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.)

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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