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Is there an open, connected subset of $\mathbb{R}^3$ with [edit: non-trivial] fundamental group that is finitely presented (I think I only need finitely generated but I'm actually interested in fp, so I'll ask the question that way) and perfect? This question here discusses the cases $\pi_1$ is not fg and perfect and finite and perfect.

FOLLOW-UP QUESTION: If $\pi_1(U)$ is fg (and hence fp, per this) and non-trivial, is it necessarily the case that $\pi_1(U)$ is free, or can there be relators? If there can be relators, what would be an example? Because, if $\pi_1(U)$ must be free if it is fg, we are done. (By the link in the first paragraph, $\pi_1(U)$ must be torsion-free.)

Jeffrey Rolland
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  • Yes, iff it is simply connected. Incidentally, for 3-manifold groups fg is the same as fp. – Moishe Kohan Jun 17 '19 at 22:04
  • Thanks so much for the reply! Where is a proof that there is no non-trivial fp perfect fundamental group of an open, connected subset of $\mathbb{R}^3$? – Jeffrey Rolland Jun 18 '19 at 02:11
  • http://pi.math.cornell.edu/~hatcher/3M/3M.pdf, Lemma 3.5. It follows that if $M$ is a compact 3-manifold then rank of the image of $H^1(M)\to H^1(\partial M)$ is $1/2$ of the rank of $H^1(\partial M)$. Thus, if $U\subset R^3$ is a connected open subset with f.g. perfect fundamental group then the boundary of the Scott compact core $M$ of $M$ is a disjoint union of spheres. From this, you conclude that $M$ is simply-connected. – Moishe Kohan Jun 18 '19 at 02:16
  • https://math.stackexchange.com/questions/679910/example-of-a-domain-in-r3-with-trivial-first-homology-but-nontrivial-fundament?rq=1 – Moishe Kohan Jun 18 '19 at 02:17
  • @TwoGun (@MoisheKohan) It's 4:00 AM and I'm just up for a minute, but I think I now understand your proof; thanks so much! It would still be interesting to see an example with $\pi_1(U)$ (non-trivial and) not free, but this absolutely answers the question. – Jeffrey Rolland Jun 18 '19 at 09:57
  • Hint: Take an open regular neighborhood of the torus embedded in $R^3$. – Moishe Kohan Jun 18 '19 at 15:06
  • Right, I did something like that with the two-holed torus; thanks so much for all your replies! – Jeffrey Rolland Jun 18 '19 at 15:54

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(From @MoiseKohan) No; see http://pi.math.cornell.edu/~hatcher/3M/3M.pdf, Lemma 3.5. It follows that if $M$ is a compact 3-manifold, then rank of the image of $H_1(M) \to H_1(\partial M)$ is 1/2 of the rank of $H_1(\partial M)$. Thus, if $U \subset \mathbb{R}^3$ is a connected open subset with f.g. perfect fundamental group, then the boundary of the Scott compact core $M$ of $U$ is a disjoint union of spheres. From this, you conclude that $M$ and $U$ are simply-connected. Incidentally, for 3-manifold groups, fg is the same as fp by the Scott compact core theorem. [This was lightly $\LaTeX$'ed and edited, all hopefully correctly.]

FOLLOW-UP ANSWER (by Jeff Rolland): Clearly, if $U$ is the subset of $\mathbb{R}^3$ with two disjoint circles of radius 1 in the $x$-$y$ plane, one centered on (0,-2,0) and the other on (0,2,0), and two lines, one $x=0$ and $y=-2$ and the other $x=0$ and $y=2$, deleted, then $U$ strong deformation retracts onto the two-holed torus $T^2 \# T^2$ and hence has non-free fundamental group.

Jeffrey Rolland
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