4
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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 $M(A)$? $\endgroup$ – YCor Dec 1 '20 at 5:06
  • $\begingroup$ $M(A)$ 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 $M(A)$. $\endgroup$ – Claudio Gorodski Dec 2 '20 at 0:29
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Claudio Gorodski Dec 1 '20 at 1:00
  • $\begingroup$ 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). $\endgroup$ – YCor Dec 1 '20 at 5:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.