Lemma 50.24.3. In Situation 50.24.1 there exists an $\mathcal{O}_ S$-module map

\[ t : Rf_*\Omega ^ n_{X/S}[n] \longrightarrow \mathcal{O}_ S \]

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

\[ Rf_*\Omega ^ p_{X/S} \otimes _{\mathcal{O}_ S}^\mathbf {L} Rf_*\Omega ^{n - p}_{X/S}[n] \longrightarrow \mathcal{O}_ S \]

given by the relative cup product composed with $t$ is a perfect pairing of perfect complexes on $S$.

**Proof.**
Let $\omega ^\bullet _{X/S}$ be the relative dualizing complex of $X$ over $S$ as in Duality for Schemes, Remark 48.12.5 and let $Rf_*\omega _{X/S}^\bullet \to \mathcal{O}_ S$ be its trace map. By Duality for Schemes, Lemma 48.15.7 there exists an isomorphism $\omega ^\bullet _{X/S} \cong \Omega ^ n_{X/S}[n]$ and using this isomorphism we obtain $t$. The complexes $Rf_*\Omega ^ p_{X/S}$ are perfect by Lemma 50.3.5. Since $\Omega ^ p_{X/S}$ is locally free and since $\Omega ^ p_{X/S} \otimes _{\mathcal{O}_ X} \Omega ^{n - p}_{X/S} \to \Omega ^ n_{X/S}$ exhibits an isomorphism $\Omega ^ p_{X/S} \cong \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\Omega ^{n - p}_{X/S}, \Omega ^ n_{X/S})$ we see that the pairing induced by the relative cup product is perfect by Duality for Schemes, Remark 48.12.6.

Uniqueness of $t$. Choose a distinguished triangle $f_*\mathcal{O}_ X \to Rf_*\mathcal{O}_ X \to P \to f_*\mathcal{O}_ X[1]$. By Lemma 50.24.2 the object $P$ is perfect of tor amplitude in $[1, \infty )$ and the triangle is locally on $S$ split. Thus $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(P, \mathcal{O}_ X)$ is perfect of tor amplitude in $(-\infty , -1]$. Hence duality (above) shows that locally on $S$ we have

\[ Rf_*\Omega ^ n_{X/S}[n] \cong R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ S}(f_*\mathcal{O}_ X, \mathcal{O}_ S) \oplus R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(P, \mathcal{O}_ X) \]

This shows that $R^ nf_*\Omega ^ n_{X/S}$ is finite locally free and that we obtain a perfect $\mathcal{O}_ S$-bilinear pairing

\[ f_*\mathcal{O}_ X \times R^ nf_*\Omega ^ n_{X/S} \longrightarrow \mathcal{O}_ S \]

using $t$. This implies that any $\mathcal{O}_ S$-linear map $t' : R^ nf_*\Omega ^ n_{X/S} \to \mathcal{O}_ S$ is of the form $t' = t \circ g$ for some $g \in \Gamma (S, f_*\mathcal{O}_ X) = \Gamma (X, \mathcal{O}_ X)$. In order for $t'$ to still determine a perfect pairing $g$ will have to be a unit. This finishes the proof.
$\square$

## Comments (2)

Comment #7502 by Hao Peng on

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