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Equivalently: can the fundamental group of a closed manifold with universal cover homeomorphic to $\mathbb{R}^n$ be generated by $k$ elements, with $k < n$?

As a bonus question: same question without the "free" requirement on the action.

(to be sure: closed = compact + without boundary).

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    $\begingroup$ arr... maybe the discrete three dimensional heisenberg group (generated by two elements) acting on $\R^3$? $\endgroup$ – fritz Nov 16 '16 at 14:54
  • $\begingroup$ I am not familiar with the construction to say if the universal cover is ever $\mathbb R^4$, but it is known that every finitely presented group is the fundamental group of compact smooth 4 manifold. You might want to look into that. $\endgroup$ – user29123 Nov 16 '16 at 16:26
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    $\begingroup$ Many closed hyperbolic 3-manifolds have fundamental group generated by $2$ elements. Indeed, if you glue two genus 2 handlebodies by a random map of their boundaries the resulting 3-manifold is highly likely to have a hyperbolic structure. $\endgroup$ – Lee Mosher Nov 16 '16 at 18:20
  • $\begingroup$ Thank you @LeeMosher for your thorough and convincing answer! If you want to put your comment as an answer I will select it. $\endgroup$ – fritz Nov 17 '16 at 14:34
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Many closed hyperbolic 3-manifolds have fundamental group generated by 2 elements. Indeed, if you glue two genus 2 handlebodies by a random map of their boundaries the resulting 3-manifold is highly likely to have a hyperbolic structure.

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Let $F$ be a free group on 2 generators. Let $N$ be a normal subgroup of index $k-1$ (e.g. the kernel of a homomorphism onto a cyclic group of order $k-1$); then $N$ is free of rank $k$. Then the 2-generated group $F/[N,N]$ is torsion-free (this is an old not-trivial fact) and virtually $\mathbf{Z}^k$ and thus acts isometrically freely properly cocompactly on the Euclidean $k$-space.

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