Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 12G (Geometry)

For a parametrized curve on the surface, the energy of a curve iswhere the first fundamental form coefficients are evaluated at $(u(t),v(t))$. A fixed-endpoint variation of a curve has coordinate variation fields $\eta ,\xi$ vanishing at $a,b$. Differentiating under the integral and integrating their derivative terms by parts givesThe fields can be chosen independently with compact support in $(a,b)$. Stationarity and the fundamental lemma of the calculus of variations therefore force both coefficients to vanish. These Euler-Lagrange equations are the geodesic equations:

The coordinate curves are assumed to be geodesics as geometric curves; use their arc-length parametrization in the geodesic equations. On $u=c$, choose increasing arc length $s$, so $\dot{u}=0$ and $\dot{v}=1/\sqrt{G}$. The transverse equation becomesThus $(F/\sqrt{G}{)}_v=G_u/(2\sqrt{G})=(\sqrt{G}{)}_u$. On $v=d$, similarly $\dot{v}=0$, $\dot{u}=1/\sqrt{E}$, and the transverse equation gives $(F/\sqrt{E}{)}_u=E_v/(2\sqrt{E})=(\sqrt{E}{)}_v$. Hence the geodesic coordinate grid identities are

The angle $\theta$ between the oriented coordinate tangent vectors satisfies $\cos \theta =F/\sqrt{EG}$. The constant-angle assumption therefore gives $F=k\sqrt{E}\sqrt{G}$ for a constant $k$. The first fundamental form is positive definite, so $EG-F^2>0$ and hence $|k|<1$. Substitution into the geodesic coordinate grid identities from part (i) givesEliminating either derivative yields $(1-k^2)(\sqrt{E}{)}_v=0$ and $(1-k^2)(\sqrt{G}{)}_u=0$. Since $E,G>0$ and $1-k^2>0$, this provesIf the angle is understood between unoriented curves, its constancy initially fixes $|F|/\sqrt{EG}$; continuity fixes the sign on each connected chart when this quantity is nonzero, and the zero case already has $F=0$. Thus the same conclusion holds. In fact constant-angle geodesic coordinate grids are locally flat, although only the displayed derivatives are needed here.