Remark

(1) Let $(\tilde{G}, \gamma)$ be the universal complexification of $G$. Let $G_0$ (resp. $\tilde{G}_0$) be the identity component of $G$ (resp. $\tilde{G}$). By the proof of Proposition 20, $(\tilde{G}_0, \gamma|G_0)$ is the universal complexification of $G_0$ and the composite morphism

defines on passing to the quotient an isomorphism of $G/G_0$ onto $\tilde{G}/\tilde{G}_0$.

(2) Suppose that $G$ is simply connected. Let $g = L(G)$, $g_C$ be the complexification of $g$, $S'$ the simply connected complex Lie group with Lie algebra $g_C$ and $\sigma$ the morphism of $G$ into $S'$ such that $L(\sigma)$ is the canonical injection of $g$ into $g_C$. We again use the notation of the proof of Proposition 20, part (a). If $H = S'$ and $\phi = \sigma$, then $\phi^* = \mathrm{Id}_{S'}$. Hence $(S', \sigma)$ is the universal complexification of $G$. Note that $\sigma$ is not in general injective (Exercise 16); however its kernel is discrete since $L(\sigma)$ is injective. On the other hand, let $\theta$ be the involution of $g_C$ defined by $g$ and let $\eta$ be the corresponding automorphism of the underlying real Lie group of $S'$; let ${S'}^\eta$ be the set of points of $S'$ invariant under $\eta$; it is a real Lie subgroup of $S'$ with Lie algebra $g$ (§ 3, no. 8, Corollary 1 to Proposition 29). By no. 1, Corollary 1 to Proposition 1, $\sigma(G)$ is a real integral subgroup of $S'$ with Lie algebra $g$ and hence $\sigma(G)$ is the identity component of ${S'}^\eta$; in particular $\sigma(G)$ is a real Lie subgroup of $S'$.