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Let $M$ be a compact Riemannian manifold and let $f:M→M$ a diffeomorphism. Let $\Lambda\subset M$ be a compact invariant subset of $M$. We say that $\Lambda $ is a hyperbolic set for $f$ when there are constants $C>0$ and $0<\lambda<1$ and an invariant decomposition $T_{\Lambda}M=E^s_{\Lambda}\oplus E^u_{\Lambda}$ such that

  1. $\|Df^n_{x}v\|\leq C\lambda^n\|v\|$ $~~\forall n\geq 0$ and $\forall v\in E^s_{x}$

  2. $\|Df^{-n}_{x}v\|\leq C\lambda^n\|v\|$ $~~\forall n\geq 0$ and $\forall v\in E^u_{x}$

  3. $Df_xE^s_x=E^s_{f(x)}$ and $Df_xE^u_x=E^u_{f(x)}$ $\forall x\in \Lambda.$

I'm currently reading basic texts on hyperbolic dynamics. An argument that appears repeatedly to these long texts says it is possible extending the hyperbolic splitting on $\Lambda$ to a neighborhood of $\Lambda $. In other words there is a neighborhood $U$ of $\Lambda$ where we get a decomposition $T_U=E^s_U\oplus E_U^u$ which still hold $(1)$ and $(2)$ (Possibly with $\lambda $ slightly higher), and possibly not holding $(3).$ Roughly speaking: "Hyperbolicity is an open property".

However all authors treat this statement as if it were obvious, is not commented on the appropriate details, which makes me think that this may be a corollary of a classical result. I wish someone would explain to me some standard or not way to do this.

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My post was to try to understand this kind of comment that constantly appear in texts –  Juan Valdez Sep 30 '13 at 18:30

1 Answer 1

up vote 4 down vote accepted

One relatively straightforward way is to use cone families. First, note that by passing to an adapted metric it is possible to assume that $C=1$. (This is a standard argument whose details should appear in the texts you are reading.) In particular, you have the one-step expansion properties $\|Df(v^u)\| \geq \lambda^{-1}\|v^u\|$ and $\|Df(v^s)\|\leq \lambda \|v^s\|$ for every $v^{u,s}\in E^{u,s}_x$.

Next, note that the hyperbolicity conditions on $\Lambda$ imply the existence of invariant cone families $K^{u,s}_x$. These can be defined by fixing $\gamma>0$ and putting $K^u_x = \{ v^u + v^s \mid v^{u,s} \in E^{u,s}_x, \|v^s\| < \gamma \|v^u\|\}$, and similarly for $K^s_x$ with the roles of $u,s$ reversed. A direct computation using the one-step expansion properties above shows that if $\gamma$ is chosen sufficiently small, then $\overline{Df(K_x^u)} \subset K_{f(x)}^u$ and $\overline{Df^{-1}(K_x^s)} \subset K_{f^{-1}(x)}^s$ for $x\in \Lambda$ and that moreover the one-step expansion holds for all $v^{u,s}\in K_x^{u,s}$ if $\lambda$ is replaced by some $\lambda'\in (\lambda,1)$.

It is straightforward to see that the existence of cones with the above properties is an open condition: given that the above conditions hold for $K_x^{u,s}$ with $x\in \Lambda$, extend the cones continuously to $x$ in a neighbourhood of $\Lambda$, and for a sufficiently small neighbourhood this continuous extension will have the same property because each of the properties above is stable under small perturbations.

Now once you have the cones on a neighbourhood $U$, you just need to get invariant subspaces $E^{u,s}_x \subset K^{u,s}_x$, and everything follows. Let's do $E^s$: the case $E^u$ is similar. Let $J = \{n\geq 0 \mid f^n(x)\in U\}$, and define $E_x^s$ to be any subspace of the appropriate dimension contained in the intersection $\bigcap_{n\in J} Df^{-n}(\overline{K_{f^nx}^s})$. Note that the sets in the intersection are compact and nested, so the intersection is nonempty even if $J=\mathbb{N}$. Take a similar intersection with $n\leq 0$ to define $E_x^u$.

Edit: Here is a clarification of how to extend the cones to a neighbourhood. Let $\tilde U$ be a small neighbourhood of some part of the attractor $\Lambda$ -- in particular, assume $\tilde U$ is small enough so that the tangent bundle $T\tilde U$ is homeomorphic to $\tilde U \times \mathbb{R}^d$. Then the subspaces $E_x^u$ on $\Lambda \cap \tilde U$ are given by a continuous function $\Lambda \cap \tilde U \to (\mathbb{R}^d)^u$, where $u$ is the dimension of the unstable subspace -- this function takes $x$ to a set of $u$ vectors forming a basis for $E_x^u$. By the Tietze extension theorem, this can be extended continuously to the neighbourhood $\tilde U$. The procedure for $E_x^s$ is analogous. This gives an extension of $E_x^{u,s}$, and hence $K_x^{u,s}$, to $\tilde U$. Cover $\Lambda$ with finitely many such neighbourhoods, enumerate them as $\tilde U_i$ for $i=1,\dots, n$, and extend first to $\tilde U_1$, then $\tilde U_2$, and so on.

This describes how to extend the cones. For the claim that the conditions on the cones extend, the key tool is the topological fact that if $\phi\colon X\to Y$ is continuous and $X'\subset X, Y'\subset Y$ are closed sets such that $\phi(X')\subset Y'$, then for every open set $Z\supset Y'$, there is an open set $V\supset X'$ such that $\phi(V) \subset Z$.

First we use this to get the expansion/contraction condition. Let $\chi(x) = \sup\{ \|Df_x(v)\| \mid v\in K_x^s, \|v\|=1\}$. Note that $\chi$ is continuous on $U$ and $\chi(x)\leq \lambda'$ for all $x\in \Lambda$. Thus for all $\lambda''\in (\lambda', 1)$, there is a neighbourhood $\hat U \subset U$ of $\Lambda$ such that $\chi(x) \leq \lambda''$ for all $x\in \hat U$. Expansion properties for $K_x^u$ follow similarly.

Finally, invariance. Extend $E_x^{u,s}$ as described above. Given $x\in U$ and $\gamma>0$ Let $K_x^u(\gamma)$ be the cone around $E_x^u$ of width $\gamma$, as in the second paragraph. Then on $\Lambda$, uniform expansion/contraction together with invariance of $E_x^{u,s}$ shows that there are $0<\gamma'<\gamma$ such that $Df(K_x^u(\gamma)) \subset K_{fx}^u(\gamma')$ for all $x\in \Lambda$. Because $Df$ and $x\mapsto K_x^u$ are continuous, and because $K_{fx}^u(\gamma)$ contains a neighbourhood of $K_{fx}^u(\gamma')$, the topological lemma shows that $\overline{Df(K_x^u(\gamma))} \subset K_{fx}^u(\gamma)$ for all $x$ in a sufficiently small neighbourhood of $\Lambda$. The invariance property for $K_x^s$ is similar.

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It is not yet clear to me, how to extend the cone field to a neighborhood of $\Lambda$ –  Juan Valdez Oct 1 '13 at 17:17
    
You say: It is straightforward to see que the existence of cones with the above properties is an open condition. And more: given the above conditions hold que for $K^{u, s}_x$ with $x\in \Lambda$, the cones extend continuously to $x$ in a neighborhood of $\Lambda$. As if it were obvious, but not obvious to me. –  Juan Valdez Oct 10 '13 at 19:06
    
I've added what I hope is a sufficient explanation -- the reason I characterised these steps as "straightforward" is because they only involve elementary arguments from point-set topology, and not many new dynamical ideas. –  Vaughn Climenhaga Oct 10 '13 at 21:03
    
Thank you for your attention, and response. –  Juan Valdez Oct 11 '13 at 2:40

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