# Thread: Gauss Map on a Regular Surface

1. ## 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!

2. Originally Posted by slevvio

$\tilde{N}(p) = f(p)N(p)$ 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.

3. 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

4. I don't think you can have that a priori, since it is equivalent to what you are trying to prove.

5. 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