Let $X$ be a smooth projective curve over a field $k$ of characteristic zero. The algebraic de Rham cohomology of $X$ is, by definition, the hypercohomology of the complex of Kähler differentials for the Zariski topology:

$H^k_{DR}(X)=\mathbb{H}(X, \mathcal{O}_X \to \Omega^1_X \to \Omega^2_X \to \cdots)$

In "Hodge cycles on abelian varieties", p. 24, Deligne claims

"For a complete smooth curve $X$ and an open affine subset, the map

$H^1_{DR}(X) \to \Gamma(U, \Omega^1_X) / d\Gamma(U, \mathcal{O}_X)$

is injective with image the set of classes represented by forms whose residues are all zero (such forms are said to be of the second kind)."

I have several questions regarding this quote:

1) How does one prove the statement?

2) How to use this to determine $H^1_{DR}(X)$?

3) I'm a bit confused with the terminology "second kind". I thought this was reserved for $H^1(X, \mathcal{O}_X)$ whereas "first kind" are differentials in $H^0(X, \Omega^1_X)$. What does it mean?

4) related to 3) how does one see the Hodge decomposition in this setting?

Any help would be appreciated. Thanks a lot!

  • $\begingroup$ Doesn't injectivity follow from $H^2=0$ because $dim(X)=1$? $\endgroup$ Jun 21, 2013 at 9:26
  • $\begingroup$ $H^2$ of what and what $H^2$? For instance $H^2_{dR}(X)=k$. $\endgroup$
    – erik
    Jun 21, 2013 at 9:40

1 Answer 1


For question 1: consider de localization long exact sequence \begin{equation} ...\to H^1_{DR,Z}(X)\to H^1_{DR}(X)\to H^1_{DR}(U)\to ... \end{equation} where $H^1_{DR,Z}(X)$ is the de Rham cohomology with support on $Z=X\setminus U$. Since $Z$ has dimension $0$ we can prove that $H^1_{DR,Z}(X)$ (which is Poincaré dual to $H^1_{DR}(Z)$) is trivial. Then you just have to note that the de Rham cohomology of $U$ is given by the cohomology of the complex $\Gamma(U,\Omega^\bullet_{U/k})$ since $U$ is affine. You will find all the details in the paper by Hartshorne: \emph{On the de Rham cohomology of algebraic varieties}, Pub. Math. IHES 1975.

For question 2: let me just note that from the above sequence you get the following exact sequence \begin{equation} 0\to H^1_{DR}(X)\to H^1_{DR}(U)\to H^0_{DR}(Z)\to H^2_{DR}(X)\to 0 \ . \end{equation}

For question 3: I don't know the answer. What I know is that the de Rham cohomology of $U$ can be computed using the de Rham complex of differentials forms on $X$ with log poles along the complement $Z$. This means that a class \begin{equation} [\omega]\in H^1_{DR}(U)=\Gamma(U,\Omega^1_{U/k})/d\Gamma(U,\Omega^0_{U/k}) \end{equation} can be represented by a 1-form $\omega\in \Gamma(U,\Omega^1_{U/k})$ that we can extend to a meromorphic 1-form on $X$ having only simple poles at any point of $Z$. In particular you can take $\omega$ to be holomorphic on $X$. From what I understand the injection of point 1 gives you all the classes $[\omega]$ such that the sum of the residues of $\omega$ at the points of $Z$ is $0$.

  • $\begingroup$ What if I try and compute the de Rham cohomology of the $\mathbb{P}^1$ by taking $Z$ to be a point, or even better, let $Z$ be several points? This argument that $H_{Z,dR}^1(X) =0$ cannot hold. $\endgroup$
    – 54321user
    Aug 5, 2016 at 18:01
  • $\begingroup$ @user251222: the first cohomology of the projective line is trivial, so in this case the boundary map from H^1(U)\to H^0(Z) is given by the residues at each point of Z and you easily deduce that \dim H^1(U)=\dimH^0(Z) -1. $\endgroup$
    – Nicola
    Mar 9, 2017 at 10:08

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