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It is known that the fundamental group of a locally path connected, path connected compact metric space is finitely presented or uncountable. Furthermore the fundamental group of every manifold is countable so the fundamental group of every compact manifold must be finitely presented.

Q1: Is there a non-compact manifold whose fundamental group is isomorphic to $\mathbb{Q}$, the group of rational numbers?

 

Q2: Or more general, for given countable group $G$, is there a manifold whose fundamental group is isomorphic to $G$?

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    $\begingroup$ Yes. Any countable group is a colimit of a sequence of finitely presentegd groups (indexed by the natural numbers). For each of these finitely presented groups, take a finite 2-complex whose fundamental group is that group. Then build the mapping telescope. That can be properly embedded in $\mathbb R^5$ (since the 2-complexes are embedable in $\mathbb R^4$), I guess, and you can take a tubular neighbourhood, which is your desired manifold. $\endgroup$ – Fernando Muro Jan 5 '15 at 20:20
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    $\begingroup$ In fact, since $\mathbb Q$ is a countable group of finite cohomological dimension one can even realize $\mathbb Q$ as the fundamental group of a manifold whose universal cover is $\mathbb R^n$, for some $n$. $\endgroup$ – Igor Belegradek Jan 5 '15 at 21:33
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    $\begingroup$ @Alain Valette: I am not sure about the smallest $n$ but some $n$ can be easily found. A process of building a desired manifold is sketched in (5) on page 25 in arxiv.org/abs/1208.5220, and at each step one can find an explicit $n$. My guess is $n=5$ for $\mathbb Q$. $\endgroup$ – Igor Belegradek Jan 5 '15 at 23:13
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    $\begingroup$ There is a 3-manifold whose fundamental group is $\mathbb{Q}$, which I learned of from a preprint of Geoffrey Mess. It is obtained as a nested union of solid tori indexed by $\mathbb{N}$, so that the $n$th solid torus wraps around the $n+1$st $n$ times. $\endgroup$ – Ian Agol Jan 5 '15 at 23:30
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    $\begingroup$ An infinitely generated abelian group is the fundamental group of a 3-manifold if and only if it is a subgroup of $\mathbb Q$. This is proved by Evans and Moser in "Solvable fundamental groups of compact 3-manifolds", which is easy to find online. They work out the example mentioned by Agol. $\endgroup$ – Igor Belegradek Jan 6 '15 at 3:25
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This was answered in the comments. The answer to both questions is "Yes". As Fernando Muro mentioned, any countable group is the fundamental group of a manifold that can be embedded in $\mathbb{R}^5$.

There is an open subset of $\mathbb{R}^3$ with fundamental group $\mathbb{Q}$, the complement of a particular solenoid.

Let $C_1$ be the complement of an unknotted solid torus. $\pi_1(C_1) = \langle x_1 \rangle$. Consider an unknotted solid torus inside the first which wraps around it twice. Let the complement of this be $C_2$. There is an inclusion $\pi_1(C_1) \hookrightarrow \pi_1(C_2) = \langle x_2 \rangle$ so that $x_1 = x_2^2$, or in additive notation, $x_1 = 2x_2$.

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Let the $n$th solid torus wrap around the inside of the $n-1$st solid torus $m_n=n$ times. Let $C_n$ be the complement of this solid torus. There is an inclusion $\pi_1(C_{n-1}) \hookrightarrow \pi_1(C_n) = \langle x_n \rangle$ so that $x_{n-1} = n x_n$.

The solenoid is the intersection of this infinite sequence of nested solid tori. The intersection of the solenoid with a meridianal disk of the first solid torus is a Cantor set. The solenoid can be viewed as a mapping cylinder of an automorphism of a Cantor set. The complement $C$ is the union of the complements. Any loop in the complement is contained in some $C_n$. The fundamental group is the direct limit, isomorphic to $\mathbb{Q}$ with $x_n = 1/n!$.

If you choose $m_n=2$ instead of $m_n=n$, the solenoid complement has a fundamental group isomorphic to the dyadic rationals. You can select other subgroups of $\mathbb{Q}$ by varying the sequence $( m_n )$.

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  • $\begingroup$ Can you please explain this line "The solenoid can be viewed as a mapping cylinder of an automorphism of a Cantor set" $\endgroup$ – Anubhav Mukherjee Jun 13 '16 at 14:30
  • $\begingroup$ @Anubhav If you choose a meridianal disk $D$, then each solid torus $T_n$ is the mapping cylinder of some automorphism on the disjoint union of discs given by $T_n \cap D$. The solenoid is $\bigcap_{n \geq 1} T_n$, and the cross-section $D \cap \bigcap_{n \geq 1} T_n = \bigcap_{n \geq 1} (T_n \cap D)$ is a Cantor set. $\endgroup$ – S. Carnahan Jun 13 '16 at 23:28
  • $\begingroup$ Thanks I got it, and can you explain me why $\pi_1(C)=\mathbb Q$... I didnt get that last night. How exactly direct limit implies this result...it seems unclear to me. $\endgroup$ – Anubhav Mukherjee Jun 14 '16 at 13:04
  • $\begingroup$ @Anubhav I will leave that as an exercise. $\endgroup$ – S. Carnahan Jun 14 '16 at 16:06

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