Geometric characterization of critical points of the Gauss map.

Question

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 $N$?

Answer 1

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.

Answer 2

The Gauss map will have a critical point if the derivative of the unit normal to the surface is zero in some direction.

This can happen in many ways for instance if the unit normal is constant along a curve. Examples would be a plane and a cylinder. For these, all points are critical. These are cases where every point on the surface has a straight-line on the surface passing through it. Such surfaces are called "ruled surfaces" Another example is a helicoid.

A different example would be a torus. Here the unit normal is constant on the top and bottom circles.

The torus exhibits a general principal which is that regions of a surface that are positively warped like a sphere or an ellipsoid are separated by critical curves(or regions) from regions where the surface is negatively warped like a saddle or the interior region of the torus (the part that is saddle shaped).

Negative and positive warping is detected by the sign of the determinant of the differential of the Gauss map. Since the differential of the Gauss map is continuous, its determinant must pass through zero along a curve that connects positively and negatively curved regions.

For smooth surfaces that are compact and have no boundary, there is a theorem of Hilbert that says that they must have a point of "positive curvature", that is, a point where the determinant of the differential of the Gauss map is positive. At the same time all such surfaces except those that are topological spheres, must also have regions of "negative curvature" as can be seen from the Gauss Bonnet theorem. So all of these surfaces must have curves along which the Gauss map is critical.

There are famous surfaces where the Gauss map has no critical points.By Hilbert's theorem, those that are "negatively curved" at every point must either have boundaries or cusps. The pseudo-sphere is a wonderful example.

I suggest that you go through a bunch of examples. Struik's book on classical Differential Geometry is a good reference.