The Classifying Space of the Discrete Heisenberg Group - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T10:43:03Z http://mathoverflow.net/feeds/question/84668 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84668/the-classifying-space-of-the-discrete-heisenberg-group The Classifying Space of the Discrete Heisenberg Group Zuriel 2012-01-01T01:48:31Z 2012-01-01T09:00:35Z <p>What is the Classifying Space of the Discrete Heisenberg Group? Which paper/book contains a detailed proof?</p> <p>Thank you for your time.</p> http://mathoverflow.net/questions/84668/the-classifying-space-of-the-discrete-heisenberg-group/84669#84669 Answer by Andy Putman for The Classifying Space of the Discrete Heisenberg Group Andy Putman 2012-01-01T02:22:33Z 2012-01-01T05:04:26Z <p>If $\Gamma$ is a finitely generated torsion-free nilpotent group, then Malcev proved that there is a connected nilpotent Lie group $G$ such that $\Gamma$ is a lattice in $G$. The Lie group $G$ is often called the Malcev completion of $\Gamma$. It is an easy exercise to show that a connected simply-connected nilpotent Lie group is homeomorphic to $\mathbb{R}^n$. It follows that $G/\Gamma$ is a classifying space for $\Gamma$.</p> <p>The baby example of this is $\Gamma = \mathbb{Z}^n$ and $G = \mathbb{R}^n$, so we obtain the usual $n$-torus $\mathbb{R}^n/\mathbb{Z}^n$ for the classifying space of $\mathbb{Z}^n$.</p> <p>Of course, the Malcev completion of the discrete Heisenberg group is the nondiscrete Heisenberg group.</p> http://mathoverflow.net/questions/84668/the-classifying-space-of-the-discrete-heisenberg-group/84676#84676 Answer by Alain Valette for The Classifying Space of the Discrete Heisenberg Group Alain Valette 2012-01-01T09:00:35Z 2012-01-01T09:00:35Z <p>As was said by Andy, the classifying space of the discrete Heisenberg group $\Gamma$ is $B\Gamma=G/\Gamma$, where $G$ is the 3-dimensional Heisenberg group over the reals. Due to the central extension $$0\rightarrow\mathbb{Z}\rightarrow\Gamma\rightarrow\mathbb{Z}^2\rightarrow 0$$ you may view $B\Gamma$ as a circle bundle over the 2-torus. Alternatively, viewing $\Gamma$ as the semi-direct product </p> <p>$\Gamma=\mathbb{Z}^2\rtimes\mathbb{Z}$, where $\mathbb{Z}$ acts by (powers of) <code>$\left(\begin{array}{cc} 1 &amp; 1 \\ 0 &amp; 1 \end{array}\right)$</code>, you can view $B\Gamma$ as the mapping torus of this automorphism of $B\mathbb{Z}^2=\mathbb{T}^2$, i.e. as a 2-torus bundle over the circle.</p>