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S Mar 17, 2023 at 14:07 history bounty ended rfauffar
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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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