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Consider the Möbius band, and take the middle circle in it (so that it deformation retracts onto it). Glue the upper boundary of a cylinder through it. This gives a space that deformation retracts into the Möbius band and into the cylinder. How can one show this doesn't embed into $\Bbb R^3$?

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    $\begingroup$ I wonder if this question will get closed for lack of effort. $\endgroup$ – Pp.. Jan 14 '15 at 20:10
  • $\begingroup$ You've done something different with your hair. $\endgroup$ – Will Jagy Jan 14 '15 at 20:14
  • $\begingroup$ @PedroTamaroff It is a joke. The post looks better as it is. $\endgroup$ – Pp.. Jan 14 '15 at 20:16
  • $\begingroup$ You can take a 3-D frame with two vectors, say $x,y$, tangent to the Moebius and one, say $z$, perpendicular to it. Let us choose it such that $y$ is perpendicular to the cylinder. We can move the frame along the Moebius (keeping all the conditions above) and without crossing the cylinder, until we return to the same point but the $z$ now points in the opposite direction. If we follow the shadow of the 2D frame $x,z$ we see it moving on the cylinder, without degenerating, but changing orientation. $\endgroup$ – Pp.. Jan 14 '15 at 20:54
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    $\begingroup$ @Laters It is not a manifold. Pick a small neighborhood of a point about the central circle of the mobius band. It's homeomorphic to T $\times I$, where by T I mean the honest to god letter T. $\endgroup$ – user98602 Jan 15 '15 at 17:50
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Assume that such an embedding exists.

Call $C\subset \Bbb R^3$ the core of the Möbius band, and $C_+\subset \Bbb R^3$ the other boundary component of the cylinder.

By compactness we may assume that the Möbius band $M$ does not intersect $C_+$ (if it does, replace $M$ with a neighbourhood of $C$ in $M$ that doesn't).

Under this condition, $\partial M$ is homologous to $2C$ in $H_1(\Bbb R^3\setminus C_+)$ (all homology groups are with integer coefficients here). Therefore:

$$\operatorname{lk}(\partial M,C_+)=0 \pmod 2$$

Now since $M$ is a Möbius band, $$\operatorname{lk}(\partial M,C)=1 \pmod 2$$

Apply now Alexander duality modulo $2$: $$\partial M\cap \operatorname{Cylinder} = \operatorname{lk}(\partial M,\partial(\operatorname{Cylinder}))=1 \pmod 2$$

The intersection number is $1$ modulo $2$, therefore there has to be at least one (geometric) intersection point between the cylinder and the boundary of the Möbius band.

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