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In the following, we let su$(N)$ denote the Lie algebra of anti-hermtian and traceless complex $N$-by-$N$ matrices, with bracket being the usual commutator.

My question is very simple, although I cannot seem to find the answer anywhere I've looked so far:

Are all solvable Lie subalgebras of su$(N)$ abelian?

As su$(N)$ is a very well-known real Lie algebra I assume that the answer to this question is known. If the answer is "yes", I would like either a proof of the statement or a reference containing a proof. If the answer is "no" I would like a counterexample (I have yet to find one, despite looking for a while).

Initially, I thought I had found a proof for the statement using arguments of simultaneous diagonalization for commuting anti-hermitian matrices, but I recently discovered that my proof was incorrect and couldn't find a way to fix the argument. I'm about to give up on the idea of figuring this one out entirely on my own, but I'm still very much interested in what the answer actually is.

Scounged
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    Subalgebras of compact Lie algebras are compact and solvable compact Lie algebras are abelian. Here I am using the definition that a Lie algebra is compact if it is the Lie algebra of a compact Lie group. – Callum May 25 '23 at 22:59
  • @Callum I see, so that means that the answer is (more or less trivially) yes. Although short, I think your comment could be turned into an acceptable answer to my question. – Scounged May 25 '23 at 23:31

1 Answers1

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As noted by Callum in a comment, this holds generally for compact Lie algebras.

Here is a more algebraic line of argument: One of the equivalent criteria for a real semisimple Lie algebra $L$ to be compact is that all its elements are $ad$-semisimple.

Now let $S \subset L$ be a solvable subalgebra and $D:=[S,S]$ its derived subalgebra. From well-known structure theory, $D$ is a nilpotent Lie algebra and hence for all $x \in D$, the map $ad_D(x): D \rightarrow D$ is nilpotent, a fortiori so is $ad_S(x):S\rightarrow S$. But as said, $ad_L(x)$ is semisimple and so must be its restriction to $S$, i.e. $ad_S(x)$, which thus is semisimple and nilpotent, i.e. zero. This implies that $D$ is abelian, and central in $S$. But that means $S$ itself is nilpotent, so we can repeat the argument for any $ad_S(x)$, $x \in S$, which are all zero, i.e. $S$ is abelian.

Torsten Schoeneberg
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