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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 an analogous of the BCH formula which $XY-YX$ in terms of the commutators $[X,Y]_\mathbf{e}$. That is, if for $\mathbf{e}=(e_1,\ldots,e_r)$ we define $\langle\mathbf{e}\rangle=e_1+\ldots+e_r$ 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 $v_n(X,Y)=\sum_{<\mathbf{e}>=n}t_\mathbf{e}[X,Y]_\mathbf{e}$ then


I would appreciate any reference about this.

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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
Have you checked Chapter 6 of the book "Analytic pro-$p$ groups" by Dixon, du Sautoy, Mann and Segal? – user91132 Mar 9 '12 at 18:21
Yes, I checked that book and it says that there is an analogous BCH formula for $\log(\exp(-x)\exp(-y)\exp(x)\exp(y))$, that is, it can be expressed as a series in terms of repetead Lie brakets (for Lie brakets I mean (X,Y)=XY-YX ) as in the BCH formula for $\Phi(X,Y)$. Now I realize that combining this with the idea on how Corollary 3 can be obtained from Corollary 2 in Chapter 6 of Segal's book, Polycyclic group, I could get a confirmation of this fact. – Diego Sulca Mar 9 '12 at 21:03

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