Is it true that any finitely presented group can be realized as fundamental group of compact 3-manifold with boundary?
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asked Nov 19, 2009 at 17:00
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18$\begingroup$ There's no need to apologize for a stupid question, and this isn't one. $\endgroup$– Tom ChurchCommented Nov 19, 2009 at 17:17
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1$\begingroup$ There is an important idea behind this question. We know that every finitely presented group is the fundamental group of a two-complex: namely, the presentation complex. Two-complexes are very low-dimensional, so it is reasonable to (try to!) embed them in a three-dimensional space. Or, equivalently, thicken them up to be three-dimensional. The various answers below (including my own) do not say why this does not work... In fact, this does work in dimension five. Every finitely presented group can be realised as the fundamental group of a compact five-manifold with boundary. $\endgroup$– Sam NeadCommented Sep 1, 2021 at 12:42
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2$\begingroup$ Dimension four is an interesting story that I'll skip. Here is an example of a two-complex that does not embed in any three-manifold. Let $K_n$ be the complete graph on $n$ vertices. Let $L$ be the two-complex with one-skeleton $K_6$ and where all three-cycles are filled in with triangles. The "vertex links" in $L$ are copies of $K_5$ (famously non-planar). This, then, is an obstruction to embedding in a three-manifold. $\endgroup$– Sam NeadCommented Sep 1, 2021 at 12:46
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1$\begingroup$ @SamNead: Thanks very much for mentioning this example! It turns out to be useful for something I'm doing right now. I suppose the easiest way to descibe $L$ is as the 2-skeleton of the 5-simplex. $\endgroup$– HJRWCommented Sep 6, 2021 at 8:48
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1$\begingroup$ @HJRW - Yes. I did not come to it that way, but that is a much tidier description. I will point out (for my future self, mostly) that L does not need to be the full two-skeleton. It is enough to fix a vertex $v$ of $K_6$ and only add the triangles meeting $v$. This is supposed to underline the "local" nature of this obstruction. $\endgroup$– Sam NeadCommented Sep 6, 2021 at 13:40
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4 Answers
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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 Autumn 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).
EDIT:
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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1$\begingroup$ Do you know if there is an algorithm to decide if a group of 3-mnfld-with-bry is trivial? $\endgroup$ Commented Nov 20, 2009 at 18:15
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2$\begingroup$ 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. $\endgroup$ Commented Nov 20, 2009 at 20:25
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$\begingroup$ 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. $\endgroup$– HJRWCommented Nov 21, 2009 at 17:51
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$\begingroup$ Could you explain a little bit how $M$ turns out to be a retract of its double $D$? $\endgroup$– MaharanaCommented Dec 27, 2009 at 7:05
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1$\begingroup$ Charlie, I don't think I understand your comment. Certainly, everything Richard has said is correct. Are you suggesting that something else isn't correct? $\endgroup$– HJRWCommented Mar 8, 2010 at 17:46
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No. The Baumslag--Solitar groups $\langle a, b | ab^m a^{-1} = b^n \rangle$ are not three-manifold groups when $m \neq n$.
See:
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)
answered Nov 19, 2009 at 17:05
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$\begingroup$ Are you sure that manifolds there can haave boundary? $\endgroup$ Commented Nov 19, 2009 at 17:22
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1$\begingroup$ Yes. These groups are never the fundamental group of any 3-manifold. $\endgroup$ Commented Nov 19, 2009 at 17:27
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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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3$\begingroup$ 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. $\endgroup$ Commented Mar 22, 2011 at 2:36
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1$\begingroup$ 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. $\endgroup$ Commented Mar 24, 2011 at 19:18
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If a finite group $G$ is (isomorphic to) the fundamental group of a three-manifold, then $G$ embeds in $\mathrm{SO}(4)$.
To see this, suppose that $M$ is a compact, connected three-manifold, possibly with boundary, having $G = \pi_1(M)$ finite. If $M$ has two-sphere boundary components, we can cap them off without changing the fundamental group. Since free products of non-trivial groups are always infinite, we deduce that $G$ is freely indecomposible. Appealing to the Poincaré conjecture [solved by Perelman], we have that $M$ is irreducible: all two-spheres in $M$ bound three-balls.
If $M$ is non-orientable, then a theorem of Livesay implies that $M$ is homeomorphic to the real projective plane crossed with a unit interval.
Thus we reduce to the case where $M$ is compact, oriented, connected, irreducible, and has finite fundamental group. All boundary components of $M$ are now oriented, and of genus at least one. Applying "one-half lives, one-half dies" we find that $M$ has no boundary components. Appealing to the elliptic part of the geometrisation conjecture [solved by Perelman] we find that the universal cover of $M$ is the three-sphere, and the deck group, and thus $G$, is (conjugate to a) subgroup of $\mathrm{SO}(4)$. QED
A quick google search finds a paper of Zimmermann giving a readable introduction to the finite subgroups of $\mathrm{SO}(4)$ - see section three of that paper. One then has to determine which of these act freely. Finally, there is another approach to this problem via spherical Seifert fibered spaces.
As a concrete example of a finite group that is not a three-manifold group, consider $\mathbb{Z}_p \times \mathbb{Z}_q$ where $p$ and $q$ are not coprime. This is a isomorphic to a subgroup of $\mathrm{SO}(4)$, but it cannot act freely. [See Theorem 9.14 in Hempel's book for a proof that, in this special case, avoids geometrisation.]