Timeline for Making images arbitrarily dense under an expanding map
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 18, 2017 at 17:27 | vote | accept | Vaughn Climenhaga | ||
| Nov 18, 2017 at 8:45 | answer | added | Anthony Quas | timeline score: 2 | |
| Nov 18, 2017 at 5:18 | answer | added | fedja | timeline score: 3 | |
| Nov 17, 2017 at 17:59 | history | edited | Vaughn Climenhaga | CC BY-SA 3.0 |
added 1046 characters in body
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| Nov 17, 2017 at 17:31 | comment | added | Vaughn Climenhaga | Anthony: I don't see how the map you describe can be uniformly expanding, since the unit disk covers itself under the map, but is also covered by the image of a disk that's far away, so surely there must be some local contraction somewhere. In any case I really wanted to consider the case where $f$ is globally expanding and is a diffeomorphism; I should have clarified this initially. Will edit the question to say this. | |
| Nov 17, 2017 at 17:28 | comment | added | Vaughn Climenhaga | Re: the second spiraling suggestion, a bounded distortion argument shows that every map of the form $f(z) = ce^{ig(|z|)}z$ with $g$ a $C^2$ function has a $\delta>0$ as in the question; I'll add details to the question to explain this. | |
| Nov 17, 2017 at 17:07 | comment | added | Anthony Quas | So here's a version (using the "lemma") with only one disk: Let your map look like $2x$ almost everywhere. Find a disk around $T^{99}(\frac 12)$ and map that linearly and conformally to a disk centred at 1/2 covering the unit disk (so that 1/2 is a period 100 fixed point) but with a slight twist by $\alpha$, say. Now $T^{100}([-1,1])$ contains $[-1,1]$ and the rotation of that line around 1/2 by $\alpha$. $T^{200}([-1,1])$ contains additionally the rotation by $2\alpha$ etc. | |
| Nov 17, 2017 at 5:43 | comment | added | user35593 | How about f (x)=ce^{(1-|x|)i}x? | |
| Nov 17, 2017 at 3:55 | comment | added | Vaughn Climenhaga | Re: the lemma about a disk and an annulus, it seems related to this question: mathoverflow.net/questions/38498/… -- one difference being that here we are requiring that everything is expanding, which wasn't asked for there. | |
| Nov 17, 2017 at 3:54 | comment | added | Vaughn Climenhaga | Anthony, I'm afraid I don't quite follow how your construction goes (with the lemma assumed). Do you take a countable sequence of disjoint disks that get smaller and smaller and converge to the origin? How can different disks all have the unit disk as their image? Or do you mean under a different number of iterates? | |
| Nov 16, 2017 at 21:30 | comment | added | Anthony Quas | (Or is there some global property of globally expanding maps that my surgery would be violating?) | |
| Nov 16, 2017 at 21:29 | comment | added | Anthony Quas | I can build a nice counterexample if I assume a lemma (whose truth I don't know): I would like to assume that if one has a disk $D$ and a disjoint annulus $A$ surrounding it, then given $C^2$ expanding maps defined on $D$ and $A$, there is a $C^2$ expanding map that extends the two pieces. Given this, you can certainly make a counterexample: take $f(x)=2x$ everywhere and use the lemma to do surgery: take little disks in the plane centred on the real axis and map them so their images are the rotated unit disk. As long as the disks avoid the "holes" created by previous surgeries, this should do. | |
| Nov 16, 2017 at 13:37 | comment | added | Vaughn Climenhaga | The spiraling map $f(z) = c e^{i|z|} z$ doesn't work as a counterexample: we need $c>1$ to be expanding, and then if $|f^n(z)| < 1$ we have $|f^k(z)| < c^{-(n-k)}$ for all $0\leq k \leq n$, so $\sum_{k=0}^n |f^k(z)| < (1-1/c)^{-1}=:L$, and we conclude that $f^n(X)$ can wrap at most $L/(2\pi)$ times around the origin. Since this is independent of $n$ it should suffice to take $\delta$ on the order of $2\pi/L$. | |
| Nov 16, 2017 at 7:05 | comment | added | user35593 | Identify $mathbb {R}^2$ with $\matbb {C}$ and take $f(x)=ce^{i|x|}x$. Then the images of the real line under $f^n$ become spirals which are more and mire dense for increasing $n$. | |
| Nov 16, 2017 at 5:13 | history | asked | Vaughn Climenhaga | CC BY-SA 3.0 |