Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 24H / a / Solution

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.