The Stacks project

50.5 Hodge cohomology

Let $p : X \to S$ be a morphism of schemes. We define the Hodge cohomology of $X$ over $S$ to be the cohomology groups

\[ H^ n_{Hodge}(X/S) = \bigoplus \nolimits _{n = p + q} H^ q(X, \Omega ^ p_{X/S}) \]

viewed as a graded $H^0(X, \mathcal{O}_ X)$-module. The wedge product of forms combined with the cup product of Cohomology, Section 20.31 defines a $H^0(X, \mathcal{O}_ X)$-bilinear cup product

\[ \cup : H^ i_{Hodge}(X/S) \times H^ j_{Hodge}(X/S) \longrightarrow H^{i + j}_{Hodge}(X/S) \]

Of course if $\xi \in H^ q(X, \Omega ^ p_{X/S})$ and $\xi ' \in H^{q'}(X, \Omega ^{p'}_{X/S})$ then $\xi \cup \xi ' \in H^{q + q'}(X, \Omega ^{p + p'}_{X/S})$.

Lemma 50.5.1. Let $p : X \to S$ be a morphism of schemes. The cup product on $H^*_{Hodge}(X/S)$ is associative and graded commutative.

Proof. The proof is identical to the proof of Lemma 50.4.1. $\square$

Given a commutative diagram

\[ \xymatrix{ X' \ar[r]_ f \ar[d] & X \ar[d] \\ S' \ar[r] & S } \]

of schemes, there are pullback maps $f^* : H^ i_{Hodge}(X/S) \longrightarrow H^ i_{Hodge}(X'/S')$ compatible with gradings and with the cup product defined above.

Comments (2)

Comment #4973 by Théo de Oliveira Santos on

Typo: de Hodge

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FM4. Beware of the difference between the letter 'O' and the digit '0'.