# Commutator Baker-Campbell-Hausdorff formula

Consider the Baker-Campbell-Hausdorff formula $\Phi(X,Y)\in\mathbb{Q}\langle\langle X,Y\rangle\rangle$ in non-commutative variables. Define $X*Y:=\Phi(X,Y)$ and $[X,Y]=(-X)*(-Y)*X*Y$, and then as usual define for any vector
$\mathbf{e}=(e_1,\ldots,e_r)\in\mathbb{N}^r$ the repeated commutator

$$[X,Y]{\mathbf{e}}:=[X,\underbrace{Y,\ldots,Y}_{e_1},\underbrace{X,\ldots,X}_{e_2},\ldots]$$ (here $[X_1,\ldots,X_r]$ is defined as $[[X_1,\ldots,X_{r-1}],X_r]$).

I think that there is a an analogous of the BCH formula on expressing $XY-YX$ in terms on the commutators $[X,Y]_\mathbf{e}$. That is, if for $\mathbf{e}=(e_1,\ldots,e_r)$ we define $<\mathbf{e}>=e_1+\ldots+e_r$ then there exist rational numbers $t_\mathbf{e}$ for all $\mathbf{e}\in\mathbb{N}^r$ and for all $r$ such that if we put $v_n(X,Y)=\sum_{<\mathbf{e}>=n}[X,Y]_\mathbf{e}$ then

$$XY-YX=\sum_{n\in\mathbb{N}}v_n(X,Y)$$.

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Use \langle x\rangle instead of <x> to obtain $\langle x\rangle$ instead of $<x>$. The spacing of < is quite different because it is interpreted as a relation. –  Mariano Suárez-Alvarez Mar 9 '12 at 17:05
1. Just to make sure: you mean that you define $[X,Y]$ as $\log(e^{-X}e^{-Y}e^Xe^Y)$, right? That $(-X)*(-Y)*X*Y$, though formally the same (?), keeps confusing me. 2. Can you prove your formula, or you expect it to be true? If the latter, did you check it up to some reasonable order, or it's just a feeling? –  Vladimir Dotsenko Mar 9 '12 at 17:06
(Similarly, you can write \mathbb Q\rangle\!\rangle X,Y\langle\!\langle which gives $\mathbb Q\rangle\!\rangle X,Y\langle\!\langle$; \newcommand is your friend :) ) –  Mariano Suárez-Alvarez Mar 9 '12 at 17:08
To add to Vladimir Dotsenko's comment, do I understand correctly that on the left hand side of your displayed equation, $[X,Y]$ means $\log(e^{-X} e^{-Y} e^X e^Y)$ but, on the right hand side, it is the standard commutator $XY-YX$? –  David Speyer Mar 9 '12 at 17:17
Yes, I used this notation because when we work in a nilpotent Lie algebra then $*$ will be the group operation which makes into a group. This question follows from Corollary 3, Chap 6, of Segal's book: Polycyclic groups, where it is proved that if $x_1,\ldots,x_s\in Tr_1(n,k)$ then $(\log x_1,\ldots,\log x_s)=\log [x_1,\ldots,x_s]+\sum_i s_i \log v_i$ where each $v_i$ is repeated group commutator of length at least $s+1$ in $x_1,\ldots,x_s$ and each $s_i$ is a universal constant dependeing only on $n$ (here $Tr_1(n,k)$ is the unipotent group of upper triangular matrices with 1s in the diagonal –  Diego Sulca Mar 9 '12 at 17:30