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Let $G$ be a linearly reductive algebraic group and $X$ be an affine $G$-variety over an algebraically closed field $\mathbb{K}$. Let $Y\subset X$ be a (closed) affine subvariety of $X$ which is also $G$-stable.

Is there any "nice" description of the invariant ring $\mathbb{K}[Y]^G$ in terms of the invariant ring $\mathbb{K}[X]^G$? Maybe something like- it's a quotient or a subring of $\mathbb{K}[X]^G$.

For my situation, $X$ is an affine space.

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  • $\begingroup$ Under your hypotheses, isn't $\mathbb K[Y]^G$ the quotient of $\mathbb K[X]^G$ by the $G$-fixed vectors in the ideal $I_{Y \subseteq X}$ of functions in $\mathbb K[X]$ vanishing on $Y$? At least this seems so as vector spaces, since linear reductivity implies that there is a $G$-module splitting of $0 \to I_{Y \subseteq X} \to \mathbb K[X] \to \mathbb K[Y] \to 0$. $\endgroup$
    – LSpice
    Jan 27, 2023 at 0:27
  • $\begingroup$ Re, assuming you meant $Y$ to be $G$-stable. Otherwise, I don't know what $\mathbb K[Y]^G$ means. $\endgroup$
    – LSpice
    Jan 27, 2023 at 0:33
  • $\begingroup$ @LSpice Thanks for your reply. Yes, I meant $Y$ is $G$-stable. For my situation, $X$ is an affine space. Also, can you give some kind of reference for the exact sequence that you've mentioned. $\endgroup$
    – It'sMe
    Jan 27, 2023 at 0:48
  • $\begingroup$ Re, for me, the exact sequence I mentioned is the definition of a closed subvariety of an affine variety. Could you say what is your definition? $\endgroup$
    – LSpice
    Jan 27, 2023 at 0:55
  • $\begingroup$ @LSpice I meant the fact about the "$G$-module splitting". Sorry, I should have been more precise. $\endgroup$
    – It'sMe
    Jan 27, 2023 at 1:10

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As we discussed in the comments, by linear reductivity of $G$, there is a $G$-module splitting of the surjection $\mathbb K[X] \to \mathbb K[Y]$; so the restriction map $\mathbb K[X]^G \to \mathbb K[Y]^G$ is a surjection. Thus $\mathbb K[Y]^G$ is a quotient of $\mathbb K[X]^G$, as both a $G$-module and an algebra; and is a $G$-submodule of $\mathbb K[X]^G$, but (even if $G$ is trivial) need not be a subalgebra.

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