# Nilradical of a Lie algebra associated to a associative algebra

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?

• You mean "let $A$ be a finite dimensional unital $K$-algebra"? – Qiaochu Yuan Sep 16 '14 at 23:06
• Yes of course thanks I have correected this issue. – Sven Wirsing Sep 16 '14 at 23:09

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 $A$.

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.

• In this case the nilradical is the whole algebra. But thanks for your additional information Dietrich. – Sven Wirsing Sep 17 '14 at 11:49
• Hi Dietrich, yes it exists, the nilpotent case of $A^{-}$ are those algebras in which every separable element is central (in finite dimension). For example, if $A/J(A)$ is separable then the complement of $J(A)$ is central. – Sven Wirsing Sep 17 '14 at 14:35
• If you ask what the nilradical of $A^{-}$ is for arbitrary associative algebras $A$, then certain Lie algebras just cannot arise as $nil(A^{-})$. – Dietrich Burde Sep 17 '14 at 14:41
• I understood the question "what is the nilradical of $A^\circ$" in the sense "how can it be descibed in terms of the structure of the associative algebra $A$, not in the sense "what is the nilradical of $A^\circ$ isomorphic to". Maybe Sven can clarify. – YCor Sep 17 '14 at 15:17
• Yes you are right YCor like the result for the case I shoed above where the nilradical is sum of the Jacobson radical and the centre. This was my intention. The extremal case is described in my answer that every separable element is central. Of course it is nice to know which algebras are appearing but my intention was to decribe the nilradical like YCor said. – Sven Wirsing Sep 18 '14 at 6:54

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.

• In this case $A$ contains a unique maximal separable subalgebra which is central in $A$. – Salvatore Siciliano Sep 18 '14 at 10:41
• Hi Salvatore:-) Yes the set of all separable elements. – Sven Wirsing Sep 18 '14 at 13:51
• Hi Sven. I hope that my previous answer about simple algebras can help you. On the other hand, as I explained, in my opinion one cannot hope to describe the nilradical for general Lie algebras coming from associative algebras. – Salvatore Siciliano Sep 18 '14 at 14:41
• Hi Salvatore thanks for your input, so maybe it helps to described it for semisimple associative algebras as well, I think as direct sums. Need to check this, too. Yes, is there maybe something in characteristics zero which is valid? In characteristics p its nice to your comment about restricted Lie-algebras, too. Is there an example in charcteric different from 2 which shows that the nilradical is not an associative subalgebra of A? – Sven Wirsing Sep 18 '14 at 16:37
• I think for semisimple associative algebras the nilradical is the direct sum of the nilradicals of the direct summands. So we have a solution for certain classes of algebras. I will investigate basic algebras now. – Sven Wirsing Sep 18 '14 at 20:56

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.).

• Sven, please could you provide an argument for your conclusion? – Salvatore Siciliano Oct 20 '14 at 14:30
• Hi Salvatore, yes of course, I will write the arguements precisely in German at the moment and then I will translate them to English and provide it here:-) – Sven Wirsing Oct 22 '14 at 7:57