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The symmetric group $\Sigma_k$ acts on $X=F({\mathbb R}^n,k)$, the ordered configuration space of $k$ points in ${\mathbb R}^n$. If $n$ is odd, the cohomology $H^*(X;{\mathbb Q})$ is a rank-one free module over ${\mathbb Q}[\Sigma_k]$.

This is checked by calculating the character of this representation. (Note also that this is obvious for $n=1$.)

Are there known (one-vector) bases of this module?

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You might look at Church and Farb's Representation theory and homological stability (, specifically section 4. – HJRW Nov 20 '10 at 18:11

This isn't exactly an answer to your question, but here's how I like to think about the fact that you quoted.

Let's assume for a minute that $n=2$, so that we can think of $X$ as the complement of the braid arrangement in $\mathbb{C}^k$. Let $G = \mathbb{Z}/2\mathbb{Z}$, which acts on $X$ by complex conjugation.

Replace $\mathbb{Q}$ with a field $F$ of characteristic $2$. The $G$-equivariant cohomology ring $H^*_G(X; F)$ is a free module over $H^*_G(pt; F) \cong F[x]$ with the property that specializing at $x=0$ gives $H^* (X; F)$ and specializing at $x=1$ gives $H^*(X^G;F)$.

Thus we have a family of $\Sigma_k$ representations over the $F$-affine line interpolating between $H^* (X; F)$ and $H^*(X^G; F)$. Since the category of $\Sigma_k$ representations is semisimple, these two representations have to be isomorphic. The fact that $H^*(X^G; F)$ is the regular representation is obvious.

This is a good way to see that $H^*(X; F)$ is isomorphic to the regular representation of $\Sigma_k$. I'm not sure how to modify this argument to get $H^* (X; \mathbb{Q})$. I'm also not sure if this will help you find a cyclic vector, since it does not give you an explicit isomorphism between $H^* (X; F)$ and $H^* (X^G; F)$.

By the way, for $n>2$ you can do something similar, where $G$ acts on $\mathbb{R}^n$ by negating the last $n-1$ coordinates.

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You prove that H^*(X;F) is the regular representation of Sigma_k, but this is false, at least for k=2, when this representation is trivial. – Semen Podkorytov Nov 22 '10 at 13:54
When $n$ is even, $H^*\!(X;\mathbb Z)$ is $2\ \text{Ind}_{\langle(1\,2)\rangle}^{S_k}1$, two copies of the representation induced from the trivial representation of a single transposition. It's when $n$ is odd that $H^*\!(X;\mathbb Z)$ is the regular representation $\text{Ind}_1^{S_k}1$. – Tom Church Nov 22 '10 at 15:51
To Tom: Even for $n$ odd, $H^*(X;\mathbb Z)$ is not the regular representation, by which I mean $\mathbb Z[\Sigma_k]$. Indeed, let $k=2$. Then $H^*=H^0\oplus H^{n-1}$ while $\mathbb Z[\Sigma_2]$ is not decomposable into a direct sum. – Semen Podkorytov Nov 25 '10 at 16:39
To Semen: When $k=2$, both the regular representation and $H^*(X; F)$ are trivial. (Recall that in my argument $F$ is supposed to have characteristic 2.) – Nicholas Proudfoot Nov 27 '10 at 14:48
Semen is correct; I should have written $H^\ast(X;\mathbb{Q})$ in both cases. – Tom Church Dec 3 '10 at 2:02

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