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Can we know if a map is hyperbolic without making any calculation, i.e without verifying if the differential of the map has or has not eigenvalues of absolute values 1?

P.S. Can we show that the spatial version of Baker map is hyperbolic only by the construction steps?

g.pomegranate
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1 Answers1

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Let $f : M \to M$ be a $C^1$ diffeomorphism on a Riemannian manifold $M$, and let $\Lambda \subset M$ be an invariant subset (i.e., $f (\Lambda) = \Lambda$).

We say that $f$ is hyperbolic on $\Lambda$ if, for each $x \in \Lambda$, there is a splitting $T_x M = E_x^u \oplus E^s_x$ with the following properties:

  1. For all $x \in \Lambda$, we have $df_x E^{u/s}_x = E^{u/s}_{f x}$.
  2. There are constants $C > 0, \lambda \in (0,1)$ such that $$ \| df^n_x |_{E^s_x} \| \leq C \lambda^n \, , \quad \text{and} \quad \| df^{-n}_x |_{E^u_x} \| \leq C \lambda^n $$ for all $x \in \Lambda$ and $n \geq 0$.

Unless the mapping in question has a very simple structure (as is the case, for instance, for the Baker's transformation) it is often somewhat challenging to find $E^{u/s}_x$ explicitly. In practice, one looks not for $E^{u/s}_x$ but rather for families of stable/unstable cones.

Theorem: $f$ is hyperbolic on $\Lambda$ iff the following holds.

For each $x \in \Lambda$ there is a splitting $T_x M = \tilde E^u_x \oplus \tilde E^s_x$ and numbers $\alpha > 0, C > 0, \lambda \in (0,1)$ with the following properties. For each $x \in \Lambda$, define \begin{gather*} \mathcal C^u_x := \{ v \in T_x M : \| \pi^s_x v\| \leq \alpha \| \pi^u_x v\|\} \\ \mathcal C^s_x := \{ v \in T_x M : \| \pi^u_x v\| \leq \alpha \| \pi^s_x v\|\} \end{gather*} For each $x \in\Lambda$,

  1. $df_x \mathcal C^u_x \subset \operatorname{Interior}(\mathcal C^u_{fx})$ and $df^{-1}_x \mathcal C^s_x \subset \operatorname{Interior}(\mathcal C^s_{f^{-1}x})$

  2. For each $v \in \mathcal C^u_x$, we have $\| df_x^n v\| \geq C^{-1} \lambda^{-n}\| v\|$ and for each $w \in \mathcal C^s_x$, we have $\| df^n_x w\| \leq C \lambda^n \| w\|$.

A good reference is the book of Brin and Stuck. Notice that in the cones definition, the subspaces $\tilde E^{u/s}_x$ need not be $df$-invariant. This makes the above alternative definition much more convenient for applications.

To get a feeling for this alternative definition, you might consider checking the above equivalence holds when $\Lambda = \{ x\}$ is just a fixed point.

An aside: Notice that I didn't reference eigenvalues or eigenspaces in either definition of hyperbolicity. The reason is that eigenvalues are only defined for endomorphisms of a single vector space, not between mappings of different vector spaces, e.g., the differential $df_x : T_x M \to T_{fx} M$. There is, however, a spectral approach to defining hyperbolicity using Mather's evolution operator formulation-- see, e.g., the book of Chicone and Latushkin.

A Blumenthal
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  • Thank you for this. I have looked at both Barreira & Valls, and Bring & Stuck. Both discuss stable/unstable cones briefly and go on to discuss how finding these cones determines hyperbolicity of the map. I was wondering if there are any results on $df_{x}$ which imply existence of these cones. For some examples, this computation is relatively easy to do by hand, but it seems to become harder as $df_{x}$ becomes larger/more complex. Is there any literature regarding what families of matrices admit stable/unstable cones? – none Nov 18 '18 at 04:01
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    @none Depends. Honestly it's not easy (nor should it be) to pass from simulations of a given dynamical system to a refined analysis of phase space properties such as the existence of nontrivial families of invariant cones. That being said much ink has been spilled on various consequences / equivalent formulations of cones conditions. A fun one that comes to mind is that of Bochi and Gourmelon, following Yoccoz. – A Blumenthal Nov 19 '18 at 02:40
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    Then there are situations when you have some kind of special structure. What comes to my mind is the theory of inertial manifolds, which is used to study certain globally well-posed dissipative parabolic PDE (some, not all of them) such as Kuramoto-Sivashinski in 1D or some reaction-diffusion equations. Again these things use the special structure of these equations (in this case, dissipation coming from the Laplacian). Note: these equations are NOT necessarily hyperbolic, but they do satisfy cones conditions! – A Blumenthal Nov 19 '18 at 02:43
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    Lastly, before I end my rant, I should also note that uniform hyperbolicity as stated in my answer is way too strong for most models. Many models which appear hyperbolic actually exhibit "cone twisting", whereby expanding directions are "twisted" into contracting directions. To put it mildly, these are hard to study, which is why there are so few instances in which mathematicians can rigorously verify asymptotic hyperbolic / chaotic regimes. Look up the Chirikov standard map or the Henon map to get a flavor of this. – A Blumenthal Nov 19 '18 at 02:45