2021-11-09

Brown Representability

Existence of semi-orthogonal decompositions and if Serre functors strongly depending on representability of functors. It is useful to have strong general criteria for such.

Here we will review a criteria for “large” categories: compactly generated triangulated categories. Next we will cover one for “small” categories, often categories of compact objects in compactly generated triangulated categories.

Recall that a functor $H: \mathcal T^{op} \to \operatorname{Ab}$ is cohomological if for any triangle $X \to Y \to Z \to X[1]$ the resulting sequence $\cdots \rightarrow \operatorname{Hom}(X,W) \rightarrow \operatorname{Hom}(Y,W) \rightarrow \operatorname{Hom}(Z,W) \rightarrow \operatorname{Hom}(X[1],W) \rightarrow \cdots$ is exact.

We say that $H$ takes coproducts to products if the natural map $H\left(\bigoplus X_a \right) \to \prod H(X_a)$ is an isomorphism for any coproduct.

Theorem. (Brown Representability) Let $\mathcal T$ be a compactly generated triangulated category and $H : \mathcal T^{op} \to \operatorname{Ab}$ a functor. Then $H$ is representable if and only if $H$ is cohomological and takes coproducts to products.

Proof. (Expand to view)

We have seen that representable functors must take triangles to long exact sequences and coproducts to products. So assume that $H$ is a cohomological functor which takes coproducts to products and let $\mathcal C$ be a set of compact generators. Set $$ U_0 := \bigcup_{C \in \mathcal C} H(C) $$ so each element is a pair $(C,u)$ with $C \in \mathcal C$ and $u \in H(C)$. Set $$ X_0 := \bigoplus_{(C,u) \in U_0} C $$ If we apply $H$ to $X_0$, we get $$ H(X_0) = \prod_{U_0} H(C) $$ We have the distinguished element $\xi_0 \in H(X_0)$ given by $u \in H(C)$ for the factor $(C,u)$. From Yoneda, we get a natural transformation $$ \xi_0 : h^{X_0} \to H $$ Now assume we have constructed $X_n$ and $\xi_n : h^{X_n} \to H$. We set $$ U_n := \bigcup_{C \in \mathcal C} \operatorname{ker} \left(\xi_n(C) : \operatorname{Hom}(C,X_n) \to H(C) \right) $$ and set $$ Y_n := \bigcup_{U_n} C $$ Since each element of $U_n$ is a map $C \to X_n$, we get a map $Y_n \to X_n$. We let $$ X_{n+1} := \operatorname{cone}(Y_n \to X_n) $$ We want to show that $\xi_n : h^{X_n} \to H$ comes via composition from some $\xi_{n+1} : h^{X_{n+1}} \to H$. If we apply $H$ to the triangle, we get $$ \cdots \leftarrow H(Y_n) \leftarrow H(X_n) \leftarrow H(X_{n+1}) \leftarrow \cdots $$ an exact sequence. The map $$ H(X_n) \to H(Y_n) = \prod_{U_n} H(C) $$ takes $\xi_n$ to the element $\prod_{(C,u)} \xi_n(C)(u) = 0$. Thus, there is some $\xi_{n+1} \in H(X_{n+1})$ with $H(f_{n+1})(\xi_{n+1}) = \xi_n$ for $f_{n+1} : X_n \to X_{n+1}$. This is equivalent to

commuting. Now we let $X = \operatorname{hocolim} X_n$. We check that we have a natural transformation $h^X \to H$ induced from the $\xi_n$. We have triangle $$ \bigoplus X_n \to \bigoplus X_n \to \operatorname{hocolim} X_n \to \bigoplus X_n [1] $$ We apply $H$ to get an exact sequence $$ \prod H(X_n) \leftarrow \prod H(X_n) \leftarrow H(\operatorname{hocolim} X_n) $$ So we need to check that $\prod \xi_n$ lie in the kernel. Recall the map $1-f: \bigoplus X_n \to \bigoplus X_n$ when restricted to the factor $X_n$ is $$ X_n \overset{\operatorname{id}_{X_n},-f_{n+1}}{\to} X_n \oplus X_{n+1} $$ Applying $H$ we have $$ H(1-f)\left(\prod \xi_n \right) = \prod (\xi_n - H(f_n)(\xi_{n-1})) $$ We saw that $\xi_n = H(f_n)(\xi_{n-1})$ so the image is $0$. Thus, we have some $\xi \in H(\operatorname{hocolim} X_n)$ giving us the desired $\xi : h^X \to H$. Let $\mathcal S$ be the full subcategory of objects $Y$ of $\mathcal T$ for which $$ \xi(Y) : \operatorname{Hom}(Y,X) \to H(Y) $$ is an isomorphism. Since both $h^X$ and $H$ take triangles to long exact sequences, we see that $\mathcal S$ is triangulated. Since both takes coproducts to products, we see $\mathcal S$ is closed under coproducts. Finally, take $C \in \mathcal C$. Then, $\xi(C)$ is surjective as $\xi_0(C)$ is surjective by construction. If $g : C \to X_n$ has $\xi(C)(g) = 0$. Then $g \in U_n$ by construction so we have a lift

Consequently $f_{n+1}g = 0$. Thus, $g: C \to X$ is the zero map. We can conclude that $C \in \mathcal S$. From [previously] (% link notes/2021_10_05.md %) we see that $\mathcal T = \mathcal S$ and $X$ represents $H$. ■

An immediate consequence of Brown Representability is a condition on existence of a right adjoint.

Corollary. Let $F: \mathcal T \to \mathcal S$ be an exact functor between triangulated categories. If $\mathcal T$ is compactly generated and $F$ commutes with coproducts, ie the natural map $\bigoplus F(X_a) \to F\left(\bigoplus X_a \right)$ is an isomorphism for coproducts, then $F$ has a right adjoint.

Proof. (Expand to view)

To construct the right adjoint is equivalent to representing the functor $$ \operatorname{Hom}_{\mathcal S}(F(-),Y) : \mathcal T^{op} \to \operatorname{Ab} $$ But, this functor is cohomological since $F$ is exact and representable functors are cohomology. It also takes coproducts to products. Thus, the previous result shows it is representable. ■