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?

For the first article on this series see Is Every Set Fieldable?.

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 GG is the product of copies of R\mathbb{R} and S1{\mathbb{S}}^1 — i.e. G≅Rn×S1×⋯×S1G \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 GG, the connected component G0G^0 of 1∈G1 \in G is itself a closed normal subgroup. In particular, if dim⁡G=n\dim G = n then G0G^0 is an nn-dimensional Lie subgroup of GG and Γ=G/G0\Gamma = G / G^0 is a 00-dimensional Lie group — i.e. a countable discrete group. We thus have an exact sequence

0⟶G0⟶G⟶Γ⟶0.0 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 0.

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

0⟶G0⟶G⟶Γ⟶0,0 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 0,

where G0G^0 is a connected Abelian Lie group — i.e. G0=Rn×S1×⋯×S1G^0 = \mathbb{R}^n \times {\mathbb{S}}^1 \times \cdots \times {\mathbb{S}}^1 — and Γ\Gamma is a countable discrete group.

We certainly know an example of such a sequence. Namely, given G0G^0 and Γ\Gamma as in the above we may take G=G0×ΓG = 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 G0G^0 and Γ\Gamma in the extremes are isomorphic to

0⟶G0⟶G0×Γ⟶Γ⟶0,0 \longrightarrow G^0 \longrightarrow G^0 \times \Gamma \longrightarrow \Gamma \longrightarrow 0,

and in particular every Abelian Lie group has the form G0×ΓG^0 \times \Gamma for some G0G^0 and Γ\Gamma as above.

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

The commutative diagram of an injective group
The commutative diagram of an injective group

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

0⟶I⟶fG⟶gK⟶00 \longrightarrow I \overset{f}{\longrightarrow} G \overset{g}{\longrightarrow} K \longrightarrow 0

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

Indeed, by taking H=IH = I and id:I→I\text{id} : I \rightarrow I for the map H→IH \rightarrow I in the definition of an injective group we get some s:G→Is : G \rightarrow I as required. As it turns out, G0G^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 DD is called divisible if given, d∈Dd \in D and n∈Zn \in \mathbb{Z}, there is some d′∈Dd ' \in D such that d=n⋅d′d = n \cdot d '.

Examples of divisible groups include R\mathbb{R} and S1{\mathbb{S}}^1. In addition, it is clear that the product of divisible groups is also divisible, so that G0=Rn×S1×⋯×S1G^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 Z\mathbb{Z}-modules — that every divisible Abelian group is injective. In particular, G0G^0 is injective and our sequence

0⟶G0⟶fG⟶gΓ⟶00 \longrightarrow G^0 \overset{f}{\longrightarrow} G \overset{g}{\longrightarrow} \Gamma \longrightarrow 0

splits in the category of Abelian groups.

We can thus find a group homomorphism s′:Γ→Gs ' : \Gamma \rightarrow G such that g∘s′=idg \circ s ' = \text{id}. Since Γ\Gamma is discrete, s′s ' is continuous and therefore our sequence splits in the category of Abelian topological groups. This implies G≅G0×ΓG \cong G^0 \times \Gamma as topological groups. Since every continuous group homomorphism between Lie groups is a smooth map, it follows that G≅G0×ΓG \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 R\mathbb{R} and S1{\mathbb{S}}^1 with a countable discrete group Γ\Gamma: G≅Rn×S1×⋯×S1×ΓG \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 g=Rn\mathfrak{g} = \mathbb{R}^n, and these are just the quotients of its simply connected form G=RnG = \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 11-dimensional and 22-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

1⟶G0⟶G⟶Γ⟶1,1 \longrightarrow G^0 \longrightarrow G \longrightarrow \Gamma \longrightarrow 1,

with G0G^0 connected and Γ\Gamma discrete, splits — which would imply G≅G0⋊ΓG \cong G^0 \rtimes \Gamma as Lie groups. The issue here is that G0G^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! 😛