4
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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.)

  1. $\mathfrak{k}$ is a real semisimple Lie algebra (Corollary 1.53).
  2. $\operatorname{ad}\mathfrak{k}\cong\mathfrak{k}$ because semisimple Lie algebras have trivial centre (Proposition 1.13). Hence, $\operatorname{Int}\mathfrak{k}$ is a compact connected Lie group with Lie algebra $\mathfrak{k}$.
  3. Since $\mathfrak{k}$ is semisimple, the universal cover of $\operatorname{Int}\mathfrak{k}$ is compact (Theorem 4.69).
  4. Hence, every connected Lie group with Lie algebra $\mathfrak{k}$ is compact.
  5. 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 $K_{\mathbb{C}}=G$?

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  • $\begingroup$ I'm curious, what does it mean to take the complexification of $K$, exactly? Are you somehow thinking of it as an algebraic variety and taking the $\mathbb{C}$ points? $\endgroup$ – freeRmodule Aug 30 '16 at 4:11
  • $\begingroup$ Complexification is the left adjoint to the forgetful functor from complex Lie groups to real Lie groups. $\endgroup$ – Qiaochu Yuan Mar 11 '17 at 22:08
  • $\begingroup$ If you take K as a maximal compact Lie subgroup of G and use the universal property of the complexification you get for free that the complexification of K injects itself in G. The proof that it is actually surjective is more involved and you can find it here: D. MONTGOMERY, Simply-connected homogeneous spaces, Proc. Amer. Math. Soc., 1 (1950), pp. 467-469 $\endgroup$ – Max Reinhold Jahnke Jan 12 at 12:45

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