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How to create normal Riemann surface for cubic?

  1. May 17, 2012 #1
    Given the function:
    [tex]w=\sqrt[3]{(z-5)(z+5)}[/tex]
    which is fully-ramified at both the finite singular points and at infinity, how does one create the normal Riemann surface for this function? It's a torus but I do not understand how to map a triple covering onto the torus so that it's fully-ramifed at -5, 5 and at infinity. Maybe though I just don't understand what I'm doing. Is that it?
     
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  3. May 17, 2012 #2

    mathwonk

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    if the riemann surface: (projectification of) w^3 - z^2 + 25 = 0 is a torus, and is totally ramified at each branch point, that means the torus maps triply onto the z line, by the projection map (z,w)-->z, with only one point over each of the branch points.

    If you are right and there are 3 ramification points of index 2 each, the total ramification index is 6, so the genus of the riemann surface is g, where 2g-2 = -6 + 6 = 0. so indeed g = 1.
     
  4. May 17, 2012 #3
    May I ask how then? That is, how exactly do I cover the torus with three copies of the complex z-plane, or rather three Riemann spheres, in such a way that a small circle (winding number one) around the places (-5,w), (5,w) or (infty,w) on the torus is mapped by the inverse mapping, to small circles around that point in each of the coverings?

    That is, should not a 2pi circle around these singular points on the torus get inverse mapped to a 6pi circuit corresponding to the three coverings?

    Or is that interpretation incorrect?
     
    Last edited: May 17, 2012
  5. May 17, 2012 #4

    mathwonk

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    you seem to have it backwards. the sphere does not cover the torus, rather it is opposite, the torus covers the sphere.

    Indeed any branched cover of two surfaces LOWERS genus.
     
  6. May 18, 2012 #5
    Ok thank you. May I ask if my following understanding of this problem correct or incorrect:

    The Riemann surface for this function is a torus. There should be a mapping then from the torus to a triple covering of the Riemann sphere. Every place on the torus (z,w) then corresponds to one point on that triple mapping. Then I should be able to segregate the torus into three (connected) domains, each of which maps a place (z,w) in each domain to a single point of it's covering of a Riemann sphere. Then I wish to know where precisely are those domains on the torus. But these domains can't be just anywhere. Rather, they have to border each other in such a way which preserves the geometry of the function, in particular, small circular contours around the three singular places (-5,w), (5,w) and (infty,w) on the torus traverse each domain only once so as to preserve the (cycle-3) ramification geometry around these points.
     
    Last edited: May 18, 2012
  7. May 18, 2012 #6

    mathwonk

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    well that is the old fashioned way of looking at them. i.e. just because there is a 3 to 1 map from the torus to the sphere does not mean it is easy to subdivide the torus into three pieces such that each piece maps 1 to 1 to the sphere. it is possible, but hard to visualize.

    it is based on the fact that an unbranched 3 to 1 cover of a simply connected set in the plane must break up into three separate pieces, each mapping one to one onto that simply connected set.

    so let T be the torus defined by your equation above and T-->C the 3:1 map from the torus to the complex plane, taking (z,w) to z.

    now this map is branched, not unbranched, so we make it un branched by removing the three branch points z = 5, -5, infinity.

    but now the target C - {5,-5} is not simply connected, so we make it simply connected by removing a line segment joining 5 and -5, and also the rest of the positive x axis in the z plane.


    Then when we also remove from the torus, the curve mapping onto this curve in the z plane, then the rest of the torus does break up into three pieces, each one mapping isomorphically onto the z plane with the interval [-5, + infinity) removed.

    if you want to picture that on the torus, you have to picture three points on the torus, joined by 6 curves. one point, over the point z=5, is an endpoint of all 6 curves, and each other point is joined to that one by three of the curves. then these 6 segments will separate the torus into three pieces, each looking like the slit z plane.

    right now i don't see it.
     
  8. May 18, 2012 #7
    Thanks for replying. May I suggest a possible covering at the risk of offending quasar who told me earlier that a horn torus is not a Riemann surface?

    Can we not allow a (normal) ring-torus approach in the limit, a horn torus? In that way we are always dealing with a Riemann surface. Suppose just for the sake of argument we can, then I believe I can cover the limiting case (a horn torus) with three Riemann spheres, each cut along two branch cuts: (infty,-5) and (5,infty). I've shows the blue, red and green surfaces for these Riemann surfaces deformed in the shapes I want below and then their coverings over the horn torus. The point infinity is at it's center and at two points along the perimeter are the points -5 and 5. It's not evident in the picture, but the three colored surfaces meet together at a vertex only at the singular points thus allowing the cycle-3 ramification at these points.

    Is this still not acceptable? Note in particular the green surface. Am I incorrect in suggesting I can (continuously) deform a Riemann sphere cut along the two branch-cuts into that shape?

    Do I absolutely need to use a ring-torus? No offense to Quasar.
     

    Attached Files:

    Last edited: May 18, 2012
  9. May 18, 2012 #8

    mathwonk

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  10. May 19, 2012 #9
    Yes, I read your notes back then and have them on my machine. Just noticed you went over y^3=1-x^2. Unfortunately for me, it's tough to understand. My fault entirely as I'm no doubt in over my head with this.

    Maybe one day if my circumstances change, I'd like to enroll in a class or two. That would help I think.

    Thanks for your help! :)

    Also, for the record, I don't think the green surface in my plots above is correct. That's ok though. I realize that sometimes you have to go through the wrong ones to get to the right one. :)
     
    Last edited: May 19, 2012
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