# Subgroups of compact Lie groups generated by a subset of nodes of the Dynkin diagram

Where can I find a reference for the following fact, or as close as possible to it?

Let $$G$$ be a semisimple compact real Lie group with rank $$r$$, let $$T$$ be a maximal torus in $$G$$, let $$\mathfrak{g}$$ and $$\mathfrak{t}$$ be their Lie algebras, let $$\Phi$$ be the root system of $$\mathfrak{g}$$ relative to $$\mathfrak{t}$$, and let $$\alpha_1,\ldots,\alpha_r \in \Phi$$ be a basis of simple roots. Then for any subset $$I \subseteq \{1,\ldots,r\}$$, the subgroup $$G_I$$ of $$G$$ associated to the Lie subalgebra of $$\mathfrak{g}$$ whose complexification is generated by $$\mathfrak{t}$$ and by the coroots $$\alpha_i^\vee$$ for $$i\in I$$, is (isogenous to?) the product of a torus of rank $$r - \#I$$ and a semisimple group (of rank $$\#I$$) described by the Dynkin diagram consisting of those nodes of the Dynkin diagram of $$\Phi$$ labeled by an element of $$I$$.

Also, how can I state this, or describe the subgroup $$G_I$$, in a less clumsy fashion? (Hopefully avoiding such phrases as “the maximal compact subgroup of the Levi factor of the parabolic subgroup defined by $$I$$ in the complexification”.)

I am looking for a description and discussion of the situation that stays as much as possible within the context of compact real Lie groups. (Essentially, how do I convince someone with little background in Lie or algebraic groups, that a subset $$I$$ of nodes of the Dynkin diagram of $$G$$ defines a subgroup of $$G$$ “as expected”?)

Edit: Let me emphasize that I'm looking for a reference, not a proof. The closest I found so far is §7–8 in chapter V of Knapp's Lie Groups Beyond an Introduction or proposition 12.6 in Malle & Testerman's Linear Algebraic Groups and Finite Groups of Lie Type, but they remain annoyingly far from the above statement.

• I think this should be in the book Parabolic geometris by Čap and Slovák. I don't have time atm to check it. Subset of nodes of the Dynkin diagram defines a disjoint union of Dynkin diagrams and hence by the classification of semi-simple Lie algebras it defines a semisimple Lie algebra. It remains to show that this semisimple Lie algebra is actually the one whose group you have described in the question (minus the central part). How exactly you do that depends on how do you go from Dynkin diagram to complex Lie algebra / compact Lie group. – Vít Tuček Jul 17 '19 at 13:41
• As for the second question, I vaguely remember that these groups should be exactly those that are centralizers of subsets of your torus $\mathfrak{t}$. – Vít Tuček Jul 17 '19 at 13:41
• Instead of your dreaded “the maximal compact subgroup of the Levi factor of the parabolic (...)”, can’t we describe $G_I$ as the intersection with $G$ of the parabolic (...)? – Francois Ziegler Jul 17 '19 at 15:47
• @FrancoisZiegler This is indeed better. – Gro-Tsen Jul 17 '19 at 15:55
• I think that Arvanitoyeorgos (2003, §§7.3–7.4) has what you want. (His $K=$ your $G_I$.) – Francois Ziegler Jul 17 '19 at 16:50

A root defines in your Lie algebra a non zero element in you algebra up to multiplication by a scalar. A collections of roots defines a subvector space in you Lie algebra.

A set of simple roots defines a set of root obtained by linear combinaisons of them with coefficients in $$\mathbb Z$$. You then obtain a linear subspace of your Lie-algebra. It is not a sub Lie algebra as its intersection with the Cartan algebra $$\mathfrak t$$ is zero.

You have a minimal and a maximal choice here. You can add the whole $$\mathfrak t$$ to obtain $$\mathfrak g_1$$ or just the linear span of the coroots associated to your chosen roots. The second option amount to take the Lie closure $$\mathfrak g_2$$ of your subspace. The subalgebra $$\mathfrak g_1$$ is the direct sum of $$\mathfrak g_2$$ and some abelian algebra $$\mathfrak h_0$$ of dimension $$r - |I|$$. The orthogonal $$\mathfrak h_1$$ of $$\mathfrak h_0$$ in $$\mathfrak h$$ is a Cartan subalgebra of $$\mathfrak g_1$$ and your choosen subset of roots is the set of roots for $$(\mathfrak g_1, \mathfrak h_1)$$.

Exponentiate the real part of this algebra in your group $$G$$ to obtain a group with Lie algebra $$\mathfrak g_1$$ or $$\mathfrak g_2$$.

• I agree with all that (except I think you mean “coroots” in the first paragraph), but I would like a reference where this is written explicitly so that I don't have to delve into explanations on this subject in a paper that's about Something Completely Different™. – Gro-Tsen Jul 17 '19 at 14:10
• I don't think you need reference, if you speak of roots and Cartan algebras, your audience should be able to follow your arguments, or at least to believe it. – InfiniteLooper Jul 17 '19 at 14:16
• The paper already contains a very annoyingly long section entitled “Recollections on compact Lie groups” containing a crash-course on the subject with no proofs but a summary of results with references. Most readers will be content to know that “a subset of Dynkin nodes defines a subgroup with the expected structure” but they will like to see a reference to be sure I'm not making this up. 😣 – Gro-Tsen Jul 17 '19 at 14:22
• sorry it wasnt clear in your question what you were asking for, will delete my answer as it not one. – InfiniteLooper Jul 17 '19 at 14:24
• No, don't delete, it's still a good explanation and there is no reason to remove it! – Gro-Tsen Jul 17 '19 at 14:28