Assume that $X$ is a compact metric space and $G$ is compact totally disconnected group. And $X$ has isometric free $G$-action i.e. $gx=x\Rightarrow g=e$.

Then the following holds $${\rm dim}\ X/G={\rm dim}\ X-{\rm dim}\ G\cdot o$$ where $o\in X$ and dimension is Hausdorff dimension ?

Thank you in advance.

  • 1
    $\begingroup$ I assume $X/G$ is equipped with the Hausdorff metric between the orbits. What metric on $G$ are you using? You can define a metric on $G$ by restricting the metric on $Gx$ and then minimizing over all $G$-orbits. Is this what you have in mind? $\endgroup$ – Moishe Kohan Apr 10 '17 at 12:55
  • $\begingroup$ Yes. That is what I intend. And I think that every orbits will have same Hausdorff dimensions. $\endgroup$ – Hee Kwon Lee Apr 10 '17 at 18:09

It may happen that $$\dim_H X/G<\dim_H X-\dim_H (G\cdot o);$$ the following example is almost identical to Example 7.8 in "Fractal geometry" by Kenneth Falconer which provides spaces $X$ and $Y$ such that $$\dim_H (X\times Y)>\dim_HX+\dim_HY.$$

Denote by $W(\varepsilon)$ the two-point space with distance $\varepsilon$ between the pair of points. Consider the $\ell_\infty$-product $$X=\prod_{n\in\mathbb{N}}W(\tfrac1{2^n}).$$ Fix a subset $\Sigma\in \mathbb{N}$ and consider the natural action of group $$G=\prod_{n\in\Sigma}\mathbb{Z}_2$$ on $X$.

The orbit can be identified with the infinite product $$G\cdot o=\prod_{n\in\Sigma}W(\tfrac1{2^n}).$$ and the quotient space can be identified with the infinite product $$X/G=\prod_{n\notin\Sigma}W(\tfrac1{2^n}).$$

Note that for appropriate choice of $\Sigma$, we have $$\dim (G\cdot o)= \dim X/G=0$$ while $$\dim X=1.$$


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