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Let $W$ be a Coxeter group and let $P_W(q) = \sum_{w \in W} q^{\ell(w)}$ be its Poincare series. When $W$ is the Weyl group of a simple algebraic group $G$ (hence $W$ is finite), $P_W(q)$ is the generating function describing the cells in the Bruhat decomposition of the flag variety $G/B$, $B$ a Borel.

What happens when $W$ is infinite, e.g. when $W$ is an affine Weyl group or a hyperbolic Coxeter group? Can $P_W$ still be associated to an infinite-dimensional "flag variety"?

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up vote 6 down vote accepted

I think that Shrawan Kumar's book "Kac-Moody groups, their flag varieties, and representation theory" will contain the flag varieties (which are really ind-varieties in the non-finite case) that you are looking for.

A crystallographic Coxeter group is one of the form $\langle s_1,\ldots,s_n| s_i^2=(s_i s_j)^{m_{ij}}=1\rangle$ where each mij is equal to 2,3,4,6 or infinity. Such Coxeter groups are precisely the Weyl groups associated to arbitrary Kac-Moody algebras. In this case, there is a corresponding Kac-Moody group, together with an associated flag variety and Schubert cells, which seems to me to be what you want.

It is this geometry that is the starting point of the geometric interpretation of Kazhdan-Lusztig polynomials in the crystallographic case. As is to be expected, the finite case is easier than the affine case, which again is easier than the arbitrary KM case.

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My understanding was that the definition of crystalographic is more subtle than this. The condition is that you can work over the integers. Your condition is necessary. It is also sufficient for finite type and affine type. My recollection is that there is a condition along the lines of: every cycle has an even number of 3's. – Bruce Westbury Jun 17 '10 at 6:31
An example to explain why crystallographic is more subtle than this: Take the Coxeter group on three generators, with $p^2=q^2=r^2=1$ and $(pq)^4=(pr)^4=(qr)^4=1$. This is a valid Coxeter group, but is NOT crystallographic. If there were an associated root system, then, for each pair of $\alpha_p$, $\alpha_q$ and $\alpha_r$, one element of the pair would have length either $1/2$ or $2$ times the other. This is impossible. – David Speyer Jun 17 '10 at 9:30
However, Peter is right that Poincare series of crystallographic Coxeter groups are always Betti numbers of Kac-Moody flag varieties, and that Kumar's book covers this very well. – David Speyer Jun 17 '10 at 9:31
@David Your example can be considered as the Weyl group associated to the Cartan Matrix a_ij=-2,2 or -1 according to whether i is less than, equal to or greater than j resp. This is not symmetrisable as your argument shows. I had thought that this flag variety construction did not require the KM group to arise from a symmetrisable Cartan datum, but I am not sufficiently versed in this subject to make a definite claim either way on the matter. – Peter McNamara Jun 19 '10 at 19:05

Coxeter groups form a very large class of groups defined by generators and relations, whose Poincare series are unlikely to have a common geometric interpretation. However, the Poincare series of an affine Weyl group has been important in work of Bott on topology of Lie groups, as well as in work of Iwahori-Matsumoto and others on p-adic groups, etc. References are given in section 8.9 of my 1990 Cambridge Press book Reflection Groups and Coxeter Groups, while the seminal 1965 IHES paper by Iwahori-Matsumoto is available online at (search for article):

Iwahori, Nagayoshi; Matsumoto, Hideya, On some Bruhat decomposition and the structure of the Hecke rings of $p$-adic Chevalley groups. Publications Mathématiques de l'IHÉS, 25 (1965), p. 5-48 Full entry | Full text djvu | pdf | Reviews MR 32 #2486 | Zbl 0228.20015

There are Kac-Moody groups in some generality, with associated "Weyl groups" and "Bruhat decompositions"; but Bruhat cells may have finite dimension in some cases, finite codimension in others. It gets complicated.

It's possible that work of Michael Davis and Ruth Charney on hyperbolic Coxeter groups would be relevant to your question, but I don't know enough about that.

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Are we assuming $W$ is crystalographic? – Bruce Westbury Jun 16 '10 at 19:48
I think the original question as well as the special cases I've mentioned all involve "crystallographic" Coxeter groups only, though the precise definition of that term in Kac-Moody theory may not be the usual one in combinatorial geometry. Other Coxeter groups, especially in the finite case, are intriguing but not obviously close to concepts coming from Lie theory. – Jim Humphreys Jun 16 '10 at 20:42
@Jim: Thanks for the answer! I was secretly hoping for a nice answer involving Kac-Moody algebras, but I guess the situation is complicated. @Bruce: Not necessarily - I'd be interested in any case for which an answer of some form exists. – Qiaochu Yuan Jun 16 '10 at 21:15

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