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.



Do cohomologically trivial line bundles affect morphisms?
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 $E.$


