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Let $(A,\cdot)$ be a finte dimensional associative algebra over $\mathbb{C}$, which is noncommutative, and $(\mathfrak{g},[\cdot,\cdot])$ be its induced Lie algebra, i.e., $\mathfrak{g}= A$ as vector spaces and $[x,y]:=x\cdot y-y\cdot x,\forall x,y\in{A}.$ If

  • the annihilator $ann(A):=\{x\in A:x\cdot y=y\cdot x=0,\forall y\in A\}$ is zero. (Note that there do exist associative algebras having non-trivial annihilator, for exmaple, nilpotent associative algebras).
  • $(\mathfrak{g},[\cdot,\cdot])$ is a reductive Lie algebra, i.e., the adjoint representation ad$:\mathfrak{g}\rightarrow\mathfrak{gl}(\mathfrak{g})$ is completely reducible.

Then is $(A,\cdot)$ necessarily a semisimple associative algebra? If not, are there any counterexamples?

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    $\begingroup$ Isn't $ker(L)=ker(R)=0$ automatically as soon as $A$ has a unit? So that e.g. $A=\mathbb{C}[x]/(x^2)$ is a counterexample? Do you maybe mean to assume that for each $a\in A$, the kernels of left and right multiplication with $a$ have trivial intersection? $\endgroup$
    – Torsten Schoeneberg
    Feb 16 at 6:21
  • $\begingroup$ Thanks, you are right. The original question has been edited. $\endgroup$
    – zhanghtam
    Feb 16 at 6:50
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    $\begingroup$ The left and right regular representations have zero kernel, as it has been said, so your first condition is weird. $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 16 at 6:57
  • $\begingroup$ The first condition is equivalent to that the annihilator of A is zero. Note that nilpotent associative algebras necessarily have non-trivial annihilator. $\endgroup$
    – zhanghtam
    Feb 16 at 9:06
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    $\begingroup$ If your algebras are not unital, you should be explicit about that. Otherwise, only confusion ensues... $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 16 at 10:00

2 Answers 2

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Any such associative product $x\cdot y$ defines a pre-Lie algebra structure on the Lie algebra $\mathfrak{g}$. This means we have $(x,y,z)=(y,x,z)$ for the associator in the algebra $(A,\cdot)$, where $A$ and $\mathfrak{g}$ have the same underlying finite-dimensional vector space, and that $[x,y]=x\cdot y-y\cdot x$. If this product $x\cdot y$ is commutative, $\mathfrak{g}$ is the zero Lie algebra. We want to exclude this.

For the reductive Lie algebra $\mathfrak{gl}_n(\Bbb C)$, these structures have been classified, see for example here, and the references therein. Then the algebra $(A,\cdot)$ is simple, see Lemma $2$.

In general, it is an open problem, which complex reductive Lie algebras admit such a pre-Lie algebra structure (or an associative structure).

The example by Torsten is the $6$-dimensional algebra $A=M_2(\Bbb C)\times A_1$, where $A_1$ has a basis $\{1,x\}$ with $1\cdot 1=1, 1\cdot x=x\cdot 1=x$ and $x\cdot x=0$. The algebra $(A,\cdot)$ is associative, but not commutative and not semisimple. It has the reductive Lie algebra $\mathfrak{sl}_2(\Bbb C)\oplus \Bbb C^3$.

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  • $\begingroup$ More Lie algebras admit pre-Lie structures than associative products inducing their products: it may well be the case that the problem of deciding which Lie algebras admit pre-Lie structures is much more difficult than the one of deciding which Lie algebras admit associative products, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Feb 16 at 22:02
  • $\begingroup$ @MarianoSuárez-Álvarez Yes, the existence problem is much harder for pre-Lie in general. Still, the associative case is also not so easy. And a classification of the associative structures might also be difficult. $\endgroup$
    – Dietrich Burde
    Feb 16 at 22:21
  • $\begingroup$ @Torsten Schoeneberg has constructed a counterexample as above. $\endgroup$
    – zhanghtam
    Feb 17 at 4:25
  • $\begingroup$ @zhanghtam Yes, thank you. I hope that the above references to pre-Lie algebra structures are still useful for you, for further questions on this topic. $\endgroup$
    – Dietrich Burde
    Feb 17 at 9:03
  • $\begingroup$ Thank you for the references @Dietrich Burde $\endgroup$
    – zhanghtam
    Feb 18 at 7:11
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To move this from comments to an answer, a nonabelian counterexample is given by

$$A = \mathbb C[x]/(x^2) \times M_2(\mathbb C) .$$

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