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I found a strange attractor which looks a lot like a solenoid. The attractor continuum is the closure of a continuous line which limits onto itself, and it is locally homeomorphic to Cantor set times Reals. It sits in the Möbius strip.

enter image description here

Does the Dyadic solenoid embed into the Möbius strip? What about solenoids in general? Could the strange attractor above actually be a solenoid?

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    $\begingroup$ If it embeds into the Möbius strip, you get a 2-fold covering, that embeds into an oriented strip (cylinder). It seems visually quite clear that it can't (embed into the plane at all), although this certainly requires a little argument. $\endgroup$ – YCor Nov 18 '18 at 15:46
  • $\begingroup$ IIRC if you follow an interval transversal to the trajectories along the flow, after one turn the interval becomes half-folded, i. e. you get a surjective non-injective self-map of it, two-to one on half of it and one-to-one on the other half. I don't see how could you possible embed such behavior in any 2-manifold... But is your question about this attractor or about a solenoid? $\endgroup$ – მამუკა ჯიბლაძე Nov 18 '18 at 17:16
  • $\begingroup$ I guess that one general statement of the form "every compact space that has a locally trivial fibration with quotient the circle and fibre a Cantor set", that embeds as a subset of the plane, is a product (circle)$\times$(Cantor), would settle everything. $\endgroup$ – YCor Nov 18 '18 at 18:55
  • $\begingroup$ But that statement is false. For example, take the suspension of an irrational rotation of the circle, which gives a foliation of the torus. Then do a Denjoy blowup (split one leaf) to make it a compact space of the type you describe embedded in the torus. @YCor $\endgroup$ – Lee Mosher Nov 18 '18 at 19:24
  • $\begingroup$ I am also pretty sure that every orientable geodesic lamination on a hyperbolic surface is of the type you describe. @YCor $\endgroup$ – Lee Mosher Nov 18 '18 at 19:26
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No solenoid can be embedded into the Mobius strip. To derive a contradiction, assume that some solenoid $S$ embeds into the Mobius strip $M$. Let $\pi:C\to M$ be a 2-fold covering map of the cylinder $C$ onto the Mobius strip. It is well-known that the solenoid $S$ contains a dense subset $D$ which is the image of the real line under a continuous map $\phi:\mathbb R\to D$ (and this image of $\mathbb R$ is called a composant of the solenoid). By the lifting property of the covering map $\pi$, there exists a continuous map $\varphi:\mathbb R\to C$ such that $\pi\circ\varphi=\phi$. Then the closure $K$ of the connected set $\phi(\mathbb R)$ in $C$ is a continuum such that $\pi(K)=\bar D=S$. Taking into account that the cylinder $C$ embeds into the plane, we conclude that $K$ is a planar continuum and hence the solenoid $S$ is a continuous image of a planar continuum.

On the other hand, by a result of Krasinkiewicz in his paper Mappings onto circle-like continua, no solenoid is a continuous image of a planar continuum (as solenoids have infinitely divisible first cohomology group whereas the first cohomology group of any planar continuum is finitely divisible). This is a desired contradiction.

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  • $\begingroup$ I still do not understand. I did know that the circle is the only solenoid which embeds into the plane. But I don't see why this implies the circle is the only one that embeds into the Möbius strip. $\endgroup$ – Forever Mozart Nov 19 '18 at 0:21
  • $\begingroup$ I also don't understand the double cover argument, or at least why it needs no further explanation (since a solenoid is obviously not Peano so doesn't satisfy the lifting criterion). But shouldn't you be able to take a transverse axis of the Mobius strip whose support on the solenoid $S$ is the Cantor Set, which can then be symmetrized across the axis to build an ad hoc double cover? Then it would remain to show that this (circle-like) double cover is a solenoid, giving the contradiction. This seems clear, but requires work, from the classification of solenoids among circle-like continua. $\endgroup$ – John Samples Nov 19 '18 at 5:55
  • $\begingroup$ @ForeverMozart I wrote more explanation in my answer. $\endgroup$ – Taras Banakh Nov 19 '18 at 7:02
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    $\begingroup$ Nice proof! Thanks! For those who stumble across this, maybe it should be said that the dense image of $\mathbb{R}$ is just one of the composants. $\endgroup$ – John Samples Nov 19 '18 at 9:59
  • $\begingroup$ @JohnSamples Thank you for the suggestion. I have added a note about composants to my answer. $\endgroup$ – Taras Banakh Nov 19 '18 at 10:21

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