The Stacks project

56.15 Special functors

In this section we prove some results on functors of a special type that we will use later in this chapter.

Definition 56.15.1. Let $k$ be a field. Let $X$, $Y$ be finite type schemes over $k$. Recall that $D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D^ b(\textit{Coh}(\mathcal{O}_ X))$ by Derived Categories of Schemes, Proposition 36.11.2. We say two $k$-linear exact functors

\[ F, F' : D^ b_{\textit{Coh}}(\mathcal{O}_ X) = D^ b(\textit{Coh}(\mathcal{O}_ X)) \longrightarrow D^ b_{\textit{Coh}}(\mathcal{O}_ Y) \]

are siblings, or we say $F'$ is a sibling of $F$ if $F$ and $F'$ are siblings in the sense of Definition 56.13.1 with abelian category being $\textit{Coh}(\mathcal{O}_ X)$. If $X$ is regular then $D_{perf}(\mathcal{O}_ X) = D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.11.6 and we use the same terminology for $k$-linear exact functors $F, F' : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$.

Lemma 56.15.2. Let $k$ be a field. Let $X$, $Y$ be finite type schemes over $k$ with $X$ separated. Let $F : D^ b_{\textit{Coh}}(\mathcal{O}_ X) \to D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ be a $k$-linear exact functor sending $\textit{Coh}(\mathcal{O}_ X) \subset D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ into $\textit{Coh}(\mathcal{O}_ Y) \subset D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$. Then there exists a Fourier-Mukai functor $F' : D^ b_{\textit{Coh}}(\mathcal{O}_ X) \to D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ whose kernel is a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ flat over $X$ and with support finite over $Y$ which is a sibling of $F$.

Proof. Denote $H : \textit{Coh}(\mathcal{O}_ X) \to \textit{Coh}(\mathcal{O}_ Y)$ the restriction of $F$. Since $F$ is an exact functor of triangulated categories, we see that $H$ is an exact functor of abelian categories. Of course $H$ is $k$-linear as $F$ is. By Lemma 56.12.3 we obtain a coherent $\mathcal{O}_{X \times Y}$-module $\mathcal{K}$ which is flat over $X$ and has support finite over $Y$. Let $F'$ be the Fourier-Mukai functor defined using $\mathcal{K}$ so that $F'$ restricts to $H$ on $ \textit{Coh}(\mathcal{O}_ X)$. The functor $F'$ sends $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ into $D^ b_{\textit{Coh}}(\mathcal{O}_ Y)$ by Lemma 56.9.5. Observe that $F$ and $F'$ satisfy the first and second condition of Lemma 56.13.2 and hence are siblings. $\square$

Remark 56.15.3. If $F, F' : D^ b_{\textit{Coh}}(\mathcal{O}_ X) \to \mathcal{D}$ are siblings, $F$ is fully faithful, and $X$ is reduced and projective over $k$ then $F \cong F'$; this follows from Proposition 56.13.4 via the argument given in the proof of Theorem 56.16.3. However, in general we do not know whether siblings are isomorphic. Even in the situation of Lemma 56.15.2 it seems difficult to prove that the siblings $F$ and $F'$ are isomorphic functors. If $X$ is smooth and proper over $k$ and $F$ is fully faithful, then $F \cong F'$ as is shown in [Noah]. If you have a proof or a counter example in more general situations, please email stacks.project@gmail.com.

Lemma 56.15.4. Let $k$ be a field. Let $X$ be a separated scheme of finite type 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 of $k$-vector spaces

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

functorial in $M$ in $D_{perf}(\mathcal{O}_ X)$. 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 56.14.2 we conclude that for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ whose support is a closed point there are isomorphisms

\[ H^0(X, M \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) = H^0(X, F(M) \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{F}) \]

functorial in $M$.

Let $x \in X$ be a closed point and apply the above with $\mathcal{F} = \mathcal{O}_ x$ the skyscraper sheaf with value $\kappa (x)$ at $x$. We find

\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(M_ x, \kappa (x)) = \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ p(F(M)_ x, \kappa (x)) \]

for all $p \in \mathbf{Z}$. In particular, if $H^ i(M) = 0$ for $i > 0$, then $H^ i(F(M)) = 0$ for $i > 0$ by Lemma 56.14.3.

If $\mathcal{E}$ is locally free of rank $r$, then $F(\mathcal{E})$ is locally free of rank $r$. This is true because a perfect complex $K$ over $\mathcal{O}_{X, x}$ with

\[ \dim _{\kappa (x)} \text{Tor}^{\mathcal{O}_{X, x}}_ i(K, \kappa (x)) = \left\{ \begin{matrix} r & \text{if} & i = 0 \\ 0 & \text{if} & i \not= 0 \end{matrix} \right. \]

is equal to a free module of rank $r$ placed in degree $0$. See for example More on Algebra, Lemma 15.74.6.

If $M$ is supported on a closed subscheme $Z \subset X$, then $F(M)$ is also supported on $Z$. This is clear because we will have $M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{O}_ x = 0$ for $x \not\in Z$ and hence the same will be true for $F(M)$ and hence we get the conclusion from Lemma 56.14.3.

In particular $F(\mathcal{O}_ x)$ is supported at $\{ x\} $. Let $i \in \mathbf{Z}$ be the minimal integer such that $H^ i(\mathcal{O}_ x) \not= 0$. We know that $i \leq 0$. If $i < 0$, then there is a morphism $\mathcal{O}_ x[-i] \to F(\mathcal{O}_ x)$ which contradicts the fact that all morphisms $\mathcal{O}_ x[-i] \to \mathcal{O}_ x$ are zero. Thus $F(\mathcal{O}_ x) = \mathcal{H}[0]$ where $\mathcal{H}$ is a skyscraper sheaf at $x$.

Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module with $\dim (\text{Supp}(\mathcal{G})) = 0$. Then there exists a filtration

\[ 0 = \mathcal{G}_0 \subset \mathcal{G}_1 \subset \ldots \subset \mathcal{G}_ n = \mathcal{G} \]

such that for $n \geq i \geq 1$ the quotient $\mathcal{G}_ i/\mathcal{G}_{i - 1}$ is isomorphic to $\mathcal{O}_{x_ i}$ for some closed point $x_ i \in X$. Then we get distinguished triangles

\[ F(\mathcal{G}_{i - 1}) \to F(\mathcal{G}_ i) \to F(\mathcal{O}_{x_ i}) \]

and using induction we find that $F(\mathcal{G}_ i)$ is a coherent sheaf placed in degree $0$.

Let $\mathcal{G}$ be a coherent $\mathcal{O}_ X$-module. We know that $H^ i(F(\mathcal{G})) = 0$ for $i > 0$. To get a contradiction assume that $H^ i(F(\mathcal{G}))$ is nonzero for some $i < 0$. We choose $i$ minimal with this property so that we have a morphism $H^ i(F(\mathcal{G}))[-i] \to F(\mathcal{G})$ in $D_{perf}(\mathcal{O}_ X)$. Choose a closed point $x \in X$ in the support of $H^ i(F(\mathcal{G}))$. By More on Algebra, Lemma 15.98.2 there exists an $n > 0$ such that

\[ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]

is nonzero. Next, we take $m \geq 1$ and we consider the short exact sequence

\[ 0 \to \mathfrak m_ x^ m \mathcal{G} \to \mathcal{G} \to \mathcal{G}/\mathfrak m_ x^ m\mathcal{G} \to 0 \]

By the above we know that $F(\mathcal{G}/\mathfrak m_ x^ m\mathcal{G})$ is a sheaf placed in degree $0$. Hence $H^ i(F(\mathfrak m_ x^ m \mathcal{G})) \to H^ i(F(\mathcal{G}))$ is an isomorphism. Consider the commutative diagram

\[ \xymatrix{ H^ i(F(\mathfrak m_ x^ m\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] \ar[d] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathfrak m_ x^ m\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \ar[d] \\ H^ i(F(\mathcal{G}))_ x \otimes _{\mathcal{O}_{X, x}} \mathcal{O}_{X, x}/\mathfrak m_ x^ n \ar[r] & \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(F(\mathcal{G})_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) } \]

Since the left vertical arrow is an isomorphism and the bottom arrow is nonzero, we conclude that the right vertical arrow is nonzero for all $m \geq 1$. On the other hand, by the first paragraph of the proof, we know this arrow is isomorphic to the arrow

\[ \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathfrak m_ x^ m\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \longrightarrow \text{Tor}^{\mathcal{O}_{X, x}}_{-i}(\mathcal{G}_ x, \mathcal{O}_{X, x}/\mathfrak m_ x^ n) \]

However, this arrow is zero for $m \gg n$ by More on Algebra, Lemma 15.100.2 which is the contradiction we're looking for.

Thus we know that $F$ preserves coherent modules. 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 $F(\mathcal{O}_ X)$ is an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ placed in degree $0$ 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 $X$. We still have to show that $f$ is an isomorphism which we will do in the next paragraph.

Let $x \in X$ be a closed point. For $n \geq 1$ denote $\mathcal{O}_{x, n}$ the skyscaper sheaf at $x$ with value $\mathcal{O}_{X, x}/\mathfrak m_ x^ n$. We have

\[ \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n}) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, F(\mathcal{O}_{x, n})) \cong \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, f^*\mathcal{O}_{x, n} \otimes \mathcal{L}) \]

functorially with respect to $\mathcal{O}_ X$-module homomorphisms between the $\mathcal{O}_{x, n}$. (The first isomorphism exists by assumption and the second isomorphism because $F$ and $F'$ are siblings.) For $m \geq n$ we have $\mathcal{O}_{X, x}/\mathfrak m^ n = \mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{O}_{x, m}, \mathcal{O}_{x, n})$ via the action on $\mathcal{O}_{x, n}$ 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. $\square$

Lemma 56.15.5. Let $X$ be a reduced scheme of finite type over a field $k$. Let $f : X \to X$ be an automorphism over $k$ which induces the identity map on the underlying topological space of $X$. Then

  1. $f^*\mathcal{F} \cong \mathcal{F}$ for every coherent $\mathcal{O}_ X$-module, and

  2. if $\dim (Z) > 0$ for every irreducible component $Z \subset X$, then $f$ is the identity.

Proof. Part (1) follows from part (2) and the fact that the connected components of $X$ of dimension $0$ are spectra of fields.

Let $Z \subset X$ be an irreducible component viewed as an integral closed subscheme. Clearly $f(Z) \subset Z$ and $f|_ Z : Z \to Z$ is an automorphism over $k$ which induces the identity map on the underlying topological space of $Z$. Since $X$ is reduced, it suffices to show that the arrows $f|_ Z : Z \to Z$ are the identity. This reduces us to the case discussed in the next paragraph.

Assume $X$ is irreducible of dimension $> 0$. Choose a nonempty affine open $U \subset X$. Since $f(U) \subset U$ and since $U \subset X$ is scheme theoretically dense it suffices to prove that $f|_ U : U \to U$ is the identity.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine, irreducible, of dimension $> 0$ and $k$ is an infinite field. Let $g \in A$ be nonconstant. The set

\[ S = \bigcup \nolimits _{\lambda \in k} V(g - \lambda ) \]

is dense in $X$ because it is the inverse image of the dense subset $\mathbf{A}^1_ k(k)$ by the nonconstant morphism $g : X \to \mathbf{A}^1_ k$. If $x \in S$, then the image $g(x)$ of $g$ in $\kappa (x)$ is in the image of $k \to \kappa (x)$. Hence $f^\sharp : \kappa (x) \to \kappa (x)$ fixes $g(x)$. Thus the image of $f^\sharp (g)$ in $\kappa (x)$ is equal to $g(x)$. We conclude that

\[ S \subset V(g - f^\sharp (g)) \]

and since $X$ is reduced and $S$ is dense we conclude $g=f^\sharp (g)$. This proves $f^\sharp = \text{id}_ A$ as $A$ is generated as a $k$-algebra by elements $g$ as above (details omitted; hint: the set of constant functions is a finite dimensional $k$-subvector space of $A$). We conclude that $f = \text{id}_ X$.

Assume $X = \mathop{\mathrm{Spec}}(A)$ is affine, irreducible, of dimension $> 0$ and $k$ is a finite field. If for every $1$-dimensional integral closed subscheme $C \subset X$ the restriction $f|_ C : C \to C$ is the identity, then $f$ is the identity. This reduces us to the case where $X$ is a curve. A curve over a finite field has a finite automorphism group (details omitted). Hence $f$ has finite order, say $n$. Then we pick $g : X \to \mathbf{A}^1_ k$ nonconstant as above and we consider

\[ S = \{ x \in X\text{ closed such that }[\kappa (g(x)) : k] \text{ is prime to }n\} \]

Arguing as before we find that $S$ is dense in $X$. Since for $x \in X$ closed the map $f^\sharp : \kappa (x) \to \kappa (x)$ is an automorphism of order dividing $n$ we see that for $x \in S$ this automorphism acts trivially on the subfield generated by the image of $g$ in $\kappa (x)$. Thus we conclude that $S \subset V(g - f^\sharp (g))$ and we win as before. $\square$

reference

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 space2 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. $\square$

Lemma 56.15.7. Let $k$ be a field. Let $X$, $Y$ be smooth proper schemes over $k$. Let $F, G : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$ be $k$-linear exact functors such that

  1. $F(\mathcal{F}) \cong G(\mathcal{F})$ for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ with $\dim (\text{Supp}(\mathcal{F})) = 0$,

  2. $F$ is fully faithful, and

  3. $G$ is a Fourier-Mukai functor whose kernel is in $D_{perf}(\mathcal{O}_{X \times Y})$.

Then there exists a Fourier-Mukai functor $F' : D_{perf}(\mathcal{O}_ X) \to D_{perf}(\mathcal{O}_ Y)$ whose kernel is in $D_{perf}(\mathcal{O}_{X \times Y})$ such that $F$ and $F'$ are siblings.

Proof. Recall that $F$ has both adjoints, see Lemma 56.8.1. In particular the essential image $\mathcal{A} \subset D_{perf}(\mathcal{O}_ Y)$ of $F$ satisfies the equivalent conditions of Derived Categories, Lemma 13.39.5. We claim that $G$ factors through $\mathcal{A}$. Since $\mathcal{A} = {}^\perp (\mathcal{A}^\perp )$ by Derived Categories, Lemma 13.39.5 it suffices to show that $\mathop{\mathrm{Hom}}\nolimits _ Y(G(M), N) = 0$ for all $M$ in $D_{perf}(\mathcal{O}_ X)$ and $N \in \mathcal{A}^\perp $. We have

\[ \mathop{\mathrm{Hom}}\nolimits _ Y(G(M), N) = \mathop{\mathrm{Hom}}\nolimits _ X(M, G_ r(N)) \]

where $G_ r$ is the right adjoint to $G$. Since $G(\mathcal{F}) \cong F(\mathcal{F})$ for $\mathcal{F}$ as in (1) we see that $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, G_ r(N)) = 0$ by the same formula and the fact that $N$ is in the right orthogonal to the essential image $\mathcal{A}$ of $F$. Of course, the same vanishing holds for $\mathop{\mathrm{Hom}}\nolimits _ X(\mathcal{F}, G_ r(N)[i])$ for any $i \in \mathbf{Z}$. Thus $G_ r(N) = 0$ by Lemma 56.14.3 and the claim holds.

Apply Lemma 56.15.6 to the functor $H = F^{-1} \circ G$ which makes sense because the essential image of $G$ is contained in the essential image of $F$ by the previous paragraph and because $F$ is fully faithful. We obtain an automorphism $f : X \to X$ and an invertible $\mathcal{O}_ X$-module $\mathcal{L}$ such that the functor $H' : K \mapsto f^*K \otimes \mathcal{L}$ is a sibling of $H$. In particular $H$ is an auto-equivalence by Lemma 56.13.3 and $H$ induces an auto-equivalence of $\textit{Coh}(\mathcal{O}_ X)$ (as this is true for its sibling functor $H'$). Thus the quasi-inverses $H^{-1}$ and $(H')^{-1}$ exist, are siblings (small detail omitted), and $(H')^{-1}$ sends $M$ to $(f^{-1})^*(M \otimes _{\mathcal{O}_ X}^\mathbf {L} \mathcal{L}^{\otimes -1})$ which is a Fourier-Mukai functor (details omitted). Then of course $F = G \circ H^{-1}$ is a sibling of $G \circ (H')^{-1}$. Since compositions of Fourier-Mukai functors are Fourier-Mukai by Lemma 56.9.3 we conclude. $\square$

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

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