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

17$\begingroup$ There's no need to apologize for a stupid question, and this isn't one. $\endgroup$– Tom ChurchNov 19, 2009 at 17:17

1$\begingroup$ There is an important idea behind this question. We know that every finitely presented group is the fundamental group of a twocomplex: namely, the presentation complex. Twocomplexes are very lowdimensional, so it is reasonable to (try to!) embed them in a threedimensional space. Or, equivalently, thicken them up to be threedimensional. 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 fivemanifold with boundary. $\endgroup$– Sam NeadSep 1, 2021 at 12:42

2$\begingroup$ Dimension four is an interesting story that I'll skip. Here is an example of a twocomplex that does not embed in any threemanifold. Let $K_n$ be the complete graph on $n$ vertices. Let $L$ be the twocomplex with oneskeleton $K_6$ and where all threecycles are filled in with triangles. The "vertex links" in $L$ are copies of $K_5$ (famously nonplanar). This, then, is an obstruction to embedding in a threemanifold. $\endgroup$– Sam NeadSep 1, 2021 at 12:46

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 2skeleton of the 5simplex. $\endgroup$– HJRWSep 6, 2021 at 8:48

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 twoskeleton. 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 NeadSep 6, 2021 at 13:40
4 Answers
A couple of extra points.
Any compact 3manifold with boundary $M$ can be doubled to give a closed 3manifold $D$. 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 BaumslagSolitar groups that Autumn mentions above) applies just as well to compact 3manifolds as closed 3manifolds.
Other classes of poison subgroups can be constructed from cohomological conditions. The KneserMilnor Theorem implies that any closed, irreducible 3manifold 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 3manifold (and hence of a compact 3manifold, by the previous paragraph).
EDIT:
Oh, and yet another source of poison subgroups comes from Scott's theorem that 3manifold 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.

1$\begingroup$ Do you know if there is an algorithm to decide if a group of 3mnfldwithbry is trivial? $\endgroup$ Nov 20, 2009 at 18:15

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 threesphere. You can use Rubinstein's algorithm to recognize the threesphere to do this. $\endgroup$ Nov 20, 2009 at 20:25

$\begingroup$ A less intelligent but more simpleminded algorithm is just to apply geometrization and deduce that the word problem is (uniformly) solvable in 3manifold groups. If you know how to solve the word problem then it's easy to tell if a group is trivial. $\endgroup$– HJRWNov 21, 2009 at 17:51

$\begingroup$ Could you explain a little bit how $M$ turns out to be a retract of its double $D$? $\endgroup$– MaharanaDec 27, 2009 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. $\endgroup$– HJRWDec 27, 2009 at 15:57
No. The Baumslag solitar groups $\langle a, b  ab^m a^{1} = b^n \rangle$ are not $3$manifold groups when $m \neq n$.
See
Heil, Wolfgang H. Some finitely presented non$3$manifold groups. Proc. Amer. Math. Soc. 53 (1975), no. 2, 497500.
(See also Peter Shalen, ThreeManifolds and BaumslagSolitar groups. Topology Appl. 110 (2001), 113118)

$\begingroup$ Are you sure that manifolds there can haave boundary? $\endgroup$ Nov 19, 2009 at 17:22

1$\begingroup$ Yes. These groups are never the fundamental group of any 3manifold. $\endgroup$ Nov 19, 2009 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 3dimensional 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 3manifold 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.)

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 3manifold 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. $\endgroup$ Mar 22, 2011 at 2:36

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. $\endgroup$ Mar 24, 2011 at 19:18
If a finite group $G$ is (isomorphic to) the fundamental group of a threemanifold, then $G$ embeds in $\mathrm{SO}(4)$.
To see this, suppose that $M$ is a compact, connected threemanifold, possibly with boundary, having $G = \pi_1(M)$ finite. If $M$ has twosphere boundary components, we can cap them off without changing the fundamental group. Since free products of nontrivial 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 twospheres in $M$ bound threeballs.
If $M$ is nonorientable, 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 "onehalf lives, onehalf 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 threesphere, 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 threemanifold 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.]