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$\DeclareMathOperator{\op}{\mathrm{op}}\DeclareMathOperator{\Ab}{\mathsf{Ab}}\DeclareMathOperator{\Vect}{\mathsf{Vect}}$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?

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    $\begingroup$ 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. $\endgroup$ Nov 24 '20 at 7:38
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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.

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  • $\begingroup$ Thanks, Kevin! I'm getting deja vu, like maybe you told me this before and I forgot... $\endgroup$
    – Tim Campion
    Nov 25 '20 at 14:44
  • $\begingroup$ @TimCampion I know what you mean! Might've just been something similar, though, like about coherentizing weak colimits of transfinite chains. Thinking about that problem was when I learned this stuff. $\endgroup$ Nov 25 '20 at 15:52
  • $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Jan 1 at 0:16
  • $\begingroup$ 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 $\varprojlim_{k<\aleph_n}$. $\endgroup$
    – Tim Campion
    Jan 1 at 0:16

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