Remark 48.27.2. Let $k$, $X$, and $\omega _ X^\bullet$ be as in Lemma 48.27.1. The identity on the complex $\omega _ X^\bullet$ corresponds, via the functorial isomorphism in part (5), to a map

$t : H^0(X, \omega _ X^\bullet ) \longrightarrow k$

For an arbitrary $K$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ the identification $\mathop{\mathrm{Hom}}\nolimits (K, \omega _ X^\bullet )$ with $H^0(X, K)^\vee$ in part (5) corresponds to the pairing

$\mathop{\mathrm{Hom}}\nolimits _ X(K, \omega _ X^\bullet ) \times H^0(X, K) \longrightarrow k,\quad (\alpha , \beta ) \longmapsto t(\alpha (\beta ))$

This follows from the functoriality of the isomorphisms in (5). Similarly for any $i \in \mathbf{Z}$ we get the pairing

$\mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \omega _ X^\bullet ) \times H^{-i}(X, K) \longrightarrow k,\quad (\alpha , \beta ) \longmapsto t(\alpha (\beta ))$

Here we think of $\alpha$ as a morphism $K[-i] \to \omega _ X^\bullet$ and $\beta$ as an element of $H^0(X, K[-i])$ in order to define $\alpha (\beta )$. Observe that if $K$ is general, then we only know that this pairing is nondegenerate on one side: the pairing induces an isomorphism of $\mathop{\mathrm{Hom}}\nolimits _ X(K, \omega _ X^\bullet )$, resp. $\mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \omega _ X^\bullet )$ with the $k$-linear dual of $H^0(X, K)$, resp. $H^{-i}(X, K)$ but in general not vice versa. If $K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$, then $\mathop{\mathrm{Hom}}\nolimits _ X(K, \omega _ X^\bullet )$, $\mathop{\mathrm{Ext}}\nolimits _ X(K, \omega _ X^\bullet )$, $H^0(X, K)$, and $H^ i(X, K)$ are finite dimensional $k$-vector spaces (by Derived Categories of Schemes, Lemmas 36.11.5 and 36.11.4) and the pairings are perfect in the usual sense.

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