Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 14 / 4 / Solution

A geodesic is an affinely parametrized curve satisfying ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$ for the Levi-Civita connection of the Riemannian metric. In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_j={\Gamma }_{ij}^k{\partial }_k$. For a vector field $V(t)=V^k(t){\partial }_k$ along the curve,Putting $V=\dot{\gamma }$ gives the geodesic equationThis is a smooth second-order ordinary differential system, so the Picard-Lindelöf theorem gives local existence and uniqueness for prescribed position and velocity.

To prove local extension of geodesic velocity, first suppose $\dot{\gamma }(0)=0$. The constant curve through $\gamma (0)$ solves this initial value problem, so uniqueness makes $\gamma$ constant locally. The zero vector field is the desired extension.

Otherwise choose a chart about $\gamma (0)$ in which the first component of the coordinate velocity is nonzero. Write the coordinate curve as $c(t)\in {\mathbb{R}}^n$ and defineAt $(0,0)$ its derivative has columns $c^{\prime }(0),e_2,\dots ,e_n$ and is invertible, because the first component of $c^{\prime }(0)$ is nonzero. The inverse function theorem therefore supplies coordinates $(t,z_2,\dots ,z_n)$ on an open neighbourhood after shrinking the parameter interval to $|t|<\delta$. Along the curve these coordinates are $(t,0,\dots ,0)$. The smooth coordinate field $\partial /\partial t$, pushed back through $\Psi$ and the original chart, agrees with $\dot{\gamma }(t)$. This proves the extension without assuming that a long geodesic has no self-intersections.

By metric compatibility, differentiation along the curve givesThe local extension above permits the ordinary vector-field version of the metric identity to be used along the arc, or the same identity follows directly from the coordinate derivative. Hence an affinely parametrized geodesic has constant speed, including the zero-speed case.

Now let $S^n$ be the unit sphere in ${\mathbb{R}}^{n+1}$. Its induced Levi-Civita derivative is the tangential projection of the ordinary ambient derivative: projection preserves the metric identity, and the commuting ambient coordinate derivatives give zero torsion. By uniqueness of the Levi-Civita connection, proved in Question 5, this is the induced connection.

Thus a curve on the sphere is a geodesic exactly when its ambient acceleration has no tangential component. The normal space is spanned by $\gamma$, so ${\gamma }^{\prime \prime }=a(t)\gamma$. Differentiating $\gamma \cdot \gamma =1$ twice gives $\gamma \cdot {\gamma }^{\prime }=0$ andwhere $c\ge 0$ is the constant speed. Therefore ${\gamma }^{\prime \prime }+c^2\gamma =0$. With initial data $\gamma (0)=p$ and ${\gamma }^{\prime }(0)=v$, all solutions areand, when $v=0$, $\boxed{\gamma (t)=p}$. The orthogonality and length conditions show directly that the displayed curve remains on the sphere, and its acceleration is $-c^2\gamma$, so it is indeed a geodesic. The nonconstant geodesics are the great circles traversed at any constant speed; arbitrary parameter intervals give their restrictions, and shifting the initial parameter simply shifts the formula. There are no other geodesics.