The Stacks project

Proposition 50.24.5. Let $S$ be a quasi-compact and quasi-separated scheme. Let $f : X \to S$ be a proper smooth morphism of schemes all of whose fibres are nonempty and equidimensional of dimension $n$. There exists an $\mathcal{O}_ S$-module map

\[ t : R^{2n}f_*\Omega ^\bullet _{X/S} \longrightarrow \mathcal{O}_ S \]

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

\[ Rf_*\Omega ^\bullet _{X/S} \otimes _{\mathcal{O}_ S}^\mathbf {L} Rf_*\Omega ^\bullet _{X/S}[2n] \longrightarrow \mathcal{O}_ S, \quad (\xi , \xi ') \longmapsto t(\xi \cup \xi ') \]

is a perfect pairing of perfect complexes on $S$.

Proof. The proof is exactly the same as the proof of Proposition 50.20.4.

By the relative Hodge-to-de Rham spectral sequence

\[ E_1^{p, q} = R^ qf_*\Omega ^ p_{X/S} \Rightarrow R^{p + q}f_*\Omega ^\bullet _{X/S} \]

(Section 50.6), the vanishing of $\Omega ^ i_{X/S}$ for $i > n$, the vanishing in for example Limits, Lemma 32.19.2 and the results of Lemmas 50.24.2 and 50.24.4 we see that $R^0f_*\Omega _{X/S} = R^0f_*\mathcal{O}_ X$ and $R^ nf_*\Omega ^ n_{X/S} = R^{2n}f_*\Omega ^\bullet _{X/S}$. More precisesly, these identifications come from the maps of complexes

\[ \Omega ^\bullet _{X/S} \to \mathcal{O}_ X[0] \quad \text{and}\quad \Omega ^ n_{X/S}[-n] \to \Omega ^\bullet _{X/S} \]

Let us choose $t : R^{2n}f_*\Omega _{X/S} \to \mathcal{O}_ S$ which via this identification corresponds to a $t$ as in Lemma 50.24.3.

Let us abbreviate $\Omega ^\bullet = \Omega ^\bullet _{X/S}$. Consider the map ( which in our situation reads

\[ \wedge : \text{Tot}(\Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]

For every integer $p = 0, 1, \ldots , n$ this map annihilates the subcomplex $\text{Tot}(\sigma _{> p} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p} \Omega ^\bullet )$ for degree reasons. Hence we find that the restriction of $\wedge $ to the subcomplex $\text{Tot}(\Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \geq _{n - p}\Omega ^\bullet )$ factors through a map of complexes

\[ \gamma _ p : \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]

Using the same procedure as in Section 50.4 we obtain relative cup products

\[ Rf_*\sigma _{\leq p} \Omega ^\bullet \otimes _{\mathcal{O}_ S}^\mathbf {L} Rf_*\sigma _{\geq n - p}\Omega ^\bullet \longrightarrow Rf_*\Omega ^\bullet \]

We will prove by induction on $p$ that these cup products via $t$ induce perfect pairings between $Rf_*\sigma _{\leq p} \Omega ^\bullet $ and $Rf_*\sigma _{\geq n - p}\Omega ^\bullet [2n]$. For $p = n$ this is the assertion of the proposition.

The base case is $p = 0$. In this case we have

\[ Rf_*\sigma _{\leq p}\Omega ^\bullet = Rf_*\mathcal{O}_ X \quad \text{and}\quad Rf_*\sigma _{\geq n - p}\Omega ^\bullet [2n] = Rf_*(\Omega ^ n[-n])[2n] = Rf_*\Omega ^ n[n] \]

In this case we simply obtain the pairing between $Rf_*\mathcal{O}_ X$ and $Rf_*\Omega ^ n[n]$ of Lemma 50.24.3 and the result is true.

Induction step. Say we know the result is true for $p$. Then we consider the distinguished triangle

\[ \Omega ^{p + 1}[-p - 1] \to \sigma _{\leq p + 1}\Omega ^\bullet \to \sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p] \]

and the distinguished triangle

\[ \sigma _{\geq n - p}\Omega ^\bullet \to \sigma _{\geq n - p - 1}\Omega ^\bullet \to \Omega ^{n - p - 1}[-n + p + 1] \to (\sigma _{\geq n - p}\Omega ^\bullet )[1] \]

Observe that both are distinguished triangles in the homotopy category of complexes of sheaves of $f^{-1}\mathcal{O}_ S$-modules; in particular the maps $\sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p]$ and $\Omega ^{n - p - 1}[-d + p + 1] \to (\sigma _{\geq n - p}\Omega ^\bullet )[1]$ are given by actual maps of complexes, namely using the differential $\Omega ^ p \to \Omega ^{p + 1}$ and the differential $\Omega ^{n - p - 1} \to \Omega ^{n - p}$. Consider the distinguished triangles associated gotten from these distinguished triangles by applying $Rf_*$

\[ \xymatrix{ Rf_*\sigma _{\leq p}\Omega ^\bullet \ar[d]_ a \\ Rf_*\Omega ^{p + 1}[-p - 1] \ar[d]_ b \\ Rf_*\sigma _{\leq p + 1}\Omega ^\bullet \ar[d]_ c \\ Rf_*\sigma _{\leq p}\Omega ^\bullet \ar[d]_ d \\ Rf_*\Omega ^{p + 1}[-p - 1] } \quad \quad \xymatrix{ Rf_*\sigma _{\geq n - p}\Omega ^\bullet \\ Rf_*\Omega ^{n - p - 1}[-n + p + 1] \ar[u]_{a'} \\ Rf_*\sigma _{\geq n - p - 1}\Omega ^\bullet \ar[u]_{b'} \\ Rf_*\sigma _{\geq n - p}\Omega ^\bullet \ar[u]_{c'} \\ Rf_*\Omega ^{n - p - 1}[-n + p + 1] \ar[u]_{d'} } \]

We will show below that the pairs $(a, a')$, $(b, b')$, $(c, c')$, and $(d, d')$ are compatible with the given pairings. This means we obtain a map from the distinguished triangle on the left to the distuiguished triangle obtained by applying $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (-, \mathcal{O}_ S)$ to the distinguished triangle on the right. By induction and Lemma 50.20.1 we know that the pairings constructed above between the complexes on the first, second, fourth, and fifth rows are perfect, i.e., determine isomorphisms after taking duals. By Derived Categories, Lemma 13.4.3 we conclude the pairing between the complexes in the middle row is perfect as desired.

Let $e : K \to K'$ and $e' : M' \to M$ be maps of objects of $D(\mathcal{O}_ S)$ and let $K \otimes _{\mathcal{O}_ S}^\mathbf {L} M \to \mathcal{O}_ S$ and $K' \otimes _{\mathcal{O}_ S}^\mathbf {L} M' \to \mathcal{O}_ S$ be pairings. Then we say these pairings are compatible if the diagram

\[ \xymatrix{ K' \otimes _{\mathcal{O}_ S}^\mathbf {L} M' \ar[d] & K \otimes _{\mathcal{O}_ S}^\mathbf {L} M' \ar[l]^{e \otimes 1} \ar[d]^{1 \otimes e'} \\ \mathcal{O}_ S & K \otimes _{\mathcal{O}_ S}^\mathbf {L} M \ar[l] } \]

commutes. This indeed means that the diagram

\[ \xymatrix{ K \ar[r] \ar[d]_ e & R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \mathcal{O}_ S) \ar[d]^{R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (e', -)} \\ K' \ar[r] & R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M', \mathcal{O}_ S) } \]

commutes and hence is sufficient for our purposes.

Let us prove this for the pair $(c, c')$. Here we observe simply that we have a commutative diagram

\[ \xymatrix{ \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p} \Omega ^\bullet ) \ar[d]_{\gamma _ p} & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p - 1} \Omega ^\bullet ) \ar[l]_-{\gamma _{p + 1}} } \]

By functoriality of the cup product we obtain commutativity of the desired diagram.

Similarly for the pair $(b, b')$ we use the commutative diagram

\[ \xymatrix{ \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p - 1} \Omega ^\bullet ) \ar[d]_{\gamma _{p + 1}} & \text{Tot}(\Omega ^{p + 1}[-p - 1] \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p - 1} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \Omega ^{p + 1}[-p - 1] \otimes _{f^{-1}\mathcal{O}_ S} \Omega ^{n - p - 1}[-n + p + 1] \ar[l]_-\wedge } \]

For the pairs $(d, d')$ and $(a, a')$ we use the commutative diagram

\[ \xymatrix{ \Omega ^{p + 1}[-p] \otimes _{f^{-1}\mathcal{O}_ S} \Omega ^{n - p - 1}[-n + p] \ar[d] & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \Omega ^{n - p - 1}[-n + p]) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{f^{-1}\mathcal{O}_ S} \sigma _{\geq n - p}\Omega ^\bullet ) \ar[l] } \]

We omit the argument showing the uniqueness of $t$ up to precomposing by multiplication by a unit in $H^0(X, \mathcal{O}_ X)$. $\square$

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