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Andy Putman
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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.

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 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.

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.

Source Link
Andy Putman
  • 44.8k
  • 14
  • 186
  • 272

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 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.