MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 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

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

as a bigraded algebra.

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 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).

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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