Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
3  
You might look at Church and Farb's Representation theory and homological stability (arxiv.org/abs/1008.1368), specifically section 4. –  HJRW Nov 20 '10 at 18:11
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

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.