Let $\mathfrak s$ be a complex semisimple Lie algebra, then $\dim _\mathbb C\mathfrak s\geq 3$. But, however, is it possible for $\mathfrak s$ to have a semisimple Lie subalgebra $\mathfrak h$ such that $\dim_\mathbb C\mathfrak s/\mathfrak h=2$?

On Lie groups level, do homogeneous spaces of the form $\mathrm{SL}(n,\mathbb C)/H$ where $H$ is a closed complex semisimple subgroup of $\mathrm{SL}(n,\mathbb C)$ have special properties, for example parallelizable, etc?


No, it's not. Indeed it's known that no simple Lie algebra of rang $\ge 3$ has a proper subalgebra of codimension 2; in rank 2, the only proper subalgebras of codimension $\le 2$ are parabolic of codimension 2, and in rank 1 the only proper subalgebras of codimension $\le 2$ are solvable (either parabolic or abelian).

So if $\mathfrak{g}$ is a semisimple Lie algebra and $\mathfrak{h}$ is a proper subalgebra of codimension $\le 2$ then there exists a direct decomposition $\mathfrak{g}=\mathfrak{g}_1\times\mathfrak{g}_2$ and $\mathfrak{h}=\mathfrak{g}_1\times\mathfrak{h}_2$ with either $\mathfrak{h}_2$ parabolic in $\mathfrak{g}_2$ (hence not semisimple), or $\mathfrak{g}_2$ has dimension 3 and $\mathfrak{h}_2$ is abelian (hence not semisimple).

Codimension 3 is possible, namely in $\mathfrak{sl}_2^2$.

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  • $\begingroup$ Do you know a reference for the fact that: A simple Lie algebra of rank $n$ doesn't have subalgebras of codimension less than $n$? Thanks! $\endgroup$ – user328669 Dec 20 '19 at 4:17
  • $\begingroup$ I know this question of yours math.stackexchange.com/questions/3003752; I guess there are printed references, but I don't know precisely. $\endgroup$ – YCor Dec 20 '19 at 7:11

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