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

  1. $g_{ii}=0$ (on each $U_i$)
  2. $g_{ij}=-\ell_{ij}g_{ji}$ (on each $U_i\cap U_j$)
  3. $g_{ij}=\ell_{ik}g_{kj}+g_{ik}$ (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?


(This should be a comment rather than an answer, but I do not have the right to make comments, since my rep is not yet established in this town.)

I personally found Atiyah's 1957 paper 'Complex analytic connections in fibre bundles' (see the proof of Proposition 2) very enlightening with respect to the problem you posed. You may want to take a look, it's always a good idea to learn from the masters.


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.

  • 1
    $\begingroup$ Why can you embed L into direct sum of line bundle of large degree? How does large degree play a role here? Would you mind explaining a bit ? Thank you! $\endgroup$ – Ben Oct 27 '15 at 23:11
  • $\begingroup$ Embedding $L$ in large degree line bundles, by dualizing is the same as finding a surjection from very negative line bundles to a line bundle and that is same as saying for a line bundle $L$, $L\otimes M$ is globally generated for large degree $M$. This of course is easy. $\endgroup$ – Mohan Oct 28 '15 at 1:26
  • $\begingroup$ I am not very familar with complex geometry. Do we need $X$ to be projective here to apply Serre twisting theorem or can we just take an injective object in which L embeds? $\endgroup$ – Ben Oct 28 '15 at 3:01
  • $\begingroup$ What do you mean by the degree of a line bundle when $\dim X>1$ ? $\endgroup$ – Heitor Fontana Dec 8 '15 at 16:36
  • $\begingroup$ As you may have guessed, I am assuming that things are projective and then I just meant large powers of an ample line bundle. $\endgroup$ – Mohan Dec 8 '15 at 18:53

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