How does the Mulmuley-Sohoni geometric approach to producing lower bounds avoid producing natural proofs (in the Razborov-Rudich sense)? The exact phrasing of the title is due to Anand Kulkarni (who proposed this site be created). This question was asked as an example question, but I’m insanely curious. I know very little about algebraic geometry, and in fact also only have a cursory, undergraduate understanding of the obstacles at play in the P/poly versus NP question (non-relativizing, non-algebrazing, likely won’t be a natural proof).

What makes algebraic geometry seem like it can bypass these sorts of obstacles? Is it just field expert intuition or do we have a really good reason to believe the approach is fundamentally more powerful than previous approachs? What weaker results have this approach been able to achieve?
[I'll answer the question as stated in the title, leaving the litany of other questions about GCT for other threads.] Proving the conjectures arising in GCT seems like it will crucially use the fact that the functions under consideration (determinant and permanent, and other related polynomials for P/poly and NP) are characterized by their symmetries.  This necessity is not a formal result, but an intuition expressed by several experts.  (Basically that in the absence of characterization by symmetries, understanding the algebraic geometry and representation theory that arises is far harder.)

This should bypass Razborov-Rudich because very few functions are characterized by their symmetries (bypassing the largeness condition in the definition of natural proofs).  Again, I have not seen a proof of this, but it is an intuition I have heard expressed by several experts.

Now, over the complex numbers, it's not clear to me that there's an analog of Razborov-Rudich.  Although most of GCT currently focuses on the complex numbers, there are analogs in the finite characteristic (promised in the forthcoming paper GCT VIII).  In finite characteristic, one might actually be able to prove a statement of the form "Very few functions are characterized by their symmetries."



[In response to Ross Snider's comment, here's an explanation of characterization by symmetries.]

First, an explanation-by-example.  For the example, define an auxiliary function $q$.  If $A$ is a permutation matrix, then $q(A)=1$ and if $A$ is diagonal, then $q(A)=det(A)$ (product of the diagonal entries).  Now, suppose $p(X)$ is a homogeneous degree $n$ polynomial in $n^2$ variables (that we think of as the entires of an $n \times n$ matrix $X$).  If $p$ has the following symmetries:


$p(X) = p(X^t)$ (transpose)
$p(AXB) = p(X)$ for all pairs of matrices $(A,B)$ such that $A$ and $B$ are each either permutation matrices or diagonal matrices and $q(A)q(B) = 1$


then $p(X)$ is a constant multiple of $perm(X)$ for all matrices $X$.  Hence we say the permanent is characterized by its symmetries.

More generally, if we have a (homogeneous) polynomial $f(x_1, ..., x_m)$ in $m$ variables, then $GL_m$ (the group of all invertible $m \times m$ matrices) acts on $f$ by $(Af)(x_1,...,x_m) = f(A^{-1}(x_1),...,A^{-1}(x_{m}))$ for $A \in GL_m$ (where we are taking the variables $x_1,...,x_m$ as a basis for the $m$-dimensional vector space on which $GL_m$ naturally acts).  The stabilizer of $f$ in $GL_m$ is the subgroup $\text{Stab}(f) = \{ A \in GL_m : Af = f\}$.  We say $f$ is characterized by its symmetries if the following holds: for any homogeneous polynomial $f'$ in $m$ variables of the same degree as $f$, if $Af' = f'$ for all $A \in \text{Stab}(f)$, then $f'$ is a constant multiple of $f$.