# equivariant cohomology of the complement to the arrangment $\cup_{i\neq j}overrightarrow{x_i} = overrightarrow{x_j}$?

Let $V=\mathbb{R}^d$ be a $d$-dimensional (Euclidian) vector space over real numbers. Let $G=SO(V)$ be a compact Lie group of linear orthogonal transformations of $V$.

Let $Conf_n(V)$ be the space of $n$-tuples $\{(\overrightarrow{x_1},\ldots,\overrightarrow{x_n}), \overrightarrow{x_i}\in V\}$ of pairwise distinct points in $V$ ($\overrightarrow{x_i} \neq \overrightarrow{x_j}$). I.e. $Conf_n(V)$ is an open space that is a complement in $V\times V \times \ldots \times V = \mathbb{R}^{n d}$ to the union of arrangements of codimension $d$: $\cup_{i,j} \{\overrightarrow{x_i} = \overrightarrow{x_j}\}$ There is a natural action of $G$ on $Conf_{n}(V)$ (namely componentwise).

Question 1): Compute the equivariant (co)homology of $Conf_{n}(V)$ with respect to this action of $SO(V)$: $$H^{\bullet}_{SO(V)} (Conf_n(V);\mathbb{R})$$

Question 2): The same question for the complex situation. I.e. $V = \mathbb{C}^d$ and the group is $G = SU(V)$. and we are interested in the description of $$H^{\bullet}_{SU(V)} (Conf_n(V);\mathbb{C})$$

Note, that in the case $V= \mathbb{C} = \mathbb{R}^2$ the answer is known to coincide with the cohomology of the open moduli spaces of curves with zero genus and $n+1$ marked points. Unfortunately for the case of $d>2$ the action of $G$ is no more the free action and the total answer should be infinite at least for $d>n\geq 2$, but I will be happy with any reasonable description, even if it will be in terms of cohomology of some finetely generated differential graded algebra.

-
If $V = \mathbb{R}^3$ and you consider not the full group $SO(3)$ but rather the circle subgroup consisting of rotations about some axis in $V$, a presentation is given by Daniel Moseley in Theorem 3.4 of arxiv.org/abs/1110.5369. Whatever the answer to your question is, it should admit lots of natural maps to Moseley's ring (one for every choice of oriented line in $V$). – Nicholas Proudfoot Apr 19 '12 at 20:40

Here is a partial answer, which at least illustrates how to attack these problems using the methods of algebraic topology.

As usual, to compute $H^\ast_G(X)=H^\ast(EG\times_G X)$ where $G$ is a compact Lie group acting on a space $X$, we examine the Leray-Serre spectral sequence of the fibration

$$X\to EG\times_G X\to BG$$

(my favourite reference for this is the book of McCleary). In the case of Question 1), $X=Conf_n(\mathbb{R}^d)$ and $G=SO(d)$ for some $n,d\ge 2$. Since $G$ is connected, $\pi_1(BG)$ is trivial. Both $X$ and $BG$ have the homotopy type of $CW$ complexes of finite type. Therefore Proposition 5.6 in McCleary applies and the SS has

$$E_2^{\ast,\ast} = H^\ast(BG;\mathbb{R})\otimes_{\mathbb{R}} H^\ast(X;\mathbb{R})$$

Now the cohomology of the classifying spaces is known to be $$H^\ast(BSO(2k);\mathbb{R})\cong \mathbb{R}[p_1,\ldots , p_{k-1},\chi],\qquad H^\ast(BSO(2k+1);\mathbb{R})\cong \mathbb{R}[p_1,\ldots , p_k],$$

where the $p_i\in H^{4i}(BSO(d);\mathbb{R})$ are Pontryagin classes and $\chi\in H^{2k}(BSO(2k);\mathbb{R})$ is an Euler class (see for instance Corollary 1.90 in this book). Meanwhile, the cohomology of configuration spaces is also known, you'll find a full description in Chapter V of the book of Fadell and Husseini (see also Section 4 of http://arxiv.org/abs/0806.4111 for the short version). The key point is that $H^\ast(Conf_n(\mathbb{R}^d);\mathbb{R})$ is generated as an algebra by elements in degree $d-1$, and hence the cohomology is concentrated in degrees $i(d-1)$ for $i=0,1,\ldots , (n-1)$.

The upshot is that the $E_2$ term is rather sparse, and this should allow you to conclude that the spectral sequence collapses in many cases (the differentials being zero for dimensional reasons). For example, if $d$ is odd then ($d$ is not a multiple of $4$ and) we have

$$H^\ast_{SO(d)}(Conf_n(\mathbb{R}^d);\mathbb{R})\cong H^\ast(BSO(d);\mathbb{R})\otimes_{\mathbb{R}} H^\ast(Conf_n(\mathbb{R}^d);\mathbb{R})$$ as algebras graded vector spaces (some extra information may be needed to conclude an isomorphism of graded algebras, see McCleary examples 1.J and 1.K).

-
Thanks a lot for your answer. There is no isomorphism on the level of algebras (as you mentioned). But one can guess about the general answer taking in mind the following equality $$H^{\bullet}_{BSO(d)}(S^{d-1})=H^{\bullet}_{BSO(d-1)}(point).$$ – Anton Khoroshkin Apr 22 '12 at 11:54

When $V$ is a complex vector space, I computed the equivariant homology of the configuration space in a paper called Operads of moduli spaces of points in $\mathbb{C}^d$, not with respect to $U(V)$, but its restriction to $U(1)$, along the diagonal embedding. I'm not sure, but I think that extending to $U(V)$ just tensors this result with $H*(BU(\dim(V)-1))$, since the answer that I got for $U(1)$ doesn't seem to allow much room for nonzero operations coming from $H_*(U\dim(V)-1)$.

The methods are closely related to those that Mark Grant sketches above, but are nicely packaged using the language of operads. If you're only interested in rational computations, I think a similar answer is obtainable in the real setting.

-
Ack! There is a danger answering these questions after not thinking about this stuff for a long while. I don't mean $U(\dim(V)-1)$ above, I mean the part of $H_*(U(V)) = \Lambda[x_1, x_3, ..., x_{2\dim(V)-1}]$ which does not come from $U(1)$ (i.e., $x_1$). – Craig Westerland Apr 21 '12 at 14:04
Thanks a lot for your answer and for the reference. My expectation coincide with yours. For SO-case the situation seems to be pretty similar, but I was wondering if the answer is already written somewhere and you give me a reference. – Anton Khoroshkin Apr 22 '12 at 11:44