1
$\begingroup$

Let $M$ be a smooth manifold, $U \subset M$ an open set, $f : U \to M$ a $C^1$ diffeomorphism onto its image and $\Lambda \in U$ a hyperbolic set for $f$.

Fix a sufficiently small $\gamma > 0$ and consider a $(\lambda, \mu)$-splitting and the family of horizontal cones $$H_x^\gamma = \{ u + v : E^u_x, v\in E^s_x, \| v \| \leq \gamma \; \| u \| \}.$$

Can someone give me a hint on how to show that $$Df_x H_x^\gamma \subset H_{f(x)}^{\lambda \mu^{-1} \gamma}?$$

P.S. This is a part of the proof for Proposition 6.4.6 from "Introduction to the Modern Theory of Dynamical Systems" - A. Katok & B. Hasselblatt.

Thank you!

$\endgroup$

1 Answer 1

0
$\begingroup$

Since $E^u, E^s$ are equivariant, the projections of $df_x (u + v)$ according to $E^u_{f x}, E^s_{fx}$ are exactly $df_x u$ and $df_x v$, respectively. So, $$ \frac{\| df_x v \|}{ \|df_x u\|} \leq \frac{\| df_x|_{E^s_x}\|}{m( df_x|_{E^u_x})} \frac{\|v\|}{\|u\|} \, , $$ where $m(A) = \min \{ \| A v \| : \| v\| =1\} = \| A^{-1} \|^{-1}$ is the minimal norm of a linear operator. The desired cone membership follows from this estimate.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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