Let $M$ be a Riemannian manifold, $U \subset M$ an open subset and $f : U \to M$ a $C^1$ diffeomorphism. I'm trying to show that a periodic orbit with period $n$ is a hyperbolic set iff $Df^n$ at any point of the orbit has no eigenvalue with absolute value $1$.

Let $\Lambda = \{p, f(p), ..., f^{n-1}(p)\}$ be a periodic orbit with period $n$. Suppose that $\Lambda$ is a hyperbolic set. Then there are constants $\lambda \in (0, 1), c > 0$ and for every $x \in \Lambda$, $T_x M = E^s_x \oplus E^u_x$ such that $$ Df_x(E^s_x) = E^s_{f(x)}, Df_x(E^u_x) = E^u_{f(x)} \\ \| Df^k_x(v) \| \leq c \lambda^k \| v \|, \text{for every} \; v \in E^s_x, k \geq 0 \\ \| Df^{-k}_x(v) \| \leq c \lambda^k \| v \|, \text{for every} \; v \in E^u_x, k \geq 0.$$

Let $x = f^i(p) \in \Lambda,\; 0 \le i \le n-1$. Suppose that $Df^n_x$ has an eigenvalue of absolute value $1$, i.e there are $\mu$ with $|\mu| = 1$ and $v \neq 0, v = v^s + v^u$, $v^s \in E^s_x, v^u \in E^u_x$, such that $Df^n_x(v) = \mu v$.

I can't obtain a contradiction. Can someone give me a hint?

  • $\begingroup$ Either $v^s\ne0$ or $v^u\ne0$. But note that $\mu$ might not be real. You need to do something about it. $\endgroup$ – John B Aug 15 '18 at 11:15
  • $\begingroup$ I will fix it, but, firstly, I want to suppose that $\mu$ is real. If $\mu$ is real, how can I obtain a contradiction? $\endgroup$ – g.pomegranate Aug 15 '18 at 11:44
  • $\begingroup$ Why do you want to prove by contradiction? If the periodic orbit of $x$ is hyperbolic then $Df^n(x)$ restricted to $E^s_x$ has spectrum inside the unit circle and $Df^n(x)$ restricted to $E^u_x$ has spectrum outside the unit circle. As $E^s_x$ and $E^u_x$ are complementary, the spectrum of $Df^n(x)$ is disjoint with the unit circle, QED. $\endgroup$ – user539887 Aug 15 '18 at 20:37

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