#### [SOLVED] Are "most" spaces aspherical?

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 $S^1$.