# Hyperbolic Fixed Point

Let $f:M\rightarrow M$ be a $C^{1}$-class diffeomorphism . Let $x\in M$ be a fixed point. I've been looking for a while on Internet for a proof of the following fact, but i couldn't find : $\lbrace x\rbrace$ is a hyperbolic set for $f$ if and only if $x$ is a hyperbolic fixed point.

The definition of hyperbolic fixed point I'm using is the following : $x$ is a fixed point of $f$ such that $d_{x}f:T_{x}M\rightarrow T_{x}M$ has no eigenvalues in the unit circle $S^{1}\subset\mathbb{C}$.

Can somebody help me ? (sketching the proof or even giving me some reference) Thank you :)

This is simply a matter of definition: a hyperbolic fixed point is defined to be a point $x$ such that $\{x\}$ is a hyperbolic set. See for example the wikipedia entry on hyperbolic sets where they use the term "hyperbolic equilibrium point" instead of "hyperbolic point".
• The notes i'm following define a hyperbolic fixed point as a fixed point of $f$ such that $T_{x}M\rightarrow T_{x}M$ has no eigenvalues on the unit circle. I would like then to prove from this definition, that $\lbrace x\rbrace$ is a hyperbolic set. Commented Apr 26, 2014 at 9:30
• In the above, I mean $d_{x}f:T_{x}M\rightarrow T_{x}M$ Commented Apr 26, 2014 at 9:37
• Using that definition, then, I would say that there is one step to the proof: apply the Jordan normal form. The Jordan blocks for eigenvalues of absolute value $>1$ give you one subspace of $T_x M$, the blocks for eigenvalues of absolute value $<1$ give you a complementary subspace, and these two subspaces give you the direct sum decomposition of $T_x M$ needed to satisfy the definition that {x} is a hyperbolic set. Commented Apr 26, 2014 at 22:03