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