# Random invariant theory bleg

So, a question I was thinking about for Cherednik algebras (more on that later) ended up reducing to the following invariant theory statement I wasn’t able to prove (or at least have been too lazy to try).

So, let G be a reductive (or if you like real groups better, compact) group acting on a complex vector space V.  Consider the polynomial functions $R=\mathbb{C}[V]$ as a module over invariant functions $R^G$.  For each irreducible representation W of G, we have a W-isotypic component $R_W\subset R$.  (Or, if you like, we can take the eigenspace decomposition for the action of class functions on G by convolution, if G is compact).  Of course, $R_W$ is an $R^G$ module in an obvious way.

Question. Is $R_W$ finitely generated as an $R^G$ module?

Note: this is well known for a finite group, since $R$ itself is a finitely generated $R^G$ module in this case.  The question above obviously fails for all of $R$ when $G$ is not finite, but the isotypic components are more “the right size.”

## 2 thoughts on “Random invariant theory bleg”

1. Allen Knutson says:

Let me call it $R_\lambda$ instead of $R_W$, where $\lambda$ is the relevant dominant weight. (Philosophically, it’s strange to have a specific irrep $W$ involved, instead of something that indexes the equivalence class. And how dare you call it $W$ anyway?) Let $\lambda^*$ denote the high weight of $W^*$.

Now consider the algebra $R \otimes \oplus_n V_{n\lambda}$, where the latter has the Cartan multiplication. This is still finitely generated. Its invariant ring is $\oplus_n R_{n\lambda}$, which being an invariant ring is finitely generated. Hence each graded piece is finitely generated over the degree $0$ piece.

Symplectic geometers know tensoring with this Cartan ring as “the shifting trick”; I don’t know what it’s called elsewhere.

2. Thanks, Allen. I should have thought of that, but I guess that’s why I have the internet.