The Stacks project

Proof. The first part of the statement follows from Derived Categories of Schemes, Lemma 36.32.8. Setting $S' = \underline{\mathop{\mathrm{Spec}}}_ S(f_*\mathcal{O}_ X)$ we get a factorization $X \to S' \to S$ (this is the Stein factorization, see More on Morphisms, Section 37.53, although we don't need this) and we see that $\Omega _{X/S} = \Omega _{X/S'}$ for example by Morphisms, Lemma 29.32.9 and 29.36.15. This of course implies that $\text{d} : f_*\mathcal{O}_ X \to f_*\Omega _{X/S}$ is zero. $\square$

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

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

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. $\square$

Proof in case $S$ is reduced. Assume $S$ is reduced. Observe that $\text{d} : R^ nf_*\Omega ^{n - 1}_{X/S} \to R^ nf_*\Omega ^ n_{X/S}$ is an $\mathcal{O}_ S$-linear map of (quasi-coherent) $\mathcal{O}_ S$-modules. The $\mathcal{O}_ S$-module $R^ nf_*\Omega ^ n_{X/S}$ is finite locally free (as the dual of the finite locally free $\mathcal{O}_ S$-module $f_*\mathcal{O}_ X$ by Lemmas 50.24.3 and 50.24.2). Since $S$ is reduced it suffices to show that the stalk of $\text{d}$ in every generic point $\eta \in S$ is zero; this follows by looking at sections over affine opens, using that the target of $\text{d}$ is locally free, and Algebra, Lemma 10.25.2 part (2). Since $S$ is reduced we have $\mathcal{O}_{S, \eta } = \kappa (\eta )$, see Algebra, Lemma 10.25.1. Thus $\text{d}_\eta $ is identified with the map

which is zero by Lemma 50.20.3. $\square$

Proof in the general case. Observe that the question is flat local on $S$: if $S' \to S$ is a surjective flat morphism of schemes and the map is zero after pullback to $S'$, then the map is zero. Also, formation of the map commutes with base change by flat morphisms by flat base change (Cohomology of Schemes, Lemma 30.5.2).

Consider the Stein factorization $X \to S' \to S$ as in More on Morphisms, Theorem 37.53.5. By Lemma 50.24.2 the morphism $\pi : S' \to S$ is finite étale. The morphism $f : X \to S'$ is proper (by the theorem), smooth (by More on Morphisms, Lemma 37.13.12) with geometrically connected fibres by the theorem on Stein factorization. In the proof of Lemma 50.24.2 we saw that $\Omega _{X/S} = \Omega _{X/S'}$ because $S' \to S$ is étale. Hence $\Omega ^\bullet _{X/S} = \Omega ^\bullet _{X/S'}$. We have

for all $p, q$ by the Leray spectral sequence (Cohomology, Lemma 20.13.8), the fact that $\pi $ is finite hence affine, and Cohomology of Schemes, Lemma 30.2.3 (of course we also use that $R^ qf'_*\Omega ^ p_{X'/S}$ is quasi-coherent). Thus the map of the lemma is $\pi _*$ applied to $\text{d} : R^ nf'_*\Omega ^{n - 1}_{X/S'} \to R^ nf'_*\Omega ^ n_{X/S'}$. In other words, in order to prove the lemma we may replace $f : X \to S$ by $f' : X \to S'$ to reduce to the case discussed in the next pargraph.

Assume $f$ has geometrically connected fibres and $f_*\mathcal{O}_ X = \mathcal{O}_ S$. For every $s \in S$ we can choose an étale neighbourhood $(S', s') \to (S, s)$ such that the base change $X' \to S'$ of $S$ has a section. See More on Morphisms, Lemma 37.38.6. By the initial remarks of the proof this reduces us to the case discussed in the next paragraph.

Assume $f$ has geometrically connected fibres, $f_*\mathcal{O}_ X = \mathcal{O}_ S$, and we have a section $s : S \to X$ of $f$. We may and do assume $S = \mathop{\mathrm{Spec}}(A)$ is affine. The map $s^* : R\Gamma (X, \mathcal{O}_ X) \to R\Gamma (S, \mathcal{O}_ S) = A$ is a splitting of the map $A \to R\Gamma (X, \mathcal{O}_ X)$. Thus we can write

where $P$ is the “kernel” of $s^*$. By Lemma 50.24.2 the object $P$ of $D(A)$ is perfect of tor amplitude in $[1, n]$. As in the proof of Lemma 50.24.3 we see that $H^ n(X, \Omega ^ n_{X/S})$ is a locally free $A$-module of rank $1$ (and in fact dual to $A$ so free of rank $1$ – we will soon choose a generator but we don't want to check it is the same generator nor will it be necessary to do so).

Denote $Z \subset X$ the image of $s$ which is a closed subscheme of $X$ by Schemes, Lemma 26.21.11. Observe that $Z \to X$ is a regular (and a fortiori Koszul regular by Divisors, Lemma 31.21.2) closed immersion by Divisors, Lemma 31.22.8. Of course $Z \to X$ has codimension $n$. Thus by Remark 50.23.7 we can consider the map

and we set $\xi = \gamma ^{0, 0}(1) \in H^ n(X, \Omega ^ n_{X/S})$.

We claim $\xi $ is a basis element. Namely, since we have base change in top degree (see for example Limits, Lemma 32.19.2) we see that $H^ n(X, \Omega ^ n_{X/S}) \otimes _ A k = H^ n(X_ k, \Omega ^ n_{X_ k/k})$ for any ring map $A \to k$. By the compatibility of the construction of $\xi $ with base change, see Lemma 50.23.10, we see that the image of $\xi $ in $H^ n(X_ k, \Omega ^ n_{X_ k/k})$ is nonzero by Lemma 50.23.11 if $k$ is a field. Thus $\xi $ is a nowhere vanishing section of an invertible module and hence a generator.

Let $\theta \in H^ n(X, \Omega ^{n - 1}_{X/S})$. We have to show that $\text{d}(\theta )$ is zero in $H^ n(X, \Omega ^ n_{X/S})$. We may write $\text{d}(\theta ) = a \xi $ for some $a \in A$ as $\xi $ is a basis element. Then we have to show $a = 0$.

Consider the closed immersion

This is also a section of a smooth morphism (namely either projection) and hence a regular and Koszul immersion of codimension $n$ as well. Thus we can consider the maps

of Remark 50.23.7. Consider the image

By Lemma 50.8.1 we have

By the Künneth formula (either Derived Categories of Schemes, Lemma 36.23.1 or Derived Categories of Schemes, Lemma 36.23.4) we see that

and

Namely, since we are looking in top degree there no higher tor groups that intervene. Combined with the fact that $\xi $ is a generator this means we can write

with $\theta _1, \theta _2 \in H^ n(X, \Omega ^{n - 1}_{X/S})$. Arguing in exactly the same manner we can write

in $H^{2n}(X \times _ S X, \Omega ^{2n}_{X \times _ S X/S}) = H^ n(X, \Omega ^ n_{X/S}) \otimes _ A H^ n(X, \Omega ^ n_{X/S})$ for some $b \in H^0(S, \mathcal{O}_ S)$.

Claim: $\theta _1 = \theta $, $\theta _2 = \theta $, and $b = 1$. Let us show that the claim implies the desired result $a = 0$. Namely, by Lemma 50.23.8 we have

By our choices above this gives

The right most equality comes from the fact that the map $\text{d} : \Omega ^{2n - 1}_{X \otimes _ S X/S} \to \Omega ^{2n}_{X \times _ S X/S}$ by Lemma 50.8.1 is the sum of the differential $\text{d} \boxtimes 1 : \Omega ^{n - 1}_{X/S} \boxtimes \Omega ^ n_{X/S} \to \Omega ^ n_{X/S} \boxtimes \Omega ^ n_{X/S}$ and the differential $(-1)^ n 1 \boxtimes \text{d} : \Omega ^ n_{X/S} \boxtimes \Omega ^{n - 1}_{X/S} \to \Omega ^ n_{X/S} \boxtimes \Omega ^ n_{X/S}$. Please see discussion in Section 50.8 and Derived Categories of Schemes, Section 36.24 for more information. Since $\xi \otimes \xi $ is a basis for the rank $1$ free $A$-module $H^ n(X, \Omega ^ n_{X/S}) \otimes _ A H^ n(X, \Omega ^ n_{X/S})$ we conclude

as desired.

In the rest of the proof we prove the claim above. Let us denote $\eta = \gamma ^{0, 0}(1) \in H^ n(X \times _ S X, \Omega ^ n_{X \times _ S X/S})$. Since $\Omega ^ n_{X \times _ S X/S} = \bigoplus _{p + p' = n} \Omega ^ p_{X/S} \boxtimes \Omega ^{p'}_{X/S}$ we may write

where $\eta _ p$ is in $H^ n(X \times _ S X, \Omega ^ p_{X/S} \boxtimes \Omega ^{n - p}_{X/S})$. For $p = 0$ we can write

by our previously given decomposition $R\Gamma (X, \mathcal{O}_ X) = A \oplus P$. Consider the morphism $(s, \text{id}) : X \to X \times _ S X$. Then $(s, \text{id})^{-1}(\Delta ) = Z$ scheme theoretically. Hence we see that $(s, \text{id})^*\eta = \xi $ by Lemma 50.23.10. This means that

This means exactly that the first component of $\eta _0$ in the direct sum decomposition above is $\xi $. In other words, we can write

with $\eta '_0 \in H^ n(P \otimes _ A^\mathbf {L} R\Gamma (X, \Omega ^ n_{X/S}))$. In exactly the same manner for $p = n$ we can write

and we can write

with $\eta '_ n \in H^ n(R\Gamma (X, \Omega ^ n_{X/S}) \otimes _ A^\mathbf {L} P)$.

Observe that $\text{pr}_1^*\theta = \theta \otimes 1$ and $\text{pr}_2^*\theta = 1 \otimes \theta $ are Hodge cohomology classes on $X \times _ S X$ which pull back to $\theta $ by $\Delta $. Hence by Lemma 50.23.9 we have

in the Hodge cohomology ring of $X \times _ S X$ over $S$. In terms of the direct sum decomposition on the modules of differentials of $X \times _ S X/S$ we obtain

Looking at the formula $\eta _0 = 1 \otimes \xi + \eta '_0$ we found above, we see that to show that $\theta _1 = \theta $ it suffices to prove that

To do this, observe that cupping with $\theta \otimes 1$ is given by the action on cohomology of the map

in the derived category, see Cohomology, Remark 20.31.2. This map is the derived tensor product of the two maps

by Derived Categories of Schemes, Remark 36.23.5. However, the first of these is zero in $D(A)$ because it is a map from a perfect complex of tor amplitude in $[n + 1, 2n]$ to a complex with cohomology only in degrees $0, 1, \ldots , n$, see More on Algebra, Lemma 15.76.1. A similar argument works to show the vanishing of $(1 \otimes \theta ) \cup \eta '_ n$. Finally, in exactly the same manner we obtain

and we conclude as before by showing that $(\xi \otimes 1) \cup \eta '_0 = 0$ in the same manner as above. This finishes the proof. $\square$

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

By the relative Hodge-to-de Rham spectral sequence

(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 precisely, these identifications come from the maps of complexes

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 (50.4.0.1) which in our situation reads

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

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

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

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

and the distinguished triangle

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 $Rf_*$

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

commutes. This indeed means that the diagram

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

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

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

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

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