Partitioning the unit ball in an ultrametric space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T08:11:23Z http://mathoverflow.net/feeds/question/78757 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/78757/partitioning-the-unit-ball-in-an-ultrametric-space Partitioning the unit ball in an ultrametric space Math-player 2011-10-21T12:10:46Z 2011-10-21T13:02:14Z <p>Consider an ultrametic space $X=K^n$ (for the norm $\| x\| =\max_{i=1,n} (\mid {x_1}\mid,\dots,\mid{x_n}\mid)$ where $K$ is an ultrametric field. Let $B(1):=\lbrace x \in X \mid \|x\| \leq 1\rbrace$ be the unit ball. Equip $X$ with Haar measure. Is it possible to partition $B(1)$ into $k$ smaller balls $B_{a_1}(r_1),\dots, B_{a_k}(r_k)$ (where $B_a(r)$ is the closed ball of radius $r$ and center $a$), $k>1$?</p> <p>Or put in a different way: is there a measure preserving bianalytic map taking $B(1)$ to the union of $B_{a_i}(r_i)$?</p> <p>Thank you</p> http://mathoverflow.net/questions/78757/partitioning-the-unit-ball-in-an-ultrametric-space/78759#78759 Answer by Gerald Edgar for Partitioning the unit ball in an ultrametric space Gerald Edgar 2011-10-21T12:24:53Z 2011-10-21T12:24:53Z <p>If you can do the case of $K$ itself, decompose into $k$ smaller balls, then for $K^n$ you can decompose into $k^n$ smaller balls. For some ultrametric fields this works. Indeed, you say "Haar measure" so you must be assuming local compactness, and it works for any such field, since they are finite extensions of $p$-adic fields.</p>