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Given a Lie group $G$ and a closed subgroup $U \subset G$:

  • Is it always possible to choose a maximal compact subgroup $K \subset G$ of $G$ such that $K \cap U$ is again a maximal compact subgroup of $U$?

  • Or equivalently: Given a maximal compact subgroup $K' \subset U$ of $U$, is it possible to extend $K'$ to a maximal compact subgroup $K$ of $G$?

Suppose the answer to the above is affirmative or reasonable criteria can be given:

  • How about several subgroups $U_1, U_2 \subset G$? Can we choose a maximal compact subgroup $K_{1,2} \subset U_1 \cap U_2$ and extensions to maximal compact subgroups $K_1 \subset U_1$ and $K_2 \subset U_2$ which then both allow for a single, common extension to a maximal compact subgroup $K \subset G$?

I have very little knowledge about Lie groups for now, so any help and any hints to helpful references are very much appreciated.

Akim Eismann
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1 Answers1

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Every compact subgroup of a connected Lie group (or actually any Lie group with finitely many connected components) is contained in a maximal one. I believe this should be proved in most basic texts on Lie groups. See, for instance, Theorem 14.1.3 on page 531 in Structure and Geometry of Lie Groups, by Hilgert and Neeb. In particular, the maximal compact subgroup of any closed subgroup is contained in a maximal one.

This answers the first two (equivalent) questions. Regarding the third: no, this is not always possible. At least I'm pretty sure. The idea is this: take two compact subgroups $U_1,U_2$ in (say) $G=\mathrm{SL}_n(\mathbb{R})$ such that $U_1 \cup U_2$ generates a non-compact subgroup. I believe this should be possible (for $n>1$). Then $K_1=U_1,K_2=U_2$ are the unique maximal compact subgroups of $U_1,U_2$ respsectively, but they do not admit a common extension.

Cronus
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  • Regarding your second paragraph, with $G=GL_n(\mathbb{R})$, I think the group generated by the union of all the compact subgroups of $G$ is the set of matrices with $\det=\pm 1$. The point is that the image of $\det$ on a compact subgroup is a compact subgroup of $\mathbb{R}\setminus {0}$, so is a subgroup of ${\pm 1}$. – Jason DeVito - on hiatus Oct 11 '20 at 17:44
  • @JasonDeVito Thank you, you are right. I changed it to SL. Although the important thing is that they don't generate a compact subgroup, of course. – Cronus Oct 12 '20 at 08:45