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Stream: learning: questions

Topic: 1-categorical vs 2-categorical limits

John Baez (Sep 19 2024 at 22:50):

Suppose we're in a 2-category $C$ and we're studying the limit of a diagram shaped like this:

$x_1 \leftarrow x_2 \leftarrow x_3 \leftarrow \cdots$

Suppose this diagram has a strict 2-limit, i.e. this diagram has a limit in the underlying 1-category of $C$. Is this strict limit also necessarily the limit in the 2-categorical sense?

John Baez (Sep 19 2024 at 22:52):

(There is some information about related topics at [[semiflexible limit]], but I'm unable to use it to answer my question.)

Mike Shulman (Sep 19 2024 at 22:54):

I guess you mean the up-to-isomorphism 2-categorical sense, a.k.a. the bilimit.

Mike Shulman (Sep 19 2024 at 22:55):

(By the way, a strict 2-limit is a bit more than a limit in the underlying 1-category: it also says says something about 2-cells into the limit corresponding to 2-cells between cones.)

Mike Shulman (Sep 19 2024 at 23:01):

I'm not sure I know the answer to that. I'm pretty sure sequential limits aren't PIE-limits, but I don't know whether they're flexible or semiflexible. I would guess not, but I don't have a counterexample right away.

Kevin Carlson (Sep 20 2024 at 00:19):

I think if you take a sequence of groups not satisfying the Mittag-Leffler condition, so that it has a nontrivial derived limit, then the naive limit is homotopically wrong. Let me try to see if I can make this more elementary and clear.

Kevin Carlson (Sep 20 2024 at 00:55):

Hmm...I know that the sequence $...\to \mathbb Z\to \mathbb Z$ where all the maps are multiplication by $2$ is supposed to have nontrivial $\lim^1$, and that ought to mean that the bilimit of either this diagram (of discrete groupoids) or its delooping is not the same as the strict limit, but I haven't been able to show this directly yet.

Mike Shulman (Sep 20 2024 at 01:05):

I think it must be the delooping.

Kevin Carlson (Sep 20 2024 at 01:12):

Right, that's where I was working too, I guess the discrete groupoid diagram has 2-limit and bilimit both empty.

Mike Shulman (Sep 20 2024 at 01:15):

Or maybe $\{0\}$?

Mike Shulman (Sep 20 2024 at 01:16):

Let's see, the Milnor exact sequence says that given a tower of fibrations of spaces $\cdots \to X_2 \to X_1\to X_0$, for each $q\in \mathbb{N}$ there is a short exact sequence

$$0 \to \lim{}^1 \pi_{q+1}(X_i) \to \pi_q(\lim X_i) \to \lim \pi_q(X_i) \to 0$$

So if we convert your delooped sequence $\cdots \to B\mathbb{Z}\to B\mathbb{Z}\to B\mathbb{Z}$ to a tower of fibrations, which doesn't change the homotopy groups, then the strict limit of the fibrations will be the homotopy limit of the original delooped tower. Taking $q=0$, then each $\pi_0(X_i) = 0$, so the right-hand object is trivial. But the sequence $\pi_1(X_i)$ on the left will be the sequence of groups $\cdots \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}$, which has nontrivial $\lim{}^1$, and thus the middle object $\pi_0(\lim X_i)$ will be nontrivial, i.e. the homotopy limit of the original delooped tower is disconnected. But certainly the strict limit of the original tower is connected (and, I think, even trivial).

Kevin Carlson (Sep 20 2024 at 01:18):

Mike Shulman said:

Or maybe $\{0\}$?

Oh yeah, yeah.

Kevin Carlson (Sep 20 2024 at 01:20):

And yeah, that argument looks right; I wasn't tracking the interaction between fibrant replacement and homotopy/strict limits. I agree the original tower has a point as its strict limit, i.e. a single object with endomorphisms given by all the elements of $\mathbb Z$ that are infinitely divisible by $2.$

Kevin Carlson (Sep 20 2024 at 01:21):

To make this a full answer to John's question we just need to remember how homotopy limits of groupoids are related to bilimits of the $(2,1)$-category; they're the same since groupoids are reflective in spaces (say in terms of $\infty$-categories.)

John Baez (Sep 21 2024 at 14:58):

Thanks so much, guys! So you're looking at the strict versus the homotopy limit of the sequence of groupoids

$B \mathbb{Z} \leftarrow B \mathbb{Z} \leftarrow \cdots$

where each arrow comes from $\mathbb{Z} \xleftarrow{2 \times -} \mathbb{Z}$. And you're saying the strict limit is the terminal groupoid, while the homotopy limit is not.

Now I'm really curious what the homotopy limit is, spelled out in the language of groupoids!

John Baez (Sep 21 2024 at 15:00):

Somehow my understanding of homological algebra never penetrated to $\lim^1$. :crying_cat:

John Baez (Sep 21 2024 at 15:07):

For some reason @Todd Trimble and @Joe Moeller and I are dealing with a diagram of Vect-enriched categories

$C_1 \leftarrow C_2 \leftarrow \cdots$

where I feel in my bones that the strict limit is the bilimit. Let me say what it is.

$C_k$ is the category of $\mathbb{N}^k$-graded vector spaces of finite total dimension, i.e. the sums of the dimensions of the grades is finite. This is a full subcategory of $\mathsf{FinVect}^{\mathbb{N}^k}$.

The functor $C_k \leftarrow C_{k+1}$ is induced by the inclusion $\mathbb{N}^k \to \mathbb{N}^{k+1}$ sending $(n_1, \dots, n_k)$ to $(n_1, \dots, n_k, 0)$.

John Baez (Sep 21 2024 at 15:10):

I guess my problem is that I don't know a good way to compute the bilimit of a big long diagram of Vect-enriched categories like

$C_1 \leftarrow C_2 \leftarrow \cdots$

even if I have a very good understanding of that diagram.

John Baez (Sep 21 2024 at 15:11):

We understand the strict limit quite well; I won't bore you by explaining it.

John Baez (Sep 21 2024 at 15:14):

The big difference between this situation and your counterexample might be that my arrows $C_k \leftarrow C_{k+1}$ are 'surjective' - more precisely, they are essentially surjective and full functors.

Mike Shulman (Sep 21 2024 at 19:54):

Are they "fibrations"? The strict limit of a tower of fibrations is generally equivalent to its homotopy limit, as long as there's a suitable model-category-like-thing around.

John Baez (Sep 21 2024 at 20:35):

Okay, good! They feel like fibrations to me. So I guess I should look for some model structure on the category of $\mathsf{Vect}$-enriched categories where they actually are.

Mike Shulman (Sep 21 2024 at 21:03):

You could try Lack's model structure that exists on any 2-category with enough limits and colimits.

John Baez (Sep 21 2024 at 21:14):

Where can I read about that? In the case of Cat, is this model structure the [[canonical model structure on Cat]]?

Mike Shulman (Sep 21 2024 at 21:17):

Homotopy-theoretic aspects of 2-monads, and yes.

Mike Shulman (Sep 21 2024 at 21:22):

Interestingly, although the notion of "2-category with enough limits and colimits" is self-dual, Lack's construction is not. So actually any 2-category with enough limits and colimits has two canonical model structures. In the case of Cat, they happen to coincide.

John Baez (Sep 22 2024 at 01:20):

Thanks!

Kevin Carlson (Sep 22 2024 at 21:07):

For taking limits, wouldn't it be fine to just look at these as categories rather than $\mathbf{Vect}$-enriched categories? Then the fibrations are just isofibrations, which these functors $C_k\leftarrow C_{k+1}$ are.

Kevin Carlson (Sep 22 2024 at 21:20):

John Baez said:

Now I'm really curious what the homotopy limit is, spelled out in the language of groupoids!

Well, an object of the homotopy limit $X$ is a choice of object in the $$n$$th copy of $B\mathbb Z$ for every $n$, so the sequence $(*_0,*_1,\ldots)$, together with an isomophism $(2\times -)(*_{n+1})\to *_{n}$ for every $n,$ so it's just a sequence $(a_n)$ of integers. And I believe a morphism $(a_n)\to (b_n)$ is another sequence $(\alpha_n)$ such that $a_n+\alpha_n=b_n+2\alpha_{n+1}.$ This is where I got stuck a few days ago though, trying to find some explicit evidence that the space of these things is not contractible. In particular, if $a_n=b_n$ then the only morphism is $0,$ so that $X$ has trivial fundamental groups at every basepoint. In context of Mike's Milnor sequence calculation, I think this is correct since $\lim \pi_1(X_i)=\lim^1\pi_2(X_i)=0.$ So we just need to show that $X$ isn't connected.
Screenshot-2024-09-22-at-2.15.23PM.png

Kevin Carlson (Sep 22 2024 at 22:01):

I'm tempted to compare the boring object $a_n=(0,0,\ldots)$ of $X$ with something that seems 2-adically spicy like $b_n(1,2,4,\ldots).$ A morphism here would thus be a sequence satisfying the recurrence $\alpha_n-2\alpha_{n+1}=2^n.$ The non-integer-valued sequence $(-\frac 23,-\frac 43,-\frac 83,\ldots),$ is one solution and the kernel of the linear operator sending $\alpha_n\mapsto \alpha_n-2\alpha_{n+1}$ is $(a_0,\frac{a_0}2,\frac{a_0}4,\ldots),$ so we'd need to find $a_0\in\mathbb Q$ such that $\frac{a_0}{2^n}-\frac 23 \cdot 2^n$ is in $\mathbb Z$ for every $n,$ or better yet, $a_0'\in \mathbb Z$ such that $a_0'-2^{2n+1}$ is a multiple of $3\cdot 2^n$ for every $n.$ But that says in particular that $a_0'-2^{2n+1}$ is 2-adic Cauchy sequence, which rules out any option but $a_0'=0,$ which is not a solution.

Kevin Carlson (Sep 22 2024 at 22:30):

So if I've got that right, the connected components of $X$ should be in some sense classes of different possible 2-adic behaviors of integer sequences, and probably that would bring us back around to $\lim^1$ if we would recall how to compute that.

Kevin Carlson (Sep 22 2024 at 22:36):

You can give an injective resolution of our sequence $S_1$ pretty easily like so:
Screenshot-2024-09-22-at-3.34.08PM.png
The $\lim^1$ is then the cokernel of $\lim(S_2)\to \lim(S_3),$ which I claim is the canonical map $\mathbb Q\to \mathbb Q/\mathbb Z\oplus (\mathbb Z/2\mathbb Z)^\mathbb N,$ i.e. the standard abelian group structure on the Cantor set $(\mathbb Z/2\mathbb Z)^\mathbb N$. I'm not sure how enlightening this is but it's fun and creepy to observe that our space $X$ has a profinite set of components, which wasn't obvious off the bat.

Kevin Carlson (Sep 22 2024 at 22:37):

I mean, I guess $X$ is just the Cantor space, up to homotopy, probably.

John Baez (Sep 22 2024 at 23:41):

Kevin Carlson said:

For taking limits, wouldn't it be fine to just look at these as categories rather than $\mathbf{Vect}$-enriched categories? Then the fibrations are just isofibrations, which these functors $C_k\leftarrow C_{k+1}$ are.

That sounds right, which is good. I just happen to be working in the 2-category of linear categories, so I want to make sure I'm taking the limit and bilimit in there.

But actually, I'm really working in the 2-category of $\mathbb{N}$-graded Cauchy complete linear categories.

(There comes a time when someone asking for help starts providing too much detail about what they're actually doing, and now is that time. You've done enough for me, now I'm just nattering.)

Alas, this '$\mathbb{N}$-grading' business. which I'd suppressed earlier in this conversation, actually makes a big difference. For example, to give a decategorified analogue, suppose $R_n$ was the free graded ring on generators of grades $2, 4, 6, \dots, 2n$. There are obvious maps $R_n \leftarrow R_{n+1}$ that kill off the generator of highest degree. The limit of the resulting diagram of rings is a lot bigger than the limit of the resulting diagram diagram of graded rings. This example actually shows up if you're studying the cohomology ring of $\mathbb{C}\mathrm{P}^\infty$ as a limit of the cohomology rings of $\mathbb{C}\mathrm{P}^n$'s. We're doing something similar but categorified.

Mike Shulman (Sep 23 2024 at 00:12):

The fibrations in Lack's model structure are the representable isofibrations. I wouldn't be surprised if in $V\text{- Cat}$ those are just the $V$-functors whose underlying ordinary functor is an isofibration, for any $V$.

Mike Shulman (Sep 23 2024 at 00:12):

Actually I think maybe I would be surprised if they weren't that.