The Stacks project

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

Shouldn't it be the case that the trace map is canonical if induced from the canonial isomorphism and canonical trace \?

I'm thinking about this because there should be a perfect pairing that is compatible with base change.

There are also:

  • 2 comment(s) on Section 50.24: Relative PoincarĂ© duality

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