Definition: Let be $f:M\to M$ a diffeomorphism of a compact manifold. We say that $A\subset M$ is an attractor when there exists a neighborhood $U\supset A$ such that $f( \overline{U})\subset int (U)$ and $$ A=\bigcap_{n\geq 0}f^n(U) $$ $U$ is called an basin of attraction of $f$.

Theorem: Let $M$ be a compact manifold and let $f:M→M$ a diffeomorphism. If $\overline{Per f}$ has hyperbolic structure, then can be partitioned into a finite number of compact, invariant and topologically transitive sets, called basic sets: $$\overline{Per(f)}=⋃_{i=1}^{m}Λ_i$$

Definition: Le be $\Lambda_i$ a basic set of $f$, then we define $$W^s(\Lambda_i)=\{x: d(f^n(x), \Lambda_i)\to 0,~n\to\infty \}$$

Question: Supose that the chain recurrent set of $f$, $\mathcal{R}(f)$ has hyperbolic structure, I would like to see that if $int (W^s(\Lambda_i))\neq \varnothing$ then the basic set $\Lambda_i$ is an attractor.

  • $\begingroup$ The case when basic set has local product structure seems to be easy. Do you know the proof in this case? $\endgroup$ – Andrey Gogolev Jul 15 '13 at 19:04
  • $\begingroup$ Seems related, from here math.byu.edu/~tfisher/documents/papers/nuthesis.pdf p.13, Theorem 1.8: "Let f:M→M be a diffeomorphism of a compact manifold M . If f has a transitive hyperbolic set Λ with nonempty interior, then Λ = M and f is Anosov" appears to be a well known folklore theorem. We could find no proof of it in the literature, so one is provided on p.52 $\endgroup$ – sigrlami Jul 16 '13 at 15:09
  • $\begingroup$ @AndreyGogolev Dear Andrey Gogolev, it is a known fact that the basic sets always have local product structure $\endgroup$ – user11178 Jul 16 '13 at 19:35
  • $\begingroup$ Juan, indeed, it seems to be clear that a hyperbolic chain recurrent set has local product structure just from definitions. Never thought about it before. $\endgroup$ – Andrey Gogolev Jul 16 '13 at 20:23

Ok, here is the answer assuming that the basic set $\Lambda$ has local product structure.

Step 1. We will prove that $\forall x\in\Lambda$ $W^u(x)\subset \Lambda$. Then any point $y$ sufficiently close to $\Lambda$ belongs to a local stable manifold of some point in $\Lambda$, hence $\Lambda$ is an attractor.

Step 2. So we need to prove that local unstable of every point in $\Lambda$ is in it. Basic set is transitive and closed; local unstable manifolds vary continuously. Therefore it sufficient to show that local unstable manifold of a point $z\in\Lambda$ is in $\Lambda$, where $z$ is a point whose tranjectory is dense.

Step 3. It is easy to show that $$ W^s(\Lambda)=\bigcup_{x\in\Lambda}W^s(x) $$ See for example Proposition 9.1 of Shub's book. (Here local product structure is essential.)

Step 4. Therefore if $p$ is in the interior of the above set. Then $\exists q\in \Lambda$ such that $p\in W^s(q)$. Take a transvesral to $W^s(q)$ that passes through $p$ and apply the $\lambda$-lemma to it to conclude that $W^u(q)$ is in $W^s(\Lambda)$. Hence $W^u(q)\subset W^s(\Lambda)$. Note that the same reasoning applies to points sufficiently close to $q$. In particular to some iterate of $z$. Hence we are done.

  • $\begingroup$ It's not clear to me who is the basin of attraction of $\Lambda$?? $\endgroup$ – user11178 Jul 16 '13 at 20:53
  • $\begingroup$ The basin is just the union of the local stables $\cup{x\in\Lambda}W^s(x,\varepsilon)$. This is an open set because we proved that $\Lambda$ is u-saturated. $\endgroup$ – Andrey Gogolev Jul 16 '13 at 21:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy