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Let $V$ be a complex vector spaces and assume that a compact group G acts linearly on $V$. Then look at the $G$-equivariant polynomial maps from $V$ to $V$. Denote this by $Mor_G(V,V)$. In the case of finite groups, $Mor_G(V,V)$ is a finitely generated module over the ring of polynomial invariants $\mathbb{C}[V]$. Does such a statement also hold for infinite (reductive) groups? If so what would be a good reference for this?

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    $\begingroup$ I'm not sure GGT is the right tag. Would you be a little more precise? $G$ acts linearly, I guess. Finite module: do you mean finitely generated module, or finite length module? Equivariant maps: do you mean polynomial maps? I guess you make it a module using addition on the target space? By the invariant ring, you mean the ring of invariant polynomials on $W$? $\endgroup$
    – YCor
    Commented Nov 17, 2015 at 0:31
  • $\begingroup$ Thanx for the remarks. I tried to be more precise and hope this helps $\endgroup$
    – Seppo
    Commented Nov 17, 2015 at 16:03

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The answer should be yes, by the following argument:

With $n := \dim(V)$, consider the polynomial ring $C[x_1..x_n,y_1..y_n] = C[V \oplus V^*]$, so G acts on the $x_i$ as on the variables of $C[V]$, but on the $y_i$ as on the variables of $C[V^*]$ with $V^*$ the dual space.

$C[x_1..x_n,y_1..y_n]^G$ is a bigraded algebra by using the x- and y-degrees. So Mor$_G(V,V)$ is the part of $C[x_1..x_n,y_1..y_n]^G$ where the y-degree is 1.

Since $G$ is reductive, $C[x_1..x_n,y_1..y_n]^G$ has a finite generating set $f_1..f_m$, which we may assume to be bigraded. Renumber the $f_i$ in such a way that $f_1..f_r$ have y-degree 0, $f_{r+1}...f_{r+k}$ have y-degree 1, and the others have higher y-degree.

Now an element from Mor$_G(V,V)$, being an invariant, can be written as a linear combination of power products of the $f_i$. But we can delete all summands whose y-degree is not 1, and so we are left with a linear combination of $f_{r+1}...f_{r+k}$ with coefficients in the subalgebra generated by $f_1..f_r$. These coefficients are all invariants.

This argument shows that $f_{r+1}...f_{r+k}$ generate Mor$_G(V,V)$ as a module over C[V]^G.

The argument generalises to Mor$_G(V,W)$,

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  • $\begingroup$ Thanx that is extremly helpful! $\endgroup$
    – Seppo
    Commented Nov 17, 2015 at 17:05

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