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?



Compact dual of a noncompact Lie group
(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 $K=G\cap U$.


