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Math Help - Gauss Map on a Regular Surface

  1. #1
    Senior Member slevvio's Avatar
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    Gauss Map on a Regular Surface

    Let S\in \mathbb{R}^3 be a connected regular surface. Let N:S \rightarrow S^2 be a Gauss map. I am trying to show that -N is the only other Gauss map.

    Let \tilde{N} :S  \rightarrow S^2 be another Gauss map. Is this argument OK?
    For p \in S,

    \tilde{N}(p) = f(p)N(p) where f(S) = \{-1,1\}, so

    \tilde{N} continuous  \implies f continuous \implies f locally constant \implies f constant by connectedness. This means \tilde{N} = +N or \tilde{N} = -N.

    Thanks for any advice on this!
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  2. #2
    Super Member Rebesques's Avatar
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    Quote Originally Posted by slevvio View Post

    \tilde{N}(p) = f(p)N(p) where f(S) = \{-1,1\}, so

    All we know is that |f(p)| will always be 1, it requires a little bit more to assert that f\equiv1 or f\equiv-1.

    To show why this is the case, consider a parametrization X(u,v) for S, with normal N. For a second parametrization X(w,z) with normal M, we will have |N|N=X_u\times X_v=JX_w\times X_z=J|M|M, where J=\frac{\partial (u,v)}{\partial (w,z)} is the Jacobian. Therefore the normal vectors to S at p, will either be equal or opposite, depending on the sign of the Jacobian. This means
    the Gauss map p\mapsto N_p\in S^2 will not alter if J>0, and will only change sign if J<0.
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  3. #3
    Senior Member slevvio's Avatar
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    When I say f(S) = \{ 1,-1\} I mean that's the image, so a priori it can jump around to different values. Thanks for your post
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  4. #4
    Super Member Rebesques's Avatar
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    I don't think you can have that a priori, since it is equivalent to what you are trying to prove.
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  5. #5
    Senior Member slevvio's Avatar
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    Well if I have a continuous unit normal field, then at each point p on the surface there are only two unit normals N(p) and -N(p) which I know in advance
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