Lemma 56.16.5. Let $k$ be a field. Let $X$ be a smooth proper scheme over $k$. Let $K \in D_{perf}(\mathcal{O}_{X \times X})$. If the Fourier-Mukai functor $\Phi _ K : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ X)$ is isomorphic to the identity functor, then $K \cong \Delta _*\mathcal{O}_ X$ in $_{perf}(\mathcal{O}_{X \times X})$.

Proof. Let $i$ be the minimal integer such that the cohomology sheaf $H^ i(K)$ is nonzero. Let $\mathcal{E}$ and $\mathcal{G}$ be finite locally free $\mathcal{O}_ X$-modules. Then

\begin{align*} H^ i(X \times X, K \otimes _{\mathcal{O}_{X \times X}}^\mathbf {L} (\mathcal{E} \boxtimes \mathcal{G})) & = H^ i(X, R\text{pr}_{2, *}(K \otimes _{\mathcal{O}_{X \times X}}^\mathbf {L} (\mathcal{E} \boxtimes \mathcal{G}))) \\ & = H^ i(X, \Phi _ K(\mathcal{E}) \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{G}) \\ & \cong H^ i(X, \mathcal{E} \otimes \mathcal{G}) \end{align*}

which is zero if $i < 0$. On the other hand, we can choose $\mathcal{E}$ and $\mathcal{G}$ such that there is a surjection $\mathcal{E}^\vee \boxtimes \mathcal{G}^\vee \to H^ i(K)$ by Lemma 56.10.1. In this case the left hand side of the equalities is nonzero. Hence we conclude that $H^ i(K) = 0$ for $i < 0$.

Let $i$ be the maximal integer such that $H^ i(K)$ is nonzero. The same argument with $\mathcal{E}$ and $\mathcal{G}$ support of dimension $0$ shows that $i \leq 0$. Hence we conclude that $K$ is given by a single coherent $\mathcal{O}_{X \times X}$-module $\mathcal{K}$ sitting in degree $0$.

Since $R\text{pr}_{2, *}(\text{pr}_1^*\mathcal{F} \otimes \mathcal{K})$ is $\mathcal{F}$, by taking $\mathcal{F}$ supported at closed points we see that the support of $\mathcal{K}$ is finite over $X$ via $\text{pr}_2$. Since $R\text{pr}_{2, *}(\mathcal{K}) \cong \mathcal{O}_ X$ we conclude by Lemma 56.12.4 that $\mathcal{K} = s_*\mathcal{O}_ X$ for some section $s : X \to X \times X$ of the second projection. Then $\Phi _ K(M) = f^*M$ where $f = \text{pr}_1 \circ s$ and this can happen only if $s$ is the diagonal morphism as desired. $\square$

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