Email from Noah Olander of Jun 8, 2020

Lemma 56.15.6. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$. 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 56.14.5 the functor $F$ is fully faithful. We claim that Lemma 56.15.4 applies to $F$. Namely, for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\dim (\text{Supp}(\mathcal{F})) = 0$ there is an isomorphism of $k$-vector spaces

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, M) = \mathop{\mathrm{Hom}}\nolimits _ X(F(\mathcal{F}), F(M)) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, F(M))$

functorial in $M$ in $D_{perf}(\mathcal{O}_ X)$. The first equality because $F$ is fully faithful. $\square$

Second proof. The proof is similar to the proof of Lemma 56.15.4 and we only sketch the arguments.

Step 1. By Lemma 56.14.5 the functor $F$ is fully faithful.

Step 2. For a closed point $x \in X$ we denote $\mathcal{O}_ x$ the skyscraper sheaf with value $\kappa (x)$ at $x$. An object $K$ of $D_{perf}(\mathcal{O}_ X)$ is isomorphic to a locally free $\mathcal{O}_ X$-module of rank $r$ placed in degree $0$ if and only if

$\dim \mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \mathcal{O}_ x) = 0 \text{ unless }i = 0\text{ and } \dim \mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \mathcal{O}_ x) = r$

Since $F(\mathcal{O}_ x) \cong \mathcal{O}_ x$ and $F$ is fully faithful we find

$\dim \mathop{\mathrm{Ext}}\nolimits ^ i_ X(K, \mathcal{O}_ x) = \dim \mathop{\mathrm{Ext}}\nolimits ^ i(F(K), \mathcal{O}_ x)$

for all $i \in \mathbf{Z}$. Thus, if $\mathcal{E}$ is a locally free $\mathcal{O}_ X$-module of rank $r$, then $F(\mathcal{E})$ is a locally free $\mathcal{O}_ X$-module of rank $r$ placed in degree $0$. In particular, we have $F(\mathcal{O}_ X) = \mathcal{L}[0]$ for some invertible $\mathcal{O}_ X$-module $\mathcal{L}$.

Step 3. Let $U \subset X$ be a dense affine open with complement $D \subset X$. Then $D = \sum _{i = 1, \ldots , r} D_ i$ is a sum of prime divisors each of which is an effective Cartier divisor, see Divisors, Lemma 31.16.7 and its proof. By Step 2 we have invertible modules $\mathcal{L}_ n$ such that $F(\mathcal{O}_ X(-nD)) = \mathcal{L}_ n[0]$. The functor $F$ transforms the maps

$\mathcal{O}_ X(-(n + 1)D) \to \mathcal{O}_ X(-nD) \to \ldots \to \mathcal{O}_ X(-D) \to \mathcal{O}_ X$

into maps $\mathcal{L}_{n + 1} \to \mathcal{L}_ n$ such that the induced map

$\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{L}_{n + 1}, \mathcal{O}_ x) \to \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{L}_ n, \mathcal{O}_ x)$

is nonzero if $x \in U$ and zero if $x \in D$. We conclude that we may write

$\mathcal{L}_ n = \mathcal{L}(\sum e_ i(n) D_ i)$

with $e_ i(n + 1) < e_ i(n) < \ldots < e_ i(0) = 0$. It follows that for $K \in D_{perf}(\mathcal{O}_ X)$ we have

\begin{align*} \Gamma (U, K) & \cong \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ X(-nD), K) \\ & \cong \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits (\mathcal{L}_ n, F(K)) \\ & \cong \mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_ X(-nD), F(K) \otimes \mathcal{L}^{\otimes -1}) \\ & \cong \Gamma (U, F(K) \otimes \mathcal{L}^{\otimes -1}) \end{align*}

The first and last isomorphisms follow from Cohomology of Schemes, Lemma 30.10.4 (details omitted). The second isomorphism holds because $F$ is fully faithful. The third isomorphism holds by comparing the system of divisors $-nD$ for varying $n$ with the system of divisors $\sum e_ i(n) D_ i$ for varying $n$.

Step 4. The functor $F$ sends $\textit{Coh}(\mathcal{O}_ X)$ to $\textit{Coh}(\mathcal{O}_ X)$. Namely, this is clear from the formula in Step 3.

Step 5. By Lemma 56.15.2 we find $F$ is a sibling to the Fourier-Mukai functor $F'$ given by a coherent $\mathcal{O}_{X \times X}$-module $\mathcal{K}$ flat over $X$ via $\text{pr}_1$ and finite over $X$ via $\text{pr}_2$. Since $F(\mathcal{O}_ X) = \mathcal{L}[0]$ we see that

$\mathcal{L} \cong F(\mathcal{O}_ X) \cong F'(\mathcal{O}_ X) \cong \text{pr}_{2, *}\mathcal{K}$

Thus by Lemma 56.12.4 there is a morphism $s : X \to X \times X$ with $\text{pr}_2 \circ s = \text{id}_ X$ such that $\mathcal{K} = s_*\mathcal{L}$. Set $f = \text{pr}_1 \circ s$. Then we have

\begin{align*} F'(M) & = R\text{pr}_{2, *}(L\text{pr}_1^*K \otimes \mathcal{K}) \\ & = R\text{pr}_{2, *}(L\text{pr}_1^*M \otimes s_*\mathcal{L}) \\ & = R\text{pr}_{2, *}(Rs_*(Lf^*M \otimes \mathcal{L})) \\ & = Lf^*M \otimes \mathcal{L} \end{align*}

where we have used Derived Categories of Schemes, Lemma 36.22.1 in the third step. Since for all closed points $x \in X$ the module $F(\mathcal{O}_ x)$ is supported at $x$, we see that $f$ induces the identity on the underlying topological space of $X$.

Step 6. Let $x \in X$ be a closed point. For $n \geq 1$ denote $\mathcal{O}_{x, n}$ the skyscaper sheaf at $x$ with value $\mathcal{O}_{X, x}/\mathfrak m_ x^ n$. Then $\text{End}_ X(\mathcal{O}_{x, n}) = \mathcal{O}_{X, x}/\mathfrak m_ x^ n$. Since $F$ and $F'$ are siblings and since $F$ is fully faithful we see that

$\text{End}_ X(\mathcal{O}_{x, n}) \to \text{End}_ X(F(\mathcal{O}_{x, n})) \cong \text{End}_ X(f^*\mathcal{O}_{x, n} \otimes \mathcal{L}) \cong \text{End}_ X(f^*\mathcal{O}_{x, n})$

are bijections. We conclude that $f^\sharp : \mathcal{O}_{X, x}/\mathfrak m_ x^ n \to \mathcal{O}_{X, x}/\mathfrak m_ x^ n$ is bijective for all $n$. Thus $f$ induces isomorphisms on complete local rings at closed points and hence is étale (Étale Morphisms, Lemma 41.11.3). Looking at closed points we see that $\Delta _ f : X \to X \times _{f, X, f} X$ (which is an open immersion as $f$ is étale) is bijective hence an isomorphism. Hence $f$ is a monomorphism. Finally, we conclude $f$ is an isomorphism as Descent, Lemma 35.22.1 tells us it is an open immersion. $\square$

[1] This often forces $f$ to be the identity, see Lemma 56.15.5.

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