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? 



Equivariant polynomial maps
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$\_G(V,W)$,


