5
$\begingroup$

Let $X$ be a compact metrizable space and $\alpha: \mathbb{Z}^d\curvearrowright X$ a continuous group action. Then it is well known that there exists a zero dimensional compact space $Y$, an action $\beta: \mathbb{Z}^d\curvearrowright Y$ and a continuous surjection $$F: Y\to X\quad\text{with}\quad \alpha_v\circ F = F\circ\beta_v\quad\text{for all}\; v\in\mathbb{Z}^d.$$ (This is even known for actions of any countable group.)

Suppose now that $X$ posesses an $\alpha$-invariant, faithful probability measure $\mu$. Then it follows from a C*-algebraic construction done in http://arxiv.org/abs/math/0604047 that there exist $Y,\beta,F$ as above with a $\beta$-invariant probability measure $\nu$ on $Y$ such that $\nu_*=\mu$, i.e. $$\int_Y f\circ F~d\nu = \int_X f~d\mu\quad\text{for all}\; f\in\mathcal{C}(X) .$$

While the result is nice, getting this only with C*-algebraic methods seems very unnatural to me.

My question is: Does there exist a direct proof of this fact, i.e. using basic topology and some measure theory? Was this maybe known before? (I have not found it yet in the literature elsewhere.)

$\endgroup$

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.