# Optional: Compact connected abelian groups are tori

## Compact connected abelian groups are tori

This video is optional.

Proposition:

A path-connected compact abelian matrix group $H$ is a torus (i.e. it's isomorphic to $U(1)^{n}$ for some $n$ ).

## Proof

Since $H$ is abelian, we have $\exp(X_{1})\exp(X_{2})=\exp(X_{1}+X_{2})\mbox{ for all }X_{1},X_{2}\in% \mathfrak{h}.$ Therefore the exponential map is a group homomorphism $\exp\colon\mathfrak{h}\to H$ where we think of $\mathfrak{h}$ as $\mathbf{R}^{n}$ with addition.

Here's how the proof will go:

1. We'll prove that exp is surjective.

2. The first isomorphism theorem for groups will then show that $H=\mathfrak{h}/\ker\exp$ .

3. Then we will identify $\ker\exp\subset\mathfrak{h}$ with $\mathbf{Z}^{n}\subset\mathbf{R}^{n}$ .

Why is exp surjective? Because $H$ is path-connected, we know from an earlier exercise that $H$ is generated by any neighbourhood of the identity matrix. In particular, $H$ is generated by the image of an exponential chart. We can always add generators to a generating set to get a bigger generating set, so $H$ is generated by $\exp(\mathfrak{h})$ . But $\exp(\mathfrak{h})$ is a subgroup, as we saw in the last video, so the subgroup generated by $\exp(\mathfrak{h})$ is $\exp(\mathfrak{h})$ itself, therefore $H=\exp(\mathfrak{h})$ , and the exponential map is surjective.

We now identify the kernel of the exponential map.

Definition:

Take $k$ linearly independent vectors $v_{1},\ldots,v_{k}$ and take all possible integer linear combinations of them to get a subgroup $\Lambda=\{\sum_{i=1}^{k}a_{i}v_{i}\ :\ a_{i}\in\mathbf{Z}\}\subset\mathbf{R}^{% n}.$ A subgroup of this form is called a lattice of rank $k$ in $\mathbf{R}^{n}$ .

Example:

$\mathbf{Z}^{n}\subset\mathbf{R}^{n}$ is a lattice, taking $v_{i}$ to be the $i$ th standard basis vector.

Lemma:

$\ker\exp\subset\mathfrak{h}$ is a lattice of rank $n$ (where $n=\dim\mathfrak{h}$ ).

Once we have established this, we are done. To see why, consider the basis of $\mathfrak{h}$ given by the $v_{i}$ s. This identifies $\mathfrak{h}$ with $\mathbf{R}^{n}$ in such a way that $\ker\exp$ is identified with $\mathbf{Z}^{n}$ . Therefore $H=\mathfrak{h}/\ker\exp$ is isomorphic to $\mathbf{R}^{n}/\mathbf{Z}^{n}=(\mathbf{R}/\mathbf{Z})^{n}=U(1)^{n}$ .

Remark:

Note that $U(1)=\mathbf{R}/\mathbf{Z}$ because we have a surjective homomorphism $\mathbf{R}\to U(1)$ , $\theta\mapsto e^{2\pi i\theta}$ , with kernel $\mathbf{Z}$ , so the first isomorphism theorem tells us $U(1)\cong\mathbf{R}/\mathbf{Z}$ .

Remark:

The fact that the rank must be $n$ is easy to see once you know that $\ker\exp$ is a lattice. If the rank were strictly less than $n$ then we would get $H\cong(\mathbf{R}/\mathbf{Z})^{k}\times\mathbf{R}^{n-k}$ , so $H$ would fail to be compact because $\mathbf{R}^{n-k}$ is not compact unless $k=n$ .

We will need to assume the following lemma:

Lemma:

A discrete subgroup of $\mathbf{R}^{n}$ is a lattice.

The proof of this lemma would take us too far away from the main business of the course. You can find it in Arnold's book "Mathematical Methods of Classical Mechanics" (on page 276 in my version of the book, where he is proving the (Arnold)-Liouville theorem on integrable systems).

We will just define discrete subgroups and prove that $\ker\exp$ is discrete.

Definition:

A subgroup $\Gamma\subset\mathbf{R}^{n}$ is discrete if, for every $X\in\Gamma$ , there exists a Euclidean ball $B\subset\mathbf{R}^{n}$ such that $B\cap\Gamma=\{X\}$ .

In other words, the points of $\Gamma$ are sufficiently spaced out that you can separate them by balls.

Example:

An example of something that is not discrete would be $\mathbf{Q}\subset\mathbf{R}$ : any ball (interval) contains infinitely many rational numbers.

By constrast, the integers $\mathbf{Z}\subset\mathbf{R}$ are discrete: an interval of length $\epsilon<1$ containing an integer contains only one integer.

Lemma:

$\ker\exp$ is a discrete subgroup of $\mathfrak{h}$ .

Proof:

The origin $0\in\mathfrak{h}$ is in $\ker\exp$ , so let's start by finding a ball $B\subset\mathfrak{h}$ such that $B\cap\ker\exp=\{0\}$ . We can take $B$ to be the domain of an exponential chart.

Remember that we have the exponential map $\exp\colon\mathfrak{h}\to H$ and that we can find neighbourhoods $U\subset\mathfrak{h}$ of the origin and $V\subset H$ of the identity such that $\exp|_{U}\colon U\to V$ is bijective. In particular, $\exp|_{U}$ is injective, so $\ker\exp\cap U=\{0\}$ (i.e. $0$ is the only possible preimage of the identity in this neighbourhood).

For a general $X\in\ker\exp$ , we will take $B=\{X+Y\ :\ Y\in U\}$ . This works because $B\cap\ker\exp=\{X+Y\ :\ Y\in U\mbox{ and }\exp(X+Y)=I\},$ but $\exp(X+Y)=\exp(X)\exp(Y)=\exp(Y)$ because $H$ is abelian and $X\in\ker\exp$ , so we deduce that $Y\in\ker\exp$ . Since $Y\in U$ , this implies $Y=0$ . Therefore $B\cap\ker\exp=\{X\}.$

## Summary

This final lemma shows that $\ker\exp$ is a discrete subgroup of $\mathfrak{h}$ , so by the previous (unproven) lemma, it is a lattice (which we saw has rank $n$ ). This implies that $H=\mathfrak{h}/\ker\exp$ is isomorphic to $U(1)^{n}$ as required.