# Example of a Lie Alegebra $\mathfrak{g}$ such that the Levi decomposion is not unique.

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?

• 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. – Matheus Manzatto Oct 2 '18 at 13:50
• Can't we just apply any non-trivial automorpism of the form $\exp(ad(x))$ to $\mathfrak{s}_1$? Then $\mathfrak{s}_2\neq \mathfrak{s}_1$. – Dietrich Burde Oct 2 '18 at 13:53
• 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 $[\mathfrak{g},\mathfrak{nr(g)}] \neq 0$. – Matheus Manzatto Oct 2 '18 at 13:59
• 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 ${\mathfrak n}_3(\mathbb C)$. – Dietrich Burde Oct 2 '18 at 14:04
• @DietrichBurde final question, I swear, what is a "suitable action"? – 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 $$\mathfrak{g}=\mathfrak{r}(\mathfrak g)\oplus \mathfrak{s_1}=\mathfrak{r}(\mathfrak g)\oplus \mathfrak{s_2}$$