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Given a locally free sheaf $M$ on $\mathbb{P}^2$ with $h^0(M)=1$. Is it true that we have $h^2(M)=0$ in this case?

I got this idea from Friedman's book "Algebraic Surfaces and holomorphic vector bundles". In Chapter 4, p.109, Ex. 4 he wrote: $h^0(Hom(V,V))=1$, by Serre duality $h^2(Hom(V,V))=h^0(Hom(V,V)\otimes K)=0$ since $K=O(-3)\subset O$.

But I don't see why $h^0(Hom(V,V)\otimes K)=0$ follows from $K=O(-3)\subset O$. I mean it could stil have dimension one or is it because $O(-3)$ has no global sections? Would $h^0(Hom(V,V)\otimes O(-i))=0$ still be true for $i=1,2$? Can one generalize to arbitrary locally free sheaves or is this only correct in this special case, i.e. $V$ stable?

${\bf Edit:}$ Given a simple sheaf $V$ on $\mathbb{P}^2$, Bjorn's answer shows $H^0(Hom(V,V)\otimes O(-i))=0$ for $i>0$ which can be written as $Hom(V,V(-i))=0$ for $i>0$. Can this be generalized?

For example given a sheaf of rings or algebras $R$ and a simple left $R$-module M, do we always have $Hom_R(M,M(-i))=0$ for $i>0$? Or do I need to be more careful in this case?

I remembered this question reading, page 8, where such an equality shows up, without further explanation, so i thought the argumentation should carry over to this more general case.

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up vote 15 down vote accepted

1) A locally free sheaf $M$ on $\mathbb{P}^2$ with $h^0(M)=1$ need not satisfy $h^2(M)=0$. For example, if $M=\mathcal{O} \oplus \mathcal{O}(-n)$ for $n>3$, then $h^0(M) = 1 + 0 = 0$, but $M^\vee = \mathcal{O} \oplus \mathcal{O}(n)$ and $K=\mathcal{O}(-3)$, so Serre duality shows that $h^2(M) = h^0(M^\vee \otimes K) = h^0(\mathcal{O}(-3)) + h^0(\mathcal{O}(n-3)) > 0$.

2) On the other hand, the statement is true when $M = \mathbf{Hom}(V,V)$ and $h^0(M)=1$. Here is why. First, $M$ is self-dual, so in this case Serre duality says that $h^2(M) = h^0(M \otimes K)$. If $V$ were a nonzero vector space over a field $k$, there would be a natural map $k \hookrightarrow \operatorname{Hom}(V,V)$ whose image consists of the multiplications by scalars; similarly, for a nonzero vector bundle $V$, there is a natural map $\mathcal{O} \hookrightarrow M = \mathbf{Hom}(V,V)$, and the global section $1$ of $\mathcal{O}$ maps to an everywhere nonvanishing global section of $M$. In your setting, $h^0(M)=1$, so this global section spans $H^0(M)$. Because this global section is everywhere nonvanishing, it cannot be in the image of $H^0(M \otimes \mathcal{O}(-i)) \hookrightarrow H^0(M)$ for any $i>0$. In particular, $H^0(M \otimes K)=0$, so $h^2(M)=0$ by the Serre duality argument.

3) The conclusion $h^2(M)=0$ can fail if you replace $V$ by a nonzero vector bundle for which $h^0(\mathbf{Hom}(V,V)) \ne 1$. For instance, if $V=\mathcal{O} \oplus \mathcal{O}(n)$ for $n \ge 3$, then $\mathbf{Hom}(V,V) = \mathcal{O} \oplus \mathcal{O}(n) \oplus \mathcal{O}(-n) \oplus \mathcal{O}$, which after twisting by $\mathcal{O}(-3)$ still has nonzero global sections.

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Thank you for this clarification and the examples. – TonyS Feb 16 '10 at 17:14

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