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 precisely, these identifications come from the maps of complexes
\[ \Omega ^\bullet _{X/k} \to \mathcal{O}_ X[0] \quad \text{and}\quad \Omega ^ d_{X/k}[-d] \to \Omega ^\bullet _{X/k} \]
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
\[ \wedge : \text{Tot}(\Omega ^\bullet \otimes _{\underline{k}} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]
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
\[ \gamma _ p : \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \longrightarrow \Omega ^\bullet \]
Using the same procedure as in Section 50.4 we obtain cup products
\[ H^ i(X, \sigma _{\leq p} \Omega ^\bullet ) \times H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \longrightarrow H_{dR}^{2d}(X, \Omega ^\bullet ) \]
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
\[ \Omega ^{p + 1}[-p - 1] \to \sigma _{\leq p + 1}\Omega ^\bullet \to \sigma _{\leq p}\Omega ^\bullet \to \Omega ^{p + 1}[-p] \]
and the distinguished triangle
\[ \sigma _{\geq d - p}\Omega ^\bullet \to \sigma _{\geq d - p - 1}\Omega ^\bullet \to \Omega ^{d - p - 1}[-d + p + 1] \to (\sigma _{\geq d - p}\Omega ^\bullet )[1] \]
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
\[ \xymatrix{ H^{i - 1}(X, \sigma _{\leq p}\Omega ^\bullet ) \ar[d]_ a \\ H^ i(X, \Omega ^{p + 1}[-p - 1]) \ar[d]_ b \\ H^ i(X, \sigma _{\leq p + 1}\Omega ^\bullet ) \ar[d]_ c \\ H^ i(X, \sigma _{\leq p}\Omega ^\bullet ) \ar[d]_ d \\ H^{i + 1}(X, \Omega ^{p + 1}[-p - 1]) } \quad \quad \xymatrix{ H^{2d - i + 1}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \\ H^{2d - i}(X, \Omega ^{d - p - 1}[-d + p + 1]) \ar[u]_{a'} \\ H^{2d - i}(X, \sigma _{\geq d - p - 1}\Omega ^\bullet ) \ar[u]_{b'} \\ H^{2d - i}(X, \sigma _{\geq d - p}\Omega ^\bullet ) \ar[u]_{c'} \\ H^{2d - i - 1}(X, \Omega ^{d - p - 1}[-d + p + 1]) \ar[u]_{d'} } \]
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
\[ \xymatrix{ \text{Tot}(\sigma _{\leq p} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \ar[d]_{\gamma _ p} & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[l]_-{\gamma _{p + 1}} } \]
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
\[ \xymatrix{ \text{Tot}(\sigma _{\leq p + 1} \Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[d]_{\gamma _{p + 1}} & \text{Tot}(\Omega ^{p + 1}[-p - 1] \otimes _{\underline{k}} \sigma _{\geq d - p - 1} \Omega ^\bullet ) \ar[l] \ar[d] \\ \Omega ^\bullet & \Omega ^{p + 1}[-p - 1] \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p + 1] \ar[l]_-\wedge } \]
and argue in the same manner.
For the pair $(d, d')$ we use the commutative diagram
\[ \xymatrix{ \Omega ^{p + 1}[-p] \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p] \ar[d] & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{\underline{k}} \Omega ^{d - p - 1}[-d + p]) \ar[l] \ar[d] \\ \Omega ^\bullet & \text{Tot}(\sigma _{\leq p}\Omega ^\bullet \otimes _{\underline{k}} \sigma _{\geq d - p}\Omega ^\bullet ) \ar[l] } \]
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 $H^0(X, \mathcal{O}_ X)$.
$\square$
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