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Consider $M = H_{\mathbb{R}}/H_{\mathbb{Z}}$, where $H_{\mathbb{R}} = \lbrace\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1\\ \end{bmatrix}|x,y,z \in \mathbb{R}\rbrace$ and $H_{\mathbb{Z}} = \lbrace\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1\\ \end{bmatrix}|x,y,z \in \mathbb{Z}\rbrace$

If we regards $M$ as a manifold (Only need to consider $M$ as topological manifold).

What is the fundamental group of $M$, $\pi_1(M)$? How to deduce that?

Thank you very much!

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  • $\begingroup$ I'm curious: did you lose interest in the question, are you not satisfied with the answer or what else is going on? $\endgroup$ – t.b. Apr 27 '12 at 3:16
  • $\begingroup$ Sorry~ because there something bother me which makes me forget to login the forum so many days. I really appreciate your help and your answer, and I can understand your answer! :) $\endgroup$ – Peter Hu Apr 29 '12 at 16:34
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If a group $G$ acts properly discontinuously and freely on a simply connected manifold $M$ then $\pi_1(M/G) \cong G$. Apply this to the present situation with $M = H_\mathbb{R}$ (note that $M$ is diffeomorphic to $\mathbb{R}^3$) and $G = H_\mathbb{Z}$ to see that $\pi_1(H_\mathbb{R}/ H_\mathbb{Z}) \cong H_\mathbb{Z}$ — Note that a discrete subgroup of a Lie group always acts properly discontinuously on the surrounding Lie group.

For a proof of this, see e.g. Corollary 4.11 on page 44 here. You should be able to find this result or closely related ones in any decent book on algebraic or differential topology.

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