3
$\begingroup$

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 :)

$\endgroup$

1 Answer 1

3
$\begingroup$

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".

$\endgroup$
11
  • $\begingroup$ 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. $\endgroup$
    – thetruth
    Commented Apr 26, 2014 at 9:30
  • $\begingroup$ In the above, I mean $d_{x}f:T_{x}M\rightarrow T_{x}M$ $\endgroup$
    – thetruth
    Commented Apr 26, 2014 at 9:37
  • 2
    $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Commented Apr 26, 2014 at 22:03
  • $\begingroup$ What would go wrong if there would be eigenvalues on the unit circle? Why couldn't we then also use Jordan decomposition? $\endgroup$ Commented Dec 18, 2014 at 21:30
  • 1
    $\begingroup$ @WillemBeek: If there are eigenvalues on the unit circle then the local dynamics are very much harder to analyze. For example, a hyperbolic fixed point has a neighborhood which contains no periodic orbits other than the fixed point itself. But if the fixed point is not hyperbolic, i.e. if there are eigenvalues on the unit circle, then one cannot draw any conclusions about the existence of nearby periodic orbits: they may or may not exist, depending on the particular example. $\endgroup$
    – Lee Mosher
    Commented Dec 19, 2014 at 16:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .