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Here is a simple question regarding the standard Lyndon basis for the free Lie Algebra. Suppose I take two lyndon words $m$ and $n$ and their standard bracketings $B(m)$ and $B(n)$ as elements in the free Lie algebra. Suppose further that $m < n$, so that $mn$ is a Lyndon word.

My question is when we express the bracket $[B(m),B(n)]$ in the Lyndon basis, is it of the form

$[B(m),B(n)] = B(mn) + \sum_{l>mn, |l| = |m|+|n|} a_{m,n}^{(l)} B(l)$

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No combinatorists out there? I think this is a basic enough question! –  Chitrabhanu Feb 5 '13 at 17:14
Yes it is basic but up to my knowledge very little is known on these structure constants. Do you know more now ? –  Duchamp Gérard H. E. Jun 7 at 9:18

2 Answers 2

up vote 2 down vote accepted

Yes it is of this form because any Lie polynomial "begins" by a Lyndon word, in particular, the least monomial of $B(l)$ is $l$. Then $$ B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v\qquad \mathbf{(LB)} $$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.


  1. $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
  2. $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
  3. But, $mn<nm$ because $mn$ is Lyndon and then $$ [B(m),B(n)]=mn+[sb1]-nm-[sb2]=mn+[sb3] $$ where the square bracket $[sb3]$ is a linear combination of monomials that are greater than $mn$ or $nm$

Hence all (monomials of $[sb3]$) are greater than $mn$ which is the form you required.

One can say a little bit more From property $\mathbf{(LB)}$, and the fact that $[B(m),B(n)]$ is a multihomogeneous Lie polynomial, one has ($\underline{w}$ being the commutative image of $w$) $$ [B(m),B(n)]=B(mn) + \sum_{mn<l\atop l\ \mathrm{Lyndon};\ \underline{l}=\underline{mn}} \gamma_{m,n}^{(l)}B(l) $$
the coefficients $\gamma_{m,n}^{(l)}$ are the structure constants of the free Lie algebra w.r.t. the Lyndon-Sirsov basis (they are integers, universal, i.e. characteristic free) and, up to my knowledge, their combinatorics is widely unknown.

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As far as I can see, your statement is equivalent to Lemma 3.5 in http://arxiv.org/pdf/0804.1254.pdf .

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Yes, equivalent up to a reversal (just to clarify in order that readers be not misled). In the statement, a Lyndon word is the (strict) minimum of a (primitive) conjugacy class as in Reutenauer's book Free Lie algebras (which I believe has become standard convention now) whereas in the paper you cite (Lemma 3.5), a Lyndon-Sirsov word is the (strict) maximum of a (primitive) conjugacy class as was used in the former "école de Lille" around Gérard Jacob. –  Duchamp Gérard H. E. Jun 8 at 7:02
See, for instance <a href="sciencedirect.com/science/article/pii/S0012365X99001247"; target="_blank">sciencedirect.com/s<wbr>cience/article/pii/…;. It seems that the question of structure constants is still widely open. –  Duchamp Gérard H. E. Jun 8 at 7:11
This, of course, does not withdraw any parcel of the merit of Sirsov who was the first to point out these words (rediscovered as explained in Bokut's paper independently by Fox and Lyndon). –  Duchamp Gérard H. E. Jun 8 at 7:22
@DuchampGérardH.E. Yes, I don't think much understood about structure constants. I feel however that the OP was asking a question about a sort of triangularity property with some structure constants, which is indeed very easy. –  Vladimir Dotsenko Jun 8 at 7:41
OK, if it is this, any of us can answer –  Duchamp Gérard H. E. Jun 8 at 8:06

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