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).

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.

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? .
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?

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 $G^{(n)} = [G^{n-1},G]$.

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    $\begingroup$ Your first question is answered by the Wikipedia article on nilmanifolds, and the answer is no. Any quotient of a nilpotent Lie group is a nilmanifold, in particular one can take a nilpotent Lie group of dimension larger than 3. $\endgroup$ – Qiaochu Yuan Nov 23 '10 at 2:01
  • $\begingroup$ those are also Heisenberg groups, no? $\endgroup$ – john mangual Nov 23 '10 at 3:19

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.

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  • $\begingroup$ @Igor: In some sense the answer is still yes, because in every nilpotency class there exists a free nilpotent group of that class with any given number of free generators. The corresponding Lie group is the "freest" in the sense that all other nil-manifolds with fund. group of that nilpotency class are factors of that Lie group. Right? The Heisenberg group is the free nilpotent group with 2 generators of class 2. $\endgroup$ – user6976 Nov 23 '10 at 2:26
  • $\begingroup$ @Mark, I do not know much about free nilpotent groups, but I do not see why any torsion-free fg nilpotent group would be a lattice in a free nilpotent Lie group. $\endgroup$ – Igor Belegradek Nov 23 '10 at 3:08
  • $\begingroup$ @Igor. I'm asking if all nilmanifolds arise as subgroups of n x n upper triangular matrices. $\endgroup$ – john mangual Nov 23 '10 at 3:24
  • $\begingroup$ I believe a dimension count should show that the $n\times n$ upper-triangular matrices are never the free $k$-step nilpotent group on $\ell$ generators except for $(n,k,\ell)=(3,2,2)$; certainly it's not true for general $n$. @John: I think you mean "arise as quotients of upper-triangular matrices". $\endgroup$ – Tom Church Nov 23 '10 at 3:32
  • $\begingroup$ @John, any finitely generated nilpotent torsion-free group is isomorphic to a subgroup of the group $UT(n, \mathbb Z)$ of upper triangular matrices with integer coefficients for some large $n$, but not as a subgroup of finite index (so the quotient space of $UT(n,\mathbb R)$ by the subgroup embedded is not compact). $\endgroup$ – Igor Belegradek Nov 23 '10 at 3:39

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