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Math Help - Something I am having trouble understanding - complex analysis

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    MHF Contributor Bruno J.'s Avatar
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    Something I am having trouble understanding - complex analysis

    Consider the function f(z)=\frac{1}{\sqrt{1-z^2}} as a multiple-valued function of the complex plane. We can construct its Riemann surface by cutting two copies of the complex plane along the line segment from -1 to 1 and pasting them together along this cut in the usual fashion. What we obtain is a cylinder, to which we could add two points at infinity (one point in each direction of the cylinder, i.e. one point on each of the two sheets); doing that, we'd obtain a surface isomorphic to the Riemann sphere, but let's stick to the non-compactified surface we had before (the cylinder).

    Now it's quite clear in view of the differential equation satisfied by w=\sin z that w has as inverse function the integral z = \int_0^w {f(t)dt}. However we all know \arcsin w is far from a single valued function (unless we consider its image to be the cylinder \mathbb{C}/2\pi \mathbb{Z}, but let's not do that now). Therefore we can expect the multi-valuedness of the function \arcsin w to be explained by the fact that the Riemann surface on which we are integrating has a non-trivial fundamental group; non homotopic paths of integration between 0 and z should yield values of w differing by an integral multiple of 2\pi. I've understood this to be true but I'm having trouble checking it computationally; I would like to check it.

    My question arises also in the context of elliptic function. I have a book in which Jacobi's \mbox{sn} function is constructed as the inverse of the elliptic integral \int_0^w \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}; the author states that integrating around the branch points in various ways results in paving the plane with the period lattice of \mbox{sn}, but does not show it by a computation.
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    MHF Contributor Bruno J.'s Avatar
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    I believe I figured it out. We fix a sheet of f(z) and we integrate f(t)dt on that sheet. The sheet has a cut from -1 to 1 and therefore the integral will be defined up to an integral number of cycles around the cut. But the integral along a simple closed curve whose interior contains the cut is equal to the integral along the cut itself, by Cauchy's theorem (since whatever branch of f(z) which we have is analytic in the region between the two curves). The integral along the cut is \int_{-1}^{1}f(t)dt + \int_{1}^{-1}(-f(t)dt) = 2 \int_{-1}^1 f(t)dt = \pm 2\pi, which is indeed a generator for the group of periods of \sin z.
    Last edited by Bruno J.; April 26th 2010 at 09:21 PM.
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