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Looking at this question and its answer

Every Lie algebra contains a maximal proper Lie subalgebra

I asked myself what is an example of a finite-dimensional Lie algebra $L$ whose only proper Lie subalgebras are trivial?

Some thoughts: Clearly the dimension of $L$ has to be greater than $1$. Moroever, $L$ cannot be abelian. I tried $\frak{sl}_2$ but that clearly doesn't work . . . same goes for any simple Lie algebra. One could try solvable . . . but the derived series produces ideals, so no go there. Thus by Levi's theorem might one that every finite dim Lie algebra of dim greater than $1$ has a non-zero Lie subalgebra?

Boris Henriques
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  • In view of the answer, one can ask whether one can classify the Lie algebras in which every proper subalgebra has dimension $\le 1$. This is obvious for $\mathfrak{g}$ of dimension $\le 2$. For 3-dimensional $\mathfrak{g}$ I guess this means the same as a non-split form of $\mathfrak{sl}_2$. Possibly this is all what's possible. – YCor Mar 21 '21 at 21:54

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For any Lie algebra $\mathfrak g$ with $\dim\mathfrak g>1$ and for every $0\neq X\in\mathfrak g$, the set $\{\lambda X\mid\lambda\in F\}$, where $F$ is the field that you are working with, is a subalgebra of $\mathfrak g$, which is neither $\{0\}$ nor $\mathfrak g$.

Torsten Schoeneberg
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José Carlos Santos
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