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What is the Classifying Space of the Discrete Heisenberg Group? Which paper/book contains a detailed proof?

Thank you for your time.

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The Heisenberg manifold, which is the quotient of the real Heisenberg group by the discrete Heisenberg (sub)group – Fernando Muro Jan 1 '12 at 2:54
up vote 11 down vote accepted

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

$\Gamma=\mathbb{Z}^2\rtimes\mathbb{Z}$, where $\mathbb{Z}$ acts by (powers of) $\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right)$, 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.

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@Alain Valette, you have mentioned that $\Gamma$ can be viewed as the semi-direct of $\mathbb{Z}^2$ and $\mathbb{Z}$; did you mean that $n\in\mathbb{Z}$ is mapped to the element $f_n\in Aut(\mathbb{Z}^2)$, such that $f_n(m_1,m_2)=m_1+nm_2$? Then why is $\Gamma$ isomorphic to the semi-direct product? It seems that the central extension $0\rightarrow\mathbb{Z}\rightarrow\Gamma\rightarrow\mathbb{Z}^2\rightarrow0$ does not split; then is there any nature semi-direct structure from this central extension? Thank you! – Zuriel Jan 2 '12 at 2:57
@ Zuriel. The central extension does not split. But you have another short exact sequence $0\rightarrow\mathbb{Z}^2\rightarrow\Gamma\rightarrow\mathbb{Z}\rightarrow 0$ (which of course splits) given as follows: give $\Gamma$ the presentation $\Gamma=<x,y,z|z=[x,y],[x,z]=[y,z]=1>$. Then the subgroup generated by $z$ and $y$ is normal and isomorphic to $\mathbb{Z}^2$, and the action of $x$ by conjugation on this subgroup is given by the 2-by-2 matrix mentioned in my answer. – Alain Valette Jan 2 '12 at 9:21

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 $\Gamma$.

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 $\mathbb{Z}^n$.

Of course, the Malcev completion of the discrete Heisenberg group is the nondiscrete Heisenberg group.

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A question: will $\Gamma$ generally be a uniform (cocompact) lattice in $G$? – Mark Jan 1 '12 at 2:40
@Mark: yes, the standard definition of the Malcev completion requires that $\Gamma$ is cocompact in $G$. This also makes Malcev completion unique. @Andy, I think you wanted to say that a simply connected nilpotent Lie group is diffeomorphic to $\mathbb R^n$. – Vitali Kapovitch Jan 1 '12 at 3:06
@Mark: all lattices in nilpotent Lie groups are cocompact. – Tom Church Jan 1 '12 at 3:57
@Tom Church Oops, I guess the standard definition of a lattice requires finite volume of the quotient. Then cocompactness is of course automatic for nilpotent groups. I've often had to work with discrete noncocompact subgroups of nilpotent Lie groups which I often also call lattices but I guess that's wrong terminology. – Vitali Kapovitch Jan 1 '12 at 4:20
@Vitali : Thanks for the correction! I've been celebrating the new year, so I'm pretty happy I managed to at least put the word "connected" in all the right places... – Andy Putman Jan 1 '12 at 5:05

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