The Stacks project

57.6 A representability theorem

The material in this section is taken from [BvdB].

Let $\mathcal{T}$ be a $k$-linear triangulated category. In this section we consider $k$-linear cohomological functors $H$ from $\mathcal{T}$ to the category of $k$-vector spaces. This will mean $H$ is a functor

\[ H : \mathcal{T}^{opp} \longrightarrow \text{Vect}_ k \]

which is $k$-linear such that for any distinguished triangle $X \to Y \to Z$ in $\mathcal{T}$ the sequence $H(Z) \to H(Y) \to H(X)$ is an exact sequence of $k$-vector spaces. See Derived Categories, Definition 13.3.5 and Differential Graded Algebra, Section 22.24.

Lemma 57.6.1. Let $\mathcal{D}$ be a triangulated category. Let $\mathcal{D}' \subset \mathcal{D}$ be a full triangulated subcategory. Let $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. The category of arrows $E \to X$ with $E \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}')$ is filtered.

Proof. We check the conditions of Categories, Definition 4.19.1. The category is nonempty because it contains $0 \to X$. If $E_ i \to X$, $i = 1, 2$ are objects, then $E_1 \oplus E_2 \to X$ is an object and there are morphisms $(E_ i \to X) \to (E_1 \oplus E_2 \to X)$. Finally, suppose that $a, b : (E \to X) \to (E' \to X)$ are morphisms. Choose a distinguished triangle $E \xrightarrow {a - b} E' \to E''$ in $\mathcal{D}'$. By Axiom TR3 we obtain a morphism of triangles

\[ \xymatrix{ E \ar[r]_{a - b} \ar[d] & E' \ar[d] \ar[r] & E'' \ar[d] \\ 0 \ar[r] & X \ar[r] & X } \]

and we find that the resulting arrow $(E' \to X) \to (E'' \to X)$ equalizes $a$ and $b$. $\square$


Lemma 57.6.2. Let $k$ be a field. Let $\mathcal{D}$ be a $k$-linear triangulated category which has direct sums and is compactly generated. Denote $\mathcal{D}_ c$ the full subcategory of compact objects. Let $H : \mathcal{D}_ c^{opp} \to \text{Vect}_ k$ be a $k$-linear cohomological functor such that $\dim _ k H(X) < \infty $ for all $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$. Then $H$ is isomorphic to the functor $X \mapsto \mathop{\mathrm{Hom}}\nolimits (X, Y)$ for some $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$.

Proof. We will use Derived Categories, Lemma 13.37.2 without further mention. Denote $G : \mathcal{D}_ c \to \text{Vect}_ k$ the $k$-linear homological functor which sends $X$ to $H(X)^\vee $. For any object $Y$ of $\mathcal{D}$ we set

\[ G'(Y) = \mathop{\mathrm{colim}}\nolimits _{X \to Y, X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)} G(X) \]

The colimit is filtered by Lemma 57.6.1. We claim that $G'$ is a $k$-linear homological functor, the restriction of $G'$ to $\mathcal{D}_ c$ is $G$, and $G'$ sends direct sums to direct sums.

Namely, suppose that $Y_1 \to Y_2 \to Y_3$ is a distinguished triangle. Let $\xi \in G'(Y_2)$ map to zero in $G'(Y_3)$. Since the colimit is filtered $\xi $ is represented by some $X \to Y_2$ with $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$ and $g \in G(X)$. The fact that $\xi $ maps to zero in $G'(Y_3)$ means the composition $X \to Y_2 \to Y_3$ factors as $X \to X' \to Y_3$ with $X' \in \mathcal{D}_ c$ and $g$ mapping to zero in $G(X')$. Choose a distinguished triangle $X'' \to X \to X'$. Then $X'' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$. Since $G$ is homological we find that $g$ is the image of some $g'' \in G'(X'')$. By Axiom TR3 the maps $X \to Y_2$ and $X' \to Y_3$ fit into a morphism of distinguished triangles $(X'' \to X \to X') \to (Y_1 \to Y_2 \to Y_3)$ and we find that indeed $\xi $ is the image of the element of $G'(Y_1)$ represented by $X'' \to Y_1$ and $g'' \in G(X'')$.

If $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$, then $\text{id} : Y \to Y$ is the final object in the category of arrows $X \to Y$ with $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$. Hence we see that $G'(Y) = G(Y)$ in this case and the statement on restriction holds. Let $Y = \bigoplus _{i \in I} Y_ i$ be a direct sum. Let $a : X \to Y$ with $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$ and $g \in G(X)$ represent an element $\xi $ of $G'(Y)$. The morphism $a : X \to Y$ can be uniquely written as a sum of morphisms $a_ i : X \to Y_ i$ almost all zero as $X$ is a compact object of $\mathcal{D}$. Let $I' = \{ i \in I \mid a_ i \not= 0\} $. Then we can factor $a$ as the composition

\[ X \xrightarrow {(1, \ldots , 1)} \bigoplus \nolimits _{i \in I'} X \xrightarrow {\bigoplus _{i \in I'} a_ i} \bigoplus \nolimits _{i \in I} Y_ i = Y \]

We conclude that $\xi = \sum _{i \in I'} \xi _ i$ is the sum of the images of the elements $\xi _ i \in G'(Y_ i)$ corresponding to $a_ i : X \to Y_ i$ and $g \in G(X)$. Hence $\bigoplus G'(Y_ i) \to G'(Y)$ is surjective. We omit the (trivial) verification that it is injective.

It follows that the functor $Y \mapsto G'(Y)^\vee $ is cohomological and sends direct sums to direct products. Hence by Brown representability, see Derived Categories, Proposition 13.38.2 we conclude that there exists a $Y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ and an isomorphism $G'(Z)^\vee = \mathop{\mathrm{Hom}}\nolimits (Z, Y)$ functorially in $Z$. For $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D}_ c)$ we have $G'(X)^\vee = G(X)^\vee = (H(X)^\vee )^\vee = H(X)$ because $\dim _ k H(X) < \infty $ and the proof is complete. $\square$


Theorem 57.6.3. Let $X$ be a projective scheme over a field $k$. Let $F : D_{perf}(\mathcal{O}_ X)^{opp} \to \text{Vect}_ k$ be a $k$-linear cohomological functor such that

\[ \sum \nolimits _{n \in \mathbf{Z}} \dim _ k F(E[n]) < \infty \]

for all $E \in D_{perf}(\mathcal{O}_ X)$. Then $F$ is isomorphic to a functor of the form $E \mapsto \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in D^ b_{\textit{Coh}}(\mathcal{O}_ X)$.

Proof. The derived category $D_\mathit{QCoh}(\mathcal{O}_ X)$ has direct sums, is compactly generated, and $D_{perf}(\mathcal{O}_ X)$ is the full subcategory of compact objects, see Derived Categories of Schemes, Lemma 36.3.1, Theorem 36.15.3, and Proposition 36.17.1. By Lemma 57.6.2 we may assume $F(E) = \mathop{\mathrm{Hom}}\nolimits _ X(E, K)$ for some $K \in \mathop{\mathrm{Ob}}\nolimits (D_\mathit{QCoh}(\mathcal{O}_ X))$. Then it follows that $K$ is in $D^ b_{\textit{Coh}}(\mathcal{O}_ X)$ by Lemma 57.5.4. $\square$

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