# Which abelian groups are $\varprojlim^1$ groups?

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 $$\mathcal D^\ast$$?

• For Question 4, can't you just take the constant inverse system with $D^{n,*}=\mathcal{D}^*$ for all $n$? – Jeremy Rickard Oct 9 '20 at 17:37
• @JeremyRickard Ah, of course you're right. Maybe there isn't really an interesting question to ask at that level of generality... – Tim Campion Oct 9 '20 at 17:40
• 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 $H_0(C),H_1(C)$ – Maxime Ramzi Oct 9 '20 at 17:52
• @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. – Tim Campion Oct 9 '20 at 17:53
• 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. – 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.
• There is a typo in the first line. Should the last Ext group be $\textrm{Ext}(\mathbb{Q}, A)$? – John Palmieri Oct 10 '20 at 2:33