Skip to main content
6 events
when toggle format what by license comment
Apr 17, 2017 at 14:49 vote accept user43198
Apr 17, 2017 at 6:10 comment added R. van Dobben de Bruyn Also, usually $(-)_x$ denotes the stalk, not the fibre. For this type of question it doesn't matter, because by Nakayama's lemma surjecting onto the stalk is equivalent to surjecting onto the fibre.
Apr 17, 2017 at 5:55 answer added R. van Dobben de Bruyn timeline score: 6
Apr 17, 2017 at 1:56 comment added R. van Dobben de Bruyn Another comment: global generation of $\mathcal E$ is completely irrelevant. Indeed, $\mathcal E \otimes \mathcal L$ is globally generated for a sufficiently ample line bundle $\mathcal L$, and $\mathcal End(\mathcal E \otimes \mathcal L) = \mathcal End(\mathcal E)$.
Apr 17, 2017 at 1:53 comment added R. van Dobben de Bruyn First counterexample: consider $\mathcal E = \mathcal O \oplus \mathcal O(1)$ on $\mathbb P^1$. Then $\mathcal End(\mathcal E) = \mathcal O(-1) \oplus \mathcal O \oplus \mathcal O \oplus \mathcal O(1)$, so this is not globally generated (even though $\mathcal E$ is). Indeed, global sections do not surject onto the stalk for any point, because the $\mathcal O(-1)$ factor does not have a nonzero global section. Of course, similar examples exist on curves of higher genera.
Apr 17, 2017 at 0:52 history asked user43198 CC BY-SA 3.0