On a compact complex manifold $X$, fix two holomorphic line bundles $L$ and $L'$. Consider a holomorphic vector bundle $V$ of rank 2 which fits in an exact sequence $$0\to L\to V\to L'\to0$$ I would like to understand why these $V$'s are parametrized by $H^1(X, (L')^\ast\otimes L)$.
Up to tensoring with $(L')^\ast$ the short exact sequence above, we can equivalently ask why, for fixed $L$, holomorphic vector bundles $V$ of rank 2 sitting in exact sequences of the form $$0\to L\to V\to \mathcal{O}_X\to0$$ are parametrized by $H^1(X,L)$.
I was thinking about reasoning with Cech cohomology. With respect to some covering $U_i$ of $X$, the transition functions of $V$ can be written in a upper-triangular matrix with the transition functions $\ell_{ij}$ of $L$ and $1$ on the diagonal and some $g_{ij}\in\mathcal{O}(U_i\cap U_j)$ at the top right entry. Thus, all the information should be encoded in the $g_{ij}$. The transition relations for the bundle then yield the following relations
- $g_{ii}=0$ (on each $U_i$)
- $g_{ij}=-\ell_{ij}g_{ji}$ (on each $U_i\cap U_j$)
- $g_{ij}=\ell_{ik}g_{kj}+g_{ik}$ (on each $U_i\cap U_j\cap U_k$)
However, when interpreting $g=(g_{ij})$ as an element of the group $C^1(\mathcal{L})$ in the Cech complex, by using the above relations I get $$(dg)_{ijk}:=g_{jk}-g_{ik}+g_{ij}=g_{jk}(1-\ell_{ij})$$ and therefore it seems that my $g$ does not even define a class in $H^1$ in general (apart from the trivial case when $L$ is trivial). Any suggestion?
Given your exact sequence, the boundary map on cohomologies gives a map $H^0(\mathcal{O}_X)\to H^1(L)$ and since the first is just $k$, the element 1 gives an element in the $H^1$. Conversely, embed $L$ in a large bundle, for example, $\oplus_{i=1}^n M_i$, where $\deg M_i$s are very large as a subbundle and let $P$ the quotient vector bundle. Again taking the long exact sequence, we get a map $H^0(P)\to H^1(L)$ and this map is onto, since $\deg M_i>>0$. Thus, given $s\in H^1(L)$, we get an element $t\in H^0(P)$ and thus a map $\mathcal{O}_X\to P$. Take the pull back to get your extension.