---
tag: JA83J040Q2
kind: definition
name: Nilmanifold
msc: ["22E25", 22F30, 53C30]
defines: 43NRBE72C6
uses: [DK87YN458D, 85TWR6E75P, HBDEPTRF11, BTYF5Y5PFY, ZZYCV5EX9J]
sources:
  - url: https://en.wikipedia.org/wiki/Nilmanifold
---

## Definition

Let $N$ be a connected [[nilpotent Lie group|DK87YN458D]] and let $H$ be a closed subgroup of $N$.
The [[homogeneous space|85TWR6E75P]] $N/H$ is a *nilmanifold*.
When $N$ is simply connected and $H = \Gamma$ is a [[lattice|BTYF5Y5PFY]], a discrete subgroup with $N/\Gamma$ compact, the space $N/\Gamma$ is a *compact nilmanifold*.

## Remarks

Example: the [[Heisenberg group|ZZYCV5EX9J]] $N$ 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|HBDEPTRF11]].
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.