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

Let $\Conf_n(V)$ be the space of $n$-tuples $\{(\vec x_1,\dotsc,\vec x_n), \vec x_i\in V\}$ of pairwise distinct points in $V$ ($\vec x_i \ne \vec x_j$). I.e. $\Conf_n(V)$ is an open space that is the complement in $V\times V \times \dotsb \times V = \mathbb{R}^{n d}$ of the union of arrangements of codimension $d$: $\bigcup_{i,j} \{\vec x_i = \vec x_j\}$. 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 longer free 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 finitely generated differential graded algebra.

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  • $\begingroup$ 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 $V$). $\endgroup$ Commented Apr 19, 2012 at 20:40

2 Answers 2

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$\DeclareMathOperator\Conf{Conf}\DeclareMathOperator\SO{SO}$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 A user's guide to spectral sequences by 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 $$H^\ast(B{\SO(2k);\mathbb{R})}\cong \mathbb{R}[p_1,\dotsc, p_{k-1},\chi],\qquad H^\ast(B{\SO(2k+1)};\mathbb{R})\cong \mathbb{R}[p_1,\dotsc, p_k],$$ where the $p_i\in H^{4i}(B{\SO(d)};\mathbb{R})$ are Pontryagin classes and $\chi\in H^{2k}(B{\SO(2k)};\mathbb{R})$ is an Euler class (see for instance Corollary 1.90 in Félix, Oprea, and Tanré - Algebraic models in geometry). Meanwhile, the cohomology of configuration spaces is also known, you'll find a full description in Chapter V of the book Geometry and topology of configuration spaces by Fadell and Husseini (see also Section 4 of Farber and Grant - Topological complexity of configuration spaces 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 $i=0,1,\dotsc, (n-1)$.

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 $$ H^\ast_{\SO(d)}(\Conf_n(\mathbb{R}^d);\mathbb{R})\cong H^\ast(B{\SO(d)};\mathbb{R})\otimes_{\mathbb{R}} H^\ast(\Conf_n(\mathbb{R}^d);\mathbb{R}) $$ 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).

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    $\begingroup$ 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 $$H^{\bullet}_{BSO(d)}(S^{d-1})=H^{\bullet}_{BSO(d-1)}(point).$$ $\endgroup$ Commented Apr 22, 2012 at 11:54
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$\DeclareMathOperator\U{U}$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 $H_*(B{\U(\dim(V)-1)})$, 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.

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  • $\begingroup$ 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., $x_1$). $\endgroup$ Commented Apr 21, 2012 at 14:04
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    $\begingroup$ 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. $\endgroup$ Commented Apr 22, 2012 at 11:44
  • $\begingroup$ @CraigWesterland, while this is on the front page, I edited to update the link to your paper, but I wasn't sure how to take into account your comment. $\endgroup$
    – LSpice
    Commented Dec 5, 2020 at 2:12

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