Timeline for Quotient of affine space by finite subgroup of SL(V) is Gorenstein
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2019 at 17:20 | comment | added | Libli | This shows that $\mathcal{O}_{V/G}[n]$ is a dualizing object on $V/G$. In particular, if you take $F =\mathcal{O}_{0}$ (structure sheaf of the point $0 \in V/G$), we get $\mathrm{Ext}^{k}(\mathcal{O}_{0}, \mathcal{O}_{V/G}) = 0$, except for $k =n$, where $\mathrm{Ext}^{k}(\mathcal{O}_{0}, \mathcal{O}_{V/G}) = \mathbb{C}^*$. Proving, as a byproduct, that $\mathcal{O}_{V/G}$ has finite injective dimension. | |
| Jan 17, 2019 at 17:15 | comment | added | Libli | Since we have $\mathrm{R} \mathcal{H}om_{V/G}^{G}(\mathcal{O}_{V}, \mathrm{L} \pi^*L) = \mathrm{L} \pi^* L$, we deduce that $\mathcal{R} \sigma_* \mathcal{H}om_{V/G}(F, \mathcal{R} \pi_*^{G} O_{V}[n]) = (\mathcal{R} \sigma_* (F \otimes \mathcal{R} \pi_*^{G} O_{V}[n]))^*$. As $\mathcal{R} \pi_*^{G} \mathcal{O}_{V} = \mathcal{O}_{V/G}$, we finally get $\mathcal{R} \sigma_* \mathcal{H}om_{V/G}(F, \mathcal{R} \pi_*^{G} O_{V}[n]) = (\mathcal{R} \sigma_* F)^*$, for all $F \in D^{b}(V/G)$ such that $\pi^* F$ is bounded above in $D^{qc}(V)$. | |
| Jan 17, 2019 at 17:03 | comment | added | Libli | Assume that $\mathrm{L} \pi^* F$ is bounded above in $D^{qc}(V)$. Then by the $G$-equivariant version of Serre duality on $V$, we have $\mathcal{R} \sigma_* \mathcal{R} \pi_*^{G} \mathrm{R} \mathcal{H}om_{V}^{G}(\mathrm{L} \pi^* F, \mathcal{O}_{V}[n]) = (\mathcal{R} \sigma_* \mathcal{R} \pi_*^{G} \mathrm{R}\mathcal{H}om_{V}^{G}(\mathcal{O}_V, \mathrm{L} \pi^* F))^*$, where $\sigma : V/G \rightarrow \mathbb{C}$. | |
| Jan 17, 2019 at 15:58 | comment | added | Libli | @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...) | |
| Jan 17, 2019 at 0:09 | comment | added | Sándor Kovács | 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. | |
| Jan 16, 2019 at 22:54 | comment | added | Libli | 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. | |
| Jan 16, 2019 at 22:49 | comment | added | Libli | 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. | |
| Jan 15, 2019 at 3:39 | comment | added | Sándor Kovács | 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)? | |
| Nov 17, 2016 at 20:58 | vote | accept | Xuqiang QIN | ||
| Oct 5, 2016 at 19:24 | history | edited | Libli | CC BY-SA 3.0 |
added 5 characters in body
|
| Oct 5, 2016 at 18:46 | history | answered | Libli | CC BY-SA 3.0 |