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Differential Geometery: Images of Gauss Maps

  1. Jun 3, 2008 #1
    1. The problem statement, all variables and given/known data

    this is a 2 part problem

    let G: T-----> (Sigma) be the gaus map of the torus T derived from its outward unit normal U. What are the image curves under G of the meridians and parallels of T? What points of (Sigma) are the image of exactly two points of T?


    Let G: M---->(Sigma) be the Gauss map of the saddle surface M: z=xy derived from the unit normal. What is the image under G of one of the straight lines, y= constant in M? How much of the sphere is covered by the entire image G(M)

    2. Relevant equations

    Part 1:

    Unit normal of z=xy

    U= [-fxU1- fyU2+U3]/ (1+ (fx)^2 + (fy)^2)^1/2

    3. The attempt at a solution

    I dont know how to do any rigorous proofs for these, but i am not sure we are suppose to.

    Ok for part 1 here is what i think.

    I think if you travel along the surface of the torus, along the meridian, you it will be mapped to a circle (in the xy plane). if you travel along the meridians you will get a circle in the xz plane.

    And the points in sigma thare are exactly 2 points in T are the North and south poles of the unit sphere.

    And for part 2 i am completley lost, am i even on the right track for part one?

    any help is appreciated.
  2. jcsd
  3. Jun 3, 2008 #2
    can anybody help with this?
  4. Jun 3, 2008 #3


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    Science Advisor
    Homework Helper

    You didn't really give us an exact notion of how stuff is oriented, but offhand, I would say meridians of the torus map to meridians of the the sphere. And parallels to parallels. I would also say that poles of the sphere are the only points which are NOT twofold images of points on T. There's an infinite number of points mapping to them. Do you see that picture?
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