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.

  • 4
    $\begingroup$ Do you have a particular model or construction in mind like randomly attaching cells of bounded dimention together? There are senses in which "random finitely presented groups" end up being aspherical, where you start with roses and attach relators, as at some point you start getting small cancellation groups. This should be closely related to your question. $\endgroup$ – Paul Plummer Apr 12 at 17:48
  • $\begingroup$ @PaulPlummer I was feeling a bit timid about actually staking out a particular model, but this gives me some hope! Are you saying that if you form the canonical 2-dimensional CW complex assoicated to a presentation of a group, this is "typically" aspherical? If so, that's amazing! For higher dimensions, I don't know a good way to write down "generators" of higher homotopy groups from a CW presentation in order to encode attaching maps, but I suppose one could pretty easily say what a "random simplicial complex" is and maybe the notion is the same in some limit. $\endgroup$ – Tim Campion Apr 12 at 18:04
  • $\begingroup$ I was being somewhat simplistic, but basically. It is actually an annoying thing that there is probably not great way to choose a model as there are a bunch of different ways to vary and relate variables to each other, so the answer depends on the model. My understanding is that for many ways to choose what a typical group is you get something aspherical. Might be interested in look at this paper and many other papers on random groups $\endgroup$ – Paul Plummer Apr 12 at 18:28
  • 1
    $\begingroup$ In dimension 2, you're saying that "most" surfaces are not spheres / projective planes, but this might depend on the way to define "most". $\endgroup$ – YCor Apr 13 at 10:02
  • 3
    $\begingroup$ Take a look at some asphericity/nonasphericity results in the 2d case: link.springer.com/content/pdf/10.1007%2Fs11856-015-1240-2.pdf. $\endgroup$ – Misha Apr 13 at 17:33

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.

  • 3
    $\begingroup$ 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. $\endgroup$ – Tim Campion Apr 13 at 5:39
  • $\begingroup$ Interesting. Oh, also I guess another class of aspherical manifolds would be tori. $\endgroup$ – cgodfrey Apr 13 at 5:57
  • 1
    $\begingroup$ 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) $\endgroup$ – Kevin Casto Apr 13 at 6:51
  • $\begingroup$ @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 $S^1$. $\endgroup$ – cgodfrey Apr 14 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.