Malcev conjugacy of Levi subalgebras

We prove the stronger assertion that the conjugating factors may all be $\exp (\mathrm{ad}x)$ with $x\in [\mathfrak{g},\mathfrak{r}]$. If this commutator is zero, then $\mathfrak{r}$ is central. Each Levi factor is a section of $\mathfrak{g}\to \mathfrak{g}\mathrm{/}\mathfrak{r}$ by [L1]. The difference of two such sections is a linear map to the central algebra $\mathfrak{r}$; the two sections preserve brackets, so that difference vanishes on the derived algebra of the quotient. The quotient is semisimple and hence perfect by [L4], so the sections, and therefore the Levi factors, coincide.

Suppose $[\mathfrak{g},\mathfrak{r}]\mathrm{\ne }0$. It is an ideal contained in the nilradical by [L3], hence is nilpotent and has nonzero center by [L5]. Jacobi shows that this center is an ideal of $\mathfrak{g}$. Choose a minimal nonzero $\mathfrak{g}$-ideal $\mathfrak{m}$ in it. Then $\mathfrak{m}$ is abelian and $0\mathrm{\ne }\mathfrak{m}\subseteq [\mathfrak{g},\mathfrak{r}]\subseteq \mathfrak{n}$.

First assume that $\mathfrak{r}$ itself contains no nonzero proper ideal of $\mathfrak{g}$. Then $[\mathfrak{g},\mathfrak{r}]=\mathfrak{r}$ by step 1.1. The ideal $[\mathfrak{r},\mathfrak{r}]$ is proper because $\mathfrak{r}$ is nonzero solvable, so minimality makes it zero. Identify both Levi factors with $\mathfrak{q}=\mathfrak{g}\mathrm{/}\mathfrak{r}$. Relative to ${\mathfrak{s}}_0$, write ${\mathfrak{s}}_1$ as the graph $x\mapsto x+h(x)$ of a linear map $h:{\mathfrak{s}}_0\to \mathfrak{r}$. Its being a subalgebra says exactly that $h$ is a $1$-cocycle for the adjoint ${\mathfrak{s}}_0$-module $\mathfrak{r}$. By [L2], $h(x)=[x,b]$ for some $b\in \mathfrak{r}$. Since $\mathfrak{r}$ is abelian, $\exp (\mathrm{ad}(-b))=1+\mathrm{ad}(-b)$ maps $x$ to $x+h(x)$. Here $-b\in \mathfrak{r}=[\mathfrak{g},\mathfrak{r}]$, proving the strengthened base case.

In the remaining case, $0\mathrm{\ne }\mathfrak{m}\mathrm{\ne }\mathfrak{r}$. The radical of $\mathfrak{g}\mathrm{/}\mathfrak{m}$ is $\mathfrak{r}\mathrm{/}\mathfrak{m}$, so it has smaller dimension. Apply the strengthened induction hypothesis in that quotient. Its conjugating elements lie in $[\mathfrak{g}\mathrm{/}\mathfrak{m},\mathfrak{r}\mathrm{/}\mathfrak{m}]=[\mathfrak{g},\mathfrak{r}]\mathrm{/}\mathfrak{m}$ and therefore lift to elements of $[\mathfrak{g},\mathfrak{r}]$. The corresponding exponentials on $\mathfrak{g}$ induce the required quotient exponentials. After applying their finite product, we may assume that ${\mathfrak{s}}_0$ and ${\mathfrak{s}}_1$ have the same image modulo $\mathfrak{m}$.

Now ${\mathfrak{s}}_0+\mathfrak{m}={\mathfrak{s}}_1+\mathfrak{m}$. As in step 2.1, ${\mathfrak{s}}_1$ is the graph over ${\mathfrak{s}}_0$ of a $1$-cocycle with values in the abelian module $\mathfrak{m}$. By [L2] it is a coboundary, and the same calculation gives one final conjugation by $\exp (\mathrm{ad}a)$ for some $a\in \mathfrak{m}\subseteq [\mathfrak{g},\mathfrak{r}]$.

This closes the induction and proves the stated claim because [L3] puts every chosen element in $\mathfrak{n}$. If $x\in \mathfrak{n}$, then ${\mathrm{ad}}_x$ maps $\mathfrak{g}$ into $\mathfrak{n}$ and its restriction to the nilpotent algebra $\mathfrak{n}$ is nilpotent; hence ${\mathrm{ad}}_x$ is nilpotent and its exponential is a finite polynomial automorphism. When $\mathfrak{r}=0$, both Levi factors equal $\mathfrak{g}$. Only finite-dimensional minimal-subspace choices occur.