Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 3 / 23H / c / Solution

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.