# Classification of Abelian Lie Groups

As a representation theorist, I’ve always been fascinated by the
innocence of classification problems:
*Now we know a couple of examples of such mathematical structure, how
about finding all possible examples?*
This is the second post in a series on fun little mathematical toy
problems I encountered during my studies and, as it turns out,
today’s question is a classification problem. Namely:
*what are all possible examples of Abelian Lie groups?*

Note that I did not say “connected Abelian Lie groups”. The
classification of connected Abelian Lie groups is a classical problem in
Lie theory and its solution is well known: every connected Abelian Lie
group
$G$
is the product of copies of
$\mathbb{R}$
and
${\mathbb{S}}^1$ — i.e.
$G \cong \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots
\times {\mathbb{S}}^1$
as Lie groups. We are interested in the classification of
*all* Abelian Lie groups, regardless of whether or not they are
connected. Nevertheless, we can use the classification of connected
Abelian Lie groups to our advantage.

Given any Lie group $G$, the connected component $G^0$ of $1 \in G$ is itself a closed normal subgroup. In particular, if $\dim G = n$ then $G^0$ is an $n$-dimensional Lie subgroup of $G$ and $\Gamma = G / G^0$ is a $0$-dimensional Lie group — i.e. a countable discrete group. We thus have an exact sequence

Hence it suffices to classify all exact sequences of Lie groups of the form

where $G^0$ is a connected Abelian Lie group — i.e. $G^0 = \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1$ — and $\Gamma$ is a countable discrete group.

We certainly know an example of such a sequence. Namely, given
$G^0$
and
$\Gamma$
as in the above we may take
$G = G^0 \times \Gamma$. But is this all we got? Surprisingly, the answer to this question is
a resounding *yes*. In other words, we claim that all short exact
sequences with
$G^0$
and
$\Gamma$
in the extremes are isomorphic to

and in particular every Abelian Lie group has the form $G^0 \times \Gamma$ for some $G^0$ and $\Gamma$ as above.

To see this, we introduce the notions of *injective* groups. An
Abelian group
$I$
is called *injective* if given an injective group homomorphism
$f : H \rightarrow G$
between Abelian groups
$G$
and
$H$
and a homorphism
$H \rightarrow I$, there is some group homomorphism
$G \rightarrow I$
such that the composition
$H \overset{f}{\to} G \to I$
is the same as the map
$H \rightarrow I$.

The reason why we are interested in injective groups is the fact that every short exact sequence

of Abelian groups with $I$ injective splits. In other words, there is some group homomorphism $s : G \rightarrow I$ such that $s \circ f = \text{id}$ — or, equivalently, there is $s ' : K \rightarrow G$ such that $g \circ s ' = \text{id}$.

Indeed, by taking
$H = I$
and
$\text{id} : I \rightarrow I$
for the map
$H \rightarrow I$
in the definition of an injective group we get some
$s : G \rightarrow I$
as required. As it turns out,
$G^0$
is an injective group, but how can we go about proving it? The answer to
this question requires us to introduce one more definition, namely the
concept of a *divisible* group: an Abelian group
$D$
is called *divisible* if given,
$d \in D$
and
$n \in \mathbb{Z}$, there is some
$d ' \in D$
such that
$d = n \cdot d '$.

Examples of divisible groups include $\mathbb{R}$ and ${\mathbb{S}}^1$. In addition, it is clear that the product of divisible groups is also divisible, so that $G^0 = \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1$ is a divisible groups. It is a well known fact from module theory — recall that Abelian groups are the same as $\mathbb{Z}$-modules — that every divisible Abelian group is injective. In particular, $G^0$ is injective and our sequence

splits in the category of Abelian groups.

We can thus find a group homomorphism $s ' : \Gamma \rightarrow G$ such that $g \circ s ' = \text{id}$. Since $\Gamma$ is discrete, $s '$ is continuous and therefore our sequence splits in the category of Abelian topological groups. This implies $G \cong G^0 \times \Gamma$ as topological groups. Since every continuous group homomorphism between Lie groups is a smooth map, it follows that $G \cong G^0 \times \Gamma$ as Lie groups.

All in all, we have just seen that every Abelian Lie group is the product of copies of $\mathbb{R}$ and ${\mathbb{S}}^1$ with a countable discrete group $\Gamma$: $G \cong \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1 \times \Gamma$. I would now like to conclude this article by saying a feel words on how one may try to generalize our proof to a classification of all Lie groups — regardless of Abelianess.

First and foremost, our proof is heavily reliant on the classification of connected Abelian Lie groups. This classification, in turn, relies on the classification of Abelian Lie algebras: to find all connected Abelian Lie groups it suffices to find all connected Lie groups whose Lie algebra is the Abelian Lie algebra $\mathfrak{g} = \mathbb{R}^n$, and these are just the quotients of its simply connected form $G = \mathbb{R}^n$ by discrete subgroups.

Since there is no general classification of finite-dimensional Lie algebras, this particular argument cannot be used to obtain a classification of all connected Lie groups. Indeed, the classification of arbitrary connected Lie groups is regarded by most as an intractable problem. Nevertheless, there are classifications of $1$-dimensional and $2$-dimensional connected Lie groups, so one could attempt a classification of low-dimensional groups.

Next there is the question of whether or not every short exact sequence of (not necessarily Abelian) Lie groups of the form

with $G^0$ connected and $\Gamma$ discrete, splits — which would imply $G \cong G^0 \rtimes \Gamma$ as Lie groups. The issue here is that $G^0$ may not be divisible in the general setting, and even if it is, it is unclear whether or not every divisible group is injective — in the category of arbitrary groups.

That about wraps it up. Hope to see you in the next post of this series! 😛