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?