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Libli
Libli
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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_*^G \omega_{\mathbb{A}^n_{\mathbb{C}}}$$\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.

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 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 $\pi_*^G \omega_{\mathbb{A}^n_{\mathbb{C}}}$ 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.

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.

Source Link
Libli
Libli
  • 7.3k
  • 25
  • 48

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 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 $\pi_*^G \omega_{\mathbb{A}^n_{\mathbb{C}}}$ 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.