# Homotopes of simple Lie algebras

Let $\mathfrak{g}$ be a complex simple Lie algebra with bracket $[x,y]$. For which $z\in \mathfrak{g}$ does the formula $$\mu(x,y)=ad (z)([x,y])=[z,[x,y]]$$ define another Lie bracket on the same vector space ? For $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$ this holds for every $z$. Explicit computation seems to show that for $\mathfrak{g}=\mathfrak{sl}(n,\mathbb{C})$ with $n\ge 3$ only $z=0$ is possible. Does this hold for all simple Lie algebras of rank at least $2$, and is there a proof somewhere ?
In the literature, often the Jacobi identity for $\mu(x,y)=[z,[x,y]]$ is studied for all $z$. Then it is known that only $\mathfrak{sl}(2,\mathbb{C})$ is possible. Another reference are simple Hom-Lie algebras, but this is different. For nonassociative algebras a new multiplication $x\circ_{z} y$ on the same vector space induced by an element $z$ is called a homotope.

What is known for simple Lie algebras ?

EdiT: I saw that $[x,y]_R:=[R(x),y]+[x,R(y)]$ is a Lie bracket iff $R$ is a classical $R$-matrix. For $R=ad(z)$ we have $[x,y]_R=[z,[x,y]]$. So the question is, for which $z$ the map $R=ad(z)$ is a classical $R$-matrix of a simple Lie algebra.

Theorem $3.2$: Let $\mathfrak{g}$ be a simple Lie algebra of rank $r\ge 2$ over an algebraically closed field $K$ of characteristic zero, and $z\in \mathfrak{g}$. Suppose that $R={\rm ad}(z)$ is a classical $R$-matrix. Then $z=0$ and $R=0$.
The proof uses the Jordan-Chevalley decomposition for $z=n+s$, the Jacobson-Morozov theorem, arguments concerning roots and algebraic group actions etc. This gives $s=0$ and ${\rm ad}(z)^3=0$, and finally $z=0$. For the case of rank $1$ we have the following result:
Proposition $3.1$: Let $\mathfrak{g}=\mathfrak{sl}(2,\mathbb{C})$. Then $R={\rm ad}(z)$ is a classical $R$-matrix for all $z\in \mathfrak{g}$. The Lie algebra $\mathfrak{g}_R$ is isomorphic to $\mathfrak{r}_{3,1}(\mathbb{C})$ for all $z\neq 0$.
Here $\mathfrak{r}_{3,1}(\mathbb{C})$ is the $3$-dimensional solvable Lie algebra given by $[e_1,e_2]=e_2$ and $[e_1,e_3]=e_3$, and $\mathfrak{g}_R$ has bracket $[x,y]_R=[z,[x,y]]$.