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I am starting to learn some Lie theory and have a few questions about Levi decomposition in a specific case.

I have a real Lie algebra $\mathfrak{g}$ of dimension $n$ with a semi-simple Lie subalgebra $\mathfrak{k}$ of dimension $n-2$. Then let $\mathfrak{s}$ be the maximal semisimple subalgebra of $\mathfrak{g}$ containing $\mathfrak{k}$.

Then by the Levi decomposition we can write $$ \mathfrak{g} = \mathfrak{s} \oplus \mathfrak{r}$$ where $\mathfrak{r}$ is the radical of $\mathfrak{g}$.

So we have three possible cases:

  1. The dimension of $\mathfrak{s} = n-2$, i.e. $\mathfrak{s} = \mathfrak{k}$,
  2. The dimension of $\mathfrak{s} = n-1$
  3. The dimension of $\mathfrak{s} = n$, i.e. $\mathfrak{s} = \mathfrak{g}$.

Are the second two cases possible?

Since semisimple Lie algebras have dimension at least $3$ it intuitively seems that one cannot have a codimension 1 (or two) semisimple subalgebra of a semisimple Lie algebra - but I am unsure how to prove this either way.

Any help or suggested references is appreciated, thanks.

Tom
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    You are correct there are no semisimple subalgebras of semisimple Lie algebras with codimension 1 or 2. I can't think of a direct proof but there is a classification of maximal semisimple subalgebras by Dynkin. – Callum Nov 08 '22 at 14:55
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    Indeed the maximum dimension of a subalgebra at all is the dimension of one of the maximal parabolic subalgebras and these can have codimension 1 or 2 only when $\mathfrak{g} $ is $\mathfrak{sl}_2$ or $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$ which you can then check by hand. Note these are not even semisimple so the dimension of a semisimple subalgebrais generally much less. – Callum Nov 08 '22 at 14:55
  • Thanks @Callum I will try to find Dynkin's classification. – Tom Nov 08 '22 at 15:42
  • Thanks @TorstenSchoenberg, I am not assuming $\mathfrak{g}$ is semisimple - I think that should only be the case in case 3. – Tom Nov 08 '22 at 15:44
  • Ah, I misunderstood the setting, sorry. – Torsten Schoeneberg Nov 08 '22 at 15:52
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    See also this post. – Dietrich Burde Nov 08 '22 at 19:14
  • Actually realised a slight inaccuracy in my earlier comment. You can have codimension 1 parabolic subalgebras in any Lie algebra of the form $\mathfrak{sl}_2 \oplus \mathfrak{g}$ by taking a parabolic subalgebra of the form $\mathfrak{b} \oplus \mathfrak{g}$. Of course in these examples the proper semisimple subalgebra of maximal dimension is $\mathfrak{g}$ in general which has codimension 3 so the broader point is still true. – Callum Nov 10 '22 at 15:03

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