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Mapping contours over normal Riemann surfaces

  1. Apr 21, 2012 #1
    Hi,

    Can someone here help me understand how to illustrate maps of analytically-continuous paths over algebraic functions onto their normal Riemann surfaces? For example, consider
    [tex]w=\sqrt{(z-5)(z+5)}[/tex]
    and it's normal Riemann surfaces which is a double covering of the complex plane onto a single Riemann sphere. Now suppose I take the figure-8 path given parametrically as
    [tex]z(t)=7 \cos(t)+7 i \sin(t)\cos(t)[/tex]

    which encircles both branch-points, and allow that path to traverse over the function in an analtyically-continuous manner. What then does the image of that path over the function look like when mapped onto it's normal Riemann surface?

    Is there already existing sofware freely available for doing this one and higher genus surfaces? My objective in all this is to understand how to compute
    [tex]\oint_{z(t)} w dz[/tex]
    and other abelian integrals.

    Thanks,
    Jack
     
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 22, 2012 #2
    I'm making progress with this guys. Very difficult though for me to visualize it. The plot below is my initial attempt. The Red points are the points at infinity for each covering. The two blue points are the singular points at -5 and 5, the blue circle is the branch-cut between coverings, and my red countour is a deformed but equivalent z(t) over one of the two possible paths of z(t) over the surface. Then I believe
    [tex]\mathop\oint\limits_{z(t)} wdz=\mathop\oint\limits_{\text{red}} wdz=2\pi ir_{\infty}=\pm 25 \pi i[/tex]

    where the [itex]r_{\infty}[/itex] is one of the two residues at infinity since the residues around the two finite singular points is zero. I think this is right but not sure. Can someone let me know if I'm on the right track?
     

    Attached Files:

    Last edited: Apr 22, 2012
  4. Apr 23, 2012 #3
    I'm new to Riemann surfaces, but if there are indeed 2 branching points at -5 and 5, and the degree of w is 2, then the genus, g, for the surface is 2 = 2(g+2-1). So, g = 0. We're definitely looking at a sphere! If you're certain the path is figure 8 and encircles the branching points, I think we indeed pick up an essential residue when we cross the branch-cut. Your figure seems to have a curve that encircles one branch point and one point at infinity, which I didn't think that's the curve you wanted? Sorry I can't be more helpful, I'm still learning this stuff too.
     
  5. Apr 24, 2012 #4
    Here's the contour over the imaginary surface of w in the z-plane. Since there are no more finite singular points, I can stretch the contour out along the purple surface towards infinity, then loop it back towards the singular point at z=5 over a [itex]4\pi[/itex] loop. That deformed contour is, I believe, equivalent to my red contour over the double-covering of the Riemann sphere which is this function's Riemann surface. But I just crudely drew that yellow contour over where I think it's going over the surface. Not sure about that.

    Surely there must be some type of transformation which maps a double-covering of the z-plane by this function onto a double-covering of it's Riemann sphere so that I can analytically compute the exact path of my contour over this Riemann surface.

    Also, the reason I'm asking is that I believe in order to evaluate abelian integrals of the form
    [tex]\mathop\oint\limits_{\gamma} \eta(z,w)dz[/tex]
    we need to determine what type of region (simply-connected, multiply-connected), the contour is enclosing over it's Riemann surface. Anyone here familiar with these integrals can tell me if that is indeed the case?

    My second version of the contour over the Riemann sphere is shown below. Can someone tell me if that red contour is equivalent to my figure-8 contour over the z-plane. Note the contour is now enclosing a simply-connected region of the Riemann surface so therefore I can write:
    [tex]\mathop\oint\limits_{\text{red}} wdz=2\pi i\left(\text{red residue}+\text{blue residue}\right)=2\pi i(12.5+0)=25\pi i[/tex]

    See what I mean guys? It makes it easy to compute the integral don't you think? May not have all this correct though. I'm just a novice although I do know the integral is [itex]\pm 25\pi i[/itex] depending on which infinity is being encircled.
     

    Attached Files:

    Last edited: Apr 24, 2012
  6. Apr 28, 2012 #5
    Here's my third attempt. This one is an actual steriographic mapping of the contour over a single Riemann sphere. I would next need to split-open the sphere along the red branch-cut, contort it into a hemisphere, do another one, then glue them together along the branch-cut with the correct polarity so that the mapping of the yellow contour reflects the actual path of the contour over this function. I'm optimistic there is already an exsiting way to do these transformations. Just need to find them. I already checked the path over an orange. Yeah, the kind you eat. It's pretty clear to me now my figure-8 contour is just a closed Jordan curve over this surface enclosing one branch-point, and one infinity point which then agrees with the residue calculations.
     

    Attached Files:

  7. May 8, 2012 #6
    Just wish to follow-up on this. The green plot below is an actual analytic mapping I created of the contour onto this double covering of the genus zero surface for [itex]w=\sqrt{z-1)(z+1)}[/itex]. The red circle is the branch-cut between coverings. Can's see much without the ability to rotate the plot. I call the mapping a linear cross-sectional squashing transformation. What I find most interesting is that a non-Jordan curve in the z-plane is transformed into a Jordan curve on the surface enclosing a simply connected region. Note also that there would be a mirror-image contour of the green contour that would be the contour over the other branch of this function.
     

    Attached Files:

    Last edited: May 8, 2012
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