# Higher homotopy groups of irreducible 3-manifolds

A 3-manifold $$M$$ is irreducible if every embedded 2-sphere bounds a 3-ball. Thanks to Papakyriakopoulos's sphere theorem, irreducibility is the same as having $$\pi_2(M)=0$$. Does irreduciblity imply that the manifold is in fact aspherical, i.e. that $$\pi_k(M)=0$$ for all $$k \geq 2$$?

(Or maybe I should say that the universal cover $$\tilde M$$ is aspherical, but the question is the same in terms of homotopy groups.)

As pointed out by @Matt Zaremsky in the comments, there is an obvious counterexample in $$S^3$$. But perhaps this is the only counterexample, or the counterexamples are easy to classify?

Given all of the tools we have about geometric classification of 3-manifolds, I expect someone would have a quick answer. I'm just not enough of an expert to make those arguments myself.

• Wait, isn't a 3-sphere a counterexample? Jan 28, 2021 at 18:55
• Right you are, I guess I didn't think about that too carefully. Let's see if I can edit to ask a better question. Jan 28, 2021 at 19:01
• This is in Hatcher's notes on 3-manifolds, which is well worth the quick read. No need for geometry.
– mme
Jan 29, 2021 at 4:19
• I would argue on the merits that it is a MathOverflow question. It might have a simple answer, but it requires a knowledge of graduate level algebraic topology to even understand the statement of the question. Feb 1, 2021 at 1:20

An irreducible 3-manifold $$M$$ is aspherical if and only if it's not a finite quotient of $$S^3$$, which in turn is equivalent to having infinite fundamental group. Essentially you've already outlined the proof: the universal cover $$\tilde M$$ is a simply-connected 3-manifold with trivial $$\pi_2$$, and so also $$H_2(\tilde M) = 0$$; if $$\tilde M$$ is not compact, then $$H_3(\tilde M) = 0$$ (because of non-compactness) and $$H_k(\tilde M) = 0$$ for higher $$k$$ (because it's a 3-manifold), so by the Hurewicz theorem $$\tilde M$$ is aspherical.

On the other hand, $$\tilde M$$ is compact if and only if $$\pi_1(M)$$ is finite. In this case, Perelman proved that $$\tilde M$$ is $$S^3$$. 3-manifolds covered by $$S^3$$ are classified, and they correspond to finite subgroups of $$SO(4)$$. I think that Scott's The geometries of 3-manifolds has a precise statement.

It's wrong for finite fundamental group, as then the universal cover is closed and has nonvanishing $$\pi_3$$ by Hurewicz. It's true for infinite fundamental group, again by Hurewicz applied to the universal cover.