complexification of maximal compact subgroup

Question

Let $K$ be a maximal compact subgroup of a Lie group $G$. Then $G/K$ is isomorphic to a Euclidean space $\mathbb R^n$. Consider the complexification $G^\mathbb C$ and $K^\mathbb C$. Is true that $G^\mathbb C/K^\mathbb C$ is isomorphic to $\mathbb C^m$?

Answer 1

No. For $d\ge 2$, take $G=\mathrm{SL}_d(\mathbf{R})$ and $K=\mathrm{SO}_d(\mathbf{R})$. Then $G^{\mathbf{C}}/K^{\mathbf{C}}=\mathrm{SL}_d(\mathbf{C})/\mathrm{SO}_d(\mathbf{C})$.

But the latter is not contractible, hence not homeomorphic to any Euclidean space. This can be seen in various ways; one is from the exact sequence $$\pi_2(H)\to\pi_2(H/L)\to\pi_1(L)\to\pi_1(H);$$ which for $H=\mathrm{SL}_d(\mathbf{C})$ and $L=\mathrm{SO}_d(\mathbf{C})$ yields, since both extreme terms vanish, an isomorphism $$\pi_2(\mathrm{SL}_d(\mathbf{C})/\mathrm{SO}_d(\mathbf{C}))\simeq\left\{ \begin{array}{rcl} \mathbf{Z}/2\mathbf{Z} & \mbox{for} & d\ge 3 \\ \mathbf{Z} & \mbox{for} & d=2. \end{array}\right. $$