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I am trying to understand Goto's proof of the following theorem (DOI: 10.32917/hmj/1206139768).

Let $g$ be a Lie algebra. If there exist nilpotent subalgebras $n_1$ and $n_2$ with $n_1 + n_2 = g$, then $g$ is solvable, and vice versa.

I am stuck at the part where he proves that any solvable Lie algebra can be written as the sum of two nilpotent subalgebras. He states that if $h$ is a Cartan subalgebra of $g$, then $g=h + g'$, where $g' = [g,g]$ is the derived subalgebra of $g$. I do not understand why this holds and how I can prove this fact.

Any help is welcome, thanks!

Mathstudent
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1 Answers1

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In fact, that step is true regardless of whether $g$ is solvable or not.

According to the result in Proof the image of a Cartan subalgebra under the quotient map is a Cartan subalgebra, if $f:L_1 \twoheadrightarrow L_2$ is a surjective Lie algebra homomorphism, and $H$ a Cartan subalgebra of $L_1$, then $f(H)$ is a Cartan subalgebra of $L_2$.

Now for any Lie algebra $g$, the derived subalgebra $g' := [g,g]$ is an ideal, so the projection map $pr: g \twoheadrightarrow g/g'$ is a homomorphism of Lie algebras. So by the above, for any CSA $h \subseteq g$, the image of $h$ in $g/g'$ is a CSA. But $g/g'$ is abelian by construction, and in any abelian (in fact, any nilpotent) Lie algebra, the only CSA is the Lie algebra itself. In other words, we have $pr(h) = g/g'$ which is equivalent to $g = h +g'$.

Torsten Schoeneberg
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