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Conditions for using Stokes' Theorem

  1. Jun 3, 2013 #1
    I'm back with more questions! :approve:

    I'm wondering what conditions must a manifold satisfy to be able to use Stokes' Theorem. I understand that it must be orientable, but does it have to necessarily be smooth?

    I tried to see if it was possible to prove Cauchy's Residue Theorem and Cauchy's Integral Formula using Stokes' Theorem, but I got stuck with results that don't make sense. Both require that an integrand can be meromorphic, so I'm not sure that Stokes' Theorem will necessarily apply to nonsmooth manifolds.
     
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  3. Jun 3, 2013 #2

    WannabeNewton

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  4. Jun 3, 2013 #3
    I didn't know. :tongue:

    I figured it out, though. I think it's cool, so I might as well share it here to explain what I was trying to do to see if there's any more information to be had.

    Let ##z\in\mathbb{C}:z=x+iy## and ##f(z)=u(x,y)+iv(x,y)## be a meromorphic function such that f is undefined for all ##z_0\in\mathcal{A}\subseteq D:\mathcal{A}=\{a_1, a_2, \cdots , a_n\}##. Then, what I wanted to say was

    $$\oint\limits_{\partial D}f(z) \, dz = \iint\limits_D d(f(z) \, dz) = \iint\limits_D \left(\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)+i\left(\frac{\partial v}{\partial y}-\frac{\partial u}{\partial x}\right)\right)dy\wedge dx$$
    which makes the Cauchy Integral Theorem rather trivially evident.

    However, this is not necessarily the most fun to use to prove the more general Residue Theorem.

    What I noted instead was that I could make the integral become

    $$\iint\limits_{D\setminus\mathcal{A}} \left(\left(\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\right)+i\left(\frac{\partial v}{\partial y}-\frac{\partial u}{\partial x}\right)\right)dy\wedge dx + \sum_{k=1}^{n}\left[\, \oint\limits_{\beta(a_k)}f(z) \, dz \,\right]$$
    where ##\beta(a_k)## traces an infinitesimally small circle around ##a_k##.

    I'm working in a similar way to get the Cauchy Integral Formula.
     
  5. Jun 4, 2013 #4

    lavinia

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    Stokes theorem applies to smooth singular chains.These exist on both oriented and non-oriented manifolds.

    The real and complex parts of a holomorphic differential are closed. This just restates the Cauchy-Riemann equations. Stokes theorem tells you that the integral of the differential must be zero around any piecewise smooth closed curve within its domain of holomorphy.

    Except for the term of degree minus 1, the terms in a Laurent series when multiplied by dz are all exact forms and so by Stokes Theorem integrate to zero around a circle. So the integral of a meromorphic function around a pole is its residue by Cauchy's theorem.
     
    Last edited: Jun 5, 2013
  6. Jun 5, 2013 #5
    I understand everything else about your post, but this is new to me. I was aware that Stokes' Theorem had something to do with chains, but I was unaware that you could use that fact to apply it to non-orientable manifolds.

    But then, I don't understand how that would work, since all of the integration I've done preserved some form of orientation (id est, ##\int_{a}^{b}f^\prime(x) \, dx = \int\limits_{[a,b]} f^\prime(x) \, dx = f(b) - f(a)##). Can you please elaborate how you would integrate over a non-orientable manifold using Stokes' Theorem?
     
  7. Jun 5, 2013 #6

    WannabeNewton

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    See chapter 4 of Spivak's Calculus on Manifolds.
     
  8. Jun 5, 2013 #7

    lavinia

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    i am sorry I was so brief. A smooth singular simplex has two natural orientations. One can always integrate a differential form over an oriented simplex whether it is in an oriented or unoriented manifold.

    A smooth singular chain is a finite formal sum of oriented smooth simplexes. The integral of a differential form over the chain is the sum of its integrals over each oriented simplex in the chain.

    When you talk about "integrating over a manifold" you mean expressing the fundamental cycle of the manifold as a smooth singular chain then integrate over that chain. An unorientable manifold does not have a fundamental cycle so there is no smooth singular chain to integrate over.
     
    Last edited: Jun 5, 2013
  9. Jun 6, 2013 #8

    lavinia

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    As an after thought here is something to think about.

    Suppose you have a function on a non-orientable manifold. Pull this function back to the orientable 2 fold cover and multiply it by any orientation form you want. Integrate this then divide by 2.
     
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