Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:M \rightarrow M$ be partially hyperbolic diffeomorphism of M with the usualy definition that at each p tangent space splits to Df invariant subspaces: $T_pM = E^s_p + E^c_p + E^u_p$ with the relevant constants on the norm of Df restricted to these spaces. I just need one which is $\mu^n < |Df^n_{E^u}|$.

I have some confusions about how people define the center-stable manifolds etc. In the case of just partially hyperbolic fixed point p in $R^n$, Robinson in his book defines the center stable manifold of the fixed point as $W^{cs}(p) = \{x | lim_{n\rightarrow \infty}\frac{1}{\mu^n}d(f^n(x),p)=0\}$ (after extending the map to whole of $R^n$ if its not defined every where already) and says that although if not unique, such center manifolds exists and are tangent to $E^c$. In the case of partially hyperbolic diffeomorphism of manifolds, I have never seen this characterization. They just define $W^{cs}(p)$ as a submanifold tangent to $E^{cs}$ basically.

My question is for a partially hyperbolic set is it not necessary that a submanifold $N(p)$ passing throgh p which satisfies $N(p) = \{x | lim_{n\rightarrow \infty}\frac{1}{\mu^n}d(f^n(x),p)=0\}$ will be tangent to $E^c$? I was able to show that the unstable manifold has to be transversal to this manifold but transversality is not enough to show that tangent space of this manifold has to be $E^c+E^s$

Why is it not preferred to define central stable behaviour in this manner for this setting? Is there also a center stable manifold theorem for this setting in the manner which says, for all p there is always a submanifold passing through p, which satisfies the convergence criteria I wrote above and is tangent to $E^s + E^c$?

Thanks

share|improve this question
add comment

Your Answer

 
discard

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

Browse other questions tagged or ask your own question.