Entropy of nested compact invariant sets - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T05:55:26Z http://mathoverflow.net/feeds/question/56892 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56892/entropy-of-nested-compact-invariant-sets Entropy of nested compact invariant sets Peng Sun 2011-02-28T12:55:24Z 2011-11-24T15:15:40Z <p>Let $f$ be a homeomorphism on a compact metric space $X$. $K_1\supset K_2\supset\cdots \supset K$ are compact subsets of $X$ such that $f(K_n)=K_n$ and $K=\bigcap K_n$. If $h(f, K_1)&lt;\infty$, do we always have $h(f,K)=\lim h(f,K_n)$?</p> <p>I can show that this is true for $C^\infty$ diffeomorphisms where the entropy map of invariant measures is upper semi-continuous.</p> <p>OK. I see the point. This is definitely false for the most general case. For example, take the union of countable hyperbolic toral automorphisms of shrinking size and add one point with Identity map on it.</p> <p>However, what if $f$ is a diffeomorphism on a compact manifold? I still expect negative answer.</p> http://mathoverflow.net/questions/56892/entropy-of-nested-compact-invariant-sets/56906#56906 Answer by Benoît Kloeckner for Entropy of nested compact invariant sets Benoît Kloeckner 2011-02-28T15:46:19Z 2011-02-28T15:46:19Z <p>The answer is negative: consider the map $\varphi:z\mapsto z^2$ acting on the unit disc $D(1)$ of $\mathbb{C}$. We know that this map has entropy $\log 2$ in restriction to every circle centered at $0$, and in fact since the dynamics is trivial transversally to this circle we have $h(\varphi,D(1))=\log 2$. Now, consider $K_n=D(1/n)$: we still have $h(\varphi,K_n)=\log 2$ since $\varphi_{|K_n}$ is conjugated to $\varphi$. But then <code>$K=\{0\}$</code> and <code>$h(\varphi,\{0\})=0$</code>!</p> <p>Ok, I cheated: $\varphi$ is not a homeomorphism. But there is an obvious way to extend the above example to an homeomorphism. First take $\psi:Y\to Y$ a positive entropy one, then construct $c\psi : cY\to cY$ its cone as follows. First, $cY$ is the topological cone over $Y$, that is $cY=(Y\times[0,1])/((x,0)\~ (y,O))$. Then $c\psi$ is the map induced on this quotient by $(x,t)\mapsto (\psi(x),t)$. Then $c\psi$ is a homeomorphism, and has the same entropy as $\psi$. But taking $K_n$ the trace on the quotient of $Y\times[0,1/n]$ you get a sequence of compacts on which the dynamics is the same than for the full map, but whose intersection is reduced to a point.</p> http://mathoverflow.net/questions/56892/entropy-of-nested-compact-invariant-sets/81809#81809 Answer by lbdl for Entropy of nested compact invariant sets lbdl 2011-11-24T15:15:40Z 2011-11-24T15:15:40Z <p>The strategy of Benoît Kloeckner fails for differentiable maps. Indeed if $K$ is a single point and $K_n$ invariant balls around $K$ it implies that $K$ is an attracting fixed point. Therefore the log of the differential of the diffeo $f$ should be close to zero near $K$ and so does the entropy.</p> <p>However conter-examples in any finite smoothness ($C^r$ maps with $1\leq r&lt;+\infty$) were given by Misisurewicz in the early seventies :</p> <p>Diffeomorphism without any measure with maximal entropy, Bull. Acad. Pol. Sci., Ser. sci. math., astr. et phys. 21 (1973), 903--910 </p>