Let $C$ be a projective curve (over an algebraically closed field, not necessarily of characteristic zero) which is smooth except for exact one node. Let $\pi:\tilde{C} \to C$ be its normalization. Let $\mathcal{V}$ be a locally free sheaf over $\tilde{C}$. Is it true that the push forward, $\pi_*(\mathcal{V})$ is a locally free sheaf? If not, are there any general conditions (for example on the rank of $\mathcal{V}$) so that this holds true?
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- 3$\begingroup$ No and yes for the rank zero vector bundle. This question does not belong on mathoverflow IMHO. $\endgroup$answer_bot– answer_bot2014-03-07 12:08:32 +00:00Commented Mar 7, 2014 at 12:08
- 1$\begingroup$ @answer_bot: I do not understand why this question does not belong to mathoverflows when there is a similar question with 6 votes: mathoverflow.net/questions/67387/… $\endgroup$user45397– user453972014-03-07 12:45:13 +00:00Commented Mar 7, 2014 at 12:45
- $\begingroup$ @answer_bot: Anyways, could you please elaborate on your answer. $\endgroup$user45397– user453972014-03-07 12:46:20 +00:00Commented Mar 7, 2014 at 12:46
- $\begingroup$ @user45397 The "similar" question is not restricted to curves and their normalizations. $\endgroup$S. Carnahan– S. Carnahan ♦2014-03-09 16:02:28 +00:00Commented Mar 9, 2014 at 16:02
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1 Answer
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1 Quoting answer_bot: "No and yes for the rank zero vector bundle." In other words, the answer is "no" except for one case: the rank zero vector bundle. That means the answer is already "no" for every rank 1 vector bundle. To user46578, I recommend that you contemplate what happens for the trivial rank 1 vector bundle on $\widetilde{C}$. If you still have trouble, perhaps you should ask on Math StackExchange.
- $\begingroup$ @Starr: Thank you very much for the answer. $\endgroup$user46578– user465782014-03-07 21:40:04 +00:00Commented Mar 7, 2014 at 21:40