Lemma 57.12.5. 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.
Email from Noah Olander of Jun 8, 2020
Proof. By Lemma 57.11.6 the functor F is fully faithful. By Lemma 57.12.4 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 57.11.4 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 57.10.2 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
Comments (0)