Talk:Baker–Campbell–Hausdorff formula

message: This article should also provide the first few terms of W for the left-multiplication variant
:e^{A+B}=e^A e^B e^W
as I'm too lazy to perform the inversion myself. I suppose, for completeness, the right-multiplication version as well.  linas 16:20, 1 March 2006 (UTC)
thread_title: Alternate form

message: I think there is misprint in the formula because ad(Y)(Y)=0
:It looks strange, yes, but I looked it up and it is intentional. I slightly rewrote the formula to get something which is clearer in my opinion, and added an explanation. -- Jitse Niesen (talk) 05:06, 22 May 2007 (UTC)
replies: text: You should have written your query as a separate question at the bottom of the page.  No, the adjoint action is fine, but you misread it. Your expression for just Y is correct and trivial. But the l.h.s. of the equation you are misquoting has the exponential of Y. Had you  concatenated  out the Hadamard rule  for all powers of Y, i.e. all powers of the trivial expression you wrote, and summed over the expansion of the exponential, you would have readily derived the braiding identity given. Cuzkatzimhut (talk) 20:25, 16 January 2018 (UTC)
thread_title: Misprint

message: What is adX?  It appears to be the adjoint representation?  If so, the definition (that comes well after the first usage) isn't consistent with that article, as far as I can tell? Petrelharp (talk) 05:03, 23 February 2011 (UTC)
replies: text: I gather it is the Adjoint endomorphism,  adX (Y) = [X,Y]. Cuzkatzimhut (talk) 12:19, 23 February 2011 (UTC)
thread_title: Definitions

message: Anyone know why this is called Hadamard's lemma?  I can find references for the lemma itself, but nowhere that it's called Hadamard's.  (Except for one dubious paper that was written after the name Hadamard's was introduced into this article...)  For example, the reference in that section doesn't call it Hadamard's.  --MOBle (talk) 16:11, 10 August 2011 (UTC)
replies: text: It is a simple lemma not necessarily linked to CBH. I grew up with that rubric, and I cite at least one such informal usage in print, on my talk page.  So, if you were researching the propriety of the labelling, I doubt Hadamard discovered it; but I stick to popular usage, e.g. like the "Runge" eponymy in the Runge-Lenz vector... Supplant a better name, if available, but I have not come across one. It is a terrible idea, however, to link CBH's label to it...     Cuzkatzimhut (talk) 12:24, 23 August 2011 (UTC)
replies: text: I edited the lemma today, reassuring the reader that Campbell proved it in print in 1897, as I just confirmed; however, this does not preclude prior implicit usage by somebody else, likely Lie... Cuzkatzimhut (talk) 21:41, 11 October 2022 (UTC)
thread_title: Hadamard's(?) Lemma

message: Happy new year.
BONFIGLIOLI and FULCI : topics in Noncommutative Algebra (Lecture Notes in Mathematics 2034),539p, SpV 2012, 
this book is well documented about CBH and Free Lie Algebra
--Guerinsylvie (talk) 09:41, 3 January 2012 (UTC)
thread_title: A good reference: LNM 2034 

message: , first few terms are wtitten in the following formula:
\begin{align}
Z(X,Y)&{}=\log(\exp X\exp Y) \\
&{}= X + Y + \frac{1}{2}[X,Y] +
\frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] \\
&{}\quad
- \frac {1}{24}[Y,[X,[X,Y]]]  \\
&{}\quad
- \frac{1}{720}([[[[X,Y],Y],Y],Y] +[[[[Y,X],X],X],X])
\\
&{}\quad +\frac{1}{360}([[[[X,Y],Y],Y],X]+[[[[Y,X],X],X],Y])\\
&{}\quad
+ \frac{1}{120}([[[[Y,X],Y],X],Y] +[[[[X,Y],X],Y],X])
+ \cdots
\end{align}
It has inconsistent format in a sense than in some terms commutators are nested towards the right, while in other terms they are nested towards the left. Also, innermost commutator is sometimes [X,Y] and sometimes [Y,X]. So I propose it to be replaced by:
\begin{align}
Z(X,Y)&{}=\log(\exp X\exp Y) \\
&{}= X + Y + \frac{1}{2}[X,Y] +
\frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] \\
&{}\quad
- \frac {1}{24}[Y,[X,[X,Y]]]  \\
&{}\quad
+ \frac{1}{720}([Y,[Y,[Y,[X,Y]]]] -[X,[X,[X,[X,Y]]]])
\\
&{}\quad -\frac{1}{360}([X,[Y,[Y,[X,Y]]]]-[Y,[X,[X,[X,Y]]]])\\
&{}\quad
+ \frac{1}{120}([Y,[X,[Y,[X,Y]]]] -[X,[Y,[X,[X,Y]]]])
+ \cdots
\end{align}
But I'm not sure if I've got all the signs right because there is a lot of it. 193.198.162.13 (talk) 09:07, 25 July 2014 (UTC)
replies: text: Well, the most important feature, beyond subjective aesthetic criteria, is that the Y - X symmetries are manifest and natural, and do not need tweaking to jump at you. So, if anything, it is the 3rd order terms that could use tweaking, and certainly not the higher, 5th,  order ones! Your fearing sign mistakes is an ipso facto illustration of the point . You might look at the following sections to appreciate the point and perhaps leave good enough alone? Why don't you get an account? Cuzkatzimhut (talk) 10:46, 25 July 2014 (UTC)
thread_title:  Better formula for first few terms 

message: The beginning of the first section at the moment talks about X,Y being small enough, while claiming to be working in an arbitrary field of characteristic zero. What norm is intended here? 
193.11.79.122 (talk) 08:27, 13 October 2014 (UTC)
replies: text: Good point. The statement "small enough" is independent of norms and fields, but the note box refers, of course, to  and  with the Hilbert-Schmidt norm for matrix groups. YohanN7 (talk) 09:33, 13 October 2014 (UTC)
thread_title:  Any field 

message: : \psi(x) ~\stackrel{\text{def}}{=} ~  \frac{x \log x}{x-1}= 1- \sum^\infty_{n=1}
{(1-x)^n \over n (n+1)}  ~, 
is not the generating function for Bernoulli numbers  — Preceding unsigned comment added by 173.48.171.226 (talk) 19:03, 18 January 2015 (UTC)
replies: text: Please sign your posts--or, better, get a WP account. Which part of footnote 2, nb2, where the Generating function involved is stated, are you disagreeing with? ψ(x) is not, but ψ(exp(y)) is, as stated. Are you conflating involvement with identity?Cuzkatzimhut (talk) 21:04, 18 January 2015 (UTC)
thread_title:  Generating function for Bernoulli numbers 

message: In "An important lemma"
--- A direct application of this idenity 
I asume there is something wrong
:e^X e^Y= e^{X+Y +\frac{1}{2} [ X, Y] } ~. 
should be:
:e^X e^Y= e^{X+Y -\frac{1}{2} [ X, Y] } ~. 
Because The Zassenhaus formula says:
: e^{t(X+Y)}= e^{tX}~ e^{tY} ~e^{-\frac{t^2}{2} [X,Y]} ~
e^{\frac{t^3}{6}(2[Y,[X,Y]]+ [X,[X,Y]] )} ~
e^{\frac{-t^4}{24}([[[X,Y],X],X] + 3[[[X,Y],X],Y] + 3[[[X,Y],Y],Y]) } \cdots
with t=0 and  
: [X [ X, Y] =0 . 
...  — Preceding unsigned comment added by ChilesheAT (talk • contribs) 12:34, 6 March 2016 (UTC)
replies: text: There is no mistake, as you may verify by consistently expanding everything in powers of X and Y... Applying the correct formula to the Zassenhaus formula to quadratic order in t you get consistency. What exactly is your argument? You must have misunderstood something, but it is not easy to reconstruct your wrong steps. Cuzkatzimhut (talk) 14:15, 6 March 2016 (UTC)
thread_title:  possible mistake In "An important lemma" 

message: Substituted at 19:47, 1 May 2016 (UTC)
thread_title: Assessment comment

message: So, am I being a complete idiot, or is is the odd part of the supersymmetric variant given by the generation function x/(1+e^x) -- i.e. Fermi-Dirac statistics, instead of Bose-Einstein statistics? The basic argument would be that Bose-Einstein -> Fock space (which is the same thing as the tensor algebra) but is for bosons, so its the symmetric algebra while for fermions it would be the exterior algebra so we have to track that extra sign through everything, and doing so would flip it to x/(1+e^x) -- finally one reinterprets x to be -E/kT to get the statistics.  I guess I should ask on mathoverflow. Or maybe just google it. 67.198.37.16 (talk) 04:12, 25 September 2016 (UTC)
thread_title:  supersymmetric variant? 

message: To prevent quasi-revelatory tweaks on the history of Lie's third fundamental theorem being presented as the CBH formula, I'd defer to the definitive chronicler of the subject, Wilfried Schmid, in Bulletin of the American Mathematical Society   6 (2) March 1982, pp 175-186. Schur's (note, not Issai S!) role and his checkered history with Lie and Engel  w.r.t. Lie's third fundamental theorem are detailed, and are adhered to in the NPOV placement in the lede.  Cuzkatzimhut (talk) 14:58, 26 September 2016 (UTC)
thread_title: Historical remarks

message: I don't think the explicit formula for X\ast Y written is correct. For instance the term [X, Y] has contributions both from n=1, r_1=s_1=1 and n=2, r_1=s_2=1, s_1=r_2=0, giving a coefficient of 1/2-1/4=1/4 instead of the supposed 1/2. Am I missing something? A (correct) version of the formula can be found in https://terrytao.wordpress.com/2011/09/01/254a-notes-1-lie-groups-lie-algebras-and-the-baker-campbell-hausdorff-formula/, exercise 17.  — Preceding unsigned comment added by 193.136.196.3 (talk) 17:53, 31 January 2018 (UTC)
replies: text: Please sign your message. Yes, you are badly misunderstanding the notation, and proposing partitions expressly excluded by it. Make a bona-fide effort to actually understand the notation, and, if need be, going to the original reference quoted from. This is not a forum. Cuzkatzimhut (talk) 18:55, 31 January 2018 (UTC)
thread_title:  Error in the explicit formula 

message: Footnote 31 is wrong. Hall has no chapter 14. Please correct.
:Apparently, what was meant is chapter 14 in Brian C. Hall, "Quantum Theorty for Mathematicians", not the (not existing) chapter 14 in Brian C. Hall "Lie Groups, Lie Algebras, and Representations: An Elementary Introduction". Can somebody please check? – Tea2min (talk) 08:32, 29 May 2020 (UTC)
thread_title:  Error in Footnote