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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?

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    $\begingroup$ I've got to run, but wouldn't it just be $d+\alpha$, where $\alpha$ is the harmonic $(0,1)$-form representing the class of $H^1(X,\mathcal{O}_X)$? $\endgroup$
    – Donu Arapura
    Commented Oct 5, 2011 at 14:55
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    $\begingroup$ would like to point out that the Hodge to de Rham spectral sequence gives an algebraic surjective map $H^1_{\rm dR}(X)\to H^1(X,{\cal O}_X)$. The inclusion $H^1(X,{\cal O}_X)\to H^1_{\rm dR}(X)$ you mention is given by the Hodge splitting, which is not algebraic. So the description of the mysterious connection must be analytic (as hinted at by Donu Arapura). On the other hand, the natural inclusion $H^0(X,\Omega_X)\to H^1_{\rm dR}(X)$ (also given by Hodge to de Rham) has a natural algebraic description in terms of connections. $\endgroup$
    – Damian Rössler
    Commented Oct 5, 2011 at 15:12
  • $\begingroup$ @Donu: thanks for your comment I don't exactly understand what you mean with $d+ \alpha$ as you need a connection on $E$. Would be glad if you could explain that to me. $\endgroup$
    – Veen
    Commented Oct 5, 2011 at 20:27

1 Answer 1

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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.

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