Are "most" spaces aspherical?

Question

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.

Answer 1

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.

Answer 2

One way to think about whether "most" spaces are aspherical is measure-theoretically. Here a few examples and non-examples of random topological spaces being aspherical.

Examples

• Presentation complexes of density random groups are aspherical for every density $d < 1/2$, and for density $d> 1/2$ these groups collapse, so this is essentially the entire interesting range of parameter.

• Random 3-manifolds. N. Dunfield and W. Thurston introduced a model for random 3-manifold using a random walk on the mapping class group to generate a random Heegaard splitting. Joseph Maher showed that these random 3-manifolds are hyperbolic with high probability, so in particular their universal cover hyperbolic space $\mathbb{H}^3$ is contractible.

• Let $Y(n,p)$ denote the Linial-Meshulam random 2-dimensional simplicial complex. This complex has vertex set $[n]$, complete $1$-skeleton, and each $2$-face appears independently with probability $p=p(n)$. Costa and Farber showed that if $p \ll n^{-1/2 - \epsilon}$, $Y(n,p)$ is nearly aspherical, in the following sense: if you delete one 2-face from every sufficiently small sphere, pinched sphere (along a vertex, or an edge), or projective plane, the resulting complex is aspherical. It is easy to check that the expected number of these local obstructions is much smaller than the expected total number of 2-faces. So you can delete one face from each one to result in an aspherical complex and have almost all the faces remaining.

• In a similar spirit, Andrew Newman and I recently showed that random 2-dimensional hypertrees (random Q-acyclic complexes) according to a certain "determinantal measure" are are aspherical, in Topology and geometry of random 2-dimensional hypertrees.

Non-examples

• If one considers the random 2-complex $Y(n,p)$ with $p \ge (\gamma n)^{-1/2}$ and $\gamma = 4^4 / 3^3$, Luria and Peled showed that $Y(n,p)$ is simply connected, so at that this point is homotopy equivalent to a bouquet of $2$-spheres, and is not aspherical. It is not "nearly aspherical" in the sense of Costa and Farber either, so there is a phase transition near $p = n^{-1/2}$ from nearly aspherical to not.

• What if we just count homotopy types of simplicial complexes on $n$ vertices? Andrew Newman showed that there are doubly exponentially many homotopy types, at least $2^{2^{0.02n}}.$ On the other hand, there are most $2^{n \choose 3}$ different fundamental groups, a much smaller number, so somehow "most" homotopy types of simplicial complexes can not be aspherical.