MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there any reasonably natural algebras whose product (bracket) almost, but does not quite, satisfy the Jacobi relation?

A priori it doesn't matter whether the bracket is anti-symmetric.

The question is deliberately vague about "almost, but does not quite", just to see if this strikes any chord. It can mean that the failure to satisfy Jacobi has a factor of epsilon, so that as epsilon goes to zero you get a Lie algebra.

share|cite|improve this question
up vote 2 down vote accepted

See Section 2.3 of the lecture notes called Geometric Models for Noncommutative Algebras by Ana Cannas da Silva and Alan Weinstein. There they define an "almost Lie algebra" to be something with an antisymmetric bracket but which does not necessarily satisfy Jacobi. In Section 3.2 they connect this to the notion of "almost Poisson manifold", which is just a manifold equipped with a bivector field. The bivector field defines a skew-symmetric bracket on smooth functions which may or may not satisfy Jacobi.

share|cite|improve this answer
Thanks to both MTS and D Burde. I hadnt thought of either answer. In a different direction (maybe) there are also L-infinity algebras (Lie algebras up to homotopy). Tks again. – Peter Mar 31 '13 at 15:15
For $L$-infinity-algebras and Hom-Lie 2-algebras, see the paper arXiv:1110.3405. – Dietrich Burde Mar 31 '13 at 21:02
In Alan Weinstein's inimitable style, the formula $[a,[b,c]]+cyc$ is called the "Jacobiator". A Lie algebra is one for which the Jacobiator vanishes. – Allen Knutson Mar 31 '13 at 23:52

There is a lot of research lately on so-called Hom-Lie algebras. A Hom-Lie Algebra is a vector space $L$ together with a bilinear skew-symmetric bracket, and a linear map $f:L \rightarrow L$ satisfying the Hom-Jacobi identity $$ [f(x),[y,z]]+[f(y),[z,x]]+[f(z),[x,y]]=0 $$ With $f=id+\epsilon g$, one obtains a Lie algebra for $\epsilon \to 0$.

share|cite|improve this answer
Is there a compatible notion of a Hom-Lie group? – Sasha Mar 30 '13 at 19:39
Good question. I have not seen Hom-Lie groups yet. There are relationships to quantum groups and quantum deformations, though. – Dietrich Burde Mar 31 '13 at 20:58
That's an interesting concept. Where does it arise? Are there natural examples? And do you know any good sources to read about them? – MTS Mar 31 '13 at 21:52
It came up first in the study of q-deformations of the Witt and Virasoro algebra. A good introduction, perhaps, is the paper and the phd-thesis of Daniel Larsson. – Dietrich Burde Apr 1 '13 at 9:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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