TITLE: Geometric characterization of critical points of the Gauss map.
QUESTION [7 upvotes]: Let $\Sigma \subset \mathbb{R}^3$ an oriented surface by Gauss map $N: \Sigma \rightarrow S^2$. How can I find a geometric characterization of critical points of $N$?

REPLY [3 votes]: Critical points of the Gauss map are exactly points where the sectional (Gaussian) curvature of $\Sigma$ vanishes. I believe this is true in general for any hypersurface in $\mathbb{R}^n$. The picture for surfaces is that critical points of the Gauss map correspond to curves in your surface which look like lines to second order (that's why they're critical points), which are exactly the curves of zero curvature; once you have such a thing through $p$ the surface will have zero curvature. 
More rigorously, this follows from the fact that the shape operator of $\Sigma$ is exactly minus the derivative of the Gauss map -- see for instance the discussion following Example 2.4 in Chapter 6 of do Carmo, Riemannian Geometry.
We can then use the Gauss equation to translate this into a statement about curvature. Suppose that the unit tangent vector $x$ is in the kernel of the derivative of the Gauss map at $p \in \Sigma$. Fix some unit tangent vector $y$ at $p$ orthogonal to $x$. Letting $\alpha(\cdot, \cdot)$ denote the second fundamental form of $\Sigma$, the Gauss equation gives the following expression for the sectional curvature $\kappa$ of $\Sigma$ at $p$:
$$
   \kappa = \langle \alpha(x, x), \alpha(y,y) \rangle - \langle \alpha(x,y), \alpha(x,y) \rangle.
$$
Since the shape operator is minus the derivative of the Gauss map and we've assumed $x$ is in its kernel, it follows that $\alpha(x, z) = 0$ for any unit tangent vector $z$ at $p$. Thus $\kappa = 0$ at a point where the Gauss map is singular.
Conversely, if $\kappa = 0$ at a point $p$, then since the sectional curvature is the product of the principal curvatures, one sees that the shape operator has nontrivial kernel, and thus the Gauss map is singular.