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Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$ and suppose that $K$ is a compact connected subgroup of $G$ with Lie algebra $\mathfrak{k}$ such that $\mathfrak{g}=\mathfrak{k}\otimes\mathbb{C}$ is the complexification of $\mathfrak{k}$.

Does this imply that $G$ is the complexification of $K$?

Here by complexification, I mean this:

Definition. A complex Lie group $G$ is a complexification of a Lie group $K\subseteq G$ if every Lie group homomorphism $K\to H$ where $H$ is a complex Lie group extends to a unique holomorphic Lie group homomorphism $G\to H$. (See Wikipedia.)

Otherwise said, I want to know if $G$ is a complex reductive group in the sense of algebraic geometry (so that we can consider GIT quotients).

It is known that if $K$ is a compact Lie group, then there is a unique complex Lie group $G$ which is the complexification of $K$. Moreover, in that case $\mathfrak{g}=\mathfrak{k}\otimes\mathbb{C}$.

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    $\begingroup$ If $K=\mathbb R/\mathbb Z$, $G=\mathbb C/(\mathbb Z+i\mathbb Z)$, and $H=\mathbb C/\mathbb Z$, with the obvious inclusions $K\subset G$ and $K\subset H$, there is no homomorphism $G\to H$ making it commute. So the answer is no (unless you modify the conditions). $\endgroup$ – user8268 Sep 19 '16 at 9:10
  • $\begingroup$ @user8268 Thanks for this counterexample. I think the missing assumption is that $G$ is a complex affine algebraic group. $\endgroup$ – Simon Parker Sep 19 '16 at 9:45
  • $\begingroup$ I think you need the inclusion $K \subset G$ to induce an isomorphism on fundamental groups for your conclusion to hold. $\endgroup$ – Sir Wilfred Lucas-Dockery Apr 17 at 20:15

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