I'm studying Lie Algebras using the book "Algebras de Lie - Luiz A. B. San Martin", and I'm stuck in this exercise.

Exercise: Find a Lie Algebra $\mathfrak{g}$, such that $\mathfrak{g}$ admits two different Levi decompositions, i.e. find a Lie Algebra $\mathfrak g=\mathfrak{r(g)}\oplus\mathfrak{s}_1 =\mathfrak{r(g)}\oplus\mathfrak{s}_2, $ where $\mathfrak{r(g)}$ is the radical of $\mathfrak{g}$ and $\mathfrak{s}_1$, $\mathfrak{s}_2$ semisimple algebras such that $\mathfrak{s}_1 \neq\mathfrak{s}_2$, $\mathfrak{s}_1$.

An immediate conclusion is that $\mathfrak{g}$ can't be semi-simple, solvable or reductive.

I saw in this question that I should consider a Lie Algebra $\mathfrak{g} = \mathfrak{r(g)\oplus s}$. And $\mathfrak{\tilde{s}} = \text{exp} (\text{ad}(z)) (\mathfrak{s})$, with $z \in \mathfrak{nr(g)}$, where $\mathfrak{nr(g)}$ is the nilradical of $\mathfrak{g}$.

I am facing two problems to solve the exercise the first one is that I was not able to find a nice example of Lie Algebra such that $\mathfrak{g}$ can be written as $\mathfrak{g} = \mathfrak{r(g) \oplus s}$, and $[\mathfrak{nr(g), s}] \neq 0$.

The second one is that I don't understand why $\mathfrak{g} = \mathfrak{r(g) \oplus} \text{exp} (\text{ad}(z)) (\mathfrak{s})$ with $z \in \mathfrak{nr(g)}$, would be a Levi's decomposition. It is not clear to me why $\mathfrak{r(g) \oplus} \text{exp} (\text{ad}(z)) (\mathfrak{s})$ generates $\mathfrak{g}$.

Can anyone help me?

  • $\begingroup$ I would like to find two different subalgebras $\mathfrak{s}_1$ and $\mathfrak{s}_2$ such that $ \mathfrak{s}_1 \neq \mathfrak{s}_2 $, and $\mathfrak{g} = \mathfrak{r(g)} \oplus \mathfrak{s}_1 = \mathfrak{r(g)} \oplus \mathfrak{s}_2 $. I will edit my question and make it clearer. $\mathfrak{s}_1$ and $\mathfrak{s}_2$ can be isomorphic, it just need to be different. $\endgroup$ – Matheus Manzatto Oct 2 '18 at 13:50
  • $\begingroup$ Can't we just apply any non-trivial automorpism of the form $\exp(ad(x))$ to $\mathfrak{s}_1$? Then $\mathfrak{s}_2\neq \mathfrak{s}_1$. $\endgroup$ – Dietrich Burde Oct 2 '18 at 13:53
  • $\begingroup$ To do this, I think that $x$ need to lies in $\mathfrak{nr(g)}$ (nilpotent radical), I'm not finding an example of a Lie Algebra such $\mathfrak{g}$ isn't solvable, semisimple or reductive such that $[\mathfrak{g},\mathfrak{nr(g)}] \neq 0$. $\endgroup$ – Matheus Manzatto Oct 2 '18 at 13:59
  • 2
    $\begingroup$ Take a semidirect product ${\mathfrak sl}_2(\mathbb C)\ltimes {\mathfrak n}_3(\mathbb C)$ where the semidirect product is by a suitable action on the Heisenberg Lie algebra ${\mathfrak n}_3(\mathbb C)$. $\endgroup$ – Dietrich Burde Oct 2 '18 at 14:04
  • $\begingroup$ @DietrichBurde final question, I swear, what is a "suitable action"? $\endgroup$ – Matheus Manzatto Oct 2 '18 at 14:06

You can get an appropriate example from parabolic subalgebras of semisimple Lie algebras. For instance, take $\mathfrak g$ to be block-upper-triangular matrices of the form $\begin{pmatrix} A & C \\ 0 & B\end{pmatrix}$ with $tr(A)=tr(B)=0$ with $A$ of size $k\times k$, $B$ of size $\ell\times\ell$ and $C$ of size $k\times\ell$. Then $\mathfrak{r}(\mathfrak g)=\mathfrak{nr}(\mathfrak g)$ are the matrices with $A=B=0$ and you get $\mathfrak{s_1}=\left\{\begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}:A\in\mathfrak{sl}_k:B\in\mathfrak{sl}_\ell\right\}$ and $ \mathfrak{s_2}=\left\{ \begin{pmatrix} A & C_0B-AC_0 \\ 0 & B\end{pmatrix}:A\in\mathfrak{sl}_k:B\in\mathfrak{sl}_\ell\right\}$ for a fixed matrix $C_0$. In these cases it is obvious that $\mathfrak{g}=\mathfrak{r}(\mathfrak g)\oplus \mathfrak{s_1}=\mathfrak{r}(\mathfrak g)\oplus \mathfrak{s_2}$

  • $\begingroup$ Your example was astonishing! $\endgroup$ – Matheus Manzatto Oct 3 '18 at 16:03

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