trace and determinant of vector bundles in cohomology

Question

Let $\mathcal{M}$ be a moduli space of stable vector bundles $E$ of fixed invariants over a smooth projective surface $S$,then there is a determinant map $\det: \mathcal{M} \rightarrow \mathbf{Pic}(S)$ mapping $E \mapsto \det(E)$. The differential at $[E] \in \mathcal{M}$ of this map is the trace map $tr: \mathbf{T}_{[E]}\mathcal{M} \rightarrow \mathbf{T}_{\mathbf{\det(E)}}\mathbf{Pic}(S)$ which is just $Ext^1(E,E) \rightarrow H^1(\mathcal{O}_S)$.
$\mathbf{Question}$: seeing an element of the LHS as an extension class $[0 \rightarrow E \rightarrow F \rightarrow E \rightarrow 0]$, is there some way to take a "trace"? I.e. what is this map on exact sequences...? For me, the trace is $tr: \mathcal{E}nd(E) \rightarrow \mathcal{O}_S$ which induces in cohomology degree 1 this map, but what is it explicitely? Thanks!!

Answer 1

Given an extension $$ 0 \to E \to F \to E \to 0, $$ you can first take its tensor product with the dual bundle $E^\vee$, thus getting $$ 0 \to E \otimes E^\vee \to F \otimes E^\vee \to E \otimes E^\vee \to 0, $$ then take its composition with the unit morphism $\mathcal{O}_S \to E \otimes E^\vee$ (applied to the last term) and with the trace morphism $E \otimes E^\vee \to \mathcal{O}_S$ (applied to the first term). As a result, you will get an extension $$ 0 \to \mathcal{O}_S \to F' \to \mathcal{O}_S \to 0 $$ whose class in $H^1(S, \mathcal{O}_S)$ is equal to the trace of the original extension multiplied by the rank of $E$.