Question 1: Let $\mathcal A$ be an abelian group. Does there exist an inverse system $(A^n)_{n \in \mathbb N} = (\cdots \to A^n \to A^{n-1} \to \cdots \to A^0)$ such that $\varprojlim^1 A^\bullet \cong \mathcal A$? If not, can we characterize the abelian groups which are $\varprojlim^1$ groups or at least say anything interesting about their isomorphism types?

The remaining questions are meant to be refinements of Question 1.

Question 2: Let $\mathcal B^0,\mathcal B^1$ be abelian groups. Does there exist an inverse system $(B^n)_{n \in \mathbb N}$ such that $\varprojlim^i B^\bullet \cong \mathcal B^i$ for $i=0,1$?

If $(C^n)_{n \in \mathbb N}$ is an inverse system, there is a canonical two-term chain complex which we'll call $\mathbf{Lim} (C^\bullet) = (\prod_{n \in \mathbb N} C^n \to \prod_{n \in \mathbb N} C^n)$, where the differential is $(c^0,c^1,\dots) \mapsto (c^0 - \gamma c^1,c^1-\gamma c^2,\dots)$ where $\gamma$ ambiguously denotes any of the linking maps for the inverse system $C^\bullet$. The point, of course, is that $H^i(\mathbf{Lim} (C^\bullet)) = \varprojlim^i(C^\bullet)$ for $i=0,1$.

Question 3: Let $\mathcal C^\ast = (\mathcal C^0 \to \mathcal C^1)$ be a two-term chain complex of abelian groups. Does there exist an inverse system $(C^n)_{n \in \mathbb N}$ of abelian groups such that $\mathbf{Lim}(C^\bullet)$ is quasi-isomorphic to $\mathcal C^\ast$?

If $(D^{n,\ast})_{n \in \mathbb N}$ is an inverse system of chain complexes of abelian groups, then define $\mathbf{Lim}(D^{\bullet,\ast})$ by by applying $\mathbf{Lim}$ levelwise to obtain a double complex, and then taking the diagonal.

Question 4: Let $\mathcal D^\ast$ be a chain complex of abelian groups. Does there exist an inverse system $(D^{n,\ast})_{n \in \mathbb N}$ of chain complexes of abelian groups such that $\mathbf{Lim}(D^{\bullet,\ast})$ is quasi-isomorphic to $\mathcal D^\ast$?

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    $\begingroup$ For Question 4, can't you just take the constant inverse system with $D^{n,*}=\mathcal{D}^*$ for all $n$? $\endgroup$ – Jeremy Rickard Oct 9 '20 at 17:37
  • $\begingroup$ @JeremyRickard Ah, of course you're right. Maybe there isn't really an interesting question to ask at that level of generality... $\endgroup$ – Tim Campion Oct 9 '20 at 17:40
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    $\begingroup$ Question 3 is equivalent to Question 2, as any chain complex of abelian groups is formal (because $\mathbb Z$ is a PID) : there is a zigzag of quasi-isomorphisms $C\to \bigoplus_n H_n(C)[n]$, so the answer to 3 is yes if and only if the answer to 2 is yes for $H_0(C),H_1(C)$ $\endgroup$ – Maxime Ramzi Oct 9 '20 at 17:52
  • $\begingroup$ @MaximeRamzi Thanks, I had some idea like that bouncing around in the back of my head, but I wasn't sure if there were finiteness / connectivity assumptions involved. $\endgroup$ – Tim Campion Oct 9 '20 at 17:53
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    $\begingroup$ Values of lim^1 are precisely cotorsion groups, i. e. ones with $Ext(\Bbb Q, G) = 0$. Values of lim^1 on a tower of f.g. groups are of the form $Ext(A, \Bbb Z)$ with A flat. $\endgroup$ – Denis T. Oct 9 '20 at 18:48

Abelian group $A$ is cotorsion if $\rm{Ext}(F, A) = 0$ for every flat $F$, or, equivalently, $\rm{Ext}(\Bbb Q, A) = 0$

Every $\varprojlim^1$ of an inverse system of abelian group is cotorsion, and, conversely, every cotorsion group is a $\varprojlim^1$. Proof can be found in Warfield, Huber. On the values of the functor $\varprojlim^1$.

If every group in the inverse system $G_i$ is f. g., then $\varprojlim^1$ is $\rm{Ext}(A, \Bbb Z)$ for $A$ torsion free countable: take $A$ equal to direct limit of $\rm{Hom}(G_i, \Bbb Z)$. In particular, that limit is always a divisible group.

Converse is also true, but version of proof I know is messy and uses structure teorem of alg. compact groups. Probably you can find something on that in Fuchs or Jensen books.

  • $\begingroup$ There is a typo in the first line. Should the last Ext group be $\textrm{Ext}(\mathbb{Q}, A)$? $\endgroup$ – John Palmieri Oct 10 '20 at 2:33
  • $\begingroup$ @JohnPalmieri Thanks! $\endgroup$ – Denis T. Oct 10 '20 at 10:30

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