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Let $\mathfrak{g}$ be a complex simple Lie algebra with bracket $[x,y]$. For which $z\in \mathfrak{g}$ defines $$ \mu(x,y)=ad (z)([x,y])=[z,[x,y]] $$ 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.

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I guess you mean "For which $z\in\mathfrak{g}$ does [formula] define another Lie bracket..." –  YCor Aug 27 '13 at 20:15
yes, for which $z$ defines the formula another Lie bracket ("does define" is perhaps better english). –  Dietrich Burde Aug 27 '13 at 20:23
this was my point (the English mistake makes the sentence pretty hard to understand) –  YCor Aug 27 '13 at 21:28

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