Section 13.34 (08TB): Derived limits

13.34 Derived limits

In a triangulated category there is a notion of derived limit.

Definition 13.34.1. Let $\mathcal{D}$ be a triangulated category. Let $(K_ n, f_ n)$ be an inverse system of objects of $\mathcal{D}$. We say an object $K$ is a derived limit, or a homotopy limit of the system $(K_ n)$ if the product $\prod K_ n$ exists and there is a distinguished triangle

\[ K \to \prod K_ n \to \prod K_ n \to K[1] \]

where the map $\prod K_ n \to \prod K_ n$ is given by $(k_ n) \mapsto (k_ n - f_{n + 1}(k_{n + 1}))$. If this is the case, then we sometimes indicate this by the notation $K = R\mathop{\mathrm{lim}}\nolimits K_ n$.

By TR3 a derived limit, if it exists, is unique up to (non-unique) isomorphism. Moreover, by TR1 a derived limit $R\mathop{\mathrm{lim}}\nolimits K_ n$ exists as soon as $\prod K_ n$ exists. The derived category $D(\textit{Ab})$ of the category of abelian groups is an example of a triangulated category where all derived limits exist.

The nonuniqueness makes it hard to pin down the derived limit. In More on Algebra, Lemma 15.86.4 the reader finds an exact sequence

\[ 0 \to R^1\mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (L, K_ n[-1]) \to \mathop{\mathrm{Hom}}\nolimits (L, R\mathop{\mathrm{lim}}\nolimits K_ n) \to \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits (L, K_ n) \to 0 \]

describing the $\mathop{\mathrm{Hom}}\nolimits $s into a derived limit in terms of the usual $\mathop{\mathrm{Hom}}\nolimits $s.

Lemma 13.34.2. Let $\mathcal{A}$ be an abelian category with exact countable products. Then

$D(\mathcal{A})$ has countable products,
countable products $\prod K_ i$ in $D(\mathcal{A})$ are obtained by taking termwise products of any complexes representing the $K_ i$, and
$H^ p(\prod K_ i) = \prod H^ p(K_ i)$.

Proof. Let $K_ i^\bullet $ be a complex representing $K_ i$ in $D(\mathcal{A})$. Let $L^\bullet $ be a complex. Suppose given maps $\alpha _ i : L^\bullet \to K_ i^\bullet $ in $D(\mathcal{A})$. This means there exist quasi-isomorphisms $s_ i : K_ i^\bullet \to M_ i^\bullet $ of complexes and maps of complexes $f_ i : L^\bullet \to M_ i^\bullet $ such that $\alpha _ i = s_ i^{-1}f_ i$. By assumption the map of complexes

\[ s : \prod K_ i^\bullet \longrightarrow \prod M_ i^\bullet \]

is a quasi-isomorphism. Hence setting $f = \prod f_ i$ we see that $\alpha = s^{-1}f$ is a map in $D(\mathcal{A})$ whose composition with the projection $\prod K_ i^\bullet \to K_ i^\bullet $ is $\alpha _ i$. We omit the verification that $\alpha $ is unique. $\square$

The duals of Lemmas 13.33.6, 13.33.7, and 13.33.9 should be stated here and proved. However, we do not know any applications of these lemmas for now.

Lemma 13.34.3. Let $\mathcal{A}$ be an abelian category with countable products and enough injectives. Let $(K_ n)$ be an inverse system of $D^+(\mathcal{A})$. Then $R\mathop{\mathrm{lim}}\nolimits K_ n$ exists.

Proof. It suffices to show that $\prod K_ n$ exists in $D(\mathcal{A})$. For every $n$ we can represent $K_ n$ by a bounded below complex $I_ n^\bullet $ of injectives (Lemma 13.18.3). Then $\prod K_ n$ is represented by $\prod I_ n^\bullet $, see Lemma 13.31.5. $\square$

Remark 13.34.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be a complex of $\mathcal{A}$. Then $\tau _{\geq -n}K^\bullet $ is an inverse system of complexes and which in particular determines an inverse system in $D(\mathcal{A})$. Let us assume that $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ exists. Then the canonical maps $c_ n : K^\bullet \to \tau _{\geq -n}K^\bullet $ are compatible with the transition maps of our inverse system. By the defining distinguished triangle of Definition 13.34.1 and Lemma 13.4.2 we conclude there exists a morphism

\[ c : K^\bullet \longrightarrow R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet \]

in $D(\mathcal{A})$ such that the composition of $c$ with the projection $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet \to \tau _{\geq -m}K^\bullet $ is equal to $c_ m$. Now the morphism $c$ may not be unique, but we claim that whether or not $c$ is an isomorphism is independent of the choice of $c$ (and of our choice of the homotopy limit). Namely, for $i \in \mathbf{Z}$ and for $m > -i$ the composition

\[ H^ i(K^\bullet ) \xrightarrow {H^ i(c)} H^ i(R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet ) \to H^ i(\tau _{\geq -m}K^\bullet ) = H^ i(K^\bullet ) \]

is the identity. Hence $H^ i(c)$ is an isomorphism if and only if the second map is an isomorphism. This is independent of $c$ and also independent of the choice of the homotopy limit (as any two choices are isomorphic).

Lemma 13.34.5. Let $\mathcal{A}$ be an abelian category with countable products and enough injectives. Let $K^\bullet $ be a complex. Let $I_ n^\bullet $ be the inverse system of bounded below complexes of injectives produced by Lemma 13.29.3. Then $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ exists, is K-injective, represents $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ in $D(\mathcal{A})$, and the following are equivalent

the map $K^\bullet \to I^\bullet $ (see proof) is a quasi-isomorphism,
the map $K^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ of Remark 13.34.4 is an isomorphism in $D(\mathcal{A})$.

Proof. The statement of the lemma makes sense as $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ exists by Lemma 13.34.3. Each complex $I_ n^\bullet $ is K-injective by Lemma 13.31.4. Choose direct sum decompositions $I_{n + 1}^ p = C_{n + 1}^ p \oplus I_ n^ p$ for all $n \geq 1$. Set $C_1^ p = I_1^ p$. The complex $I^\bullet = \mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ exists because we can take $I^ p = \prod _{n \geq 1} C_ n^ p$. Fix $p \in \mathbf{Z}$. We claim there is a split short exact sequence

\[ 0 \to I^ p \to \prod I_ n^ p \to \prod I_ n^ p \to 0 \]

of objects of $\mathcal{A}$. Here the first map is given by the projection maps $I^ p \to I_ n^ p$ and the second map by $(x_ n) \mapsto (x_ n - f^ p_{n + 1}(x_{n + 1}))$ where $f^ p_ n : I_ n^ p \to I_{n - 1}^ p$ are the transition maps. The splitting comes from the map $\prod I_ n^ p \to \prod C_ n^ p = I^ p$. We obtain a termwise split short exact sequence of complexes

\[ 0 \to I^\bullet \to \prod I_ n^\bullet \to \prod I_ n^\bullet \to 0 \]

Hence a corresponding distinguished triangle in $K(\mathcal{A})$ and $D(\mathcal{A})$. By Lemma 13.31.5 the products are K-injective and represent the corresponding products in $D(\mathcal{A})$. It follows that $I^\bullet $ represents $R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet $ (Definition 13.34.1). Since $R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet \cong R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ as derived limits are defined on the level of the derived category, we see that $I^\bullet $ represents $R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $. Moreover, the complex $I^\bullet $ is K-injective by Lemma 13.31.3. By the commutative diagram of Lemma 13.29.3 and since $K^ i = (\tau _{\geq -n}K^\bullet )^ i$ for $n \gg 0$ we see that we get a unique map $\gamma : K^\bullet \to I^\bullet $ such that the diagrams

\[ \xymatrix{ K^\bullet \ar[r] \ar[d]_\gamma & \tau _{\geq -n} K^\bullet \ar[d] \\ I^\bullet \ar[r] & I_ n^\bullet } \]

commute. It follows that $\gamma $ is a map of complexes which represents the map $c : K^\bullet \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K^\bullet $ of Remark 13.34.4 in $D(\mathcal{A})$. In other words, the diagram

\[ \xymatrix{ K^\bullet \ar[r]_-c \ar[d]_\gamma & R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K^\bullet \ar[d]^{\cong } \\ I^\bullet \ar[r]^-{\cong } & R\mathop{\mathrm{lim}}\nolimits I_ n^\bullet } \]

is commutative in $D(\mathcal{A})$. The lemma follows. $\square$

Lemma 13.34.6. Let $\mathcal{A}$ be an abelian category having enough injectives and exact countable products. Then for every complex there is a quasi-isomorphism to a K-injective complex.

Proof. By Lemma 13.34.5 it suffices to show that $K \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K$ is an isomorphism for all $K$ in $D(\mathcal{A})$. Consider the defining distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K \to \prod \tau _{\geq -n}K \to \prod \tau _{\geq -n}K \to (R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K)[1] \]

By Lemma 13.34.2 we have

\[ H^ p(\prod \tau _{\geq -n}K) = \prod \nolimits _{p \geq -n} H^ p(K) \]

It follows in a straightforward manner from the long exact cohomology sequence of the displayed distinguished triangle that $H^ p(R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n}K) = H^ p(K)$. $\square$

Comments (3)

Comment #8295 by ZL on April 06, 2023 at 08:40

Can you explain what is the canonical map $K^{\bullet}\rightarrow R \lim \tau_{\geq -n} K^{\bullet}$ in tag 13.34.5? I imagine that it comes from the exact sequence $\mathrm{Hom}(K^{\bullet},R \lim \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})\rightarrow \mathrm{Hom}(K^{\bullet}, \prod \tau_{\geq -n} K^{\bullet})$ , since the canonical map $K^{\bullet}\rightarrow \prod \tau_{\geq -n} K^{\bullet}$ in the second term will be mapped to $0$ in the third term. But I can not get a unique map from $K^{\bullet}$ to $R \lim \tau_{\geq -n} K^{\bullet}$ .

Comment #8296 by Stacks Project on April 06, 2023 at 11:28

This is a good question, thank you! The answer is that it is constructed in the proof. But let me explain a more general canonical map.

I claim: given an abelian category $\mathcal{A}$ with countable products and enough injectives, an inverse system $\ldots \to K_3^\bullet \to K_2^\bullet \to K_1^\bullet$ of bounded below complexes of $\mathcal{A}$ , an (unbounded) complex $K^\bullet$ , and maps of complexes $K^\bullet \to K_n^\bullet$ compatible with the transition maps in the inverse system, there is a canonical map $\gamma : K^\bullet \to R\lim K_n^\bullet$ in $D(\mathcal{A})$ .

To construct $\gamma$ we argue as in the proof of tag 13.34.5. Namely, we find an inverse system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ of bounded below complexes of injectives with termwise surjective transition maps and quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$ compatible with transition maps. Then $R\lim K_n$ is represented by the complex $I^\bullet$ where $I^p = \lim I_n^p$ as in the proof of 13.34.5. The map $\gamma$ is defined to be the one corresponding to the obvious map of complexes $K^\bullet \to I^\bullet$ . Of course we have to show that this map is independent of choices...

If the transition maps in the system $\ldots \to I_3^\bullet \to I_2^\bullet \to I_1^\bullet$ are not termwise surjective, we can still get a unique map easily. Namely, then the Rlim is represented by $C[-1]$ where $C$ is the cone on $\prod I_n^\bullet \to \prod I_n^\bullet$ , see references given in proof of 13.34.5. But now you can use that $K^\bullet \to \prod I_n^\bullet \to \prod I_n^\bullet$ is zero on the nose, to get a canonical map into $C[-1]$ , see Lemma 13.9.3 and its proof. The non-uniqueness that your are pointing out corresponds, in the proof of Lemma 13.9.3, to the choice of the homotopy witnessing the vanishing of $g \circ f$ ; however, since in our case $g \circ f = 0$ it makes sense to take the $0$ homotopy and this does indeed give a well defined map into the cone. Again, one has to show that the map $\gamma$ so obtained does not depend on the choice of the inverse system $I_n^\bullet$ and the quasi-isomorphisms $K_n^\bullet \to I_n^\bullet$ ... but this does not present any problems (as far as I can see at the moment).

The next time I go through all the comments I will make the corresponding edits. Please feel free to complain if you think this doesn't work.

Comment #8922 by Stacks project on April 10, 2024 at 17:01

OK, I fixed this in the opposite way, because actually to prove the "canonicity" would be quite annoying. So I fixed it by making the map maximally noncanonical. See somewhat large set of changes here.

The Stacks project

13.34 Derived limits

Comments (3)

Post a comment