Let $H \subset G$ be an inclusion of reductive groups over an algebraically closed field $k$ of char $0$. For simplicity, let's assume that $G$ is split and $H$ contains a maximal torus for $G$. Then the map $k[G]^G \rightarrow k[H]^H$, where invariants are for the conjugation action, is injective. As is well known, it coincides with the map of representation rings induced by restriction, $R(G) \rightarrow R(H)$.

Now suppose I'm interested in $k[G^n]^G \rightarrow k[H^n]^H$, where the $G$-action on $G^n$ is via diagonal conjugation, and similarly for $H$. This should certainly not be injective for all $n$. What can one say about the kernel? In particular, is there a representation-theoretic description?

For example, one can give algebra generators for $k[G^n]^G$ in terms of representations, as of matrix coefficients for $n$-fold tensor powers of irreducible representations for $G$. Can one describe the kernel as an ideal in terms of these generators in some clean way?


1 Answer 1


Even in the case $n=1$, the map $k[G^n]^G\to k[H^n]^H$ is not injective.

Example: let $G=GL_2(\mathbb{C})$ and $H=SL_2(\mathbb{C})$. Then $\mathbb{C}[H]^H\cong\mathbb{C}[t]$ where $t$ corresponds to the trace, and $\mathbb{C}[G]^G\cong\mathbb{C}[t,1/d]$ where $d$ corresponds to the determinant. The map $\mathbb{C}[G]^G\to \mathbb{C}[H]^H$ restricts the determinant and trace to $H$. This map sends $d-1$ to 0 (hence it is not injective).

In general, the main theorem in On invariants of a set of matrices by E. B. Vinberg states that the natural map $H^n/\!/H\to G^n/\!/G$, corresponding to the map on coordinate rings $k[G^n]^G\to k[H^n]^H$, is a finite morphism.

Therefore, the image of $H^n$ in $G^n/\!/G$ is closed. So if we let $k[H^n]^G$ be the coordinate ring of this image, denoted $H^n/\!/G$, then we can say that the kernel of $k[H^n]^G\to k[H^n]^H$ is trivial (and more interestingly this inclusion is an integral extension).

As with the example, the general kernel is basically just the polynomials that cut $H$ out of $G$ (if we can do so with conjugate invariant polynomials).

Remark: Adam Sikora and I use Vinberg's theorem to discuss coordinate rings of $G$-character varieties in Varieties of Characters, in particular addressing the situation when $G=SL_m$ and $H$ is a subgroup.

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    $\begingroup$ In your (counter)example, $H$ does not contain a maximal torus in $G$, as supposed in the statement. $\endgroup$
    – LSpice
    Jul 6, 2018 at 0:05
  • $\begingroup$ Thanks for pointing that out. I tried to address the general question posed, without the "for simplicity" assumption. In the case that H contains a maximal torus T of G then the GIT quotients for n=1 are T modulo the respective Weyl groups. So in that case the map is injective. This doesn't appear to make any difference for the general n>1 situation. $\endgroup$ Jul 6, 2018 at 20:22

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