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Let $T$ be the isometry of a metric compact with a dense orbit. Prove that the topological entropy $T$ is zero.

What happened now: $T$ is the isometry of a metric compact, that is, for any two points of a compact with metric d, we have $d(T(a), T(b)) = d(a, b)$. Since by the condition with a dense orbit, then $T$ is a topologically transitive map. We have that $d(T^ i (a), T^i (b)) = . . . = d(T ^2 (a), T^2 (b)) = d(T(a), T(b)) = d(a, b) $. Help to prove. Happy New Year)

Alp Uzman
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gkndy
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1 Answers1

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Hint: There are multiple ways of calculating/defining topological entropy, in particular there are ways of calculating it via the so-called Bowen metrics (I mentioned these metrics in a previous answer here: https://math.stackexchange.com/a/4326733/169085). Say we want to use the spanning numbers w/r/t Bowen metrics. We'll need to take the logarithm of the $(n,\epsilon)$-spanning number; divide it by $n$ and then take limits. But observe that since the transformation is an isometry these spanning numbers are independent of $n$, as you've observed.

(Also, this statement is true without the point transitivity assumption.)

Alp Uzman
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  • Solve $N_n(\epsilon)$ - the number of points of the minimum epsilon network for $(X, d_n)$. Then the entropy has the form $h(T) = lim_{ε\to0} lim sup_{n\to∞} \frac{ln(N(\epsilon ))}{n}$ Because the transformation is an isometry, independent of n. Therefore, with n go to infinity, the entropy is 0. – gkndy Dec 30 '21 at 13:48
  • It seems to be solved this way – gkndy Dec 30 '21 at 14:05
  • Add: $d_{n}(a, b)=\max \left{d\left(T^{i} (a), T^{i} (b)\right), i=0,1, \ldots ., n-1\right}$ – gkndy Dec 30 '21 at 14:22