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A theorem of Knop states that if $G$ is semisimple and connected acting on a vector space $V$ over a field $K$ of characteristic 0, then the degree of the Hilbert series of $K[V]^G$ is less than or equal to $-\operatorname{dim}( K[V]^G)$.

This in turn implies the following: If $f_1,\dots,f_r$ is a homogeneous system of parameters for $K[V]^G$, then $K[V]^G$ is generated as a ring by elements of degree at most $\operatorname{max}(d_1+\cdots+d_r-r,d_1,d_2,\dots,d_r)$ where $d_i=\operatorname{deg}(f_i)$ and $r$ is the Krull dimension.

Is this still true if $V$ is replaced by a $G$-variety $X$? Or at least a sufficiently nice $G$-variety? It seems likely as we are simply moving to a quotient ring of $K[V]^G$ but I would like to know for sure.

Edit: To add more information about the situation I am working with: I am trying to find a degree bound for a specific action $G$ acting on $V$. I'm looking at a quotient ring $K[V]^G/I$ which I know can be realized as $K[X]^G$ for a certain subvariety $X$ of $V$. I am viewing both of these rings with the standard grading.

I know quite a bit about both $K[X]^G$ and $I$ and can thus say some things about $K[V]^G$. Actually, since posting this question, I have found a set of generators for $K[X]^G$. However, in general, I am still interested in knowing if a degree bound for $K[X]^G$ can be found if the a set of equations defining the null cone are known/a homogeneous system of parameters of $K[X]^G$ is known.

But I must clarify that I am assuming that $K[X]^G$ is a specific quotient ring of $K[V]^G$ with the standard gradings.

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This is true, but in a limited sense.

Every $G$-variety $X$ can be equivariantly embedded into a $G$-module $V$ (i.e., $G$ acts linearly and morphically on $V$). This can be derived from the local finiteness of the coordinate ring $K[X]$, see somewhere at the beginning of Mumford's "Geometric Invariant Theory". It can also be turned into an explicit procedure. Since $G$ is semisimple and char$(K) = 0$, $G$ is linearly reductive. This implies that the epimorphism $K[V] \to K[X]$ remains surjective when restricting to invariants. So a generating set of $K[V]^G$ maps to a generating set of $K[X]^G$.

This translates to a degree bound as follows: Embedding $X$ into a finite-dimensional vector space is the same as choosing a finite generating set for the algebra $K[X]$. Now if $K[V]^G$ is generated in degree at most $d$, this means that there is a generating set in $K[X]^G$ such that every invariant from this set can be expressed as a polynomial in the chosen generators of $K[X]$ whose total degree does not exceed $d$.

Unfortunately this bound is somewhat unnatural, as it depends on the choice of an embedding $X \subseteq V$. But I don't think anything stronger can be inferred.

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  • $\begingroup$ I actually have the reverse problem. I'm trying to find a bound on $K[V]^G $ and $X $ is a specific subvariety of $V $ chosen because it's much easier to work with and I have a way to turn a degree bound on $K[X]^G $ to a degree bound on $K[V]^G $. I just don't quite have a degree bound for $K [X]^G $ yet. I know a set of homegeneous invariants that cut out the null cone. If $X $ was a vector space, then there are of course many techniques to get a degree bounds from that info. However, $X $ is not a vector space. I was hoping some of the techniques might still work. $\endgroup$ – batconjurer Nov 22 '15 at 14:38
  • $\begingroup$ Now with this comment you are changing the ground rules completely. It would be helpful if you posed the question in more exact terms. You seem to make the tacit assumption that $K[X]$ is graded. Is it standard graded? Are you seeking a generalization of Knop's result and the degree bound implied by it to other graded situations? $\endgroup$ – Gregor Kemper Nov 22 '15 at 15:43
  • $\begingroup$ Sorry, you are correct. I have edited this question to hopefully clarify what I am looking for. But I am assuming $K[X]$ has the standard grading and is a quotient ring of some previously chosen $K[V]$ with $G$ acting on $V$. $\endgroup$ – batconjurer Nov 22 '15 at 17:48

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