# Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$?

$$\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$$Question 1: What is an example of a sequence $$(X_\alpha)_{\alpha<\kappa}$$ of abelian groups such that $$\varprojlim^2_{\alpha < \kappa} X_\alpha \neq 0$$?

Here $$\varprojlim^2_{\alpha<\kappa}$$ is the second derived functor of the limit functor. Necessarily $$\kappa$$ will be uncountable, and of uncountable cofinality. I suspect it should be possible to give an example where the $$X_\alpha$$ are vector spaces and $$\kappa = \omega_1$$ is the first uncountable ordinal.

This question likely sounds funny -- usually one only discusses $$\varprojlim^n_{\alpha<\kappa}$$ for $$n\geq 2$$ in more exotic abelian categories than $$\Ab$$ or $$\Vect$$. This is because one usually only deals with the case where $$\kappa = \omega$$ or at least has cofinality $$\omega$$, in which case the functor $$\varprojlim_{\alpha < \kappa}^n : \Ab^{\kappa^{\op}} \to \Ab$$, i.e. the $$n$$th derived functor of the limit functor, vanishes for $$n \geq 2$$. The usual proof uses a very natural 2-step resolution, which only works when $$\kappa = \omega$$, or by extension when $$\kappa$$ is of countable cofinality.

But when it comes to longer sequences, this resolution is not available. In fact, I've only seen $$\varprojlim_{\alpha<\kappa}^n$$ discussed for $$\kappa$$ of uncountable cofinality in Neeman's Triangulated Categories, appendix A, which contains methods of constructing resolutions in $$\Ab^{\kappa^{\op}}$$, but the resolutions are not of finite length.

Another way of saying that $$\varprojlim^2_{n<\omega} = 0$$ in abelian groups is that $$\varprojlim^1_{n<\omega}$$ is right exact. So a closely related question is:

Question 2: What is an example of an epimorphism $$(X_\alpha \to Y_\alpha)_{\alpha<\kappa}$$ of inverse systems of abelian groups such that the induced map $$\varprojlim^1_{\alpha<\kappa} X_\alpha \to \varprojlim^1_{\alpha<\kappa} Y_\alpha$$ is not an epimorphism?

And by the way,

Question 3: What is the global dimension of the category $$\Ab^{\kappa^{\op}}$$ of $$\kappa$$-indexed inverse systems of abelian groups, for a given regular cardinal $$\kappa$$? How about $$\Vect^{\kappa^{\op}}$$, where $$\Vect$$ is the category of vector spaces over your favorite field?

• There's also a very natural uncountably-indexed inverse system of abelian groups for which the question of vanishing of Rlim^i runs into delicate matters of set theory independent of ZFC. This is the system indexed on the poset of functions f: N --> N (where f\leq g means f(n)\lleq g(n) for all n) whose f^{th} term is \oplus_{n,m:m\leq f(n)} Z and transition maps given by the projections. See people.vcu.edu/~cblambiehanso/higher_limits.pdf for the most recent and rather impressive contribution. Nov 24 '20 at 7:38

A great survey on this and some related topics is Osofsky's "The subscript of $$\aleph_n$$, projective dimension, and the vanishing of $$\varprojlim^{(n)}$$." As far as I am aware, this 1974 paper still describes the state of the art on the matter.
Osofsky (exposing material from Mitchell's book on rings with many objects) gives the following example of a sequence such as you want. Fix any ring $$R$$ and let $$\Delta R$$ be the constant functor at $$R$$ indexed by $$\aleph^{\mathrm{op}}_1$$. Then $$\Delta R$$ can be shown to have homological dimension $$2$$ via a technical argument involving its nice ordered basis (see section 5 of Osofsky.) Now considering any projective resolution $$\cdots\to P_2\to P_1\to P_0\to \Delta R\to 0,$$ if $$K$$ is the kernel of $$P_1\to P_0$$ then the length-2 extension $$0\to K \to P_1\to P_0\to \Delta R\to 0$$ must be nontrivial, or else the kernel $$K'$$ of $$P_0\to \Delta R$$ would be a summand of $$P_1$$, making $$\Delta R$$ of projective dimension at most $$1$$.
That is, $$\mathrm{Ext}^2(\Delta R, K)=\varprojlim^{(2)}_{\aleph_1^{\mathrm{op}}} K$$ is nontrivial whenever $$K$$ is the kernel of the map between the first two projectives resolving $$\Delta R$$, for any ring $$R$$. The same arguments go through when $$\aleph_1$$ is replaced by $$\aleph_n$$ to get nontrivial $$n$$th derived functors of lim. So the answer to your question 3 is "$$n+1$$, if $$\kappa=\aleph_n$$, and $$\infty$$, if it's larger." These holds, passing to cofinalities, even when $$\kappa$$ is replaced with an arbitrary directed set, and I think even an arbitrary filtered category.
• Regarding Question 3, I think I agree that the dimension of the particular functor $\varprojlim : Ab^{\aleph_n^{op}} \to Ab$ is $n+1$, but I think this is strictly smaller than the global dimension of the category $Ab^{\aleph_n^{op}}$, at the very least when $n=0$. For instance the functor $F: Ab^{\omega^{op}} \to Ab$, $X_\bullet \mapsto Hom(\mathbb Z/p^\infty, \varprojlim_{k<\omega} X_k)$ has $Ext(\mathbb Z/p^\infty, \varprojlim^1_{k<\omega} X_k) \subseteq R^2 F(X_\bullet)$ by the Grothendieck spectral sequence. Jan 1 at 0:16
• The values of the functor $\varprojlim^1_{k<\omega}$ are the cotorsion groups, and there are many of these (e.g. $A = \mathbb Z_p$) such that $Ext(\mathbb Z/p^\infty, A) \neq 0$. So it seems that $dim(Ab^{\omega^{op}}) \geq 2$. I suspect that in general $dim(Mod_R^{\aleph_n^{op}}) \overset ? = dim(R) + n+1$, but I'm not sure. At any rate, the global dimension of $R$ has to be a lower bound, so $dim(Mod_R^{\aleph_n^{op}})$ is at least sometimes bigger than $n+1$, even though that's the dimension of the particular functor $\varprojlim_{k<\aleph_n}$. Jan 1 at 0:16