Let $U$ be a smooth algebraic variety defined over $k \hookrightarrow \mathbb{C}$. Let $\mathcal{E}$ be a locally free sheaf on $U$ equipped with an integrable connection

$\nabla: \mathcal{E} \to \mathcal{E} \otimes_{\mathcal{O}_U} \Omega^1_U$.

Let $X$ be a smooth variety including $U$ as the complement of a divisor $D$ with simple normal crossings.

Definition. $(\mathcal{E}, \nabla)$ has regular singularities along $D$ if there exists a coherent $\mathcal{O}_X$-module $\mathcal{E}_X$ and a logarithmic integrable connection $\nabla_X: \mathcal{E}_X \to \mathcal{E}_X \otimes \Omega^1_X(\log D)$ extending $(\mathcal{E}, \nabla)$, i.e. there exists an isomorphism $(\mathcal{E}_X, \nabla_X)_{|U} \cong (\mathcal{E}, \nabla)$.

For such connections, one can form the de Rham complex $$ DR(\mathcal{E}_X)=[\mathcal{E}_X \stackrel{\nabla_X}{\longrightarrow} \mathcal{E}_X \otimes \Omega^1_X(\log D) \to \cdots] $$ and take the hypercohomology of $X$ with values in this complex. Let me call it

$H_{dR}^i(U, (\mathcal{E}, \nabla)):=\mathbb{H}^i(X, DR(\mathcal{E}_X))$

By a fundamental theorem of Deligne, $H^i_{dR}(U, (\mathcal{E}, \nabla)) \otimes \mathbb{C}$ is isomorphic to $H^i(U^{an}, \ker(\nabla^{an}))$, the cohomology of $U^{an}$ with values in the local system of complex vector spaces $ker(\nabla^{an})$.

That's for the general theory. My question aims to understand what do these groups actually compute when $\mathcal{E}$ is the Gauss-Manin connection, which has regular singularities by a theorem of Griffiths. To fix the ideas, let $f: Y \to S$ be a smooth proper family of algebraic varieties and look at the (locally free) sheaves of relative cohomology $\mathcal{H}^i(Y/S)$. One has the Gauss-Manin connection

$\nabla^i_{GM}: \mathcal{H}^i(Y/S) \to \mathcal{H}^i(Y/S) \otimes \Omega^1_S$.

Question: What do the groups $H^j_{dR}(S, (\mathcal{H}^i(Y/S), \nabla^i_{GM}))$ compute? For instance, what is the relation between these groups and the de Rham cohomology of one particular fiber $Y_s$?

Somehow, I will expect a relation between $H^0_{dR}(S, (\mathcal{H}^i(Y/S), \nabla^i_{GM}))$, that is the global sections horizontal with respect to the connection and $H^i_{dR}(Y_s)$, but I don't know how to precise the relation.

I will be happy to understand the case of curves and abelian varieties. Do things get simpler? For instance, if $f: Y \to S$ is a family of curves over a curve S=$\bar{S}$ minus a finite set of points, does the first cohomology group of the connection on $\mathcal{H}^1(Y/S)$ vanishes?

Thanks for your help!

  • 2
    $\begingroup$ It's probably obvious, but since the Gauss-Manin connection on $\mathcal{H}^i(Y/S)$ is obtained as the D-module pushforward of $\mathcal{O}_Y$ to $S$ and the de Rham cohomology is the pushforward to a point, there is a spectral sequence relating $H^j_{dR}(S, (\mathcal{H}^i(Y/S), \nabla^i_{GM}))$ and the de Rham cohomology of $Y$ itself. $\endgroup$ Jan 11, 2013 at 19:44

1 Answer 1


The module with integrable connection $(\mathcal{H}^i, \nabla^i_{GM})$ corresponds to the local system $Rf^{an}_* \mathbb{C}_{Y^{an}}$ on $S^{an}$, so the cohomology of the de Rham complex of $(\mathcal{H}^i, \nabla^i_{GM})$ will compute cohomology $H^j(S^{an}, R^if^{an}_* \mathbb{C}_{Y^{an}})$. By definition, the stalks of $R^if^{an}_* \mathbb{C}_{Y^{an}}$ are (singular) cohomology groups of the fibers of $f^{an}$. There is a Leray spectral sequence $$H^j(S^{an}, R^if^{an}_* \mathbb{C}_{Y^{an}}) \Rightarrow H^{i+j}(Y^{an}, \mathbb{C}_{Y^{an}})$$ which these isomorphisms identify with $$H^j(S, (\mathcal{H}^i, \nabla^i_{GM})) \Rightarrow H^{i+j}_{dR}(Y/\mathbb{C}).$$

Fix $s\in S$, then $\pi_1(S, s)$ homotopically acts on $Y^{an}_s = (f^{an})^{-1}(s)$, hence you get an action on $H^i(Y^{an}_s, \mathbb{C})$, called the monodromy action. For $j=0$, you get should get $$H^0(S^{an}, \mathcal{H}^{i, an})^{\nabla^{i, an}_{GM}} = H^0(S^{an}, (\mathcal{H}^{i, an}, \nabla^{i, an}_{GM})) = H^0(S^{an}, R^i f^{an}_* \mathbb{C}_{Y^{an}}) = H^i(Y^{an}_s, \mathbb{C})^{\pi_1(S, s)}.$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .