Power of minimal system is transitive?

88 Views Asked by At

Let $(X,f)$ be a minimal dynamical system. What is the best thing we can deduce about $(X,f^n)$, some $n\in \mathbb{N}$? In particular, do we know that it is at least topologically transitive? Even more weakly, does there exist one $n\in \mathbb{N}$ for which $(X,f^n)$ is transitive?

My motivation for asking is because I know that a transitive system is not necessarily totally transitive.

1

There are 1 best solutions below

4
On BEST ANSWER

You should be more precise about the notions involved in your question, especially when speaking of (topological) transitivity - if you define it as existence of a dense orbit then you trivially get that $(X,f)$ minimal implies $(X,f)$ transitive assuming minimality means that all orbits are dense. Also, in general, it's not the same whether you consider one-sided orbits or two-sided orbits. Anyway, in the following example, it doesn't really matter and besides the trivial YES for $n=1$ we get NO for $n>1$.

Consider a special instance of odometer - you can define $X$ as the product space of the sequence of the discrete spaces $A_k:=\{0,1,\dots,p_k-1\}$ where $p_k$ denotes the $k$-th prime number - in fact, you can take just the sequence of all positive integers as well but the example is probably more evident with the prime numbers. Topologically, the space $X$ is the Cantor set and we define the continuous map $f$ as the addition of one from left to right - you can find all the necessary definitions eg. in https://sciendo.com/article/10.2478/amsil-2022-0010. Now consider $f^n$ for any $n>1$. To see that it's not transitive, just take the least prime factor $p=p_k$ of $n$ and look at how the point represented by the sequence of zeros is mapped under the iterates of $f^k$. Or, even better, imagine how the $k$-level cylinders of $X$ are mapped under the iterates of $f^k$ - there are exactly $p_1 \cdots p_k$ such cylinders.

If the space is connected, then all the iterates are minimal - at least in case of compact metric $X$ and continuous $f$. It holds in more general situations as well but one must be careful with exact definitions of the notions involved.