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While trying to deal with a problem involving BCH (Baker-Campbell-Hausdorff) formula, I've noticed something strange. Everywhere in the literature I've managed to fetch (for example: this and this paper, also Wikipedia), fourth order term is simply just:

\begin{equation} -\frac{1}{24}[Y,[X,[X,Y]]] \end{equation}

I understand that some commutators that might seem different are actually the same, for example $[X,[Y,[X,Y]]]$ and $[Y,[X,[X,Y]]]$, which can be simply shown:

\begin{equation} [X,[Y,[X,Y]]]-[Y,[X,[X,Y]]]= [\mathrm{ad}_X, \mathrm{ad}_Y][X,Y]=\mathrm{ad}_{[X,Y]}[X,Y]=0 \end{equation}

Therefore:

\begin{equation} -\frac{1}{24}[Y,[X,[X,Y]]]=-\frac{1}{24}[X,[Y,[X,Y]]] \end{equation}

But what about the terms with $[X,[X,[X,Y]]]$ and $[Y,[Y,[X,Y]]]$? Why are they not present? I fail to see how they cancel or can be written as a part the previously mentioned term.

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Danijel
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1 Answers1

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I am only a physicist, and thus lack deep understanding of the combinatoric symmetry underlying Bernoulli numbers, namely the recondite reason $B_{2n+1}=0$ for integers $n>0$. (Note added : actually the evenness of the shifted generating function does it.)

In your case, $B_3=0$ leads to the absence of $[X,[X,[X,Y]]]$ from the expansion, cf. W Miller's book (Symmetry Groups and their Applications, Academic Press, NY 1972, pp 159-161), or my crib notes for physics applications, Application Algorithm 1, the most popular.

Namely, $$ \log (\exp X ~ \exp (tY)) |_{t=1}= X +\int_0^1 dt~ \psi ( e^{\operatorname{ad}X} e^{t\operatorname{ad}Y}) ~ Y ~, $$
where $\psi (e^{-y})=\sum_{n=0}^\infty B_n y^n/n!$ is the generating function of the Bernoulli numbers.

So, you see, in your case, for terms linear in Y, we are merely looking at $\psi(e^{\operatorname{ad}_X}) ~ Y$ and the cubic, quintic,... terms in X are missing. There should be a less "just-so" symmetry argument, to be sure.

The obverse argument for X - Y interchange follows by the antisymmetry of even powers of generators' terms.

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Cosmas Zachos
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