Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 3 / 23H (Differential Geometry)

For an oriented surface normal $N$ and a unit-speed curve tangent $T$, take signed geodesic curvature $k_g=(dT/ds)\cdot (T\times N)$. This convention uses boundary orientation with the region on the right; reversing to the usual left-hand boundary convention reverses both orientation and the corresponding curvature convention. With positive orientation consistent this way, the Gauss-Bonnet theorem for a region $D$ with piecewise smooth boundary isFor a disk with smooth boundary there are no corner terms and $\chi (D)=1$. Stating the signed convention is necessary to reproduce the sign in part (ii).

Let $T,n,B$ be the Frenet frame of the original space curve, with $T^{\prime }=kn$, $n^{\prime }=-kT+\tau B$, $B^{\prime }=-\tau n$. Put $w=\sqrt{k^2+{\tau }^2}$. The normal-indicatrix arc-length satisfies $d\bar{s}/ds=w>0$, and its unit tangent is $\overline{T}=(-kT+\tau B)/w$. On the unit sphere its outward normal is $n$, and $\overline{T}\times n=(-kB-\tau T)/w$.

Differentiate $\overline{T}$ in $s$. The derivative of $1/w$ is parallel to $\overline{T}$ and contributes nothing to its projection onto $\overline{T}\times n$. The remaining derivative is $(-k^{\prime }T+{\tau }^{\prime }B-(k^2+{\tau }^2)n)/w$, givingWith the alternative left-normal signed-curvature convention the displayed scalar changes sign, while the area conclusion is unchanged.

Because $k>0$ and the original smooth closed curve has periodic Frenet data, $\arctan (\tau /k)$ is a continuous real-valued periodic function. Therefore $\int k_{g}d\bar{s}=-\oint d\arctan (\tau /k)=0$. The simple closed normal curve divides the sphere into two disks by the Jordan theorem. Apply Gauss-Bonnet theorem to either disk, choosing its consistent boundary orientation. Its Gaussian curvature is one and its total boundary curvature remains zero. Thus each region has area $2\pi$, proving the two regions have equal area.