Lemma 50.20.1. Let $k$ be a field. Let $X$ be a nonempty smooth proper scheme over $k$ equidimensional of dimension $d$. There exists a $k$-linear map

$t : H^ d(X, \Omega ^ d_{X/k}) \longrightarrow k$

unique up to precomposing by multiplication by a unit of $H^0(X, \mathcal{O}_ X)$ with the following property: for all $p, q$ the pairing

$H^ q(X, \Omega ^ p_{X/k}) \times H^{d - q}(X, \Omega ^{d - p}_{X/k}) \longrightarrow k, \quad (\xi , \xi ') \longmapsto t(\xi \cup \xi ')$

is perfect.

Proof. By Duality for Schemes, Lemma 48.27.1 we have $\omega _ X^\bullet = \Omega ^ d_{X/k}[d]$. Since $\Omega _{X/k}$ is locally free of rank $d$ (Morphisms, Lemma 29.34.12) we have

$\Omega ^ d_{X/k} \otimes _{\mathcal{O}_ X} (\Omega ^ p_{X/k})^\vee \cong \Omega ^{d - p}_{X/k}$

Thus we obtain a $k$-linear map $t : H^ d(X, \Omega ^ d_{X/k}) \to k$ such that the statement is true by Duality for Schemes, Lemma 48.27.4. In particular the pairing $H^0(X, \mathcal{O}_ X) \times H^ d(X, \Omega ^ d_{X/k}) \to k$ is perfect, which implies that any $k$-linear map $t' : H^ d(X, \Omega ^ d_{X/k}) \to k$ is of the form $\xi \mapsto t(g\xi )$ for some $g \in H^0(X, \mathcal{O}_ X)$. Of course, in order for $t'$ to still produce a duality between $H^0(X, \mathcal{O}_ X)$ and $H^ d(X, \Omega ^ d_{X/k})$ we need $g$ to be a unit. Denote $\langle -, - \rangle _{p, q}$ the pairing constructed using $t$ and denote $\langle -, - \rangle '_{p, q}$ the pairing constructed using $t'$. Clearly we have

$\langle \xi , \xi ' \rangle '_{p, q} = \langle g\xi , \xi ' \rangle _{p, q}$

for $\xi \in H^ q(X, \Omega ^ p_{X/k})$ and $\xi ' \in H^{d - q}(X, \Omega ^{d - p}_{X/k})$. Since $g$ is a unit, i.e., invertible, we see that using $t'$ instead of $t$ we still get perfect pairings for all $p, q$. $\square$

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