Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary?

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    $\begingroup$ There's no need to apologize for a stupid question, and this isn't one. $\endgroup$ – Tom Church Nov 19 '09 at 17:17

A couple of extra points.

Any compact 3-manifold with boundary $M$ can be doubled to give a closed 3-manifold $D$. As $M$ is a retract of $D$, it follows that $\pi_1(M)$ injects into $\pi_1(D)$. Therefore, any "poison subgroup" (such as the Baumslag--Solitar groups that Richard mentions above) applies just as well to compact 3-manifolds as closed 3-manifolds.

Other classes of poison subgroups can be constructed from cohomological conditions. The Kneser--Milnor Theorem implies that any closed, irreducible 3-manifold with infinite fundamental group is aspherical. It follows that any freely indecomposable infinite group with cohomologial dimension greater than 3 cannot be a subgroup of a closed 3-manifold (and hence of a compact 3-manifold, by the previous paragraph).


Oh, and yet another source of poison subgroups comes from Scott's theorem that 3-manifold groups are coherent, meaning that every finitely generated subgroup is finitely presented. This rules out subgroups like $F\times F$ (where $F$ is a free group), which is not coherent.

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    $\begingroup$ Do you know if there is an algorithm to decide if a group of 3-mnfld-with-bry is trivial? $\endgroup$ – Anton Petrunin Nov 20 '09 at 18:15
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    $\begingroup$ There is an algorithm to determine if $\pi_1(M)$ is trivial for compact $M$: If the boundary has a component that is not a sphere, then $M$ will have nontrivial homology. If the boundary is a union of spheres, then cap them off with balls to get a closed manifold $N$. Then, by the Poincare conjecture, which we now know, the question is whether or not $N$ is the three-sphere. You can use Rubinstein's algorithm to recognize the three-sphere to do this. $\endgroup$ – Autumn Kent Nov 20 '09 at 20:25
  • $\begingroup$ A less intelligent but more simple-minded algorithm is just to apply geometrization and deduce that the word problem is (uniformly) solvable in 3-manifold groups. If you know how to solve the word problem then it's easy to tell if a group is trivial. $\endgroup$ – HJRW Nov 21 '09 at 17:51
  • $\begingroup$ Could you explain a little bit how $M$ turns out to be a retract of its double $D$? $\endgroup$ – Maharana Dec 27 '09 at 7:05
  • $\begingroup$ Maharana, sure. D = M<sub>1</sub> U M<sub>2</sub>, where the M<sub>i</sub> are copies of M. Now the identifications M<sub>i</sub>->M agree on their boundaries, so extend to a map D->M. This is the retraction. $\endgroup$ – HJRW Dec 27 '09 at 15:57

No. The Baumslag solitar groups $\langle a, b | ab^m a^{-1} = b^n \rangle$ are not $3$-manifold groups when $m \neq n$.


Heil, Wolfgang H. Some finitely presented non-$3$-manifold groups. Proc. Amer. Math. Soc. 53 (1975), no. 2, 497--500.

(See also Peter Shalen, Three-Manifolds and Baumslag-Solitar groups. Topology Appl. 110 (2001), 113--118)

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  • $\begingroup$ Are you sure that manifolds there can haave boundary? $\endgroup$ – Anton Petrunin Nov 19 '09 at 17:22
  • $\begingroup$ Yes. These groups are never the fundamental group of any 3-manifold. $\endgroup$ – Autumn Kent Nov 19 '09 at 17:27

I recently heard of a result due to Aitchison and Reeves which shows that any finitely presented group arises as the fundamental group of a 3-dimensional orbifold (where fundamental group means the topological and not the orbifold fundamental group). In fact, they say that the orbifold can be taken to be the quotient of a closed oriented hyperbolic 3-manifold by an isometric involution with isolated fixed points, all modelled on $x\mapsto -x$.

(I'm certainly no expert on this topic, just passing on what I heard.)

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    $\begingroup$ I asked Aitchison, he only can make it to be fundamental group of $M^3/\mathbb Z_2$ where $M^3$ is closed orientable 3-manifold and $\mathbb Z_2$ acts on $M$ with isolated fixed points. The question if $M$ can be made hyperbolic and $\mathbb Z_2$ action isometric is not yet resolved. $\endgroup$ – Anton Petrunin Mar 22 '11 at 2:36
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    $\begingroup$ Ah, that's good to know - I must have misunderstood, although I thought I was quite insistent on knowing about the hyperbolic case. Still, that's quite a while ago now, probably I've misremembered things. $\endgroup$ – Joel Fine Mar 24 '11 at 19:18

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