I was wondering if some concepts of Lie algebras with two multiplications exist in the litterature. By this I mean an object $(A,+, \times , \star)$ such that $(A,+,\times)$ and $(A,+,\star)$ are Lie algebras, and there exist some equationnal affinities between the two products (probably something which looks like the Jacobi identity).

For example : $x\times(y \star z) =(x \times y) \star z + y \star ( x\ \times z) $.

Does anybody has ever heard or read about something similar ?

  • 1
    $\begingroup$ What does motivate this question? $\endgroup$ – Berci Jun 12 at 15:22
  • $\begingroup$ denoting the null lie bracket $(x\times y) = 0$ and denoting $x\star y$ any other lie bracket that you can define in this same lie algebra, they satisfy this relation. But well, a little bit uninteresting $\endgroup$ – math.h Jun 12 at 16:59
  • $\begingroup$ The motivation is that there exists a categorical description for Lie algebras. So I want to apply the same characterisation to algebras with two multiplications and see what comes out. What I expect is some kind of Lie algebras where the brackets talk to each other. $\endgroup$ – Sov Jun 13 at 12:02

Post-Lie algebras and post-Lie algebra structures have two Lie algebra structures on a vector space $V$, say a Lie bracket $[x,y]$ and a Lie bracket $\{x,y\}$, together with a bilinear product $x\cdot y$ such that

\begin{align} x\cdot y -y\cdot x & = [x,y]-\{x,y\} \label{post5}\\ [x,y]\cdot z & = x\cdot (y\cdot z) -y\cdot (x\cdot z) \label{post6}\\ x\cdot \{y,z\} & = \{x\cdot y,z\}+\{y,x\cdot z\} \label{post7} \end{align} for all $x,y,z \in V$. This is related to geometric structures on Lie groups, Rota-Bater operators, homology of partition posets, Koszul operads, isospectral flows, Lie-Butcher series and many other topics. For a reference see for example the articles Post-Lie algebra structures on pairs of Lie algebras and Rota-Baxter operators and post-Lie algebra structures on semisimple Lie algebras.

  • $\begingroup$ Thank you for the references ! But here it is more likely to have three multiplications, and the third one makes speak the other two together. $\endgroup$ – Sov Jun 13 at 11:59
  • $\begingroup$ We also obtain conditions without the third one $x\cdot y$, which you could consider then, e.g., $\{[x,y],z\}+\{[y,z],x\}+\{[z,x],y\}=0$ for all $x,y,z$. There are several ways to consider triples $(V,[,],\{,\})$. $\endgroup$ – Dietrich Burde Jun 13 at 12:03

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.