TITLE: factorization of the cohomology of configuration space
QUESTION [6 upvotes]: 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.)

REPLY [3 votes]: 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.