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

Let $G \subset SL(V, \mathbb{C})$ be a finite group and $R=(\operatorname{Sym}\[V\])^G$ is the ring of polynomial invariants, $W$ some irreducible complex representation of $G$. I want to know is there any methods (or at least examples) of computing generators and relations of the $R$ module $M=(\operatorname{Sym}\[V\] \otimes_\mathbb{C} W)^G$?

share|improve this question
    
$M$ is often free. It is locally free, so if $R$ has no nontrivial vector bundles, say if $R$ is a free ring, then it is free. This gets you the answer instantly in a lot of cases. –  Will Sawin Nov 9 '12 at 17:12
    
@Will $M$ is not locally free even for $\operatorname{dim} V=2$. In this case $M$ is reflexive module over Kleinian singularity. Moreover, projective dimension of such module is infinite in this example. –  Sasha Pavlov Nov 9 '12 at 18:47
add comment

1 Answer

up vote 2 down vote accepted

The polynomial ring Sym(V) is naturally graded: $Sym(V) = \oplus Sym(V)_d$ Suppose you have can compute the isotypic decomposition of these graded components

Sym(V$)_d$ =

${\oplus_{\chi \in A_d} U_\chi}$

where the $U_\chi$ are irreducible representations of $G$. Then $M = \oplus_d \oplus_{\chi \in A_d} (U_\chi \otimes W)^G$. Now $(U_\chi \otimes W)^G = 0$ unless $W \cong U_\chi^*$ in which case $(U_\chi \otimes W)^G$ is one dimensional. Thus your problem really amounts to decomposing $Sym(V)$ as a $G$-representation. So far none of these depends on $G$ being finite (but it should be reductive).

$M$ is called the module of $W$-covariants of $V$. The Hilbert Series of $M$ may be computed using Molien's Theorem. A minimal generating set for $M$ contains only elements of degree less than or equal to the order of $G$. Lots of other things are known but I suggest you read about it. One good reference for this is {\it Invariants of Finite Groups and Their Applications to Combinatorics} R. Stanley Bull. A.M.S. 1979.

share|improve this answer
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.