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I Contour integration over Riemann surface

  1. Jul 23, 2016 #1
    Cauchy integral theorem states that the contour integration of a complex harmonic function along a closed simply connected path=0. What if this simply connected path is drawn over a Riemann surface of function like ##f(z)=\sqrt z##. Will that be possible in the first place? and will the integration over that path yield zero?
     
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  3. Jul 27, 2016 #2

    FactChecker

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    As long as the winding number of the path around the point z=0 is 0, the integral will be 0.
     
  4. Jul 27, 2016 #3
    Is the winding number really 0? If we move counterclockwise over Riemann surface representing the real part of ##f(z)=\sqrt z## we will end on an infinite looping around ##z=0## switching from the positive to the negative roots.
     
  5. Jul 27, 2016 #4

    FactChecker

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    It is not guaranteed that the winding number is 0. That is essential. For closed paths that do not wind around zero, the integral is 0.
     
  6. Jul 27, 2016 #5
    I don`t get it. If a closed path is taken around ##z=0## in the complex flat plane, the integral of ##f(z)=\sqrt z## vanishes. But in my case, what is the value of the integration with a closed path winding around ##z=0## in Riemann surface that describe the same function?
     
  7. Jul 27, 2016 #6

    andrewkirk

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    I think the reference to a 'closed path' is to a closed path in the Riemann surface, not in the domain of the 'function' (by the way ##z\mapsto \sqrt z## is not a function because it doesn't give unique results). I am not sure that there is any closed path on the Riemann surface in ##\mathbb C^2## generated by the relation ##z_2{}^2=z_1## (which is a more careful way of expressing the square root notion), whose ##z_1## coordinate winds around 0.

    If that's correct then the integral of any sufficiently nice function along a closed path on that Riemann surface will be zero.
     
  8. Jul 28, 2016 #7

    Ssnow

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    Hi, I agree with the ideas of @andrewkirk because a closed path that contains the zero on the riemann surface ##f=z^2## seems not viable. I want to be precise, the ##0## must be in the interior of the support of the path and I don't think this is possible.
    In the end the case where ##0## is not inside the path was explained before ...
     
  9. Jul 28, 2016 #8

    andrewkirk

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    It occurs to me that the meaning of the integral is not clear in the question. The Riemann surface ##R## is a one-dimensional complex manifold that is embedded in ##\mathbb C^2##. We can define a function ##f:R\to\mathbb C## and we can define a simple, closed curve in ##R## as a function ##\gamma:[0,1]\to R## that is continuous when we interpret ##R## as a two-dimensional subset of 4D Euclidean space, such that ##\gamma(0)=\gamma(1)##. But I cannot see any obvious interpretation of the notion of a 'line integral' of ##f## along ##\gamma##. Note that ##f## takes two complex arguments, rather than one.

    For a loop integral of ##g:\mathbb C\to\mathbb C## around a loop ##\eta## in the complex plane we have the simple formulation
    $$\oint_\eta g(z)\,dz=\int_0^1 g(\eta(t))\eta'(t)\,dt$$

    There is no immediately obvious (to me) equivalent of this when the function to be integrated has a domain in ##\mathbb C^2## rather than ##\mathbb C##. Some form of complex Jacobean may be needed. But then I have never studied complex manifolds. Perhaps there is a standard definition of a line integral of a function on a complex manifold. Presumably it would use the metric on the manifold. What would the standard metric on ##\mathbb C^2## be? Would it just be
    $$ds^2=dz_1{}^2+dz_2{}^2$$
     
  10. Jul 29, 2016 #9

    Ssnow

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    Sorry I correct myself ##f=\sqrt{z}##
     
  11. Jul 29, 2016 #10

    vanhees71

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    That's tricky ;-). If you integrate over a closed path in the Riemann surface, it gives 0. To demonstrate that on your example take the usual Riemann surface for the square-root function by cutting along the negative real axis. Now a closed curve around 0 must run over the two sheets ("double cover" of the complex plane). Let's take a circle. The first sheet is starting from ##\text{arg} z=-\pi## and runs to ##\text{arg} z=\pi##. On this sheet you have along the unit circle
    $$\sqrt{z}=\sqrt{\exp(\mathrm{i} \varphi)}=\exp(\mathrm{i} \varphi/2), \quad \varphi = \text{arg} z \in (-\pi,\pi)$$
    Then you enter the 2nd sheet of the Riemann surface, where you have
    $$\sqrt{z}=-\sqrt{\exp(\mathrm{i} \varphi)}=-\exp(\mathrm{i} \varphi/2) \quad \varphi =\text{arg} z \in (-\pi,\pi)$$
    Only then you close the circle and thus you get trivially 0.

    Equivalently you can also integrate ##\exp(\mathrm{i} \varphi/2)## along ##\varphi \in (-\pi,3 \pi)##, which covers also the two branches of the square-root function, and it gives of course also 0.
     
  12. Jul 30, 2016 #11
    My understanding is that Riemann conceived of his idea in order to devise a coordinate system in which multi-valued functions become single-valued so that the foundation of Complex Analysis of single-valued functions could be applied to multivalued functions. The normal Riemann surface for the square root is a Riemann sphere with a double covering of the complex plane mapped onto it conformally with every point on the sphere equal to ##(z,\sqrt{z})##. And therefore for the function ##f(z)=\sqrt{z}## over this new coordinate system, we have the simple linear function ##f(z,w)=w## which is an entire function zero over any closed contour over its Riemann surface. Contrast this with ##f(z)=\sqrt{z}+\frac{k}{z}##. What do you suppose the integral of this function over a closed contour (winding number of one) about the point ##(0,\infty)## over it's Riemann surface is? ##4k\pi i##? What about ##f(z)=\sqrt[n]{z}+\frac{k}{z}##? ##2nk\pi i##? What might the Riemann surface of ##\sqrt[n]{z}## even look like? I imagine it like one of those cantelopes with deep furrows with each section between the furrows corresponding to one copy of the z-plane. Once this perspective is taken, we can then re-cast the Residue Theorem for multivalued functions over their normal Riemann surfaces, then Laurent's Theorem and then all the other theorems for single-valued functions.
    For example, the Laurent series for ##\sqrt{z}## is ##\sqrt{z}##, the series for ##\sqrt{z}+\frac{1}{z}## is ##\frac{1}{\left(\sqrt{z}\right)^2}+\sqrt{z}##. That was easy. What would the Laurent series for ##\frac{1}{\sqrt{z(z-1)(z-2)}}## be about the point ##(0,0)## over its Riemann surface? But if you know that all the theorems of single-valued functions apply to multivalued functions over their Riemann surface, how might you calculate the coefficients for this series expansion? Use Laurent's Theorem.
     
    Last edited: Jul 30, 2016
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