11
$\begingroup$

Let $G$ be a compact, connected and simply connected Lie group. 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. However, if $n>1$, one gets a complicated space. Any ideas?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ You mean singular cohomology groups of the quotient space $T^n/W$, no? How can $T/W$ be contractible, being a closed manifold? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 7, 2012 at 19:42
  • 4
    $\begingroup$ @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). $\endgroup$
    – Jason DeVito - on hiatus
    Commented Feb 7, 2012 at 21:02
  • 1
    $\begingroup$ 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 $n$. $\endgroup$
    – José Manuel Gómez
    Commented Feb 7, 2012 at 22:25
  • 2
    $\begingroup$ @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. $\endgroup$
    – Vitali Kapovitch
    Commented Feb 8, 2012 at 17:24

2 Answers 2

Reset to default
6
$\begingroup$

In the case $G=\mathrm{U}(m)$, this is the character variety $\mathrm{Hom}(\mathbb{Z}^n, \mathrm{U}(m))/\mathrm{U}(m)$.

See my paper here for general results on its topology (and similar results for other compact and complex reductive $G$).

In particular, the rational cohomology ring is worked out at the end. So from this you can at least get the free part of the integral cohomology, as discussed in this answer.

Also see this paper for the computation of its fundamental group.

Examples:

  1. For $n=1$ and $G$ simply connected you always get closed ball. See Remark 6.2 here.

  2. For $n=2$ and $G=\mathrm{SU}(2)$ you get a 2-sphere. Here is a 3D print of it. It is the boundary of $\mathrm{Hom}(F_2,\mathrm{SU}(2))/\mathrm{SU}(2)$. See Example 3.14 here.

  3. For $n=3$ and $G=\mathrm{SU}(2)$ you get 3-dimensional orbifold with 8 isolated singularities. See the discussion on page 23 here for a more detailed discussion. Also, there is this video giving a visualization of this orbifold.

Improve this answer
$\endgroup$
4
$\begingroup$

The case n=2 is well studied and has beautiful anwser. This quotient space is a weighted projective space (and for $SU(n)$ it is $CP^{n-1}$). See E. Looijenga, Root systems and elliptic curves, Invent. Math. 38 (1976), 17–32. And I.N. Bernshtein and O.V. Shvartsman, Chevalley’s theorem for complex crystallographic Cox- eter groups, Funct. Anal. Appl. 12 (1978), 308–310. As pointed out in Sean's answer these quotient spaces are moduli spaces of flat $G$ connections in the trivial $G$ bundle over the torus. It is interesting to extend this story to non-trivial $G$ bundles. A wonderful book about some of what goes on "Almost commuting elements in compact Lie groups" by Borel, Friedman and Morgan.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .