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There was given the following definition of a Procesi bundle in this paper:

let $V$ denote a symplectic $\mathbb C$-vector space with a finite group $\Gamma$ of symplectic linear automorphisms and let $f: X \to V/\Gamma$ be a conical symplectic resolution.

Then a vector bundle $\mathcal F$ over $X$ is called a Procesi bundle if

(i) $\operatorname{End}_{\mathcal O_X} \mathcal F$ is $\mathbb C^*$-equivariantly isomorphic to $\mathbb C[V]\#\mathbb C[\Gamma]$;

(ii) $Ext^j(\mathcal F, \mathcal F) = 0$ for j >0.

Is it true that any fiber of Procesi bundle should be isomorphic as a $\Gamma$-module to the regular representation of $\Gamma$?

If so, how to prove this fact?

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I think it is easy to prove by semi-continuity. Namely, generically on $X$ the statement is obvious (it is obvious on the open subset of $X$ where the map to $V/\Gamma$ is an isomorphism). Now, the multiplicity of some irreducible representation $\pi$ of $\Gamma$ in the fiber $\mathcal F_x$ of $\mathcal F$ at a point $x\in X$ is a semi-continuous function of $x$, thus for all $x$ it cannot be smaller than what happens generically (i.e. it can't be smaller than $\dim \pi$). If for some $x$ it jumps, then automatically the dimension of the space $\mathcal F_x$ jumps, which contradicts the assumption that $\mathcal F$ is a vector bundle.

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  • $\begingroup$ Well, this statement isn't obvious for me generically on $X$. $\endgroup$
    – Rgdn Dznrbx
    Commented Sep 28, 2021 at 19:23
  • $\begingroup$ Of course, if one establishes the fact that $$\operatorname{End}f_*\mathcal F = \operatorname{End}\mathcal F,$$ the rest of the proof is obvious. Namely, we can just notice that on some open set $U$ the isomorphism $$\operatorname{End}(\mathcal F(x)) = \operatorname{End}(\mathcal f_*F(y)) = (\mathcal{End}(\mathcal f_*\mathcal F))(y)$$ (for $x \in U$, $y = f(x)$) holds. And the equality $\operatorname{End}((\mathcal f_*F)(y)) = \operatorname{End}(\mathbb C[\Gamma])$ of $\Gamma$-modules (which holds since $\Gamma(V/\Gamma, f_*\mathcal F)$ is $\mathbb C[V]\#\Gamma$) finishes the proof. $\endgroup$
    – Rgdn Dznrbx
    Commented Sep 28, 2021 at 19:23
  • $\begingroup$ But how to prove that $\operatorname{End}f_*\mathcal F = \operatorname{End}\mathcal F$? Equivalently, one can try to prove that $\operatorname{rk}\mathcal F = |\Gamma|$ (since we have an obvious morphism $\operatorname{End}\mathcal F \to \operatorname{End}f_*\mathcal F$ and matrix algebra has only one irreducible representation). But I can't solve this problem. $\endgroup$
    – Rgdn Dznrbx
    Commented Sep 28, 2021 at 19:23
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    $\begingroup$ The point is that the isomorphism in (i) is assumed to commute with $\mathbb C[V/\Gamma]$, so generically on $V/\Gamma$ the two endomorphism algebras are the same -- that is all you need. $\endgroup$
    – Alexander Braverman
    Commented Sep 30, 2021 at 22:29

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