Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

One of the motivation of the theory of Lie Algebras is that every associative algebra $A$ is a LA when the bracket is defined by $[a,b]=ab-ba$ : this is skew-symmetric and satisfies the Jacobi identity $[[a,b],c]+[[b,c],a]+[[c,a],b]=0$. Conversely, every abstract LA can be embedded into an associative algebra (its envelopping algebra). And for some good reason, one is really interested in sub-LAs rather than sub-algebras. A similar attitude, with different motivation lead to the notion of Jordan algebras.

If $A$ is an associative algebra, one may consider instead the ternary bracket $$[a,b,c]_3=abc+bca+cab-acb-cba-bac.$$ Does $[.,.,.]_3$ satisfy non-trivial identities, besides skew-symmetry? Is there any theory of abstract objects, vector spaces endowed which a ternary skew-symmetric product satisfying these identities?

More generally, we may consider a $d$-bracket, which bears the name of standard non-commutative polynomial in $d$ non-commuting variables. For $d=2$, it is nothing but the standard bracket. When $d=2p$, the $d$-bracket does satisfy non-trivial identies, for instance $$\sum_{i\in\frak A_7}[[a_{i_1},a_{i_2},a_{i_3},a_{i_4}],a_{i_5},a_{i_6},a_{i_7}]=0,\qquad\forall a_1,\ldots,a_7\in A.$$ I don't know if something non-trivial exists when $d$ is odd.

share|improve this question
    
Have you checked if these are $L_\infty$ structures? –  Jim Conant May 10 '12 at 16:50
    
(Ginzburg and Berger have studied a family of algebras, called antisymmetrizer algebras. The thirds one is the quotient of the free algebra on three generators $a$, $b$, $c$ modulo what you write as $[abc]_3$; the algebra is encodes then properties of triads of elements which "ternary-Lie-commute". The algebra is $3$-Koszul in the sense of Berger, so quite special: maybe you can extract information from that fact) –  Mariano Suárez-Alvarez May 10 '12 at 18:21
    
If you remove $bca-acb$ (that is, the only two terms with $c$ in the middle) then you get a Lie triple system jstor.org/discover/10.2307/… en.wikipedia.org/wiki/Triple_system#Lie_triple_systems –  Fernando Muro May 10 '12 at 19:40
    
You might be interested in my answer to a previous MO question concerning n-Lie algebras: mathoverflow.net/questions/49437/… –  José Figueroa-O'Farrill May 12 '12 at 9:54
add comment

4 Answers 4

up vote 13 down vote accepted

"Identities for the ternary commutator" by Bremner classifies all such identities up to degree 7. A recent exposition can also be found in "Ternutator Identities" by C. Devchand, D. Fairlie, J. Nuyts, G. Weingart. Similar identities for n-ary commutators are proven in "Multi-operator brackets acting thrice" by T. Curtright, X. Jin, L. Mezincescu. I don't know if there is an accepted definition for what a Lie n-ary algebra should be for $n\geq 3$ (not to be confused with Lie n-algebras from nlab). For $n=3$ the most standard object one encounters are Lie triple systems.

share|improve this answer
4  
Ternutators is a painful neologism... –  Mariano Suárez-Alvarez May 10 '12 at 18:17
    
Thanks! This is definitely the answer I was looking for. I'm impressed by the complexity of such a short paper. Page 618 is impressive! The referee must have had hard time... –  Denis Serre May 10 '12 at 20:10
3  
"Ternutation" = the act of neezing. Compare "Sternutation". –  Noam D. Elkies May 11 '12 at 2:06
    
Which gives éternuer in French. –  Chandan Singh Dalawat May 11 '12 at 9:29
add comment

The proper setting is the theory of operads, which allows to deal with any number of generators at a single stroke.

The Lie operad is the sub-operad of the associative operad generated by 12-21.

There exists some n-ary Lie operads, but I do not remember whether they are exactly what you ask.

share|improve this answer
add comment

One way to define a Lie structure on a vector space $V$, is as a map $\wedge^2V\to V$ such that its natural extension to $d\colon\wedge^k V\to \wedge ^{k-1}V$ satisfies $d^2=0$. This exactly gives the Jacobi identity. Similarly one can define a Lie infinity structure to be any map $d\colon \wedge V\to \wedge V$ with square $0$. In your case $[\cdot,\cdot,\cdot]_3$ gives a map $\wedge^k V\to\wedge^{k-2}V$ and the question becomes whether it squares to $0$, which I haven't had time to work out. But if it does, then you have a L-infinity structure.

share|improve this answer
add comment

You might want to look at the paper "On Lie k-Algebras" by P. Hanlon by M. Wachs (http://www.sciencedirect.com/science/article/pii/S0001870885710389). They consider algebras satisfying the generalized Jacobi identity you specify. I wanted to leave this as a comment but I don't have enough reputation.

share|improve this answer
add comment

Your Answer

 
discard

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.