Recently I did some explicit computations that involved the BCH series, $\log(e^x e^y)$. Here $x$ and $y$ are non-commuting variables, and the BCH series lives in the graded completion $FL(x,y)$ of the free Lie algebra generated by $x$ and $y$.

Mostly by chance I found that when BCH is written in the Lyndon basis of $FL(x,y)$, the number of Lyndon words that occur in its degree $n$ piece is {2, 1, 2, 1, 6, 5, 18, 17, 55, 55, 186, 185, 630, 629, 2181, 2181, 7710, 7709, 27594, 27593, 99857, 99857}, for $n$ running from 1 to 22.

There is an obvious pattern in this sequence - it seems that the odd-numbered terms are almost equal to the even-numbered terms that follow them, with a decline of one in 2/3 of the times, and with precise equality in the remaining 1/3 of the times. I have no idea why this is so. Perhaps you do?

Why care? The truth is that I'm curious but I don't care much; I just stumbled upon this by chance. Yet Lyndon words are a very effective tool for computations in free Lie algebras, and the BCH formula appears in many of these computations. The fact that there is some unexpected symmetry in the Lyndon word description of BCH suggests that BCH contains less information than one might think, possibly leading to some computational advantage. Though in (my) reality, the computational bottlenecks are anyway elsewhere.

Some further details and observations are at http://drorbn.net/AcademicPensieve/2012-12/nb/BCH-Lyndon_Question.pdf.

  • $\begingroup$ For comparison, I wonder if you could remind us the total number of Lyndon words in each of these spaces? I assume it's growing much faster than the sequence you computed, but I could imagine that BCH essentially saturates some natural subspace of of FL(x,y), say the subspace with some obvious symmetry, and that the pattern is exactly the pattern of dimensions of those spaces. $\endgroup$ Commented Dec 12, 2012 at 6:03
  • $\begingroup$ Ah, I see that you answer (at least) some of my question in the linked pdf, which I have only just started to read. $\endgroup$ Commented Dec 12, 2012 at 6:04
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    $\begingroup$ It looks to me (from the data in the linked pdf) like there are two different perhaps not-so-related things that are happening: at n odd, BCH gets all possible Lyndon words except one of them when n=6k+3 (k>0), in which case it misses x^{4k+1}yx^{2k}y, and at n=2k, the Lyndon words that appear in BCH are exactly those obtainable from the Lyndon words of length 2k-1 by prepending x, except that x^{2k-1}y does not appear. I have no explanations, though. $\endgroup$ Commented Dec 12, 2012 at 10:11
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    $\begingroup$ Let me mention that the Lyndon words of length n obtained by prepending an x, form a basis of the subspace of FL_n of the form [x,FL_{n-1}]. This seems as if it could be helpful for showing that only terms of this form appear in the n-even case. $\endgroup$ Commented Dec 14, 2012 at 19:39
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    $\begingroup$ I tool a liberty to add this sequence to the OEIS as oeis.org/A220587 $\endgroup$ Commented Dec 16, 2012 at 19:11

1 Answer 1


This pattern of vanishing coefficients in the expansion of the Baker-Campbell-Hausdorff formula in the basis of Lyndon words is explained in section IV.C of the article An efficient algorithm for computing the Baker–Campbell–Hausdorff series and some of its applications of Fernando Casas and Ander Murua.


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