Baker-Campbell-Hausdorff formula: prime divisors of denominators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T01:59:54Z http://mathoverflow.net/feeds/question/43445 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43445/baker-campbell-hausdorff-formula-prime-divisors-of-denominators Baker-Campbell-Hausdorff formula: prime divisors of denominators Vipul Naik 2010-10-24T23:41:50Z 2010-10-27T22:50:00Z <p>Consider the Baker-Campbell-Hausdorff formula (<a href="http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula" rel="nofollow">Wikipedia page</a>):</p> <p>$$Z(X,Y) := X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]] + \dots$$</p> <p>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.</p> <p>For a prime <em>p</em> and a natural number <em>n</em>, denote by $f(p,n)$ the largest <em>k</em> 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 $n$. </p> <p>It's pretty easy to see that $f(p,n) = 0$ for $n &lt; p$, and is nonzero for $n \ge p$.</p> <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.</p> <p>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.</p> <p>UPDATE: Chapter 3 of the Springer Lecture Notes by Klass, Leedham-Green, and Plaskett (<a href="http://www.springerlink.com/content/y454145107303034/" rel="nofollow">access online if you have an a university subscription</a>) 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.</p>