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In Brin Stuck, the following Corollary 5.6.6 is left as an exercise, that of which I'm not sure on how to prove. The stable-manifold theorem states that there exists $\epsilon >0$ small enough such that local stable manifolds $W^s_\epsilon(x) = \{ y \in M: d(f^n(x), f^n(y)) < \epsilon \quad \forall n \geq 0\}$ is a $C^1$ embedded manifold for every $x \in \Lambda$ in the hyperbolic set. Now we're interesting in the global stable manifold (defined below). Proposition 5.6.5 is just a set theoretic result, while the Corollary that follows seems to be a differential geometry question. Note that there is a typo here, authors should have written that $W^s(x)$ is an immersed $C^1$ submanifold (not embedded). I'm trying to find a prove for this claim. As a matter of fact, one can see that $W^s(x)$ is defined as a increasing union of embedded $C^1$-manifolds ! I posted a question here where I asked whether or not an increasing sequence of embedded submanifold is still an embedded submanifold. That is not the case, but at the time I thought that the stable manifold was to be a embedded submanifold (following the typo). Therefore, the question remains open as to whether or not such an increasing sequence can end up being an immersed submanifold !

Maybe that's not the way to do it...

Anyway, thanks for the help !

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Noam Eluar
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  • A duplicate of Global stable manifold always an embedded submanifold? Typo or misreading?. – user539887 Jan 12 '19 at 12:33
  • It's not a duplicate. I'm asking for a proof/ref of Corollary 5.6.6. – Noam Eluar Jan 12 '19 at 12:37
  • Please look at the comments and the answer. – user539887 Jan 12 '19 at 12:48
  • It seems to me these are discussions on the notion of "$C^1$ embedded manifold" used by the authors. In the answer, it says that it means that "there exists a $C^1$-smooth injective immersion from the open disk (whose image is in the stable manifold)". If this is the case, there's nothing to prove. I'll bet that the user meant to say "...(whose image is the stable manifold)", not the "in". That's the regular definition of a immersed submanifold. We need the all image of the disk to be $W^s(x)$, not just a part of it...If you have a proof, I'll be very thankful if you could provide it ! – Noam Eluar Jan 12 '19 at 13:05
  • I don't have the book at my disposal, but from the discussion under the other question it follows that Corollary 5.6.6 is wrong, provided that the topologies on $W^s(x)$ and $W^u(x)$ are relative topologies inherited from $M$. And this choice of topologies is perfectly natural in a textbook on dynamics. The global manifolds are homeomorphic to the unit balls, but when considered with the submanifold topologies (I cannot give a reference now). From the dynamicist's (as opposed to the topologist's) point of view the submanifold topology is rather too fine to be interesting. – user539887 Jan 12 '19 at 13:24
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    Regarding a reference on your last question, see Morton Brown's PAMS paper The Monotone Union of Open $n$-Cells is an Open $n$-Cell. – user539887 Jan 12 '19 at 13:29
  • Thanks a lot for taking the time to answer both of my questions, even though I'm still a little be confused: but then why do we call it the global stable manifold if it's not a manifold at all? I've read in different places that it was a immersed submanifold, so I'm a little bit confused by what you're saying... – Noam Eluar Jan 12 '19 at 13:38
  • For instance (https://arxiv.org/pdf/math/0506292.pdf) clearly state that the global stable manifold is immersed. – Noam Eluar Jan 12 '19 at 13:52
  • And this is a standard definition, I think. – user539887 Jan 12 '19 at 13:55
  • Sorry for spamming, but that also the case in Shub : Global stability of dynamical systems (1967) (see 6.6) – Noam Eluar Jan 12 '19 at 13:55
  • @LutzL $n$ is just the iteration of the dynamic $f$. $W^s(x)$ is the intersection of the preimage of the $\epsilon-$ball around $f^n(x)$ for all $n$ ! – Noam Eluar Jan 12 '19 at 14:47
  • Oh sorry ! It's for all $n \geq 0$, not just one specific $n$! I'll edit – Noam Eluar Jan 12 '19 at 15:07

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