Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 24H (Differential Geometry)

Parametrize the regular curve by arclength $s$. Its unit tangent is $T={\gamma }^{\prime }$, and its curvature is $\kappa =|T^{\prime }|$. Where $\kappa >0$, define the principal normal $N=T^{\prime }/\kappa$ and binormal $B=T\times N$. The torsion of a space curve is $\tau =-B^{\prime }\cdot N$. Orthonormality and differentiation give the Frenet-Serret formulasFor a general parameter, $\kappa =|\dot{\gamma }\times \ddot{\gamma }|/|\dot{\gamma }{|}^3$ and $\tau =\det (\dot{\gamma },\ddot{\gamma },\stackrel{...}{\gamma })/|\dot{\gamma }\times \ddot{\gamma }{|}^2$. The normal and torsion formulas require nonzero curvature; at zero-curvature points this particular Frenet frame need not be defined.

The planar planar isoperimetric inequality is $4\pi |\Omega |\le |\partial \Omega {|}^2$, with equality exactly for a disk. For one boundary component of length $L$, take its unit-speed periodic parametrization $\gamma =(x,y)$ and subtract its mean position. Green's area formula and the Cauchy-Schwarz inequality giveApply the periodic Wirtinger inequality to each mean-zero coordinate: $\int |\gamma {|}^2\le (L/2\pi {)}^2\int |{\gamma }^{\prime }{|}^2$. Since unit speed gives the final integral $L$, it follows that $|A_{\mathrm{signed}}|\le L^2/(4\pi )$. For several boundary components, orient each with the domain on its left. Green's formula, followed by the triangle inequality, givesThis proves the stated inequality, including disconnected domains or holes. Equality permits only one component; equality in the two integral inequalities makes its coordinates first sine-cosine harmonics with the corresponding orthogonal rotation relation. Unit speed then makes the curve a circle, so its domain is a disk. Conversely disks attain equality.

Let $c$ be the center of the containing disk and take arclength around the closed curve. The periodic function $\Phi (s)=|\gamma (s)-c{|}^2/2$ has a maximum. At a maximum $s_0$,By the Cauchy-Schwarz inequality and the containing radius, $1\le |\gamma (s_0)-c||{\gamma }^{\prime \prime }(s_0)|\le r|k(s_0)|$. ThereforeClosedness provides an interior maximum on the periodic parameter domain. Without it, the straight segment $\gamma (s)=(s,0)$ for $-r<s<r$ lies in the disk but has signed curvature identically zero, giving the required counterexample.