Complexification of a Lie group

Remarks

By the universal property, a complexification is unique up to a unique isomorphism compatible with $\gamma$; it always exists.

Example: $G = \R$ with $G_\C = \C$ and $\gamma$ the inclusion. A continuous homomorphism $\varphi \colon \R \to H$ is $\varphi(t) = \exp(tX)$ for some $X$ in the Lie algebra of $H$. Then $\psi(z) = \exp(zX)$ is a holomorphic homomorphism extending $\varphi$. It is the only one, since two holomorphic maps on $\C$ that agree on $\R$ are equal.

Non-example: $G = \R$ with $G_\C = \C^\times$ and $\gamma(t) = e^t$ fails clause 2. Take $H = \C$ and $\varphi(t) = t$. For 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 $\psi \circ \gamma \ne \varphi$.

The map $\gamma$ need not be injective. For 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. For a compact Lie group $G$, the map $\gamma$ is injective, and $G$ is a maximal compact subgroup of $G_\C$.