Computation of metric Entropy by another metric which is induced by a homeomorphism

Question

First let me explain problem in general case

If $T:\mathbb R^n\to \mathbb R^n$ and $S:\mathbb R^n\to \mathbb R^n$ are two conjugated linear Dynamical Systems which means there exist homeomorphism $h:\mathbb R^n\to \mathbb R^n$ such that $ h(S(x))=T(h(x))$

Now we know $h$ induced another metric $d_h$ on $\mathbb R^n$ which is topological equivalent to euclidean metric now we raise a question. no we can compute metric entropy via two metrics

1. $H_d(T)=?$

2. $H_{d_{h}}(T)=?$

by pesin formula we know that $$H_d(T)=\sum_{{\mu\in \{Spec(T)\cap{|\mu|\gt1}}\}} log{|\mu|}$$

How compute metric entropy with new metric ??

W.L.O.G we can simplify this problem and take$n=1$ and $$T(x)=2x$$ $$S(x)=3x$$ are two linear systems that are conjugate .

we can contract question in following form Compute entropy of $T(x)=2x$ by following metric??

$$d_1(x,y)=|x-y|$$ $$d_2(x,y)=|logx-logy|$$

$$d_3(x,y)=|x^c-y^c|$$

I started to compute generating set or separated set but working with these was so though for i thought should be some short cut

Thanks for any hints

Answer 1

"Metric entropy" always refers to a measure-theoretic entropy. What you call a metric entropy actually is a topological entropy. Right formula for the topological entropy of T is $$h_{d}(T) = \sum\limits_{|\lambda|>1}\log |\lambda|,$$ where the sum is taken over all the eigenvalues of $T$ and $d$ is Euclidean distance. This formula belongs to R. Bowen, see his paper. Here the topological entropy for maps on general metric spaces was introduced for the first time. For non-compact metric spaces the topological equivalence of two metrics isn't sufficient for the coincidence of the topological entropies. So, in general situation the problem of the computation of the topological entropy depends on the choice of metric. But in your special case it is easy.

For $d_1$ use Bowen's formula or just act like in the case of $d_3$

For $d_2$ notice that $T$ is an isometry, so $H_{d}(T)=0$.

For $d_3$ notice that $d_3(Tx,Ty)=2^{c}d_3(x,y)$. Now just compute the minimal number of Bowen balls required to cover $[0,1]$. At the end you'll get that $H_d(T)=c\log2$.

But i still don't get why you need the second dynamical system if you just want to compute the topological entropy in new metric.

UPDATE: how to compute $H_d(T)$, where $d=d_3$.

Define $\rho_{m}(x,y) := \max\limits_{0\leq j \leq m-1} d(T^{j}x,T^{j}y)$. Choose a real $\varepsilon>0$, a positive integer $m$ and a closed interval $K$. We have to find the smallest cardinality of a set which $(\varepsilon,m)$-spans $K$. This is equivalent to find the smallest number of open balls of radius $\varepsilon$ in metric $\rho_{m}$ (such a ball is called Bowen's ball) which can cover $K$. So, using the noticed property we have that $\rho_{m}(x,y) < \varepsilon$ iff $2^{c (m-1)}d(x,y)<\varepsilon$. Thus, we have to cover $K$ by open balls of radius $\varepsilon' =\varepsilon 2^{-c(m-1)}$ in metric $d(x,y)$. Denote by $H_{d}(T,K)$ the topological entropy of $T$ on $K$. It is easy to see that if there is $L>0$ such that the inequality \begin{equation} d(x,y) \leq L |x-y| \end{equation} holds for all $x,y \in K$, then $H_{d}(T,K) \leq c \log 2$. Analogously, the reverse inequality \begin{equation} d(x,y) \geq L |x-y| \end{equation} for all $x,y \in K$ implies $H_{d}(T,K) \geq c \log 2$.

Notice that, by the mean value theorem, both inequalities hold if $K$ is separated from $0$, so in this case $H_{d}(T,K)=c\log 2$. Now let $K=[-\delta,\delta]$ where $\delta>0$ is small enough.

if $c \geq 1$ then (use the mean value theorem) we have the first inequality and thus, $H_{d}(T,K) \leq c \log 2$. So, taking the supremum over all the compact $K$-s, we have $H_{d}(T)=c\log 2$.

If $0<c<1$ then we have only $H_{d}(T,K) \geq c \log 2$. So we have to investigate this case more accurately. Let's take a look at the special case $c=\frac{1}{2}$. It seems that $B_{d}(x,\varepsilon) \approx B_{|.|}(x,2\frac{1}{\sqrt{|x|}}\varepsilon)$. But $\frac{1}{|x|}$ grows not so fast, so integral $\int\limits_{-\delta}^{\delta}\frac{1}{\sqrt{|x|}}dx$ converges. We have to use this somehow.