Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question came up (unexpectedly) in a problem I was working on a few years ago. It may not be too difficult but I never got around to figuring out the answer, because all I needed at that time was a constant multiple of Haar measure.

Let $k\ge 2$ and suppose that $p_1,\ldots , p_k$ are distinct primes. Let $\mu$ be the Haar probability measure on $\mathbb{Z}_{p_1}\times\cdots\times\mathbb{Z}_{p_k}$, and let $\mathrm{H}^k$ denote $k-$dimensional Hausdorff measure on the same space with respect to the metric $d$ defined by $$d(\mathbf{x},\mathbf{y})=\max_{1\le i\le k}\{|x_i-y_i|_{p_i}\}.$$ It is easy to verify that $\mathrm{H}^k$ is finite, regular, and translation invariant and therefore $\mathrm{H}^k=c\mu$ for some real constant $c$. Furthermore if $k=2$ then it is not too difficult to show that $c=1$. Is it always true that $c=1$?

share|improve this question
    
Hausdorff measure is unfortunately not multiplicative for metric spaces in general. But in this case (self-similar factors) we have more information. So Anthony's proof has a chance of working. –  Gerald Edgar Dec 16 '11 at 15:12
    
@Gerald sorry, you are quite right. –  Vitali Kapovitch Dec 16 '11 at 15:30
    
@Vitali Ok thanks for the update, I was starting to think that maybe I had overlooked something obvious. –  Alan Haynes Dec 16 '11 at 15:49
add comment

1 Answer

up vote 2 down vote accepted

I think the answer is yes. Let $X=\mathbb Z_{p_1}\times \ldots \times \mathbb Z_{p_k}$. Notice that the metric on $X$ is an ultrametric, so that if $A$ is a subset of $X$ of diameter $d$ and $a\in A$, then the ball of radius $d$ about $a$ is a superset of $A$ that has the same diameter. Denote this ball by $B(A)$.

Writing $a=(a_1,\ldots,a_k)$ in $X$. The ball of radius $r$ about $a$, $B_r(a)$, is the product $B_r(a_1)\times\ldots\times B_r(a_k)$, where $B_r(a_i)$ is the ball of radius $r$ about $a_i$ in $\mathbb Z_{p_i}$. Of course $B_r(a_i)$ is equal to $B_{p_i^{-s_i}}(a_i)$ where $s_i$ is the largest integer such that $p_i^{-s_i}\le r$. The diameter of the ball is $d=\max_i p_i^{-s_i}$. The volume of the ball with respect to Haar measure is $\prod_i p_i^{-s_i}\le d^k$. In general, we have for any ball, $\mu(B)\le \text{diam}(B)^k$.

Now if $A_1,\ldots,A_n$ is a cover of $X$, then so is $B(A_1),\ldots,B(A_n)$. These sets have the same diameters by the first observation. Now we have $$ \sum\text{diam}(A_i)^k = \sum\text{diam}(B(A_i))^k \ge \sum \mu(B(A_i)) \ge 1. $$

Hence $c\ge 1$.

On the other hand, by considering the map $\phi\colon t\mapsto (t\bmod \log p_1,\ldots,t\bmod\log p_k)$ (using density of $\phi(\mathbb R)$ in the $k$-dimensional torus), we can find a sequence $t_n\to\infty$ such that $\phi(t_n)$ is a vector with all coordinates lying in $(\log p_j-\epsilon_n,\log p_j)$ with $\epsilon_n\to 0$. Write $m^i_n=\lceil t_n\log p_i\rceil$. Then $e^{-t_n}$ lies between $p_i^{-m^i_n}$ and $p_i^{-m^i_n}e^{\epsilon_n}$.

For balls of diameter $e^{-t_n}$, the measure of the ball is close (within a factor $e^{k\epsilon_n}$) to the $k$th power of the diameter. Hence $c=1$.

share|improve this answer
    
Yes, I think that solves it. Thank you! –  Alan Haynes Dec 16 '11 at 23:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.