Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $M$ be a compact, finite-dimensional Riemannian manifold, let $T: M \rightarrow M$ be an Anosov diffeomorphism, and let $\mathcal{R} = \{ R_1,\dots,R_n \}$ be a Markov partition. Because of structural stability, a small $C^1$ perturbation $T'$ of $T$ will also be Anosov. Is there a well-defined Markov partition $\mathcal{R}'$ for $T'$ that can be regarded as a perturbation of $\mathcal{R}$?

share|improve this question
add comment

1 Answer

The answer is yes (don't know why I didn't see this before).


Let $T'$ be a small perturbation of $T$ and $U$ the corresponding topological conjugacy, viz. $U^{-1}T'U = T$. Now the key is to show that

$UW^s_{(T)}(x,R_j) = W^s_{(T')}(Ux,UR_j)$

and similarly for the unstable manifolds.

Let $x' \in W^s_{(T)}(x,R_j)$, i.e. $x' \in R_j$ and $d(T^nx,T^nx') \downarrow 0$. Then $Ux' \in UR_j$ and $d(U^{-1}T'^n(Ux),U^{-1}T'^n(Ux')) \downarrow 0$. By continuity, $d(T'^n(Ux),T'^n(Ux')) \downarrow 0$. So $Ux' \in W^s_{(T')}(Ux,UR_j)$.

Now $UW^s_{(T)}(x,R_j) = W^s_{(T')}(Ux,UR_j)$ implies that

$T'UW^s_{(T)}(x,R_j) = T'W^s_{(T')}(Ux,UR_j) \subset UW^s_{(T)}(U^{-1}T'Ux,R_k) = W^s_{(T')}(T'Ux,UR_k)$.

Again, the unstable case follows similarly. So $U\mathcal{R}$ is a Markov partition for $T'$. $\square$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.