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I am trying to understand uniformly hyperbolic dynamical systems from the definition given here.

I understand Smale's horseshoe with expansion and contraction that is very clear to see, but I don't see how the derivative gives us this expansion and contraction. How does the derivative express the squeezing and stretching on the original set?

Dan M
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1 Answers1

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The distance along a path $\gamma\colon[0,1]\to M$ between two points $p,q\in M$ is given by $$ d(p,q)=\int_0^1\|\gamma'(t)\|\,dt. $$ Therefore, for a diffeomorphism $f\colon M\to M$ we have $$ d(f(p),f(q))=\int_0^1\|(f\circ\gamma)'(t)\|\,dt\le\sup\|df\|\int_0^1\|\gamma'(t)\|\,dt=\sup\|df\|d(p,q). $$ In particular, if the derivative contracts, then the length of $\gamma([0,1])$ decreases when we apply $f$.

In other words, lengths decrease when the derivative contracts. Similarly, lengths increase when the derivative expands.

That's the beginning of hyperbolicity theory, which generalizes this simple example in various ways.

John B
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  • My thinking is that the tangent space expands and contracts, so I'm guessing you pick the directional derivative for $\gamma'(t_0)$ for all the points in the path so you have one vector rather than all of the tangent space for each, then this is either in the stable or unstable subspace, but the subspaces depend continuously so you'll have all of the derivatives being (un)stable and then that gives you the length of the image path, and because of the above the length is contracted or expanded, is that correct? – Dan M Jun 11 '20 at 13:56
  • No, but I have already replied to your question. – John B Jun 11 '20 at 16:36
  • @JohnB thank you for the reply, I am happy with the response but I suppose I am unhappy with the definition of $\gamma'(t)$. For some $p \in M$ where $M$ is the manifold, it seems $\gamma'(p)$ could be defined as any vector in $T_p M$. How does one make this choice? If the tangent space has a stable and unstable subspace this choice should be important unless there is some way that this is independent of the choice but I cannot see this. – math Jun 11 '20 at 19:39