The Stacks project

50.5 Hodge cohomology

Let $p : X \to S$ be a morphism of schemes. We define the de 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 (1)

Comment #4973 by Théo de Oliveira Santos on

Typo: de Hodge


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