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Stokes theorem is easy to prove

  1. Jul 23, 2004 #1

    mathwonk

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    i thought stokes theorem (green's thm) was hard after reading it in spivak, who calls it trivial nonetheless. however lang showed it is indeed trivial in his analysis I.

    the same proof occurs in courant. I.e. the point is that the theorem is easy for a rectangle, where it follows immediately from fubini and the fundamental theorem of calculus. the same argument works on a circle.

    then one discusses the concept of a "parametrized" plane region, i.e. the image of a rectangle or circle under a mapping. there is a concept also of "pulling back" a vector field or covector filed, under the mapping.

    Then the pull back of the curl is the curl of the pullback, the pull back of the integral is the integral of the pullback, etc...

    the upshot is that every term in the theorem pulls back faithfully under the parameter map so that once the theorem is proved for a rectangle or circle it is also true for every region which can be parametrized by a rectangle or circle, i.e. essentially any convex region or deformation of one. that does it.

    i apologize for posting a new thread on this question asked elswhere but I could not find that thread again.
     
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  3. Jul 23, 2004 #2
    From the development you mentioned, it seems that it is quite similar to the proof outlined in Spivak's "Calculus on Manifolds".

    I think what Spivak referred to as "trivial" is Stokes' Theorem on chains? He explained that he thinks it is trivial because the theorem is a consequence of good definitions (e.g. pullbacks, forms...).
     
  4. Jul 24, 2004 #3

    mathwonk

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    Wong, you are entirely correct. Thinking back, the IDEA of the proof is exactly the same in all three places I cited. The distinction for me was the presentation. After stating it was trivial for the reasons you gave, Spivak presented it in the general case in n dimensions, and my eyes were spinning from the many ellipses and subscripts.

    Lang on the other hand just wrote it out for n = 2, where it was extremely clear, and then left it to me to generalize. I found this very helpful.

    I learned a lot from Spivak's book though and recommend it highly as a short treatment of several vbl calc.
     
  5. Jul 24, 2004 #4

    mathwonk

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    Wong, Ironically I see now that it was your post in another thread asserting that Stokes is difficult to prove that I was trying to respond to. I got the impression you had been put off, as had I, by the abstract presentation in books like Spivak.
     
  6. Jul 26, 2004 #5
    Oh, it's been hard for me to get to the internet lately...

    Mathwonk, I agree with what you said. But in fact I do not have Lang's book and do not know what he did. Did his just prove "the divergence theorem" and "the curl theorem" in 3D or did he prove the more general Stokes' theorem about the integration on forms?

    As to the proof, I think what is easy is the *steps* of the proof about the *Stoke's Theorem on chains* (what's proved in Spivak's Chapter four), as it involves only some moderate manipulations and the fundamental theorem of calculus. However, I think what is most difficult for people to accept is that a "form" is somehow associated with the concept of volume of a manifold (and thus related to the concept of integration). Once a person gets over that, the rest is more or less a natural consequence.

    As to the general Stoke's Theorem (on manifolds), I think the proof involves more technical points, and that is why I referred to it as "difficult" in another thread. As you said, it's true that one is able to piece together the local result (i.e. Stoke's Theorem on Chains) to form the global one (i.e. Stoke's Theorem on Manifolds), but one has to use a "partitions of unity". As far as I know, what Spivak did in Chapter five of "Calculus on Manifolds" is to generalise the theorem to m-dimensional manifolds embedded in R^n and forms with compact supports. The existence of partitions of unity is natural in this case because the embedded manifold may be considered as a subset of R^n and the existence of partition of unity for the manifold is guaranteed by the existence of partitions of unity of an arbitrary set subbordinate to a certain open cover in R^n, which is proved somewhere in chapter three of Spivak's.

    Besides partitions of unity, one needs to define the notion of a manifold and the tangent bundle, and to extend the definition of tensors and forms, and to show that certain operations are co-ordinate independent.

    So yes, I think technically the proof of Stoke's Theorem is quite tedious. And I will not recommend it to people who just want to apply it... for their health.
     
    Last edited: Jul 26, 2004
  7. Jul 29, 2004 #6

    mathwonk

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    my point is just that if one works through the proof of the ordinary green's thm on a rectangle, the rest is just formalities that enable you to transfer the same proof to a parametrized setting. Emphasizing the formalities (chains, forms, partitions of unity) over the substance (fubini, plus FTC) makes it look harder than it is.
    the proof in 2 dimensions already shows the full idea that is used in higher dimensions.
     
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