A question on cohomology of vector bundles on a surface

Let $X$ be a smooth projective surface. Let $K$ be the canonical bundle. Let $F$ be a vector bundle (of rank $2$) such that the kernel of the map $tr_{2} : Ext^2(F,F) \to H^2(\mathcal O_X)$ which is sometimes denoted by $Ext^2_0(F,F)=0$. Let $End(F) : = F \otimes F^*$.

Then can we say that there exists an isomorphism between $H^0( End(F) \otimes K) $ and $H^0(K)$?

From injectivity of $tr_2$ we have $(H^0((End(F))^* \otimes K))^* \hookrightarrow (H^0(K))^*$. I do not see how to deduce the desired isomorphism from here.

Do we need to assume some more conditions? e.g $F$ is slope stable with respect to some ample divisor of $X$?

Solution 1:

Assume for simplicity that characteristic of the base field is not 2. Then $$ \mathcal{E}nd(F) \cong \mathcal{E}nd_0(F) \oplus \mathcal{O}_X,\tag{*} $$ the splitting is given by the identity and trace morphisms. It follows that the kernel of the trace morphism on $H^2$ is equal to $$ H^2(X,\mathcal{E}nd_0(F)), $$ which therefore vanishes. By Serre duality it follows that $$ H^0(X,\mathcal{E}nd_0(F) \otimes K_X) = 0. $$ Now tensoring (*) by $K_X$ and taking $H^0$, one obtains the isomorphism $$ H^0(X,\mathcal{E}nd(F) \otimes K_X) = H^0(X,K_X). $$