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Let $G$ be a compact semisimple group and let $\Gamma$ be a finite subgroup of $G$. I am interested, for $(\pi,V)\in \widehat G$ (irred rep of $G$), in a formula for $\mathrm{dim} V^\Gamma$, the dimension of the invariant space of $V$ by $\Gamma$. When $\Gamma$ is the trivial group, Weyl dimension formula says that $$ \mathrm{dim}(V) = \prod_{\alpha\in\Phi^+} \frac{\langle\lambda+\rho,\alpha\rangle}{\langle\rho,\alpha\rangle}, $$ where $\lambda$ is the highest weight of $\pi$ (I think the notation is usual).

One idea: let $P:V\to V$ given by $$ P(v)=\frac1{|\Gamma|} \sum_{\gamma\in\Gamma} \pi(\gamma)(v), $$ then $P$ is surjective onto $V^\Gamma$, thus $\mathrm{dim}(V^\Gamma)= \mathrm{Tr}(P) = |\Gamma|^{-1} \sum_{\gamma\in\Gamma} \chi_\pi(\gamma)$, where $\chi_\pi$ is the character of $\pi$. Hence, by applying the Weyl character formula, we obtain a formula for $\mathrm{dim}(V^\Gamma)$. However, I expect that there is a more explicit formula (like Weyl dimension formula, without a sum over the Weyl group) in terms of the root system of $\mathfrak g$.

Question: Is there a formula for $\mathrm{dim}(V^\Gamma)$ similar to Weyl Dimension Formula?

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What makes you think such a formula is possible? How are discrete subgroups of compact Lie groups connected to root systems? –  Vít Tuček Aug 14 '13 at 20:00
    
After Jim Humphreys and Dan Petersen's answers, it seems that such formula does not exists in general. However, I still think that it's possible when $\Gamma$ is inside the maximal torus (in particular $\Gamma$ is abelian). –  emiliocba Aug 15 '13 at 10:16
    
OK. Just to let you know. A similar problem in different setting is considered in the eighth chapter of Huang, Pandzic - Dirac operators in representation theory. –  Vít Tuček Aug 15 '13 at 23:45
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2 Answers

up vote 5 down vote accepted

Probably there is no explicit formula of the type you want. In any case, it's important to look first at the most accessible special cases (even though Weyl's formulas may be kept in the background). For example, consider $G= \mathrm{SU}(2)$ and its finite subgroups, using both the McKay correspondence and the Lie theoretic information about representations of $G$ afforded by symmetric powers of the natural 2-dimensional module. In this situation you are asking for the Poincare series of the resulting algebra of invariants of a given finite subgroup $\Gamma$.

More generally you could consider the multiplicities of all irreducible representations of $\Gamma$ in the various irreducible representations of $G$. This has been studied in detail by Kostant in the mid-1980s, with emphasis on the interplay of Lie theory and McKay's theory. Even in this relatively small case, the results are fairly complicated and therefore cautionary for the general question you are raising. The root system and Weyl group (along with its Coxeter elements) do however come directly into the picture. Here are references to the announcement and full article:

On finite subgroups of $\mathrm{SU}(2)$, simple Lie algebras, and the McKay correspondence. Proc. Nat. Acad. Sci. U.S.A. 81 (1984), no. 16, Phys. Sci., 5275–5277.

The McKay correspondence, the Coxeter element and representation theory. The mathematical heritage of Elie Cartan (Lyon, 1984). Asterisque 1985, Numero Hors Serie, 209–255.

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Your idea seems to have already been implemented, see Section 2 of Chenevier and Renard's "Level one algebraic cusp forms of classical groups of small ranks". But it appears to be not as straightforward as just applying the ordinary Weyl character formula - you need a certain "degenerate" version of it.

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It's a long paper but might contribute to the solution of the problem here. A preprint version is at front.math.ucdavis.edu/1207.0724 –  Jim Humphreys Aug 14 '13 at 22:14
    
Excelent reference! I think that they applied the "degenerate" Weyl character formula to give a better computer algorithm. –  emiliocba Aug 15 '13 at 10:10
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@emiliocba I don't think its just an issue of a better algorithm. For an element which is not regular WCF gives the expression $\frac 0 0$ which needs to somehow be interpreted correctly. –  Dan Petersen Aug 15 '13 at 10:30
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