2019-04-12 17:17:11 8 Comments

There's a heuristic idea that "most" closed manifolds $M$ are aspherical (i.e. $\pi_{\geq 2}(M) = 0$). Does this heuristic extend usefully to all spaces -- or at least to all finite CW complexes?

To make this question more precise, I should say something about in what sense "most" manifolds are aspherical. I don't know a lot about this heuristic, but here's where I'm coming from:

It's true in low dimensions: trivially in 0 or 1 dimensions, and by classification of surfaces in 2 dimensions. In 3 dimensions, I've heard it said that part of the upshot of Thurston's Geometrization Conjecture is that "most" 3-manifolds are hyperbolic, and in particular aspherical.

There's some discussion of this heuristic in this survey article of Luck (at the end).

How do things look if we think about CW complexes? Well, every 0 or 1-dimensional CW complex is aspherical. And the Kan-Thurston theorem tells us that every space is homology-equivalent to an aspherical space. But it's really not clear to me whether I should think of "most" spaces as being aspherical.

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## 1 comments

## @cgodfrey 2019-04-13 03:34:40

As of the writing of Peter May's

A concise introduction to algebraic topology(where I saw this statement):One theorem that helps me to understand

whythis is the case:Theorem(Serre): Let $X$ be a simply-connected finite CW complex with non-0 reduced homology $\tilde{H}_*(X) \neq 0$. Then for any $N \in \mathbb{N}$ there's a $i > N$ with $\pi_i (X) \neq 0$.One reference: Mosher and Tangora,

Cohomology operations and applications in homotopy theory.Your comment about curvature does seem relevant. For instance, one can use the above theorem to show that if $M$ is a compact Riemannian manifold with positive curvature, then $M$ has infinitely many non-0 higher homotopy groups.

To see this, observe that when $M$ has positive curvature its universal cover $\tilde{M}$ is compact, simply-connected, non-contractible (e.g. because $H_{\dim \tilde{M}}(\tilde{M}; \mathbb{Z}) \neq 0$) and has the homotopy type of a finite complex (by Morse theory, for instance). Furthermore the covering map $\tilde{M} \to M$ induces isomorphisms $\pi_i(\tilde{M}) \simeq \pi_i(M)$ for $i> 1$.

In some ways it's easier for an infinite dimensional complex to be aspherical: for instance, when $G$ is a (non-trivial) finite group the classifying space $BG$ (a.k.a. $K(G, 1)$) is necessarily infinite dimensional, since using group cohomology one can show $H_i(BG; \mathbb{Z}) \neq 0$ for infinitely many $i$.

So, while this is by no means a complete answer to your question, we can see that for a finite, non-contractible CW complex $X$ to be aspherical, $\pi_1(X)$ must be infinite.

## @Tim Campion 2019-04-13 05:39:31

I absolutely agree. In fact, I'm so used to thinking along these lines that I found myself scratching my head the other day when I heard the phrase "finite aspherical space" in a talk (and then slapping my forehead when I remembered "duh, think about hyperbolic manifolds"!). So my sense is that the spaces which homotopy theorists tend to think of as "typical" are in some sense a very biased sample, just as Luck's survey indicates that the manifolds we think of as "typical" are actually very special.

## @cgodfrey 2019-04-13 05:57:38

Interesting. Oh, also I guess another class of aspherical manifolds would be tori.

## @Kevin Casto 2019-04-13 06:51:20

Surely it's much easier/weaker that "most" spaces have infinite $\pi_1$ than that "most" spaces are aspherical? (e.g., Gromov's work on random groups)

## @cgodfrey 2019-04-14 05:20:59

@KevinCasto I'm not familiar with Gromov's work on random groups but you're definitely right. All I'm saying is that infinite $\pi_1$ is a necessary condition for a non-contractible finite complex to be aspherical. There are certainly way more such complexes with infinite $\pi_1$, e.g. take the product of any finite CW complex with $S^1$.