I have found the following theorem in Onishchik & Vinberg "Lie Groups and Algebraic Groups": "On any compact Lie subgroup $K$ there exists a unique real algebraic group structure and the complex algebraic group $K(\mathbb{C})$ is reductive." (In this book, they define a complex algebraic group to be reductive if its tangent algebra is reductive, coinciding with the notion of reductive Lie algebras.)

Since we know furthermore that any algebraic group defines a Lie group (at least where the underlying field is $\mathbb{R}$ or $\mathbb{C}$), the authors point out that we can think of the tangent algebra to the algebraic group as the Lie algebra to the Lie group. So in particular, $K(\mathbb{C})$ has a (complex) reductive Lie algebra. Let $H$ denote the Lie group underlying $K(\mathbb{C})$. We furthermore know that any complex reductive Lie group arises as the complexification of a compact Lie group. Suppose now that $K$ is a compact real Lie group. Letting $G$ be the complexification of $K$, is it true that $H = G$ or at least $H\cong G$?


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