Nilmanifold

Remarks

Example: the Heisenberg group $N$ of real matrices $\begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}$ and the subgroup $\Gamma$ of those with $x, y, z \in \Z$. The group $N$ is nilpotent, since its commutator subgroup is the center $\set{x = y = 0}$. It is diffeomorphic to $\R^3$, so it is simply connected. The subgroup $\Gamma$ is discrete, and each element of $N$ is an element with $x, y, z \in [0, 1)$ times an element of $\Gamma$, so $N/\Gamma$ is compact.

Non-example for the compact case: let $\Z$ be the integer points on the first axis of $\R^2$. Then $\R^2/\Z$ is a nilmanifold, but $\Z$ is not a lattice in $\R^2$, and the cylinder $\R^2/\Z$ is not compact.

Each nilmanifold is a solvmanifold. Many texts use the word nilmanifold only for compact nilmanifolds. By a theorem of Malcev, a simply connected nilpotent Lie group has a lattice exactly when its Lie algebra has a basis with rational structure constants.