# Left- (right-) multiplications of an algebra that are derivations

Let us say that $$A$$ is a (finite-dimensional) algebra over a field of characteristic zero. We can assume commutativity but not associativity, if that makes it easier. Indeed, I am mostly interested in the case of complex Jordan algebras.

Question: What is known about left-multiplication operators $$L_a:A\to A$$, $$L_ax=ax$$, that are derivations of $$A$$, in the sense that $$L_a(xy)=(L_ax)y+x(L_ay)$$ for all $$x$$, $$y\in A$$? What about algebras such that all left-multiplications are derivations?

I think these are never semisimple algebras. An obvious remark is that $$A^3=0$$ is a sufficient condition for $$L(A)\subset\mathrm{Der}(A)$$.

(Of course, Lie algebras fit that category.)

• Maybe you could define explicitly $M(A)$ as the set of $a$ such that $L_a$ is a derivation. This is a subspace. I don't know whether it's a subalgebra in general, I guess not (but unsure esp. in the commutative case). What are you asking exactly about $M(A)$?
– YCor
Dec 1 '20 at 5:06
• $M(A)$ certainly does not look like a subalgebra, I do not know exactly what to expect from $M(A)$. I was just doing some calculations and I was wondering whether there was some underlying structure that could simplify them. As it turns out, from this short discussion now I feel like we cannot expect too much from $M(A)$. Dec 2 '20 at 0:29

## 1 Answer

An algebra whose (left) multiplications are derivations is referred to as a (left) Leibniz algebra (or Loday algebra). There is a large literature about this class of non-associative algebras. See e.g. the following survey by Joerg Feldvoss: https://arxiv.org/abs/1802.07219.

• Thanks, it is a very interesting reference! I have only browsed it, but I will read it more carefully. It still remains the first question: what about algebras in which some left multiplications are derivations? Perhaps some ideas in Feldvoss' paper can also help understand this question. Dec 1 '20 at 1:00
• Note that a commutative left-Leibniz algebra (over a ring with $2$ invertible) is somewhat degenerate: it's the same as an algebra (module + bilinear law) $A$ satisfying $A^3=0$ (so it's also associative, Jordan, etc).
– YCor
Dec 1 '20 at 5:07