How to find sub-algebras of $\mathfrak{su}(n,\mathbb{R})$?

Question

I'm still a beginner on the field of Lie algebras. I understand that a simple Lie algebra $\mathfrak{g}$ does not contain any ideal other than the trivial ideals $0$ and itself. Is there a way to check if $\mathfrak{g}$ contains any sub-algebras at all, i.e. sub-algebras that are not necessarily ideals?

Edit: I found the book "The Lie Algebras su(N)" by Walter Pfeifer (it is publicly available on his website). There the author claims that for for each $n$, the algebra $\mathfrak{su}(n,\mathbb{R})$ is a subalgebra of $\mathfrak{su}(n+1,\mathbb{R})$ (c.f. page 87). Unfortunately there is no citation for that statement. Is there any source where I can look up the proof? Can this statement be made even stronger, i.e. the only subalgebras of $\mathfrak{su}(n)$ are $\mathfrak{su}(n')$ with $n'<n$?

Answer 1

There are many publications classifying subalgebras of simple or semisimple Lie algebras. In general this is very complicated.

As an example, the subalgebras of $\mathfrak{sl}(3,\Bbb R)$ and $\mathfrak{sl}(3,\Bbb C)$ have been classified up to conjugacy by Pavel Winternitz in $2004$, and the table in the appendix is already rather long.

More recently, Douglas and Repka classified the subalgebras of all semisimple Lie algebras of rank two. The paper was published in Linear Multilinear Algebra $66$ in $2018$.

Answer 2

As has been discussed, the answer in general is very complicated but to give you some ideas, here are a few important types of examples.

A semisimple Lie algebra always contains Cartan subalgebras. These are (in this scenario) maximal abelian subalgebras containing only semisimple elements. Think the subalgebra of diagonal matrices in $\mathfrak{sl}_n$

In a complex semisimple Lie algebra, we can find a maximal solvable subalgebra called a Borel subalgebra $\mathfrak{b}$. Think the subalgebra of upper triangular matrices in $\mathfrak{sl}_n$. We can also find subalgebras which contain $\mathfrak{b}$ called parabolic subalgebras. These ideas extend to real forms of $\mathfrak{g}$ as long as it isn't compact (although we lose some of these unless the real form is split).

In a compact semisimple Lie algebra every element is semisimple so the $1$-dimensional subalgebras are only differentiated by their rank but in other real semisimple Lie algebras we also have nilpotent elements and ones which are neither. So even the 1-dimensional subalgebras have lots of different possibilities.

We could also find for example $\mathfrak{so}(n,\mathbb{R}) \leq \mathfrak{sl}(n,\mathbb{R})$ as a maximal compact subalgebra, or maybe $\mathfrak{sl}_m \leq \mathfrak{sl}_n$ for $m\leq n$ as a semisimple subalgebra.

Moving away from the semisimple ones a solvable Lie algebra $\mathfrak{g}$ always contains its derived subalgebra $\mathfrak{g}' = [\mathfrak{g},\mathfrak{g}]$ which is always nilpotent. Indeed everything on the derived series or the lower central series will be a subalgebra. Assuming it wasn't abelian to start with this gives a collection of proper, non-trivial subalgebras.

If it were abelian then any subspace is an abelian subalgebra but these are all basically the same up to dimension.