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Question 1. Given a topological space $X$ and two metrics $a$ and $b$ on it, compatible with the topology, what conditions should I impose on them so that box-counting (or other, for example Hausdorff) dimensions of $(X,a)$ and $(X,b)$ are equal?

Question 2. Is there a notion of a dimension for topological (as opposed to metric) spaces which can assume non-integral values?


My motivation

Let $G$ be a finitely generated group and let $p$ be a prime number. Consider the compact topological space $X:=\prod_G \mathbb Z/p$, infinite product of copies of the cyclic group of order $p$, indexed by elements of $G$. Let $T$ be an element of the integral group ring of $G$. Note that $T$ gives a map $X\to X$ in a natural way (in this context $T$ is sometimes called cellular automaton). Define $Y$ to be the subset of those points $x$ of $X$ such that $T(x)=0$.

I want to measure "how big" $Y$ is.

If we choose a generating set for $G$ then we get a metric $d$ on $G$, and we also get a metric on $X$: two sequences $x_i$ and $y_i$ of $X$ are $p^{-|B(1,k)|}$ apart if they agree on the ball $B(1,k)$ of radius $k$ around the neutral element $1$ of $G$, but they don't agree on any larger ball. It's straightforward to see that box counting dimension of $X$ is $1$.

Unfortunately the metric we just defined depends on the generators chosen for $G$, so I'd like to know if I have a chance to get any kind of "intrinsic size of $Y$" this way.

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@Greg, why did you remove dimension-theory tag? –  Łukasz Grabowski Oct 19 '10 at 13:24
    
@Lukasz I only meant to remove the differential geometry tag. I don't know how the dimension-theory tag disappeared, but to make amends I put it back. –  Greg Kuperberg Oct 19 '10 at 15:05
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2 Answers

up vote 5 down vote accepted

Q2. If you want that dimension for topological spaces to agree with Hausdorff dimension (for example) in case of metric space, then NO. For any $0 \le s \le \infty$, there is a metric on the Cantor set so that the Hausdorff dimension is $s$.

Another topological result. Let $X$ be a separable metrizable space. The infimum of the Hausdorff dimensions of all metric spaces homeomorphic to $X$ is the topological dimension of $X$. (Integer valued.)

for the My Motivation comment. See the notion "topological entropy".

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Thanks for your answer. I just had a quick look at "topological entropy" in wikipedia. It's defined for a map. Do you have an intuition to share what TE will measure for the map T in my question? –  Łukasz Grabowski Oct 19 '10 at 22:12
    
@Lukasz For instance, suppose that you have an alphabet (such as $\mathbb{Z}/p$) with a measure and then the product measure on sequences. Then the measure-space entropy of the shift map is the Shannon entropy, while the topological entropy, the maximum over all invariant measures, is the log of the number of symbols. In your case, if $T = g \in G$, $g$ acts by a symbolic shift map separately on each orbit, so the topological entropy is infinite but $\log p$ on each orbit. –  Greg Kuperberg Oct 20 '10 at 1:47
    
@Gerald The infimum is equal to which topological dimension? –  Greg Kuperberg Oct 20 '10 at 1:48
    
@Greg: I said separable metrizable space, since in that case they are all equal: covering dimension, small inductive dimension, large inductive dimension. –  Gerald Edgar Oct 20 '10 at 3:55
    
Greg: thanks, I'll think about. Although the case $T=g$ is in a sense a trivial one - what I want to do is, really, give a meaning to dimension over F_p of the kernel of T, which is 0 in the case $T=g$. Generally, over complex numbers people know how to do it (it's called Von Neumann dimension), people also know it when a group is amenable (by Orstein-Weiss lemma); it's a huge open question whether meaningful dimensions exist for kernels of elements of the group ring of more general groups. –  Łukasz Grabowski Oct 21 '10 at 18:05
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The so-called "box" dimension is best interpreted as a cheap approximation to Hausdorff dimension. Hausdorff dimension is much more robust, but it is not preserved by homeomorphisms. All Cantor sets in the line $\mathbb{R}$ are equivalent under ambient homeomorphism, but their Hausdorff dimension takes all values in the interval $[0,1]$. So there you have metrics on $\mathbb{R}$ that give you different answers, moreover metrics that are all the same metric in different positions. By taking products of Cantor sets, an abstract Cantor set can have any Hausdorff dimension in $[0,\infty]$.

Things are better for bi-Lipschitz changes of metric. By a simple computation, a bi-Lipschitz change of metric has a bounded effect on Hausdorff measure, and therefore cannot change Hausdorff dimension. Unfortunately, although your different metrics are mutually bi-Lipschitz on $G$, as written they are not bi-Lipschitz on $X$.

Meanwhile there are various definitions of topological dimension, which are all integer-valued and generally agree on well-behaved topological spaces. The Wikipedia article on dimension lists some of them. The bad news is that $X$ is a Cantor set and its topological dimension is zero by any common definition.

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