Assume we have a locally free sheaf $R$ of associative $O_S$-algebras of rank $r^2$, possibly noncommutative, where $S=\mathbb{P}^2$ over $\mathbb{C}$. Furthermore given a locally free $O_S$-module $E$ also of rank $r^2$, which is a simple $R$-module, i.e. $End_R(E)=\mathbb{C}$.

We also know that all bundles come from a cyclic degree $r$ covering $f:X\rightarrow S$ ramified on a smooth curve of degree $nr$, i.e. $E$ is also a locally free $O_X$-module of rank $r$, and $R$ also an $O_X$-algbera of rank $r$ and $E$ is an $R$-module.

Now if $L$ is a cohomologically trivial $O_X$-line bundle, i.e. $H^i(X,L)=0$ for all $i$, what can we say about $Hom_R(E,L\otimes_{O_X} E)$? Does it vanish? Is it isomorphic to $\mathbb{C}$? Can it have bigger dimension?

If this is too broad, here is the situation im especially interested in: $X$ is a $K3$-surface with two disjoint (-2)-curves $C_1,C_2$, constructed as a double cover of $\mathbb{P}^2$ ramified on a sextic $C$. The line bundle $L$ is given by $O_X(C_1-C_2)$ and $R$ and $E$ are rank 2 locally free sheaves on $X$.

I was hoping that there are no such morphisms: my idea was to construct a map $\alpha: L\otimes_{O_X}E\rightarrow E$, such that if there were a map $\beta: E \rightarrow L\otimes_{O_X} E$, one could look at $\alpha\circ \beta=c id_E$ by simplicity. Then one had to show that a multiple of the identity can not factor in this way. But i couldn't make this argument work.

Another idea was to use the determinant bundle: if i had a map $E\rightarrow L\otimes_{O_X} E$, this would give me a map $det(E)\rightarrow L^2\otimes_{O_X} det(E)$, so a map $O_X\rightarrow L^2$, but i couldn't see if there can be such maps. If $L$ is cohomologically trivial, especially $H^0(L)=0$ does this imply $H^0(L^2)=0$?

Maybe there can be nontrivial elements in $Hom_R(E,L\otimes_{O_X} E)$? Or are there other ways to attack this problem? Any idea is welcome.

Edit: We may restrict to those bundles $E$ with first Chern class $c_1(E)=C_1-C_2+f^*(nH)$ on $X$, with $H$ a line in $\mathbb{P}^2$, where $n=0$ and $n=1$ would be enough, so $c_1(E)=C_1-C_2$ and $c_1(E)=C_1-C_2+f^*H=C_1-C_2+C_2+f(C_2)=C_1+f(C_2)$. We may also assume $E$ is a semistable $O_X$-module.


1 Answer 1


If you are specifically interested in the case where $L=\mathcal{O}_{X}(C_{1}-C_{2}),$ all that is needed for your determinant bundle argument to work is the vanishing $H^{0}(\mathcal{O}_{X}(2C_{1}-2C_{2}))=0.$ But this follows from taking cohomology in the exact sequences $$0 \rightarrow \mathcal{O}_{X}(-2C_{2}) \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(C_{1}-2C_{2}) \rightarrow 0$$ $$0 \rightarrow \mathcal{O}_{X}(C_{1}-2C_{2}) \rightarrow \mathcal{O}_{X}(2C_{1}-2C_{2}) \rightarrow \mathcal{O}_{C_{1}}(2C_{1}-2C_{2}) \rightarrow 0 $$

EDIT: Up to now we have only established that the elements of ${\rm Hom}(E,E \otimes L)$ have rank at most 1 on each fiber. If the restriction of $E$ to $C_{1}$ is $\mathcal{O}_{C_1}(a) \oplus \mathcal{O}_{C_1}(b)$ for $a,b \in \mathbb{Z}$ satisfying $|a-b| \leq 1,$ then the simplicity of $E$ and the sequence $$0 \rightarrow E \otimes E^{\ast}(-C_2) \rightarrow E \otimes E^{\ast} \otimes L \rightarrow E \otimes E^{\ast} \otimes \mathcal{O}_{C_1}(-2) \rightarrow 0 $$ give you what you want; I don't see right away how to proceed without extra assumptions on $E.$

  • 1
    $\begingroup$ Ah, thanks, so one needs two sequences to show the vanishing of $H^0(L^2)$. I added some extra assumptions one can impose on $E$ at the end of the question, maybe that will help? $\endgroup$
    – TonyS
    Feb 6, 2012 at 19:51

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