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Hi all

I thought that the following was true but was unable to think of a proof and after browsing the internet now I am not so sure. My understanding of graded rings and modules is not great so I'm thinking this may be obvious to some people but not me.

In the category of coherent sheaves on a projective variety $X$, is a locally free sheaf a projective (in the categorical sense)? I'm mainly interested when $X$ is a nonsingular curve but would be curious as to the general answer.

Thanks.

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This is never true whenever $X$ has positive dimension. Let $L$ be ample on $X$ and let $E$ be a nonzero coherent sheaf on $X$. Let $P$ be any point of $X$ at which $E$ has a nonzero fiber, so we get a surjection $\mathcal{O}_X \to \mathcal{O}_P$ ($\mathcal{O}_P$ being the skyscraper sheaf at $P$). We can also find a $k>0$ such that the sheaf $\mathrm{Hom}(E, E\otimes L^{-k})= \mathrm{End}(E)\otimes L^{-k}$ has no nonzero global sections. Now tensor the surjection $\mathcal{O}_X\to \mathcal{O}_P$ by $E$ and $E\otimes L^{-k}$, getting surjections $a:E\to E_P$ and $b:E\otimes L^{-k}\to (E\otimes L^{-k})_P = E_P$. We cannot lift $a$ along $b$ because by assumption on $k$ there are no nonzero maps $E\to E\otimes L^{-k}$.

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What is true, is that a locally free sheaf is locally projective (of course) --- you can lift morphisms locally. –  Sasha Oct 12 '12 at 6:02
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It is also true that some things that would normally be done with projective resolutions, can in this category be done with locally free resolutions. –  Charles Staats Oct 12 '12 at 13:07
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If you are trying to lift a specific homomorphism, or to analyse extensions of vector bundles on curves, notice the following (almost obvious) observation:

Given a short exact sequence of vector bundles $$ 0\to E\to W\to F\to 0, $$ a vector bundle homomorphism $f:V\to F$ lifts to a homomorphism $V\to W$ if and only if the extension class $\delta(W)$ satisfies $$\delta(W)\in \ker\left(H^1(X,\underline{Hom}(F,E))\to H^1(X,\underline{Hom}(V,E)) \right). $$ This is Lemma 3.1 in Narasimhan - Ramanan.

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