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The Stacks project

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


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