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Jun 15, 2020 at 7:27 history edited CommunityBot
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May 19, 2019 at 13:12 comment added Anthony Quas The point is that for most $x$ and $y$, $y$ does not lie in the unstable manifold of $x$ (otherwise the unstable manifold would be of the wrong dimension). This implies (as I showed above) that there is no $t$ lying in the intersection of the unstable manifolds.
May 19, 2019 at 11:56 comment added Adam I am really interested to know what is mathematic reason in nonlinear Baker, two unstable manifold intersect each other.
May 19, 2019 at 11:55 comment added Adam @AnthonyQuas: It doesn't mean they intersect each other, does it? because maybe they are very close to each other.
May 19, 2019 at 5:14 history edited მამუკა ჯიბლაძე CC BY-SA 4.0
might be confusing - it is not anybody's name
May 19, 2019 at 0:33 comment added Anthony Quas If $t\in W^u(x)\cap W^u(y)$, then $d(f^{-n}t,f^{-n}x)\to 0$ and $d(f^{-n}t,f^{-n}y)\to 0$, so that $d(f^{-n}x),f^{-n}y)\to 0$. Of course this is not the case for typical nearby $x$ and $y$.
May 18, 2019 at 23:43 comment added Adam @Mike Miller. You're right. I modified the question. I'm looking for a map that defines on $ S^1 \times [-1, 1]$
May 18, 2019 at 23:40 history edited Adam CC BY-SA 4.0
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May 18, 2019 at 22:19 comment added mme The link by Lee Mosher is for a map on the unit disc, not some arbitrary Riemannian manifold. You're asking for that specific map?
May 18, 2019 at 22:08 history edited Adam CC BY-SA 4.0
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May 18, 2019 at 15:47 comment added Adam @LeeMosher : I understood stable and unstable manifold intersect each other but I would like to know why unstable manifolds intersect each other. I will appreciate if you help me or introduce some references.
May 18, 2019 at 15:45 comment added Adam @LeeMosher : Yes, But I asked why unstable manifold $W_{x}^{u}$ and $W_{y}^{y}$ intersect each other. And then, why it is unique?
May 18, 2019 at 15:43 history edited Adam CC BY-SA 4.0
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May 18, 2019 at 15:40 comment added Lee Mosher Perhaps you can provide a definition or a link to a "Baker map". I know what the Baker's map is, but presumably that is not what you mean because its stable and unstable manifolds do indeed have a local product structure.
S May 18, 2019 at 15:39 history edited Adam CC BY-SA 4.0
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S May 18, 2019 at 15:39 history suggested Michal CC BY-SA 4.0
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May 18, 2019 at 15:37 review Suggested edits
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May 18, 2019 at 15:32 comment added Adam @AnthonyQuas : Thanks for your comment. No, I know when we have local product structure stable and unstable manifold intersect each other . But as I said I want to show that unstable manifolds intersect each other in Baker map. We see they intersect each other when we draw it but I do not know how I can prove it.
May 18, 2019 at 15:30 comment added Adam @LeeMosher : Thanks for your comment. I forgot to write map is Baker. When we draw picture we will see their unstable manifold intersect each other but I can not prove it
May 18, 2019 at 10:12 history edited Adam CC BY-SA 4.0
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S May 18, 2019 at 10:01 history suggested user64494 CC BY-SA 4.0
The title is improved.
May 18, 2019 at 9:59 review Suggested edits
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May 18, 2019 at 9:56 history edited Adam CC BY-SA 4.0
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May 18, 2019 at 1:16 comment added Anthony Quas Is the result you want that (under some conditions) the stable manifold of $x$ meets the unstable manifold of $y$?
May 18, 2019 at 0:43 comment added Lee Mosher It's not true. In fact if we assume that $f$ is Anosov then locally the stable manifolds form a decomposition, hence the only way that that $W^u_x$ and $W^u_y$ can have nonempty intersection is if they are equal. Are you sure you wrote the question correctly?
May 18, 2019 at 0:23 history asked Adam CC BY-SA 4.0