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 $\mathbb C^m$?

  • $\begingroup$ what do you mean by "isomorphic"? $\endgroup$ – YCor May 16 '17 at 23:50
  • $\begingroup$ biholomorphic equivalent $\endgroup$ – Ronald May 17 '17 at 1:58
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    $\begingroup$ not in your first sentence... $\endgroup$ – YCor May 17 '17 at 9:10
  • $\begingroup$ yes sure, the first is a diffeomorphism $\endgroup$ – Ronald May 17 '17 at 22:27

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 $$\pi_2(\mathrm{SL}_d(\mathbf{C})/\mathrm{SO}_d(\mathbf{C}))\simeq\left\{ \begin{array}{rcl} \mathbf{Z}/2\mathbf{Z} & \mbox{for} & d\ge 3 \\ \mathbf{Z} & \mbox{for} & d=2. \end{array}\right. $$

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  • $\begingroup$ Thanks YCore! I am wondering now if $G$ was complex Lie group then $G/K^\mathbb C$ is contractible?? $\endgroup$ – Ronald May 17 '17 at 22:31
  • $\begingroup$ What do you mean? if $G$ is a complex Lie group then $K^{\mathbf{C}}=G$. $\endgroup$ – YCor May 17 '17 at 23:43
  • $\begingroup$ Why should it be? Maybe $G$ is not reductive ! $\endgroup$ – Ronald May 17 '17 at 23:49
  • $\begingroup$ in general complexification of a Lie group is not well-defined. But if you mean $G$ a real algebraic group, then $K^{\mathbf{C}}$ will be a Levi factor and the quotient will indeed be biholomorphic to the unipotent radical, and hence to some $\mathbf{C}^k$. $\endgroup$ – YCor May 17 '17 at 23:57

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