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Let $M$ be compact manifold. suppose $f:M\rightarrow M$ is $C^{2}$.

There is a continuous splitting of the tangent bundle $TM=E^{ss}+E^{s}+E^{u}$ invariant under the derivative $Df$ of the diffeomorphism $f$ and where strong expansion of the unstable bundle $E^{u}$, strong contraction $E^{ss}$ and weak contraction $E^{s}$.

Special case, you can consider $M=S^{1}\times \mathbb{D}$ so f is solenoid map such that $$(x,y,z)\rightarrow (mx mod1, \lambda y +u(x), \mu z+v(x))$$ where $m=2$ and $\mu<\lambda<\frac{1}{2}$. $\Lambda$ is attractor.

If A,B are two nearby embedded disks transverse to $W^{u}$ then there is a $\mathit{holonomy}$ map defined on subset of $A \cap \Lambda $ by $p\rightarrow W^{u}(p)\cap B$ whenever this make sense. In other words, From A to B along unstable leaves. The holonomies are always continuous and in fact Holder continuous.There woild be no need these holonomies are always Lipschitz continuous.

Question Why does Lipschitz property holonomies fails, when stable leaves $W^s(x)$ inside the leaves $W^{ss}(x)?$

In fact,The question is paragraph of a paper(Dimension and product structure of hyperbolic measures)by Barreira,Pesin and Schmeling, unfortunately they did not explain at all.They used this result in another paper(Dimension product structure of hyperbolic sets)by Hasselbalt and Schmeling.They did not explain,again.

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  • $\begingroup$ $W^{ss}$ has codimension 1 along $W^{s}$ then as we know codimension 1 is Lipschitz $\endgroup$ – Michal Nov 1 '18 at 23:45
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I think before discussing about the regularity of holonomies of $W^s$, one needs to discuss the existence of the foliation $W^s$.

In the above setting there are algebraic examples where the distibution $E^s$ is not even integrable: see for instance Example 6.1 of the book of Y. Pesin "Lectures on Partial Hyperbolicity and Stable Ergodicity", which is in fact a construction due to S. Smale

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  • $\begingroup$ Thank you very much for your answer.That is true we have to talk more about existence of the foliation$W^{s}$ but i assumed they are exist because other wise the problem should be much harder. $\endgroup$ – Michal Nov 1 '18 at 23:43

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