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Is a fattened Möbius Spiral Band homeomorphic to a Torus?

(Due to the same Euler Characteristic $\chi$ ?)

Are both non-orientable?

Following (3D printable plastic) Torus has a square section that rotates around a circle. Edges can be four closed helical geodesics. Made one such model in steel by twisting and welding ends for topological interest.

EDIT1:

Two continuous surfaces on the fat Möbius Spiral Band make it orientable. However if one of them is ignored making it a thin Möbius Spiral Band, orientability is lost. Is that how the thin band should be viewed as a special case of the fat/thick orientable Möbius Spiral Band?

Möbius spiral

Narasimham
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    Since they are closed surfaces same Euler characteristic means they are homeomorphic, as soon as they are both orientable or non-orientable. In particular there is one orientable surface with $\chi=0$ (the torus) and a non-orientable surface with $\chi=0$ (the Klein bottle).

    In particular the torus can be embedded in the Euclidean three-dimensional space, while the Klein bottle cannot: hence since your "Moebius spiral band" can be embedded in the three space (since you can 3D-print it without nasty overlapping), you can deduce that it is actually a torus (and hence it is orientable).

     – Dario Nov 24 '14 at 16:52
  • @Dario:3D Space immersion of Klein bottle (wikipedia) was given by Robert Israel. Is it not correct? Also the fattened Moebius band is logically and topologically an orientable torus. Is that correct? – Narasimham Nov 25 '14 at 05:22
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    Klein bottle can be immersed in $\mathbb{R}^3$, not embedded: the usual immersion has self-intersection... Anyway your Moebius spiral is a torus and is orientable. – Dario Nov 25 '14 at 08:32

1 Answers1

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They are homeomorphic, and they are both orientable.

It’s not clear whether you’re considering just the surfaces of these objects, or the whole solid objects, but in either case, the answer is the same. I’ll assume in the following that you mean the surfaces, but it adapts straightforwardly to the solid bodies as well. I’m leaving a lot of details out here; if you would like me to elaborate on anything, let me know.

To get the homeomorphism, first pick nice parametrisations of the torus and the fattened Möbius band. $\newcommand{\x}{\mathbf{x}}\newcommand{\y}{\mathbf{y}}\newcommand{\R}{\mathbb{R}}$The torus can be parametrised by pairs $(\theta,\x)$, where $\theta \in [0,2 \pi)$, and $\x \in \R^2$, $|\x| = 1$, i.e. $x$ is a point on the standard unit circle. Here $\theta$ represents the “major” angle, i.e. the angle on the large-diameter circle round the origin, and $\x$ represents the point on the small-diameter circle.

The fattened band is exactly the same idea, but a little tricker to write down. Let $Q_\theta$, for $\theta \in [0,2\pi]$, be the unit square in $\R^2$ rotated by $\theta/4$ radians. So as $\theta$ varies from $0$ to $2\pi$, $Q_\theta$ rotates by a quarter-turn, and ends up back where it started. Now the fattened band can be parametrised by pairs $(\theta,y)$, where $y \in Q_\theta$.

Now one direction of the homemorphism, from the band to the torus, is described easily by sending $(\theta,\y)$ to $(\theta, \frac{\y}{|\y|})$.

Once one has the homeomorphism, it follows that since the torus is orientable, so is the fattened Möbius band.

One take-home here is the Möbius band is very different from the fattened Möbius band; the non-orientability of the former doesn’t imply anything about the latter.

Peter LeFanu Lumsdaine
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    To emphasize your last comment, the boundary of a compact region in $\Bbb R^3$ (i.e., compact $3$-manifold with boundary) is always orientable. :) – Ted Shifrin Nov 25 '14 at 14:56