Let $G$ be a compact connected semisimple Lie group and let $\mathfrak g$ denote its Lie algebra. Is the following result true? Does it follows directly from Dynkin's classification of maximal Lie subalgebras?:

There are $\mathfrak h_1,\dots,\mathfrak h_r$ proper Lie subalgebras of $\mathfrak g$ such that: (1) for any Lie subalgebra $\mathfrak a$ of $\mathfrak g$ there is $a\in G$ such that $\textrm{Ad}_a(\mathfrak a)\subset \mathfrak h_i$ for some $i$; (2) if a Lie subalgebra $\mathfrak a$ of $\mathfrak g$ contains properly $\mathfrak h_i$ for some $i$, then $\mathfrak a=\mathfrak g$; (3) for each $i$, the only connected Lie subgroup $H_i$ of $G$ with Lie algebra $\mathfrak h_i$ is closed in $G$.

I am confident that the answer is affirmative, except in the finiteness of the number of Lie subalgebras $\mathfrak h_i$. I hope there is a precise (modern) reference for it.

Some related questions:

Modern reference for maximal connected subgroups of compact Lie groups

Maximal subgroups of semisimple Lie groups

  • $\begingroup$ Thank you for your suggestion to clarify the question. I am in fact interested in both questions. $\endgroup$ – emiliocba Jul 19 '19 at 23:18
  • 1
    $\begingroup$ I think they are two questions in one: are there finitely many maximal subalgebras up to conjugation, and do they correspond to closed subgroups. I think that the answer (to both) is positive and does not rely on the classification. $\endgroup$ – YCor Jul 19 '19 at 23:22

It's true, and does not rely on classification of maximal subalgebras.

The first part of the statement is that there are finitely maximal subalgebras up to conjugation. It follows from the fact that in a semisimple real Lie algebra there are finitely many semisimple subalgebras up to conjugation, and that every maximal subalgebra in a compact semisimple real Lie algebra, if not abelian (modulo its core, see edit), is product of a semisimple subalgebra with its centralizer, and if abelian (modulo its core) has to be a maximal torus (in the quotient by the core) and there's only a single such conjugacy class.

For being closed: if $\mathfrak{h}$ is a maximal subalgebra and $H$ its corresponding analytic subgroup, I claim that $H$ is closed. Otherwise, $H$ is dense (since otherwise its Lie algebra would be larger), and hence $\mathfrak{h}$ is $G$-invariant, hence an ideal, hence not maximal.

Edit: Recall that the core is the largest ideal in the subalgebra. There are finitely many possibilities, since any semisimple Lie algebra has finitely many ideals. Actually, even if I didn't use it, it is not hard to prove that the quotient by core of a maximal subalgebra in a semisimple Lie algebra is either simple, or direct product of two isomorphic simple Lie algebras, in such a way that the maximal subalgebra modulo its core is the graph of an isomorphism between the factors. Given this, everything boils down to the case when $\mathfrak{g}$ is simple.

  • $\begingroup$ Nice answer! The proof is much simpler than I expected. Thank you so much! $\endgroup$ – emiliocba Jul 19 '19 at 23:42
  • $\begingroup$ Why is an ideal not maximal? Couldn't it have codimension 1? $\endgroup$ – LSpice Jul 20 '19 at 1:42
  • 2
    $\begingroup$ @LSpice no because $\mathfrak{g}$ is semisimple. So a proper ideal $\mathfrak{n}$ is direct factor of its centralizer, and then adding a 1-dimensional subalgebra of its centralizer yields a larger proper ideal. $\endgroup$ – YCor Jul 20 '19 at 7:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.