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