Lemma 57.4.1. Let $k$ be a field. Let $X$ be a proper scheme over $k$ which is Gorenstein. Consider the complex $\omega _ X^\bullet$ of Duality for Schemes, Lemmas 48.27.1. Then the functor

$S : D_{perf}(\mathcal{O}_ X) \longrightarrow D_{perf}(\mathcal{O}_ X),\quad K \longmapsto S(K) = \omega _ X^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} K$

is a Serre functor.

Proof. The statement make sense because $\dim \mathop{\mathrm{Hom}}\nolimits _ X(K, L) < \infty$ for $K, L \in D_{perf}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.7. Since $X$ is Gorenstein the dualizing complex $\omega _ X^\bullet$ is an invertible object of $D(\mathcal{O}_ X)$, see Duality for Schemes, Lemma 48.24.4. In particular, locally on $X$ the complex $\omega _ X^\bullet$ has one nonzero cohomology sheaf which is an invertible module, see Cohomology, Lemma 20.52.2. Thus $S(K)$ lies in $D_{perf}(\mathcal{O}_ X)$. On the other hand, the invertibility of $\omega _ X^\bullet$ clearly implies that $S$ is a self-equivalence of $D_{perf}(\mathcal{O}_ X)$. Finally, we have to find an isomorphism

$c_{K, L} : \mathop{\mathrm{Hom}}\nolimits _ X(K, L) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ X(L, \omega _ X^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} K)^\vee$

bifunctorially in $K, L$. To do this we use the canonical isomorphisms

$\mathop{\mathrm{Hom}}\nolimits _ X(K, L) = H^0(X, L \otimes _{\mathcal{O}_ X}^\mathbf {L} K^\vee )$

and

$\mathop{\mathrm{Hom}}\nolimits _ X(L, \omega _ X^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} K) = H^0(X, \omega _ X^\bullet \otimes _{\mathcal{O}_ X}^\mathbf {L} K \otimes _{\mathcal{O}_ X}^\mathbf {L} L^\vee )$

given in Cohomology, Lemma 20.50.5. Since $(L \otimes _{\mathcal{O}_ X}^\mathbf {L} K^\vee )^\vee = (K^\vee )^\vee \otimes _{\mathcal{O}_ X}^\mathbf {L} L^\vee$ and since there is a canonical isomorphism $K \to (K^\vee )^\vee$ we find these $k$-vector spaces are canonically dual by Duality for Schemes, Lemma 48.27.4. This produces the isomorphisms $c_{K, L}$. We omit the proof that these isomorphisms are functorial. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FY8. Beware of the difference between the letter 'O' and the digit '0'.