Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 24H (Differential Geometry)

The Theorema Egregium states that the Gaussian curvature of a surface in ${\mathbb{R}}^3$ is determined entirely by its first fundamental form; in particular a local isometry preserves it.

Let $r(u^1,u^2)$ be a regular parametrization, $g_{ij}=r_i\cdot r_j$, $n$ a unit normal, and $b_{ij}=r_{ij}\cdot n$ the second fundamental form. Differentiating the metric and decomposing $r_{ij}$ into tangent and normal parts gives the Gauss formulaDifferentiating $n\cdot r_j=0$ gives $n_i=-b_i{}^k{r}_k$, the Weingarten formula. Compute two successive ambient derivatives of a tangent vector and subtract them. Ambient derivatives commute, so their tangential parts yield the Gauss equationwhere $S$ is the shape operator and $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ is the intrinsic curvature of the connection with the displayed Christoffel symbols. Taking $X={\partial }_1,Y={\partial }_2,Z={\partial }_2$ and the inner product with ${\partial }_1$ givesBut $K=\det S=\det (b)/\det (g)$, soBoth the connection and its curvature on the right are computed from $g$ and its derivatives alone. This proves the theorem, rather than merely using its name as a substitute for the argument.

Write the plane as $\{x:\nu \cdot x=c\}$ for a fixed unit vector $\nu$, and put $h(x)=\nu \cdot x-c$. Suppose $X$ misses the plane. Choose a point of $X$ and a number $A$ larger than its distance to the plane. The hypothesis $D_n\to \infty$ makes $\{x\in X:|h(x)|\le A\}$ bounded, and closedness of $X$ makes it compact. Thus $|h|$ attains its global minimum at some $p\in X$, with $|h(p)|>0$.

Locally the sign of $h$ is constant, and $h$ has a local minimum or maximum at $p$. Its surface gradient vanishes, so the tangent plane of $X$ there is parallel to the given plane and its normal is $\pm \nu$. For a surface tangent vector, the intrinsic Hessian of $h$ is the normal component of the second fundamental form, namely ${\mathrm{Hess}}_{X}h=\pm b$ at $p$. The Hessian is semidefinite because this is a local extremum. Since $X$ is a minimal surface, ${\mathrm{tr}}_{g}b=2H=0$. A semidefinite symmetric form with zero trace must be zero, so $b(p)=0$. This makes $p$ a planar point, contradicting the assumption. Therefore $\boxed{X\cap \Pi \ne \varnothing }$.

The boundaryless hypothesis ensures that the minimizing point is an interior surface point. The condition on $D_n$ ensures that the infimum is attained; disjointness from a plane alone would not do so.