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A classical result from vector calculus says that if a vector field $F: \mathbb{R}^3 \to \mathbb{R}^3$ satisfies curl$(F) = 0$, then $F$ is conservative, i.e. $F = \nabla f$ for some $f: \mathbb{R}^3 \to \mathbb{R}$. I am curious about the following rough sketch of a proof of this fact, using Stokes's theorem:

"Proof:" Recall that $F$ is conservative if and only if $\int_C F \cdot dr = 0$ for every closed curve $C$. To show the latter condition, let $C$ be any curve, and choose an oriented surface $S$ with $\partial S = C$(!). By Stokes's theorem,

$\int_C F \cdot dr = \int_S \text{curl}(F) \cdot dS = \int_S 0 \cdot dS = 0.$

Thus $F$ is conservative.

I am wondering how the formal details of (!) would go: namely, how does one show that any closed curve $C$ is the boundary of some oriented surface $S$? This seems intuitively obvious. I would be happy with a proof of this fact even under very nice assumptions on $C$, say if $C$ is a simple, closed, smooth curve.

I haven't been able to find a reference for this, other than a parenthetical mention in Stewart's calculus, which says 'This can be done, but the proof requires advanced techniques.'

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Look for "Seifert surface" in any knot-theory reference.

A quick-and-easy version is this (assuming C is smooth enough):

  1. Project $C$ to the plane so that the image has no cusps or triple points or other weirdness (yeah, that needs Sard's theorem or something to guarantee such a projection exists -- - that's the tough part).

  2. Mark all "crossings" as "overcrossings" or "undercrossings" -- look at any knot-diagram to see what this means.

  3. At each crossing that looks like this:

    ^    ^
    |    |
     \  /
      `/
      / \
     /   \
    

(You can rotate others to look that way) "smooth out" the crossing to look like this:

  ^     ^
  |     |
   \   /
    | |
   /   \
  /     \
  1. The end result is a collection of simple closed curves in the plane. Fill in the innermost one with a disk, then remove it from your drawing; repeat until they're all filled in.

  2. Now stack up those disks, one atop the other, each a little higher off the plane than the one below it, and glue in little "half-twisted ribbons" to put back in the crossings.

The end result is a surface whose boundary is like the curve $C$.

There's a problem: that surface may be nonorientable. And that's where a proper proof of the existence of Seifert surfaces comes in. :)

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  • $\begingroup$ Cool idea! I didn't expect knot theory to enter in here. $\endgroup$ – J Richey Nov 21 '17 at 23:41
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You don't need to prove that it is the boundary of a surface. In fact, most calculus courses won't define what "being a surface" means, which is one of the reasons for which I personally think it would be more pedagogical to focus on "homotopy" terminology and not on "surface" terminology on calculus courses, since what you will generally see is an homotopy (and homotopies are much easier to define and understand elementarily and rigorously at the same time).

That personal output aside, what you need is to build a (smooth) map $H:I \times I \to \mathbb{R}^3$ such that $H(t,0)=c_1(t/2)$, $H(t,1)=c_1(1-t/2)$. This is trivial by convexity of $\mathbb{R}^3$ (note that this point of view also directly points out the reason as to why the topology of the ambient space is relevant for the result to be true) - just pick $H(t,s)=(1-s)c_1(t/2)+sc_1(1-t/2)$.

Then you have two options: prove Stokes' theorem for homotopies, which is a nice exercise through an explicit calculation: Define $I(s)=\int_{H(s,\cdot)} F$, calculate its derivative and use the fundamental theorem of calculus...

...or you can apply the "conventional" Stokes' theorem to the pull-back of the form you are integrating to $I \times I$, which is essentially the same thing but in more generality and using the traditional Stokes' theorem for manifolds with boundaries.

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    $\begingroup$ @JRichey It is not esoteric. The intuition of a surface as a "curve moving through space" is natural. The explicit parametrizations via this point of view makes it also computationally good for a calculus course, meanwhile explaining where the formulas for parametrizations come from (for instance, the parametrization of the sphere is just rotating a curve etc). "Surface" is... what, for an undergraduate? I don't think they are able to answer that satisfactorily. (... continuing) $\endgroup$ – Aloizio Macedo Nov 21 '17 at 23:59
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    $\begingroup$ (...continuing) This is also not an empty rant: I've seen multiple very reasonable doubts stemming from the fact that people simply didn't have a definition. To give an example, I've seen students wondering why they "couldn't apply Stokes' theorem to the disk minus a point: it is a surface, with boundary being the circle!". And they are not wrong. And now you have the job to either explain: "well, the form is not compactly supported..." or shrug away and say that "we are restricting to closed surfaces", and then you have a new term to define (closed, which is not trivial for an undergrad)... $\endgroup$ – Aloizio Macedo Nov 22 '17 at 0:02
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    $\begingroup$ or you could simply say, for instance, that the top curve and bottom curve are not homotopic, and they have a pretty clear visualization of the phenomenom. End of story. $\endgroup$ – Aloizio Macedo Nov 22 '17 at 0:03
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    $\begingroup$ I see! As you say, I suppose the notion of homotopy plays directly into 'parameterized surfaces.' Maybe I will try that the next time I teach this material... $\endgroup$ – J Richey Nov 22 '17 at 0:07
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    $\begingroup$ Not at all. Well-justified mathematical passion should never be put aside! $\endgroup$ – J Richey Nov 22 '17 at 0:14

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