Solvable Lie groups

Question

I've read the following fact:

An arbitrary connected solvable Lie group is diffeomorphic to $\mathbb{R}^n\times T^m$.

Can anyone give me a reference on it? I'm not very familiar with Lie groups and Lie algebras. Thanks.

Answer 1

There is a really good series of books on Lie groups and Lie algebras by Onishchik and Vinberg (although very pricey). In their third volume, in second chapter, they have a section on the topology of solvable Lie groups.

Here are two results that should help you see why this is the case:

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Theorem 3.1 (page 50-51)

Let $G$ be a connected real Lie group.

(i) If there is a connected normal Lie subgroup $G_1$ of $G$ of codimension 1, then there exists a 1-dimensional Lie subgroup $C$ of $G$ such that $C \ltimes G_1,$ the semidirect product of $C$ with $G_1$.

(ii) If the Lie group $G$ is solvable, then it has a connected normal Lie subgroup of codimension 1.

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And then here is part of the succeeding corollary.

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Part of a Corollary (page 51)

Let $G$ be a connected solvable real Lie group. Then there exists one-parameter subgroups $C_i$ of $G$ for $1 \leq i \leq n = dim\, G$ such that: $G = C_1 \cdot C_2 \cdots C_n$ where each element of $g\in G$ is uniquely represented in the form $g = g_1 \cdot g_2 \cdots g_n$ where each $c_i \in C_i$.

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This corollary follows from induction on the dimension of the Lie group. A 1-dimensional Lie group has this form. By way of induction, assume $G$ has dimension $n$, and this result holds for dimension less than $n$. By Theorem 3.1, we know that $G = C \ltimes G_1$ where $C$ is 1-dimensional and $G_1$ has dimension less than $n$. Hence, the corollary.

Now, since each element has a unique representation, we can define an explicit diffeomorphism with each element of $G$ with the product of the 1-parameter subgroups $C_i$ i.e. $G \simeq \prod C_i$. Notice that every 1-parameter subgroup is isomorphic to $\mathbf{R}$ or $\mathbf{T}$ and hence diffeomorphic as well. Hence, the claimed diffeomorphism.