Timeline for Commutator Baker-Campbell-Hausdorff formula
Current License: CC BY-SA 3.0
13 events
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| Apr 1, 2015 at 21:55 | history | edited | Diego Sulca | CC BY-SA 3.0 |
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| Mar 9, 2012 at 21:03 | comment | added | Diego Sulca | 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. | |
| Mar 9, 2012 at 18:21 | comment | added | user91132 | Have you checked Chapter 6 of the book "Analytic pro-$p$ groups" by Dixon, du Sautoy, Mann and Segal? | |
| Mar 9, 2012 at 18:08 | comment | added | John Jiang | @Diego, you could adopt the angle bracket suggestions of Mariano. | |
| Mar 9, 2012 at 17:36 | history | edited | Diego Sulca | CC BY-SA 3.0 |
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| Mar 9, 2012 at 17:32 | comment | added | Diego Sulca | Yes, $[X,Y]$ means $\log(e^{-X}e^{-Y}e^Xe^Y)$ and $XY-YX$ is the standard commutator in non-commutative variables. | |
| Mar 9, 2012 at 17:30 | comment | added | Diego Sulca | 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 | |
| Mar 9, 2012 at 17:17 | comment | added | David E Speyer | 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 $XY-YX$? | |
| Mar 9, 2012 at 17:14 | history | edited | Diego Sulca | CC BY-SA 3.0 |
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| Mar 9, 2012 at 17:08 | comment | added | Mariano Suárez-Álvarez |
(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 :) )
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| Mar 9, 2012 at 17:06 | comment | added | Vladimir Dotsenko | 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? | |
| Mar 9, 2012 at 17:05 | comment | added | Mariano Suárez-Álvarez |
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.
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| Mar 9, 2012 at 16:59 | history | asked | Diego Sulca | CC BY-SA 3.0 |