Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 25G (Differential Geometry)

Choose the orientation of a surface represented by a smooth Gauss map $N:S\to {\mathbb{S}}^2$. The shape operator at $p$ isIf its principal curvatures are $k_1,k_2$, then by definition

Let $E(x)=Ax+b$ be a proper Euclidean motion of Euclidean three-space, so $A\in SO(3)$. Orient $\tilde{S}=E(S)$ by transporting the normal:The chain rule gives the orthogonal conjugacyThe determinant and trace are invariant under similarity, so the Euclidean invariance of the shape operator yields

If the surface orientation is reversed, then $N$ and $S_p$ are multiplied by $-1$. Therefore the orientation reversal of surface curvature says

A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.