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 


$[B(m),B(n)] = B(mn) + \sum\_{l>mn, |l| = |m|+|n|} a\_{m,n}^{(l)} B(l)$



Bracket of lyndon words?
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 $mn$. 


Computationally


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 $mn$

- $$
B(n)B(m)=nm+[\sum\_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum\_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2]
$$
all the monomials within square brackets are of same length and strictly greater than $nm$. 

- 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. 


