Email from Noah Olander of Jun 8, 2020

Lemma 57.13.5. Let $k$ be a field. Let $X$ be a proper scheme over $k$ which is regular. Let $F : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ X)$ be a $k$-linear exact functor. Assume for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\dim (\text{Supp}(\mathcal{F})) = 0$ there is an isomorphism $\mathcal{F} \cong F(\mathcal{F})$. Then there exists an automorphism $f : X \to X$ over $k$ which induces the identity on the underlying topological space1 and an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ such that $F$ and $F'(M) = f^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{L}$ are siblings.

Proof. By Lemma 57.12.6 the functor $F$ is fully faithful. By Lemma 57.13.4 the essential image of the identity functor is contained in the essential image of $F$, i.e., we see that $F$ is essentially surjective. Thus $F$ is an equivalence. Observe that the quasi-inverse $F^{-1}$ satisfies the same assumptions as $F$.

Let $M \in D_{perf}(\mathcal{O}_ X)$ and say $H^ i(M) = 0$ for $i > b$. Since $F$ is fully faithful, we see that

$\mathop{\mathrm{Hom}}\nolimits _ X(M, \mathcal{O}_ x[-i]) = \mathop{\mathrm{Hom}}\nolimits _ X(F(M), F(\mathcal{O}_ x)[-i]) \cong \mathop{\mathrm{Hom}}\nolimits _ X(F(M), \mathcal{O}_ x[-i])$

for any $i \in \mathbf{Z}$ for any closed point $x$ of $X$. Thus by Lemma 57.12.4 we see that $F(M)$ has vanishing cohomology sheaves in degrees $> b$.

Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. By the above $F(\mathcal{F})$ has nonzero cohomology sheaves only in degrees $\leq 0$. Set $\mathcal{G} = H^0(F(\mathcal{F}))$. Choose a distinguished triangle

$K \to F(\mathcal{F}) \to \mathcal{G} \to K[1]$

Then $K$ has nonvanishing cohomology sheaves only in degrees $\leq -1$. Applying $F^{-1}$ we obtain a distinguished triangle

$F^{-1}(K) \to \mathcal{F} \to F^{-1}(\mathcal{G}) \to F^{-1}(K')[1]$

Since $F^{-1}(K)$ has nonvanishing cohomology sheaves only in degrees $\leq -1$ (by the previous paragraph applied to $F^{-1}$) we see that the arrow $F^{-1}(K) \to \mathcal{F}$ is zero (Derived Categories, Lemma 13.27.3). Hence $K \to F(\mathcal{F})$ is zero, which implies that $F(\mathcal{F}) = \mathcal{G}$ by our choice of the first distinguished triangle.

From the preceding paragraph, we deduce that $F$ preserves $\textit{Coh}(\mathcal{O}_ X)$ and indeed defines an equivalence $H : \textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ X)$. By Functors and Morphisms, Lemma 56.7.8 we get an automorphism $f : X \to X$ over $k$ and an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ such that $H(\mathcal{F}) = f^*\mathcal{F} \otimes \mathcal{L}$. Set $F'(M) = f^*M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{L}$. Using Lemma 57.11.2 we see that $F$ and $F'$ are siblings. To see that $f$ is the identity on the underlying topological space of $X$, we use that $F(\mathcal{O}_ x) \cong \mathcal{O}_ x$ and that the support of $\mathcal{O}_ x$ is $\{ x\}$. This finishes the proof. $\square$

[1] This often forces $f$ to be the identity, see Varieties, Lemma 33.32.1.

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