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Given two semisimple complex Lie algebras $\frak{g}$ and $\frak{n}$ such that the root system of $\frak{n}$ arises as a sub-root system of the root system of $\frak{g}$, does this then imply that $\frak{n}$ arises as a sub-Lie algebra of $\frak{g}$?

A second less elementary question is if the subalgefbras of $\frak{g}$ corresponding to the isotropy subgroups for compact symmetric spaces can be understood in terms of sub-root systems?

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  • $\begingroup$ Do you mean for some of the n's to be g's? $\endgroup$ Commented Aug 31, 2024 at 16:59
  • $\begingroup$ This is hardly research level. For the first question - the subalgebra of $\mathfrak g$ generated by root vectors of the subsystem will be isomorphic to $\mathfrak n$. For the second - rather the compact symmetric spaces are described in terms of involutions (order two automorphisms) of $\mathfrak g$, or equivalently in terms of $\mathbb Z/2\mathbb Z$-gradings of $\mathfrak g$. In other words only those subalgebras correspond to symmetric spaces which are fixed points of an involution, or equivalently zero degree parts of some $\mathbb Z/2\mathbb Z$-grading $\endgroup$ Commented Aug 31, 2024 at 18:13
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    $\begingroup$ @მამუკაჯიბლაძე That's true for subsystems closed under addition, but not all subsystems. For instance, $C_2$ has two subsystems of type $A_1\times A_1$, but only one gives rise to a subalgebra $\mathfrak{sl}_2\times \mathfrak{sl_2} \subset \mathfrak{sp}_4$. $\endgroup$ Commented Aug 31, 2024 at 18:27
  • $\begingroup$ @GrantB. Thanks, I should be more careful! Still I think if the root system of $\mathfrak n$ embeds into the root system of $\mathfrak g$ then one can also embed it in such a way that its root vectors would generate a subalgebra of $\mathfrak g$ isomorphic to $\mathfrak n$, no? $\endgroup$ Commented Aug 31, 2024 at 18:44

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My immediate instinct was "no", and I tried the system of short roots in $\mathsf C_2$. This is almost a counterexample, but not quite: there's no semisimple subalgebra of $\mathfrak{sp}_4$ normalised by "the" Cartan subalgebra whose roots are the short roots, but $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ is a subalgebra whose root system has the same type as the system of short roots. (This was mentioned by @GrantB. in a comment.) Similarly for $\mathfrak g_2$: there's no semisimple subalgebra normalised by "the" Cartan subalgebra whose roots are the short roots, but the subalgebra generated by the long root spaces is $\mathfrak{sl}_3$, which is of the same type.

I am visiting Jeff Adler and mentioned your question to him, and he pointed out that I was just being too conservative with rank: if instead of $\mathsf C_2$ one looks at $\mathsf C_3$, then the subsystem of short roots is of type $\mathsf D_3$, but there is no $\mathfrak{so}_6$ subalgebra of $\mathfrak{sp}_6$. (For example, one can verify this using Borel–de Siebenthal theory.) Since this is Jeff's answer, I have made it community wiki to avoid reputation. If you like it, then please go up-vote one of his questions or answers.

If, however, you require additionally that the sub-root system be closed, in the sense that any integral linear combination of roots of $\mathfrak n$ that is a root of $\mathfrak g$ is also a root of $\mathfrak n$, then, as mentioned by @მამუკაჯიბლაძე and @GrantB., the subalgebra of $\mathfrak g$ generated by the appropriate root spaces with respect to a fixed Cartan subalgebra of $\mathfrak g$ is a semisimple subalgebra of the desired type. (I mention, by the way, that it's very difficult for me to think of an algebra denoted by $\mathfrak n$ as semisimple; my brain can't but see it as the nilradical of a parabolic subalgebra!)

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    $\begingroup$ Thanks a lot for the great answer. With hindsight $\frak{n}$ is quite a poor choice to denote a semisimple Lie algebra . . . but it is probably too late to change it now . . . $\endgroup$ Commented Aug 31, 2024 at 21:42

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