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.

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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

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.

  • $\begingroup$ 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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