The Stacks project

Lemma 115.13.3. Let $X$ be a proper scheme over a field $k$ which is regular.

  1. Let $F : D_{perf}(\mathcal{O}_ X)^{opp} \to \text{Vect}_ k$ be a $k$-linear cohomological functor such that

    \[ \sum \nolimits _{n \in \mathbf{Z}} \dim _ k F(E[n]) < \infty \]

    for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $F$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.

  2. Let $G : D_{perf}(\mathcal{O}_ X) \to \text{Vect}_ k$ be a $k$-linear homological functor such that

    \[ \sum \nolimits _{n \in \mathbf{Z}} \dim _ k G(E[n]) < \infty \]

    for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $G$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(K, E)$ for some $K \in D_{perf}(\mathcal{O}_ X)$.

Proof. This follows from Derived Categories of Varieties, Theorem 57.6.3 and Lemma 57.6.4. We also give another proof below.

Proof of (1). The derived category $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums, is compactly generated, and $D_{perf}(\mathcal{O}_ X)$ is the full subcategory of compact objects, see Derived Categories of Schemes, Lemma 36.3.1, Theorem 36.15.3, and Proposition 36.17.1. By Derived Categories of Varieties, Lemma 57.6.2 we may assume $F(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. Then it follows that $K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Lemma 115.13.2.

Proof of (2). Consider the contravariant functor $E \mapsto E^\vee $ on $D_{perf}(\mathcal{O}_ X)$, see Cohomology, Lemma 20.50.5. This functor is an exact anti-self-equivalence of $D_{perf}(\mathcal{O}_ X)$. Hence we may apply part (1) to the functor $F(E) = G(E^\vee )$ to find $K \in D_{perf}(\mathcal{O}_ X)$ such that $G(E^\vee ) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$. It follows that $G(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E^\vee , K) = \mathop{\mathrm{Hom}}\nolimits _ X(K^\vee , E)$ and we conclude that taking $K^\vee $ works. $\square$


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