Timeline for Can one construct the GIT quotient of a projective bundle?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2016 at 16:57 | comment | added | t3suji | P.S. Come to think of it, the morphism need not be injective. | |
| Sep 11, 2016 at 14:03 | comment | added | t3suji | Yes, for any $G$-equivariant sheaf $F$ on $X$, you can consider $F':=(\pi_*(F))^G$ (invariants in the direct image), it is a sheaf on $X//G$. If $F$ descends, then $F'$ is its descent: $F\simeq\pi^*F'$. In general, there is a morphism $\pi^*F'\to F$; it is injective, but not surjective. | |
| Sep 11, 2016 at 9:43 | history | edited | Bernie | CC BY-SA 3.0 |
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| Sep 11, 2016 at 9:39 | comment | added | Bernie | @t3suji : Ah, thanks. I think misunderstood that the whole time. So it makes sense to talk about the the invariant direct image of $End(E)$ on $X//G$ although this sheaf itself does not descend to $X//G$? The last point just means that there is no sheaf $F$ on the quotient such that $\pi^{*}F\cong End(E)$ as $G$-linearized sheaves? | |
| Sep 11, 2016 at 5:38 | comment | added | t3suji | End(E)^G makes no sense as a sheaf on X: G-invariance is not defined for local sections when G acts non-trivially on X. | |
| Sep 10, 2016 at 22:35 | comment | added | Bernie | @t3suji : No. I am saying $End(E)^G$, the sheaf of invariants, descends. G acts nontrivially via conjugation on End(E) with the finite stabilizers, so that this bundle will not descend. | |
| Sep 10, 2016 at 16:47 | comment | added | t3suji | Why do you say that End(E) descends? Are you assuming that the stabilizers of points act trivially (or at least by scalars) on E's fibers? | |
| Sep 10, 2016 at 16:04 | history | asked | Bernie | CC BY-SA 3.0 |