Manifold with a nilpotent fundamental group.

Question

I am looking for a manifold with a nonabelian nilpotent fundamental group. I know the above terms, but I couldn't find out an example of that.

score 5 · Accepted Answer · answered Jun 08 '20 at 13:14

5

The Klein bottle has a fundamental group with a normal $\mathbb Z \oplus \mathbb Z$ subgroup and a $\mathbb Z / 2 \mathbb Z$ kernel.

answered Jun 08 '20 at 13:14

Lee Mosher

120,280

Thank you. I was looking for something like that. – Bartek Jun 09 '20 at 12:58

Bartek · Answer 2 · 2020-07-08T16:24:30.070

4

Probably an another example is here: The Heisenberg manifold, because the Heisenberg group is a nilpotent group.

edited Jul 08 '20 at 16:24

answered Jun 09 '20 at 13:05

Bartek

197

Yes, that's a good example. – Lee Mosher Jun 09 '20 at 14:12

score 3 · Answer 3 · answered Jun 08 '20 at 13:23

3

Let $M$ be a compact homogeneous flat pseudo-Riemannian manifold. Then the fundamental group of $M$ is $2$-step nilpotent, and there are examples where it is non-abelian.

answered Jun 08 '20 at 13:23

Dietrich Burde

130,978

Is $\mathbb{R}^n/\Gamma$ (where $\Gamma$ is a certain group of isometries of $\mathbb{R}^n$ without fixed points) an example? – Bartek Jun 09 '20 at 12:56
1

See here. – Dietrich Burde Jun 09 '20 at 13:00

Manifold with a nilpotent fundamental group.

3 Answers3