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  "body": "\n## Definition\n\nLet $G$ be a connected real or complex [[Lie group|7JBRGCVJRB]].\nIts [[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 $h, h' \\in H$.\nThe 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]].\n\n## Remarks\n\nA 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 $0$.\n\nExample: the group $G$ of affine maps $x \\mapsto a x + b$ of $\\R$ with $a \u003e 0$ and $b \\in \\R$, written $(a, b)$, with $(a, b)(a', b') = (a a', a b' + b)$.\nIt is connected, being homeomorphic to $(0, \\infty) \\times \\R$.\nThe 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 $(1, b)$.\nThen $G^{(2)} = \\set{e}$, and $G$ is solvable.\n\nNon-example: $G = SL(2, \\R)$.\nIts 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.\nSo the derived series of the Lie algebra never reaches $0$, and $G$ is not solvable.\n",
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