I believe the following statement is true, and I've even seen it referenced here. Could someone point me to a proof?

The fundamental group of a closed hyperbolic 3-manifold is not a free product.

  • 3
    $\begingroup$ I think the point is that if a 3-manifold has $\pi_1$ a free product, it is a connected sum (Stallings?) and hence has $\pi_2\ne 0$ (sphere theorem) and thus isn't hyperbolic. $\endgroup$
    – Paul
    Mar 9, 2011 at 3:32
  • 1
    $\begingroup$ Your question is answered in my answer to mathoverflow.net/questions/46874/… $\endgroup$ Mar 9, 2011 at 13:46
  • $\begingroup$ Also by Robert Bell's answer to the same question... $\endgroup$
    – HJRW
    Mar 9, 2011 at 15:37

2 Answers 2


If $M$ is a closed $3$-manifold and $\pi_1(M) \cong A \ast B$ with $A$ and $B$ nontrivial, then Kneser's conjecture (which is a theorem -- the proof can be found in Hempel's book on 3-manifolds) says that we can write $M = M' \sharp M''$ where $M'$ and $M''$ are closed 3-manifolds with $\pi_1(M')=A$ and $\pi_1(M'')=B$. In particular, $M$ contains an embedded $2$-sphere which does not bound a ball (namely, the sphere from the connect sum decomposition). However, all embedded $2$-spheres in hyperbolic 3-manifolds bound balls, as can be seen by lifting to the universal cover.

  • $\begingroup$ This is the proof that was communicated orally to me, but I was looking for the name or to see it in a book somewhere. This is just what I needed. Thanks! $\endgroup$
    – JeremyKun
    Mar 9, 2011 at 22:27

One can also see it using the theory of ends. If $\pi_1M$ were freely decomposable, then it would follow from the easy direction of Stallings' Ends Theorem that $\pi_1M$ had two or infinitely many ends. On the other hand, by the Svarc--Milnor Lemma, $\pi_1M$ is quasi-isometric to hyperbolic 3-space, which has one end. Because ends are a quasi-isometry invariant, we are done.

This argument shows that this isn't really a 3-manifold fact. Indeed, it proves:

If $M$ is any closed manifold with universal cover homeomorphic to $\mathbb{R}^n$ for $n>1$ then $\pi_1M$ is freely indecomposable.

  • $\begingroup$ That's a nice observation! $\endgroup$ Mar 9, 2011 at 4:58
  • $\begingroup$ As Robert Bell points out in the question that Igor Belegradek links to above, in fact any contractible manifold of dimension at least two is one-ended. So the same result applies whenever $M$ is aspherical. $\endgroup$
    – HJRW
    Mar 9, 2011 at 15:46

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