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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.

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Lazard in Michel Lazard: "Groupes analytiques $p$-adiques", Publ. IHES, 26 has results on the p-adic convergence of the BCH formula. –  Torsten Ekedahl Oct 25 '10 at 4:11
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The definition given for $f(p,n)$ doesn't involve $p$. Did you mean for denominators divisible by $p^k$? –  Gerry Myerson Oct 25 '10 at 20:50
    
Gerry: thanks. I somehow hadn't noticed your comment, but I just noticed the error independently and fixed it. Probably a lot of people reading the question guessed I was referring to $p^k$ or else the statements following it are not true. –  Vipul Naik Oct 27 '10 at 23:05
    
The best I know is that the multiplicity of $p$ in the denominators of the terms of length $\leq n$ is at most $(n-1)/(p-1)$. The proof follows from the elementary observation that if $v_p(x)\geq1/(p-1)$ then $v_p(e^x-1)\geq1/(p-1)$ and $\log(1+x)\geq1/(p-1)$. See the above-mentioned paper of Lazard. I don't know any intelligent lower estimates of the multiplicity, however –  Pavol S. Feb 14 '11 at 21:36
    
Thanks Trial! I learned about that upper bound through some other source (outside Math Overflow) using the method you suggested, shortly after posting the question, and that was precisely what I needed, so I forgot about the question still being on Math Overflow. –  Vipul Naik Feb 15 '11 at 2:05

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