I am looking for a proof of the following fact:
If $G$ is a finite subgroup of $SL_n(\mathbb{C})$ acting on $\mathbb{A}_{\mathbb{C}}^n$, then the resulting quotient scheme is Gorenstein.
Thanks.
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1$\begingroup$ Just FYI, the canonical sheaf being a line bundle is not enough for being Gorenstein. E.g., a cone over an abelian surface has a trivial canonical sheaf, but it is not Gorenstein, because it is not Cohen-Macaluay. $\endgroup$– Sándor KovácsCommented Jan 17, 2019 at 0:11
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2 Answers
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You probably want to show that the quotient scheme $\mathbb{A}^n_{\mathbb{C}}/G$ is Gorenstein.
A proof can be made along the following line. Since $G \subset SL_n(\mathbb{C})$, the canonical bundle of $\mathbb{A}^n_{\mathbb{C}}$ is $G$-equivariant and satisfy the following : for all $x \in \mathbb{A}^n_{\mathbb{C}}$ which is fixed by some subgroup $H \subset G$, the group $H$ acts trivially on $\omega_{\mathbb{A}^n_{\mathbb{C}}} \otimes \mathbb{C}(x)$. Hence, by a classical criterion for descent, there is a line bundle $L$ on $\mathbb{A}^n_{\mathbb{C}}/G$ such that $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \pi^*L$, where $\pi : X \rightarrow \mathbb{A}^n_{\mathbb{C}}/G$ is the quotient map.
Now we know that $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}}$ and that $\pi_*(\mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}})^G = \mathcal{O}_{\mathbb{A}^n_{\mathbb{C}}/G}$. Using Grothendieck-Serre duality for the projective morphism $\pi$, it's easy to see that $\pi_*(\omega_{\mathbb{A}^n_{\mathbb{C}}})^G$ is a (up to a shift) a dualizing object on $\mathbb{A}^n_{\mathbb{C}}/G$. Using the projection formula and the equality $\omega_{\mathbb{A}^n_{\mathbb{C}}} = \pi^*L$, one finds that $L$ is a dualizing object (up to a shift) on $\mathbb{A}^n_{\mathbb{C}}/G$. Hence $\mathbb{A}^n_{\mathbb{C}}/G$ is Gorenstein.
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$\begingroup$ Could you explain how to conclude in your proof that $L$ has finite injective dimension (which is one of the requirements of being a dualizing object)? $\endgroup$ Commented Jan 15, 2019 at 3:39
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$\begingroup$ If you worry that my proof could also be phrased in positive characteristic (while it is well-known that the result is false in positive characteristic) the issue is not there. The descent lemma, I am referring to uses the Reynolds operator, so it does't work in positive charcateristic. $\endgroup$– LibliCommented Jan 16, 2019 at 22:49
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$\begingroup$ As far as the finite injective dimension is concerned, we know that $L$ is a line bundle which is a dualizing object on $A_{\mathbb{C}^{n}/G$, so it has finite injective dimension. $\endgroup$– LibliCommented Jan 16, 2019 at 22:54
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$\begingroup$ You are using a circular argument. Having finite injective dimension is a requirement for being a dualizing object. So if you want to play it this way, then my question is, how do you know that $L$ is a dualizing object? More to the point, $L$ having finite injective dimension is equivalent to $\mathscr O$ having finite injective dimension which is equivalent to being Gorenstein. The fact that L satisfies the duality part of the definition (see 3) in stacks.math.columbia.edu/tag/0A7B) is trivial. The real question is 1), which you seem to suggest follows from 3), but that's not true. $\endgroup$ Commented Jan 17, 2019 at 0:09
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$\begingroup$ @SándorKovács I don't think my argument is circular, I guess you just don't understand it. Let me give you some again some extra details. Let $F \in D^b(V/G)$, we have $\mathrm{R} \mathcal{H}om_{V/G}(F, \mathcal{R} \pi_{*}^{G} \mathcal{O}_{V}[n]) = \mathcal{R} \pi_*^{G} \mathrm{R} \mathcal{H}om_{V}^{G}(\mathrm{L} \pi^*F, \mathcal{O}_{V}[n])$, by the $G$-equivariant projection formula. Since $\omega_{V}$ is $G$-equivariantly locally trivial, the object $\mathcal{O}_{V}[n]$ is a $G$-equivariant dualizing object on $V$ (continued...) $\endgroup$– LibliCommented Jan 17, 2019 at 15:58
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- By the Hochster-Roberts Theorem it is Cohen-Macaulay
- The canonical sheaf of $\mathbb A^n$ is $G$-invariant (because the elements of $G$ have det=1) and hence it descends to the canonical sheaf of the quotient, which is then a line bundle.
answered Oct 5, 2016 at 18:47