The Stacks project

Proof. By Duality for Schemes, Lemma 48.27.1 we have $\omega _ X^\bullet = \Omega ^ d_{X/k}[d]$. Since $\Omega _{X/k}$ is locally free of rank $d$ (Morphisms, Lemma 29.34.12) we have

Thus we obtain a $k$-linear map $t : H^ d(X, \Omega ^ d_{X/k}) \to k$ such that the statement is true by Duality for Schemes, Lemma 48.27.4. In particular the pairing $H^0(X, \mathcal{O}_ X) \times H^ d(X, \Omega ^ d_{X/k}) \to k$ is perfect, which implies that any $k$-linear map $t' : H^ d(X, \Omega ^ d_{X/k}) \to k$ is of the form $\xi \mapsto t(g\xi )$ for some $g \in H^0(X, \mathcal{O}_ X)$. Of course, in order for $t'$ to still produce a duality between $H^0(X, \mathcal{O}_ X)$ and $H^ d(X, \Omega ^ d_{X/k})$ we need $g$ to be a unit. Denote $\langle -, - \rangle _{p, q}$ the pairing constructed using $t$ and denote $\langle -, - \rangle '_{p, q}$ the pairing constructed using $t'$. Clearly we have

for $\xi \in H^ q(X, \Omega ^ p_{X/k})$ and $\xi ' \in H^{d - q}(X, \Omega ^{d - p}_{X/k})$. Since $g$ is a unit, i.e., invertible, we see that using $t'$ instead of $t$ we still get perfect pairings for all $p, q$. $\square$

Proof. Since $X$ is smooth over $k$ it is geometrically reduced over $k$, see Varieties, Lemma 33.25.4. Hence $H^0(X, \mathcal{O}_ X) = \prod k_ i$ is a finite product of finite separable field extensions $k_ i/k$, see Varieties, Lemma 33.9.3. It follows that $\Omega _{H^0(X, \mathcal{O}_ X)/k} = \prod \Omega _{k_ i/k} = 0$ (see for example Algebra, Lemma 10.158.1). Since the map of the lemma factors as

by functoriality of the de Rham complex (see Section 50.2), we conclude. $\square$

Proof. It is tempting to think this follows from a combination of Lemmas 50.20.2 and 50.20.1. However this doesn't work because the maps $\mathcal{O}_ X \to \Omega ^1_{X/k}$ and $\Omega ^{d - 1}_{X/k} \to \Omega ^ d_{X/k}$ are not $\mathcal{O}_ X$-linear and hence we cannot use the functoriality discussed in Duality for Schemes, Remark 48.27.3 to conclude the map in Lemma 50.20.2 is dual to the one in this lemma.

We may replace $X$ by a connected component of $X$. Hence we may assume $X$ is irreducible. By Varieties, Lemmas 33.25.4 and 33.9.3 we see that $k' = H^0(X, \mathcal{O}_ X)$ is a finite separable extension $k'/k$. Since $\Omega _{k'/k} = 0$ (see for example Algebra, Lemma 10.158.1) we see that $\Omega _{X/k} = \Omega _{X/k'}$ (see Morphisms, Lemma 29.32.9). Thus we may replace $k$ by $k'$ and assume that $H^0(X, \mathcal{O}_ X) = k$.

Assume $H^0(X, \mathcal{O}_ X) = k$. We conclude that $\dim H^ d(X, \Omega ^ d_{X/k}) = 1$ by Lemma 50.20.1. Assume first that the characteristic of $k$ is a prime number $p$. Denote $F_{X/k} : X \to X^{(p)}$ the relative Frobenius of $X$ over $k$; please keep in mind the facts proved about this morphism in Lemma 50.19.5. Consider the commutative diagram

The left two horizontal arrows are isomorphisms as $F_{X/k}$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The right square commutes as $\Theta _{X^{(p)}/X}$ is a morphism of complexes and $\Theta ^{d - 1}$ is zero. Thus it suffices to show that $\Theta ^ d$ is nonzero (because the dimension of the source of the map $\Theta ^ d$ is $1$ by the discussion above). However, we know that

is surjective and hence surjective after applying the right exact functor $H^ d(X^{(p)}, -)$ (right exactness by the vanishing of cohomology beyond $d$ as follows from Cohomology, Proposition 20.20.7). Finally, $H^ d(X^{(d)}, \Omega ^ d_{X^{(d)}/k})$ is nonzero for example because it is dual to $H^0(X^{(d)}, \mathcal{O}_{X^{(p)}})$ by Lemma 50.20.1 applied to $X^{(p)}$ over $k$. This finishes the proof in this case.

Finally, assume the characteristic of $k$ is $0$. We can write $k$ as the filtered colimit of its finite type $\mathbf{Z}$-subalgebras $R$. For one of these we can find a cartesian diagram of schemes

such that $Y \to \mathop{\mathrm{Spec}}(R)$ is smooth of relative dimension $d$ and proper. See Limits, Lemmas 32.10.1, 32.8.9, 32.18.4, and 32.13.1. The modules $M^{i, j} = H^ j(Y, \Omega ^ i_{Y/R})$ are finite $R$-modules, see Cohomology of Schemes, Lemma 30.19.2. Thus after replacing $R$ by a localization we may assume all of these modules are finite free. We have $M^{i, j} \otimes _ R k = H^ j(X, \Omega ^ i_{X/k})$ by flat base change (Cohomology of Schemes, Lemma 30.5.2). Thus it suffices to show that $M^{d - 1, d} \to M^{d, d}$ is zero. This is a map of finite free modules over a domain, hence it suffices to find a dense set of primes $\mathfrak p \subset R$ such that after tensoring with $\kappa (\mathfrak p)$ we get zero. Since $R$ is of finite type over $\mathbf{Z}$, we can take the collection of primes $\mathfrak p$ whose residue field has positive characteristic (details omitted). Observe that

for example by Limits, Lemma 32.19.2. Similarly for $M^{d, d}$. Thus we see that $M^{d - 1, d} \otimes _ R \kappa (\mathfrak p) \to M^{d, d} \otimes _ R \kappa (\mathfrak p)$ is zero by the case of positive characteristic handled above. $\square$

Proof. By the Hodge-to-de Rham spectral sequence (Section 50.6), the vanishing of $\Omega ^ i_{X/k}$ for $i > d$, the vanishing in Cohomology, Proposition 20.20.7 and the results of Lemmas 50.20.2 and 50.20.3 we see that $H^0_{dR}(X/k) = H^0(X, \mathcal{O}_ X)$ and $H^ d(X, \Omega ^ d_{X/k}) = H_{dR}^{2d}(X/k)$. More precisesly, these identifications come from the maps of complexes

Let us choose $t : H_{dR}^{2d}(X/k) \to k$ which via this identification corresponds to a $t$ as in Lemma 50.20.1. Then in any case we see that the pairing displayed in the lemma is perfect for $i = 0$.

Denote $\underline{k}$ the constant sheaf with value $k$ on $X$. Let us abbreviate $\Omega ^\bullet = \Omega ^\bullet _{X/k}$. Consider the map (50.4.0.1) which in our situation reads

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

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

We will prove by induction on $p$ that these cup products via $t$ induce perfect pairings between $H^ i(X, \sigma _{\leq p} \Omega ^\bullet )$ and $H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet )$. For $p = d$ this is the assertion of the proposition.

The base case is $p = 0$. In this case we simply obtain the pairing between $H^ i(X, \mathcal{O}_ X)$ and $H^{d - i}(X, \Omega ^ d)$ of Lemma 50.20.1 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 $\underline{k}$-modules; in particular the maps $\sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p]$ and $\Omega ^{d - p - 1}[-d + p + 1] \to (\sigma _{\geq d - p}\Omega ^\bullet )[1]$ are given by actual maps of complexes, namely using the differential $\Omega ^ p \to \Omega ^{p + 1}$ and the differential $\Omega ^{d - p - 1} \to \Omega ^{d - p}$. Consider the long exact cohomology sequences associated to these distinguished triangles

By induction and Lemma 50.20.1 we know that the pairings constructed above between the $k$-vectorspaces on the first, second, fourth, and fifth rows are perfect. By the $5$-lemma, in order to show that the pairing between the cohomology groups in the middle row is perfect, it suffices to show that the pairs $(a, a')$, $(b, b')$, $(c, c')$, and $(d, d')$ are compatible with the given pairings (see below).

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

Hence if we have $\alpha \in H^ i(X, \sigma _{\leq p + 1}\Omega ^\bullet )$ and $\beta \in H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet )$ then we get $\gamma _ p(\alpha \cup c'(\beta )) = \gamma _{p + 1}(c(\alpha ) \cup \beta )$ by functoriality of the cup product.

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

and argue in the same manner.

For the pair $(d, d')$ we use the commutative diagram

and we look at cohomology classes in $H^ i(X, \sigma _{\leq p}\Omega ^\bullet )$ and $H^{2d - i}(X, \Omega ^{d - p - 1}[-d + p])$. Changing $i$ to $i - 1$ we get the result for the pair $(a, a')$ thereby finishing the proof that our pairings are perfect.

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