# Associative algebras whose induced Lie algebras are reductive.

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?

• 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? Feb 16 at 6:21
• Thanks, you are right. The original question has been edited. Feb 16 at 6:50
• The left and right regular representations have zero kernel, as it has been said, so your first condition is weird. Feb 16 at 6:57
• The first condition is equivalent to that the annihilator of A is zero. Note that nilpotent associative algebras necessarily have non-trivial annihilator. Feb 16 at 9:06
• If your algebras are not unital, you should be explicit about that. Otherwise, only confusion ensues... Feb 16 at 10:00

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 $$\mathfrak{sl}_2(\Bbb C)\oplus \Bbb C^3$$.

• 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? Feb 16 at 22:02
• @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. Feb 16 at 22:21
• @Torsten Schoeneberg has constructed a counterexample as above. Feb 17 at 4:25
• @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. Feb 17 at 9:03
• Thank you for the references @Dietrich Burde Feb 18 at 7:11

To move this from comments to an answer, a nonabelian counterexample is given by

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