# Question about maximal compact subgroups of Lie groups

Let $$G$$ be a compact (connected) semisimple Lie group. Let $$G_\mathbb{C}$$ be the complexification of $$G$$.

Is $$G$$ a maximal compact subgroup of $$G_\mathbb{C}$$?

• $\mathfrak g$ is a maximal subalgebra of its complexification on which the Killing form is negative definite, so at least $G$ is a maximal connected compact subgroup. Commented Nov 4, 2022 at 14:05
• @LSpice maximal compact subgroups of connected Lie groups are connected (the inclusion is even a homotopy equivalence).
– YCor
Commented Nov 4, 2022 at 14:11
• @LSpice For $\mathfrak{g}$ not simple, it is not true that $\mathfrak{g}$ is maximal in $\mathfrak{g}_\mathbf{C}$. For $\mathfrak{g}$ semisimple the larger subalgebras correspond to non-compact subgroups anyway, so it's fine. In the general case one should deal with the center too.
– YCor
Commented Nov 4, 2022 at 14:13
• @LSpice ah, the formulation "maximal on which..." is ambiguous.
– YCor
Commented Nov 4, 2022 at 14:14
• @LSpice no, I missed the assumption (I edited to streamline)
– YCor
Commented Nov 4, 2022 at 14:20

$$\DeclareMathOperator\Lie{Lie}\newcommand\g{\mathfrak g}\newcommand\C{{\mathbb C}}\g = \Lie(G)$$ is maximal among subalgebras of $$\g_\C = \Lie(G_\C)$$ on which the Killing form is negative definite, so $$G$$ is a maximal connected, compact subgroup of $$G_\C$$. According to @YCor, maximal compact subgroups of connected groups are connected, so $$G$$ is also a maximal compact subgroup.