If ##M^{2} \subset \mathbb{R}^{3}## is a surface with given normal field, we define the Gauss (normal) map(adsbygoogle = window.adsbygoogle || []).push({});

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

by

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

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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##.

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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

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

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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?

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# A Degree of Gauss normal map

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