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I've been watching Professor Ted Shifrin's brilliant lectures on Stokes's Theorem, and I had a few questions that I don't think were answered:

  1. Rectangles in the plane are not manifolds with boundary as the lectures define them, so I don't think Stokes's Theorem (as discussed in the lectures) would apply to them. But Professor Shifrin said that Stokes's Theorem was a generalization of Green's Theorem, which was grounded in rectangles in the lectures. How do these two ideas hold at once?

Professor Shifrin provided an example of a region on which the integral of a differential form on which wasn't parametrized "exactly," but the integral over the small segment canceled out. Then, he asserted that a similar cancellation would not occur when integrating over a Möbius strip. (The relevant portion of the lecture starts here: https://youtu.be/5k13cowATAw?t=644)

  1. Why doesn't a similar cancellation occur when integrating over a Möbius strip? I would think that the segment would have opposite orientation when parametrized each time.

  2. Why does the integral on that segment even matter? (I know this is probably a stupid question.) On first glance, my thought was that this boundary region had 2d volume zero, so it wouldn't matter in the grand scheme of integration.

  3. Professor Shifrin said that the Möbius strip could be parametrized locally but not globally. Why don't we define the integral over the whole manifold to be the sum of the integrals along each coordinate patch, similar to what we do for piecewise smooth manifolds?

Thank you so much for your help!!

RHyp
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    @TedShifrin care to take this one? You have a fan – Ninad Munshi Mar 25 '23 at 03:54
  • This is a bad answer (I guess that's why it's a comment), but the answer to 1 is basically that the corner points don't matter. You either amend the theorem statement to allow for piecewise smooth curves, or you have to do four separate integrals. Alternatively, conceptually, you can modify a rectangle by an arbitrarily small amount to smooth out the corners, and, since your functions/vector fields are well-behaved, no issues arise. (More generally you can explore the whole topic of manifolds with corners, but to be honest this probably won't be enlightening to delve into) – pancini Mar 25 '23 at 04:20
  • See Areas of polygons in the the language of Riemannian geometry for an elaboration of the previous comment, to address your (1). For the others, probably better to wait for Ted to respond since he knows what he said (and I haven't seen this particular lecture when I studied this stuff, and don't feel like watching now) :) – peek-a-boo Mar 25 '23 at 04:31
  • Please ask one question at a time. – Shaun Mar 25 '23 at 05:09

1 Answers1

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I think somewhere in there I did mention that Stokes's Theorem applies more generally to manifolds with corners but that we would not worry about proving those cases. There are certainly lots of examples that show up (like a hemisphere with a disk attached to its boundary).

The Möbius strip can certainly be parametrized globally (just as a circle can be). Indeed, one exercise in the text gives the parametrization. What I believe I said was that it could not be oriented globally, even though it can be oriented if you remove an appropriate line segment. Although we can easily define the integral of a $k$-form over a parametrized $k$-manifold by pulling back, to integrate a $k$-form over a general $k$-dimensional manifold requires orientation; if you check the definitions carefully, the different parametric patches (or charts, if you prefer) have to be compatibly oriented in order for the partition of unity argument to give you something well-defined.

A question worth thinking about (which I don't believe I raised in lecture) is this: Suppose you have a vector field defined on $\Bbb R^3$ (just for simplicity). Does its flux across a Möbius band make sense? What about finding its work done around the boundary of the Möbius strip (suitably oriented)? Then in what sense can you say that the flux of its curl across the Möbius strip gives that work? Indeed, what happens when you cut the Möbius strip and get an orientable surface? But, as you suggested in your second question, doesn't the cut show up twice (rather than cancelling out) when you do the line integral?

Ted Shifrin
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  • Thanks for the answer! You mentioned manifolds with corners later, and you clearly answered my fourth question. However, I still don't understand why we don't define an integration over a Möbius strip using an infinitesimally good parametrization similar to the concentric circle example (or would the coordinate patches not work out here, either?). Additionally, while I know that the integral over the Möbius strip double-counts a portion (rather than cancelling out), I don't understand how this is different from the concentric circle example. This might be the key to the question you posed. – RHyp Mar 25 '23 at 18:56
  • Also, I had some more questions after watching a later lecture. Could I email you about those? – RHyp Mar 25 '23 at 19:05
  • You can define the integral of a function over a non-orientable surface provided you still know what $dA$ should mean, but you need an orientation to define flux or the integral of a $2$-form. Ultimately, even for a parametrized $k$-dimensional submanifold, we can define the integral by pullback, but why is that well-defined independent of the parametrization? Here you will run into trouble with something like the Möbius strip. Great questions! Yes, you may email (or — if they are of general interest — post here). – Ted Shifrin Mar 25 '23 at 19:48
  • I think I'll continue with this thread here and start another with questions related to a different lecture. I think you can integrate a differential form over the Möbius strip as long as you exclude an infinitesimally small portion of it. Why can't we define the integral of a differential form over the Möbius strip this way? Related to this, you justified the integral of a 2-form in the concentric circle example by using a cancellation argument. Why does the non-orientability of the Möbius strip prevent such an argument? Why does the integral double and not cancel? – RHyp Mar 25 '23 at 20:22
  • Go back to the definition of integration of forms and understand why you need orientation for this to make sense. Of course, you can orient a piece of a Möbius strip, but you cannot orient the whole thing. – Ted Shifrin Mar 25 '23 at 20:49
  • After rereading section 8.1 of the book, I now understand why manifolds $\textbf{must}$ be oriented in order for integration to make sense. Now, can you tell whether a manifold is orientable from a parametrization? – RHyp Mar 25 '23 at 22:45
  • Nope. It’s compatibility of different parametrizations (defined on open subsets of $\Bbb R^k$, not on arbitrary subsets). This turns out to be a huge point, so that we have overlaps (hence compatibility) on open sets. – Ted Shifrin Mar 25 '23 at 23:22
  • That is fascinating. Thank you so much!! – RHyp Mar 26 '23 at 01:08