Lemma 50.24.4. In Situation 50.24.1 the map $\text{d} : R^ nf_*\Omega ^{n - 1}_{X/S} \to R^ nf_*\Omega ^ n_{X/S}$ is zero.

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

$\text{d} : H^ n(X_\eta , \Omega ^{n - 1}_{X_\eta /\kappa (\eta )}) \longrightarrow H^ n(X_\eta , \Omega ^ n_{X_\eta /\kappa (\eta )})$

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

$R^ qf_*\Omega ^ p_{X/S} = \pi _*R^ qf'_*\Omega ^ p_{X/S'}$

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

$R\Gamma (X, \mathcal{O}_ X) = A \oplus P$

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

$\gamma ^{0, 0} : H^0(Z, \Omega ^0_{Z/S}) \longrightarrow H^ n(X, \Omega ^ n_{X/S})$

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

$\Delta : X \to X \times _ S X$

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

$\gamma ^{p, q} : H^ q(X, \Omega ^ p_{X/S}) \longrightarrow H^{q + n}(X \times _ S X, \Omega ^{p + n}_{X \times _ S X/S})$

of Remark 50.23.7. Consider the image

$\gamma ^{n - 1, n}(\theta ) \in H^{2n}(X \times _ S X, \Omega ^{2n - 1}_{X \times _ S X})$

By Lemma 50.8.1 we have

$\Omega ^{2n - 1}_{X \times _ S X} = \Omega ^{n - 1}_{X/S} \boxtimes \Omega ^ n_{X/S} \oplus \Omega ^ n_{X/S} \boxtimes \Omega ^{n - 1}_{X/S}$

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

$H^{2n}(X \times _ S X, \Omega ^{n - 1}_{X/S} \boxtimes \Omega ^ n_{X/S}) = H^ n(X, \Omega ^{n - 1}_{X/S}) \otimes _ A H^ n(X, \Omega ^ n_{X/S})$

and

$H^{2n}(X \times _ S X, \Omega ^ n_{X/S} \boxtimes \Omega ^{n - 1}_{X/S}) = H^ n(X, \Omega ^ n_{X/S}) \otimes _ A H^ n(X, \Omega ^{n - 1}_{X/S})$

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

$\gamma ^{n - 1, n}(\theta ) = \theta _1 \otimes \xi + \xi \otimes \theta _2$

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

$\gamma ^{n, n}(\xi ) = b \xi \otimes \xi$

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

$\gamma ^{n, n}(\text{d}(\theta )) = \text{d}(\gamma ^{n - 1, n}(\theta ))$

By our choices above this gives

$a \xi \otimes \xi = \gamma ^{n, n}(a\xi ) = \text{d}(\theta \otimes \xi + \xi \otimes \theta ) = a \xi \otimes \xi + (-1)^ n a \xi \otimes \xi$

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

$a = a + (-1)^ n a \Rightarrow a = 0$

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

$\eta = \eta _0 + \eta _1 + \ldots + \eta _ n$

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

\begin{align*} H^ n(X \times _ S X, \mathcal{O}_ X \boxtimes \Omega ^ n_{X/S}) & = H^ n(R\Gamma (X, \mathcal{O}_ X) \otimes _ A^\mathbf {L} R\Gamma (X, \Omega ^ n_{X/S})) \\ & = A \otimes _ A H^ n(X, \Omega ^ n_{X/S}) \oplus H^ n(P \otimes _ A^\mathbf {L} R\Gamma (X, \Omega ^ n_{X/S})) \end{align*}

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

$\xi = (s, \text{id})^*\eta = (s^* \otimes \text{id})(\eta _0)$

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

$\eta _0 = 1 \otimes \xi + \eta '_0$

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

\begin{align*} H^ n(X \times _ S X, \Omega ^ n_{X/S} \boxtimes \mathcal{O}_ X) & = H^ n(R\Gamma (X, \Omega ^ n_{X/S}) \otimes _ A^\mathbf {L} R\Gamma (X, \mathcal{O}_ X)) \\ & = H^ n(X, \Omega ^ n_{X/S}) \otimes _ A A \oplus H^ n(R\Gamma (X, \Omega ^ n_{X/S}) \otimes _ A^\mathbf {L} P) \end{align*}

and we can write

$\eta _ n = \xi \otimes 1 + \eta '_ n$

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

$\theta _1 \otimes \xi + \xi \otimes \theta _2 = \gamma ^{n - 1, n}(\theta ) = (\theta \otimes 1) \cup \eta = (1 \otimes \theta ) \cup \eta$

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

$\theta _1 \otimes \xi = (\theta \otimes 1) \cup \eta _0 \quad \text{and}\quad \xi \otimes \theta _2 = (1 \otimes \theta ) \cup \eta _ n$

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

$(\theta \otimes 1) \cup \eta '_0 = 0$

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

$(P \otimes _ A^\mathbf {L} R\Gamma (X, \Omega ^ n_{X/S}))[-n] \xrightarrow {\theta \otimes 1} R\Gamma (X, \Omega ^{n - 1}_{X/S}) \otimes _ A^\mathbf {L} R\Gamma (X, \Omega ^ n_{X/S})$

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

$\theta : P[-n] \to R\Gamma (X, \Omega ^{n - 1}_{X/S}) \quad \text{and}\quad 1 : R\Gamma (X, \Omega ^ n_{X/S}) \to R\Gamma (X, \Omega ^ n_{X/S})$

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

$b \xi \otimes \xi = \gamma ^{n, n}(\xi ) = (\xi \otimes 1) \cup \eta _0$

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. $\square$

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