The simplest non-Abelian example

Share this page

It is worth considering the simplest non-abelian example more closely. The integer Heisenberg group $H_{\mathbb Z}$ is the simplest non-trivial example of a nilpotent group. It is generated by two elements $u,v$ with the relations that the commutator $w = uvu^{-1}v^{-1}$ commutes with $u$ and $v$. Explicitly, it is the group of $3\times 3$ upper-triangular matrices with integer entries and diagonal entries 1: take $u$ and $v$ to be the matrices

  \[ u = \left( \begin{array}{ccc} 1 &  1 &  0 \\ 0 &  1 &  0 \\ 0 &  0 &  1 \end{array} \right),\   v = \left( \begin{array}{ccc} 1 &  0 &  0 \\ 0 &  1 &  1 \\ 0 &  0 &  1 \end{array} \right), \]    

so that

  \[ w = \left( \begin{array}{ccc} 1 &  0 &  1 \\ 0 &  1 &  0 \\ 0 &  0 &  1 \end{array} \right). \]    

Any element of $H_{\mathbb Z}$ can be written uniquely in the form $u^ k v^ l w^ m$ for some integers $k,l,m$.

The group $H_{\mathbb Z}$ sits inside the more usual 3-dimensional Heisenberg group $H$ consisting of the strictly upper-triangular matrices with real entries

  \[ \left( \begin{array}{ccc} 1 &  x &  z \\ 0 &  1 &  y \\ 0 &  0 &  1 \end{array} \right). \]    

By analogy with the case of $\mathbb Z^2$ in the plane, you might expect the group $H_{\mathbb Z}$ to converge to $H$ under re-scaling. As a manifold, the group $H$ is just $\mathbb R^3$, so you would predict that $H_{\mathbb Z}$ has cubic polynomial growth just like the abelian group $\mathbb {Z}^3$. But actually it has quartic growth. This is easy to see: because $vu = uvw$ we have $v^ m u^ n = u^ n v^ m w^{mn}$ for any $m,n$, and so any of the $n^4$ elements $u^ a v^ b w^ c$ with $0 \leq a \leq n$, $0 \leq b \leq n$, and $0 \leq c \leq n^2$ can be obtained as the product of a string of at most $n$ copies of $u$ and at most $n$ copies of $v$ arranged in a suitable order. Why do we get quartic growth from such a 3-dimensional group? The answer takes us to a fascinating piece of geometry.

Although $H$ is identified with $\mathbb R^3$, we find that if we want left-multiplications in the group to be isometries we must warp the usual metric of $\mathbb R^3$ somewhat. In fact the length of a path must be defined as the integral of

  \[ \{ dx^2 \  + \  dy^2 \  + \  (dz - xdy)^2 \} ^{1/2}, \]    

which means that each plane perpendicular to the $x$-axis has been sheared in the $y$-direction by an amount proportional to $x$. As we re-scale the word-metric on $H_{\mathbb Z}$ we must re-scale the metric on $\mathbb R ^3$ too, multiplying $x$ and $y$ by $\lambda = 1/r_ i$, and $z$ by $\lambda ^2$. In the limit, this means that the only paths of finite length are those whose direction at each point lies in the plane given by $dz - xdy = 0$. Let us call them the allowable paths. They give us a new metric on $\mathbb R^3$ in which the distance from one point to another is the length of the shortest allowable path joining them. This metric defines the usual topology on $\mathbb R^3$, but, unlike the taxi-cab metric, it is far from equivalent — even locally — to the usual metric. Nevertheless, it is a metric which those of us who have tried to park a car know all too well. To move a car just a little bit sideways we must take it along a disproportionately long path. This is because the position of a car sitting on an expanse of tarmac is described by three coordinates $(\xi ,\eta ,\theta )$, where $(\xi ,\eta )$ is the position of the mid-point of the front wheels, and $\theta $ is the angle in which the axis of symmetry of the car is pointing. When we want to move the car we can move $\xi $ and $\eta $ any way we like, within reason, but the change in the third coordinate is constrained by the differential relation $d\theta = \cos \theta \  d\eta \  - \  \sin \theta \  d\xi ,$ assuming the car is of unit length. (A simple change of coordinates puts this in the form $dz = ydx$ which we found for the group $H$.)

What is remarkable about this new metric is that on one hand it defines the usual topology of $\mathbb R^3$, but on the other hand it defines a metric space of Hausdorff dimension four. Hausdorff dimension is a concept defined only for metric spaces. To say it is four means, essentially, that the number of balls of radius $r/n$ needed to cover a ball of radius $r$ grows like $n^4$ as $n \to \infty $. In the new metric on $\mathbb R^3$ this is the case because a small $\varepsilon $-ball for the new metric looks (in the usual coordinates) like a very flat ellipsoid with axes $(\varepsilon , \varepsilon , \varepsilon ^2)$. This explains the quartic growth rate of the group $H_{\mathbb Z}$. More generally, for any metric space the topological dimension is bounded above by the Hausdorff dimension, and so the finite-dimensionality of the limit space $Y$ follows from the polynomial growth of the group.

Back to the main article

  • Want facts and want them fast? Our Maths in a minute series explores key mathematical concepts in just a few words.

  • What do chocolate and mayonnaise have in common? It's maths! Find out how in this podcast featuring engineer Valerie Pinfield.

  • Is it possible to write unique music with the limited quantity of notes and chords available? We ask musician Oli Freke!

  • How can maths help to understand the Southern Ocean, a vital component of the Earth's climate system?

  • Was the mathematical modelling projecting the course of the pandemic too pessimistic, or were the projections justified? Matt Keeling tells our colleagues from SBIDER about the COVID models that fed into public policy.

  • PhD student Daniel Kreuter tells us about his work on the BloodCounts! project, which uses maths to make optimal use of the billions of blood tests performed every year around the globe.