The self contained version of this question is: "Given a Möbius strip $M$ do there exist two curves $\gamma_1,\gamma_2$ in $M$ such that $M\setminus\gamma_1$, $M\setminus\gamma_2$ are orientable and with different area?". The area form on $M\setminus\gamma_i$ is the natural one as a submanifold of $\mathbb R^3$. For more context see below.

I was trying to answer this question with a response along the lines of "there is no global area form on a Möbius strip, and therefore no way to define a total area". The obvious objection is that if I take a slip of paper with a certain area, the total area along which one can move on the slip of paper is simply the area of the front of the slip plus the area of the back. This is essentially the notion that OP suggests in asking whether the Möbius strip will have area $4\pi$.

I wanted to make the further counter-argument that this is not an area in the normal sense, since we are actually adding areas of a pair of orientable submanifolds of our choosing to obtain a number which we'd call the "total area" of the Möbius strip (in fact we are adding the area of the same submanifold twice since a mathematical strip does not have thickness, but this doesn't change the underlying question). My differential geometry is a bit rusted, so I consulted the internet myself to be able to give a more precise answer. I eventually found exactly what I was looking for here. So now I only needed to concoct a curve along which to cut the strip so as to get an area different than $2\cdot(\text{one side of the square without identifying})$. But I'm stumped!

  • $\begingroup$ If you cut a mobius band certainly it have an area so the question is about the definition of area that you are using. $\endgroup$ – Masacroso Sep 26 '16 at 23:04
  • $\begingroup$ @Masacroso The definition of the area of a cut $M\setminus\gamma$ is the area given by the inherited metric as a submanifold of $\mathbb R^3$ with the Euclidean metric. The question is whether or not two different curves exist with the respective cuts having different total area, as a priori it depends on $\gamma$ as suggested in the linked MO post. $\endgroup$ – GPerez Sep 26 '16 at 23:12
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    $\begingroup$ Any piecewise-$C^1$ curve you take in the Möbius strip will have $\mathscr H^2$-measure zero, and the area of the Möbius strip, after all, is its $\mathscr H^2$ measure. ($\mathscr H^2$ is $2$-dimensional Hausdorff measure.) If you take a "space-filling curve" on the Möbius band, then all bets are off. $\endgroup$ – Ted Shifrin Sep 26 '16 at 23:53
  • $\begingroup$ @TedShifrin Certainly, but I could do that on an orientable surface as well. I would like to exploit the non-orientability of the strip. $\endgroup$ – GPerez Sep 27 '16 at 7:42
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    $\begingroup$ Well, I gave you one answer with Hausdorff measure. Another is to think of the area density given by taking the absolute value of an area $2$-form. Again, I reiterate that I believe that it is with fluxes that you'll find the problem. $\endgroup$ – Ted Shifrin Sep 27 '16 at 17:12

Just to close the circle, I think the best answer (from a differential-geometric point of view) is one that Ted Shifrin mentioned in comments. A nonorientable manifold like the Möbius strip has a perfectly good notion of area; you just can't compute it using differential forms. Instead, what's needed is a density, which can be integrated on any manifold. For a manifold (or manifold with boundary) with a Riemannian metric (such as any submanifold of $\mathbb R^n$ with the induced metric), the appropriate object is the Riemannian density, which is the unique positive density that yields the value $1$ whenever it's applied to an orthonormal basis. This is explained in my Introduction to Smooth Manifolds (2nd ed.), pp. 427-434.

  • $\begingroup$ That book was a treasure when I studied Differential Geometry, I'll definitely go back to it. Thanks for the response! $\endgroup$ – GPerez Sep 27 '16 at 23:37
  • $\begingroup$ I also realized that the pretext of of the MO post was to (say, obstinately) use a differential form defined on a submanifold (or many). In this case, surely we can make the computed area depend on the choice of submanifold? $\endgroup$ – GPerez Sep 27 '16 at 23:44

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