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A Degree of Gauss normal map

  1. Nov 30, 2016 #1
    If ##M^{2} \subset \mathbb{R}^{3}## is a surface with given normal field, we define the Gauss (normal) map

    $$n:M^{2} \rightarrow \text{unit sphere}\ S^{2}$$

    by

    $$n(p) = \textbf{N}(p), \qquad \text{the unit normal to $M$ at $p$}.$$

    -------------------------------------------------------------------------------------------------------------------------------------------------------

    It can be shown that

    $$n^{*}\text{vol}^{2}_{S}=K\text{vol}^{2}_{M},$$

    where ##\text{vol}^{2}_{S}## is the volume form in ##S^{2}##, ##\text{vol}^{2}_{M}## is the volume form of ##M^{2}## and ##K## is the Gauss curvature.

    This means that

    1. the Gauss map is a local diffeomorphism in the neighbourhood ##U## of any ##p \in M^{2}## at which ##K(p)\neq 0## (alternatively, ##u \in M## is a regular point for the Gauss map provided ##K(u)\neq 0##)
    2. if ##U## is positively oriented then ##n(U)## will be positively oriented on ##S^2## iff ##K>0##.

    -------------------------------------------------------------------------------------------------------------------------------------------------------

    Let ##y \in V## be a regular value of ##\phi: M^{n} \rightarrow V^{n}##; that is, ##\phi_{*}## at ##\phi^{-1}(y)## is onto. For each ##x \in \phi^{-1}(y)##, ##\phi_{*}:M_{x}\rightarrow V_{y}## is also ##1:1##; that is, ##\phi_{*}## is an isomorphism. Put

    $$\text{sign}\ \phi(x) := \pm 1$$

    where the ##+## sign is used iff ##\phi_{*}:M_{x}\rightarrow V_{y}## is orientation-preserving. Then

    $$\text{deg}(\phi)=\sum_{x \in \phi^{-1}(y)} \text{sign}\ \phi(x)$$

    -------------------------------------------------------------------------------------------------------------------------------------------------------

    1. Why does ##n^{*}\text{vol}^{2}_{S}=K\text{vol}^{2}_{M}## mean that the Gauss map is a local diffeomorphism in the neighbourhood ##U## of any ##p \in M^{2}## at which ##K(p)\neq 0##?

    2. Why does ##n^{*}\text{vol}^{2}_{S}=K\text{vol}^{2}_{M}## mean that ##u \in M## is a regular point for the Gauss map provided ##K(u)\neq 0##?

    3. Why does ##n^{*}\text{vol}^{2}_{S}=K\text{vol}^{2}_{M}## mean that, if ##U## is positively oriented then ##n(U)## will be positively oriented on ##S^2## iff ##K>0##?

    4. How can we use these three facts to evaluate the Brouwer degree of the Gauss normal map?
     
  2. jcsd
  3. Nov 30, 2016 #2

    lavinia

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    Science Advisor

    The Gauss curvature is the determinant of the Gauss map. This and the Inverse Function Theorem is all you need for 1-3.

    The Brouwer degree sums the number of points in the inverse image of a regular value with orientation taken into account. When the Gauss curvature is positive the orientation sign is +1. When the Gauss curvature is negative the sign is -1



    Assume that the surface is compact without boundary. Then convince yourself that:

    The pull back of any 2-form that integrates to 1 over the 2-sphere integrates over the surface to the Brouwer degree of the Gauss map.

    In particular the pull back of 1/4π times the volume form of the unit sphere integrates to the Brouwer degree.

    The the Brouwer degree is independent of the regular value.
     
    Last edited: Dec 1, 2016
  4. Dec 1, 2016 #3
    Thanks!

    Can you help me with another thing?

    Let's say that you have a map from ##M^n## to ##S^n## such that the vector field ##\bf v## on ##M^n## can be extended to be a non-vanishing vector field on all of the interior region ##U^{n+1}## of ##M^n##.

    I was wondering why this means that the Brouwer degree of the map from ##M^n## to ##S^n## is ##0##

    Any ideas?
     
  5. Dec 1, 2016 #4

    lavinia

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    Not sure what you are asking. Are you saying that the manifold is the boundary of a 1 higher dimensional manifold?
     
  6. Jan 13, 2017 at 4:06 PM #5

    mathwonk

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    if lavinia's guess is correct, and i believe it is, then you should be able to use stokes' theorem.
     
  7. Jan 13, 2017 at 7:39 PM #6

    lavinia

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    Science Advisor

    Also try the Implicit function Theorem.
     
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