Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 23H (Differential Geometry)

For an oriented surface with unit normal $N$, the Gauss map sends $p$ to $N(p)\in S^2$. The shape operator is $-dN$ on the tangent plane; its eigenvalues are the principal curvatures $k_1,k_2$. The Gaussian curvature is $K=k_1{k}_2$ and the mean curvature is $H=(k_1+k_2)/2$. Reversing orientation reverses both principal curvatures and $H$, but preserves $K$. The Theorema Egregium says that Gaussian curvature is determined by the first fundamental form, so is invariant under local isometries.

A minimal surface has $H=0$, so $k_2=-k_1$ and $\boxed{K=-k_1^2\le 0}$. For a nonflat example take the catenoid parametrization $X(u,v)=(\cosh v\cos u,\cosh v\sin u,v)$. Its first fundamental form has $E=G={\cosh }^{2}v$ and $F=0$. Choosing the normal appropriately, its second fundamental form has coefficients $e=-1$, $f=0$, $g=1$. Thus its principal curvatures are $\pm {\mathrm{sech}}^{2}v$, proving $H=0$ and $\boxed{K=-{\mathrm{sech}}^{4}v<0}$.

There is no compact boundaryless minimal surface in ${\mathbb{R}}^3$. For an immersion $X$ of a minimal surface, the coordinate functions are harmonic: ${\Delta }_{S}X=0$. Consequently ${\Delta }_S|X{|}^2=2|{\nabla }_{S}X{|}^2=4$, because the immersion has a two-dimensional orthonormal tangent frame. Integrating over a compact surface without boundary gives zero on the left by the divergence theorem and $4\mathrm{Area}(S)>0$ on the right, a contradiction. If surfaces with boundary are allowed, a closed planar disk is a compact minimal example; the boundaryless convention matters.