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 $G_\mathbb{C}$?
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$\begingroup$ $\mathfrak g$ 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. $\endgroup$– LSpiceCommented Nov 4, 2022 at 14:05
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2$\begingroup$ @LSpice maximal compact subgroups of connected Lie groups are connected (the inclusion is even a homotopy equivalence). $\endgroup$– YCorCommented Nov 4, 2022 at 14:11
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$\begingroup$ @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. $\endgroup$– YCorCommented Nov 4, 2022 at 14:13
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2$\begingroup$ @LSpice ah, the formulation "maximal on which..." is ambiguous. $\endgroup$– YCorCommented Nov 4, 2022 at 14:14
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1$\begingroup$ @LSpice no, I missed the assumption (I edited to streamline) $\endgroup$– YCorCommented Nov 4, 2022 at 14:20
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1 Answer
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$\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.
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2$\begingroup$ I think this statement (that maximal compact subgroups of connected Lie groups are connected) is due to Iwasawa around 1950. $\endgroup$– YCorCommented Nov 4, 2022 at 15:36
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$\begingroup$ Thank you very much @LSpice and YCor for your help! $\endgroup$– MiraCommented Nov 4, 2022 at 16:04
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1$\begingroup$ @YCor there is question about that statement here mathoverflow.net/questions/140622/… $\endgroup$– MiraCommented Nov 4, 2022 at 16:05
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2$\begingroup$ @asma, awesome! I have been trying for a long time to remember that comment by @abx, so thank you for indirectly pointing me to it! $\endgroup$– LSpiceCommented Nov 4, 2022 at 17:51