By Tim Campion

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.


@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):

There is no simply-connected, non-contractible finite CW-complex all of whose homotopy groups are known.

One theorem that helps me to understand why this 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 $S^1$.

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