Timeline for Is there an elementary proof that distal maps are invertible?
Current License: CC BY-SA 4.0
15 events
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| May 25, 2021 at 19:24 | comment | added | Random | Recently I took an introductory course on topological dynamics, and this was (mistakenly) one of the exercises there. The intended "proof" was precisely this, so at least you're not the first person to make this error :) | |
| May 25, 2021 at 17:53 | history | edited | Nicolast | CC BY-SA 4.0 |
As pointed out in the comments, the argument is completely wrong. I edited the post accordingly
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| May 25, 2021 at 17:45 | comment | added | Nicolast | As pointed out in the comments, I was too quick to assert that $\delta$ is continuous. So there is really nothing to save... | |
| May 25, 2021 at 12:35 | comment | added | Benjamin Steinberg | @VilleSalo your edited description is clearer | |
| May 25, 2021 at 12:34 | comment | added | Benjamin Steinberg | For equicontinuous to imply distal you also need T injective I suppose | |
| May 25, 2021 at 12:31 | comment | added | Ville Salo | This system is obviously distal (since it can be written as a disjoint union of rotations, which are distal minimal systems), and it is not equicontinuous because two points that are very close but on different circles diverge slowly. | |
| May 25, 2021 at 12:26 | comment | added | Ville Salo | No not a Sturmian. You have a rotation on each individual concentric circle of radius $r$, and the point is that these circles rotate at different speeds. This is $(r, \theta) \mapsto (r, \theta+r)$ in polar coordinates. (Sorry, my description was misleading, you have to rotate the angles by the distance.) | |
| May 25, 2021 at 12:23 | comment | added | Benjamin Steinberg | @isnt this essentially a Sturmian shift? The argument about $\delta$ being continuous would be fine if the powers of T are equicontinuous but apparently distal is weaker than equicontinuous. | |
| May 25, 2021 at 12:21 | comment | added | Ville Salo | Right. Well, actually I suppose the most basic example of rotating points on a disk by their distance from the center is a counterexample. Using the unit disk of $\mathbb{C}$ as the model, $\delta(x, y) = 2$ for $x,y$ on the boundary, but nearby it can be arbitrarily small, just take two points on different circles. | |
| May 25, 2021 at 12:16 | comment | added | Benjamin Steinberg | @VilleSalo, hopefully the OP will chime in on which interpretation was meant and if the former, explain why. I’m not a dynamics person so I don’t know distal flows to check for counterexamples. Symbolic dynamics is my dynamical limit. | |
| May 25, 2021 at 12:11 | comment | added | Ville Salo | The question is, does the sentence parse as "(This function is continuous and vanishes on the diagonal) by distality." or as "This function is continuous and (vanishes on the diagonal by distality)."? @BenjaminSteinberg's example is for the latter reading. | |
| May 25, 2021 at 12:09 | comment | added | Benjamin Steinberg | The Cantor set $2^{\mathbb N}$ with its usual metric saying two infinite words are close if they have a long common prefix and the shift map $T$ (eliminating the first letter of an infinite word) is a counterexample to $\delta$ being continuous. let $z$ be a fixed infinite word over the alphabet $\{0,1\}$ and consider the sequences of infinite words $u_n=0^nz$ and $v_n=1^nz$. Then $u_n$ converges to $0^{\infty}$ and $v_n$ converges to $1^{\infty}$. Clearly $\delta(u_n,v_n)=0$ all $n$ since $T^n(u_n)=T^n(v_n)$ but $\delta(0^\infty,1^\infty)\neq 0$. | |
| May 25, 2021 at 10:56 | comment | added | Nate River | Hmm, maybe I’m missing something - is it obvious that $\delta$ is continuous? Naively this is a pointwise infimum of continuous functions which need not be continuous.. | |
| May 25, 2021 at 10:54 | vote | accept | Nate River | ||
| May 25, 2021 at 10:55 | |||||
| May 25, 2021 at 8:18 | history | answered | Nicolast | CC BY-SA 4.0 |