# On maximal closed connected subgroups of a compact connected semisimple Lie group?

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:

Maximal subgroups of semisimple Lie groups

• Thank you for your suggestion to clarify the question. I am in fact interested in both questions. – emiliocba Jul 19 '19 at 23:18
• 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. – YCor Jul 19 '19 at 23:22

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.
• @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. – YCor Jul 20 '19 at 7:00