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I am trying to solve Exercise 2.5.7 from Introduction to Dynamical Systems by Brin and Stuck:

Prove that the topological entropy of a continuously differentiable map of a compact manifold is finite.

Definitions needed: we have $$ d_n(x,y) := \max_{0\leq k\leq n-1} d(f^k(x), f^k(y)) $$ The topological entropy of a map $f$ is defined as $$ \lim_{\epsilon\downarrow 0} \limsup_{n\to\infty} \tfrac1n \log\text{cov}(n,\epsilon,f), $$ where $\text{cov}(n,\epsilon,f)$ is the minimum cardinality of a covering of by sets of $d_n$-diameter less than $\epsilon$.

My ideas for this exercise: since this manifold is compact we know there exist $N,M$ such that $ \|f(x)\|\leq M $ and $ \|Df(x)\|\leq N. $ Compactness also definitely will give that for every $\epsilon$ the expression $$ \limsup_{n\to\infty} \tfrac1n \log\text{cov}(n,\epsilon,f), $$ exists and is finite.

Jolien
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1 Answers1

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In fact you can prove a more general result: If $(X,d)$ is a metric space, and $f:X\to X$ is a continuous Lipschitz map, with Lipschitz constant $L(f)$. Let $D(X)$ be the ball dimension of $X$, which is defined by $$D(X):=\overline{\lim}_{\epsilon\to 0}\frac{\log b(\epsilon)}{|\log\epsilon|},$$ where $b(\epsilon)$ is the minimum of the number of elements of a covering of $X$ by balls of radius $\epsilon$. Then we have that $h_{top}(f)\leq D(X)\max(0,\log L(f)).$ In your setting, $L(f)$ is $\sup_x\|Df_x\|$, so we have the finite entropy property, since for a compact smooth manifold, the ball dimension is equal to its topological dimension. You can check that the ball dimension is finite for any set in a bounded subset of a finite dimensional Euclidean space. Then note that a smooth manifold is locally Euclidean, and the space is compact, and the ball dimension is invariant passing to a biLipchitz-equivalent metric. Here I omit some details, but I think it is not too hard to check.

To prove this, let $L>\max(1,L(f))$. Then $d(f(x),f(y))\leq Ld(x,y)$ for all $x,y\in X$. So $f^m(B_d(x,L^{-n}\epsilon))\subset B_d(f(x),\epsilon)$ when $0\leq m\leq n$ and hence $B_d(x,L^{-n})\subset B_{d_n}(x,\epsilon)$ for all $x\in X$, $n\in\mathbb{N}$, and $\epsilon>0$. Thus $$S(f,\epsilon,n)\leq b(L^{-n}\epsilon).$$ Note that $|\log(L^{-n}\epsilon)|=n\log L-\log\epsilon$, so $$n=\frac{|\log(L^{-n}\epsilon)|}{\log L}(1+\frac{\log\epsilon}{|\log(L^{-n}\epsilon)|})=\frac{1}{\log L}|\log(L^{-n}\epsilon)|(1+O(\frac{1}{n})),$$ so $$\overline{\lim}_{n\to\infty}\frac{\log S(f,\epsilon,n)}{n}\leq\overline{\lim}_{n\to\infty}\frac{\log b(L^{-n}\epsilon)}{n}=\log L\overline{\lim}_{n\to\infty}\frac{\log b(L^{-n}\epsilon)}{|\log(L^{-n}\epsilon)|}=D(X)\log L$$ and $h_{top}(f)\leq D(X)\log L<\infty$. And so the proof is completed.

Siming Tu
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  • Thanks for your answer. Could you explain me why $D(X)$ is finite and why $n$ is equal to that product? If I expand the right side of the first equality I get $-n$ plus something with $\epsilon$. – Jolien Nov 25 '15 at 09:09
  • @jolien I add some details in the answer with respect to your question. If there is something wrong, thanks for telling me. – Siming Tu Nov 25 '15 at 09:56
  • Thanks for adding details. I understand your reasoning, but I'm still stuck on the ball dimension. If I understand correctly we would like to show that $b(\epsilon)\approx \epsilon^{-D(X)}\epsilon$ for small $\epsilon$. Could you perhaps give me another hint? – Jolien Nov 25 '15 at 10:30
  • @jolien You can first try first to prove that in $\mathbb{R}^n$ for a open ball of radius $1$ (in fact, it is n), the ball dimension is finite. The for a smooth compact manifold we can find a finite cover with which piece smooth homeomorphic to a open ball of radius $1$ in some $\mathbb{R}^n$, so you have that the ball dimension of a compact smooth manifold. – Siming Tu Nov 25 '15 at 11:12