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Thread: Gauss Map on a Regular Surface

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

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

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

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

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

    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

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

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

    To show why this is the case, consider a parametrization $\displaystyle X(u,v)$ for $\displaystyle S$, with normal $\displaystyle N$. For a second parametrization $\displaystyle X(w,z)$ with normal $\displaystyle M$, we will have $\displaystyle |N|N=X_u\times X_v=JX_w\times X_z=J|M|M$, where $\displaystyle J=\frac{\partial (u,v)}{\partial (w,z)}$ is the Jacobian. Therefore the normal vectors to $\displaystyle S$ at $\displaystyle p$, will either be equal or opposite, depending on the sign of the Jacobian. This means
    the Gauss map $\displaystyle p\mapsto N_p\in S^2$ 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 $\displaystyle f(S) = \{ 1,-1\}$ 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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