TITLE: Example of an uncountable sequence of abelian groups with nonvanishing $\varprojlim^2$?
QUESTION [8 upvotes]: $\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$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?

REPLY [10 votes]: 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.