Let $X$ be a smooth projective variety over the complex numbers. One has the Hodge-Decomposition


$H^1\_{DR}(X) \simeq \Omega^1(X) \oplus H^1(X,\mathcal O\_X)$ (here consider the underlying manifold).
With $H^1\_{DR}(X)$ I denote the first De-Rham cohomology group, which is also the first hypercohomology group of the complex 
$\mathcal O\_X \rightarrow \Omega^1\_X \rightarrow \Omega^2\_X \rightarrow...$


given by exterior derivation.


Furthermore one knows that the following isomorphisms hold


1) $H^1\_{DR}(X) \simeq Ext^1\_{Conn}(\mathcal O\_X, \mathcal O\_X)$, where the right group is the group of extensions of locally free modules on $X$ equipped with an integrable connection, i.e. an element of it is an exact sequence


$0\rightarrow \mathcal O\_X \rightarrow E \rightarrow \mathcal O\_X \rightarrow 0$,


where $E$ is locally free equipped with integrable connection and the sequence is compatible with the connections of the three sheaves (here $ \mathcal O\_X$ is equipped with the trivial connection $d$).


2) $Ext^1(\mathcal O\_X,\mathcal O\_X) \simeq H^1(X,\mathcal O\_X)$, where now the left group is the same as before, but you don't regard the sheaves with connection, i.e. just exact sequences of modules.


With 1) and 2) and the inclusion $H^1(X,\mathcal O\_X) \subset H^1\_{DR}(X)$ one has an inclusion


$Ext^1(\mathcal O\_X,\mathcal O\_X) \subset Ext^1\_{Conn}(\mathcal O\_X, \mathcal O\_X)$.


But this means that for any exact sequence


$0\rightarrow \mathcal O\_X \rightarrow E \rightarrow \mathcal O\_X \rightarrow 0$


of modules, where $E$ is locally free, one finds canonically an integrable connection on $E$, which makes the sequence an exact sequence of modules with connection.


Now: can one describe this mysterious connection in some way which is not just by abstract arrows as in the above procedure? Where does it come from?



Ext groups with connection and Hodge Decomposition
Sorry, my comment yesterday was rushed and this will be only slightly less so. Thus it is more
of a hint. It is perhaps easier to view $E$ etc. as a $C^\infty$ bundle equipped with a $\bar\partial$ operator. Then the exact sequence
$$0\to \mathcal{O}\_X\to E\to \mathcal{O}\_X\to 0$$
admits a $C^\infty$ splitting, with respect to which
$$\bar \partial\_E= \begin{pmatrix} \bar \partial&\alpha\\\ 0 &\bar \partial\end{pmatrix}$$
where $\alpha$ is a $(0,1)$-form. Integrability should force this to be a closed form. Factoring out the dependence on the splitting should
give the Dolbeault class. A $C^\infty$ integrable connection on $E$, compatible with the sequence, can expressed similarly with $\bar \partial$ replaced by $d$ and $\alpha$ now just a closed $1$-form. So you see the
inclusion naturally from this perspective.


