Timeline for Differential of a specific morphism to a Grassmannian
Current License: CC BY-SA 4.0
14 events
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S Mar 17, 2023 at 14:07 | history | bounty ended | rfauffar | ||
S Mar 17, 2023 at 14:07 | history | notice removed | rfauffar | ||
Mar 17, 2023 at 14:07 | vote | accept | rfauffar | ||
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Mar 17, 2023 at 7:38 | answer | added | Rina | timeline score: 1 | |
Mar 13, 2023 at 3:10 | history | edited | rfauffar | CC BY-SA 4.0 |
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S Mar 10, 2023 at 16:11 | history | bounty started | rfauffar | ||
S Mar 10, 2023 at 16:11 | history | notice added | rfauffar | Draw attention | |
Mar 9, 2023 at 17:41 | history | edited | rfauffar | CC BY-SA 4.0 |
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Mar 8, 2023 at 15:06 | comment | added | rfauffar | @Libli Thanks for your nice example. The specific situation I have is the following: let $G\leq\mathrm{Aut}(X)$ be a finite group such that $X/G\simeq\mathbb{P}^d$, and let $Q$ be the linear system $\pi^*|\mathcal{O}(1)|$ where $\pi:X\to\mathbb{P}^d$ is a quotient map. | |
Mar 7, 2023 at 22:34 | comment | added | Libli | for any $Q \subset H^0(X,\mathcal{M}) = \mathbb{C}^{r+1}$ the map $\Phi$ can't be a local embedding. | |
Mar 7, 2023 at 22:26 | comment | added | Libli | I think it would be helpful you'd me slightly more specific about the example you hae in mind. In general, there is no chance that this map is a local ismorphism. Take $X = \mathbb{P}^r$, $\mathcal{M} = \mathcal{L} = \mathcal{O}_{\mathbb{P}^r}(1)$. We have (up to a finite quotient whose kernel acts trivially on $X$) $Aut^0(X) = \mathrm{SL}_{r+1}$ which is of dimension $(r+1)^2-1 = r^2+2r$. Since $\mathcal{M}$ is $\mathrm{SL}_{r+1}$-equivariant, one checks that $\{0\} \times \mathrm{SL}_{r+1} \subset \mathrm{Div}_{\geq 0}(X) \times_{\mathrm{Pic}(X)} Aut^0(X)$. An easy dimension count shows that | |
Mar 7, 2023 at 18:41 | history | edited | rfauffar | CC BY-SA 4.0 |
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Mar 7, 2023 at 17:10 | history | edited | rfauffar | CC BY-SA 4.0 |
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Mar 7, 2023 at 16:26 | history | asked | rfauffar | CC BY-SA 4.0 |