6
$\begingroup$

This question is a follow-up to my previous question factorization of the regular representation of the symmetric group, which was answered in a very satisfactory way.

Let $\operatorname{Conf}(n,\mathbb{R}^3)$ be the configuration space of $n$ labeled points in $\mathbb{R}^3$, and consider the cohomology $B_n := H^*(\operatorname{Conf}(n,\mathbb{R}^3))$, which is a graded representation of $S_n$. Let $W_n := V[n] + q^2 V[n-1,1]$ be the graded representation that is the 1-dimensional trivial representation in degree 0 and the $(n-1)$-dimensional irreducible permutation representation in degree 2.

Question: Does there exist a graded representation $M_n$ such that $M_n\otimes W_n\cong B_n$?

If we forget about the grading, then $B_n$ is just the regular representation and $W_n$ is the vector representation $\mathbb{C}^n$; the existence of a representation whose tensor product with the vector representation is isomorphic to the regular representation is explained in the aforementioned post. So I am now asking a more refined version of the question in which the grading is being taken into account.

I will also note that I "know" that the answer is positive. That is, I can define a graded representation $M_n$, I conjecture that there is an isomorphism $M_n\otimes W_n\cong B_n$, and I've checked this conjecture on a computer up to $n=10$. What I really want to know is whether the graded representation $B_n$ is already known to factor in this way. If so, then I would really like to understand why this is the case, as I believe that it will help me to understand the representation $M_n$ that I am interested in.

(In case anyone would like to know, my representation $M_n$ that conjecturally solves this problem is the intersection cohomology of the hypertoric variety associated with the braid arrangement.)

$\endgroup$
  • $\begingroup$ Is this configuration space $S_n$-equivariantly the total space of a bundle such that either the fiber or the base is homotopy equivalent to a wedge of $2$-spheres, or something like it? $\endgroup$ – Qiaochu Yuan Jan 13 '16 at 5:48
  • $\begingroup$ Yes, but then you have to break symmetry. You can project to $\operatorname{Conf}(n-1,\mathbb{R}^3)$ by forgetting the last point; this is a fiber bundle that behaves like a product on the level of cohomology, and it tells you that $B_n \cong B_{n-1}\otimes W_n$. But this only holds $S_{n-1}$-equivariantly, since you have chosen which point to forget. This does tell you that, if $M_n$ is a solution to my problem, then the restriction of $M_n$ to $S_{n-1}$ is isomorphic to $B_{n-1}$. $\endgroup$ – Nicholas Proudfoot Jan 13 '16 at 6:11
3
$\begingroup$

Here's a geometric construction of a factorization that works for points in $\mathbb R^2$ (and not any other dimension). Given the close relationship between the cohomology rings of configuration spaces of points in $\mathbb R^d$ for varying $d$ (cf my answer to Cohomology of configuration space as a representation of the symmetric group) I'd expect that one can extract a factorization in arbitrary dimension from this construction, even though the geometric part breaks down.

Note first that the affine group $G = \mathbb C \rtimes \mathbb C^\times$ is the subgroup of Möbius transformations fixing the point $\infty \in \mathbb P^1$. It follows that the quotient $\mathrm{Conf}(\mathbb C,n)/G$ is equal to the moduli space $M_{0,n+1}$ of $(n+1)$ points on the projective line modulo symmetries. This makes $\mathrm{Conf}(\mathbb C,n)$ homotopic to a trivial circle bundle over $M_{0,n+1}$, and so $$ H^\bullet(\mathrm{Conf}(\mathbb C,n)) \cong H^\bullet(M_{0,n+1}) \oplus H^{\bullet-1}(M_{0,n+1}).$$

So one only needs to construct such an $S_n$-equivariant factorization for the cohomology $H^\bullet(M_{0,n+1})$. Now there is an $S_n$-equivariant fiber bundle $M_{0,n+1} \to M_{0,n}$ by forgetting the last marking. Each fiber $F$ is $\mathbb P^1$ minus $n$ points, and so $H^0(F) \cong V[n]$, $H^1(F) \cong V[n-1,1]$ as $S_n$-representations. The Leray-Serre spectral sequence degenerates, and this gives the claimed factorization.

$\endgroup$
  • $\begingroup$ Thanks, that's a nice argument! I agree that the fact that such a factorization exists for $\mathbb{R}^2$ provides more evidence that it should exist for $\mathbb{R}^3$. Furthermore, it's nice that you get an interesting description of the factor. $\endgroup$ – Nicholas Proudfoot Jan 13 '16 at 21:09
  • $\begingroup$ I don't why I didn't realize this earlier, but your argument actually becomes easier when you replace $\mathbb{R}^2$ with $\mathbb{R}^3$. You can replace the group of M\"obius transformations by $SU(2) \cong S^3$ acting on itself by left multiplication, and replace $G$ by the trivial group. What's making this case easier is the fact that $S^3$, unlike $\mathbb{C}P^1$, admits a free transitive action by a group (namely $SU(2)$). $\endgroup$ – Nicholas Proudfoot Feb 29 '16 at 21:17
  • $\begingroup$ Oh! That's very nice. $\endgroup$ – Dan Petersen Mar 1 '16 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.