Baker-Campbell-Hausdorff formula: prime divisors of denominators

2010-10-24T23:41:50Z

Consider the Baker-Campbell-Hausdorff formula (Wikipedia page):

$$Z(X,Y) := X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] + \dots$$

Many sources, including the Wikipedia page, have an explicit expression for the terms, so the question I'm asking should be answerable just from that expression.

For a prime p and a natural number n, denote by $f(p,n)$ the largest k such that, if we truncate the formula to terms that involve products of length at most $n$, then one or more of the denominators is divisible by $p^k$. Note that this truncation is valid when we are working in a situation of nilpotency class $n$.

It's pretty easy to see that $f(p,n) = 0$ for $n < p$, and is nonzero for $n \ge p$.

My question: Is there a direct explicit expression for $f(p,n)$ (or a sandwiching of it between two fairly close expressions)? For instance, inspection of the first few terms suggests that $f(2,n) = n - 1$, but I'm not sure how to derive this from the general expression.

Analogue: In the power series for the exponential $e^x$, the analogue to $f(p,n)$ is the sum $[n/p] + [n/p^2] + [n/p^3] + \dots$ where $[]$ denotes the greatest integer function.

UPDATE: Chapter 3 of the Springer Lecture Notes by Klass, Leedham-Green, and Plaskett (access online if you have an a university subscription) contains some estimates. However: (i) I'm not sure all the numerical calculations there are correct, since they don't agree with others I have seen, (ii) the authors aren't concerned about the precise growth of $f(p,n)$ -- they only care that it grows slowly enough that the series converges under certain conditions.

Baker-Campbell-Hausdorff formula: prime divisors of denominators - MathOverflow

Baker-Campbell-Hausdorff formula: prime divisors of denominators