---
tag: V3XM2N7K0X
kind: definition
name: Solvable Lie group
msc: ["22E25", "22E15", "20F16"]
defines: 9CNSVFQ801
uses: [7JBRGCVJRB, 0FXJ7YFMMX, 0Q5G5DJ4P2]
sources:
  - url: https://en.wikipedia.org/wiki/Solvmanifold
---

## Definition

Let $G$ be a connected real or complex [[Lie group|7JBRGCVJRB]].
Its [[derived series|0Q5G5DJ4P2]] is $G^{(0)} = G$ and $G^{(k+1)} = [G^{(k)}, G^{(k)}]$ for $k \ge 0$, where $[H, H]$ is the subgroup generated by the commutators $h h' h^{-1} h'^{-1}$ with $h, h' \in H$.
The group $G$ is a *solvable Lie group* if $G^{(k)} = \set{e}$ for some $k \ge 0$, that is, if $G$ is a [[solvable group|0FXJ7YFMMX]].

## Remarks

A connected Lie group is solvable if and only if its Lie algebra $\mathfrak{g}$ is a solvable Lie algebra, that is, its derived series $\mathfrak{g} \supseteq [\mathfrak{g}, \mathfrak{g}] \supseteq \cdots$ reaches $0$.

Example: the group $G$ of affine maps $x \mapsto a x + b$ of $\R$ with $a > 0$ and $b \in \R$, written $(a, b)$, with $(a, b)(a', b') = (a a', a b' + b)$.
It is connected, being homeomorphic to $(0, \infty) \times \R$.
The commutator of $(a, 0)$ and $(1, t)$ is $(1, (a - 1)t)$, and every commutator has first entry $1$, so $G^{(1)}$ is the abelian group of translations $(1, b)$.
Then $G^{(2)} = \set{e}$, and $G$ is solvable.

Non-example: $G = SL(2, \R)$.
Its Lie algebra $\mathfrak{sl}(2, \R)$ satisfies $[\mathfrak{sl}(2, \R), \mathfrak{sl}(2, \R)] = \mathfrak{sl}(2, \R)$, since the basis elements $H, E, F$ with $[H, E] = 2E$, $[H, F] = -2F$, $[E, F] = H$ are all brackets.
So the derived series of the Lie algebra never reaches $0$, and $G$ is not solvable.