Let $k$ be a field and $X$ a topological space.

Write $\mathrm{Sh}(X)$ for the category of sheaves of vector spaces on $X$, and $\mathrm{Loc}(X)$ for the subcategory of local systems of finite dimensional $k$-vector spaces.

The category not local systems is an abelian category, so we can form the derived category $D(Loc(X))$. This is the category of complexes of local systems on $X$ with quasi-isomorphisms inverted.

We can also consider the subcategory $D_{\mathrm{Loc}}(X)$ of $D(X):=D(\mathrm{Sh}(X))$ consisting of complexes whose cohomology sheaves are local systems on $X$.

I have two questions:

  1. Is $D_{\mathrm{Loc}}(X)$ a triangulated subcategory of $D(X)$? More precisely is it closed under taking mapping cones?
  2. Under what hypotheses (if any) are $D_{\mathrm{Loc}}(X)$ and $D(\mathrm{Loc}(X))$ equivalent?
  • 7
    $\begingroup$ $S^2$ is a counterexample to both questions because you can use a class in $H^2$ to make a mapping cone between two shifted constant sheaves, but the derived category of local systems is just the the derived category of vector spaces. There might be a positive answer to 2 for $K(\pi,1)$s, I don't know. $\endgroup$
    – Will Sawin
    May 19, 2020 at 2:24
  • 3
    $\begingroup$ I believe that 2. is essentially the definition of what it means for a (reasonably nice) topological space $X$ to be/become a $K(\pi,1)$ "after the derived functor of $k$-proalgebraic completion of the fundamental group is taken". To give an example, if $k$ is a field of characteristic $0$, the following property 2'. holds if and only if $X$ is a "rational $K(\pi,1)$". Notice that a $K(\pi,1)$ space $X$ need not be a rational $K(\pi,1)$ if $X$ is not a nilpotent space (and of course a rational $K(\pi,1)$ space need not be a $K(\pi,1)$). $\endgroup$ May 19, 2020 at 3:21
  • 3
    $\begingroup$ 2' is as 2 in the question, but with the category $\operatorname{Loc}(X)$ of local systems of finite-dimensional vector spaces replaced by its subcategory $\operatorname{Loc}_{un}(X)\subset \operatorname{Loc}(X)$ of unipotent local systems. Here a local system is called unipotent if the related representation of the fundamental group has a filtration with trivial representations in the successive quotients. $\endgroup$ May 19, 2020 at 3:24
  • 1
    $\begingroup$ @WillSawin By the way, $S^2$ is not a counterexample to Question 1. It is a counterexample to Question 2. The answer to Question 1 is always positive, for any reasonable (locally connected and locally simply connected may be sufficient) space $X$. This is so because the kernel and cokernel of any sheaf morphism between two local systems are again local systems, and any sheaf extension of two local systems is a local system. $\endgroup$ Jun 22, 2020 at 2:37
  • 1
    $\begingroup$ Restricting to finite dimensional objects almost never preserves exts / produces a full triangulated subcategory. The most famous example is representations of $sl_2$. Finite dimensional objects are semi-simple, so there are no exts. Whereas in the full category, the self-exts of the trivial representation are big, the cohomology of $SU(2)$. Similarly, Margulis's supperrigidity theorem roughly says that finite dimensional representations of $SL_3Z$ are the same as reps of $SL_3R$ and thus roughly semisimple. But the group has cohomology, coming from infinite dimensional representations. $\endgroup$ Jun 22, 2020 at 17:14

1 Answer 1


UPD: I didn't notice that you're asking about finite dimensional local systems, so this answer doesn't really answer your question. The easy way to fix that is to consider the category $D^b_f(Loc(X))$ of complexes of representations of $\pi_1$ with finite dimensional cohomology; in order to treat the case of an honest category of finite dimensional representations, it seems, the use of (derived) algebraic completion is unavoidable.

I think that both statements are true if $X$ is sufficiently nice and $K(\pi, 1)$, and if you consider bounded derived categories. Sufficiently nice here means that Exts between two local systems in the category of local systems and in the category of sheaves coinside with each other. I think it is enough to assume that $X$ is locally contractible; from this assumption follows the existence of the universal cover $\tilde X$.

The cone of two bounded complexes with locally constant cohomology has locally constant cohomology, because on a locally contractible space $X$ the category of locally constant sheaves is a so called Serre subcategory of the category of sheaves --- abelian subcategory closed under extensions. Then the long exac sequence of cohomology objects shows you that a cone of two complexes with locally constant cohomology has locally constant cohomology as well.

There are at least two ways to show that Exts between two local systems in two categories are the same. One can show that there are enough local systems, that are projective (resp., injective) as local systems, which are adapted to the functor $\mathrm{Ext}^*(-, V)$ (resp. $\mathrm{Ext}^*(V, -)$), where $V$ is also a local system. For projective local systems we can just take the regular representation of $\pi_1(X)$, let's call it $P$. For an injective, one can take the representation $\prod_{g \in G} \mathbb{Q}g$, where $\mathbb{Q}$ is some injective hull of your base ring. The $G$-action here is by multiplying to $g^{-1}$ from the right. Let's denote this module by $Q$. I think, by using the rule $v \mapsto \prod_{g \in G} g\otimes gv$ one can embed any $G$-module $V$ into $Q \hat\otimes V$, that is, infinite product of $Q$ ($V$ in the tensor product is considered with trivial $G$-action).

Now, both $P$ and $Q$ come from $\tilde X$, that is, if $\pi: \tilde X \longrightarrow X$ is the universal cover, then $P = \pi_! \underline{\mathbb{Z}}_{\tilde X}$ (where $\mathbb{Z}$ is the base ring) and $Q = \pi_*\underline{\mathbb{Q}}_{\tilde X}$. Since $\pi$ is a covering, there are no higher derived functors of $\pi_*$ and $\pi_!$.

Now one can use the adjointness $\mathrm{Ext}_{X}(F, Q) = \mathrm{Ext}_{\tilde X}(\pi^*F, \underline{\mathbb{Q}}_{\tilde X})$. Since $\mathbb{Q}$ is injective, $\tilde X$ is contractible and $\pi^*F$ is a locally constant (hence constant) sheaf, this vanishes in higher degrees. Or, you can show that $P$ is adapted by using the fact that for a covering map $\pi_!$ is left adjoint to $\pi^*$ (a fact which I think I know how to prove, but I was unable to find a reference for it in the case of an infinite covering. You also need to assume $X$ to be locally compact to use this, i suppose).

Now, using the induction on the length of a complex, you can show that the inclusion functor from $D^b(Loc(X))$ into $D^b(X)$ is a fully faithful embedding. And since local systems generate $D^b_{Loc}(X)$ as a triangulated subcategory of $D^b(X)$, it is precisely the essential image of this functor. This is more or less standart; i think that an appendix to this Positselski's paper has a good treatment of that stuff.

Sadly, I don't know what will happen if one wants to consider unbounded categories.

  • $\begingroup$ Also, among related questions, mathoverflow shows mathoverflow.net/a/163698/43309 this answer by Julian Holstein, whose papers give another proof for the statement, since if $X$ is aspherical, the chains on $\Omega X$ are quasi-isomorphic to the group algebra of $\pi_1$ $\endgroup$ Jun 21, 2020 at 23:52
  • $\begingroup$ This is a convincing answer to a question somewhat different from the one asked by OP above. In the question above, OP denotes by $\operatorname{Loc}(X)$ the category of local systems of finite-dimensional $k$-vector spaces on $X$. This changes the meaning of the question, and also changes the answer, as compared to the case when the local systems are allowed to have infinite-dimensional fibers (as the argument in your answer necessarily presumes, in case $\pi_1(X)$ is infinite). In case infinite-dimensional local systems are allowed, I agree with your answer. $\endgroup$ Jun 22, 2020 at 0:28
  • $\begingroup$ Yes, I only just noticed that the question was about finite-dimensional local systems. $\endgroup$ Jun 22, 2020 at 0:30
  • 1
    $\begingroup$ That's the reason why I mentioned unipotent local systems in my comments to the question above. If one wants to restrict oneself to finite-dimensional local systems, one may as well go all the way and restrict to finite-dimensional unipotent local systems. These are conceptually similar restrictions (at least on some level of abstraction), but the one with unipotent local systems is simpler and related to more familiar concepts. $\endgroup$ Jun 22, 2020 at 0:53
  • $\begingroup$ I understand now, thank you! Well, this argument at least shows that the derived category of complexes with finite dimensional locally constant cohomology is equivalent to the derived category of complexes of representations with finite dimensional cohomology also. Do you, by the way, have a reference about the unipotent case? $\endgroup$ Jun 22, 2020 at 1:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .