding.",
"I've edited, saying explicitly that the choices of $x$ and $y$ are simultaneous.",
"Why should P2 win? Notice that the choices are done independently (simultaneously, if you want), so that P2 cannot choose a suitable $y$ such that $xy\\in W$, because (s)he does not know $x$.",
"I'm confused. If $W$ is the whole group, player 1 wins, otherwise player 2 wins. Right?"
] | 4 |
Science
| 0 |
130
|
mathoverflow
|
Characterising critical points of $E(f)=\int_{M}| \bigwedge^2 df|^2 \text{Vol}_{M}$
|
$\newcommand{\id}{\operatorname{Id}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\TM}{\operatorname{TM}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{\Cof}{\operatorname{Cof}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\tr}{\operatorname{tr}}$ $\newcommand{\TM}{\operatorname{T\M}}$ $\newcommand{\TN}{\operatorname{T\N}}$ $\newcommand{\TstarM}{T^*\M}$
Let $\M,\N$ be $d$-dimensional oriented Riemannian manifolds. Let $\tilde E:C^{\infty}(M,N) \to \R$ be defined by
$$\tilde E(f)=\frac{1}{2}\int_{\M} | \bigwedge^2 df|^2 \text{Vol}_{\M}.$$ $\tilde E(f)$ measures the mean action of $f$ on 2$D$-cubes. (How $f$ affects areas of surfaces, locally).
Note that $\bigwedge^2 df\in \Omega^2\big(\M,\Lambda_2(f^*{\TN})\big)$. Let $\nabla^{\Lambda_2(f^*{\TN})}$ be the induced connection on $\Lambda_2(f^*{\TN})$ and let $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} $ be the adjoint of the covariant exterior derivative $d_{\nabla^{(f^*{\TN})}}$.
> **Question:** Does every critical point of $\tilde E$ satisfy $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$?
Of course, by counting degrees of freedom, this doesn't look reasonable.
The converse direction follows from the Euler-Lagrange equation of $\tilde E$: I proved [here](https://mathoverflow.net/a/271264/46290) that the E-L equation is $$h_{f^*\TN} \big( \tr_{\TM}\big( df \otimes \delta_{\nabla^{\Lambda_2(f^*{\TN})}}(\bigwedge^2 df)\big) \bigg)=0,$$
where $h_{f^*\TN}:f^*\TN \otimes \Lambda_2(f^*\TN) \to f^*\TN$ is the linear extension of
$$ \tilde w \otimes (w_1 \wedge w_2) \to \langle \tilde w,w_2 \rangle w_1-\langle \tilde w,w_1 \rangle w_2.$$
Thus $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$ implies $f$ is critical.
**Edit 1:**
A natural way to produce critical points of a functional is to look at its symmetries. $\tilde E$ is conformally-invariant exactly at dimension $4$. So, in dimension $4$ conformal maps are critical. However, I proved that they also satisfy the stronger equation $\delta_{\nabla^{\Lambda_2(f^*{\TN})}} \big( \bigwedge^2 df \big) =0$.
(In fact I proved that a conformal map $\M^d \to \N^d$ satisfies $\delta_{\nabla^{\Lambda_k(f^*{\TN})}} \big( \bigwedge^k df \big) =0$ if and only if $d=2k$ or it is a homothety).
**Edit 2:**
As I shall prove below (see "Edit 3"), in dimension $d=2$ the answer is positive. (Note that in this case the "number of constraints" is the same). Focusing on the simplest next case, we consider $d=3,\M=\N=\R^3$ with the flat metrics.
**First, let's try to answer a more degenerate version of the question (forgetting $h_{f^*\TN}$):**
> Is there a smooth map $f:\mathbb{R}^3 \to \mathbb{R}^3$, which satisfy
>
> $$\delta \big( df \wedge df \big) \neq 0, \, \text{and } \, \tr \big( df \otimes \delta(df \wedge df) \big)=0, $$
If there is such an $f$ then it cannot be a local diffeomorphism; its Jacobian must vanish somewhere.
Here $ \delta \big( df \wedge df \big)\in \Omega^1\big(\R^3;\Lambda_2(\R^3)\big)$ is a one-form on $\R^3$ with values in $\Lambda_2(\R^3)$, and
$$ \delta \big( df \wedge df \big)(X)=\sum_i \nabla_{e_i}(df \wedge df)(e_i,X)=\sum_i (\nabla_{e_i} df)(e_i) \wedge df(X) + df(e_i) \wedge (\nabla_{e_i} df)(X)$$ $$ =\Delta f \wedge df(X)+\sum_i df(e_i) \wedge (\nabla_{e_i} df)(X).$$
We can try to look first for harmonic counter-examples, but so far I failed doing even that.
**Edit 3:**
> The answer is positive for $d=2$, so we need to restrict our attention to $d \ge 3$.
Indeed, in this case $E(f)=\int_{\M}| \bigwedge^2 df|^2 \text{Vol}_{\M}=\int_{\M}(\Det df)^2 \text{Vol}_{\M}$. We shall prove a map $f$ is $E$-critical if and only if its determinant is constant.
Since $\Det df$ is constant $\iff$ $\nabla (\bigwedge^2 df)=0 \iff \delta_{\nabla^{\Lambda_2(f^*\TN)}} \big( \bigwedge^2 df \big)=0$, we are done.
The Euler-Lagrange equation of $E$ can be written as
$$ \delta (\Det df \cdot \Cof df) =0, \tag{1}$$ where $\Cof df:\TM \to f^*(\TN)$ is the _cofactor map_ of $df$ defined by $$ \Cof df= \star_{f^*TN}^{d-1} (\wedge^{d-1} df) \star_{TM}^1, $$
and $\delta$ is the adjoint of the pullback connection on $f^*\TN$.
(The Euler-Lagrange equation of the Jacobian functional $E(f)=\int_{\M} \Det df \text{Vol}_{\M}=\int_{\M} f^* \text{Vol}_{\N}$ is $\delta (\Cof df)=0$. Details can be found in lemma 2.9 in my paper [here](https://arxiv.org/abs/1701.08892)).
Expanding equation $(1)$ we get $$ 0=\delta(\Det df \cdot \Cof df )= \Det df \delta(\Cof df ) - \tr_{g}(d \Det df \otimes \Cof df). $$
We now use the fact the Jacobian functional is a null-Lagrangian, i.e. every smooth map is a critical point, or equivalently $\delta (\Cof df)=0$. (This is essentially Stokes theorem, you can see lemma 2.5 [here](https://arxiv.org/abs/1701.08892)).
So, the E-L equation $(1)$ reduces to $$\tr_{g}(d \Det df \otimes \Cof df)=0. \tag{2}$$
Let $f$ be a map satisfying equation $(2)$. We shall prove $\Det df$ is constant; suppose that $\Det df_p \neq 0$ for some $p \in \M$. This implies $ \Cof (df_p)$ is invertible, so by equation $(2)$ $d\Det df_p=0$.
Now [we observe](https://math.stackexchange.com/a/2489068/104576) that any $C^1$ function $g : \M \to \mathbb R$ on a connected manifold, satisfying $g(p) \ne 0 \implies dg_p = 0$ is constant.
|
https://mathoverflow.net/questions/272355/characterising-critical-points-of-ef-int-m-bigwedge2-df2-textvol
|
[
"ap.analysis-of-pdes",
"riemannian-geometry",
"calculus-of-variations",
"critical-point-theory"
] | 22 | 2017-06-16T06:50:03 |
[
"For the local problem, it may be useful to assume that $M,N$ are both flat in which case you can get rid of the geometry and concentrate on the multilinear algebra."
] | 1 |
Science
| 0 |
131
|
mathoverflow
|
Smooth thickenings of non-smoothable manifolds
|
It is known that any closed topological manifold is homotopy equivalent to an open smooth manifold.
**Question 1.**_What can be said about the**smallest** dimension of a smooth manifold that is homotopy equivalent to a given closed topological manifold?_
The following somewhat heavy-handed argument yields a smooth manifold of roughly twice the dimension, namely, use
* [ West's solution ](http://www.maths.ed.ac.uk/~aar/papers/west.pdf) of Borsuk's conjecture that any compact ANR of dimension $n$, and in particular any closed $n$-manifold, is homotopy equivalent to a finite polyhedron of dimension $\max\\{n, 3\\}$;
* [ Stallings-Dranishnikov-Repovs's ](http://www.pef.uni-lj.si/repovs/clanki/1993/BullAustralMathSoc.pdf) embedding up to homotopy type theorem that any finite $n$-dimensional polyhedron is homotopy equivalent to a finite $n$-dimensional subpolyhedron of $\mathbb R^{2n}$, so its regular neighborhood is the desired open smooth manifold.
Here is a specific question that shows the state of my ignorance on this matter:
**Question 2.** _Is there a closed $n$-manifold which is**not** homotopy equivalent to a smooth $(n+1)$-manifold? _
The naive idea to look at the product of a non-smoothable manifold of dimension $\ge 5$ with $\mathbb R$ fails, because such a product is also non-smoothable (by the topological product structure theorem of Kirby-Siebenmann).
**Edit:** Misha kindly corrects me that a $5$-manifold is smoothable if and only if its Kirby-Siebenmann invariant vanishes; in particular, this apples to products of a $\mathbb R$ and a closed $4$-manifold $M$. Thus $M\times\mathbb R$ is smoothable iff $M$ has zero KS invariant. Smooth $4$-manifolds have zero KS invariant, but amazingly so do some non-smoothable ones.
|
https://mathoverflow.net/questions/97598/smooth-thickenings-of-non-smoothable-manifolds
|
[
"gt.geometric-topology"
] | 22 | 2012-05-21T12:43:39 |
[
"@Misha: certainly, not all $(n+1)$-dimensional thickenings of closed $n$-manifolds are proper homotopic to $I$-bundles. One can start with an $I$-bundle, and then attach a one-sided $h$-cobordism along the boundary; there are general methods to construct those in higher dimensions. One can also take boundary connected sums with (possibly noncompact) contractible manifolds with boundary, and this way one gets tons of examples when $n\\ge 3$ whose ends are quite complicated. ",
"@Igor: Concerning KS invariant for 4-manifolds: It vanishes iff $M^4\\times {\\mathbb R}$ is smoothable. In dimension 4 there are non-smoothable manifolds with both zero and non-zero KS invariants since there are other, gauge-theoretic, obstructions to smoothability. ",
"@Igor: Actually, I realized that even for n=3, a tame 4-dimensional thickening of a hyperbolic 3-manifold need not be properly homotopy-equivalent to an interval bundle. ",
"@Misha: I am familiar with examples of closed non-smoothable aspherical manifolds by Davis-Januszkiewicz and Davis-Hausmann, in fact, my question was motivated by the desire to see how much the action dimension of arxiv.org/abs/math/0010141 would change if one defines it using smooth (!) properly discontinuous actions.",
"@Mark: corrected; I hope you won't insists on accents in Repovs. @Misha: Do you know a $4$-dimensional example? In higher dimensions there is an old preprint of Galewski-Hollingworth that implies that there is no $(n+1)$-dimensional compact smooth thickening $V$ of an $n$-dimensional Poincare complex $X$, provided both of them are orientable; compactness of $V$ can be replaced by the assumption that an image of $X$ separates a neighborhood in $V$. Showing that $V$ is properly h.e. to an $\\mathbb R$-bundle seems tricky.",
"@Igor: 1. Sometimes, the product with ${\\mathbb R}$ is smoothable. 2. You can avoid Stallings' theorem if you use immersion to ${\\mathbb R}^{2n}$ instead of an embedding. 3. Try Davis-Januszkiewicz argument, see the reference here and see if this gives a counter-example at least for a tame smoothing. (Try to argue that your open smooth (n+1)-manifold is properly homotopy-equivalent to an open interval bundle over a non-smoothable n-manifold.) ",
"Should be Dranishnikov?"
] | 7 |
Science
| 0 |
132
|
mathoverflow
|
On certain representations of algebraic numbers in terms of trigonometric functions
|
Let's say that a real number has a _simple trigonometric representation_ , if it can be represented as a product of zero or more rational powers of positive integers and zero or more (positive or negative) integer powers of $\sin(\cdot)$ at rational multiples of $\pi$ (the number of terms in the product is assumed to be finite, the empty product is taken to be $1$).
_Examples:_
* $\sqrt[3]{\sin\\!\left(\frac\pi3\right)}$ has a simple trigonometric representation, because $\sqrt[3]{\sin\\!\left(\frac\pi3\right)}=\frac{3^{1/6}}{2^{1/3}}$.
* $\sqrt{\sin\\!\left(\frac\pi{10}\right)}$ has a simple trigonometric representation, because $\sqrt{\sin\\!\left(\frac\pi{10}\right)}=\frac{2^{1/2}}{5^{1/4}}\,\sin\\!\left(\frac\pi5\right)$.
* $\pi$ does not have a simple trigonometric representation, because it is not an algebraic number.
_Questions:_
* Do $\sqrt{\sin\\!\left(\frac\pi5\right)},$ $\sqrt{\sin\\!\left(\frac\pi8\right)},$ $\sqrt{\sin\\!\left(\frac\pi{12}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{15}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{20}\right)},$ $\sqrt{\sin\\!\left(\frac\pi{24}\right)}$ have simple trigonometric representations?
* Is there an algorithm that, given a rational power of $\sin(\cdot)$ at a rational multiple of $\pi$, would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?
* More generally, is there an algorithm that, given a real algebraic number (in some explicit form, e.g. as its minimal polynomial and a rational isolating interval), would determine if it has a simple trigonometric representation? If so, could you give (or outline) a concrete example of such an algorithm (efficient, if possible)?
|
https://mathoverflow.net/questions/194181/on-certain-representations-of-algebraic-numbers-in-terms-of-trigonometric-functi
|
[
"nt.number-theory",
"algorithms",
"algebraic-number-theory",
"computability-theory",
"galois-theory"
] | 22 | 2015-01-17T20:50:44 |
[
"You mean solve all minimal polynomials with $\\sin(\\pi\\times rational)$?",
"This is related I guess: mathworld.wolfram.com/TrigonometryAngles.html"
] | 2 |
Science
| 0 |
133
|
mathoverflow
|
Infinitely many planets on a line, with Newtonian gravity
|
(I previously asked essentially this [on physics.stackexchange](https://physics.stackexchange.com/questions/56843/infinitely-many-planets-on-a-line-with-newtonian-gravity), but was actually
hoping for answers with something closer to a proof than what I got there.)
Suppose we have a unit mass planet at each integer point in 1-d space. $\:$ As described in that answer, the sum of the forces acting on any particular planet is absolutely convergent. $\;\;$ Suppose we move planet_0
to point $\epsilon$, where $\: 0< \epsilon< \frac12 \:$. $\;\;$ For similar reasons, those sums will still be absolutely convergent.
Now we let Newtonian gravity apply. $\:$ What will happen?
If it's unclear what an answer might look like, you could consider the following more specific questions:
Will there be a positive amount of time before any collisions occur?
(As opposed to, for example, a collision at time $\frac1n$ for each positive integer $n$.)
"Obviously" (at least, I hope I'm right), planet_0 will collide with planet_1. $\:$ Will that be the first collision?
planet_0 will start out moving right, and all of the other planets will start out moving to the left.
Will there be a positive amount of time before any of them turn around?
How long will it be before there are any collisions? $\:\:$ (perhaps just an approximation for small $\:\epsilon\:$)
|
https://mathoverflow.net/questions/128796/infinitely-many-planets-on-a-line-with-newtonian-gravity
|
[
"mp.mathematical-physics",
"differential-equations"
] | 22 | 2013-04-25T20:31:29 |
[
"@Anthony Quas and Jon: I think this is not what he means. He does not \"live in 1D world\" but in 3-space. Just the planets happen to be on the line. The gravity law must be inverse squares. Otherwise the questions make no sense.",
"@AnthonyQuas is correct. The Poisson equation in one dimension in this case is $\\phi''(x)=\\delta(x)$ that has as a solution $\\theta(x) x$. So, the potential between two bodies is not decaying in this case.",
"Excellent question! Perhaps You should start as in thermodynamics with a large number $N$ of particles located at $-N,\\dotsc,-1, \\epsilon, 1,\\dotsc, N$ and see what happens then. At least in this case the total energy is finite, which is not the case when infinitely many particles are present.",
"point masses $\\:$",
"Are the planets point masses, or some radius $r \\lt \\frac{1}{2}$?",
"Yes, particles clump together, typically forming smaller systems first. This is studied extensively in cosmology, both analytically and numerically. Gravitational effects are easy to model, and affect dark matter. youtube.com/watch?v=8C_dnP2fvxk However, dissipative effects such as the inelastic contraction of gas clouds are important, too. ",
"I think that as long as each planet is at most $\\delta$ from its nearest integer, the total force on each planet is $O(\\delta)$. This can be used to prove rigorously that there's a positive $\\tau>0$ before any collision can occur. ",
"Of course if you really live in a 1D world, gravitational force presumably doesn't decay at all?",
"I read recently about a very similar problem that appeared in a 1949 letter from Ulam to von Neumann. (In that case the particles started at points of $\\mathbb Z$ with each node being occupied with probability 1/2). He showed(?) that something analogous to the universe happens: nearby groups of particles come together; and then those \"solar systems\" form galaxies etc."
] | 9 |
Science
| 0 |
134
|
mathoverflow
|
Are there "chain complexes" and "homology groups" taking values in pairs of topological spaces?
|
Throughout this question, notation of the form $(X,A)$ denotes a sufficiently nice pair of topological spaces. I think for most of what I'm saying here, it is enough to assume that the inclusion $A \hookrightarrow X$ is a [cofibration](http://ncatlab.org/nlab/show/Hurewicz+cofibration), or that $(X,A)$ is a neighborhood deformation retract ([NDR](http://ncatlab.org/nlab/show/neighborhood+retract)) pair. All maps $f:(X,A) \to (Y,B)$ are continuous and satisfy $f(A) \subset B$.
Consider a sequence $\mathcal{S}$ given by: $$\ldots \to (X_{n-1},A_{n-1}) \stackrel{d_{n-1}}{\to} (X_n,A_n) \stackrel{d_n}{\to} (X_{n+1},A_{n+1}) \to \ldots$$ so that $d_n \circ d_{n-1}(X_{n-1}) \subset A_{n+1}$ for each $n$.
> Has this object been defined and studied? If so, where?
One can associate to this sequence of space-pairs a "homology" which takes values in the category of space-pairs. That is, define $$\mathcal{HT}_n(\mathcal{S}) = \left(d_n^{-1}(A_{n+1}),d_{n-1}(X_{n-1})\right).$$ The construction is functorial if one considers the obvious analogue of "chain maps" in the category which contains objects like $\mathcal{S}$. I'm wondering if basic definitions and properties, etc. of this or related constructions have been set down somewhere.
|
https://mathoverflow.net/questions/134443/are-there-chain-complexes-and-homology-groups-taking-values-in-pairs-of-topo
|
[
"at.algebraic-topology",
"chain-complexes",
"reference-request",
"gn.general-topology"
] | 22 | 2013-06-21T17:21:30 |
[
"@ViditNanda This is two years old by now. Did you make any progress?",
"The pair should be $(X,A)$ of course. ",
"@Vidit: I've sent an email on this. One point is that a pair $(X,S)$ defines filtrations $E_n(X,A)$ which are a base point in dim $0$, $A$ in levels $1$ to $n-1$ and $X$ thereafter. So if $X$ already has a filtration $X_*$ then we get two filtrations for a given $n$. ",
"Ronnie, I'm not sure I see an \"obvious\" bifiltration anywhere in this question. Could you explain why that would be the natural thing to do? I am happy to discuss this via email if things get too intricate for comment boxes.",
"@Vidit: Reading your question again, I would be inclined to go for defining invariants of bifiltered spaces, rather than invariants in spaces. Such bifiltered spaces in truncated form can be regarded as special cases of $n$-cubes of spaces, to which my work with Loday applies. I'm slowly writing up something on the bifiltered case, and my exposition [60] ``Triadic Van Kampen theorems and Hurewicz theorems'', Algebraic Topology, Proc. Int. Conf. March 1988, Edited M.Mahowald and S.Priddy, Cont. Math. 96 (1989) 39-57. (pdf on my web site) gives an idea. Also I missed the word \"discrete\"!",
"Thank you, Ronnie. I already have your book(s), being a fan if the groupoid point of view, so I will take a look.",
"@Vidit: You mention a \"filtered cell complex\". The book \"Nonabelian algebraic topology\" (2011) see my web page, pages.bangor.ac.uk/~mas010/nonab-a-t.html , builds algebraic topology directly from filtered spaces, via homotopically defined functors. Problem 16.1.17 is about relating these methods to Morse Theory. ",
"Johannes: somewhere in between. I'd call it concretely-motivated curiosity. My thesis work involved simplifying a filtered cell complex via discrete Morse theory in a way that preserved algebraic topological invariants (persistent homology groups). As such, I find the process of constructing massive chain groups and then taking a huge quotient somewhat inefficient, and was wondering if there is a \"compact representation\" (both words used non-technically) of some object in the category of topological spaces itself from which one could derive the invariants whenever necessary.",
"Where did you meet this structure? In a concrete problem or just out of curiosity?"
] | 9 |
Science
| 0 |
135
|
mathoverflow
|
Fake CM elliptic curves
|
Suppose one has an elliptic curve $E$ over $\mathbb{Q}$ with conductor $N < k^3$ for some (large) positive $k$, with the property that its Fourier coefficients satisfy $$ a_p=0, \; \mbox{ for all } \; p \equiv -1 \mod{4}, \; \; \; k/3 < p < k. $$ Can one prove unconditionally that $E$ has CM? This follows from work of Serre (under GRH) or Elkies (under something like Szpiro's conjecture), since otherwise we'd have a surplus of supersingular primes. It does not appear to be a consequence of, say, Serre's argument without additional hypotheses (though I'd be happy to be wrong on this score).
|
https://mathoverflow.net/questions/49937/fake-cm-elliptic-curves
|
[
"nt.number-theory",
"elliptic-curves"
] | 22 | 2010-12-20T00:09:05 |
[
"Dear BCrd : I've pondered that, but I really don't know. My admittedly uneducated guess would be \"probably not\" with the level of precision I'm after....",
"Dear Mike: Can one make \"effective\" (in the spirit of your question) any information coming from the knowledge that Sato-Tate holds in the non-CM case? (Sorry to answer a question with another question...)."
] | 2 |
Science
| 0 |
136
|
mathoverflow
|
Is the equivariant cohomology an equivariant cohomology?
|
Suppose a finite group $G$ acts piecewise linearly on a polyhedron $X$. Then there are two kinds of equivariant cohomology (or homology).
$\bullet$ With coefficients in a $\Bbb Z G$-module $M$. A reference is K. Brown's "Cohomology of groups". Namely, $H^\ast_G(X;M)=H^\ast(Hom_{\Bbb Z G}(C_\ast,M))$, where $C_\ast$ is the simplicial chain complex of a suitable invariant triangulation of $X$. Note that $H^\ast_G(X;M)\simeq H^\ast_G(X\times EG;M)$. Generalized cohomology (including stable cohomotopy) of $X$ with coefficients in a module can be defined via generalized cohomology of $X\times_G EG$ with local coefficients, which in turn is defined in the last chapter of the Buoncristiano-Rourke-Sanderson book.
$\bullet$ Representation-graded. A reference is "Equivariant homotopy and cohomology theory" by J. P. May et al. (but to simplify matters, I'm thinking of trivial coefficients, as in [Segal](http://www.mathunion.org/ICM/ICM1970.2/Main/icm1970.2.0059.0064.ocr.pdf) and [Kosniowski](http://resolver.sub.uni-goettingen.de/purl?GDZPPN002309874) rather than coefficients in a Mackey functor). Instead of defining $H^*_G$ let me only say that there is a Hurewicz homomorphism $\omega^*_G(X)\to H^*_G(X)$, and the equivariant stable cohomotopy $\omega_G^{V-W}(X)$ can be defined as the equivariant homotopy set $[S^{W+U}\wedge X,S^{V+U}]_G$, where $U$ is a sufficiently large (with respect to the partial order by inclusion) finite-dimensional $\Bbb RG$-submodule of $\Bbb R G\oplus\Bbb RG\oplus\dots$, and the "representation sphere" $S^U$ is the one-point compactification of $U$ (viewed as a Euclidean space with an action of $G$).
What relations are there between these two kinds of (e.g. ordinary) equivariant cohomology? More specifically,
> is cohomology with coefficients in a $\Bbb ZG$-module a special case of representation-graded cohomology?
A related (equivalent?) question is whether cohomology with coefficients in a (nontrivial!) module is representable in some reasonable sense.
For instance, by reducing both sides to non-equivariant cohomology, I know that $H^m_{\Bbb Z/2}(X;I^{\otimes m})\simeq H_{\Bbb Z/2}^{mT}(X_+)$, where $\Bbb Z/2$ acts freely on $X$, $X_+=X\sqcup$(basepoint), $I$ is the augmentation ideal of $\Bbb Z[\Bbb Z/2]$ (so $I^{\otimes m}$ is $\Bbb Z$ if $m$ is even and $I$ if $m$ is odd) and $T$ is the augmentation ideal of $\Bbb R[\Bbb Z/2]$ (so $mT$ is $\Bbb R^m$ with the antipodal involution). Something of this kind can also be done for stable cohomotopy. However this all depends on a twisted Thom isomorphism which I don't know for more general modules.
|
https://mathoverflow.net/questions/68330/is-the-equivariant-cohomology-an-equivariant-cohomology
|
[
"at.algebraic-topology",
"equivariant-homotopy",
"equivariant",
"group-cohomology"
] | 22 | 2011-06-20T17:42:25 |
[
"When the action is free, $H^n_G(X;M)\\simeq [X,K(M,n)]_G$; the relative case looks a bit more complicated, $H^n_G(X,Y;M)\\simeq [(X,Y),(K(M,n)\\times BG,BG)]_G$ (see VI.3.11 in the Goerss-Jardine book and projecteuclid.org/euclid.ijm/1256052280). On the other hand, for every $G$-spectrum $E$, $h^{V-W}_G(X):=[S^W\\wedge X,S^V\\wedge E]_G$ is an equivariant cohomology theory. So we want to find for an $M$ e.g. a $V$ such that $H^{dim V+1}_G(S^V\\wedge B|M|)\\simeq M$, where $|M|$ is $M$ with the trivial action of $G$. This is indeed possible for $G=\\Bbb Z/2$ but not in general.",
"Another reference for generalized cohomology with local coefficients is K. Brown's paper, ams.org/journals/tran/1973-186-00/S0002-9947-1973-0341469-9"
] | 2 |
Science
| 0 |
137
|
mathoverflow
|
bound on the genus of a fiber of the Albanese map of a surface with $h^1({\mathcal O})=1$?
|
This is maybe more an open problem than a question, since I have seriously thought about it and asked several people working on algebraic surfaces with no success. I hope somebody here can suggest an approach different from the standard arguments in surface theory.
BACKGROUND: let $X$ be a smooth minimal complex projective surface of general type. An _irrational pencil_ is a morphism with connected fibers $f\colon X\to B$, with $B$ a smooth curve of genus $b>0$.
For $b>1$, $X$ has at most finitely many pencils of genus $b$, having such a pencil is a topological property and it is possible to bound explicitly the genus of a general fiber of $f$ in terms of $K^2_X$ (Arakelov' theorem).
For $b=1$, namely for _elliptic pencils_ , things are very different in general: a surface can have infinitely many such pencils, the genus of the general fibers of these pencils can be unbounded, and it is possible that a surface with an elliptic pencil deforms to a surface without elliptic pencils.
However, if $h^1({\mathcal O}_X)=1$, then the Albanese map $a\colon X\to Alb(X)$ is an elliptic pencil, and for fixed $K^2$ the genus of a general fiber of $a$ is bounded, since the moduli space of surfaces with fixed $K^2$ is quasiprojective.
QUESTION: can one give a bound for the genus of the general fiber of the Albanese pencil of a minimal surface of general type $X$ with $h^1({\mathcal O}_X)=1$ in terms of $K^2_X$? Such a bound would be very interesting in the fine classification of surfaces of general type.
|
https://mathoverflow.net/questions/49169/bound-on-the-genus-of-a-fiber-of-the-albanese-map-of-a-surface-with-h1-mathc
|
[
"ag.algebraic-geometry",
"algebraic-surfaces"
] | 22 | 2010-12-12T12:12:39 |
[
"If $K_S-F$ is effective, than of course I have a bound since $K_S(K_S-F)\\ge 0$ and $K_SF=2g(F)-2$. And by the same argument, I would have a bound if I could determine explicitly an $m$ such that $mK_S-F$ is effective, but that's precisely what I don't know how to do.",
"I GUESS in some sense you can make it. For example, you can consider the canonical map of the surface, which will work if $K_S^2$ is not so small. Assume that the fiber of the Albanese map is $F$, then consider whether $\\mathscr{O}(K_S-F)$ has global section or not. If it has a global section, then after computing some intersection number, maybe you can have a bound of $g(F)$."
] | 2 |
Science
| 0 |
138
|
mathoverflow
|
Row of the character table of symmetric group with most negative entries
|
The row of the character table of $S_n$ corresponding to the trivial representation has all entries positive, and by orthogonality clearly it is the only one like this.
Is it true that for $n\gg 0$, the row of the character table of $S_n$ corresponding to the sign representation has the most negative entries of any row (about half of them)?
Note that for $S_4$, in addition to the sign representation row there are two other rows which also have 2 negative entries.
Possibly the fact that row sums are positive is relevant here.
**EDIT** : see also <https://realopacblog.wordpress.com/2025/05/02/maximizing-negative-entries-in-symmetric-group-character-tables/>
|
https://mathoverflow.net/questions/420865/row-of-the-character-table-of-symmetric-group-with-most-negative-entries
|
[
"co.combinatorics",
"gr.group-theory",
"rt.representation-theory",
"symmetric-groups",
"characters"
] | 21 | 2022-04-22T09:54:56 |
[
"@PerAlexandersson: I don't see any obvious connection to that conjecture.",
"Is this related perhaps to the conjecture that about half of the partition numbers, p(n) are even, as n goes to infinity?",
"@DenisSerre: the sum of the row corresponding to the sign representation is oeis.org/A000700. This quantity grows much slower than $p(n)$, hence why I said “about half.”",
"Is it clear/known that \"about half of conjugacy classes have negative signature\" ?",
"Some questions of a somewhat similar flavor that have been considered lately include whether most entries in the character table are highly divisible, or in fact are equal to $0$ (see e.g. arxiv.org/abs/2010.12410).",
"@FedorPetrov: each conjugacy class. The character table is a $p(n)\\times p(n)$ table, where $p(n)$ is the number of partitions of $n$.",
"Do we count each permutation once or each conjugacy class once?"
] | 7 |
Science
| 0 |
139
|
mathoverflow
|
Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?
|
_(Copied[from MSE](https://math.stackexchange.com/questions/4472850/can-a-4d-spacecraft-with-just-a-single-rigid-thruster-achieve-any-rotational-v). Offering four bounties over time, I got no response, other than twenty-nine upvotes.)_
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](https://space.stackexchange.com/a/59430) to [Could this three-thruster spherical spacecraft de-tumble itself with zero final velocity?](https://space.stackexchange.com/questions/53590) 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 at [Chiral Anomaly's answer](https://physics.stackexchange.com/a/471197/247726) to [Rigid body dynamics derivation from Newton's laws for higher dimensions](https://physics.stackexchange.com/questions/471106). 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 $\\{M,\Omega(t)\\}$ changes by a positive multiple of $T$.
* * *
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).
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 is $L(t)=\\{M,\Omega(t)\\}$. Its 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: $(m_1+m_2)\omega_{12}\,f\,\tau_{34}+\dotsb$ .)
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 terms of $L$ instead of $\Omega$ (with $\ell_{ij}=(m_i+m_j)\omega_{ij}$), those constants are
$$ \ell_{12}^2+\ell_{13}^2+\ell_{23}^2+\ell_{14}^2+\ell_{24}^2+\ell_{34}^2, $$$$ \left(\tfrac1{m_1+m_2}\right)\ell_{12}^2+\left(\tfrac1{m_1+m_3}\right)\ell_{13}^2+\left(\tfrac1{m_2+m_3}\right)\ell_{23}^2+\left(\tfrac1{m_1+m_4}\right)\ell_{14}^2+\left(\tfrac1{m_2+m_4}\right)\ell_{24}^2+\left(\tfrac1{m_3+m_4}\right)\ell_{34}^2, $$$$ \ell_{12}\ell_{34}-\ell_{13}\ell_{24}+\ell_{23}\ell_{14}. $$
We can assume $m_1>m_2>m_3>m_4>0$. Taking the squared magnitude of $L$, and subtracting $(m_1+m_3)$ or $(m_2+m_4)$ times the doubled energy, we get
$$ \left(\tfrac{m_2-m_3}{m_1+m_2}\right)\ell_{12}^2-\left(\tfrac{m_1-m_2}{m_2+m_3}\right)\ell_{23}^2-\left(\tfrac{m_3-m_4}{m_1+m_4}\right)\ell_{14}^2-\left(\tfrac{m_1-m_2+m_3-m_4}{m_2+m_4}\right)\ell_{24}^2-\left(\tfrac{m_1-m_4}{m_3+m_4}\right)\ell_{34}^2, $$$$ \left(\tfrac{m_1-m_4}{m_1+m_2}\right)\ell_{12}^2+\left(\tfrac{m_1-m_2+m_3-m_4}{m_1+m_3}\right)\ell_{13}^2+\left(\tfrac{m_3-m_4}{m_2+m_3}\right)\ell_{23}^2+\left(\tfrac{m_1-m_2}{m_1+m_4}\right)\ell_{14}^2-\left(\tfrac{m_2-m_3}{m_3+m_4}\right)\ell_{34}^2, $$
both being constant. Notice the signs of the coefficients; the first expression has only one positive term, so (if the expression is $0$) it defines a cone, or a pair of opposite cones, around the $\ell_{12}$ axis in 6D. And the second expression defines a pair of opposite cones around the $\ell_{34}$ axis. If the torque $T$ is inside one of these cones, and the initial angular momentum $L_0$ is in the same cone, then $L$ stays in that cone, and there's no solution with the final angular momentum $L_1$ outside of the cone. Thus, we must take $T$ to be outside of all four of these cones. That is, the first expression (with $\tau_{ij}$ in place of $\ell_{ij}$) must be negative, and the second expression 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 $L'=L\odot L$.)
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
$$\sum^6\mathbf a^4=((\mathbf a\mathbf a)\mathbf a)\mathbf a+(\mathbf a(\mathbf a\mathbf a))\mathbf a+2(\mathbf a\mathbf a)(\mathbf a\mathbf a)+\mathbf a((\mathbf a\mathbf a)\mathbf a)+\mathbf a(\mathbf a(\mathbf a\mathbf a))$$ $$\newcommand{\aaaa}[3]{\mathbf a\underset{#1}\odot\mathbf a\underset{#2}\odot\mathbf a\underset{#3}\odot\mathbf a} =\aaaa{1}{2}{3}+\aaaa{2}{1}{3}\begin{matrix}{}+\aaaa{1}{3}{2} \\\ {}+\aaaa{2}{3}{1}\end{matrix}+\aaaa{3}{1}{2}+\aaaa{3}{2}{1}.$$
(In the $1$-dimensional algebra, the product is associative, and the power series simplifies to $\sum_nt^na^{n+1}=a/(1-at)$.)
This converges absolutely, 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://mathoverflow.net/questions/448773/can-a-4d-spacecraft-with-just-a-single-rigid-thruster-achieve-any-rotational-v
|
[
"ca.classical-analysis-and-odes",
"ds.dynamical-systems",
"mp.mathematical-physics",
"differential-equations"
] | 21 | 2023-06-13T06:58:24 |
[
"Re, I had not meant to make any adverse changes, and didn't even realise that it made a difference in the rendering. My apologies. I have changed back to back-to-back $$ $$ environments.",
"@LSpice - Now the displayed equations are too closely spaced."
] | 2 |
Science
| 0 |
140
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mathoverflow
|
Reference request: deforming a G-local system to a variation of Hodge structure
|
Let $X$ be a smooth connected quasiprojective variety over $\mathbb{C}$ and let $G$ be a complex reductive group. Let $$\iota: G\to GL_N$$ be a representation and let $$\rho: \pi_1(X(\mathbb{C}))\to G(\mathbb{C})$$ be a homomorphism. I am looking for a reference for the following fact:
> $\rho$ is deformation-equivalent to a homomorphism $$\rho': \pi_1(X(\mathbb{C}))\to G(\mathbb{C})$$ such that the local system on $X$ associated to $\iota\circ \rho'$ underlies a polarizable complex variation of Hodge structure.
If $X$ is projective, this is due to Simpson [1, Theorem 3]; in the quasiprojective case some special cases have been written down by T. Mochizuki, e.g. if $G=GL_N$ [2, Theorem 10.5] or if $\rho$ is rigid with Zariski-dense image [2, Lemma 10.13].
I think this can probably be extracted with some difficulty from existing literature, but I'm hoping the statement has been written down explicitly somewhere.
EDIT: If it helps, I'm primarily interested in the case $G=PGL_n$.
[1] _Simpson, Carlos T._ , [**Higgs bundles and local systems**](http://dx.doi.org/10.1007/BF02699491), Publ. Math., Inst. Hautes Étud. Sci. 75, 5-95 (1992). [ZBL0814.32003](https://zbmath.org/?q=an:0814.32003).
[2] _Mochizuki, Takuro_ , [**Kobayashi-Hitchin correspondence for tame harmonic bundles and an application**](http://www.arxiv.org/abs/math/0411300), Astérisque 309. Paris: Société Mathématique de France (ISBN 978-2-85629-226-6/pbk). viii, 117 p. (2006). [ZBL1119.14001](https://zbmath.org/?q=an:1119.14001).
|
https://mathoverflow.net/questions/413301/reference-request-deforming-a-g-local-system-to-a-variation-of-hodge-structure
|
[
"ag.algebraic-geometry",
"complex-geometry",
"hodge-theory"
] | 21 | 2022-01-06T11:59:59 |
[
"@SashaP: actually I think this should probably be false over infinite finitely generated fields; I have an idea for a counterexample conditional on Fontaine-Mazur, but it’s maybe too involved for an MO comment.",
"@SashaP: This is a nice question! I think I see an argument for the analogous statement over finite fields (i.e. working in a component of the rigid generic fiber of the deformation ring of a given residual rep of the geometric fundamental group). Plausibly I see an argument over p-adic fields as well, but I need to check details...send me an email and let's discuss further? I don't yet see how to do it over finitely generated infinite fields, which is I think the most interesting case...",
"Apologies for an off-topic comment, but do you know if the analogous statement (say, at least for $GL_n$) is true for 'arithmetic' in place of 'underlies a polarizable VHS'?",
"@MoisheKohan: No worries!",
"I see, I just did not read the OP carefully.",
"@MoisheKohan: As I remark in the question, Mochizuki has proven a quasiprojective version, just not for arbitrary $G$...",
"I remember that Simpson proved this for projective varieties. I do not remember seeing a quasiprojective version."
] | 7 |
Science
| 0 |
141
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mathoverflow
|
A multiple integral
|
Let us consider the multiple integral $$I_{n}=\int_{-\infty }^{\infty }ds_{1}\int_{-\infty}^{s_{1}}ds_{2}\cdots \int_{-\infty }^{s_{2n-1}}ds_{2n}\;\cos {(s_{1}^{2}-s_{2}^{2})}\;\cdots \cos {(s_{2n-1}^{2}-s_{2n}^{2})}.$$
Using an indirect method (Landau-Zener formula from the theory of molecular scattering), we had shown in <http://arxiv.org/abs/1201.1975> that $$I_n=\frac{2}{n!}\left(\frac{\pi}{4}\right)^{\,n}.$$ I'm very interested how to get this result by direct mathematical (preferably simple) methods, without using the Landau-Zener formula. Can anyone help?
Actually, I asked similar question an year ago [Multiple Integral (American Mathematical Monthly problem 11621 and its generalization)](https://mathoverflow.net/questions/125237/multiple-integral-american-mathematical-monthly-problem-11621-and-its-generali) but have not got an answer.
|
https://mathoverflow.net/questions/172278/a-multiple-integral
|
[
"ca.classical-analysis-and-odes",
"real-analysis",
"integration"
] | 21 | 2014-06-20T04:15:04 |
[] | 0 |
Science
| 0 |
142
|
mathoverflow
|
Bounding failures of the integral Hodge and Tate conjectures
|
It is well know that the integral versions of the Hodge and Tate conjectures can fail. I once heard an off hand comment however that they should only fail by a "bounded amount". My question is what this means and whether this can be made more precise.
> > Let $\pi: X \to B$ be a smooth projective morphism of smooth varieties over $\mathbb{C}$. Assume that the (usual rational) Hodge conjecture holds for the fibre $B_x$ over every point $x \in B(\mathbb{C})$. Does there exist an integer $d$ such that for any integral Hodge class $h$ on $B_x(\mathbb{C})$, the class $dh$ is represented by an algebraic cycle for all $x \in B(\mathbb{C})$?
Of course the point of this question is whether $d$ can be chosen uniformly with respect to the family.
As a brief remark, it is quite easy to prove this when $B={\mbox{Spec}}\ \mathbb{C}$. Indeed, by assumption the algebraic classes form a subgroup of finite index inside the Hodge classes!
I am also similarly interested in the analogous question for the integral Tate conjecture for families of varieties over number fields and finite fields.
|
https://mathoverflow.net/questions/168997/bounding-failures-of-the-integral-hodge-and-tate-conjectures
|
[
"ag.algebraic-geometry",
"nt.number-theory",
"arithmetic-geometry",
"hodge-theory"
] | 21 | 2014-06-04T01:36:23 |
[
"It seems unlikely that any of the known counterexamples to the integral Hodge conjecture would give a negative answer to your question. On the other hand, there seems to be no reason to expect a positive answer since the dimension of the space of Hodge classes could jump over infinitely many subvarieties of $B$. Interesting!"
] | 1 |
Science
| 0 |
143
|
mathoverflow
|
Is the mapping class group of $\Bbb{CP}^n$ known?
|
In his paper ["Concordance spaces, higher simple homotopy theory, and applications"](https://pi.math.cornell.edu/~hatcher/Papers/ConcordanceSpaces.pdf), Hatcher calcuates the smooth, PL, and topological mapping class groups of the $n$-torus $T^n$. This requires an understanding of the surgery structure set of the $n$-torus, as well as the space of self-equivalences (easy, because the torus is aspherical and a Lie group).
At least $\pi_0$ and $\pi_1$ of $\text{hAut}(\Bbb{CP}^n)$, the space of self-homotopy equivalences, are known: the former is $\Bbb Z/2$ because $[\Bbb{CP}^n, \Bbb{CP}^n] = [\Bbb{CP}^n, \Bbb{CP}^\infty] = \Bbb Z$, and the only invertible elements are $\pm 1$; further the homology of this space seems to have been calculated by Sasao, who gives that $\pi_1 \text{hAut} = \Bbb Z/(n+1)$.
The best references I know for the surgery structure set of $\Bbb{CP}^n$ are Wall's book on surgery and the Madsen-Milgram book on Top/PL cobordism groups "The classifying spaces for surgery and cobordism of manifolds", neither of which seem to give a completely explicit description.
From this, it's not immediately clear to me whether enough is known to run Hatcher's argument for $T^n$ on complex projective space.
Has anybody computed the various mapping class groups of $\Bbb{CP}^n$ when $n \geq 3$? Maybe in any specific examples, or when $n$ is very large? If not, is it out of reach with current technology?
(I would also be interested in other infinite families of high-dimensional manifolds, like real projective spaces and lens spaces, but I imagine those require strictly more work to understand, given the presence of non-trivial fundamental group without being aspherical like the torus is.)
|
https://mathoverflow.net/questions/349319/is-the-mapping-class-group-of-bbbcpn-known
|
[
"reference-request",
"at.algebraic-topology",
"gt.geometric-topology",
"differential-topology",
"mapping-class-groups"
] | 21 | 2019-12-29T07:40:43 |
[
"Regarding your last comment: Wall (MR0156354, MR0177421) and Kreck (MR0561244) have computed various mapping class groups of highly connected manifolds up to extension problems.",
"Cool, I wasn't aware of that paper!",
"@skupers Kreck points out to me that the oriented smooth mapping class group $\\pi_0 \\text{Diff}^+(\\Bbb{CP}^3)$ was calculated by Brumfiel to be $\\Bbb Z/4$, all of whose elements are represnted by diffeomorphisms supported in a ball. I did not check but presumably $\\pi_0 \\text{Diff}$ is the dihedral group on 8 elements.",
"Kreck and Su have announced a paper containing the case n=3, see Remark 1.4 of arxiv.org/abs/1907.05693. You could try asking one of them.",
"Only small homotopy groups seem relevant in Hatcher's computation, though, so I hope that doesn't cause too much trouble. Thanks for the reference!",
"There are spectral sequences developed by Schultz, and Becker-Schultz that converge to homotopy groups of the identity component of the space of self-maps of $CP^n$. See section 6 in arxiv.org/abs/0912.4874 and references therein. The computational difficulties are substantial, e.g., in the linked paper we only manage to get partial information on $\\pi_7$."
] | 6 |
Science
| 0 |
144
|
mathoverflow
|
Cauchy matrices with elementary symmetric polynomials
|
$\newcommand{\vx}{\mathbf{x}}$ Let $e_k(\vx)$ denote the [elementary symmetric polynomial](http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial), defined for $k=0,1,\ldots,n$ over a vector $\vx=(x_1,\ldots,x_n)$ by
\begin{equation*} e_k(\vx) := \sum_{1 \le l_1 < l_2 < \cdots < l_k \le n} x_{l_1}x_{l_2}\cdots x_{l_k}. \end{equation*}
To avoid messy conjugates in what follows, let me limit $x$ to be a real vector.
[**EDIT**]: Changed notation below to highlight that size of $M_k$ is independent of size of the vectors being used to define it.
> **Definition** (GC-matrix) Let $(\vx^1,\ldots,\vx^m)$ be vectors in $\mathbb{R}^{n}$ such that each component $|x^p_j| < 1$ (for $1\le p \le m$, $1\le j \le n$). Denote by $\vx^i \circ \vx^j$ the vector formed by taking elementwise products; also, let $\mathbf{1}$ denote the length $n$ vector of all ones.
>
> Define now the $m\times m$ _Generalized Cauchy (GC) matrix_ $M_k$, for $1\le k\le n$, as \begin{equation*} M_k := \left[\frac{1}{e_k(\mathbf{1}-\vx^i\circ \vx^j)}\right]_{i,j=1}^m. \end{equation*}
**Example** For $e_1(\vx)=\sum_lx_l$, the GC matrix above essentially (up to scaling) boils down to a standard [Cauchy matrix](http://en.wikipedia.org/wiki/Cauchy_matrix), of the form \begin{equation*} \left[\frac{1}{1-\lambda_i\lambda_j}\right], \end{equation*} for suitable scalars $\lambda_i$; while for $e_n(\vx)=\prod_l x_l$, the GC matrix reduces to a Hadamard product of ordinary Cauchy matrices.
> **Question.** The above example shows that $M_1$ and $M_n$ are positive semidefinite (since the involved Cauchy matrices are). So my question is there a slick representation for $e_k$ that I could use to conclude positive semidefiniteness of $M_k$?
* * *
**PS:** I suspect the answer follows immediately from well-known (in the correct circles) results; but I need some MO help here to put me rapidly in the correct direction!
|
https://mathoverflow.net/questions/114101/cauchy-matrices-with-elementary-symmetric-polynomials
|
[
"matrices",
"rt.representation-theory",
"linear-algebra",
"co.combinatorics"
] | 21 | 2012-11-21T13:01:30 |
[
"Unfortunately, this question has seems to have a negative answer (as proved to me very recently by one of the mathematicians with whom I discussed it; will update this question once I get a chance.)"
] | 1 |
Science
| 0 |
145
|
mathoverflow
|
p-groups as rational points of unipotent groups
|
Is it true that every finite p-group can be realized as the group of rational points over $\mathbb{F_p}$ of some connected unipotent algebraic group defined over $\mathbb{F_p}$? For abelian p-groups, the answer is yes via Witt vectors, but is it true in general?
|
https://mathoverflow.net/questions/69397/p-groups-as-rational-points-of-unipotent-groups
|
[
"p-groups",
"algebraic-groups"
] | 21 | 2011-07-03T06:46:19 |
[
"Sorry, after a bit more thought, it occurs to me that, since $p$-groups have centres, it's OK to have the Abelian group as sub-object. Then Theorem 1.8(c) of the paper by Kumar and Neeb shows that $\\operatorname{Ext}^1_{\\text{alg}}(G, \\mathbb G_a) \\cong H^2(\\mathfrak g, \\mathfrak{gl}_1)^{\\mathfrak g}$; but I still can't see my way through to showing that the necessary map is surjective.",
"L. Spice, do you want to replace $GL_1$ with $\\mathbb{G}_a$?",
"By using a filtration of your $p$-group with all successive quotients $\\mathbb F_p$, you can reduce the problem to showing that $\\operatorname{Ext}^1_{\\text{alg}}(\\operatorname{GL}_1, G) \\to \\operatorname{Ext}^1_{\\mathbb Z}(\\mathbb F_p, G(\\mathbb F_p))$ is surjective for all connected, unipotent $\\mathbb F_p$-groups $G$. A quick Google search turned up “Extensions of algebraic groups” by Kumar and Neeb (#48 at math.unc.edu/Faculty/kumar), but the Abelian group by which you're extending there is the subobject, not the quotient."
] | 3 |
Science
| 0 |
146
|
mathoverflow
|
Deciding whether a given power series is modular or not
|
The degree 3 modular equation for the Jacobi modular invariant $$ \lambda(q)=\biggl(\frac{\sum_{n\in\mathbb Z}q^{(n+1/2)^2}}{\sum_{n\in\mathbb Z}q^{n^2}}\biggr)^4 $$ is given by $$ (\alpha^2+\beta^2+6\alpha\beta)^2-16\alpha\beta\bigl(4(1+\alpha\beta)-3(\alpha+\beta)\bigr)^2=0, $$ where $\alpha=\lambda(q)$ and $\beta=\lambda(q^3)$. This has a very simple rational parametrization $$ \alpha=\frac{p(2+p)^3}{(1+2p)^3}, \qquad \beta=\frac{p^3(2+p)}{1+2p} $$ (with $p$ ranging from 0 to 1 as as $q$ changes in the range).
By certain heuristical reasons (which are hidden behind analysis in my [recent joint work](http://arxiv.org/abs/1012.3036)), I expect that modularity can occur in some other similar parameterizations. In particular, the expansions $$ \mu(q) = 4096q - 294912q^2 + 12238848q^3 - 379846656q^4 $$ $$ \+ 9737920512q^5 - 217011585024q^6 + 4333573472256q^7 - 79091807551488q^8 $$ $$ \+ 1337378422542336q^9 - 21157503871942656q^{10} + 315428695901356032q^{11} $$ $$ \- 4455786006742302720q^{12} + 59885350975571779584q^{13} + O(q^{14}) $$ and $$ p = 4q + 12q^2 - 48q^3 + 156q^4 - 12q^5 - 6576q^6 + 78144q^7 - 607812q^8 $$ $$ \+ 3017364q^9 + 156q^{10} - 208502832q^{11} + 2876189520q^{12} - 24837585384q^{13} + O(q^{14}) $$ (which, of course, can be further extended) satisfy $$ \mu(q)=\frac{p(4+p)^5}{(1+4p)^5} \quad\text{and}\quad \mu(q^5)=\frac{p^5(4+p)}{1+4p}. $$ (These relations define $p$ and $\mu$ in a unique way.)
Is there any way to identify $\mu(q)$ with a known modular function, or to show that $\mu(q)$ is not modular at all?
Thanks! Best wishes already from 2011.
|
https://mathoverflow.net/questions/50804/deciding-whether-a-given-power-series-is-modular-or-not
|
[
"nt.number-theory",
"modular-forms"
] | 21 | 2010-12-31T05:11:02 |
[
"@Kevin, thanks for this hint. Because I have no guess about the index of the underlying group in $\\Gamma(1)$, I am not sure that $O(q^{1000})$ would be enough. I didn't try to expand so far (the coefficients grow extremely fast), but what I did (up to $O(q^{50})$) was verifying a possible algebraic relation between $\\mu(q)$ and the classical $j$-invariant. None was found...",
"If $\\mu$ were modular then presumably there would be an algebraic relation between $\\mu(q)$ and $\\mu(q^n)$ for all positive integer values of $n$, the relation being of degree something like the index of $\\Gamma_0(n)$ in $SL(2,Z)$ (but perhaps this isn't precisely right---the exact degree would depend on the level of $\\mu$). So you could expand $\\mu$ out to $O(q^1000)$ and then it would be easy to search for these relations. If it doesn't work out, i.e. if $n=5$ is OK but the others don't seem to come out, then this is evidence to suggest that $\\mu$ isn't modular."
] | 2 |
Science
| 0 |
147
|
mathoverflow
|
Schemification (schematization?) of locally ringed spaces
|
**Motivation:**
Say $F: D \to Sch$ is a diagram in the category of schemes, and we're interested in whether it has a [colimit](http://en.wikipedia.org/wiki/Limit_\(category_theory\)#Colimits_2) (gluings, pushouts, and "categorical" quotients are all examples of colimits). Its colimit $Q$ in the category of locally ringed spaces always exists, and a scheme $Y$ is the scheme-colimit of $F$ iff there is a map $Q\to Y$ that is "initial among maps to schemes", i.e. any other map from $Q$ to a scheme factors through it.
Thus the problem of when colimits of schemes exists can be answered if we know when a locally ringed space admits a "schemification".
**Examples:**
1) The process of turning a classical variety $V$ into a "variety with generic points" $V^s$ is a schemification. Proving this boils down the classical affine case. If $R$ is a finite-type reduced $k=\overline{k}$-algebra, and we map $V=mSpec(R)$ to a scheme $Y$, first cover $Y$ by affines $Spec(B_i)$, and then pull back this cover to $V$ and refine it by principal opens $mSpec(R_{f_{ij}})$, so we have maps $B_i \to R_{f_{ij}}$ that determine the map $V\to Y$ (by the adjunction of $Spec({\cal O}(-))$ to the inclusion $AffSch\hookrightarrow LRS)$. But these define the desired map $V^s\to Y$ because $V^s$ can be obtained by gluing $Spec(R_{f_{ij}})$.
2) For $k=\overline{k}$, if we let $\mathbb{G_m}=\mathbb{A}^1_k\setminus 0$ act on $\mathbb{A}^1_k$, then the coequalizer in schemes of the action and the projection $G_m \times_k \mathbb{A}^1_k \rightrightarrows \mathbb{A}^1_k$ (i.e. the "categorical quotient") is $Spec(k)$, hence the locally ringed space coequalizer $Q$, which has two points and is not a scheme, has $Spec(k)$ as its schemification.
3) Some locally ringed spaces have no schemification; for example, two affine lines glued along their generic points. This is precisely because the gluing diagram has no coequalizer in schemes, as per BCnrd and Anton Geraschenko's [answer here](https://mathoverflow.net/questions/9961/colimits-of-schemes/23966#23966).
> When does a locally ringed space $X$ admit a "schemification", i.e. a map to a scheme $X^s$ through which any other map to a scheme must factor?
**EDIT:** (Response to Martin's answer)
4) Any locally ringed space $X$ with exactly one closed point $x$ has its "affinization", $X^a:=Spec({\cal O}_X(X))$, as a schemification. This is because in a map $F:X\to Y$ with $Y$ a scheme, every point maps to a generization of $f(x)$, so and since open sets are closed under generization, $f$ factors through an affine neighborhood of $f(x)$. But any map from $X$ to an affine factors uniquely through $X^a$, so we're done.
5) Schemification commutes with disjoint unions, in that if $X= \coprod X_i$ then $X^s$ exists iff all $X_i^s$ exist, and in that case $X^s = \coprod X_i^s$. So _if $X$ is locally connected_ , since it then decomposes into a disjoint union of connected clopen components, we might as well assume it is connected.
|
https://mathoverflow.net/questions/60524/schemification-schematization-of-locally-ringed-spaces
|
[
"ag.algebraic-geometry",
"ct.category-theory"
] | 21 | 2011-04-03T23:40:12 |
[
"@Andrew: I'm thinking of something like a \"generalized GAGA\", but I'm not really sure what conditions you need on $(X,A)$. I think something like what I'm talking about is covered in chapter 8 of that book, but I really don't know the specifics.",
"@Buschi, Chapter IV seems to be about forming locally ringed topoi (or spaces) from ringed topoi (or spaces). If I understand, this construction is right adjoint to the inclusion $LRT \\to RT$. @Harry, Could you say more specifically what result you're talking about? I'm dubious, because I'd expect any ringed topos to be a relative scheme over itself...",
"@Andrew: There is some stuff in the next chapter (chapter 5) that might be useful to you (if you can show that your locally ringed topos is a relative scheme for some other ringed topos $(X,A)$, you have a universal scheme associated to it by way of a crazy adjunction induced by the universal map $(X,A)\\to (Set,Z)$), but I don't think it really addresses your question in any sort of concrete way.",
"I haven't. Should I? I just skimmed through a copy and didn't see anything about turning espaces annelés en anneaux locaux into schemas.",
"HAve you read CAP.IV of \"Topos Annelés et schemas relatifs\" by M. Hakim , Ergebn. Math. Grenzgeb. , 64 ? "
] | 5 |
Science
| 0 |
148
|
mathoverflow
|
Closed connected additive subgroups of the Hilbert space
|
It is a classical result that a closed and connected additive subgroup of $\mathbb{R}^n$ is necessarily a linear subspace. However, this is no longer true in infinite dimension: a very easy example is the subgroup $L^2(I,\mathbb{Z})$ of the real Hilbert space of all $L^2$ real valued functions on the unit interval $I:=[0,1].$
Indeed, any element $\phi$ of $L^2(I,\mathbb{Z})$ is connected to the origin by the path $\gamma:I\ni t\mapsto \phi\chi_{[0,t]}\in L^2(I,\mathbb{Z}),$ where $\chi_{[0,t]}$ is the characteristic function of the interval $[0,t].$ Actually, up to a reparametrization, this path is also $1/2$-Hölder continuous. (Indeed, if $\sigma:[0,1]\to\big[0,\ 1+ \|\phi\| _2^2\, \big]$ is the strictly increasing, surjective continuous map $t\mapsto t+\int_0^t \phi^2dx$, then $\|\gamma(t)-\gamma(t')\|_2\le|\sigma(t)-\sigma(t')|^{1/2}$, meaning that $\gamma\circ\sigma^{-1}$ is $1/2$-Hölder continuous).
So we may say that $L^2(I,\mathbb{Z})$ is even $1/2$-Hölder-path-connected, though it is certainly not a linear subspace.
It is also not hard to see that the Hölder exponent $1/2$ is critic: any closed subgroup $G$ of a Hilbert space $H$, which is connected by $\alpha$-Hölder paths, with $\alpha > 1/2,$ is necessarily a linear space. (Reason: as a consequence of the generalized parallelogram identity, it turns out that the lattice generated by $n$ vectors $g_1,\dots,g_n$ in $H$, with norms $\|g_k\|\leq r,$ is a $rn^{1/2}$-net in their linear span. In particular, if $\gamma:[0,1]\to G$ is an $\alpha$-Hölder path, for any $n\in\mathbb{N},$ the $n$ elements $g_{k,n}:=\gamma(\frac{k+1}{n})-\gamma(\frac{k}{n})\in G,\quad k=0,\dots,n-1$ are a $Cn^{1/2 - \alpha }$-net in their linear span. Since $G$ is closed this implies that it is a cone, hence a linear subspace).
I find this quite nice, but at this point some questions arise quite naturally. Let $H$ be the infinite dimensional real separable Hilbert space.
* Let $0 < \alpha < 1/2.$ Are there closed additive subgroups of $H$ which are connected by $\alpha$-Hölder paths, but not by $\beta$-Hölder paths for any $\beta >\alpha \, $?
* More generally: connected / non-connected w.r.to paths with given modulus of continuity? Are there closed, connected, not path-connected additive subgroups?
* Are these objects just pathologies/curiosities of the mathematical Zoo, or did anybody gave an application of them to functional analysis?
|
https://mathoverflow.net/questions/45322/closed-connected-additive-subgroups-of-the-hilbert-space
|
[
"hilbert-spaces",
"topological-groups",
"fa.functional-analysis"
] | 21 | 2010-11-08T08:48:41 |
[] | 0 |
Science
| 0 |
149
|
mathoverflow
|
Almost everywhere differentiability for a class of functions on $\mathbb{R}^2$
|
A while ago, I came across the following problem, which I was not able to resolve one way or the other.
> Let $f,g\colon\mathbb{R}^2\to\mathbb{R}$ be continuous functions such that $f(t,x)$ and $g(t,x)$ are Lipschitz continuous with respect to $x$ and increasing in $t$. Is it necessarily true that the partial derivative $f_x=\partial f/\partial x$ exists almost everywhere with respect to the measure $\iint\cdot\,d_tg(t,x)dx$?
The notation $\int\cdot\,d_tg(t,x)$ refers to the Lebesgue-Stieltjes integral with respect to $g(t,x)$, for each fixed $x$. At first sight, this looks like a basic real-analysis problem for which it should be possible to either prove true or construct a simple counterexample. However, after playing around with this for a while, I was none the wiser.
The reason for the interest in this question comes from attempts to construct a particular decomposition of stochastic processes. Supposing that $X$ is a continuous martingale, it can be useful to decompose $f(t,X_t)$ into a martingale term and a 'drift' component, $$ \begin{align} f(t,X_t)=\int_0^tf_x(s,X_s)\,dX_s+A_t.&&{\rm(1)} \end{align} $$ For this expression to be well-defined, it is necessary that $f_x$ exists outside of a null set. That is, $\int1_{\\{f_x(s,X_s)\rm\ doesn't\ exist\\}}dX_s$ should be zero. However, letting $g(t,x)=\mathbb{E}[\vert X_t-x\vert]$, it can be shown that the expected value of the square of this integral is $\iint1_{\\{f_x\rm\ doesn't\ exist\\}}d_tg(t,x)dx$. So a positive answer to my question would mean that decomposition (1) is well-defined. Furthermore, it can then be shown that $A$ has zero [quadratic variation](http://en.wikipedia.org/wiki/Quadratic_variation), so that $f(t,X_t)$ is what is sometimes referred to as a _Dirichlet process_ (although this term does also have a completely different meaning). Decomposition (1) was used to prove some results about the distribution of martingales, but I also published it as a stand-alone paper ([here](http://dx.doi.org/10.1214/09-AOP476)). However, not knowing an answer to the question above, this did require imposing the annoying additional condition that the left and right hand derivatives of $f(t,x)$ with respect to $x$ exist everywhere.
Some further points are in order.
* Lebesgue's differentiation theorem tells us that the partial derivative $f_x$ exists almost everywhere with respect to the Lebesgue measure. That does not help here though, because $\iint\cdot d_tg(t,x)dx$ need not be absolutely continuous. In fact, even restricting to the case where $g(t,x)$ is convex in $x$, it is possible to construct nontrivial examples where $\iint\cdot d_tg(t,x)dx$ is supported on the graph $\\{(t,\gamma(t))\colon t\in\mathbb{R}\\}$ of a (discontinuous) function $\gamma\colon\mathbb{R}\to\mathbb{R}$. One such example is given by $g(t,x)=\mathbb{E}[(M_t-x)_+]$ where $M$ is the martingale constructed in [this blog post](https://almostsure.wordpress.com/2016/10/05/a-martingale-which-moves-along-a-deterministic-path/). Another (similar) example is the function $f(t,x)$ constructed [here](https://almostsure.wordpress.com/2016/10/12/do-convex-and-decreasing-functions-preserve-the-semimartingale-property-a-possible-counterexample/).
* If the left and right hand partial derivatives of $f(t,x)$ with respect to $x$ exist everywhere, then the question has a positive answer.
* If the left and right hand partial derivatives of $g(t,x)$ with respect to $x$ exist everywhere, then $f_x$ exists in a weak sense. Given any bounded measurable $\theta\colon\mathbb{R}^2\to\mathbb{R}$ with compact support,$$\iint h^{-1}(f(t,x+h)-f(t,x))\theta d_tgdx\to\iint f_x\theta d_tgdx$$as $h\to0$.
* Rather than monotonicity in $t$, I am really interested in the weaker condition that $\iint\cdot\vert d_tf(t,x)\vert dx$ is locally finite ($\int\cdot\vert d_tf(t,x)\vert$ is the variation of $\int\cdot d_tf(t,x)$), and similarly for $g$. If I find a positive answer to this, then I think I would update my papers on arXiv to reflect this.
|
https://mathoverflow.net/questions/77957/almost-everywhere-differentiability-for-a-class-of-functions-on-mathbbr2
|
[
"ca.classical-analysis-and-odes",
"real-analysis"
] | 21 | 2011-10-12T14:15:06 |
[] | 0 |
Science
| 0 |
150
|
mathoverflow
|
Is the Dieudonne module actually a cohomology group?
|
One often times thinks of the Dieudonne module $M(X)$ of a $p$-divisible group (say over $k$, a perfect characteristic $p$ field) as being some sort of cohomology theory
$$M:\left\\{p\text{- divisible groups}/k\right\\}\to \left\\{F\text{-crystals }/k\text{ with slopes in }[0,1]\right\\}.$$
This is supported by the famous observation of Mazur-Messing-Oda that if $A/k$ is an abelian variety, then
$$M(A[p^\infty])=H^1_\text{crys}(A/W(k))$$
as $F$-crystals.
There are two natural questions that come from this, in my opinion. First, can one make sense of $M(X)$ as a crystalline cohomology group (in a literal sense) if $X$ 'comes from a scheme' (e.g. is the $p^\infty$-torsion of some group scheme). I've asked this [here](https://mathoverflow.net/questions/219015/extension-of-messing-mazur-oda-to-general-groups) as well as asking when $M$ (an $F$-crystal) 'comes from a scheme'.
But, the other question, and the one I am asking here is the following. Is there some way to interpret the understanding of the Dieudonne module 'as cohomology' in a rigorous way? For example, is there a natural site over $X$ (perhaps with underlying category $p$-divisible groups over $X$) such that $M(X)$ is cohomology on this site? I would even be interested in knowing if there is a rigorous way of considering $M(X)/pM(X)$ as a cohomology theory (in the Mazur-Messing-Oda setup it's $H^1_\text{dR}(A/k)$).
The obvious guess is to create a crystalline/infinitesimal site on $X$, but I don't really know how to make this precise.
|
https://mathoverflow.net/questions/235186/is-the-dieudonne-module-actually-a-cohomology-group
|
[
"arithmetic-geometry",
"crystalline-cohomology"
] | 21 | 2016-04-03T05:24:29 |
[
"Doesn't Mazur-Messing interpret the Dieudonne module of a p-divisible group G as rigidified extensions? This suggests that defining a category of G-invariant D-modules should be possible. I would naively guess that the resulting cohomology would be \"generated in degree 1\", just as for abelian varieties, though.",
"One wants not a site on $X$ but a site whose \"fundamental group\" is in some way \"dual\" to $X$. The simplest picture of this seems like it should be the quotient of a point by the Cartier dual of $X$."
] | 2 |
Science
| 0 |
151
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mathoverflow
|
Density of first-order definable sets in a directed union of finite groups
|
This is a generalization of the following [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) by John Wiltshire-Gordon.
Consider an inductive family of finite groups: $$ G_0 \hookrightarrow G_1 \hookrightarrow \ldots \hookrightarrow G_i \hookrightarrow \ldots $$
We may view each group as a subgroup of the next. Let $G$ be the directed union of the $G_i$: $$ G := \bigcup_{i=0}^\infty \ G_i $$
Now, $G$ is a countable group. Suppose we're asked a question like, "What proportion of the elements of $G$ are commutators?", or "What proportion of pairs $(g,h) \in G^2$ satisfy $g^2h^2=1$ but $g^2 \neq 1$?" We try to make sense of such questions in terms of _density_ on $G^n$, which is not defined for all subsets. For $E \subseteq G^n$, let $$ d(E) := \lim_{i \to \infty} \frac{|E \cap G_i^n|}{|G_i^n|}, $$ if the limit exists.
My question is: If $E$ is a subset of $G^n$ that is first-order definable in the language of groups, does the density of $E$ necessarily exist? John's [question](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) covers the (still unresolved) special case of when $E$ is defined by an atomic formula.
A first-order definable subset $E \subseteq G^n$ is the set of $n$-tuples of group elements where a particular first-order formula in $n$ free variables is true. Such a formula is a finite string of symbols and may involve multiplication, inversion, the identity element, equality, logical connectives ($\wedge$, $\vee$, $\neg$, $\implies$), quatifiers ($\exists$ and $\forall$), and parentheses. For example, the following first-order formula defines the set of commutators $g \in G$:
$$ (\exists x)(\exists y)(g = xyx^{-1}y^{-1}). $$
On the other hand, if you try to define the commutator subgroup by such a formula, you run into trouble. You might want to say, "$g$ is a commutator or $g$ is a product of two commutators or $g$ is a product of three commutators or ...," but infinite disjunctions are not allowed.
Please note that the counterexample to a similar question in a comment by Vipul Naik [here](https://mathoverflow.net/questions/39798/in-an-inductive-family-of-groups-does-the-probability-that-a-particular-word-is) is _not_ a counterexample to this question. In the inductive family $$ A_3 \hookrightarrow S_3 \hookrightarrow \ldots \hookrightarrow A_{2i-1} \hookrightarrow S_{2i-1} \hookrightarrow A_{2i+1} \hookrightarrow S_{2i+1} \hookrightarrow \ldots, $$ the groups alternate between having all the elements as commutators and half the elements as commutators. However, in the directed union, which is $A_\infty$, all the elements are commutators. The upshot is that quantifiers in a first-order formula may demand that you "look ahead" in your inductive family.
|
https://mathoverflow.net/questions/43900/density-of-first-order-definable-sets-in-a-directed-union-of-finite-groups
|
[
"gr.group-theory",
"lo.logic",
"combinatorial-group-theory",
"pr.probability",
"finite-groups"
] | 21 | 2010-10-27T16:10:17 |
[
"I agree, cool question!",
"This is a great question.",
"It may also not be true that you will get a density, i.e. that there will be a limit law. I definitely don't know, but Vardi might. Gerhard \"Ask Me About System Design\" Paseman, 2010.11.02",
"@Gerhard: It's not true that you'll always get density zero or one, if that's what you mean. For the union $S_\\infty$ of the symmetric groups $$ S_1 \\hookrightarrow S_2 \\hookrightarrow \\ldots \\hookrightarrow S_n \\hookrightarrow \\ldots, $$ the set of commutators is $A_\\infty$, and its density is $1/2$. An even more basic example would be the set of elements squaring to one in the union of the groups $\\mathbb{Z}/3\\mathbb{Z} \\times (\\mathbb{Z}/2\\mathbb{Z})^i$, which has density $1/3$.",
"This shouts \"zero-one law\" at me. Perhaps looking at limit laws, and works of Vardi, Fagin, and others on finite structures may give you what you want. Gerhard \"Ask Me About System Design\" Paseman, 2010.10.27"
] | 5 |
Science
| 0 |
152
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mathoverflow
|
If $X\times Y$ is homotopy equivalent to a finite-dimensional CW Complex, are $X$ and $Y$ as well?
|
Is there a space $X$ that is not homotopy equivalent to a finite-dimensional CW complex for which there exists a space $Y$ such that the product space $X\times Y$ is homotopy equivalent to a finite-dimensional CW complex? If so, how might we construct an example?
A first consideration could be where $X$ has infinitely many nontrivial homology groups, in which case $X$ is not homotopy equivalent to a finite-dimensional CW complex. Yet, in this case it follows from the Künneth Formula that no suitable $Y$ exists. Of course, this only rules out spaces with infinitely many nontrivial homology groups.
Another consideration could be where $X$ is an Eilenberg–Maclane space for which $\pi_1(X)$ has non-trivial torsion, in which case $X$ is, again, not homotopy equivalent to a finite-dimensional CW complex. In this case, if $\pi_1(X)$ is abelian then $X$ is homotopy equivalent to $L\times Z$ for some other space Z and an infinite-dimensional lens space $L$, and hence $X$ has infinitely many nontrivial homology groups, which implies, again, that no suitable $Y$ exists. This has led me to wonder if there exists a space $Y$ and a noncommutative group $G$ with non-trivial torsion such that $Y\times K(G,1)$ is homotopy equivalent to a finite-dimensional CW complex, where $ K(G,1)$ denotes an Eilenberg–Maclane space.
Any insight into approaching this problem is greatly appreciated.
|
https://mathoverflow.net/questions/267669/if-x-times-y-is-homotopy-equivalent-to-a-finite-dimensional-cw-complex-are-x
|
[
"at.algebraic-topology",
"homotopy-theory",
"cohomology",
"classifying-spaces"
] | 21 | 2017-04-19T14:52:52 |
[
"There are examples by R. Bing and many other mathematicians when a product of a non-manifold B and a real line gives a euclidean space, see e.g. the following papers and references there: ams.org/journals/bull/1958-64-03/S0002-9904-1958-10160-3/… ams.org/journals/proc/1961-012-01/S0002-9939-1961-0123303-2/… However, I do not know whether such space B has the homotopy type of a CW-complex.",
"A postscript to Igor Belegradek's comment: The argument in Proposition A.11 of my book shows that a space dominated by a CW complex of dimension $n$ is homotopy equivalent to a CW complex of dimension $n+1$, where the extra dimension arises from a mapping telescope construction. The argument, which is due to J.H.C.Whitehead if I remember correctly, uses only elementary homotopy theory. Using cohomology and Wall's work one obtains the stronger result that the increase in dimension is not really necessary, at least in dimensions 3 and greater.",
"Just to add references to Tom's answer. The last sentence it is theorem E (p.63) of C.T.C. Wal, \"Finiteness conditions for CW complexes\", see math.uchicago.edu/~shmuel/tom-readings/…. The statement that a homotopy retract of a CW-complex is a CW complex is e.g. in Hatcher's \"Algebraic Topology\", proposition A.11.",
"If $Y$ is nonempty then $X$ is a homotopy retract of a CW complex, therefore homotopy equivalent to a CW complex. If $X$ is a homotopy retract of a finite-dimensional CW complex, then $X$ has no cohomology above some dimension $d$, for any coefficient system. This in turn implies (if $d\\ge 3$) that $X$ is weakly, therefore strongly equivalent, to a $d$-dimensional complex.",
"Not exactly what you wanted, but there are pointed CW complexes $X$ that are not finite, but such that $S^1\\wedge X$ is finite, for example $BG$ for $G$ an infinite acyclic group.",
"A triviality: Y can be the empty space"
] | 6 |
Science
| 0 |
153
|
mathoverflow
|
Are the eigenvalues of the Laplacian of a generic Kähler metric simple?
|
It is a [theorem of Uhlenbeck](http://www.jstor.org/stable/2374041) that for a generic Riemannian metric, the Laplacian acting on functions has simple eigenvalues, i.e., all the eigenspaces are 1-dimensional. (Here "generic" means the set of such metrics is the complement of a [meagre set](http://en.wikipedia.org/wiki/Meagre_set) in the space of all metrics on a given manifold.)
I would like to know whether this property is also true for a generic Kähler metric. More precisely, given a compact complex manifold with a fixed choice of Kähler class, is it true that for the generic representative of this class, the Laplacian acting on functions has simple eigenvalues?
(To be honest, on a first reading it seems as if one might be able to apply similar arguments to those of Uhlenbeck to prove this result. I guess I would like to know if someone has already done this, or if there is a cunning counter-example that I'm missing, before I commit the time and energy to try!)
|
https://mathoverflow.net/questions/32810/are-the-eigenvalues-of-the-laplacian-of-a-generic-k%c3%a4hler-metric-simple
|
[
"dg.differential-geometry",
"complex-geometry",
"riemannian-geometry",
"fa.functional-analysis"
] | 21 | 2010-07-21T08:52:36 |
[] | 0 |
Science
| 0 |
154
|
mathoverflow
|
Homotopy flat DG-modules
|
A right DG-module $F$ over an associative DG-algebra (or DG-category) $A$ is said to be homotopy flat (h-flat for brevity) if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $F\otimes_A M$ is acyclic. Homotopy projective and injective DG-modules are defined in the similar way, e.g., a left DG-module $P$ over $A$ is h-projective if for any acyclic left DG-module $M$ over $A$ the complex of abelian groups $Hom_A(P,M)$ is acyclic. Homotopy adjusted DG-modules are used to define derived functors on the derived categories of DG-modules.
My question is: can one describe the class of h-flat DG-modules in any way alternative to the definition? Here are examples of the kind of description I have in mind:
1. A DG-module is h-projective if and only if it is homotopy equivalent to a DG-module obtained from the free DG-module $A$ over $A$ using the operations of shift, cone, and infinite direct sum. (Similarly, a DG-module is h-injective if and only if it is homotopy equivalent to a DG-module obtained from the cofree DG-module $Hom_{\mathbb Z}(A,\mathbb Q/\mathbb Z)$ using the operations of shift, cone, and infinite product.)
2. A module over a noncommutative ring is flat if and only if it is a filtered inductive limit of projective (or even projective and finitely generated, or free and finitely generated) modules (the Govorov-Lazard theorem).
The class of h-flat DG-modules is closed with respect to the operations of shift, cone, filtered inductive limit, and passage to a homotopy equivalent DG-module. It also contains the free DG-module $A$ over $A$. Can all h-flat DG-modules be obtained from that one DG-module using these four operations?
References: 1. Keller's paper "Deriving DG-categories"; 2. my preprint arXiv:0905.2621, section 1.
|
https://mathoverflow.net/questions/40036/homotopy-flat-dg-modules
|
[
"ct.category-theory",
"homological-algebra",
"differential-graded-algebras",
"flatness"
] | 21 | 2010-09-26T09:53:57 |
[
"I think the paper arxiv.org/abs/1607.02609 answers your question.",
"The following paper solves a closely related problem: L.W. Christensen and H. Holm, \"The direct limit closure of perfect complexes\", Journ. of Pure and Appl. Algebra 219, #3, p.449-463, 2015. arxiv.org/abs/1301.0731"
] | 2 |
Science
| 0 |
155
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mathoverflow
|
Is there a "direct" proof of the Galois symmetry on centre of group algebra?
|
Let $G$ be a finite group, and $n$ an integer coprime to $|G|$. Then we have the following map, which is clearly not a morphism of groups in general: $$g\mapsto g^n.$$
This induces a linear automorphism of $\mathbb{Z}[G]^G$, the algebra of $G$-invariant functions on $G$ under convolution, and surprisingly, this induced map is also an algebra automorphism, as can be seen by passing to $\mathbb{C}$ and noting that this is the Galois action on characters, which permutes the set of primitive idempotents of this algebra.
My question is whether this surprising fact can be explained directly, without using character theory? I would be interested in both a high-concept explanation of this symmetry, or a generators and relations argument for why there exists, for any $g,h$ in $G$, an $a,b\in G$ with: $$(gh)^n=ag^n a^{-1}bh^n b^{-1}.$$
|
https://mathoverflow.net/questions/415174/is-there-a-direct-proof-of-the-galois-symmetry-on-centre-of-group-algebra
|
[
"gr.group-theory",
"rt.representation-theory",
"finite-groups",
"characters"
] | 21 | 2022-02-01T13:38:45 |
[
"By a generators and relations argument I would think some symbolic argument that holds in any torsion group of exponent coprime to $n$, building $a,b$ from $g,h$ using these assumptions and the assumed $n$th root function. If one could give a group of this type where the result fails, it would rule out such an argument, which would also be very interesting.",
"What do you mean by \"a generators-and-relations argument for... $(gh)^n=ag^na^{-1}bh^nb^{-1}$\"? I doubt there exist such $a,b$ in the free group over $g,h$. For $n=2$ there are none: any $x,y,z$ in a free group satisfying $x^2y^2=z^2$ must commute (it follows there are none whenever $n$ is even).",
"A slightly different (but definitely character theoretic) proof of this interesting fact uses the structure constants $a_{KLM}=\\frac{|K||L|}{|G|}\\sum_{\\chi\\in\\mathrm{Irr}(G)}\\frac{\\chi(x_K)\\chi(x_L)\\overline{\\chi(x_M)}}{\\chi(1)}.$ If $K^\\prime$ etc. are the classes containing the respective $n^{th}$ powers and if $\\sigma$ is Chris' automorphism then $a_{K^\\prime L^\\prime M^\\prime}=\\frac{|K||L|}{|G|}\\sum_{\\chi\\in\\mathrm{Irr}(G)}\\frac{\\chi^\\sigma(x_K)\\chi^\\sigma(x_L)\\overline{\\chi^\\sigma(x_M)}}{\\chi(1)}=a_{KLM}^\\sigma,$ so $a_{K^\\prime L^\\prime M^\\prime}=a_{KLM}^\\sigma=a_{KLM}.$",
"There are a number of similar facts which seem non-obvious from a group-theoretic point of view, but are easy to prove with characters: for example, the number of times $ x \\in G$ is expressible as a commutator is unchanged if we replace $x$ by a different generator of $\\langle x \\rangle.$",
"@spin ah, thanks",
"@FedorPetrov: Note the assumption $\\gcd(n, |G|) = 1$. So if $g^n = 1$, then $g = 1$.",
"The Frobenius group of order $20$ is an example, trivial outer automorphism group, with nonrational characters, take $n=3$.",
"Basic question: What is an example of a finite group $G$ and an integer $n$ prime to $|G|$ such that there is no automorphism of $G$ which induces the permutation $\\text{Conj}(g) \\mapsto \\text{Conj}(g^n)$ on conjugacy classes?",
"Wow, that's a beautiful find. I am getting a slightly cohomological whiff from it (as, e.g., in the proof of the transfer's homomorphism property).",
"There is a theory of products and powers of conjugacy classes; see Beltrán, Felipe, and Melchor - Some problems about products of conjugacy classes in finite groups for what appears to be a recent survey. It seems to be in a different direction, but I wonder if that might be the place to look.",
"Its true, and follows from the fact that $g\\mapsto g^n$ is an algebra automorphism of $\\mathbb{Z}[G]^G$, by expanding the product of the classes of $g$ and $h$. I don't know of a non character theoretic proof of why this map is actually an algebra automorphism however.",
"Is the quoted equality about $(g h)^n$ true, but you don't know how to prove it, or you're not sure if it's true?"
] | 12 |
Science
| 0 |
156
|
mathoverflow
|
monoidal (∞,1)-categories from weakly monoidal model categories
|
In _Higher Algebra_ section 4.1.7, Jacob Lurie proves that the underlying $(\infty,1)$-category of a monoidal model category is a monoidal $(\infty,1)$-category.
Dominic Verity and Yuki Maehara have (independently) defined the lax Gray tensor product for model categories $S$ and $T$ for two different models of $(\infty,2)$-categories and proved that their tensor product is "weakly monoidal" in various senses:
* Maehara provides functors $\otimes_n \colon T^{\times n} \to T$ for each $n \geq 2$ that define the left adjoint parts of $n$-variable Quillen adjunctions. But these functors are associative in a weak sense, eg with a zig zag of natural weak equivalences
$$ X \otimes_2 (Y \otimes_2 Z) \to \otimes_3(X,Y,Z) \leftarrow (X \otimes_2 Y) \otimes_2 Z.$$
* Verity constructs two binary tensor products $\otimes, \otimes' \colon S \times S \to S$ that are naturally weakly equivalent, one of which is a closed left Quillen bifunctor and the other of which is associative up to isomorphism.
Morally, this weakness should not matter for the underlying $(\infty,1)$-category, but has anyone verified that a monoidal structure on the underlying $(\infty,1)$-category can be produced under these weaker hypotheses?
Alternatively, would someone like to take this on? I'd be happy to help publicize a result along these lines whenever it appears.
(FYI: Andrea Gagna, Yonatan Harpaz, Edoardo Lanari have a third version of the lax Gray tensor product that defines a monoidal model category in the usual sense. But I'm still interested in the question of about weak monoidal model categories, which is why I've set this aside.)
|
https://mathoverflow.net/questions/370172/monoidal-%e2%88%9e-1-categories-from-weakly-monoidal-model-categories
|
[
"model-categories",
"monoidal-categories",
"infinity-categories"
] | 20 | 2020-08-26T09:02:52 |
[
"There was further discussion in chat. To sum up: the suggestion from my second comment was misguided because Lurie really only deals in reflective localizations, and so can't say much about the localization from a model category to its associated $\\infty$-category. But Rune points out that Hinich has results on how DK localization interacts with cocartesian fibrations which could do the trick if generalized to the locally cocartesian case.",
"@TimCampion Excellent!",
"@AlexanderCampbell Great, that's good to know! Regarding pre-complicial sets, Verity indeed showed that the Gray tensor product is monoidal biclosed on them, and Yuki, Chris Kaupulkin and I observed in Theorem 1.33 here that the usual model structures are monoidal with respect to it. Relatedly, in that same preprint, we construct a monoidal biclosed lax Gray tensor product on a category of marked cubical sets (without having to futz around with the category a la precomplicial sets) and show it's a monoidal model category.",
"@TimCampion Yuki's structure is indeed an example of a lax monoidal category (which is the standard name in the literature). Lurie's \"corepresentable operads\" correspond to (symmetric) colax monoidal categories. Of course, the opposite of a lax monoidal category is a colax monoidal category, so I agree that this a promising approach.",
"In the second situation, isn't the full subcategory of pre-complicial sets a genuine monoidal model category?",
"Hiya Emily, nice to see you here. I think that this question was explored in the paper by Heuts, Hinich, and Moerdijk that fixes all of the mistakes from Moerdijk-Weiss. I'm not sure if they verified transport of structure to the associated ∞-category, but it's another data point: arxiv.org/abs/1305.3658",
"Localization is a product-preserving functor from relative categories to $\\infty$-categories. That means it's compatible with the enrichment of each category in itself, which should give you enough structure to say a functor of 2-categories to the 2-category of relative categories gives after localization one to the $(\\infty,2)$-category of $\\infty$-categories, which would give another approach to the first part. Unfortunately this kind of naturality for enrichments from module structures hasn't been worked out yet, however, though I believe Hadrian Heine is working on this.",
"It's conceivable that Lurie's framework might even be helpful here -- if you're interested mostly in the $\\infty$-category presented by your model category, then maybe some of Lurie's general results on localization for $\\infty$-operads could be used to say things like \"there is a monoidal structure on the presented $\\infty$-category\".",
"Ignoring the model structure, I would like to call the situation of the first bullet a \"lax monoidal category\" (or maybe \"oplax\"?). I've only seen lax monoidal categories (or rather their dual) in one source, namely Higher Algebra Definition 6.2.4.3 where they're called \"corepresentable ($\\infty$-)operads\" (of course, it's jazzed up to be $\\infty$ everything). I think another example of a (op)lax monoidal category (interacting nicely with a model structure) would be Boardman's handicrafted smash products of spectra."
] | 9 |
Science
| 0 |
157
|
mathoverflow
|
Is every positive integer the rank of an elliptic curve over some number field?
|
For every positive integer $n$, is there some number field $K$ and elliptic curve $E/K$ such that $E(K)$ has rank $n$?
It's easy to show that the set of such $n$ is unbounded. But can one show that _every_ positive integer is the rank of some elliptic curve over a number field?
The analogous question for a fixed number field is expected to have a negative answer (c.f. e.g., [this question](https://mathoverflow.net/questions/137970/is-it-expected-that-every-natural-number-is-the-rank-of-some-elliptic-curve-over)) but is still conjectural. But I wonder if one might be able to prove a positive answer to the question I asked above.
|
https://mathoverflow.net/questions/334853/is-every-positive-integer-the-rank-of-an-elliptic-curve-over-some-number-field
|
[
"nt.number-theory",
"arithmetic-geometry",
"elliptic-curves"
] | 20 | 2019-06-26T07:04:40 |
[
"The \"trick\" here is we can use $q \\to\\infty$ results because the question lets us pass to any finite extension. A similar result would follow over number fields from heuristics for the distribution of $L$-functions. In particular, it requires our heuristics to be true only in the 100% setting, and not in the power savings setting rqeuired by the boundedness of rank arguments over the integers.",
"So for $q$ sufficiently large we can show the existence of an elliptic curve $E$ and discriminants $D_1,\\dots, D_k$ such that of the $2^k$ twists generated by $D_1,\\dots, D_k$, $n$ have analytic rank $1$and the rest have rank $0$. Then one just has to use the analytic rank $1$ part of the BSD conjecture (which I think is now known in full generality over function fields?) and deduce that $n$ of these twists have algebraic rank $1$, hence the pullback to the composite of quadratic extensions has rank $n$.",
"It should be possible to prove the analogous statement over function fields. For a quadratic twist of a fixed elliptic curve $E$, Katz proved that $1/2$ the twists have analytic rank $0$ and half the twists have analytic rank $1$, in the large $q$ limit. It should be possible to check by explicitly examining monodromy that the root numbers of different quadratic twists are suitably independent.",
"@VesselinDimitrov Yes it's also natural to consider the Tate-Shafaverich group, from the point of view of the BSD conjecture and analytic class number formula. Both are special cases of the Tamagawa number conjecture, which uses Galois cohomology. The class number arises as the boundary in the Poitou-Tate localization sequence. The ideal class group and the Mordell-Weil group are both Picard groups but for Dedekind domains which are of somewhat different nature.",
"@FrançoisBrunault: That rel Picard over $\\mathbb{Q}$, though, kills the much more mysterious Shafarevich-Tate part; the Selmer group I think should be the true analogy. It is very easy to construct many $E/K$ of rank zero, much harder to construct number fields of class number one (apart from finitely many known cases).",
"@FrançoisBrunault: I didn't think of this, thanks for noting it! It makes sense, then. I assumed the class group was similar to a Selmer group, which includes only an upper bound on the rank. Still it looked to me that Stanley's question should be harder than the current one, but I could be off.",
"@VesselinDimitrov The Mordell-Weil group of $E$ is isomorphic to the relative Picard group of $E/\\mathbb{Q}$, so this is a natural analogue. By the way it is known that every abelian group is the Picard group of some Dedekind domain (Claborn 1966). One may ask the same thing while restricting the class of abelian groups and/or Dedekind domains considered.",
"@StanleyYaoXiao: Your question should be infinitely harder than this one. I don't think there is much of a similarity between a class number and a Mordell-Weil rank.",
"This other question I asked may be of relevance: mathoverflow.net/questions/88175/…",
"Sorry, I realized that my comment sounds less than clear, so let me add that apart from the remark in the first sentence, I do regard there both $K$ and $E$ as varying. Basically I am saying that a version of Silverman's specialization theorem can be used $r$ times over an $r$-dimensional char. $0$ base $T$ assuming that there exists an elliptic scheme $A/T$ such that $A(T) / \\mathrm{tors} \\cong \\mathbb{Z}^n$. From this point of view the problem is geometric, and I am guessing that an explicit such $A/T$ is given by $A_{1,n+1} / A_{1,n}$ (but any other construction would suffice).",
"@Wojowu: I am merely pointing out the existence of the phenomenon 'idle curiosity'. If you would label Goldbach's conjecture as such, that is up to you. Anyway, I wanted to give my reasons for my downvote, not enter into a discussion about the value of mathematics, so I will leave it here.",
"One may ask more generally: given a representation of a finite group $\\rho : G \\to \\mathrm{GL}_n(\\mathbb{Q})$, does $\\rho$ arise as the Galois module $E(L) \\otimes \\mathbb{Q}$ for some elliptic curve $E/K$ and some extension $L/K$ with Galois group $G$. We can also fix the number field $K$. If I recall correctly, if one replaces \"arises\" by \"appears inside\" then for function fields $K=\\mathbb{F}_p(t)$ these kinds of things are known.",
"@RP: What would be the use of knowing the answer of any question on this website? Unless you're only in it for cryptographic applications, this question is useful because of human curiosity.",
"@RP_ I would -1 your comment if I could. Since when does usefulness of the result dictate whether the question is good for this site? Which ranks are possible over a fixed number field is similarly useless, and yet it's a very interesting open problem.",
"-1. I think this question is a little silly. What could possibly be the use of knowing the answer? I mean, if you could reduce the Goldbach conjecture for $2n$ to the existence of an elliptic curve of rank $n$ over some number field, then I would see the point. As the matter stands I'm afraid I don't.",
"I am not sure whether it is fair to say that the fixed number field case should admit a negative answer; of course that comes down to a famous unsolved problem. For the existence of $E/K$, an iterated application of Silverman's specialization theorem reduces us to proving that the group of sections of the elliptic scheme $A_{1,n+1} \\to A_{1,n}$ has rank $n$, generated by the $n$ 'point doubling' sections $(x_1, x_2,\\ldots,x_n) \\mapsto (x_1,x_1;x_2,\\ldots,x_n)$, etc. (Here $A_{g,n}$ is the space of $n$-pointed abelian varieties of dimension $g$). This should be true."
] | 16 |
Science
| 0 |
158
|
mathoverflow
|
Non-rigid ultrapowers in $\mathsf{ZFC}$?
|
_Originally[asked and bountied at MSE](https://math.stackexchange.com/questions/4293034/non-rigid-ultrapowers-in-mathsfzfc):_
> **Question** : Can $\mathsf{ZFC}$ prove that for every countably infinite structure $\mathcal{A}$ in a countable language there is an ultrafilter $\mathcal{U}$ on $\omega$ such that the ultrapower $\mathcal{A}^\mathcal{U}$ has a nontrivial automorphism?
"Obviously" the answer is yes! _(... right?)_
Note that this is trivial in $\mathsf{ZFC+CH}$: $\mathsf{ZFC}$ alone proves that any nontrivial ultrapower is countably saturated, which in the presence of $\mathsf{CH}$ lets us run a back-and-forth argument to show that $\mathcal{A}^\mathcal{U}$ is non-rigid whenever $\mathcal{A}$ is countable and $\mathcal{U}$ is a nonprincipal ultrafilter on $\omega$. Separately, note that if we drop the requirement that the ultrafilter be on $\omega$ a positive answer follows from the Keisler-Shelah theorem, but [$\mathsf{ZFC}$ can't bring down KS to $\omega$ as one might expect](https://shelah.logic.at/files/130108/vive.pdf), so that ultimately doesn't seem like a promising direction.
I strongly suspect that I'm missing a very simple argument, but at present I'm not seeing it.
_As a minor aside, I'm especially interested in a proof which avoids using results about first-order logic (techniques from the model-theory of first-order logic are fine). This comes from the original motivation for this question, which was a weakness in[an MSE answer of mine](https://math.stackexchange.com/questions/41469/non-axiomatisability-and-ultraproducts/3569736#3569736) where the whole point was to avoid some standard model theory. However, in retrospect this seems to be jumping the gun here, so I'll relegate this aspect of the question from the question itself to this mere side comment._
|
https://mathoverflow.net/questions/408178/non-rigid-ultrapowers-in-mathsfzfc
|
[
"set-theory",
"lo.logic",
"model-theory",
"ultrafilters",
"ultrapowers"
] | 20 | 2021-11-10T08:01:04 |
[
"@PaulLarson \" An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?\" I'm not certain, I'm not too familiar with Scott processes.",
"The proof of Harrington's theorem on the Scott ranks of counterexamples to Vaught's Conjecture should show that separating points characterizes rigidity for structures of size aleph_1 too.",
"A countable structure is rigid if and only if its Scott process separates points. An ultrapower of a structure of infinite Scott height by an ultrafilter on omega shouldn't have its points separated by its Scott process. Should that make it nonrigid?",
"@PaulLarson Ooh, fun question (although not directly related to this one unless I'm missing something?) - I can't think of anything off the top of my head, but this is well outside my normal sphere so that doesn't mean much.",
"Is there anything known about rigid structures whose countable elementary substructures are all non-rigid?",
"@FarmerS Good question! :P The best I can say is that Shelah's Vive la difference series probably has some relevant info, per Douglas Ulrich's comment, but I haven't worked through those papers to see if there's an answer there.",
"Maybe this is easy, but do we know there is a countably infinite structure and some non-principal ultrafilter on $\\omega$ such that the ultraproduct is rigid?"
] | 7 |
Science
| 0 |
159
|
mathoverflow
|
A spin extension of a Coxeter group?
|
Consider a Coxeter group $W$ with simple generators $S$ and Coxeter matrix $\left( m_{s,t}\right) _{\left( s,t\right) \in S\times S}$.
Let $\mathfrak{M}$ be the set of all pairs $\left(s, t\right) \in S^2$ satisfying $s \neq t$ and $m_{s,t} < \infty$.
For every $\left( s,t\right) \in \mathfrak{M}$, let $c_{s,t}$ be an element of $\left\\{ 1,-1\right\\} $. Assume that:
* We have $c_{s,t}=c_{s^{\prime},t^{\prime}}$ for any two elements $\left( s,t\right) $ and $\left( s^{\prime},t^{\prime}\right) $ of $\mathfrak{M}$ for which there exists some $q \in W$ satisfying $s^\prime = qsq^{-1}$ and $t^\prime = qtq^{-1}$.
* We have $c_{s,t}=c_{t,s}$ for each $\left( s,t\right) \in \mathfrak{M}$.
Let $W^{\prime}$ be the group with the following generators and relations:
_Generators:_ the elements $s\in S$ and an extra generator $q$.
_Relations:_ \begin{align*} s^{2} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\\ q^{2} & =1;\\\ qs & =sq\ \ \ \ \ \ \ \ \ \ \text{for every }s\in S;\\\ \left( st\right) ^{m_{s,t}} & =1\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=1;\\\ \left( st\right) ^{m_{s,t}} & =q\ \ \ \ \ \ \ \ \ \ \text{for every }\left( s,t\right) \in \mathfrak{M} \text{ satisfying } c_{s,t}=-1. \end{align*}
There is clearly a surjective group homomorphism $\pi:W^{\prime}\rightarrow W$ sending each $s\in S$ to $s$, and sending $q$ to $1$. There is also a group homomorphism $\iota:\mathbb{Z}/2\mathbb{Z} \rightarrow W^{\prime}$ which sends the generator of $\mathbb{Z}/2\mathbb{Z}$ to $q$.
> **Question.** Is $\iota$ injective? Equivalently, is the sequence \begin{equation} 1\longrightarrow\mathbb{Z}/2\mathbb{Z}\overset{\iota}{\longrightarrow}W^{\prime }\overset{\pi}{\longrightarrow}W \longrightarrow1 \end{equation} exact? Equivalently, is $\left\vert \operatorname*{Ker}\pi\right\vert =2$ ?
**Background.** This would generalize at least one of the two "spin symmetric groups" to the situation of any Coxeter group. It would explain one of the results (Theorem 2.3 **(b)**) in [Alexander Postnikov and Darij Grinberg, _Proof of a conjecture of Bergeron, Ceballos and Labbé_](http://nyjm.albany.edu/j/2017/23-70.html), and prove a generalization of this result (Conjecture 6.1 **(b)**).
I have tried generalizing the [standard approach to constructing the spin symmetric groups by embedding them in the Hecke-Clifford algebra](https://arxiv.org/abs/1110.0263), but to no avail so far. Nor has the existing literature on central extensions of Coxeter groups been particularly helpful ([Howlett's _On the Schur multipliers of Coxeter groups_](http://onlinelibrary.wiley.com/doi/10.1112/jlms/s2-38.2.263/abstract) counts the extensions abstractly, but doesn't help understanding whether a given one exists; [Burichenko's _On extensions of Coxeter groups_](http://www.tandfonline.com/doi/abs/10.1080/00927879508825315?journalCode=lagb20) gives a criterion that I don't seem to properly understand, as it gives me wrong answers).
|
https://mathoverflow.net/questions/285263/a-spin-extension-of-a-coxeter-group
|
[
"algebraic-combinatorics",
"coxeter-groups",
"hecke-algebras"
] | 20 | 2017-11-04T12:46:36 |
[
"@DavidESpeyer: Thanks -- I've answered the Question positively in the meantime (a year ago), but only in a month or so will probably be able to write up my proof. Meanwhile, I did look into Howlett's and others' Schur-multiplier papers, but never found myself able to get something out of it that wasn't obviously wrong; it's too much of a foreign language to me.",
"Morris, \"Projective representations of finite reflection groups. III.\", Comm. Alg. (2004), is doing something very like this, although I haven't unwound his notation to see whether it exactly matches your formulation or not. The title of that paper has the word finite in it, but most of the results don't seem to need it. See also Howlett, \"On the Schur multipliers of Coxeter groups\", JLMS (1988), which is less concrete but is better about not including unneeded finiteness hypotheses.",
"This is a quotient of the Tits group (MSN) in a natural way. It seems like maybe Théorème 2.5 there will give you what you want, although I can't see it right away.",
"@VictorProtsak: Thanks. I've reverted to the $\\mathfrak{M}$ notation from the paper.",
"Sure, $s\\ne t$. Actually, your last relation involving $q$ should also have restriction $s\\ne t$.",
"@VictorProtsak: I take it you want $s \\neq t$, since otherwise picking $s=t$ and $s' = t'$ to be two non-conjugate simple reflections would break it. Anyway, nice question! Perhaps what we did in the proof of Claim 1 in the proof of Lemma 4.0.3 in Alex's and my paper could be of use.",
"Darij, very interesting question! Clearly your two conditions are necessary: conjugate elements have the same order. But are they sufficient to guarantee that if $st$ and $s't'$ have the same order in $W$, then $c_{s,t}=c_{s',t'}$ ? More precisely, do you know whether the following statement is true for general $W$? $$ $$ If $st$ and $s't'$ are conjugate in $W$ then there is $q\\in W$ simultateously conjugating either $s$ to $s'$ and $t$ to $t'$ or $s$ to $t'$ and $t$ to $s'$.",
"Jim, for any finite dihedral group (finite Coxeter group of rank 2) $\\langle s,t\\rangle\\simeq D_m$ of type $I_2(m)$, in the nontrivial case $c_{s,t}=-1$ this construction produces the dihedral group $D_{2m}$ of type $I_2(2m)$ with double the size. So starting from $W(G_2)=D_6$ you would get $D_{12}$.",
"@Gro-Tsen: Interesting idea; but this would only give two \"spin extensions\" (and I have no idea how to precisely find their generators and relations), whereas my $c_{s,t}$ in general allow for much more freedom. (Not saying that the approach is broken; could be my extensions are all equivalent.)",
"@darij: This is an intriguing area which I haven't gone into, but I wonder for example what would happen in the frequent case that -1 is in a given finite irreducible Coxeter group $W$? (This is a special case of the longest element of $W$, which has been well studied using the classification.) A small example is the $G_2$ case, though symmetric groups of rank > 1 aren't examples.",
"I had in mind that every Coxeter group can be realized as isometries of a certain \"canonical\" symmetric bilinear form (not necessarily definite), so that it embeds in $\\mathit{O}(p,q)$: wouldn't its inverse image in $\\mathit{Pin}^+(p,q)$ or $\\mathit{Pin}^-(p,q)$ answer your question? Or am I talking nonsense?"
] | 11 |
Science
| 0 |
160
|
mathoverflow
|
Isoperimetric inequality and geometric measure theory
|
The following version of the isoperimetric inequality can be easily deduced from the Brunn-Minkowski inequality:
> **Theorem.** _If $K\subset\mathbb{R}^n$ is compact, then $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}\mu_+(K), $$ where $\omega_n$ is the volume of the unit ball and
> $$ \mu_+(K)=\liminf_{h\to 0} \frac{|\\{x:\, 0<{\rm dist}\, (x,K)\leq h\\}|}{h} $$ is the Minkowski content._
If $K$ is the closure of a bounded set with $C^2$ boundary, then $\mu_+(K)=H^{n-1}(\partial K)$ (Hausdorff measure) so we have $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K). $$ However, the above inequality is true for _any_ compact set without assuming anything about regularity of the boundary. My question is:
## Is there a simple proof of the following result?
> **Theorem.** _If $K\subset\mathbb{R}^n$ is compact, then $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}H^{n-1}(\partial K). $$_
This result can be proved using the machinery of the geometric measure theory. If $H^{n-1}(\partial K)=\infty$, the inequality is obvious. If $H^{n-1}(\partial K)<\infty$, then $K$ has finite perimeter and the isoperimetric inequality for sets of finite perimeter yields $$ |K|^{\frac{n-1}{n}}\leq n^{-1}\omega_n^{-1/n}P(K)= n^{-1}\omega_n^{-1/n} H^{n-1}(\partial^* K) \leq n^{-1}\omega_n^{-1/n} H^{n-1}(\partial K), $$ where $P(K)$ is the perimeter of $K$ and $\partial^*K\subset\partial K$ is the reduced boundary. For details see [1].
Unfortunately, this argument is very far from being elementary.
**[1] Ambrosio, L., Fusco, N., Pallara, D.:** _Functions of bounded variation and free discontinuity problems._ Oxford Mathematical Monographs. The Clarendon Press, Oxford University Press, New York, 2000.
|
https://mathoverflow.net/questions/294882/isoperimetric-inequality-and-geometric-measure-theory
|
[
"geometric-measure-theory",
"isoperimetric-problems",
"hausdorff-measure"
] | 20 | 2018-03-10T13:32:46 |
[] | 0 |
Science
| 0 |
161
|
mathoverflow
|
Hahn-Banach and the "Axiom of Probabilistic Choice"
|
Stipulate that the Axiom of Probabilistic Choice (APC) says that for every collection $\\{ A_i : i \in I \\}$ of non-empty sets, there is a function on $I$ that assigns to $i$ a finitely-additive probability measure $\mu_i$ on $A_i$.
Then Hahn-Banach (HB) implies APC, since HB is equivalent to the existence of a finitely-additive probability measure on every boolean algebra, and hence implies the existence of such a measure on the direct sum of the powerset algebras $P(A_i)$.
APC is non-trivial in that it implies the Banach-Tarski paradox and the existence of nonmeasurable sets (the proof Foreman and Wehrung use to show that HB implies Banach-Tarski works).
**Question:** Does APC imply HB?
(When I think about this, I find I keep on wanting to use Stone representation, but of course to do that would be to assume Boolean Prime Ideal, which is stronger than HB or APC.)
|
https://mathoverflow.net/questions/281142/hahn-banach-and-the-axiom-of-probabilistic-choice
|
[
"set-theory",
"axiom-of-choice",
"hahn-banach-theorem"
] | 20 | 2017-09-14T07:57:09 |
[
"1. D. Pincus, The strength of Hahn–Banach's Theorem, in: Victoria Symposium on Non-standard Analysis, Lecture notes in Math. 369, Springer 1974, pp. 203-248. Implies Hahn-Banach theorem is weaker than the ultrafilter lemma",
"@Fedor: You might have remembered that HB+SKM (Strong Krein-Milman) imply AC.",
"@Wojowu ah, indeed, I remembered not well",
"@FedorPetrov Not at all. It is very much weaker. It follows from the ultrafilter lemma (and at the same time doesn't imply it).",
"If I remember well, HB is equivalent to AC.",
"Oh, I wasn't implying it's trivial. I was hoping for someone to complete the proof. :)",
"... sorry, I don't see the \"so\"...",
"Well, using Bartle integrals you can get linear functionals on $\\ell^\\infty(A_i)$ which do not come from $\\ell^1(A_i)$. And HB is equivalent to \"Every Banach space has a nontrivial functional\" (continuous or otherwise). So..."
] | 8 |
Science
| 0 |
162
|
mathoverflow
|
Looking for an effective irrationality measure of $\pi$
|
Most standard summaries of the literature on [irrationality measure](http://mathworld.wolfram.com/IrrationalityMeasure.html) simply say, e.g., that $$ \left| \pi - \frac{p}{q}\right| > \frac{1}{q^{7.6063}} $$ for all sufficiently large $q$, without giving any indication of how large qualifies as "sufficiently large." It would occasionally be useful (for example, [for my answer here](https://math.stackexchange.com/a/1313626/30836)) to be able to come up with explicit bounds on the size of $q$.
I don't have access to Salikhov's paper where the exponent $7.6063$ is obtained, but Hata's paper which obtained an exponent of $8.0161$ is online [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa63/aa6344.pdf). Looking through it, the proof seems to be effective (all the bounds he's using appear to be defined explicitly), but it is definitely not written to be clear about what those bounds are.
Has anyone extracted explicit bounds on the size of the denominator from any irrationality measure calculation for $\pi$ more recent than Mahler's proof (referenced in the Hata paper) that $$ \left| \pi - \frac{p}{q}\right| > \frac{1}{q^{42}} $$ unrestrictedly?
|
https://mathoverflow.net/questions/210509/looking-for-an-effective-irrationality-measure-of-pi
|
[
"nt.number-theory",
"diophantine-approximation"
] | 20 | 2015-06-07T11:02:16 |
[
"Salikhov's paper is available here: mathnet.ru/links/93c99ab6587bdc1bf0912ec96b1d9f4f/rm9175.pdf",
"I wasn't aware this was migrated, I'll delete my answer in that case.",
"The exponent -42 got by Mahler was quite extraordinary in his time (\"Striking inequality\", Alan Baker). It is valid for all rational p/q with q > 1 and also indicates that rationals \"may not be very closed to $\\pi$\". I guess all improvement of the exponent -42 is valid just \"for all sufficiently large q\" in the ordinary meaning of this expression without explicity giving of bounds. Maybe possible but certainly quite difficult to get them.",
"@muaddib: Nope, you must be thinking of some other Micah...",
"I know this is random but by any chance did you go to New College?"
] | 5 |
Science
| 0 |
163
|
mathoverflow
|
Is the determinant of cohomology a TQFT?
|
If $M$ is an oriented $d$-manifold, let $D(M)$ denote the top exterior power of $H^*(M,\mathbf{C})$. Then $D(M_1 \amalg M_2) = D(M_1) \otimes D(M_2)$. Is there a good recipe for a map $D(M) \to D(N)$ induced by a cobordism from $M$ to $N$?
In some dimensions, there is a natural identification $D(M) = \mathbf{C}$ and you could take every cobordism to the identity map. But such a TQFT would not be completely trivial when $d = 4$, I think. For example it would distinguish $S^1 \times \mathbf{C}P^2$ from the mapping torus of complex conjugation.
If there's no problem defining it for $(d+1)$-manifolds, can it be "extended down" any distance, i.e. associate something to a $(d-1)$-manifold?
|
https://mathoverflow.net/questions/241807/is-the-determinant-of-cohomology-a-tqft
|
[
"at.algebraic-topology",
"dg.differential-geometry",
"topological-quantum-field-theory"
] | 20 | 2016-06-09T10:31:23 |
[
"I would have thought that the $d$-dimensional theory (or maybe I mean $d\\pm 1$) would assign $\\mathrm{pt} \\mapsto \\mathbb C$-as-an-$E_d$-algebra. But that defines the trivial (framed) theory. Perhaps there's a nontrivial choice of how to descend from a framed theory to an oriented one, but I'm not seeing it.",
"I'm sure the answer is \"yes, $D$ is part of an invertible TFT\", but I'm not seeing totally immediately how to write it down. From my point of view, you want to look at the \"free topological boson on $M$\", which is the theory whose field is a map $\\phi: M \\to \\mathbb C$ and whose EOM is $\\mathrm d \\phi = 0$. This has a \"cotangent quantization\" in the Costello-Gwilliam language, which up to some degree shift is $D$. If you can realize it as a fully extended TFT, you win.",
"This TFT would presumably be invertible, so can in principle be classified using the cohomology of Madsen-Tillman spectra. There's a known invertible 2d TFT which assigns to a closed oriented surface the number $\\lambda^{\\chi(X)}$ where $\\chi$ is the Euler characteristic and $\\lambda \\in \\mathbb{C}^{\\times}$, and this TFT is a categorification of that. Maybe this paper of Freed will be helpful: arxiv.org/abs/1406.7278"
] | 3 |
Science
| 0 |
164
|
mathoverflow
|
Homeomorphisms of the sphere mapping a geodesic triangulation to another one
|
Consider the standard Riemannian 2-sphere $S$, equipped with a geodesic triangulation $T$. Let $L(S,T)$ be the space of homeomorphisms of $S$ which map $T$ to a geodesic triangulation. What is the homotopy type of $L(S,T)$? A similar question has been solved by [E. Bloch, R. Connelly, D. Henderson, Topology 23 (1984), 161-175](http://www.sciencedirect.com/science/article/pii/0040938384900375), when $S$ is a simplicial disc in the plane, and it is a "tour de force"!
This question is on behalf of Jean Cerf, my former advisor, who is interested in S. Cairns' work.
|
https://mathoverflow.net/questions/262048/homeomorphisms-of-the-sphere-mapping-a-geodesic-triangulation-to-another-one
|
[
"gt.geometric-topology",
"riemannian-geometry",
"triangulations"
] | 20 | 2017-02-12T11:27:31 |
[
"I agree with the change of title. It is better, but I thought of (S,T) as a structure, not an object; it was ambiguous.",
"I have improved the title",
"The question in the edited title doesn't seem to be the same as the one in the body (and is much easier)."
] | 3 |
Science
| 0 |
165
|
mathoverflow
|
Polynomials with roots in convex position
|
Let $\mathcal P_n$ denote the set of all monic polynomials of degree $n$ with real or complex coefficients such that $P\in\mathcal P_n$ if for all $k\in\lbrace 0,1,\dots,n-2\rbrace$ the $n-k$ roots of $P^{(k)}=\left(\frac{d}{dx}\right)^kP$ are in strictly convex position (ie are the $n-k$ vertices of a convex polygon with $n-k$ extremal vertices).
The set $\mathcal P_n$ is clearly an open subset of all monic polynomials of degree $n$ over $\mathbb R$ or $\mathbb C$.
What is the geometry and topology of $\mathcal P_n$?
Facts:
(1) Over the reals, $\mathcal P_n$ has at least $2^{n-1}$ connected components: Indeed, consider a very fast decreasing sequence (I guess $n\longmapsto 1/((1+n)^{(1+n)^{1+n}})!$ will probably work) of strictly positive reals $\alpha_0 > \alpha_1 > \cdots$ and a sequence of signs $\epsilon_0,\epsilon_1,\ldots,\epsilon_{n-2}\in\lbrace \pm 1\rbrace$. Then
$$x^n+\sum_{k=0}^{n-2}\epsilon_k\alpha_k x^k\in \mathcal P_n(\mathbb R)$$ and different choices of signs correspond to different connected components. (Proof: Up to translation, we can assume that all roots sum up to $0$. If two polynomials of the above form are connected by a continuous path of real polynomials with roots summing up to $0$ , then all derivatives except the very last one do not have a root at $0$ (since it is otherwise in the interior of the convex hull of all roots). All coefficients up to degree $d-2$ are thus necessarily non-zero and different signs correspond to different connected components). Are there other connected components?
(2) Over the complex numbers, all the polynomials described in (1) are in the same connected component. Choosing $\epsilon_i$ on the complex unit circle suggests however that $\pi_1(\mathcal C_n)$ might be $\mathbb Z^{n-1}$ where $\mathcal C_n$ denotes the connected component of $x^n+\sum_{k=0}^{n-2}\alpha_kx^k$ in $\mathcal P_n(\mathbb C)$. Do we have $\mathcal C_n=\mathcal P_n(\mathbb C)$?
(3) We have the equalities $\mathcal P_{n-1}=\lbrace P'/n \ |\ P\in\mathcal P_n\rbrace$.
|
https://mathoverflow.net/questions/26799/polynomials-with-roots-in-convex-position
|
[
"gt.geometric-topology",
"cv.complex-variables",
"polynomials"
] | 20 | 2010-06-02T01:21:03 |
[
"I have sketched a proof in order to adress the the comment of Denis Serre.",
"I am not sure of why all these real polynomials are in different CCs. Do you pretend that if a coefficient of degree $\\le n-2$ is zero, then $P$ is not in ${\\mathcal P)_n$ ? That seems wrong for the constant coefficient.",
"Ī̲ have some ideas how to prove $\\mathcal C_n=\\mathcal P_n(\\mathbb C)$ and, similarly, to make a retraction of the real case to components described by the OP. Still searching for a suitable compactification of $\\mathcal P_n$ because Ī̲ detest to construct ε–δ proofs."
] | 3 |
Science
| 0 |
166
|
mathoverflow
|
On random Dirichlet distributions
|
Fix a dimension $d\ge2$.
* Let $Q_d$ denote the positive quadrant of $\mathbb{R}^d$, that is, $Q_d$ is the set of points $\mathbf{x}=(x_i)_i$ in $\mathbb{R}^d$ such that $x_i>0$ for every $i$.
* For every $\mathbf{x}$ in $Q_d$, let $|\mathbf{x}|=x_1+\ldots+x_d$.
* Let $\Delta_d$ denote the set of points $\mathbf{x}$ in $Q_d$ such that $|\mathbf{x}|=1$.
* For every $\mathbf{a}$ and $\mathbf{b}$ in $Q_d$, define $\mathbf{a}\cdot \mathbf{b}$ in $Q_d$ by $(\mathbf{a}\cdot \mathbf{b})_i=a_ib_i$ for every $i$.
* For every $\mathbf{a}$ in $Q_d$, let $\mathrm{Dir}(\mathbf{a})$ denote the Dirichlet distribution of parameter $\mathbf{a}$.
**The problem in a nutshell**
Fix $\mathbf{a}$ and $\mathbf{b}$ in $Q_d$. Choose a random parameter $\mathbf{u}$ in $\Delta_d$ with distribution $\mathrm{Dir}(\mathbf{a})$. Then choose a random point $\mathbf{X}$ in $\Delta_d$ with distribution $\mathrm{Dir}(\mathbf{b}\cdot \mathbf{u})$.
> My aim is to understand the (absolute) distribution of $\mathbf{X}$.
**Some more notations**
For every $\mathbf{a}=(a_i)_i$ in $Q_d$, $\mathrm{Dir}(\mathbf{a})$ is the absolutely continuous probability measure on $\Delta_d$ whose density $f(\ |\mathbf{a})$ at $\mathbf{x}$ is proportional to $x_1^{a_1-1}\cdots x_d^{a_d-1}$. More precisely, $$ f(\mathbf{x}|\mathbf{a})=\Gamma(|\mathbf{a}|)\mathbf{x}^{\mathbf{a}-1}/\Gamma(\mathbf{a}), $$ with the following shorthands: $$ \Gamma(\mathbf{a})=\Gamma(a_1)\cdots\Gamma(a_d),\quad \mathbf{x}^{\mathbf{a}-1}=x_1^{a_1-1}\cdots x_d^{a_d-1}. $$ The density $f_{\mathbf{a},\mathbf{b}}$ of the distribution of $\mathbf{X}$ is $$ f_{\mathbf{a},\mathbf{b}}(\mathbf{x})=\int_{\Delta_d} f(\mathbf{x}|\mathbf{b}\cdot \mathbf{u})f(\mathbf{u}|\mathbf{a})\mathrm{d}u_1\cdots\mathrm{d}u_{d-1}. $$
**Some special cases**
If $a_i=b_i=1$ for every $i$, $\displaystyle f_{\mathbf{1},\mathbf{1}}(\mathbf{x})\propto\int_{\Delta_d} \frac{\mathbf{x}^{\mathbf{u}-1}}{\Gamma(\mathbf{u})}\mathrm{d}u_1\cdots\mathrm{d}u_{d-1}.$
The case $d=2$ yields $$ f_{\mathbf{1},\mathbf{1}}(x,1-x)\propto\int_0^1\frac{x^{w-1}(1-x)^{-w}}{\Gamma(w)\Gamma(1-w)}\mathrm{d}w=\frac1{\pi x}\int_0^1\left(\frac{x}{1-x}\right)^{w}\sin(\pi w)\mathrm{d}w, $$ hence $$ f_{\mathbf{1},\mathbf{1}}(x,1-x)=\frac1{x(1-x)}\frac1{\pi^2+(\log[x/(1-x)])^2}. $$ Writing $\mathbf{X}=(X_1,X_2)$ with $X_1\ge0$, $X_2\ge0$ and $X_1+X_2=1$, this can be rewritten as the fact that, for every $x$ in $(0,1)$, $$ P(X_1\le x)=P(X_2\le x)=\frac12+\frac1\pi\arctan\left(\frac1\pi\log\left(\frac{x}{1-x}\right)\right). $$ Are there other cases where the density $f_{\mathbf{a},\mathbf{b}}$ is (reasonably) explicit? Or, for example, where the moments $E(\mathbf{X}^\mathbf{n})$ of $\mathbf{X}$ with $\mathbf{n}=(n_1,\ldots,n_d)$ any $d$-uplet of integers, are (reasonably) explicit?
|
https://mathoverflow.net/questions/55147/on-random-dirichlet-distributions
|
[
"pr.probability",
"probability-distributions"
] | 20 | 2011-02-11T09:23:01 |
[] | 0 |
Science
| 0 |
167
|
mathoverflow
|
Etale fundamental group of a curve in characteristic $p$
|
Let $C$ be a connected, smooth, proper curve of genus $g$ over an algebraically closed field $k$ of characteristic $p>0$. Let $\pi_1(C)$ be the etale fundamental group of $C$ - I only care about this as an abstract profinite group, so I omit base points.
It is well known by Grothendieck that $\pi_1(C)$ is topologically generated by $2g$ elements. Formally, there is a surjection $\phi:\widehat{F_{2g}}\rightarrow \pi_1(C)$, where the first group is the profinite completion of the free group on $2g$ elements. Moreover, if we let $G^{(p)}$ be the prime-to-$p$ quotient of $G$, then $$\pi_1(C)^{(p)}\cong\widehat{F_{2g}/w}^{(p)},$$ where $w=\prod_{i=1}^g[x_i,y_i].$ I.e., you get what you expect from the characteristic 0 case.
Recall that the prime-to-p quotient is the quotient $G/K$, where $K$ is the intersection of the kernels of all the maps from $G$ to finite prime-to-$p$ groups.
**Question: Is there any case of a curve $C$ of genus at least 2 for which its etale fundamental group is known?**
I phrase the question in this way because the Etale fundamental group does depend on the curve in question. In fact, even its abelianization will depend on the $p$-rank of the Jacobian.
|
https://mathoverflow.net/questions/186687/etale-fundamental-group-of-a-curve-in-characteristic-p
|
[
"ag.algebraic-geometry",
"fundamental-group"
] | 20 | 2014-11-09T23:09:03 |
[
"I think $\\pi_1$ is not known for any hyperbolic curve in any reasonable sense. For instance, just knowing for each prime-to-$p$ cover the $p$-rank of its Jacobian seems very hard - it is not at all obvious that there is a usable finite description of this data.",
"Take $G_1$ an infinitely generated free $p$-group and $G_2$ the product of $G_1$ with $\\mathbb Z/p$. Then they have the same finite quotients but are distinct. This is not so great an example, because group cohomology distinguishes them.",
"@Niels: Not to belabour the point, but i think knowing whcih groups $G$ arise suffices to figure out the cardinality of the set of $G$-corvers. The point being that if there are $k$ non-conjugate homomorphisms to $G$, then there is a homomorphism $\\pi_1\\to G^k$, and the image $H$ of this homomorphism has $k$ non-conjugate homomorphisms to $G$. So we can detect whether theres at least $k$ $G$-covers by asking whether such a group $H$ occurs as a finite quotient. Probably even this is not enough to determine the profinite group. Perhaps I'll ask this a separate question. Thanks for your help!",
"@jacob: for an affine curve, we just know if a given finite group $G$ arises as a quotient of $\\pi_1$ or not and nothing more. In particular the cardinality of the set of $G$-covers is not known for general $G$. And this data would be probably insufficient to recover the $\\pi_1$ anyway. To sum up: no the $\\pi_1$ is not known for any curve (complete or not) over any field of positive characteristic except for the trivial cases of $\\mathbb P^1$ and elliptic curves.",
"@Niels: You said Joels observation of knowing the finite quotients does not tell us, for a fixed finite group, the cardinality of the set of $G$-covers.My third sentence was meant to say that,if we are working over a countable field, it does. Because the answer is either countable infinity,or some finite number,either of which can be detected on the level of finite quotients. As you say, this is irrelevant if we are working with proper curves, but if we talk about affine curves this starts mattering. So, my question is: If the $\\pi_1$ of a hyperbolic affine curve over a countable field known?",
"@jacob I am not sure I understand your third sentence. About the last sentence : the $\\pi_1$ of a proper curve is invariant by algebraically closed field extension. So fixing a specific base field does not change the question for proper curves. And the answer is no, the $\\pi_1$ is not known. By specialization theory we just know it is topologically of finite type. For affine curves the situation is even worse since the $\\pi_1$ is not topologically of finite type.",
"Aha! That is illuminating. But over a countable field like $\\mathbb{F}_{p^{\\infty}}$ we know that the set of $G$-covers is at most countably infinite. So in that context, at least, one can answer Niels's question simply by knowing the finite quotients. Am I correct that even in this restricted setting, we do not know the $\\pi_1$?",
"@Joël I agree \"completely wrong\" was a bit strong :) but I disagree with your interpretation of the question. Clearly the question is : \"is the $\\pi_1$ known up to isomorphism ?\". An answer to your weak version wouldn't answer the following question : \"given a fixed finite group $G$, what is the cardinality of the set of $G$-covers\" ?",
"@jacob I think I remember you can find an example in Fried-Jarden Field Arithmetic.",
"@jacob. The finite quotient of $\\mathbb Z_p^I$ for $I$ infinite are all the abelian $p$-group. Now take two infinite sets $I$, one enumerable say and another say larger than the continuum, and you have two profinite groups non-isomorphic (because they don't have the same cardinality) having the same set of finite quotients. So Niels is right that there is more in understanding a $\\pi^1$ than its understanding it's finie quotient. Now in algebraic geometry one could have the point of view that it is the finite quotients, corresponding to finite cover, that are what we want to understand.",
"@Niels, Ah okay, interesting. But \"completely wrong\" is perhaps not the right term (or possibly I am more wrong that you explain in your comment): one at least knows all the finite quotients of this profinite group (the $pi^1$ of a non-complete curve), which is already something important, and sufficient for many applications. Then it depends of what we mean by \"know\", and apparently the OP was not completely sure about that. But, do we know the complete list of finite quotient of the $\\pi^1$ for a curve of genus $\\geq 2$ without any point removed ?",
"@Niels: Actually, I was wondering about this. Is there an example of 2 non-isomorphic profinite groups with the same finite quotients? Perhaps this is worth posting as a separate question...",
"@Joël : what you wrote is completely wrong. Abyankhar's conjecture describes the set of finite quotients of the $\\pi_1$, but since in the affine case the $\\pi_1$ is not topologically of finite type, this is not enough information to get back the $\\pi_1$. For instance, the $\\pi_1$ of $\\mathbb A^1$ is not known in positive characteristic, even if one knows that its finite quotients are the quasi-$p$ groups. In fact to my knowledge there is no example of an hyperbolic curve whose $\\pi_1$ is \"known\" in positive characteristic.",
"Of course (you certainly know this), if you remove $t$ points to your curve, then the problem is solved, provided than $t>0$. It is Abyankhar's conjecture, now a theorem of Raynaud and Harbater.",
"Tamagawa proved that for all profinite groups $G$, there are only finitely many smooth proper curves over $\\bar{\\mathbb F_p}$ (up to isomorphism) with etale fundamental group isomorphic to $G$. Maybe you can find something useful in his article; see Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, J. Algebraic Geom. 13 (2004), 675–724."
] | 15 |
Science
| 0 |
168
|
mathoverflow
|
Finiteness of etale cohomology for arithmetic schemes
|
By an _arithmetic scheme_ I mean a finite type flat regular integral scheme over $\mathrm{Spec} \, \mathbb{Z}$.
> Let $X$ be an arithmetic scheme. Then is $H_{et}^2(X,\mathbb{Z}/n\mathbb{Z})$ finite for all $n \in \mathbb{N}$?
Remarks: Let $j:U \to X$ be the open subset given by removing the fibres of $X \to \mathrm{Spec} \, \mathbb{Z}$ lying above those primes dividing $n$.
1. I can easily show that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z})$ is finite. For example, it follows from the Leray spectral sequence that $H_{et}^1(X,\mathbb{Z}/n\mathbb{Z}) \to H_{et}^1(U,\mathbb{Z}/n\mathbb{Z})$ is injective, and the latter group is finite by a special case of [1, Proposition II.7.1].
2. If $n$ invertible on $X$, then the answer is _yes_ and this is again a special case of [1, Proposition II.7.1]. So I'm interested in the case where $n$ is not invertible on $X$. (In particular, this shows that $H_{et}^2(U,\mathbb{Z}/n\mathbb{Z})$ is finite).
3. If $\dim X = 1$ then the answer if _yes_ by [1,Theorem II.3.1].
4. Another application of the Leray spectral sequence argument shows that is suffices to show that $H^0(X, R^1j_* \mathbb{Z}/n\mathbb{Z})$ is finite, but I don't see why this should be the case.
5. The analogous question concerning finiteness of $H_{et}^i(X,\mathbb{Z}/n\mathbb{Z})$ is also interesting, but for my application I only need $i = 2$.
References:
[1] Milne - Arithmetic duality theorems.
|
https://mathoverflow.net/questions/284354/finiteness-of-etale-cohomology-for-arithmetic-schemes
|
[
"ag.algebraic-geometry",
"nt.number-theory",
"arithmetic-geometry",
"etale-cohomology"
] | 20 | 2017-10-25T04:51:27 |
[
"@MartinBright: Thanks for the references! I have had a look at the paper of Sato, but it is written in the language of derived categories which are not my forte.... Could you please be more specific which result is most relevant? He seems to have many different purity results in his paper.",
"A couple of references: the purity result of Gabber referred to is here arxiv.org/abs/1207.3648 Exposé XVI, and indeed assumes the torsion order is invertible. For p-torsion there is some kind of purity theorem in this article of Sato arxiv.org/abs/math/0610426 but I don't know whether it is any use to you.",
"@JasonStarr: as you know, there's always an excision sequence relating the cohomology of a scheme to that of an open subscheme with the extra term being cohomology with supports along the closed complement (rather than cohomology on that complement), so the Artin-Schreier stuff doesn't directly intervene (but there's likely no purity theorem for this situation with torsion-orders not invertible; with reasonable torsion-orders I think Gabber has proved purity theorems for abstract regular schemes).",
"I have one more comment. For every flat, regular, affine $\\mathbb{Z}$-scheme $X$ whose reduction $X_p$ over $\\mathbb{Z}/p\\mathbb{Z}$ is regular of dimension $\\geq 1$, the cohomology $H^1_{\\text{et}}(X_p,\\mathbb{Z}/p\\mathbb{Z})$ is infinite, because of the many different Artin-Schreier covers. If there were a long exact sequence, presumably this would imply that $H^2_{\\text{et}}(X,\\mathbb{Z}/p\\mathbb{Z})$ is also infinite. That might be evidence that there is no long exact sequence, or it might be evidence that the cohomology is infinite.",
"The first paper of Esnault that I know including such a Gysin sequence (the one with the joint appendix with Deligne) is \"Deligne's Integrality Theorem in Unequal Characteristic and Rational Points over Finite Fields\", Ann. of Math. 164 (2006), pp. 715--730. A second paper, where she uses alterations, is \"Coniveau over $\\mathfrak{p}$-adic Fields and Points over Finite Fields\", C. R. Acad. Sci. Paris, Ser. I 345 (2007), pp. 73--76.",
"@Jason Starr: Which paper of Esnault are you referring to?",
"Esnault's uses alterations (in the strong form of a $G$-Galois cover that has a regular modification) and the map from the cohomology of the original scheme to the $G$-invariants of the cohomology of the $G$-cover to deal with the regularity properties. You are completely correct that she needs to invert several integers.",
"@JasonStarr: The purity theorems (such as in Milne's notes that you mention) require the closed complement to satisfy regularity properties (which $X \\bmod p$ may not) and more importantly assume torsion-orders for the sheaf are invertible on the scheme being considered (and avoiding the latter assumption is the main thrust of the question here).",
"@JasonStarr: In the papers of Deligne and/or Esnault surely the relevant torsion sheaves have torsion orders invertible on that base (and section 6 of Weil II assumes properness). With \"invertible\" torsion-orders one can use Deligne's \"generic base change\" expose from SGA 4.5 to see that the higher direct images on ${\\rm{Spec}}(\\mathbf{Z})$ are constructible, and it is a general theorem (perhaps due to some combination of Artin, Mazur, Zink?) that the cohomology of any constructible abelian sheaf (no hypothesis on torsion-orders) on the ring of $S$-integers of a number field is finite.",
". . . Esnault cites Equation 6 in Section 3.6, p. 389 of \"La Conjecture de Weil, II\" by Pierre Deligne. Deligne considers a finite type flat scheme over a DVR, but he seems to assume that it has equicharacteristic. However, in the papers of Esnault (including the paper with a joint appendix with Deligne), the exact sequence is applied when the DVR has mixed characteristic . . .",
"The version in Milne is Theorem 16.1 of \"Lectures on Etale Cohomology\", but he only formulates there the theorem for finite type schemes over a field. The purity theorem over a Dedekind domain base has been used by Esnault in her work on congruences for rational points . . .",
"There is a purity theorem that is really a Gysin sequence relating etale cohomology of $U$, etale cohomology of $X$, and etale cohomology with supports in the closed complement of $U$ (which then gets turned into etale cohomology of the complement via purity). There is a reference in Milne. I will write it in a moment."
] | 12 |
Science
| 0 |
169
|
mathoverflow
|
Infinitely generated non-free group with all proper subgroups free
|
Is there any example of group $G$ satisfying the following properties?
1. $G$ is non-abelian, infinitely generated (i.e. it is not finitely generated) and not a free group.
2. $H< G$ implies that $H$ is a free group.
Clearly such an example should be torsion-free and countable.
* * *
(Added, from comments) For completeness, here's why it has to be countable. First, it is enough to construct a subgroup of index 2. Indeed, a theorem of Swan asserts that a torsion-free group with a free finite-index subgroup is free. Next, if a group $G$ has no subgroup of index 2 (i.e. every element is product of squares), it is easy to construct a nontrivial countable subgroup $H$ with the same property). In the current setting, $G$ is uncountable so $H$ is a proper subgroup, and hence has to be free, contradiction.
|
https://mathoverflow.net/questions/351295/infinitely-generated-non-free-group-with-all-proper-subgroups-free
|
[
"gr.group-theory",
"examples",
"free-groups"
] | 20 | 2020-01-27T14:45:56 |
[
"Since a partly equivalent question was just asked, I have added the short argument to discard the uncountable case.",
"@YCor You're probably right I should not have have said \"clearly\" but in my mind it was a very direct application of some known facts (to me at least) in infinite group theory. Swan Theorem says that if your group is torsion-free and it is a finite extension of a free group, then it must be free itself.",
"@W4cc0 thanks; I wouldn't have said \"clearly\"! I didn't know Swan's theorem, what does it say exactly? Here's an argument to show residual finiteness: it's enough to check that there is a proper subgroup of index 2. Indeed, the latter being free by assumption, it has to be res. finite and hence so is the whole group. Now, if by contradiction there is no subgroup of index 2, then every element is a finite product of squares. From this, it is easy to construct a nontrivial countable subgroup in which every element is a finite product of squares, and hence this is a non-free countable subgroup.",
"@YCor such a group must be countable because if it is not, then it would be residually finite, but then a very well known theorem of Swan would imply that it is free",
"@SantanaAfton -- Mark explained how to construct a non-free group in which every finitely generated subgroup is free. This contradicts your intuition that since \"every finitely-generated subgroup is both proper and free\" ... the group must itself be free.",
"@W4cc0 on the other hand do you maintain your claim that such a group has to be countable? or do you want to require the group to be countable? (I guess the countable case and uncountable case are quite distinct questions.)",
"@YCor, I meant $G$ non-abelian (even if it makes sense in this case, it is easy to see that there are no abelian such groups), of course it would have no sense in the other case; but unfortunately I always write the opposite of what I would like to, sorry.",
"Higman constructed non-free groups of cardinality $\\aleph_1$ in which every countable subgroup is free. But every proper subgroup being free sounds hard.",
"@SantanaAfton It's overly simplistic, if not senseless, to think of a free group as a group with no relations.",
"@W4cc0 no, you don't: there are always free abelian subgroups. So the question was correctly stated, and wouldn't if \"non-abelian\" were added. Also as already mentioned, every group with your property has a non-abelian free subgroup.",
"@SantanaAfton Every finitely generated group of the rationals/Hawaiian earring is free but these are not free",
"@HJRW Are you referring to the HNN construction? In that example, if you take the relation $a^t=ab$ and the group generated by all letters involved, you don’t get a proper subgroup. This never happens in OP’s situation.",
"@SantanaAfton -- see Mark Sapir's comment above.",
"It feels as though no such group exists. Since every finitely-generated subgroup is both proper and free by assumption, how could the group have any nontrivial relations?",
"I should have been more clear, I meant free non-abelian.",
"@BenWieland a torsion-free abelian group in which every proper subgroup is cyclic, is itself cyclic, by an easy argument.",
"@MarkSapir $\\mathbf{Z}[3/2]=\\mathbf{Z}[1/2]$, so it's not a proper subgroup. Actually, the only unital subrings of $\\mathbf{Z}[1/2]$ are itself and $\\mathbf{Z}$. Also, every subgroup of $\\mathbf{Z}[1/2]$ is $\\{0\\}$, infinite cyclic, or has finite index and isomorphic to $\\mathbf{Z}[1/2]$ (namely equals $k\\mathbf{Z}[1/2]$ for some unique positive odd $k$.)",
"@MarkSapir Oh, yeah, but probably better to call the subgroup $3\\cdot\\mathbb Z[\\frac12]$",
"@BenWieland: $\\mathbb{Z}[1/2]$ is the derived subgroup of the Baumslag-Solitar group $BS(1,2)$, it contains non-free subgroup $\\mathbb{Z}[3/2]$",
"and this londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/…",
"$\\mathbb Z[\\frac12]$",
"arxiv.org/pdf/1903.03334.pdf",
"Sorry. I missed the infinitely Generated.",
"About the countability issue, I think I'm not aware of a non-free group in which every countable subgroup is free. (By analogy, $\\mathbf{Z}^\\mathbf{N}$ is not free abelian but all its countable subgroups are. Also, the Hawaiian earring group is locally free but has non-free countable subgroups.)",
"@PaulPlummer: I do not think it is clear (or even true) that these groups must be countable,",
"@MarkSapir and YCor Thanks, for some reason my comment sounded like a better idea when typing it, even though I knew of locally free nonfree groups.",
"@PaulPlummer: Locally free but non-free groups are easy to construct. Take a proper ascending HNN extension of the free group $F_k$, and the normal subgroup $N$ generated by $F_k$. For example $\\langle a,b,t \\mid a^t=ab, b^t=ba>$. The normal closure $N$ of $a,b$ is locally free but not free. The OP does not want this example. He wants all proper subgroups to be free, not just finitely generated subgroups.",
"@PaulPlummer I don't really understand your first comment. A free group itself has plenty of generating families under which it is not free, so using non-freeness of a given generating family will not help much.",
"Also why must such groups be countable? I am not seeing it immediately",
"@BenjaminSteinberg To complement Mark's comment, groups with Olshanskii's property are generated by every non-commuting pair."
] | 30 |
Science
| 0 |
170
|
mathoverflow
|
Characteristic subgroups and direct powers
|
Solved question: Suppose _H_ is a characteristic subgroup of a group _G_. Is it then necessary that, for every natural number _n_ , in the group $G^n$ (the external direct product of $G$ with itself $n$ times), the subgroup $H^n$ (embedded the obvious way) is characteristic?
Answer: No. A counterexample can be constructed where $G = \mathbb{Z}_8 \times \mathbb{Z}_2$ (here $\mathbb{Z}_n$ is the group of integers modulo $n$) with $H$ the subgroup
$$\\{ (0,0), (2,1), (4,0), (6,1) \\}$$
This subgroup sits in a weird diagonal sort of way and just happens to be characteristic (a quirk of the prime $2$ because there isn't enough space). We find that $H \times H$ is not characteristic in $G \times G$.
(ASIDE: The answer is _yes_ , though, for many important characteristic subgroups, including fully invariant subgroups, members of the upper central series, and others that occur through typical definitions. Since for abelian groups of odd order, all characteristic subgroups are fully invariant, the answer is yes for abelian groups of odd order, so the example of order $2^4$ has no analogue in odd order abelian groups.)
My question is this:
1. Strongest: Is it true that if _H_ is characteristic in _G_ and $H \times H$ in $G \times G$, then each $H^n$ is characteristic in $G^n$ [NOTE: As Marty Isaacs points out in a comment to this question, $H \times H$ being characteristic in $G \times G$ implies _H_ characteristic in _G_ , so part of the condition is redundant -- as explained in (2)]?
2. Intermediate: Is there some finite $n_0$ such that it suffices to check $H^n$ characteristic in $G^n$ for $n = n_0$? Note that if $H^n$ is not characteristic in $G^n$ for any particular _n_ , then characteristicity fails for all bigger $n$ as well. I'd allow $n_0$ to depend on the underlying prime of _G_ if we are examining $p$-groups.
3. Weakest: Is there a test that would always terminate in finite time, that could tell, for a given _H_ and _G_ , whether $H^n$ is characteristic in $G^n$ for all _n_? The "try all _n_ " terminates in finite time if the answer is _no_ , but goes on forever if the answer is _yes_. In other words, is there a finite characterization of the property that each direct power of the subgroup is characteristic in the corresponding direct power of the group?
ADDED: My intuition, for what it's worth, is that those subgroups _H_ of _G_ that can be characterized through "positive" statements, i.e., those that do not make use of negations or $\ne$ symbols, would have the property that $H^n$ is characteristic in $G^n$. On the other hand, those whose characterization requires statements of exclusion (_not a ..._) would fail because negative statements are difficult to preserve on taking direct powers. But I don't know how to make this rigorous.
|
https://mathoverflow.net/questions/35701/characteristic-subgroups-and-direct-powers
|
[
"gr.group-theory"
] | 20 | 2010-08-15T18:10:41 |
[
"Do you have a reference for \"if $H$ is fully invariant in $G$ then $H \\times H$ is fully invariant in $G \\times G$? Thanks in advance.",
"In the case of finite abelian groups, is it true that $\\operatorname{Aut}(G \\times G \\times G)$ is generated by automorphisms stabilizing one copy of $G$ ? This holds at least when $G=(\\mathbf{Z}/p^n\\mathbf{Z})^k$.",
"@Marty Isaacs: I have added this note to Question 1. I had already incorporated a similar observation into Question 2, as you probably noticed.",
"There is a redundancy in Question 1 since if $H \\times H$ is characteristic in $G \\times G$, then $H$ is automatically characteristic in $G$. This is because every automorphism of $G$ extends to an automorphism of $G \\times G$ stabilizing the first component.",
"No reference to a paper or book, but here's an online proof I wrote up: groupprops.subwiki.org/wiki/… It's rather long. What fails for the prime 2 is the first half of the proof: \"it is the direct sum of its intersections with the direct summands\" because in that part we use that the doubling map is an automorphism. The second half is bookkeeping that takes homomorphisms between cyclic subgroups and converts them to automorphisms by adding the identity map.",
"That odd prime statement sounds handy, do you have a reference for \"characteristic subgroups of abelian groups of odd order are fully invariant\"?"
] | 6 |
Science
| 0 |
171
|
mathoverflow
|
Large values of characters of the symmetric group
|
For $g$ an element of a group and $\chi$ an irreducible character, there are two easy bounds for the character value $\chi(g)$: First, the bound $|\chi(g)|\leq \chi(1)$ by the dimension of the representation and, second, the bound $|\chi(g) |\ \leq \sqrt{ |Z(g)|}$ by the centralizer arising from the formula $\sum_{\chi} |\chi(g)|^2= |Z(g)|$.
I am interested on upper bounds for characters of the symmetric group that improve slightly on the second bound.
More specifically, fix $\delta> 0$ small. I want to know for which $g\in S_n$ there exists an irreducible character $\chi$ with $|\chi(g)| > \sqrt{ |Z(g)|} e^{- \delta \sqrt{n}}$.
Recall that for a permutation $g\in S_n$ with $m_k$ cycles of size $k$ for all $k$, so that $\sum_k m_k k =n$, we have $|Z(g)| = \prod_k (k^{m_k} m_k!)$.
For example, known upper bounds for the dimension of representations of the symmetric group imply there does not exist such a character for $g$ the identity. On the other hand, such a character does exist if $g$ is an $n$-cycle, since the right side is less than $1$. I suspect and hope that this can only happen for $g$ containing some relatively large cycles.
I found works in the literature giving bounds for $\chi(g)$ that improve on the $|\chi(g)|\leq \chi(1)$ by a multiplicative factor ([Asymptotics of characters of symmetric groups related to Stanley character formula by Féray and Śniady](http://doi.org/10.4007/annals.2011.173.2.6)) or a power ([Characters of symmetric groups: sharp bounds and applications by Michael Larsen and Aner Shalev](https://doi.org/10.1007/s00222-008-0145-7)) but it's not obvious if it's possible to transform them into a bound of the form I need.
|
https://mathoverflow.net/questions/441235/large-values-of-characters-of-the-symmetric-group
|
[
"co.combinatorics",
"rt.representation-theory",
"symmetric-groups",
"characters"
] | 19 | 2023-02-20T06:22:56 |
[
"But now suppose we consider a permutation with $n/{3k}$ $k-1$-cycles, $k$-cycles, and $k+1$-cycles, or something like that. The square root of the centralizer is now $\\sqrt{ (n/3k)!^3 (k^3-k)^{n/3k} }$ which is smaller by a factor of roughly $\\sqrt{ 3^{n/k} e^{ - \\pi \\sqrt{2n/3}}}$. So the upper bound you desire should imply that slightly changing the size of the strips gives us vastly fewer border strip tableaux than before, which seems unlikely (instead of merely more cancellation, which is more feasible.)",
"@TimothyChow Actually, consider the following. If we consider a permutation with $n/k$ $k$-cycles, its centralizer has size $(n/k)! k^{n/k}$. This is the sum of the squares of the character values and the number of character values is $e^{ \\pi \\sqrt{2n/3}}$ so one character value has to be at least $\\sqrt{ (n/k)! k^{n/k} e^{ - \\pi \\sqrt{2n/3}}}$. So the number of border strip tableaux is at least that number as well.",
"@SamHopkins Yes, good point, and an \"unsigned combinatorial description for the square of the characters\" as ruled out by this paper would certainly be very convenient for proving the bound I want if it was possible.",
"Of course this does not in any way preclude bounds of the form you are interested in, but it might also be worth pointing out on the negative side of things that there have been recent dramatic advances which show that symmetric group character values are computationally hard: arxiv.org/abs/2207.05423",
"@TimothyChow For $n=4$ if $g$ is a $3$-cycle and a $1$-cycle and the character corresponds to the partition $3+1$ then $\\sqrt{ | Z(g)|} =\\sqrt{3}<2$ and the number of border-strip tableaux is either $2$ or $0$ depending on the order of the cycles. So maybe the truth or falsity depends on the order. Another evidence is that this bound is true is if all the cycles are the same size so that there is no choice of order, by Corollary 9 of the paper A bijection proving orthogonality of the characters of $S_n$ by Dennis E White mentioned in a deleted answer.",
"My first instinct is to try Murnaghan-Nakayama, but I don't know if that will work. Best-case scenario would be that the number of border-strip tableaux is already $\\le \\sqrt{|Z(g)|}$ (so that you don't even have to worry about sign-cancellation), but maybe this is false?",
"@PerAlexandersson Thanks! It seems to me that one of these formulas will be helpful if there exists a relatively simple combinatorial proof of the bound $\\chi(g)^2 \\leq |Z(g)|$, because then one can try to study the proof more carefully to see when it can be sharp. So far I don't see how to do that for any of them, but it could be possible...",
"Some various combinatorial formulas for Sn-characters can be found here: symmetricfunctions.com/murnaghanNakayama.htm"
] | 8 |
Science
| 0 |
172
|
mathoverflow
|
What algebraic properties are preserved by $\mathbb{N}\leadsto\beta\mathbb{N}$?
|
Given a binary operation $\star$ on $\mathbb{N}$, we can naturally extend $\star$ to a semicontinuous operation $\widehat{\star}$ on the set $\beta\mathbb{N}$ of ultrafilters on $\mathbb{N}$ as follows:
$$\mathcal{U}\mathbin{\widehat \star}\mathcal{W}:=\\{A:\\{k: \\{a: a\star k\in A\\}\in\mathcal{U}\\}\in\mathcal{W}\\}.$$
It's a standard (and [quite useful!](https://web.williams.edu/Mathematics/lg5/Hindman.pdf)) fact that associativity is preserved under this transformation: if $\star$ is associative, then so is $\widehat{\star}$. On the other hand, commutativity [is _not_ preserved](https://math.stackexchange.com/questions/87109/addition-on-ultrafilters-is-non-commutative). Say that an equational sentence $\sigma$ involving a single binary function symbol is a **$\beta$ -property** iff we have $$(\mathbb{N};\star)\models\sigma\quad\implies\quad (\beta\mathbb{N};\widehat{\star})\models\sigma$$ for every binary operation $\star$ on $\mathbb{N}$.
> **Question** : Which equational sentences are $\beta$-properties?
Of course there is no need to restrict attention to equational properties per se _(which is why I've used "$\implies$" rather than "$\iff$" in the definition of $\beta$-property)_, and we can similarly transform arbitrarily many arbitrary-arity operations on $\mathbb{N}$; however, already this question seems difficult. I recall seeing a partial(?) answer to this question which involved the way variables were/were not _reordered_ in the left vs. right hand sides of $\sigma$, but at the moment I can't track it down.
* * *
Incidentally, the following question is at least of a related _flavor_ , and may involve relevant ideas (and be easier to tackle at first): what is $(\beta\mathbb{N};\widehat{+})$ _(or $(\beta\mathbb{Z};\widehat{+})$ using the analogous definition)_ like from a universal-algebraic perspective? For example, is it congruence modular?
|
https://mathoverflow.net/questions/427135/what-algebraic-properties-are-preserved-by-mathbbn-leadsto-beta-mathbbn
|
[
"set-theory",
"lo.logic",
"gn.general-topology",
"model-theory",
"universal-algebra"
] | 19 | 2022-07-22T17:24:11 |
[
"@BenjaminSteinberg xy=zw is also preserved. :P",
"One thing that is preserved of course is either xy=x or xy=y",
"Yes. I think chapter 6 talks about free subsemigroups. Of course it doesn't looks at nonassociative magmas. My impression is that few general semigroup identities are preserved.",
"@BenjaminSteinberg Is that book Algebra in the Stone-Cech compactification?",
"There are some results in Hindman's book on conditions guaranteeing the Stone-Cech compactification of a semigroup contains a free semigroup on 2 generators and hence satisfies no semigroup identity. If is of course possible for the original semigroup to satisfy an identity and so this identity is not preserved. Commutativity is such and identity and so are Malcev's nilpotent identities.",
"Notice also many quasiidentities like cancellativity are not preserved."
] | 6 |
Science
| 0 |
173
|
mathoverflow
|
Mumford-Tate conjecture for mixed Tate motives
|
Let $X$ be a (not necessarily smooth or proper) variety over a number field $k$. Suppose we are given
1. A subquotient $V_{dR}$ of the algebraic de Rham cohomology $H_{dR}^i(X)$ (defined in the non-smooth case via a thickening, as in e.g. Hartshorne's algebraic de Rham cohomology paper),
2. For each embedding $\iota: k\to \mathbb{C}$, a subquotient $V_\iota$ of $$H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})$$ respecting the $\mathbb{Q}$-mixed Hodge structure, and
3. For some fixed prime $\ell$ and algebraic closure $\bar k$ of $k$, a subquotient $V_\ell$ of $$H^i(X_{\bar k, \text{ét}}, \mathbb{Q}_\ell)$$ respecting the action of $\text{Gal}(\bar k/k)$, such that
4. For each $\iota$, the corresponding comparison isomorphism $$H^i_{dR}(X)\otimes_k \mathbb{C} \to H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})\otimes \mathbb{C}$$ sends $V_{dR}$ to $V_\iota$, and
5. For each $\iota$, the corresponding comparison isomorphism $$H^i(X_{\mathbb{C}}^{an}, \mathbb{Q})\otimes \mathbb{Q}_\ell \to H^i(X_{\bar k, \text{ét}}, \mathbb{Q}_\ell)$$ sends $V_\iota$ to $V_\ell$, and
6. The $V_\iota$ is of mixed Tate type (i.e. all $h^{p,q}$'s vanish for $p\not=q$). Or equivalently $V_\ell$ is an iterated extension of powers of the cyclotomic character.
(If you'd like, you can think of $V$ as the realization of a mixed Tate motive over $k$ under any of the various formalisms for mixed Tate motives over number fields -- I think that a priori the conditions above are slightly weaker but expected to be equivalent.)
> I'd like to know -- is the analogue of the Mumford-Tate conjecture known to be true for such $V$? Explicitly, one expects that the Lie algebra of the image of $$\text{Gal}(\bar k/k)\to GL(V_\ell)$$ is the extension of scalars to $\mathbb{Q}_\ell$ of the Lie algebra of a $\mathbb{Q}$-group, given as the Tannaka dual of the subcategory of mixed Tate Hodge structures over $k$, generated by the Hodge realization $(V_{dR}, V_{\iota_1}, V_{\iota_2}, \cdots)$.
My feeling is that one can probably extract this from the literature by comparing ranks of various Ext groups (in one's favorite category of mixed Tate motives and in the $\ell$-adic and Hodge settings) but I am hoping it is written explicitly somewhere.
**EDIT:** Fixed some errors in the formulation of the Hodge realization.
|
https://mathoverflow.net/questions/379972/mumford-tate-conjecture-for-mixed-tate-motives
|
[
"ag.algebraic-geometry",
"nt.number-theory",
"galois-representations",
"mixed-hodge-structure"
] | 19 | 2020-12-29T09:18:09 |
[
"@DonuArapura It seems like I might have to, since I have an application in mind!",
"Hi Daniel. I'm not sure it is written down anywhere, so it would be nice if you did."
] | 2 |
Science
| 0 |
174
|
mathoverflow
|
Is there a classification of reflection groups over division rings?
|
I asked a [version](https://math.stackexchange.com/questions/3491953/reflection-groups-of-division-rings) of this question in Math StackExchange about a week ago but I've received no feedback so far, so following the advice I received on [meta](https://meta.mathoverflow.net/questions/4416/would-this-question-be-appropriate-for-mathoverflow) I decided to post it here.
* * *
## Details
The irreducible finite [complex reflection groups](https://en.wikipedia.org/wiki/Complex_reflection_group) were classified by Shephard and Todd. The list consists of a three-parameter family of imprimitive groups and 34 exceptional cases. If $\mathbb{K}$ is any field of characteristic zero, it is known (see e.g. Section 15-2 [here](https://www.springer.com/gp/book/9780387989792)) that if a group has a representation as a $\mathbb{K}$-reflection group then it also has a representation as a complex reflection group, so we get no new examples for fields. I'm interested to know if an analogue of this classification exists when we allow $\mathbb{K}$ to be a division ring of characteristic zero.
I have only found partial results. The irreducible finite quaternionic reflection groups were classified in [this paper](https://www.sciencedirect.com/science/article/pii/0021869380901489) by Cohen. Finite $\mathbb{K}$-reflection groups of rank $1$, or equivalently finite subgroups of division rings of characteristic $0$, were classified in [this paper](https://www.jstor.org/stable/1992994?seq=1) by Amitsur. The classification for rank $2$ can perhaps be extracted from the [classification of finite subgroups of $GL(2,\mathbb{K})$](https://core.ac.uk/download/pdf/82740228.pdf) by Banieqbal using a case-by-case check.
In view of the complex and quaternionic lists, I would expect the full classification (if there is one) to follow a form similar to this:
* An infinite family $G_n(M,P,\alpha)$ of imprimitive reflection groups of rank $n$, where $M$ is a finite subgroup of a division ring $\mathbb{K}$ (i.e. an Amitsur group) and $[M,M]\le P \trianglelefteq M$, possibly with some extra data $\alpha$ in low rank. Algebraically it should correspond to something like the group of generalized permutation matrices with entries in $M$ whose determinant is in $P$, like in the cases of fields and quaternions (note that order doesn't matter when computing the determinant, since $[M,M]\le P$).
* A family or families of examples in rank $2$ (or perhaps in rank $\le m$ where $m^2$ is the dimension of $\mathbb{K}$ as a division algebra over its center). In the quaternionic case they are the primitive reflection groups whose complexification is imprimitive, and are all constructed from certain $2$-dimensional primitive complex reflection groups; it's not clear to me how this construction could generalize to arbitrary $\mathbb{K}$.
* A number of exceptional cases of small rank. These are the ones I'm most interested in. In the quaternionic case these are precisely the primitive reflection groups whose complexification is also primitive; in the general case they might correspond to primitive reflection groups which remain primitive after tensoring the representation with a splitting field.
* * *
## Question
My main question is thus:
> Is there a classification of groups representable as a $\mathbb{K}$-reflection group over some division ring $\mathbb{K}$ of characteristic zero?
I would also appreciate any references dealing with this problem or with particular cases. If the classification turns out to be intractable, or currently out of reach, I would ask if at least an example can be found of a new exceptional reflection group of rank $\ge 3$ (see details above).
* * *
### Update
After much searching, I finally found a brief reference to this problem in the literature. The mention occurs at the end of Section 3 in [this 1981 paper](https://darkwing.uoregon.edu/%7Ekantor/PAPERS/GenLinearGps.pdf) by Kantor, which I quote here for convenience:
> [...] For example, consider the problem of determining all finite primitive reflection groups $G$ in $GL(n,D)$, for $D$ an arbitrary noncommutative division ring of characteristic $0$. If $n=1$, this is just the famous problem solved by Amitsur (1955) (and independently and almost simultaneously by J. A. Green). If $n=2$ and $G$ is solvable, the problem seems to involve even more difficult number theory than Amitsur used. But if $n\ge 3$, and if simple group classification theorems are thrown at the problem, no new nonsolvable examples arise. [...]
If the last sentence is true, it means that the new examples of "exceptional groups" I asked for must necessarily be solvable. However, the author does not provide any in-text citation for that statement, and I haven't been able to find which result is being alluded to.
|
https://mathoverflow.net/questions/349809/is-there-a-classification-of-reflection-groups-over-division-rings
|
[
"gr.group-theory",
"rt.representation-theory",
"division-rings",
"reflection-groups"
] | 19 | 2020-01-06T00:40:10 |
[] | 0 |
Science
| 0 |
175
|
mathoverflow
|
xkcd's "Unsolved Math Problems", straight lines in random walk patterns
|
STEM student's favourite source of amusement posted a comic titled "Unsolved Math Problems" one of which looks like something that could actually be tackled.
> If I walk randomly on a grid, never visiting any square twice, placing a marble every N steps, on average how many marbles will be in the longest line after NK steps?
[Original comic (with ilustration of the problem)](https://xkcd.com/2529)
I know there are some results in form of power laws (critical exponents) like the 0.587597... constant for SARW in 3D.
Can we say anything interesting about the process described in the comic?
|
https://mathoverflow.net/questions/406469/xkcds-unsolved-math-problems-straight-lines-in-random-walk-patterns
|
[
"random-walks"
] | 19 | 2021-10-18T01:49:48 |
[
"@Keba if you get boxed in discard that walk and begin again. A better way to think of the question might be: consider ALL possible SAWs of NK steps with a marble placed at every Nth step. What is the average number of marbles on the longest line?",
"@ZachTeitler, re, a (literally) bold new notation for the $5$-adics?",
"@Gro-Tsen It's a \"conjection\" whatever that is. (Also: I wonder what $\\mathbb{5}$ is supposed to mean.)",
"@SamHopkins A horizontal or vertical line through the origin will have $\\sqrt{K/N}$ marbles on average and I would wildly guess that the correct answer is within $(KN)^\\epsilon$ of this. It looks conceivable to prove this by bounding for each line the expectation of the $\\ell$th power of the number of marbles on the line and summing over all lines passing through at least a few points within a distance $NK$ of the origin.",
"If we consider the analogous process for a random walk which is not self-avoiding, is this understood? (SAWs are much harder than regular random walks, from what I understand…)",
"@Gro-Tsen: $\\epsilon$ cannot be $\\lt 0$.",
"I believe deeplinking to images on XKCD is discouraged, so I changed the link to point to the comic.",
"What I very much want to know is: is the Euler field manifold hypergroup isomorphic to a Gödel-Klein meta-algebreic $\\epsilon<0$ quasimonoid connection under Sondheim calculus? (Also, what in Apollo's name is going on with this curve?)",
"@MartinHairer Is this 'dropping a marble' part redundant then? Is it roughly equivalent to how many collinear gridpoints were visited?",
"@Keba either seems interesting but I think the original intention was point-like marbles.",
"The self-avoiding random walk with $N$ steps is usually simply defined as the uniform measure on the set of all self-avoiding walks of $N$ steps. It doesn't have any kind of Markov property, but is conjectured in $2D$ to converge to an SLE curve with self-similarity exponent $3/4$. Rigorously, almost nothing non-trivial is known about its large-$N$ behaviour.",
"I feel like we are getting nerd sniped again...",
"Also, how large are these marbles? Do they fill the whole square? Are they just points?",
"I'm no expert on random walks so I probably miss something easy, but why is this process well-defined? I might corner myself in?"
] | 14 |
Science
| 0 |
176
|
mathoverflow
|
Is there a simpler proof of the key lemma in the paper by Hiroshi Iriyeh and Masataka Shibata on the 3D Mahler conjecture?
|
In this [remarkable paper](https://arxiv.org/abs/1706.01749) 30 pages are occupied by the proof of the following innocently looking lemma:
Let $K$ be an origin-symmetric convex body in $\mathbb R^3$. There exist three planes through the origin splitting $K$ into $8$ parts of equal volume and such that each two of these planes split the cross-section of $K$ by the third one into $4$ parts of equal area.
I cannot shake off the feeling that there must be a half-page proof of this statement though I don't have one yet. I also know that MO is swarming with good topologists. Anybody up to the challenge?
|
https://mathoverflow.net/questions/271972/is-there-a-simpler-proof-of-the-key-lemma-in-the-paper-by-hiroshi-iriyeh-and-mas
|
[
"gt.geometric-topology",
"convex-geometry"
] | 19 | 2017-06-11T18:05:22 |
[
"it might be of interest: in the first part of the proof of Theorem 1 keithmball.files.wordpress.com/2014/11/… Ball constructs a similar map to simplex as Makeev does but he uses Browers Fixed Point Theorem (BFPT) unlike Makeev. So it quite might be that one needs to apply BFPT in a proper way which I don't see yet how.",
"@fedja yes, the claim is indeed bit different, but the topological argument is still the same",
"@FedorPetrov Except in our case the measures to partition depend on the planes. However the idea that if the solution is unique in some generic position, then there is an odd number of solutions in every generic position is amazing and certainly very promising. It should be rather standard, of course, but the good side of ignorance (mine) is the possibility to get surprised with the facts everybody else considers routine :-).",
"(continued) Makeev partitions two measures which are absolutely continuous w.r.t. Lebesgue measure, but actually this absolute continuity is not essential.",
"Roman Karasev (private communication) suggests that this essentially follows from the results of [V.V. Makeev. Equipartition of a continuous mass distribution. Journal of Mathematical Sciences, 140:4 (2007), 551--557, mathnet.ru/links/a1722f058ed63ed36cb13d9e67fd255f/znsl299.pdf, theorems 5 and 6]",
"Now one can hope that geometric considerations will show that (in this \"generic\" situation) the intersection class in $H^6((\\mathbb{RP}^2)^3,\\mathbb{Z}/2\\mathbb{Z})$ will be nontrivial, and that furthermore that there is no contribution from some degenerate case e.g. two planes coinciding. Then it probably still remains necessary to pass to some perturbation of $K$, and to run this argument there instead. So even if this sketch can be made to work, I doubt that a fully rigorous proof would fit in a half-page...",
"Some naive thoughts: We can try and derive this from an intersection theory statement on $(\\mathbb{RP}^2)^3$. Points correspond to triples of planes, and the condition that three planes split $K$ into $8$ parts of equal volume will give a codimension $3$ cycle for \"generic\" $K$. (We get codimension $3$ instead of codimension $7$ because diametrically opposite regions have the same area, by reflection symmetry of $K$). Similarly each of the conditions that two planes split a cross-section into four parts of equal area should lead to a codimension $1$ cycle."
] | 7 |
Science
| 0 |
177
|
mathoverflow
|
Checking Mertens and the like in less than linear time or less than $\sqrt{x}$ space
|
Say you want to check that $|\sum_{n\leq x} \mu(n)|\leq \sqrt{x}$ for all $x\leq X$. (I am actually interested in checking that $\sum_{n\leq x} \mu(n)/n|\leq c/\sqrt{x}$, where $c$ is a constant, and in finding the best $c$ such that this is true in a given range, such as $3\leq x\leq X$, say; all of these problems are evidently related.)
The natural algorithm (the same you can find in section 4 of [1], say) has running time $O(X \log \log X)$ and takes space $O(\sqrt{X})$ (where we think of integers as taking constant space). I've coded it with some optimizations (for $\sum_{n\leq x} \mu(n)/n$, using interval arithmetic) and should have a result for $x=10^{14}$ in about a fortnight; $x=10^{12}$ takes an afternoon. (When I first coded this, less carefully and on worse hardware, it took a week.)
Is it possible to do things in either less time or less space? Space is important here in practice - ideally you would want to keep everything in cache, and it is rare to have more than 4MB per processor core - that is, enough for $1.6\cdot 10^7$ values of $\mu$, or $5\cdot 10^5$ large integers.
Notice that I am asking for a check for all $x\leq X$, and not just for a single $x=X$.
* * *
Further remarks: I am aware that there are algorithms for computing a single value of $\sum_{n\leq x} \mu(n)$ in time $O(x^{2/3} \log \log x)$ [2], or even (in what looks like a more sophisticated but less practical way that may have never been coded) in time $O(x^{1/2+\epsilon})$ ([3]; see also [Mertens' function in time $O(\sqrt x)$](https://mathoverflow.net/questions/95726/mertens-function-in-time-o-sqrt-x)). Once can of course use such an algorithm to compute the sum for values of $x$ that are about $c\sqrt{x}$ apart, and then use what I've called the "natural" algorithm to deal with intervals in which such a computation shows that the statement to be verified could be violated. This results in a running time of $O(x^{7/6} \log \log x)$, or $O(x^{1+\epsilon})$, and while memory usage could be decreased by applying the "natural" algorithm on intervals of length smaller than $\sqrt{x}$, this would result in a longer running time.
The same is true, of course, of the "natural" algorithm itself: one could store a list of primes $p\leq \sqrt{x}$ in the main memory in $\sqrt{x}$ bits, and compute $\mu(m)$ in blocks of size $M$ at a time; this would require working in memory $O(M)$ (to be stored in the cache) for the most part, but the running time would be $O(x^{3/2}/M)$. Or is there a way around this? Can one, say, select the primes $p\leq \sqrt{x}$ that might divide integers in an interval of length $M \lll \sqrt{x}$, and do so in time less than $\sqrt{x}$ or $\sqrt{x}/\log x$? Or is there any other way to take space substantially less than $O(\sqrt{x})$ while still having essentially linear running time? Or a way to have better than linear running time?
PS. I could also ask about parallelising this, but I would not like to risk making the discussion too hardware-specific here.
[1] _François Dress_ , MR 1259423 [**Fonction sommatoire de la fonction de Möbius. I. Majorations expérimentales**](http://projecteuclid.org/euclid.em/1048516214), _Experiment. Math._ **2** (1993), no. 2, 89--98.
[2] _Marc Deléglise and Joël Rivat_ , MR 1437219 [**Computing the summation of the Möbius function**](http://projecteuclid.org/euclid.em/1047565447), _Experiment. Math._ **5** (1996), no. 4, 291--295.
[3] _J. C. Lagarias, and A. M. Odlyzko_ , MR 0890871 [**Computing $\pi(x)$: an analytic method**](http://dx.doi.org/10.1016/0196-6774%2887%2990037-X), _J. Algorithms_ **8** (1987), no. 2, 173--191.
|
https://mathoverflow.net/questions/237308/checking-mertens-and-the-like-in-less-than-linear-time-or-less-than-sqrtx-s
|
[
"nt.number-theory",
"analytic-number-theory",
"computational-complexity",
"computation"
] | 19 | 2016-04-25T12:22:32 |
[
"Well, numbers of size about $X$ without prime factors between $X^(1/3)$ and $X^{1/2}$ are very roughly as common as the primes - but the sum of the reciprocals of the primes diverges. Or rather - the sum $\\sum_n 1/n$ over all $X\\leq n<2X$ without such prime factors is about a constant times $1/\\log X$, i.e., much larger than $1/\\sqrt{X}$. The same happens if you work over much smaller intervals. So I'm afraid this cannot work as stated.",
"You might consider \"faulty\" versions of the natural algorithm, where one computes the wrong value on some of the values, or forgets to store the primes above cube root of X. You might still be able to show the error is small enough that the inequalities you want to check are preserved. Gerhard \"Faulty Computing Isn't Computationally Wrong\" Paseman, 2016.04.25.",
"And yes, the running time is as you said and I said. I'm looking for something better.",
"Gerhard: that's exactly what I mean by \"the natural algorithm\".",
"the simplest algorithm for checking $M(x)$ not for every $x \\le X$ but for every $x = \\lfloor X/n\\rfloor$ is by using $\\sum_{n=1}^x M(x/n) = 1$, and computing $M(k)$ for every different $k \\in \\{ \\lfloor X/n \\rfloor \\}$, it takes $\\mathcal{O}(X)$ additions and $\\mathcal{O}(\\sqrt{X}\\log X)$ in space",
"It may be that I am talking about the \"natural algorithm\" already, which is a variant of the one at mathoverflow.net/a/50691 for determining distinct prime factors of integers up to X in a serial fashion. However I still wonder what parameter you are talking about for space usage. Gerhard \"Space Usage: The Final Frontier\" Paseman, 2016.04.25.",
"Do you mean $\\sqrt{x}$ or $\\sqrt{X}$? If the latter, there is a simple sieving process that computes the sum on the fly, and you can save extremal values for later processing and check that the inequality is satisfied. You can even sift just the squarefree values for processing. The memory storage is $O(\\log X)$ bits for each prime at most $\\sqrt{X}$. The runtime should be not much longer than $O(X\\log\\log X)$. Gerhard \"Sometimes Has Xtreme Case Sensitivity\" Paseman, 2016.04.25."
] | 7 |
Science
| 0 |
178
|
mathoverflow
|
Which manifolds decompose into pants?
|
In this [nice paper](http://arxiv.org/abs/math/0205011) Mikhalkin uses certain (more geometrical than algebraic) aspects of tropical geometry to prove that every complex projective hypersurface in $\mathbb C \mathbb P ^n$ decomposes as a smooth manifold into some _$(n-1)$-dimensional pair-of-pants_.
Following Mikhalkin, a $k$-dimensional pair-of-pants is a particular real $(2k)$-manifold with corners. It is obtained by removing $k+2$ generic hyperplanes from $\mathbb C \mathbb P^k$. For instance, a $1$-dimensional pair-of-pants is a sphere minus 3 points (as everybody knows), while a $2$-dimensional pair-of-pants is $\mathbb C \mathbb P^2$ minus 4 generic lines.
Boundaries and corners arise when you take the compact version of these pair-of-pants, i.e. you remove an open regular neighborhood of the $k+2$ generic hyperplanes:
* When $k=1$ you get the genuine compact pair-of-pants $P$ with 3 boundary components.
* When $k=2$ you get a real 4-manifold with boundary and corners: the boundary consists of four compact 3-manifolds homeomorphic to $P\times S^1$ (corresponding to the four lines), glued together along six tori (corresponding to the six intersections between the lines). The tori are the corners.
The 4-dimensional manifold with corners we get with $k=2$ is a nice object, which may look intriguing to a low-dimensional topologist.
As everybody knows, not only complex hypersurfaces (i.e. curves) in $\mathbb C \mathbb P^2$ decompose into pair-of-pants: every oriented surface of negative Euler characteristic does! It would be then natural to ask the following:
> Which compact 4-manifolds decompose into pair-of-pants?
The set of course includes all complex hypersurfaces in $\mathbb C \mathbb P^3$.
**Edit:** It is not true that every curve in $\mathbb C \mathbb P^2$ decomposes into pants: a line or a cubic give $S^2$ and $T^2$ and they of course do not decompose into pants. Reading more carefully Mikhalkin's paper, it seems to me that he also allows to collapse the natural fibering of the boundaries of the blocks: for instance, by collapsing the boundaries of a two-dimensional pair-of-pants we get $S^2$.
|
https://mathoverflow.net/questions/113875/which-manifolds-decompose-into-pants
|
[
"gt.geometric-topology",
"tropical-geometry",
"4-manifolds"
] | 19 | 2012-11-19T13:53:05 |
[
"I think that the following holds: let $X$ be obtained by gluing $n$ Pairs-of-Pants, and suppose that $X'$ is obtained by the same gluing rules, except that you add a Dehn twist in one of the gluing faces. Then, in analogy with the Lickorish-Wallace theorem, $X'$ should be obtained from $X$ by a single surgery along a (probably homologically essential, zero-square) torus.",
"Thinking a bit more about your comment, I realized that Mikhalkin probably also admits the collapsing of the natural fibrations of the boundaries of the pants: the situation is slightly more complicated than I guessed, I have edited the text accordingly.",
"Hi Marco. I would put no restriction at all (I can't see any reasonable restriction...). You glue along diffeomorphisms of blocks of type $P\\times S^1$: every such diffeomorphism must preserve the fiber, but there are infinitely many non-isotopic choices. You are absolutely right about $\\chi$, I must admit I didn't think about it. It might be that you get only finitely many manifolds for each $\\chi$, but since you might choose among infinitely many gluing maps between the $P\\times S^1$ faces, this is not clear (to me).",
"Hi Bruno! What sort of restrictions do you put on the gluing maps between (not necessarily different?) Pairs-of-Pants? For example, how would you obtain $\\mathbb{CP}^2$ in this picture? Is it explained in Mikhalkin's paper? In any case, it looks like a necessary condition is to have $\\chi>0$, since your building blocks all have $\\chi=1$, and both the submanifolds you glue them along and the corners have $\\chi=0$. In particular, $\\chi$ of your manifolds seems to determine how many PoPs you need."
] | 4 |
Science
| 0 |
179
|
mathoverflow
|
Can a number be palindromic in more than 3 consecutive number bases?
|
_$2017:$ Was initially asked on [MSE](https://math.stackexchange.com/questions/2234587/can-a-number-be-palindrome-in-4-consecutive-number-bases)_ \- but wasn't solved or updated there since.
**Update $2019$:** I've returned to this problem, made some progress and updated the post here.
(I've basically rewritten this entire post here)
* * *
## **Introduction and problem**
An $k$-palindrome is a number palindromic in $k$ consecutive number bases. Here, I'm wondering about the existence of $k=4$. Note that if it does not exits, $k\gt 4$ can't exist by definition.
Also note that if a number is even length (number of digits is even) palindrome in base $b$, then it is divisible by $b+1$ and thus can't be palindromic in that number base. So we can look for odd length cases of digits in the first (last, depending on how you define it) base only.
> Can a natural number be _nontrivially_ [palindromic](https://en.wikipedia.org/wiki/Palindromic_number) in **more than** $3$ consecutive number bases?
>
> _Nontrivially_ means that I'm not counting one-digit palindromes.
Smallest number $N$ which is nontrivially palindromic in $k$ consecutive number bases:
$$ \begin{array}{|c|} \hline k& N & \text{Palindrome} \\\ \hline 1& 3 & 11_2 \\\ 2& 10 & 101_3=22_4\\\ 3& 178 & 454_6 =343_7 = 262_8\\\ 4& ?& \\\ \hline \end{array} $$
Where the index denotes the number base representation. For example $11_2$ is three in binary.
Note that $k=1$ is not special as those are just palindromes and easily can be constructed.
It is not hard to see $k=2,3$ have infinitely many examples. But finding all of them is hard.
> I conjecture solution for $k=4$ does not exist. That is, there are no $4$-palindromes.
My **question** here is:
> Can you provide arguments (help) or direction (on proving) why (why not) would this be true?
My ideas are presented below:
**First approach to proving this**
One idea to prove this is proving $(1)$ and $(2)$ which would imply this. The $(1)$ is now proven. The $(2)$ will be much harder to prove. The details are included below. Here are the claims:
* $(1)$ A $3$ digit (when written in palindromic bases) number can't be a $4$-palindrome.
* $(2)$ To be a $4$-palindrome, a number must have $3$ digits (when written in palindromic bases).
The second claim is based on the fact that _almost-all_ "almost $4$-palindromes" have $3$ digits. That is, numbers palindromic in three out of four consecutive bases (that is, for example in bases $b,b+3$ and either $b+1$ or $b+2$).
_Almost-all_ since the only two known "almost $4$-palindromes" that do not have $3$ digits are $71240, 1241507$ which have $5$ digits (when talking about digits, they are counted in the "pivot" number base among the consecutive bases in which the number is palindromic - "pivot" being either the smallest or largest base).
> Proving $(2)$ is equally hard as proving $(3)$ from alternate approach below, since all cases of $d$ must be resolved. The $(1)$ specifically asked to resolve $d=3$ which was accomplished.
**Second approach to proving this**
> $(3)$ I actually conjecture I have found all $3$-palindromes. If this is true, then a $4$-palindrome does not exist since neither of my $k=3$ solutions can be extended to a fourth consecuitve number base.
More details are linked at the end of the next section:
## **" $k$-problem system" results and conjectures**
What I was able to prove both computationally and by hand:
> Expanding $(1):$ In short, I have proven I have all $3$ digit $3$-palindromes, and nontrivial $d\le 3$ solutions don't exist. Now it is easy to see now that a $3$ digit $3$-palindrome can't be palindromic in fourth consecutive base, which means $4$-palindromes with $d\le 3$ digits do not exist. (Digits counted when written in palindromic number bases).
>
> More details about this are linked at the end of this section, as well.
Expanding $(3):$ What I was able to conjecture is presented below:
In short, we can write the problem of finding $k$-palindromes in form of solving linear diophantine equations whose order grows when solving for longer palindromes of length $d$ (number of digits). Lets call that system needed to be solved "$k$-problem system".
There is the "general", and there is the "non-general" case of this problem system.
The non-general $k$-problem system is related to finding $d=2l+1$ (WLOG $d$ is odd) digit solutions, where all representations in consecutive bases have exactly $d$ digits.
The general $k$-problem system allows different lengths of digits in those $k$ consecutive number bases, where "pivot" number base among these $k$ consecutive number bases, usually smallest or largest, is taken to have $d$ digits.
The $k=2$ case has infinitely many solutions for every case of $d$. It is hard to find them all.
The $k=3$ is what will be the case of the "problem-system" from now on:
> Expanded on $(3)$ in this context:
>
> For the non-general case of the "problem-system", I have computationally solved (proved) I have all solutions for two smallest cases, $d=3,5$. I also strongly conjecture same is true for $d=7$. I also conjecture based only on computation, that a $3$-palindrome solution for $d\ge 9$ does not exist. That is, I conjecture I have all solutions.
>
> For the general case of the "problem-system", I have computationally solved (proved) I have all solutions for smallest case, $d=3$. I also conjecture I have all solutions for $d\ge 5$ since another conjecture I strongly believe is that the general case is equivalent to the non-general case, except for having an extra finite set of solutions for sufficiently small bases $b$ which I have collected and computed all (conjectured).
* * *
_Summing this all up_ : the important thing is I have all $3$ digit $3$-palindromes and that they can't be $4$-palindromes. I have all conjectured solutions for $3$-palindromes, but unable to prove the fact that large $d$ can't have solutions for $k=3$.
* * *
Note that I was trying to find counterexamples to these conjectures for a long time now but couldn't.
The more details as promised: A post mainly focusing on non-general "problem-system" [can be read on MSE](https://math.stackexchange.com/questions/3305696/hard-system-in-integers-related-to-natural-number-representations) if you are interested in more details, or if you want to see the "all $3$-palindromes" solutions for the non-general case (the extra solutions from general case are easily re-computable if my conjectures are true).
I'll also include all _infinite families_ (the non-sporadic solutions) for $k=3$ given there, here:
Given $k\in\mathbb N_0 = \mathbb N \cup \\{0\\}$, "infinite families" giving $3$-palindromes:
$$ \begin{array}{l,l,l,l} d & (a_i)&(c_i)k & b\\\ d=3 & (2,6)&(1,1)k & 2k+8\\\ d=5 & (31,32,0)&(3,2,1)k & 4k+47 \\\ d=7 & (34,50,10,74)&(1,1,1,1)k & 2k+76 \\\ d=7 & (8,33,0,41)&(1,3,1,3)k & 6k+58 \\\ d=7 & (112,15,0,36)&(4,0,1,0)k & 6k+175 \\\ d=7 & (227,160,187,200)&(5,3,5,3)k & 6k+280 \\\ d=7 & (5,23,6,14)&(2,6,5,0)k & 12k+39 \\\ d=7 & (93,78,30,50)&(10,6,7,0)k & 12k+119 \\\ d=7 & (47,150,249,26)&(2,6,11,0)k & 12k+291 \\\ \end{array} $$
Where these give digits of $3$-palindromes in base $b$, which are also palindromic in $b-1,b-2$.
That is, for example, observing the second row (family), the only $d=5$ case infinite family, means $(31+3k,32+2k,0+1k,32+2k,31+3k)$ are digits of a $3$-palindrome in base $4k+47$, for every $k$. We can convert this to a decimal value easily.
> $(3):$ If we can prove these are all of the solutions for $k=3$ (and that sporadic solutions exist only for sufficiently small bases as I conjectured), we have solved the $k=4$ problem, the $4$-palindrome problem: There does not exist a number palindromic in four consecutive number bases.
The main thing needing proving is that $d\ge 9$ can't have solutions. Or can you find a counterexample, and actual solution? That would make this much more interesting.
The $d=9$ specifically for example will be very surprising if it had a solution (I found the period of solutions if they exist, must be larger than $500$ in this case, which is unlikely given $d=3,5$ have periods $2,4$ and $d=7$ has periods $2,6,12$). Period being families producing a solution every period amount of bases.
|
https://mathoverflow.net/questions/268590/can-a-number-be-palindromic-in-more-than-3-consecutive-number-bases
|
[
"nt.number-theory",
"diophantine-equations",
"palindromes"
] | 19 | 2017-04-29T08:32:32 |
[
"@GerryMyerson Regarding a relevant sequence, I think A279093 is updated with all known 3-palindromic solutions.",
"Loosely related: oeis.org/A214425",
"@joro Update: Now it is known that the $3$ digit solution does not exist. I will have to update this post.",
"@JoseBrox If we have $n$ and $n+3$, and we also include either $n+1$ or $n+2$ and \"observe almost palindromic in four consecutive\", then surprisingly, only 3 digit palindromes seem to have a chance to be palindromic in four consecutive bases ~ Updated the question",
"@Vepir Yes, sorry, I meant $n$ and $n+3$. I was just trying to understand if the obstruction for the existence of consecutive palindromes for $k=4$ (in case it exists) is of an algebraic/number-theoretic nature or rather of an analytic/combinatoric nature (which is what the existence of examples for $n$ and $n+3$ suggests).",
"@JoseBrox Why $n, n+4$? There are a lot of examples, smallest one being $7=111_2=11_6$. You can easily modify the linked python code to generate them. If you meant $n, n+3$ there are again a lot of examples, smallest being $31=11111_2=111_5$. I only found interesting the palindromes which share consecutive bases. Smallest example for $k=1,2,3,4,\\dots$ consecutive bases: $3(=11_2), 10(=11_3=101_4),178(= 454_6 =343_7 = 262_8), ? (k=4)$... The solution for four is either very large or does not exist.",
"@Vepir Do you know of any example of palindrome simultaneously for bases $n$ and $n+4$?",
"@joro: That doesn't seem to be proven. At MSE, Vepir searched up to $10^7$ and found a one parameter family which does not extend to $4$ bases, and a single sporadic example, but there is no proof these are the only $3$ digit solutions.",
"Is it known that three digit solution doesn't exist?",
"This is perfectly tweetable mathematics and thus certainly not offtopic on mathovertweet. I vote Reopen",
"@Andy, the question was asked at m.se two weeks ago. If it hasn't been answered to Vepir's satisfaction, no reason why it shouldn't be posted here. And no reason why anyone here has to accept Vepir's suggestion to answer there. I'm sure that if someone posts an answer here, Vepir will survive the trauma.",
"I voted to close. MO is not intended to advertise questions on math.se, so if you want answers there, you should just ask there."
] | 12 |
Science
| 0 |
180
|
mathoverflow
|
A Linear Order from AP Calculus
|
In teaching my calculus students about limits and function domination, we ran into the class of functions
$$\Theta=\\{x^\alpha (\ln{x})^\beta\\}_{(\alpha,\beta)\in\mathbb{R}^2}$$
Suppose we say that $g$ weakly dominates $f$, and write $f\preceq g$, if
$$\lim_{x\to\infty}\frac{f(x)}{g(x)} \hspace{3 mm} \text{is finite}$$
We can then readily see that $(\Theta,\preceq)$ is a total order isomorphic to the lexicographic order on $\mathbb{R}^2$.
But we can get more complicated total orders with, say
$$\Theta_n=(x^{\alpha_0}(\ln{x})^{\alpha_1}(\ln\ln{x})^{\alpha_2}\cdots(\ln^{n-1} x)^{\alpha_{n-1}})_{\vec{\alpha}\in\mathbb{R}^{n}}$$
$$\Phi=\\{e^{p(x)}\\}_{p(x)\in\mathbb{R}[x]}$$
which are isomorphic as total orders to the lexicographic orders on $\mathbb{R}^n$ and $\operatorname{List}\mathbb{R}$
All of these complicated orders live inside what I'd call "the AP Calc linear order" $(\Omega,\preceq)$ defined as:
$$\Omega_0=\\{f\in\mathscr{C}^0((\lambda,\infty))\\}_{\lambda\in\mathbb{R}}$$
$$f\preceq g\Longleftrightarrow \max\left\\{\left|\liminf_{x\to\infty} \frac{f(x)}{g(x)}\right|,\left|\limsup_{x\to\infty} \frac{f(x)}{g(x)}\right|\right\\}<\infty$$
$$\Omega=\Omega_0/\simeq \hspace{5 mm} \text{where} \hspace{5 mm} f\simeq g \Leftrightarrow \left[f\preceq g \text{ and } g\preceq f\right]$$
where the refinement on $\preceq$ is made so as to avoid problems with things like $\sin{x}$.
This seems to be a very complicated linear order, as it includes as a suborder things like
$$\Psi=\\{p_0(x)e^{p_1(x)}e^{e^{p_2(x)}}\cdots\exp^{n-1}(p_{n-1}(x))\\}_{p_i(x)\in\mathbb{R}[x]\forall i}$$
My question is the following: is there any combinatorial description or universal construction, i.e. as a colimit, of the isomorphism type of $(\Omega,\preceq)$?
|
https://mathoverflow.net/questions/215798/a-linear-order-from-ap-calculus
|
[
"co.combinatorics",
"ct.category-theory",
"real-analysis",
"order-theory",
"linear-orders"
] | 19 | 2015-08-27T09:12:46 |
[
"This is essentially just a question involving Landau notation, no?",
"With Anthony Quas, I suggest you look up \"Hardy field\", and also \"o-minimal structure\" whose germs at infinity produce Hardy fields. The question as it stands suffers from problems noted by Eric Wofsey.",
"@DmitryV: That's not what I was talking about (I was also thinking like a category theorist and not caring about that). If $f/g$ oscillates between approaching $0$ and approaching $\\infty$, then $f\\not\\leq g$ and $g\\not\\leq f$. Actually, if $f$ and $g$ don't have to be positive, you could just have an unbounded set where $f$ is zero and $g$ is not and an unbounded set where $g$ is zero and $f$ is not.",
"@EricWofsey you're right, I was being too much a category theorist and forgetting to mod out by isomorphisms, turning this pre-order into a total order. I'll edit it now.",
"A side comment: The class $\\Theta$ is a very small sub-class of the logarithmico-exponential functions studied by Hardy (see also Boshernitzan's work). These have very nice properties including no oscillation: for any two distinct functions in a Hardy Field, eventually $f$ is bigger than $g$ or vice versa.",
"This is not a total order; $f$ and $g$ can oscillate between $f\\gg g$ and $g\\gg f$.",
"Related mathoverflow.net/questions/29624/…"
] | 7 |
Science
| 0 |
181
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mathoverflow
|
A question in Fontaine--Laffaille theory
|
Let $K$ be finite unramified extension of $\mathbf{Q}_p$ with ring of integers $W$. Let ${\rm MF}$ be the category of strongly divisible $W$-modules $M$ with ${\rm Fil}^0M=M$ and ${\rm Fil}^{p-1}M=0$. Let ${\rm Crys}(G_K)$ be the category of finite free $\mathbf{Z_p}$-representations of $G_K$ which become crystalline over $\mathbf{Q}_p$. There is a functor $$A:{\rm MF} \to {\rm Crys}(G_K), \qquad M \mapsto {\rm Hom}(M, A_{\rm crys}),$$ where the ${\rm Hom}'s$ are required to preserve the filtration and the Frobenius action, which is fully faithful by Fontaine-Laffaille theory.
Given $T$ in the image of the above functor, one can construct a finite free $W$-module with filtration and Frobenius, namely $B(T)={\rm Hom}_{G_K}(T, A_{\rm crys})$. Note that $B(A(M))[1/p]$ is isomorphic to $M[1/p]$ by the theory of crystalline representations. I believe that $B$ is not the inverse to $A$ in general, however.
Questions: 1) What is an example of a $T$ for which $B(T)$ is not the strongly divisible module associated to $T$? 2) For which $T$ is $B(T)$ the strongly divisible lattice associated to $T$?
|
https://mathoverflow.net/questions/60134/a-question-in-fontaine-laffaille-theory
|
[
"nt.number-theory",
"galois-representations"
] | 19 | 2011-03-30T15:31:30 |
[] | 0 |
Science
| 0 |
182
|
mathoverflow
|
The oriented homeomorphism problem for Haken 3-manifolds
|
Haken famously described an algorithm to solve the homeomorphism problem for [the 3-manifolds that bear his name](http://en.wikipedia.org/wiki/Haken_manifold) (fleshed out by many others, including Hemion and Matveev who fixed some gaps). But it's often natural to consider 3-manifolds _equipped with an orientation_ , and so my question is:
> **Question 1:** Is there an algorithm _in the literature_ that determines whether or not a pair of oriented Haken 3-manifolds M and N are _orientedly_ homeomorphic?
By Haken's Algorithm, we can reduce to the case in which M and N are (unorientedly) homeomorphic, and so the question can be reduced to:
> **Question 2:** Is there an algorithm _in the literature_ that determines whether or not an orientable Haken 3-manifold M admits an orientation-reversing self-homeomorphism?
**Notes:**
1. Note the words 'in the literature'. I don't really doubt that this algorithm exists, but the details seem quite complicated, particularly in the non-geometric case, so I hope that someone will be able to supply a reference. A very short argument would also be nice. Answers of the form 'I don't know, but I'm sure it can be done if you work out each geometric case individually and then think hard enough about the JSJ decomposition', while appreciated, would be less useful.
2. The hyperbolic case can be deduced from Sela's solution to the homeomorphism problem, which also computes the automorphism group. I've no doubt the other non-Haken cases can be handled similarly. Hence my interest in the Haken case.
3. It may be that a simple modification of Haken's Algorithm handles the oriented case. This isn't at all clear to me, but I would be very interested to hear if it's true.
4. **(added later)** I would also be happy to confirm that this question is `open' (in the sense that an answer is not in the literature, rather than that the answer is in doubt)---I very much suspect that this is the case. An authoritative pronouncement by someone who knows the literature very well would therefore be a valid answer.
|
https://mathoverflow.net/questions/107573/the-oriented-homeomorphism-problem-for-haken-3-manifolds
|
[
"3-manifolds",
"gt.geometric-topology",
"reference-request",
"open-problems"
] | 19 | 2012-09-19T07:37:20 |
[
"That's a very nice observation, Ian!",
"Certain special cases can be deduced from the literature. If you look at Theorem 6.1.6 in Matveev's book, it says that there is an algorithm to tell if two Haken manifolds with boundary pattern are homeomorphic taking boundary pattern to boundary pattern. If you have two oriented knot complements, then you can tell if they are orientation preserving homeomorphic by choosing a boundary pattern which is not preserved under mirror image (such as a single closed (1,1) curve). However, this is also appealing to the knot complement problem. springerlink.com/content/978-3-540-45898-2",
"Actually, Ian, this is precisely my motivation for asking the question. As far as I can tell, the treatments of the homeomorphism problem for reducible manifolds in the literature miss this point.",
"If you connect sum with a manifold that doesn't admit an orientation reversing isometry, then check homeomorphism, then this checks if the manifolds are orientation preserving homeomorphic. However, I'm not sure if the homeo. problem for reducible (or even connect sums of Haken) manifolds is in the literature explicitly. "
] | 4 |
Science
| 0 |
183
|
mathoverflow
|
Reference request: Parallel processor theorem of William Thurston
|
Sometime in the 1980's or 1990's, Bill Thurston proved a theorem regarding the existence of a universal parallel processing machine, using a certain class for such machines having finite deterministic automata at each node. There is no such published paper on MathSciNet, and I have no record of this theorem other than my own faulty memory.
I'd like to ask whether there exists a preprint somewhere, or some other reference.
Although I cannot quite reproduce the theorem's exact statement, here's what I roughly remember. Caveat: please take this all with a grain or three of salt, because I am sure my memory has gaps and/or errors.
The parallel processors of this class are built on an underlying framework consisting of a connected graph with finite valence at each vertex. At each node of this graph one places a finite deterministic automaton. This automaton has some special capabilities. After each computation, it can send one of finitely many outputs along each adjacent edge. Also, before each computation it can receive inputs along each adjacent edge (which have previously been sent by adjacent automata, or have been set up as part of an initialization) and it uses those inputs together with the current state to compute its next state and its next outputs.
I believe there are strong regularity constraints. The graph should have an automorphism group which is transitive on vertices. In fact the entire setup, including the automata themselves and how their inputs and outputs are associated to the adjacent edges, should have an automorphism group that is transitive on vertices. The reason for these constraints is so that the entire network can be "finitely described" or "finitely programmed", and perhaps that is the true constraint, however it may be formalized.
One programs or initializes the processor by choosing initial states and initial inputs. I believe there must be some finiteness condition, for example outside of some bounded region the initial states and initial inputs are perhaps in some default position. Then one lets the processor run until some point (I don't remember the halting conditions), at which time one reads off the terminal states and outputs.
I remember Thurston proved two theorems about universal machines in this class (subject to some restrictions about how to encode or program one machine within another, which I totally do not remember; but this argues for some strong regularity constraints in the class of machines).
One theorem is that a processor of this type whose underlying graph is the Cayley graph of a free group cannot be universal. Roughly speaking the reason one might expect this to be a universal processor is because of exponential growth of nodes, but the trouble is that there is too much bottlenecking at each node.
The main theorem is that there exists a processor of this type whose underlying graph is any Cayley graph of a lattice in the 3-dimensional solvable Lie group. For example, one such lattice is the Cayley graph of the semidirect product $$\mathbb{Z}^2 \rtimes_M \mathbb{Z} $$ using the action of $\mathbb{Z}$ on $\mathbb{Z}^2$ generated by the matrix $M = \pmatrix{2 & 1 \\\ 1 & 1}$. Here the rough idea is that one still has exponential growth of nodes, but the bottlenecking phemonenon has been avoided.
**Added:** I'm convinced now that so far this is just a description of computable functions in the ordinary sense. But this jogs a really dim memory that what's missing is a way to constrain or measure how the programming process is allowed to compress information. I'm still hoping that someone will have a more accurate record of this theorem.
|
https://mathoverflow.net/questions/213181/reference-request-parallel-processor-theorem-of-william-thurston
|
[
"reference-request",
"graph-theory",
"computational-complexity",
"computer-science",
"geometric-group-theory"
] | 19 | 2015-08-06T08:29:11 |
[
"Maybe you could try theoretical computer science, cstheory.stackexchange.com",
"@GerryMyerson: No, it's not in that paper, although the issues of mesh partitions discussed in that paper remind me of the \"bottlenecking\" issues that are at the heart of this parallel processing model I'm asking about, and this may have been around the same time as well.",
"@IgorRivin: I did ask Matt, and although he remembers working on somewhat similar problems around the same (vague) time, this particular theorem is not something he remembers.",
"@GerryMyerson: I wondered the same, but it seems far from Cayley graphs and solvable Lie groups...",
"Is it possible we are talking about the paper, Miller, Teng, Thurston, and Vavasis, Automatic mesh partitioning, available at cs.cmu.edu/~glmiller/Publications/Papers/MiTeThVa93.pdf ?",
"@Algernon: You might be right, I could be overlooking some kind of space-or-time compression restrictions. But I cannot remember.",
"My recollection that this is actually due to Matt Grayson, so you might want to ask him...",
"Your model is what people call a cellular automaton. It is easy to simulate a Turing machine with a \"one-dimensional\" cellular automaton (i.e., the underlying graph is a bi-infinite path, a Cayley graph of the one-generator free group!), with reasonable finiteness constraint. On a free group with a larger generator set, you can simply ignore all but one of the generators and simulate the Turing machine along the remaining generator. I suspect your memory is putting the emphasis of whatever Thurston's result was on the wrong place."
] | 8 |
Science
| 0 |
184
|
mathoverflow
|
Coarse moduli spaces of stacks for which every atlas is a scheme
|
Let $X = [P/G]$ be a smooth finite type separated DM-stack over $\mathbb C$ given as the quotient of a smooth quasi-projective scheme $P$ by the action of a smooth (finite type separated) reductive group scheme $G$. Since $X$ has finite inertia, the coarse space $X^c$ of $X$ exists as an algebraic space.
At _atlas_ of $X$ is an étale morphism $U\to X$ with $U$ an algebraic space.
I sm interested in the converse of the following statement:
**Thm.** If the coarse moduli space of $X$ is a scheme,then every atlas $U$ of $X$ is a scheme.
_Proof._ The map from an atlas $U$ to the coarse moduli space $X^c$ of $X$ is quasi-finite separated. Therefore $U$ is a scheme by Knutson Cor. II.6.16., p. 138 QED
So the converse would read as follows:
**Q.** Suppose that every atlas $U$ of $X$ is a scheme. Is the coarse moduli space of $X$ a scheme?
I expect the answer to be negative, but can't find a good example. Note that a counterexample can not be an algebraic space, as an algebraic space with the property that every atlas is a scheme is itself a scheme (the identity morphism being an atlas).
Edit: I have an application in mind of the above question, and in the context of my application the stack $X$ is even generically a scheme. Not sure if that helps...
|
https://mathoverflow.net/questions/186379/coarse-moduli-spaces-of-stacks-for-which-every-atlas-is-a-scheme
|
[
"ag.algebraic-geometry",
"complex-geometry",
"stacks",
"algebraic-stacks",
"algebraic-spaces"
] | 19 | 2014-11-06T08:42:34 |
[
"Yes it looks like the example from Corollary 6 does the job! After taking a quick look at the proof, it seems like the condition that the coarse space is a scheme should be equivalent to something like the following: for each point $x \\in P$, there exists a linearized ample line bundle $L$ such that $x$ is $L$-semistable. I'm not sure how this interacts with your atlas condition though.",
"@DoriBejleri I think (but I'd have to double-check) that Kollar's paper arxiv.org/pdf/math/0501294.pdf contains such examples. Do you agree? (BTW, this question is seven years old, and I somehow decided to revive it now, but I don't even remember the \"application in mind\" at the moment.)",
"Do you have an example, without assuming your condition on the atlases, where $P$ is quasi-projective and $G$ is reductive, $[P/G]$ is separated DM, but the coarse space of $[P/G]$ is not a scheme?"
] | 3 |
Science
| 0 |
185
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mathoverflow
|
The cofinality of $(\mathbb{N}^\kappa,\le)$ for uncountable $\kappa$?
|
For a partially ordered set $P$, a set $A\subseteq P$ is _cofinal_ if for each element of $P$ there is a larger element in $A$. The _cofinality_ of $P$, ${\rm cof}(P)$, is the minimal cardinality of a cofinal family in $P$.
Let $\kappa$ be an infinite cardinal number. Consider the set $\mathbb{N}^\kappa$ of all functions from $\kappa$ to $\mathbb{N}$ (equivalently, $\kappa$-sequences of natural numbers), with the pointwise partial order: $f\le g$ if $f(\alpha)\le g(\alpha)$ for all $\alpha<\kappa$.
Let $$f(\kappa):={\rm cof}(\mathbb{N}^\kappa,\le).$$ The cardinal $f(\aleph_0)$ is the well-understood dominating number $\mathfrak{d}$, which is consistently any cardinal of uncountable cofinality that is not larger than the continuum (see [Blass](http://www.math.lsa.umich.edu/~ablass/hbk.pdf)).
Basic facts include:
1. For cardinal numbers $\kappa\le\mu$, we have that $f(\kappa)\le f(\mu)$: A projection of a cofinal family in $\mathbb{N}^\mu$ on the first $\kappa$ corrdinates is cofinal in $\mathbb{N}^\kappa$.
2. $\kappa<f(\kappa)$: Given $\\{h_\alpha:\alpha<\kappa\\}\subseteq\mathbb{N}^\kappa$, define $h(\alpha):=h_\alpha(\alpha)+1$ for all $\alpha<\kappa$.
3. The value of $f(\kappa)$ with "$\le$" replaced by eventual dominance remains the same (see [Comfort](http://dml.cz/bitstream/handle/10338.dmlcz/106682/CommentatMathUnivCarol_029-1988-4_5.pdf)).
Some deeper results:
1. [Jech-Prikry](http://journals.cambridge.org/download.php?file=%2F4622_1176E3359F6DC1B50E531EBE349A0A5F_journals__PSP_PSP95_01_S0305004100061272a.pdf&cover=Y&code=f33a33db8489693def0a6e06d5ccf9fa): Assume that $2^{\aleph_0}$ is regular and smaller than $2^{\aleph_1}$. If $f(\aleph_1)=2^{\aleph_0}$, then there is an inner model with a measurable cardinal.
2. If $\kappa^{\aleph_0}=\kappa$, then $f(\kappa)=2^\kappa$ (see [Hathaway](http://arxiv.org/abs/1401.7948)).
I did not conduct an extensive literature search. I would appreciate your pointing out relevant results that are not mentioned here.
**Question:** Are there additional cardinals $\kappa$ for which $f(\kappa)$ can be evaluated in ZFC?
(The question also applies to large cadrinals. I do not request that the cardinals provably exist.)
**Update.** Intially, I also asked whether $\kappa^{\aleph_0}=\kappa$ could be weakened to ${\rm cof}([\kappa]^{\aleph_0},\subseteq)=\kappa$. Todd Eisworth reminds in the comments below that the latter hypothesis is true in ZFC for $\aleph_1$.
Strangely, I couldn't find an explicit reference for the following.
**Question.** Is it consistent (modulo suitable large cardinals) that $f(\aleph_1)<2^{\aleph_1}$? Assuming that (See Assaf Rinot's comment below), is there an explicit reference for that?
**Motivation:** The values $f(\kappa)$ pop up whenever I study (usually, jointly with colleagues) the character and other cardinal invariants of topological groups. They are unavoidable.
(Comment: By no means do I mean that the symbol $f(\kappa)$ should be reserved for the function thus defined. This is just a short, temporary name for brevity.)
|
https://mathoverflow.net/questions/221236/the-cofinality-of-mathbbn-kappa-le-for-uncountable-kappa
|
[
"set-theory",
"gn.general-topology",
"topological-groups",
"foundations"
] | 19 | 2015-10-18T10:35:04 |
[
"The Komjath-Shelah model does not involve large cardinals. Just force to add $\\aleph_{\\omega_1}$ many Cohen reals over a model of $2^{\\aleph_0}=\\aleph_1, 2^{\\aleph_1}=\\aleph_2, 2^{\\aleph_2}=\\aleph_{\\omega_1+1}$ and $(\\aleph_{\\omega_1})^{\\aleph_0}=\\aleph_{\\omega_1}$. E.g., force with $Add(\\aleph_2,\\aleph_{\\omega_1+1})*Add(\\omega,\\aleph_{\\omega_1})$ over a model of GCH.",
"@ToddEisworth I had such a feeling from what I did see in the literature. This is absurd that such a \"simple\" question is still open. I wonder if Shelah knows about it (probably not, for otherwise it should have been solved :) ). I'll have a chance to mention/remind/motivate it to him and others next week, in my lecture at the YST. Would appreciate any futher info about this problem that I can mention then.",
"@BoazTsaban: Jech's Chapter 29 mentions that the question of whether $f(\\aleph_1)<2^{\\aleph_1}$ is consistent is open, and I seem to recall that the question was also mentioned in the original \"green version\" of his book from 1978.",
"@Todd: Oops. I should have considered these immediate cases. Thanks for pointing this out. Will update the question accordingly.",
"Should have said \"stronger implication\" as the assumption is weaker...",
"Just observe that the cofinality of $([\\aleph_1]^{\\aleph_0},\\subseteq)$ is $\\aleph_1$, as witnessed by initial segments. Same holds for all the $\\aleph_n$ by induction.",
"In the particular case where $\\kappa=\\aleph_1$, one may show that not only CH, but also $2^{\\aleph_0}=\\aleph_2$, is enough to imply that $f(\\aleph_1)=2^{\\aleph_1}$. See Jech's book for a reference (Chapter 29).",
"A wild guess concerning the consistency of $f(\\aleph_1)<2^{\\aleph_1}$: have a careful look at Komjath-Shelah model of [F1445].",
"@ToddEisworth could you provide some details, or a reference?",
"If the weakened implication in (2) holds then we would have $f(\\aleph_1)=2^{\\aleph_1}$ in ZFC."
] | 10 |
Science
| 0 |
186
|
mathoverflow
|
Homotopy type of the affine Grassmannian and of the Beilinson-Drinfeld Grassmannian
|
The affine Grassmannian of a complex reductive group $G$ (for simplicity one can assume $G=GL_n$) admits the structure of a complex topological space. More precisely, the functor $$X\mapsto |X^{an}|$$ that associates to a scheme locally of finite type the underlying topological space of its analytification ([SGA1, XII, 1]) can be left-Kan-extended to a functor $PSh(Sch)\to Top$, and therefore applied to the ind-scheme $Gr_G$.
Some topological properties of this space are known: its connected components are in bijection with the topological fundamental group $\pi_1(G)$ ([Beilinson-Drinfel'd, _Quantization of Hitchin's integrable system and Hecke eigensheaves_ , Proposition 4.5.4]). The étale fundamental group (which is the profinite completion of the topological fundamental group) of each orbit in $Gr_G$ under the left action of $L^+G$ is trivial (Richarz, _Affine Grassmannians and the Geometric Satake Equivalences_ , proof of Proposition 3.1).
Also, $|Gr_G^{an}|$ is known to be homotopy equivalent to a polynomial loop space ([Zhu, An introduction to the affine Grassmannian and the Geometric Satake Equivalence] and [Pressley/Segal, Loop Groups, Sec. 8.3]).
However, my knowledge essentially stops here. For example, the homotopy groups of this polynomial loop space are unknown to me.
Has anyone ever written or thought about the topology of the affine Grassmannian in further detail? Relevant questions to me are the following:
1. What are the fundamental groups of the connected components of $|Gr_G^{an}|$?
2. The cohomology of the affine Grassmannian is much more studied and used than its homotopic properties. Is there any chance that the connected components of $|Gr_G^{an}|$ satisfy conditions like "being a simple space", which often allows homotopical issues to be translated into cohomological issues, by means of the homological Whitehead theorem?
Needless to say, the answer to such questions for the Beilinson-Drinfel'd Grassmannian will probably be more difficult. However, any ideas or references in this more sophisticated context would be highly appreciated.
Thank you in advance.
|
https://mathoverflow.net/questions/364818/homotopy-type-of-the-affine-grassmannian-and-of-the-beilinson-drinfeld-grassmann
|
[
"ag.algebraic-geometry",
"at.algebraic-topology",
"homotopy-theory",
"geometric-representation-theory"
] | 19 | 2020-07-04T04:17:06 |
[] | 0 |
Science
| 0 |
187
|
mathoverflow
|
support of the coupling between two probability measures
|
Given two Borel probability measures $\mu$ and $\nu$ on $\mathbb{R}$, let $\Pi(\mu, \nu)$ denote all couplings between them, i.e., all Borel probability measures on $\mathbb{R}^2$ such that the marginal distribution of the first and second coordinate are $\mu$ and $\nu$ respectively. Can we describe the set of all possible support of the joint law:
$S(\mu, \nu) = \\{E \subset \mathbb{R}^2: \exists \lambda \in \Pi(\mu, \nu), s.t. \lambda(E) = 1\\}$.
[Strassen](http://projecteuclid.org/euclid.aoms/1177700153) (Theorem 11) characterized this collection: $C \in S(\mu, \nu)$ if and only if for all $U$ open in $\mathbb{R}$, $\nu(U) \leq \mu(U^C)$, where $U^C = \\{x: \exists y \in U, s.t. (x,y) \in C\\}$. But this condition seems not easy to verify! For instance, it is probably not true to only consider open interval $U$.
Let us focus on the normal distribution $\mu=\nu=N(0,1)$. Can we characterize $S' = S(N(0,1), N(0,1))$ **more explicitly** in this case, i.e., what kind of subset in the plane admits a probability measure whose marginals are standard normal? For instance we can ask the following question: for what pair of $a < b$ do we have $\\{(x,y): a \leq |x-y| \leq b\\} \in S'$, that is, can we construct a joint distribution on this strip that has standard normal marginals. It seems a non-trivial question. For $b = \infty$, one can show that the largest possible $a$ is between $\sqrt{\pi/2}$ and $3/2$. I am thinking is there a systematic way to do it in the simple case of real line. Note that Strassen's result work for any Polish space.
|
https://mathoverflow.net/questions/56968/support-of-the-coupling-between-two-probability-measures
|
[
"pr.probability"
] | 19 | 2011-02-28T21:29:35 |
[] | 0 |
Science
| 0 |
188
|
mathoverflow
|
Does this variant of a theorem of Hasse (really due to Gauss) have an "elementary" proof?
|
BACKGROUND
Here are 3 theorems of varying difficulty. Let $M$ be the $Z/2$ subspace of $Z/2[[x]]$ spanned by $f^k$, with the $k>0$ and odd, and $f=x+x^9+x^{25}+x^{49}+\cdots$. For $g$ in $M$, let $S(g)$ consist of the primes, $p$, for which the coefficient of $x^p$ in $g$ is 1. Note that each $p$ in $S(f^k)$ is congruent to $k$ mod 8.
T1.----- If $k=3 {\rm\ or\ } 5$, $S(f^k)$ consists of the $p$ that are $k$ mod 8
T2.----- $S(f^7)$ consists of the $p$ that are 7 mod 16
T3.----- If $k=19 {\rm\ or\ } 21$, then $S(f^k)$ consists of the $p$ that are $k$ or $k+8$ mod 32.
To prove T1 when $k=3$, we write $f^k$ as $f*f^2$ and use the fact that if $p$ is 3 mod 8, then $p$ is uniquely the sum of a square and twice a square. When $k=5$ we argue similarly using Fermat's two square theorem.
As I indicated in a comment on a recent MO question of Joel Bellaiche, "[Primes and x^2+2y^2+4z^2](https://mathoverflow.net/questions/100701/primes-and-x22y24z2)" ,T2 follows from a result of Hasse on the class number of $Q(\sqrt{-2p})$, using Gauss' theorem that the number of representations of $2p$ as a sum of 3 squares is 12*(this class number). Hasse's proof is an application of the Gauss theory of genera and ambiguous forms.
T3 is thornier. Because $f$ is the mod 2 reduction of (the Fourier expansion of) the normalized weight 12 cusp form for the full modular group, each $g$ is the mod 2 reduction of a modular form of integral weight. A profound result of Deligne, relating Hecke eigenforms to Galois representations, then shows that $S(g)$ is a "Frobenian set". Nicolas, Serre and Bellaiche, continuing in this vein, developed a theory of level 1 modular forms in characteristic 2 that led to more precise results. Their investigations motivated me to try to determine $S(f^k)$ empirically for small $k$, and I was led to conjecture T3. Joel then applied his methods to give a proof. But this is very hard, and so I ask:
QUESTION
Does there exist an "elementary proof" of T3, using the theory of binary quadratic forms, along the lines of the Hasse-Gauss argument?
EDIT: Motivated by my recent simple proof of T2 (see my answer to the question of Joel cited above), I've found arguments that ought to reduce the proof of T3 to Sage calculations. The point is that forms of weight 2 are easier to deal with than forms of weight 3/2, so one should work with quadratic forms in 4 variables rather than in 3, even when the genera that arise have more than 1 class in them. Here's the idea of my argument for f21.
Let p be a prime that is 5 mod 8. Writing $f^{21}$ as $(f)(f^2)(f^2)(f^{16})$ we find that if $R$ is (1/16)*(the number of representations of $p$ by $G_1=x^2+2y^2+2z^2+16t^2$ with $x,y,z$ and $t$ all odd), then $p$ is in $S(f^21)$ if and only if $R$ is odd. Now since $p$ is 5 mod 8, in any representation of $p$ by $G_1$, $x,y$ and $z$ must be odd. So if we set $G_2=x^2+2y^2+2z^2+64t^2$ then $R=(N1-N2)/16$, where $N_1$ and $N_2$ are the numbers of representations of $p$ by $G_1$ and $G_2$ respectively. Now write $p$ as $a^2+4b^2$ with $a$ and $b$ congruent to 1 mod 4. Computer calculations indicate:
Conjecture 1. $N_1=p+1+2a$
Conjecture 2. $N_2=((p+1)/2)+a+4b$
If these conjectures hold then $R=(p+1+2a-8b)/32$. The numerator here is $4(b-1)^2 +(a+3)(a-1)$, which mod 64 is $(a+3)(a-1)$. So $R$ has the same parity as $(a+3)(a-1)/32$ and is odd just when $a$ is $5$ or $9$ mod $16$. Now mod $32$, $p=a^2+4$. So $R$ is odd just when $p$ is $29$ or $85$ mod $32$, and so the conjectures imply Joel's result for $S(f^21)$.
How does one attack the conjectures? The theta series attached to $G_1$ and $G_2$ are modular forms for $\Gamma_0 (64)$ and $\Gamma_0 (256)$ respectively. If the conjectures are to hold it seems that each of these theta series should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(i)$. It should be possible, using Sage, to get an explicit formulation of this, and prove the conjectures.
My proposed treatment of $S(f^{19})$ is entirely similar. Suppose $p$ is $3$ mod $8$. Writing $f^{19}$ as $(f)(f)(f)(f^{16})$ and arguing as above we find that if we take $H_1$ and $H_2$ to be $x^2+y^2+z^2+16t^2$ and $x^2+y^2+z^2+64t^2$ respectively, and let $N_1$ and $N_2$ be the number of representations of $p$ by $H_1$ and $H_2$, then $p$ is in $S(f^{19})$ just when $R=(N_1-N_2)/16$ is odd. Now Jacobi's 4 square theorem, (see the argument in my answer to Joel's question), shows that $N_1$ is $2(p+1)$. Write $p$ as $a^2+2b^2$ with $a =1$ or 3 mod 8. The computer suggests:
Conjecture 3. $N_2=p+1+4a$
So if the conjecture holds, $R=(p+1-4a)/16$, and one sees easily that this is odd just when $p$ is 19 or 27 mod 32. Once again the theta series attached to H2 is a modular form for $\Gamma_0 (256)$. The conjecture indicates that it should be a linear combination of Eisenstein series and cusp forms attached to Grossencharaktere for $\mathbb Q(\sqrt{-2})$; all this should admit a proof using Sage.
|
https://mathoverflow.net/questions/106267/does-this-variant-of-a-theorem-of-hasse-really-due-to-gauss-have-an-elementar
|
[
"nt.number-theory",
"quadratic-forms",
"modular-forms",
"power-series"
] | 19 | 2012-09-03T12:27:05 |
[] | 0 |
Science
| 0 |
189
|
mathoverflow
|
Is there some way to see a Hilbert space as a C-enriched category?
|
The inner product of vectors in a Hilbert space has many properties in common with a hom functor. I know that one can make a projectivized Hilbert space into a metric space with the Fubini-Study metric and then follow Lawvere in considering it a $[0,\infty)$-enriched category; it's particularly nice because the distance between two vectors is a function of the inner product.
Is there a way of defining a monoidal structure on $\mathbb{C}$ such that the inner product is the hom so that we can think of a (perhaps projectivized) Hilbert space as $\mathbb{C}$-enriched?
|
https://mathoverflow.net/questions/47644/is-there-some-way-to-see-a-hilbert-space-as-a-c-enriched-category
|
[
"hilbert-spaces",
"ct.category-theory",
"enriched-category-theory"
] | 19 | 2010-11-28T21:30:13 |
[
"You might also be interested in John Baez's paper on the subject: arxiv.org/abs/q-alg/9609018",
"Related question: mathoverflow.net/questions/476/…"
] | 2 |
Science
| 0 |
190
|
mathoverflow
|
are there high-dimensional knots with non-trivial normal bundle?
|
Does there exist a smooth embedding $\varphi\colon S^k\to S^n$ such that $\varphi(S^k)$ has non-trivial normal bundle? I looked at some of the old papers by Kervaire, Haefliger, Massey, Levine but I couldn't find an answer. There are various related results, for example it is shown by Kervaire et al that in many dimensions the normal bundle is trivial. Furthermore Kervaire showed that there exists an immersion, such that the pullback of the normal bundle is non-trivial.
|
https://mathoverflow.net/questions/382047/are-there-high-dimensional-knots-with-non-trivial-normal-bundle
|
[
"at.algebraic-topology",
"gt.geometric-topology"
] | 19 | 2021-01-24T03:39:42 |
[
"@StefanFriedl Thanks for the corrections.",
"@archipelago thanks, you put me on the right track!",
"Namely Danny Ruberman writes: \"Jerry Levine's paper \"A classification of differentiable knots\" (Annals 82 (1965) 15-50) determines (see proposition 6.2) the possible normal bundles in some range of codimensions in terms of maps in various exact sequences. It's hard to summarize the results, but he gives a table at the end dealing with relatively low dimensions, where calculations can be made explicit. The case n=11 and k=6 contains Haefliger's example; in fact there are exactly 5 possible normal bundles among all embeddings.\"",
"more googling gives an answer in mathoverflow.net/questions/237384/….",
"@archipelago but for this conversation it's more important that by accident you misquoted their comment, they write \"since Haefliger (unpublished) has shown that S^11 embeds in R^17 with a +++++non-trivial++++ normal bundle.\" That's a very good find and answers my question with a shaky reference. It would be great if there was a reference with details",
"@archipelago Since Levine was my advisor I should point out that it's a paper by Hsiang, Levine and Szcarba",
"@archipelago: Of course, thanks for pointing that out.",
"@archipelago thanks for your comments! But from you and I unearthed so far in the literature it could in principle be that the normal bundle is always trivial, isn't it? You'd think that this question was answered in the 60's.",
"According to Hsiang, Lee, and Szczarba, it is an unpublished result of Haefliger that $S^{11}$ embedds into $S^{17}$ with trivial normal bundle (see 'On the normal bundle of a homotopy sphere embedded in Euclidean space'). This would show that Kervaire's bound is sharp.",
"@MichaelAlbanese Any normal bundle arising from such an embedding is stably trivial, since the tangent bundle of $S^k$ is.",
"Do you know if Kervaire's example $\\nu$ is stably trivial? Maybe I'm oversimplifying, but if you have an immersion $f : S^k \\to S^n$ with $k < n$, then it factorises as $i_n\\circ f_0$ where $i_n : \\mathbb{R}^n \\to S^n$ is the standard inclusion and $f_0 : S^k \\to \\mathbb{R}^n$. The map $h : S^k \\to S^{n+k+1}$ given by $i_{n+k+1}\\circ(f_0, \\iota)$ is an embedding where $\\iota: S^k \\to \\mathbb{R}^{k+1}$ is the standard embedding. If I'm not mistaken, the normal bundle of $h$ is $\\nu\\oplus\\varepsilon^{k+1}$ which is trivial if and only if $\\nu$ is stably trivial.",
"Using this, one sees that all embeddings with $k\\le 6$ have trivial normal bundle. I believe it is an open question whether there is an embedding with nontrivial normal bundle for k=7 and n=11.",
"If k<2(n-k)-1 then the normal bundle is trivial by a result of Kervaire (Theorem 8.2 in his \"An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant\"). If n=k+1,k+2,k+3 it is trivial as well."
] | 13 |
Science
| 0 |
191
|
mathoverflow
|
About the equivariant analogue of $G_n/O_n$
|
Let $BO_n$ and $BG_n$ be the classifying spaces for rank $n$ vector bundles and for spherical fibrations with fiber $S^{n-1}$, respectively, and let $G_n/O_n$ be the homotopy fiber of $BO_n\to BG_n$.
There is an interesting old fact, that the stabilization map $G_n/O_n\to G_{n+1}/O_{n+1}$ is $(2n-4)$-connected, about twice as good as the corresponding maps $BO_n\to BO_{n+1}$ and $BG_n\to BG_{n+1}$. Equivalently, the map $G_n/O_n\to G/O$ is $(2n-4)$-connected. I was wondering about the history of this result.
But what I was really wondering about was the following equivariant analogue. Let $\Gamma$ be a finite group. For a real representation $V$ of $\Gamma$, let $O(V)^\Gamma$ be the group of linear (isometric) automorphisms respecting the $\Gamma$-action and let $G(V)^\Gamma$ be the topological monoid of equivariant homotopy equivalences from the unit sphere $S(V)$ to itself. Write $G(V)^\Gamma/O(V)^\Gamma$ for the homotopy fiber of the map of classifying spaces, and write $G^\Gamma/O^\Gamma$ for the colimit of this over all $V$ in a complete universe (or the sequential colimit over direct sums of copies of the regular representation).
I think maybe with some work I could get a good connectivity statement for the map $G(V)^\Gamma/O(V)^\Gamma\to G^\Gamma/O^\Gamma$, but I wonder if someone has already done so, and more generally if anyone has any insights to offer about any of this.
Note that $BO^\Gamma$ is the product, over irreducible real representations $W$, of either $BO$, $BU$, or $BSp$ depending on the type of $W$, while $G^\Gamma$ is the union of the invertible components of $(\Omega^\infty S^\infty)^\Gamma$. In particular, $\pi_1(BG^\Gamma)$ is the group of units of the Burnside ring of $\Gamma$.
|
https://mathoverflow.net/questions/425108/about-the-equivariant-analogue-of-g-n-o-n
|
[
"at.algebraic-topology",
"homotopy-theory"
] | 18 | 2022-06-20T06:53:22 |
[] | 0 |
Science
| 0 |
192
|
mathoverflow
|
The free complete lattice on three generators, beyond ZF
|
_This was originally[asked at MSE](https://math.stackexchange.com/q/4261226/28111); although it is still under bounty it seems unlikely to be answered there._
[$\mathsf{ZF}$ proves that there is no free complete lattice on three generators](https://www.jstor.org/stable/1993166?seq=1#metadata_info_tab_contents) since any such lattice would have to surject onto the ordinals. This isn't a problem in the context of set theories which admit a universal set, such as [$\mathsf{NFU}$](https://plato.stanford.edu/entries/settheory-alternative/#NewFounRelaSyst) or [$\mathsf{GPK}_\infty^+$](https://plato.stanford.edu/entries/settheory-alternative/#SystGPKiOlivEsse). However, in such theories recursive constructions (which are utterly unproblematic in $\mathsf{ZF}$) become more complicated. So the status of the free complete lattice on three generators in such a theory is unclear to me:
> Is the existence of a free complete lattice on three generators consistent with either of $\mathsf{NFU}$ or $\mathsf{GPK_\infty^+}$?
Of course $\mathsf{NFU}$ and $\mathsf{GPK_\infty^+}$ are not the only two set theories with a universal set, and I'd be interested in an answer for any such theory. My focus on these two is due to $(i)$ my impression that $\mathsf{NFU}$ is the best-understood such set theory by far at the moment and $(ii)$ my personal liking of $\mathsf{GPK_\infty^+}$.
**EDIT** : I forgot to include exactly what "free complete lattice on three generators" means for me in this context (given that such notions are often more fragile in non-$\mathsf{ZF}$ contexts). While to a certain extent I'm interested in any reasonable interpretation of the phrase, I'm specifically interested in the following precisiation: a complete lattice $L$ with distinguished elements $a,b,c$ such that for every complete lattice $M$ and every $x,y,z\in M$ there is exactly one complete lattice homomorphism $L\rightarrow M$ sending $a$ to $x$, $b$ to $y$, and $c$ to $z$.
|
https://mathoverflow.net/questions/405269/the-free-complete-lattice-on-three-generators-beyond-zf
|
[
"set-theory",
"lo.logic",
"universal-algebra",
"foundations",
"new-foundations"
] | 18 | 2021-10-01T10:56:33 |
[
"Given the results in the Crawley-Dean paper, I suspect that if the free complete lattice with $3$ generators exists in a model of $\\mathsf{NFU}$ with choice, then it would actually have the same with any number of generators, including $|V|$.",
"@JamesHanson In fact, I can't even show that \"the free complete lattice on $V$ generators\" doesn't exist! (That is, a pair $(X,f)$ where $X$ is a complete lattice and $f:\\{x:x= x\\}\\rightarrow X$ is such that for every other complete lattice $A$ and every map $g:\\{x:x =x\\}\\rightarrow A$ there is exactly one complete lattice homomorphism $h: X\\rightarrow A$ such that $h\\circ f=g$.)",
"@JamesHanson Yes, I don't see how to get Dedekind-MacNeille completions to work in such settings either. (On that note I forgot to include my definition of freeness - fixed!)",
"I think you can show that in $\\mathsf{NFU}$ there is no complete lattice generated by three elements that satisfies the conditions in Crawley and Dean's definition 2. The issue I'm having with applying this to the definition of freeness in your MSE question is actually that it seems likely that not all lattices have completions in $\\mathsf{NFU}$. The Dedekind-MacNeille completion of a lattice naturally lives one type higher than the lattice itself, which makes it seem like in general you only know that completions exist for lattices on Cantorian sets."
] | 4 |
Science
| 0 |
193
|
mathoverflow
|
Automorphic forms and coherent cohomology
|
Why is it (and what does it mean) that automorphic forms do not contribute in the coherent cohomology of Siegel modular varieties parametrizing abelian varieties of dimension $d>2$ (see section 7 of the [thesis](https://tel.archives-ouvertes.fr/tel-02940906/document) of Nguyen or section 1.4.1 of this [paper](https://arxiv.org/pdf/1812.09269.pdf) of Boxer, Calegari, Gee, and Pilloni)?
Apparently this comes from the fact that the Hodge-Tate weight of the associated motive is neither regular nor weakly regular, i.e. the same number appears more than twice in the Hodge-Tate weights, but why does this mean that we cannot find them in the coherent cohomology? Apparently this is also true for the Betti cohomology of the associated locally symmetric space. Does the theorem of Franke (generalization of Eichler-Shimura) have anything to say about this?
|
https://mathoverflow.net/questions/375826/automorphic-forms-and-coherent-cohomology
|
[
"nt.number-theory",
"arithmetic-geometry",
"automorphic-forms",
"galois-representations"
] | 18 | 2020-11-06T10:32:18 |
[
"A good reference for coherent cohomology of Shimura varieties is the paper of Michael Harris, cited as [Har90b] in the BCGP paper. In particular, Proposition 4.3.2 there gives a formula for the infinitesimal character associated to the coherent cohomology of an automorphic vector bundle. You might also find the paper of Gross cited as [Gro16] in BCGM useful: it explains the role played by the endomorphism ring of the abelian variety.",
"I also agree that a short answer with some reference could be useful, not only for the OP but for other people interested in this question as well. Even if it is not perfect, it would be better than nothing.",
"@naf: I think I understand now why the motives of abelian varieties of dimension d>2 cannot come from the coherent cohomology of a Shimura variety: The former is never regular, since HT weights are d 1's and d 0's. The latter however are required to be regular. Could you give me a reference for coherent cohomology that discusses this?",
"@YemonChoi: Coherent cohomology actually corresponds to non-degenerate limits of discrete series. Without knowing anything about what the OP is willing to assume, it is not possible (for me) to write a short answer which would also be useful.",
"(Alternatively, is there a representation-theoretic explanation of why certain representations are not contained in the discrete series for the relevant group, without going through Hodge-Tate weights?)",
"@naf I am neither a number theorist nor an algebraic geometer, but perhaps the OP would find it useful to see an expanded version of the statement \"Another way to see this...\" that you refer to. I don't know if \"computing the possible infinitesimal characters\" is supposed to be a standard procedure in this setting, but perhaps an outline of how one does this for a toy example could be helpful for the OP?",
"@Tim: The explanation in the paper of BGGP is fairly detailed and seems quite clear to me. What I said corresponds to the statement in brackets beginning with \"Another...\" in the second paragraph there. If there is something specific you do not understand you should ask a more focused question.",
"@naf If you know the correct statement then do you also know an answer? It might be a bit more constructive to maybe expand on this a bit more :-)",
"Thank you, and sorry for getting it wrong. I would be happy to edit the question, but could you elaborate more or give me some references to learn more about this?",
"The question as written doesn't make sense (or is wrong...): the correct statement is that the motive of an abelian variety of dimension $>2$ with endomorphism ring the integers does not occur in the (motive associated to) the coherent cohomology of (automorphic vector bundles on) a Shimura variety."
] | 10 |
Science
| 0 |
194
|
mathoverflow
|
Mysterious sum equal to $\frac{7(p^2-1)}{24}$ where $p \equiv 1 \pmod{4}$
|
Consider a prime number $p \equiv 1 \pmod{4}$ and $n_p$ denotes the remainder of $n$ upon division by $p$. Let $A_p=\\{ a \in [[0,p]] \mid {(a+1)^2}_p<{a^2}_p\\}$.
I Conjecture
$$\sum_{n \in A_p } n=\dfrac{7(p^2-1)}{24}$$
The question has already been asked in this [thread](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4?noredirect=1#comment9793873_4639883). If it is a repetition, please vote to close this thread.
**Addition** After have [answer](https://math.stackexchange.com/questions/4639883/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4/4640608#4640608) Numerically, I find this equality : for every odd integer p ≥ 3 $$\sum_{n \in A_p} n=\dfrac{7p^2 -12p+5}{24}+\dfrac{1}{p}\sum_{n=1}^{p-1}{n^2}_p$$ Perhaps, it is useful to demonstrate in a simple way the fact that [the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p))](https://math.stackexchange.com/questions/4639930/even-more-mysterious-sum-equaling-to-frac7p2-124-h-p-where-p-equ).
|
https://mathoverflow.net/questions/440995/mysterious-sum-equal-to-frac7p2-124-where-p-equiv-1-pmod4
|
[
"nt.number-theory",
"prime-numbers"
] | 18 | 2023-02-16T06:48:42 |
[
"In the same vein, see also in Franz Lemmermeyer's book \"Reciprocity Laws\", especially chapter 1 and exercise 1.32 (Gauss).",
"@SB1729 Good question",
"@Pascal I did some simple computations, this problem is so interesting feels like doing more, so I think we have $\\sum_{a\\in A_p}a^2=\\frac{(p-1)(5p^2+7p-10)}{24}$. In general I think it is interesting to know if there is a degree $d$ polynomial $P_d$ such that $\\frac{24}{p-1}\\sum_{a\\in A_p}a^d=P_d(p)$ for any prime $1$ modulo $4$ and if there are some patterns for the polynomials $P_d$.",
"I have added a numerical result that I think is interesting",
"Thank you very much , I received a proof in the other thread that I had reported here",
"This has been answered in your MSE post. Generally it's best practice to leave a reasonable amount of time (at least a week) between crossposts to avoid situations like this.",
"I believe this has nothing to do with primes. Try for instance powers of primes 1 mod 4 (the pattern is immediate for powers 2,3,4 for instance), or products of primes 1 mod 4, where the same formula holds (and probably also with class numbers).",
"Let $(a+1)^2=a^2+2a+1\\equiv\\alpha\\pmod{p}$ and $a^2\\equiv \\beta\\pmod{p}$. Then $\\beta-\\alpha\\equiv-2a-1\\pmod{p}$. Also for $a\\in A_p$ we have $\\beta-\\alpha\\geq1$. But according to the size of $a$, I think we should get $\\beta-\\alpha=2p-2a-1$. Not sure though.",
"To test that I understand correctly, for $p=5$ we are talking about $3+4$, because $0^2_5 < 1^2_5 < 2^2_5 = 3^2_5 > 4^2_5 > 5^2_5$, right?",
". . . and as already noted at math.stackexchange.com/questions/4639930 when p is -1 mod 4 the sum differs from 7(p^2-1)/24 by the class number of Q(sqrt(-p)) ! This too is yet unproved.",
"$[0.p]\\cap \\mathbb N$",
"What is $[[0,p]]$?"
] | 12 |
Science
| 0 |
195
|
mathoverflow
|
Two curious series for $1/\pi$
|
On Jan. 18, 2012 I conjectured that for any prime $p>3$ we have $$R_p^2\equiv\frac1{10}\left(512\left(\frac{10}p\right)-27\left(\frac{-15}p\right)-475\right)\pmod p,$$ where $(\frac{\cdot}p)$ denotes the Legendre symbol and $$R_p:=\frac1p\sum_{n=0}^{p-1}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}.$$ Motivated by this I believed in 2011 that $$\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{c}{\pi}$$ for some algebraic number $c$, but I could not figure out the exact value of $c$ at that time.
On August 19, 2020 I learned from Dr. Deyi Chen that Maple has a function to justify the form of an algebraic number if we know enough digits of this number. Inspired by this, here I pose the following conjecture.
**Conjecture 1.** We have $$\begin{aligned}&\sum_{n=0}^{\infty}\frac{6n+1}{(-1728)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}\\\&\qquad=\frac{24}{25\pi}\sqrt{375+120\sqrt{10}}. \end{aligned}\tag{1}$$
On June 16, 2011 I conjectured that for any odd prime $p\not=5$ we have $$\sum_{n=0}^{p-1}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\equiv 0\pmod p.$$ Motivated by this I formulated the following conjecture on August 20, 2020.
**Conjecture 2.** We have $$\begin{aligned}&\sum_{n=0}^{\infty}\frac{4n+1}{(-160)^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-20)^{n-k}\\\&\qquad=\frac{\sqrt{30}}{5\pi}\cdot\frac{5+\root 3\of{145+30\sqrt6}}{\root 6\of{145+30\sqrt6}}. \end{aligned}\tag{2}$$
I have checked $(1)$ and $(2)$ for the first $100$ decimal digits.
**Question**. How to prove the identities $(1)$ and $(2)$?
Note that my two other similar conjectural identities $$\sum_{n=0}^\infty\frac{357n+103}{2160^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k(-324)^{n-k}=\frac{90}{\pi}\tag{3}$$ and $$\sum_{n=0}^\infty\frac{n}{3645^n}\binom{2n}n\sum_{k=0}^n\binom nk\binom{n+2k}{2k}\binom{2k}k486^{n-k}=\frac{10}{3\pi}\tag{4}$$ found in 2011-2012 and published in [this paper](http://maths.nju.edu.cn/%7Ezwsun/157b.pdf) also remain open.
|
https://mathoverflow.net/questions/369569/two-curious-series-for-1-pi
|
[
"nt.number-theory",
"sequences-and-series",
"combinatorial-identities",
"congruences",
"ramanujan"
] | 18 | 2020-08-19T07:20:37 |
[
"Thank you, this is very interesting!",
"You may look at my talk available from maths.nju.edu.cn/~zwsun/CNT-talk3.pdf and my recent paper available from dx.doi.org/doi:10.3934/era.2020070.",
"What is the relation between the congruence and the identity for pi? Is there an explanation for why we would expect such things?"
] | 3 |
Science
| 0 |
196
|
mathoverflow
|
Cycles in algebraic de Rham cohomology
|
Let $F$ be a number field, $S$ a finite set of places, and $X$ a smooth projective $\mathscr{O}_{F,S}$-scheme with geometrically connected fibers. For each point $t\in \text{Spec}(\mathscr{O}_{F,S})$, there is a Chern class map $$c_{1,t}:\text{Pic}(X_t)\to H^2_{dR}(X_t),$$ where $H^2_{dR}$ denotes algebraic de Rham cohomology. (See below for one construction of $c_1$.)
> Let $c\in H^2_{dR}(X_F)$ be an element so that there exists a non-zero integer $N$ such that $Nc$ specializes to an element of the image $c_{1,t}\otimes k(t)$, for almost all closed points $t\in \text{Spec}(\mathscr{O}_{F,S})$. Is $c$ in the image of $$c_{1, F}: \text{Pic}(X_F)\otimes F\to H^2_{dR}(X_F)?$$ Here $c_{1,F}$ means $c_{1, \eta}$, for $\eta$ the generic point of $\text{Spec}(\mathscr{O}_{F,S})$, and $k(t)$ is the residue field of $t$.
**Is the answer to the above question (and its evident analogue for higher degree Chern classes/cycle class maps) known in general? Does the question have a name, and is it discussed somewhere?**
In general, given a cohomology theory, there's some conjectural description of the image of the cycle class map (the Hodge conjecture for Betti/de Rham cohomology, the Tate conjecture for $\ell$-adic cohomology) -- my motivation for asking the question is that I realized I hadn't seen such a conjecture for algebraic de Rham cohomology.
* * *
I'll now give a brief description of $c_1$, for completeness. Let $\Omega^\bullet_{dR}$ be the algebraic de Rham complex of some variety $Y$. There's a natural map $$d\log: \mathscr{O}_Y^*[-1]\to \Omega^{\bullet}_{dR}$$ sending a nowhere-vanishing function $f$ to $df/f$. Applying $H^2$ gives the desired map $$c_1:\text{Pic}(Y)=H^1(Y, \mathscr{O}_Y^*)=H^2(Y, \mathscr{O}_Y^*[-1])\overset{d\log}{\longrightarrow} H^2_{dR}(Y).$$
By construction, it factors through $F^1H_{dR}^2(Y)$, the first piece of the Hodge filtration.
* * *
Some remarks, added September 6, 2022:
* François Charles has explained to me how to give a positive answer to the above question for divisors on Abelian varieties, using Bost's work on foliations. I think a similar argument handles products of curves.
* One natural strategy is to try to find a prime where the Picard rank does not jump. This will not work, unfortunately; there are varieties where the Picard rank jumps at _every_ prime, e.g. the square of a non-CM elliptic curve.
* The question can be reduced to the case of surfaces, if I'm not mistaken.
|
https://mathoverflow.net/questions/429644/cycles-in-algebraic-de-rham-cohomology
|
[
"ag.algebraic-geometry",
"arithmetic-geometry",
"algebraic-cycles",
"chern-classes",
"derham-cohomology"
] | 18 | 2022-09-02T07:26:35 |
[
"@PeterScholze: Thanks! This does indeed seem quite related, and indeed the discussion on page 81 of \"Une Introduction aux Motifs\" is closely related to the argument François Charles sketched for the case of Abelian varieties.",
"Related in spirit, but I think a bit different, is the Ogus conjecture, discussed at mathoverflow.net/questions/389391/… (see in particular the reference to Yves Andre's \"Une Introduction aux Motifs\"). Ogus' conjecture is the only one I'm aware of in the neighborhood of what you are proposing.",
"@R.vanDobbendeBruyn: yes, I was hedging because otherwise specialization isn’t defined everywhere. But since it’s an almost-everywhere condition it’s fine to take N=1 if you don’t mind specialization not being defined somewhere.",
"You don't lose generality by taking $N = 1$, right? Because $N$ is invertible in $F$ and in $\\kappa(t)$ for almost all $t$, and all maps are maps of vector spaces.",
"@DanPetersen: Good point. To be honest my actual motivation was: Simpson's conjecture says that a local system underlies an integral VHS iff it arises from geometry; this is the non-abelian analogue of the Hodge conjecture. The Fontaine-Mazur conjecture says a local system is arithmetic iff it arises from geometry; this is the non-abelian analogue of the Tate conjecture. Now there is a conjecture that a flat vector bundle has almost all p-curvatures nilpotent iff it comes from geometry (the p-curvature conjecture is closely related) -- I was wondering what the \"abelian\" analogue of this is.",
"An analogue of Hodge and Tate conjecture with respect to algebraic de Rham cohomology is the Grothendieck period conjecture: if $X$ is a smooth projective variety over $\\overline{\\mathbf{Q}}$, then conjecturally the image of the cycle class map is given by the intersection $ H^{2k}_{\\mathrm{sing}}(X^{\\mathrm{an}},\\mathbf Q(k)) \\cap H^{2k}_{\\mathrm{dR}}(X/{\\overline{\\mathbf{Q}}})$.",
"Clearly, this is a special case of the Litt conjecture. :)"
] | 7 |
Science
| 0 |
197
|
mathoverflow
|
Čech functions and the axiom of choice
|
A **Čech closure function** on $\omega$ is a function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ such that (i) $X\subseteq\varphi(X)$ for all $X\subseteq\omega$, (ii) $\varphi(\emptyset)=\emptyset$, and (iii) $\varphi(X\cup Y)=\varphi(X)\cup\varphi(Y)$ for all $X,Y\subseteq\omega$; in other words, it obeys the Kuratowski closure axioms except possibly the idempotent law $\varphi(\varphi(X))=\varphi(X)$.
In 1947 E. Čech asked if there exists such a closure function (on any set, not necessarily $\omega$) which is also **surjective** and **nontrivial** (not the identity map). Čech's question has been answered in the affirmative [under various set-theoretic assumptions](https://www.sciencedirect.com/science/article/pii/0166864182900608), including ZFC. [The ZFC example](https://pdfs.semanticscholar.org/8870/c3c8a574f24aa1ee411f82283a1076c53303.pdf) is rather complicated and makes heavy use of the axiom of choice; it seems unlikely that one could construct such a thing without choice.
**Question.** Is it consistent with ZF that there is no nontrivial surjective Čech closure function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$?
That is, there is no function $\varphi:\mathcal P(\omega)\to\mathcal P(\omega)$ satisfying the conditions (i)-(iii) above and also (iv) for every $Y\subseteq\omega$ there exists $X\subseteq\omega$ such that $\varphi(X)=Y$, and (v) $\varphi(X)\ne X$ for some $X\subseteq\omega$. (Condition (ii) is now redundant, as it follows from (i) and (iv) that $\varphi(X)=X$ if $X$ is finite.)
|
https://mathoverflow.net/questions/361038/%c4%8cech-functions-and-the-axiom-of-choice
|
[
"gn.general-topology",
"set-theory",
"axiom-of-choice"
] | 18 | 2020-05-20T03:03:19 |
[
"@Haim Looks to me like they both use choice. My guess is, if you can construct such a function without choice, it will be by a completely different method.",
"@bof Out of Theorem 1 and Theorem 2 in the Galvin-Simon paper, which one is using AC in its proof?",
"@YCor: I got a partial answer to this in this question.",
"@YCor: Are you asking for a definable class (the class of functions such that blah blah, defined by formula) whose elements are individually not definable (so that you cannot write down such $F$'s to pick out individual elements)?",
"@bof I'd be happy with a weaker definition of \"constructible\". I'm just wondering whether the slogan \"(exists in ZF)=(can be constructed)\" has any foundation.",
"Is it correct to think as \"in ZF there's such a function\" as \"one can construct such a function\"? More precisely, is it true (from some general principle) that if \"there is such a function\" is a theorem of ZF, then there exists $f_0\\in P(\\omega)^{P(\\omega)}$ satisfying the given condition, and a formula of set theory $F(x)$ with only parameter $x$, such that for every $f\\in P(\\omega)^{P(\\omega)}$, $F(f)$ holds iff $f=f_0$?",
"I think it would be a good point in time to migrate this to MathOverflow. I don't have an obvious solution at hand, and I have quite a lot of things on my hands right now (which is why I might be missing something obvious).",
"Also a candidate: the Feferman–Levy model where the reals are a countable union of countable sets, or the Truss model (although if this holds there, I'd imagine the Solovay model would already capture this).",
"I could imagine several candidate models: Maybe Cohen's model is a suitable example with the Dedekind-finite set of reals \"spoiling things\", but more likely would be Solovay's model where there are no MAD families, etc."
] | 9 |
Science
| 0 |
198
|
mathoverflow
|
Is the Frog game solvable in the root of a full binary tree?
|
_This is a cross-post from[math.stackexchange.com](https://math.stackexchange.com/q/3800570/318073)$^{[1]}$, since the bounty there didn't lead to any new insights._
* * *
For reference,
> The **Frog game** is the generalization of the _Frog Jumping (see it on[Numberphile](https://www.youtube.com/watch?v=X3HDnrehyDM))_ that can be played on any graph, but by convention, we restrict the game to tree graphs. The _Frog Jumping_ is equivalent to the case on "paths (tree graphs with exactly 2 leaves)" of the **Frog game** , which was resolved in the linked video.
The mechanics of the Frog game are in short,
> Given a tree graph and one frog on each vertex, the goal of the game is for all frogs to host a party on a single vertex of the graph. All of the $f$ frogs on some vertex can "jump" to some other vertex if and only if both vertices have at least one frog on them and both vertices are exactly $f$ edges apart from each other.
>
> If a sequence of "jumps" exists such that all frogs end up on a single vertex, then the party is successfully hosted on that vertex and we call that vertex a "lazy toad" (or just "lazy") because the frog that started there never jumped during the game.
Trivially, if $|V|=n$ is the number of vertices of some graph $G=(V,E)$, then every game on $G$ must end in $n-1$ moves (or less if the party isn't hosted, e.g. if the sequence of jumps is suboptimal and we can no longer make moves or if none of the vertices are lazy).
* * *
Here, I'm trying to resolve the Frog game on "full binary trees".
Let $T_h=(V,E)$ be a full binary tree of height $h$. Denote its layers with $0,1,2,\dots,h$ where the root $r\in V$ is on the layer $0$ ("bottom layer"), where layer $i=1,2,\dots,h$ contains vertices $v_{i,j}$ labeled with $j=1,2,\dots,2^i$.
That is, we have $|V|=2^{h+1}-1$. Let $h\in \mathbb N$ because $h=0$ is a single vertex which is trivial.
> **Conjecture.** For all $h\ge 4$, every vertex of $T_h$ is a "lazy toad" (is "lazy").
I couldn't prove the above, and for simplicity, my question is just about the root vertex:
> Let $h\in\mathbb N, h\ne 2$. Can we prove that the root $r$ of every such $T_h$ is a "lazy toad" (is "lazy")?
However, if the approach that can solve the root can be generalized to all other vertices, that's a big bonus.
* * *
**Solving the root**
**Reduction.** We say that $T_h$ can be reduced to $T_s,s\le h-2$, if there exists a sequence of moves that can bring all frogs from the top $s$ layers $h,h-1,\dots,h-s+1$ to the root vertex.
I've managed to reduce $T_h$'s for all $h\in(2,20)$ by hand pretty easily. At most, I needed to consider only $31$ vertices or less to find a pattern that allows a reduction of entire layers, which is very efficient if you consider that $T_{19}$ has over a million vertices. It looks like that maybe an inductive argument would be able to solve the problem, but I did not manage to find a sufficient set of move sequences.
I'll summarize some immediate results below.
> **Trivial reduction.** If $h=2^t+t-2,t\ge 2$ and $T_{t-1}$ is solvable in root, then $T_h$ can be reduced to $T_{h-t}$.
In the above reduction, we are simply (at the top layers) solving $T_{t-1}$ subtrees in root and then jumping from that root to the root of $T_h$.
> **Necessary minimal reduction condition.** Let $h\gt2,a\ge2$. If reduction from $T_h$ to $T_{h-a}$ is possible, then there exists $b\in[0,h-a+1]$ and a partition of the number $(2^a-1)2^b$ into the layers $H=\\{h,h-1,h-2,\dots,h-a+1\\}$ where $(h-l)\in H$ is used at most $(2^{a-l-1})2^b$ times.
In the above condition, we say that "a partition is solvable" if we can bring the corresponding amounts of frogs to the corresponding layers. I.e. the above condition simply says that we must be able to partition the total number of frogs from the top layers (that we are reducing) into the corresponding vertices from which they can jump to the root.
It remains to show (or find an explicit set of move sequences that will imply) that at least one partition from the above condition is solvable. Surprisingly, for all $h\in(2,20)$ that I've solved so far, the solution was always available in the smallest or one of the smallest partitions ($a\le 5,b\le 3$) of no more than $31$. The $h=20$ is the first case that requires considering more than $31$ vertices.
I still do not see how to construct explicit move sequences for larger $h$ without tackling each one individually. Alternatively, could there be a way to prove that the root is lazy, without finding explicit move sequences?
|
https://mathoverflow.net/questions/370694/is-the-frog-game-solvable-in-the-root-of-a-full-binary-tree
|
[
"graph-theory",
"trees",
"induction",
"binary-tree"
] | 18 | 2020-09-02T09:53:11 |
[] | 0 |
Science
| 0 |
199
|
mathoverflow
|
Orientation-reversing homotopy equivalence
|
If there is an orientation-reversing homotopy equivalence on a closed simply-connected orientable manifold is there an orientation-reversing homeomorphism?
It is not true, for instance, that if there is an orientation-reversing homeomorphism there must be an orientation-reversing diffeomorphism (consider exotic 7-spheres).
|
https://mathoverflow.net/questions/373348/orientation-reversing-homotopy-equivalence
|
[
"dg.differential-geometry",
"at.algebraic-topology",
"gt.geometric-topology",
"homotopy-theory"
] | 18 | 2020-10-05T06:15:17 |
[
"This can be heuristically reduced to finding manifolds that are homotopy equivalent but not homeomorphic, which should show you how to organize thinking about this. If $X$ does not admit an orientation-reversing homotopy equivalence and $Y$ is homotopy equivalent to $X$ but not homeomorphic, then the connected sum of $X$ and the opposite orientation of $Y$ is a manifold that admits an orientation-reversing homotopy equivalence, but no orientation-reversing homeo. $X=\\mathbb C\\mathbb P^4$ and $Y$ with different Pontrjagin classes. This isn't any easier than my previous example, though.",
"Correction: there is a smooth manifold homotopy equivalent to $S^4\\times S^4$ with nonzero $p_2$ and thus no orientation reversing homeomorphism, but I don't see an easy way to construct it. However, linear sphere bundles have nontrivial $p_1$ and so do provide counterexamples to the more natural question of whether a particular homotopy equivalence is homotopic to a homeomorphism. That is, the homotopy equivalence which reverses orientation on the base sphere is not homotopic to a homeomorphism, even though the homotopy equivalence which reverses orientation on fiber is realized by a homeo.",
"@ConnorMalin I'm not really sure what you're asking, but the first topic is called surgery theory and the second topic is called smoothing theory (or triangulation theory). For most purposes, I recommend the two volumes \"Surveys on Surgery Theory,\" ed Cappell-Ranicki-Rosenberg. But if you really want a foundational reference for the topological category, I think the only option is Kirby-Siebenmann.",
"@BenWieland Do you know of a standard reference for the obstructions to homotoping a homotopy equivalence to a homemorphism? And for isotopy between a homeomorphism and a diffeomorphism?",
"The difference between homeomorphism and diffeomorphism is subtle (torsion). The difference between homotopy equivalence and homeomorphism is crude (rational), the same as the difference between homotopy equivalence and diffeomorphism. Examples involving exotic spheres are difficult and overkill. There are easy smooth examples using Pontrjagin classes. There are linear $S^4$ bundles over $S^4$ homotopy equivalent to $S^4\\times S^4$ with nonzero $p_2$. Obviously smoothly chiral. Given the difficult but blackbox theorem that $p_2$ is a rational homeomorphism invariant, topologically chiral.",
"According to arxiv.org/pdf/0907.5283.pdf (Section 2, 4th bullet point), the lens space $L_5(1,1)$ has an orientation reversing homotopy equivalence, but no orientation reversing homeomorphism. Of course, it is not simply connected."
] | 6 |
Science
| 0 |
200
|
mathoverflow
|
Are these local systems on $\mathscr{M}_{g,1}$ motivic?
|
Let $(\Sigma_g, x)$ be a pointed topological surface of genus $g$, and let $MCG(g,1)$ be the mapping class group of this pointed surface. Then $MCG(g,1)$ has a natural action on $\pi_1(\Sigma_g, x)$ $$MCG(g)\to \text{Aut}(\pi_1(\Sigma_g))$$ and hence acts on the set of irreducible characters of $\pi_1(\Sigma_g)$: $$\gamma: MCG(g,1)\to \text{Aut}(\text{Char}^{\text{irr}}(\pi_1(\Sigma_g))).$$
Suppose $$\rho: \pi_1(\Sigma_g)\to GL_n(\mathbb{C})$$ is an irreducible representation of $\pi_1(\Sigma_g)$ which is fixed by this action, i.e. for every element $\tau$ of $MCG(g,1)$, there exists a matrix $\varphi_\rho(\tau)$ conjugating $\rho^\tau$ to $\rho$ as a representation of $\pi_1(\Sigma_g)$. Because $\rho$ is irreducible, $\varphi_\rho(\tau)$ is well-defined up to scaling, and so we obtain a natural projective representation $$\varphi_\rho: MCG(g,1)\to PGL_n(\mathbb{C}).$$ Choosing a representation of $PGL_n(\mathbb{C})$ (say, the adjoint representation), we obtain a representation of $MCG(g,1)$ and hence a local system on the moduli space of pointed curves $\mathscr{M}_{g,1}$.
> Are the local systems obtained this way motivic?
I know how to produce several examples of such $\rho$ fixed by $MCG(g,1)$ (e.g. using the Parshin trick) and in these examples the local systems in question are indeed motivic.
Of course this question is too hard as stated in this generality -- so it would be better to ask:
> Do the local systems thus obtained underly $\overline{\mathbb{Z}}$-variations of Hodge structure?
This question is naturally divided into three parts:
> 1. Is $\varphi_\rho$ always defined over $\overline{\mathbb{Q}}$? I.e. is it conjugate to a representation into $PGL_n(\overline{\mathbb{Q}})$?
>
> 2. Is $\varphi_\rho$ always integral, i.e. is it conjugate to a representation into $PGL_n(\overline{\mathbb{Z}})$?
>
> 3. Does $\text{ad}_{PGL_n}\circ \varphi_\rho$ always underly a $\overline{\mathbb{Q}}$-variation of Hodge structure?
>
>
The only method of proof I can think of at the moment would be to show that $\varphi_\rho$ is _rigid_ (which would imply (1) and (3), using non-abelian Hodge theory) or _cohomologically rigid_ (which would imply (1-3) via recent results of Esnault and Groechenig). I haven't had any luck showing this thus far.
* * *
Some brief remarks on this old question (9/11/22):
1. As written, the answer is "no": I know how to make counterexamples for $g\leq 2$. Indeed, for $g\leq 2$ one may construct such local systems which do not have quasi-unipotent mondoromy at infinity. It's possible that the answer is "yes" in $g\leq 2$ if one additionally assumes this.
2. Such examples are impossible for $g\geq 3$, as any local system on $\mathscr{M}_{g,n}$ automatically has quasi-unipotent monodromy at infinity in this case by a result of Aramayona-Souto.
3. The answer is "yes" for representations of rank $r<\sqrt{g+1}$ with finite orbit under $MCG(g,1)$ by one of the results of [this paper](https://arxiv.org/abs/2205.15352) of mine and Landesman. This is because such representations necessarily have finite image.
|
https://mathoverflow.net/questions/345955/are-these-local-systems-on-mathscrm-g-1-motivic
|
[
"ag.algebraic-geometry",
"gt.geometric-topology",
"mapping-class-groups"
] | 18 | 2019-11-13T08:17:12 |
[
"@BenWieland: (1) is an interesting idea -- I'll play with it a bit and see what I get. Thanks!",
"Oops, my (2) is not reducible as an $M_{g,1}$ representation. That was the whole point, yet it ruins it. The whole point is that it is the degeneration of an extension to a split extension, but such degenerations are trivial away from the split locus.",
"Yes, expand . . $M_2$ is virt $M_{0,6}$, so admits non-rigid reps. So ask $g\\geqslant3$, but why stop there? Why did you think that 2 was enough? I don't really know any non-rigid reps, but here are two ideas. (1) a Burau-ish rep of $M_g$ on the homology of the abelian cover, as a $\\mathbb Z[t_1,...t_{2g}]$-mod. (2) The rep of of $M_{g,1}$ on the Lie algebra of the 2-step nilpotent quotient of the fundamental group of the curve: shouldn't a 2-step nilpotent Lie algebra have a defm towards splitting, natural enough that it carries the rep? What happens when you restrict $M_{g,1}$ to $\\pi_1S_g$?",
"@BenWieland: I should have assumed $g\\geq 2$ -- the hope was that this construction rigidifies things, and it's motivated by some arithmetic considerations I omitted from the question, but which I can expand on if you'd like. Do you know of an example of a non-rigid representation of $\\pi_1(M_{g,n})$ for some $g\\geq 2$? Re: reps of $\\pi_1(M_{g, 1})$, the Parshin trick gives examples, and there are others coming from TQFT's...",
"This seems very arbitrary to me. Why not ask if all representations of $M_{g,n}$ are motivic? But they're not all rigid. If you had a nonrigid representation of $M_{g,1}$, you could restrict to $\\pi_1(\\Sigma_g)$ and check if it remained irreducible. But I don't know any representations of $M_{g,1}$. For example, the Burau representation of $M_{0,n}^1$ is not rigid and probably remains irreducible when restricted to $\\pi_1(P^1-n))$. (But it sounds motivic.)",
"@WillSawin: It seems to me that such reps will always be rigid as projective reps of the mapping class group, so at least (1) and (3) should be OK. And I think (2) should be OK by Esnault-Groechenig if the mapping class group action on the tangent space to $\\rho$ has no invariants...",
"It might be an easier project to show that, if $\\rho$ is rigid as a representation satisfying this additional condition, then $\\varphi_{\\rho}$ is motivic, and if $\\rho$ is not rigid then $\\varphi_\\rho$ is not (always) motivic. Determining whether $\\rho$ is / isn't always rigid could be a separate matter.",
"You’re right, fixed. Do you know how to make such a nontrivial family?",
"Your construction of a representation does not really work since you are trying to act via $Aut(\\pi_1(\\Sigma))$ instead of $Out(pi_1(\\Sigma))$. What you are actually getting here are projective representations of $MCG(g,1)$. I am sure, these are non-motivic in general and there are nontrivial families of (equivalence classes ) such representations. One more thing: Each fixed point $[\\rho]$ does get you a linear representation of $MCG(g)$ but in a different dimension: You act on the Zariski tangent space to the character variety at the point $[\\rho]$.",
"I agree with everything you’ve written—I tried the strategy in your second comment at some point but didn’t have any luck there either...some things are known in the case of 2-dimensional reps IIRC (due to Goldman and collaborators, for example), but not much.",
"With regards to showing at least the first couple of your three points, might it make sense to show that $\\rho$ has these properties first? I think if $\\rho$ is algebraic or integral then $\\varphi_{\\rho}$ is algebraic or integral as well. For algebraicity at least a plausible avenue of proof is to show that $\\rho$ has no 1-parameter deformations that respect this invariance property.",
"In the $\\ell$-adic world, if $\\rho$, $\\pi_1(\\Sigma_g) \\to GL_n( \\overline{\\mathbb Q}_\\ell)$ factors continuously through the profinite completion of $\\Sigma_g$, and this factorized representation is stable under the action of the arithmetic fundamental group of the moduli space of curves, then I think by the same construction we get a projective arithmetic representation. Of course by the grand theory of motives we expect variations of Hodge sructures and arithmetic representations should appear at the same time."
] | 12 |
Science
| 0 |
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