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.

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    $\begingroup$ Wait, isn't a 3-sphere a counterexample? $\endgroup$ Jan 28, 2021 at 18:55
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    $\begingroup$ 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. $\endgroup$ Jan 28, 2021 at 19:01
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    $\begingroup$ This is in Hatcher's notes on 3-manifolds, which is well worth the quick read. No need for geometry. $\endgroup$
    – mme
    Jan 29, 2021 at 4:19
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    $\begingroup$ 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. $\endgroup$
    – Ben Webster
    Feb 1, 2021 at 1:20

2 Answers 2


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.


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