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Every subalgebra of a solvable Lie algebra is solvable. So a Lie algebra $ \mathfrak{g} $ containing a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ cannot be solvable.

Is the converse true? In other words, is it true that a Lie algebra $ \mathfrak{g} $ contains a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ if and only if $ \mathfrak{g} $ is not solvable?

Ian Gershon Teixeira
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    If you think a bit, you will easily identify a counter-example. But there are only few of these: Only one among simple real Lie algebras. – Moishe Kohan Feb 11 '22 at 20:22
  • oh wow as coudy says $ \mathfrak{sl}_2(\mathbb{R}) $ is an obvious counterexample. that's actually really embarrassing I can't believe I posted that. Are you saying though that this is the only counter example? In other words, every non solvable lie algebra not containing $ \mathfrak{so}_3(\mathbb{R}) $ is an extension of $ \mathfrak{sl}_2(\mathbb{R}) $ by a solvable Lie algebra? – Ian Gershon Teixeira Feb 11 '22 at 20:24
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    Well, among semisimple Lie algebras, you arte also missing directs sums of $sl(2,R)$'s. – Moishe Kohan Feb 11 '22 at 20:35
  • ah sure so does extension of direct sums of $ sl_2(R) $ by solvable exhaust all of them? – Ian Gershon Teixeira Feb 11 '22 at 20:38
  • Yes, direct sums of those and solvable algebras are the only examples. You should really pick up a textbook in Lie algebras and Lie groups and do some serious reading if you are to succeed in your project. – Moishe Kohan Feb 11 '22 at 21:03
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    Thanks for the advice! – Ian Gershon Teixeira Feb 11 '22 at 21:09
  • I really mean it. At this point, you do not need to know the proofs, but a detailed collection of key theorems. Vinberg and coauthors have several papers on this in Encyclopedia of Mathematics. – Moishe Kohan Feb 11 '22 at 22:09
  • That's actually a crazy coincidence I just checked out Vinberg's Lie Groups and Lie Algebras II from the library 2 days ago (first math book I've ever checked out!)! I was reading the section about lattices in abelian/nilpotent/solvable groups just yesterday and it really is lovely. You seem like the most knowledgeable person on Lie groups and related subjects that I've ever met. Is the Vinberg series where you got a lot of your knowledge? – Ian Gershon Teixeira Feb 13 '22 at 02:20
  • This was my first source. Then I have read much more. You are overestimating me: YCor is more knowledgeable in this regard. – Moishe Kohan Feb 13 '22 at 03:57

2 Answers2

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As coudy already said, the answer to your question is no, and as pointed out in the comments by Moishe Kohan, if you modify your question to also consider $\mathfrak{sl}(2,\mathbb{R})$ then the answer becomes yes. So we have

Theorem: A real finite-dimensional Lie algebra is not solvable if and only if it has a subalgebra isomorphism to either $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.

Here's a proof sketch: You already proved one direction. For the other direction, note that by the Levi decomposition, every real finite-dimensional Lie algebra is a semidirect product of a solvable Lie algebra and a semisimple Lie algebra. But every real semisimple Lie algebra has a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.

You can complete the proof sketch without the full power of the classification. For example, the complexification of your real semisimple Lie algebra will be a complex semisimple Lie algebra, and this complex semisimple Lie algebra will have a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{C})$ which is preserved by the complex conjugation map. So your original real semisimple Lie algebra will contain a real form of $\mathfrak{sl}(2,\mathbb{C})$, and there are only two of these: $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{so}(3)$.

Max
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  • You still have to argue after your last sentence. I do not see how to do it on the cheap without discussing root systems. – Moishe Kohan Feb 13 '22 at 03:54
  • Yes, I had in mind that you use root spaces to find the $\mathfrak{sl}(2,\mathbb{C})$ subalgebra. I just wanted to point out that you don't have to classify root systems to see that the theorem holds. – Max Feb 13 '22 at 11:04
  • Ok, so you found that the Lie algebra $g$ contains $sl(2,R)$. But then what? The result is that if $g$ is simple, it contains $so(3)$ or it is isomorphic to $sl(2,R)$. – Moishe Kohan Feb 13 '22 at 14:15
  • Yes, I am only trying to sketch a proof of this weaker statement. It was too long for a comment. I can remove it if you want. – Max Feb 13 '22 at 16:01
  • it is interesting to note that the original property I suggested (the Lie algebra has no $ so(3) $ subalgebra) is actually equivalent to the corresponding simply connected Lie group being torsion free. https://math.stackexchange.com/questions/4045532/torsion-free-lie-groups – Ian Gershon Teixeira Feb 16 '22 at 16:40
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No, the Lie algebra of $SO_{2,1}$ is not solvable. The Tits alternative may help you decide if a group is virtually solvable or not.

Let G be a finitely generated linear group over a field. Then two following possibilities occur:

either G is virtually solvable (i.e. has a solvable subgroup of finite index)

or it contains a nonabelian free group (i.e. it has a subgroup isomorphic to the free group on two generators).

coudy
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