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Given a smooth affine variety $X$ over $\mathbb C$ with an algebraic symplectic form $\omega$ and a finite group $G$ acting on $X$ by symplectomorphisms, then

  1. Is it true that $X$ is trivially a symplectic singularity in the sense of Definition 2.1 in Kaledin's paper Symplectic singularities from the Poisson point of view?

  2. Is it true that the canonical projection $X\to X/G$ is a symplectic resolution of the symplectic singularity $X/G$?

  3. If $X$ is trivially a symplectic singularity, are the connected components of the isotropy types $X_H$, $H\leq G$, a stratification satisfying the hypothesis in Theorem 2.3 in Kaledin's paper? In particular, does the product decomposition (2.1) hold true?

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    $\begingroup$ I wouldn't say $X$ is a symplectic singularity but it certainly has symplectic singularities by that definition. $\endgroup$
    – Will Sawin
    Commented Feb 24, 2023 at 17:09
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    $\begingroup$ I made a mistake. I meant to write "symplectic variety", not "symplectic singularity". So, you confirm that $X$ is a symplectic variety, right? It is obvious to me, too. But since there is no symplectic resolution of $X$ other than $X$ itself (if we are allowed to do that), I was wondering if I don't want too much. $\endgroup$
    – Flavius Aetius
    Commented Feb 24, 2023 at 17:18
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    $\begingroup$ You are certainly allowed to take a variety to be a resolution of itself. I don't think $X \to X/G$ is a symplectic resolution because it is not a resolution since it is (usually) not birational. $\endgroup$
    – Will Sawin
    Commented Feb 24, 2023 at 18:16
  • $\begingroup$ This is a very good point! Thank you. What about the orbit type strata $X_H^i$ which are simply the connected components of the isotropy types? We know that they define a canonical stratification of $X$. But are they the canonical stratification in Kaledin's theorem $2.3$? $\endgroup$
    – Flavius Aetius
    Commented Feb 24, 2023 at 23:54

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