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Let $\mathfrak{g}_0$ be a noncompact simple Lie algebra, and fix a Cartan involution $\theta$ of $\mathfrak{g}_0$, which gives a Cartan decomposition $\mathfrak{g}_0=\mathfrak{k}_0+\mathfrak{p}_0$. Since the automorphism group $G=\mathrm{Aut}\mathfrak{g}_0$ has finite center ($\textbf{Right?}$), there is an automorphism $\Theta$ of $G$ whose differential is $\theta$, and the subgroup $K$ of the fixed points of $\Theta$ on $G$ is a maximal compact subgroup of $G$. Moreover, $G=Ke^{\mathfrak{p}_0}$ as a polar decomposition of $G$.

Let $\mathfrak{g}=\mathfrak{g}_0\oplus\sqrt{-1}\mathfrak{g}_0$ be the complexification of $\mathfrak{g}_0$, and denote by $G_\mathbb{C}$ the universal complexification of $G$. Set $\mathfrak{u}_0=\mathfrak{k}_0+\sqrt{-1}\mathfrak{p}_0$ which is a compact Lie algebra. Let $U$ be the subgroup of $G_\mathbb{C}$ corresponding to Lie algebra $\mathfrak{u}_0$.

$\textbf{QUESTION}$

1) Is $U$ a compact dual of $G$?

2) Does the polar decomposition $U=Ke^{\sqrt{-1}\mathfrak{p}_0}$ hold? In particular, does $K\subseteq G\cap U$ hold?

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  • $\begingroup$ What do you mean by the universal complexification of $G$? $\endgroup$ – Mikhail Borovoi Mar 29 '17 at 16:58
  • $\begingroup$ @MikhailBorovoi Let $G$ be a real Lie group. A universal complexification of $G$ is a complex Lie group $G_\mathbb{C}$ and a continuous group homomorphism $i:G\rightarrow G_\mathbb{C}$ with the property that, if $f$ is an arbitrary continuous group homomorphism from $G$ to a complex Lie group $H$, then there exists a unique continuous group homomorphsim $\tilde{f}:G_\mathbb{C}\rightarrow H$ such that $f=\tilde{f}\circ i$. $\endgroup$ – Hebe Mar 30 '17 at 9:23
  • $\begingroup$ I think @MikhailBorovoi's point may have been that what you are calling the universal complexification is usually just called the complexification, with no modifier (although your terminology is eminently logical). $\endgroup$ – LSpice Apr 4 '18 at 2:04
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(0) The group ${\rm Aut}(\mathfrak{g}_0)$ does not have to be connected (even over $\mathbb C$), take $\mathfrak{g}_0=\mathfrak{su}(2,2)$ as a counter-example. So let $G={\rm Inn}(\mathfrak g_0)$, that is, the identity component (over $\mathbb C$) of ${\rm Aut}(\mathfrak{g}_0)$. Then right, $G$ has trivial center.

(1) Since $\mathfrak u_0$ is a compact semisimple Lie algebra, yes, $U$ is a compact semisimple Lie group. Since $\mathfrak u_0$ is a real form of $\mathfrak g$, yes, $U$ is a real form of $G_{\mathbb C}$. Thus yes, $U$ is a compact real form of $G_{\mathbb C}$.

(2a) No, this polar decomposition does not hold. Indeed, otherwise the compact group $U$ would be homeomorphic to the direct product of $K$ and $\mathfrak p_0$, hence it would be noncompact, which is a contradiction.

(2b) However, it is true that $K=G\cap U$. Indeed, we have $K\subset G$. Let us show that $K\subset U$. Let $\rho\colon G_{\mathbb C}\to G_{\mathbb C}$ denote the complex conjugation corresponding to $G$, that is, such that $G=(G_{\mathbb C})^\rho$. Then $\theta$ commutes with $\rho$. We have $$ K=G^\theta=\{g\in G_{\mathbb C}\ |\ \rho(g)=g,\ \theta(g)=g.\} $$ On the other hand $$ U=(G_{\mathbb C})^{\theta\circ\rho}=\{g\in G_{\mathbb C}\ |\ \theta(\rho(g))=g.\} $$ We see that $K\subset U$, as required. Thus $K\subset G\cap U$. Since $G\cap U$ is a compact subgroup of $G$ containing the maximal compact subgroup $K$ of $G$, we conclude that $K=G\cap U$.

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  • $\begingroup$ Thank you for your answer! I have two questions. (1) I understand that $\mathrm{Aut}\mathfrak{g}_0$ is not necessary connected, but since $\mathrm{Inn}\mathfrak{g}_0$ has trivial center and the quotient group $\mathrm{Aut}\mathfrak{g}_0/\mathrm{Inn}\mathfrak{g}_0$ is finite, it is follows that $\mathrm{Aut}\mathfrak{g}_0$ has finite center. Right? $\endgroup$ – Hebe Mar 30 '17 at 9:13
  • $\begingroup$ (2) I do not quite understand the complex conjugation $\rho$. Of course, on the complex Lie algebra $\mathfrak{g}$, one may define the complex conjugation $\rho$ corresponding to the real form $\mathfrak{g}_0$. However, we do not assume that $G_\mathbb{C}$ is simply connected, in which way is $\rho$ lifted from $\mathfrak{g}$ to $G_\mathbb{C}$? Thank you. $\endgroup$ – Hebe Mar 30 '17 at 9:15
  • $\begingroup$ @Hebe: We have $G_{\mathbb C}={\rm Inn}(\mathfrak g)\subset{\rm Aut}(\mathfrak g)\subset {\rm GL}(\mathfrak g)$. The complex conjugation on $\mathfrak g$ induces a complex conjugation on $\rm GL$, $\rm Aut$ and $\rm Inn$. $\endgroup$ – Mikhail Borovoi Mar 30 '17 at 9:42
  • $\begingroup$ @Hebe: Your argument is correct, but actually ${\rm Aut}(\mathfrak g_0)$ has trivial center. $\endgroup$ – Mikhail Borovoi Mar 30 '17 at 9:45
  • $\begingroup$ Thank you! Now I understand the two points! Finally, now that $\mathrm{Aut}\mathfrak{g}_0$ has trivial center, if we set $G=\mathrm{Aut}\mathfrak{g}_0$, do your arguments for (1) and (2b) in your answer still work? Or in which process is connectedness needed? $\endgroup$ – Hebe Mar 30 '17 at 10:03

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