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Let $\mathfrak{g}$ be – for the sake of simplicity – a complex Lie Algebra. Then we define the antisymmetric linear transformations as $$ \mathfrak{A(g)} =\big\{\left.\,\alpha \in \mathfrak{gl(g)}\,\right|\;[\alpha(X),Y]+[X,\alpha(Y)]=0\text{ for all }X,Y\in \mathfrak{g}\,\big\} $$ which is again a Lie algebra, and via $X.\alpha :=[\operatorname{ad}X,\alpha]$ a $\mathfrak{g}$-module.

It can be shown by direct calculation, that $\mathfrak{A(g)}=0$ for semisimple Lie algebras $\mathfrak{g}$, and $\mathfrak{A(g)}\neq 0$ for solvable Lie algebras $\mathfrak{g}$, where the one dimensional ideal is used.

As $\mathfrak{A(g)}=\mathfrak{gl(g)}$ for abelian Lie algebras $\mathfrak{g}$, the complexity of $\mathfrak{A(g)}$ somehow measures the complexity of $\mathfrak{g}$ in reverse, which makes it interesting for solvable Lie algebras without center. Being a Lie algebra itself it also allows all combinations like a semidirect product with $\mathfrak{g}$, iterations of $\mathfrak{A}(.)$, or other funny things like defining $(X,Y)\longmapsto \alpha([X,Y])$ as new Lie product.

Where can I find more about this?

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    $\begingroup$ You mean (nonzero) finite-dimensional solvable Lie algebras? For $\infty$-dimensional solvable Lie algebra there's not always a 1-dimensional ideal. Also strange remark, but the definition doesn't make use of the Lie bracket of the source. So it makes sense inside linear maps $\mathfrak{v}\to\mathfrak{g}$ for an arbitrary vector space $\mathfrak{v}$ (but for $\mathfrak{v}=\mathfrak{g}$ indeed one gets a Lie subalgebra of the Lie algebra of linear endomorphisms). $\endgroup$
    – YCor
    Commented Mar 18, 2020 at 17:53
  • $\begingroup$ Yes, I meant finite dimensional cases. What do you mean by "doesn't make use of the Lie bracket of the source"? It is defined by the Lie bracket of the source $\mathfrak{g}$. It is an invariant of $\mathfrak{g}$, which is somewhere in between $0$ and $\mathfrak{gl(g)}$. As the structure constants of $\mathfrak{g}$ define the linear equation system which determines $\mathfrak{A(g)}$, there is a direct connection between them. E.g. if $\mathfrak{g}$ is the matrix algebra of first row only matrices, then $\mathfrak{g}\cong \mathfrak{A(g)}$. $\endgroup$
    – Marius S.L.
    Commented Mar 18, 2020 at 18:24
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    $\begingroup$ An extension of your assertion on semisimple Lie algebras: Let $W(\mathfrak{g})$ be the set of $x$ such that $B(x,[y,z])=0$ for all $y,z\in\mathfrak{g}$ and every invariant symmetric bilinear form (ISBF) on $\mathfrak{g}$. Then for every $\alpha\in\mathfrak{A}(\mathfrak{g})$ we have $\alpha$ valued in $W(\mathfrak{g})$. In general $W(\mathfrak{g})$ contains the center, but in case $\mathfrak{g}$ has a non-degenerate ISBF (there are many examples beyond semisimples), then it's an equality, and in particular if in addition the center is trivial then $\mathfrak{A}(\mathfrak{g})=0$. $\endgroup$
    – YCor
    Commented Mar 18, 2020 at 18:43
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    $\begingroup$ Beware that Borel subalgebras of semisimple Lie algebras are very non-typical solvable Lie algebras (and their nilpotent radical is very non-typical as nilpotent Lie algebras). Still, observations: every $\alpha\in\mathfrak{A}(\mathfrak{g})$ preserves the ascending central series. Also, whenever it preserves an ideal, it preserves its centralizer (and more generally its centralizer modulo another $\alpha$-invariant ideal). $\endgroup$
    – YCor
    Commented Mar 18, 2020 at 19:26
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    $\begingroup$ Of course deriv(...) is worse, I never suggested it. There are plenty of words that could be used (transfiguration, transmutation, mutation, metamorphosis, transference, etc– just browse a dictionary of synonyms, e.g., here); I also thought of "paraderivation". The problem with 'antisymmetric' is that whenever you specify to a Lie algebra which has a scalar product, you get a clumsy terminology. $\endgroup$
    – YCor
    Commented Mar 19, 2020 at 10:27

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