Are all Nilmanifolds quotients of Heisenberg Group

Question

I've been reading some wonderful blog entries where Terry Tao and Ben Green prove some generalizations of Weyl Equidstribution using a "higher" Fourier Analysis. Unfortunately, all the information I can find about nilmanifolds is embedded for some difficult papers (at least for non-Harmonic analysts. also this one).

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Weyl Equidistribution says you can take remainders (mod 1) of $\{n \alpha\}$ for 0 < n < N and the odds of it lying in any interval approaches the uniform distribution. (I think it's even known how quickly it converges.)

You can tell this sequence apart from purely random numbers in other ways. The "gaps" in $\{ \{ n \alpha\}: 0 < n \leq N \}$ take up to three values while random numbers on the circle have Poisson distributed gaps.

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The only example of a Nilmanifold I'll mention is the Heisenberg nilmanifold.
$$ \left( \begin{array}{ccc} 1 & \mathbb{R} & \mathbb{R} \\\\ 0 & 1 & \mathbb{R}\\\\ 0 & 0 & 1 \end{array} \right) \mod \left( \begin{array}{ccc} 1 & \mathbb{Z} & \mathbb{Z} \\\\ 0 & 1 & \mathbb{Z}\\\\ 0 & 0 & 1 \end{array} \right) $$ This quotient space is the unit cube, but the quotient map is funny. I think it's $(x,y,z) \equiv ( \{ x\}, \{y\}, z - \lfloor z - x\lfloor y \rfloor \rfloor )$. It's not even clear to me the third coordinate lies in [0,1]. This can be extended to n by n matrices. Are all Nilmanifolds quotients of the Heisenberg group in this way? .

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To prove equidistribution, you show the average value $e^{ i n \alpha }$ as 0 < n < N approaches 0 as N gets large. If you move around a circle enough, it's kinda of intuitive that your average location is in the center.

With these nilmanifolds (which the circle S¹ is also an example) you can get equidistribution for "bracket polynomials" like $\alpha n \lfloor \beta n \rfloor $ mod 1. What are the analogues of the Fourier coefficients here?

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Clarification, I'm using "Heisenberg group" to mean the group of upper-triangular matrices with real entries above the diagonal. I guess I'm trying to ask if groups like $$ \left( \begin{array}{cccc} 1 & \mathbb{R} & \mathbb{R} & \mathbb{R} \\\\ 0 & 1 & \mathbb{R} & \mathbb{R} \\\\ 0 & 0 & 1 & \mathbb{R} \\\\ 0 & 0 & 0 & 1 \end{array} \right) \mod \left( \begin{array}{cccc} 1 & \mathbb{R} & \mathbb{Z} & \mathbb{Z} \\\\ 0 & 1 & \mathbb{Z} & \mathbb{Z} \\\\ 0 & 0 & 1 & \mathbb{Z} \\\\ 0 & 0 & 0 & 1 \end{array} \right) $$ are nilmanifolds. [there's no typo] And if all of them look like that? It seems one way to build a nilpotent lie group abstractly is to take a lie group and quotient out $G^{(n)} = [G^{n-1},G]$.

Answer 1

The answer to the question in the title is emphatic no in higher dimensions. In dimension $3$ nilmanifolds (that are not tori) are indeed quotients of the Heisenberg group.

There are lots of nilmanifolds in each dimension $>2$, in fact nilpotent Lie algebras are not classified (and probably not classifiable as there are too many of them with no apparent structure), and as long as all structure constants of the Lie algebra are rational the corresponding Lie group has a torsion free lattice, whose quotient is a nilmanifold.

Topologically nilmanifolds are precisely the iterated principal circle bundles, e.g. in dimension $3$ any principal circle bundle over a $2$-torus is a nilmanifold. Such circle bundles are classified by the Euler number which can take any value in $\mathbb Z$, so there are countably many $3$-dimensional nilmanifolds. In dimensions $1$ and $2$ the only nilmanifolds are tori.