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



*

*Why use a unit sphere to define the Gauss normal map?

*$n: M^{2} \rightarrow S^{2}$ is a map from points on $M^2$ to points on $S^2$, so $n(p)$ ought to be a point. But, the text defines $n(p)$ as a normal vector. Am I missing something here?

 A: *

*This is analogous to the situation for curves. In the theory of curves, you attach a special orthonormal frame (called the Frenet-Serret) at each point of the curve that was constructed naturally from the curve and by differentiating the frame, you get interesting invariants (curvature, torsion). For surfaces in $\mathbb{R}^3$, you don't have a natural choice of a basis for each tangent plane but you do have (up to a sign) a natural choice for the basis of $T_pM^{\perp}$ at each point. This is the vector $\mathbf{N}$ and by differentiating it you'll get the second fundamental form which will give you interesting invariants of the surface (principal curvatures, Gauss and mean curvature, etc). The fact that $\mathbf{N}$ is normalized to be of unit length also allows you to interpret $d\mathbf{N}|_{p}$ as a map from the tangent plane $T_pM$ to itself.

*If you wear your "internal" glasses and forget that $S^2$ sits inside $\mathbb{R}^3$ then $\mathbf{N} \colon M \rightarrow S^2$ is a map between two surfaces which map points to points as you don't have any notion of a "vector" on $S^2$. From the external point of view, $S^2$ sits inside $\mathbb{R}^3$ which is a vector space and so you can interpret $N(p)$ as a vector in $\mathbb{R}^3$ and this point of view is relevant with the identification of $d\mathbf{N}|_p$ as a map of signature $T_pM \rightarrow T_pM$ and not as a map of the signature $T_pM \rightarrow T_{N(p)} S^2$ (as you can identify $T_{\mathbf{N}(p)} S^2 = \mathbf{N}(p)^{\perp} = T_pM$ as subspaces of $\mathbb{R}^3$). Of course, if $M$ wasn't sitting in $\mathbb{R}^3$ in the first place (for example, if it was a surface in $\mathbb{R}^4$ or an abstract surface) you wouldn't be able to talk of the normal to $M$ in the first place.

