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  "body": "\n## Definition\n\nLet $G$ be a real [[Lie group|7JBRGCVJRB]].\nA *complexification* (or universal complexification) of $G$ is a pair $(G_\\C, \\gamma)$ with two properties.\n\n1. $G_\\C$ is a complex Lie group and $\\gamma \\colon G \\to G_\\C$ is a continuous homomorphism.\n2. For each complex Lie group $H$ and each continuous homomorphism $\\varphi \\colon G \\to H$, there is exactly one holomorphic homomorphism $\\psi \\colon G_\\C \\to H$ with $\\varphi = \\psi \\circ \\gamma$.\n\n## Remarks\n\nBy the universal property, a complexification is unique up to a unique isomorphism compatible with $\\gamma$; it always exists.\n\nExample: $G = \\R$ with $G_\\C = \\C$ and $\\gamma$ the inclusion.\nA continuous homomorphism $\\varphi \\colon \\R \\to H$ is $\\varphi(t) = \\exp(tX)$ for some $X$ in the Lie algebra of $H$.\nThen $\\psi(z) = \\exp(zX)$ is a holomorphic homomorphism extending $\\varphi$.\nIt is the only one, since two holomorphic maps on $\\C$ that agree on $\\R$ are equal.\n\nNon-example: $G = \\R$ with $G_\\C = \\C^\\times$ and $\\gamma(t) = e^t$ fails clause 2.\nTake $H = \\C$ and $\\varphi(t) = t$.\nFor a holomorphic homomorphism $\\psi \\colon \\C^\\times \\to \\C$, the map $z \\mapsto \\psi(e^z)$ is a holomorphic homomorphism $\\C \\to \\C$, so $\\psi(e^z) = cz$ for a constant $c$, and $c \\cdot 2\\pi i = \\psi(1) = 0$ gives $\\psi = 0$, so $\\psi \\circ \\gamma \\ne \\varphi$.\n\nThe map $\\gamma$ need not be injective.\nFor the universal cover $\\widetilde{\\mathrm{SL}}_2(\\R)$, the complexification is $\\mathrm{SL}_2(\\C)$, and $\\gamma$ is the covering map onto $\\mathrm{SL}_2(\\R)$, with infinite kernel.\nFor a compact Lie group $G$, the map $\\gamma$ is injective, and $G$ is a maximal compact subgroup of $G_\\C$.\n",
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