# A Degree of Gauss normal map

1. Nov 30, 2016

### spaghetti3451

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

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

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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?

2. Nov 30, 2016

### lavinia

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
3. Dec 1, 2016

### spaghetti3451

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?

4. Dec 1, 2016

### lavinia

Not sure what you are asking. Are you saying that the manifold is the boundary of a 1 higher dimensional manifold?

5. Jan 13, 2017 at 4:06 PM

### mathwonk

if lavinia's guess is correct, and i believe it is, then you should be able to use stokes' theorem.

6. Jan 13, 2017 at 7:39 PM

### lavinia

Also try the Implicit function Theorem.

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