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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 :)

thetruth
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1 Answers1

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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".

Lee Mosher
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  • 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. – thetruth Apr 26 '14 at 09:30
  • In the above, I mean $d_{x}f:T_{x}M\rightarrow T_{x}M$ – thetruth Apr 26 '14 at 09:37
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    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. – Lee Mosher Apr 26 '14 at 22:03
  • What would go wrong if there would be eigenvalues on the unit circle? Why couldn't we then also use Jordan decomposition? – Willem Beek Dec 18 '14 at 21:30
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    @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. – Lee Mosher Dec 19 '14 at 16:58
  • @LeeMosher: Thank you. That sounds horrible, when there is no neighborhood which contains no other periodic orbits..

    But still I don't see what would precisely go wrong in the proof-strategy you described earlier. Couldn't we now decompose the $T_xM$ in the subspace given by Jordan blocks of absolute value $ = 1$ and jordan blocks of absolute value $\neq 1$?

    Thank you again!

     – Willem Beek Dec 19 '14 at 17:28
  • @WillemBeek: You may certainly decompose $T_x M$ in that manner. However, that decomposition violates the requirements for the definition of a hyperbolic set. – Lee Mosher Dec 19 '14 at 17:32
  • @Lee Mosher Your first comment is correct, up to the point that it is for the complexification of the tangent space. Note that in general the eigenvalues may not be real, and so we need to take the real parts (or imaginary parts) of the complexification. – John B Jan 20 '16 at 00:18
  • @LeeMosher Let $E^s$ and $E^u$ be the two subspaces you mentioned in the first comment, i.e. $E^s$ = span ${v : v$ eigenvector of $df_x$ with eigenvalue $\lambda, |\lambda| < 1 }$ and $E^u$ = span ${v : v$ eigenvector of $df_x$ with eigenvalue $\lambda, |\lambda| > 1 }$. How do we know that $T_x M = E^s \bigoplus E^u$? Can you give me some ideas on how to check this fact, please? Thank you! – g.pomegranate Nov 30 '17 at 19:34
  • @g.pomegranate Generally speaking, if you have a new question about a three-and-a-half year old question and its answer, you should simply post your own new question. Burying new questions in the comments to the answers of very old questions is bad practice on any question-and-answer web site. – Lee Mosher Nov 30 '17 at 19:38
  • Ok. Thank you for your observation. I'm going to post it in a new question. – g.pomegranate Nov 30 '17 at 19:40