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.

## 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

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

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

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

Given a commutative diagram

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.

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## Comments (2)

Comment #4973 by Théo de Oliveira Santos on

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