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

Let $G$ be a compact, connected and simply connected. Let $T\subset G$ be a maximal torus and let $W$ be the corresponding Weyl group. Then we have the diagonal action of $W$ on $T^{n}$ for $n\ge 0$. I would like to know if the cohomology groups $H^{*}(T^{n}/W;\mathbb{Z})$ have been computed or if anything is known about them. The case $n=1$ is particularly simple as $T/W$ is contractible when $G$ simply connected but when $n>1$ one gets a complicated spaces. Any ideas?

share|improve this question
1  
You mean singular cohomology groups of the quotient space $T^n/W$, no? How can $T/W$ be contractible, being a closed manifold? –  Mariano Suárez-Alvarez Feb 7 '12 at 19:42
3  
@Mariano: $T/W$ isn't, in general, a closed manifold since $W$ doesn't act freely on $T$. For example, when $G = SU(2)$, $W = \mathbb{Z}/2\mathbb{Z}$ acts on the circle as complex conjugation. The quotient is homeomorphic to a closed interval (and is contractible). –  Jason DeVito Feb 7 '12 at 21:02
    
I mean the cohomology of the honest quotient $T^{n}/W$. When $n=1$ one can show that a closed alcove is homeomorphic to $T/W$ and thus it is contractible. For $n>1$ on gets a complicated quotient. For example when $G=SU(2)$ and $n=2$ one has $T^{2}/W$ is homeomorphic to $\mathbb{S}^2$ and it gets more complicated for higher values of $n$. –  José Manuel Gómez Feb 7 '12 at 22:25
    
@José Manuel Gómez I'm confused by this. I think the quotient is always contractible. The Weyl group is generated by reflections and the Weyl chamber walls project to the boundary of the quotient $T/W$. In particular for $G=SU(3)$ the quotient is a flat triangle, not a sphere, topologically. –  Vitali Kapovitch Feb 8 '12 at 17:24
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.