Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 26J (Differential Geometry)

For a unit-speed space curve, setThese form the Frenet trihedron. With the signed torsion convention compatible with the question, the Frenet-Serret formulas areFor an oriented surface with unit normal $N$, the signed geodesic curvature is

On the unit sphere $N=\alpha$. Since $\alpha \cdot t=0$, differentiation givesWrite $\alpha =An+Bb$. Then $A=-1/\kappa$, and differentiating this expression and comparing the $n$ coefficient with $\dot{\alpha }=t$ givesThus, wherever the displayed quantities are defined,Finally,which is the stated curvature decomposition for a spherical curve.

For a positively oriented simple closed curve bounding a disc $D$ on the unit sphere, the local Gauss-Bonnet theorem givesHere $K=1$, soAt fixed length $L$, maximizing the enclosed area $A$ is therefore exactly the same as minimizing ${\int }_0^L{\kappa }_{g}ds$.

From part (a),As $\alpha$ is orthogonal to $t$,To compute $\dot{\beta }\cdot \dot{\alpha }=\dot{\beta }\cdot t$, only its $t$ component matters. Differentiating $-\phi t$ contributes $-\dot{\phi }t$, while differentiating $-{\kappa }^{-1}\dot{\phi }n$ contributes $+\dot{\phi }t$ through $\dot{n}=-\kappa t-\tau b$. Every other term is normal to $t$. Hence

At an area critical point the assumed first-variation identity holds for every smooth $\phi$ compactly supported where ${\kappa }_g\ne 0$:The fundamental lemma of the calculus of variations gives
$({\kappa }_g^{-1}{)}^{\prime }=0$ on each component of $\{{\kappa }_g\ne 0\}$. Thus ${\kappa }_g$ is constant there. A nonzero constant component cannot have a boundary point where continuity makes ${\kappa }_g=0$. Since the closed curve is connected, either ${\kappa }_g\equiv 0$ or it is one nonzero constant everywhere. Therefore every area-maximizing curve has constant geodesic curvature.

For a unit-speed spherical curve,the normal component is $-\alpha$ because
$\alpha \cdot \dot{t}=-1$, and the tangential normal component is the definition of ${\kappa }_g$. If ${\kappa }_g$ is constant, thenMoreover $e\cdot \alpha ={\kappa }_g$. Hence the whole curve lies in the fixed planeso it is planar.

The preceding parts show that an area maximizer among curves of fixed length is a plane section of the sphere, hence a circle. If its smaller cap has angular radius $0\le \theta \le \pi /2$, thenFor this circle,No curve of the same length encloses more area than the maximizing circle. Equivalently every curve enclosing the smaller area $A$ satisfies the spherical isoperimetric inequalitywith equality precisely for circles.