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Given an infinitely-presented group $G$, can I get a smooth manifold $M$ whose fundamental group $\pi_1(M)\cong G$?

This is true if $G$ is finitely-presented and $n\geq 4$, for $n=3$ this post gives some hints on why it is false.

I know there are some restrictions, for given a compact connected smooth manifold, it admits a Riemannian metric such that pulling it back via the universal covering, its fundamental group acts geometrically. In particular, we can prove that $\pi_1(G)$ is finitely-presented.

SubGui
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    The group $G=\bigoplus_{i=1}^{\infty} \Bbb Z$ does not arise as a fundamental group of a smooth manifold. – Dietrich Burde Jun 19 '25 at 19:19
  • Obviously no big enough group can be a fundamental group of a manifold (smooth or not) As in: group with cardinality above $2^c$. – freakish Jun 19 '25 at 19:51
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    You need a countable group for this, presentation is irrelevant. I think it is a duplicate question though: https://math.stackexchange.com/questions/3526063/manifold-with-infinitely-many-based-loops-but-only-two-unbased-loops-up-to-homo/3527567#3527567 – Moishe Kohan Jun 19 '25 at 20:39
  • I'm particularly interested in a finitely generated, but not finitely-presented group, the Grigorchuk group. Is it possible to construct a smooth manifold having it as its fundamental group? – SubGui Jun 20 '25 at 00:13
  • Can you prove that every finitely generated group is countable? If you can, read the linked question. – Moishe Kohan Jun 23 '25 at 20:16

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