Nilradical of a Lie algebra associated to a associative algebra

Question

Let $A$ be a finite dimensional (unital) $K$-Algebra. By $A^{\circ}$ we denote the associated $K$-Lie-algebra of $A$ with respect to the product $a\circ b:=ab-ba$. In addition, we denote by $rad(A^{\circ})$ the largest nilpotent ideal of $A^{\circ}$ and by $J(A)$ the largest nilpotent ideal of $A$ (the Jacobson-radical).

If $A/J(A)$ is separable and commutative there exist a complement $H$ of $J(A)$ in $A$ by the Wedderburn-Malcev-Theorem. I could proove that in this case $J(A)+Z(A)=rad(A^{\circ})$ holds ($Z(A)$ is the center of $A$.).

What is the nilradical (in terms of structure properties of the associative algebra) of $A^{\circ}$ for an arbitrary finite dimensional (unital) associative algebras $A$? What is the solution for certain associative classes like division algebras, simple, semisimple or basic algebras?

Answer 1

If $A$ is a finite-dimensional simple algebra over a field of characteristic not 2 then the nilradical of the Lie algebra $A^-$ coincides with the center $Z(A)$ of $A$. This follows by combining some results in the first chapter of the Herstein's book "Topics in Ring Theory". Such a result fails in characteristic 2. Indeed, if $A$ is the algebra of 2x2 matrices over a field of characteristic 2, then in this case $[A,A]= sl(2,F)$ is nilpotent (being isomorphic to a 3-dimensional Heisenberg algebra) and it coincides with the nilradical of $A^-$. This example also shows that the nilradical of $A^-$ need not be a subring of $A$ and so, in general, one cannot expects that the nilradical of a Lie algebra arising from an associative algebra $A$ "can be described in terms of the associative structure of $A$". I think it is also worth to mention that if $A$ is defined over a field of characteristic $p>0$ then $A^-$ has a natural structure of restricted Lie algebra via the $p$-map given by the ordinary $p$-exponenentation, and it is not difficult to see that the $p$-radical (that is, the maximal $p$-nilpotent restricted ideal) of this restricted Lie algebra coincides with the Jacobson radical $J(A)$ of $A$.

Answer 2

If the associated Lie algebra $A^{-}$ is nilpotent, then the question what $rad(A^{-})=A^{-}$ can be, seems related to the question, which nilpotent Lie algebras admit an associative algebra structure (where the Lie bracket is given by $[x,y]=x\cdot y-y\cdot x$ with some associative bilinear product $x\cdot y$ on the vector space). Let $G$ be a connected and simply connected Lie group. If $G$ admits a left-invariant and right-invariant affine structure, then its Lie algebra $\mathfrak{g}$ admits an associative algebra structure. If $G$ admits a left-invariant affine structure, then $\mathfrak{g}$ admits a so-called left-symmetric algebra structure.

It is known that not all nilpotent Lie algebras admit an associative algebra structure. In fact, not even all nilpotent Lie algebras admit a left-symmetric algebra structure. Such examples have been given by Y. Benoist in his paper Une nilvariété non affine. See also the paper Affine structures on nilmanifolds.

Answer 3

The case $rad(A^{\circ}) = A$ (in other words that $A$ is as Lie-algebra nilpotent) can be described that every separable element is central. This follows from the determination of Cartan-subalgrbas of $A^{\circ}$ based on a theorem by Salvatore Siciliano.

I think the following theorem holds but I have to go through the proof again: If we take a perfect field of characteristik not equal to $2$ then the nilradical is exactly the sum of the Jacobson radical and the centre of the associative algebra.

Answer 4

If $A/rad(A)$ is separable and $char(K)\ne 2$ then the nilradical of $A^{\circ}$ is exactly $rad(A)+Z(A)$.The steps for the proff are as follows (which I will refine later): 1.) Proof the theorem in the solvable case. 2.) Use Hersteins results to extend them to simple $K$-Algebras. 3.) Use 2.) and extend them to semisimple algebras by using diagonals. 4.) The general case will be reduced to 1.) and 3.).