Let $G$ be a connected complex semisimple Lie group, i.e. the Lie algebra $\mathfrak{g}$ of $G$ is a complex semisimple Lie algebra. A well-known theorem states that $\mathfrak{g}$ has a compact real form, i.e. we can write $\mathfrak{g}=\mathfrak{k}\otimes\mathbb{C}$ where $\mathfrak{k}$ is a real subalgebra of $\mathfrak{g}$ (by viewing $\mathfrak{g}$ as a real Lie algebra of twice the dimension), such that $\operatorname{Int}\mathfrak{k}$ is compact. Here $\operatorname{Int}\mathfrak{k}$ denotes the connected subgroup of $\operatorname{GL}(\mathfrak{k})$ with Lie algebra $\operatorname{ad}\mathfrak{k}$. The reference I am using for these statements and definitions is Knapp's book Lie groups Beyond an Introduction.
Question. Is there a compact subgroup $K$ of $G$ such that $G$ is equal to the complexification $K_{\mathbb{C}}$ of $K$?
In the event of an affirmative answer, I am more interested in the proof than the statement itself.
Here is what I know so far: (The references are all in Knapp's book mentioned above.)
- $\mathfrak{k}$ is a real semisimple Lie algebra (Corollary 1.53).
- $\operatorname{ad}\mathfrak{k}\cong\mathfrak{k}$ because semisimple Lie algebras have trivial centre (Proposition 1.13). Hence, $\operatorname{Int}\mathfrak{k}$ is a compact connected Lie group with Lie algebra $\mathfrak{k}$.
- Since $\mathfrak{k}$ is semisimple, the universal cover of $\operatorname{Int}\mathfrak{k}$ is compact (Theorem 4.69).
- Hence, every connected Lie group with Lie algebra $\mathfrak{k}$ is compact.
- By 4, the unique connected subgroup of $G$ with Lie algebra $\mathfrak{k}$ is compact. Call it $K$.
So now we know that $G$ has a compact subgroup $K$ such that the complexification $\mathfrak{k}_{\mathbb{C}}$ of its Lie algebra is equal to $\mathfrak{g}$. Can we conclude from this that $K_{\mathbb{C}}=G$?