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It is well known that cutting a Möbius strip "in half" down the middle results in a band with two twists, homeomorphic to a cylinder. See this question for example.

If instead, one begins the cut one third of the way along the breadth of the Möbius strip, and maintains this distance, the resulting shape is composed of an interlinked Möbius strip and doubly-twisted cylinder. See this question for example.

The familiar argument proceeds by identifying the opposite sides of a square, to obtain a Möbius strip, and trisecting said square. enter image description here

This explains the Möbius strip's dissection into a cylinder and a Möbius strip. It does not explain why the resulting surfaces are interlinked. It is clear that this property is dependent on the ambient space being $\mathbb{R}^3$.

How does one prove that the resulting surfaces are interlinked?

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  • $\begingroup$ The figure in the second link seems to have shown the interlink. Cut one third along the breadth, and the strip with a twin is caught in the middle of the two connected strips. As a result of this, there should be a interlink. Do I miss anything? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ @JCQ I do not understand. There are only two strips, which strip has a twin? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ What I am referring to is described the same as on the figure. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ Are you presenting a formal argument? I am fine with an argument from a diagram, but I think some delicacy is required to do so rigourously. Why must the internal strip be caught? $\endgroup$ Commented 14 hours ago

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I can't furnish a formal proof, but perhaps this animation can help you "see" why the two bands must be interlinked. Notice how the "marginal" purple band essentially goes through the loop created by the central band.

enter image description here

I'm not a topologist but upon further reflection, can offer the following idea. In the parametrization $$\vec m (u,v) = \begin{bmatrix} (1 + u \cos \tfrac{v}{2}) \cos v \\ (1 + u \cos \tfrac{v}{2}) \sin v \\ u \sin \tfrac{v}{2} \end{bmatrix}$$ for $(u,v) \in [-1,1] \times (-\pi, \pi]$, consider the curves $$\gamma_1 = \vec m(0,v), \quad v \in (-\pi, \pi]$$ which is the unit circle in the $xy$-plane, and $$\gamma_2 = \vec m(1,v), \quad v \in (-2\pi, 2\pi]$$ which is the boundary of the Mobius band. Since the central portion of the band contains $\gamma_1$, and the purple "marginal" band contains $\gamma_2$, then the latter must be interlinked with the former if $\gamma_2$ intersects the $xy$-plane in exactly two points, once inside the unit disk, and once outside. To this end, we solve for the cases where $z = 0$ for $\gamma_2$: i.e., $\sin \frac{v}{2} = 0$, or $v = 2\pi k$ for some integer $k$. Thus the intersection points are $(0,0,0)$ for $k$ odd, and $(2,0,0)$ for $k$ even, and since these are unique, it follows that the purple band is interlinked with the blue.

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  • $\begingroup$ Thank you for your answer, the animation is pretty. I will not mark this as accepted, because I would prefer a more formal proof. I think it is intuitive that the band's are interlinked, indeed one can verify this physically with paper and scissors, but I would like to put this on formal footing. $\endgroup$ Commented 14 hours ago
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    $\begingroup$ @Kepler'sTriangle While there are some minor gaps, I have amended the above to outline a line of reasoning that might be more persuasive. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ Thank you for the further comments. That is an elegant argument, and indeed it is my favourite among those I have seen here. $\endgroup$ Commented 14 hours ago
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We can define two curves: the boundary $C_1$ of the original Möbius strip, and the line $C_2$ down the middle of both the original Möbius strip and the new smaller Möbius strip. To determine whether the two pieces are linked, it's enough to determine whether $C_1$ and $C_2$ are linked.

We can take any diagram of a Möbius strip (I modified the one photographed by David Benbennick on Wikipedia) and draw $C_1$ and $C_2$ on top of it, taking care only to keep track of which strand is on top at every crossing, and obtain a link diagram:

Möbius strip from Wikipedia, with C1 and C2 drawn on top of it

I admit that the MS Paint method of generating the link diagram is somewhat informal. I can imagine a formal but tedious method, which is to take a parameterization of some specific embedding of the Möbius in $\mathbb R^3$, which gives us parameterizations of the two curves, and then to project it down onto a plane from a point; at each crossing, we could compute which piece of a curve is closer to the point of projection.

To determine that the diagram represents a Hopf link in particular, we could perform some Reidemeister moves on this diagram to reduce it to a standard diagram of the Hopf link with two crossings. But to determine that $C_1$ and $C_2$ are linked somehow, it's enough to check that this diagram is not tricolorable: it is impossible to color each strand of the diagram, using at least two colors, such that at every crossing, either one or three colors are present. (There are $8$ strands total in the diagram above, so this is not too hard to check by a mix of deduction and brute force.)

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    $\begingroup$ I think you can obtain the link diagram by starting with the 2-d picture OP posted and then drawing paths from the seconds that are linked together. From there you can use your argument. This doesn't require any specific embedding in 3-d. $\endgroup$ Commented 4 hours ago

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