📂Geometry

Uniqueness Theorem of Geodesics

Theorem¹

Let’s say point $p$ is on the surface $M$ and let $\mathbf{X} \in T_{p}M$ be the unit tangent vector at point $p$. Then, there exists a unique geodesic $\boldsymbol{\gamma} : (-\epsilon, \epsilon) \to M$ that satisfies the following initial value condition.

$$\boldsymbol{\gamma} (0) = p \quad \text{and} \quad \boldsymbol{\gamma}^{\prime}(0) = \mathbf{X}$$

Description

This theorem states that, at least locally, there exists a straight line of shortest distance connecting two points on a surface.

In global areas, the existence of a shortest-distance geodesic cannot be assured. A simple example can be seen by considering a unit circle that does not contain the origin. In this case, as shown in the following image, we can understand that there is no straight line of shortest distance going from point $a$ to $b$.

Proof

Strategy: In differential geometry, theorems about existence and uniqueness are almost always proved by the Picard’s theorem, which asserts the existence of a solution to the ODE system.

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Let $\mathbf{x}$ be a coordinate chart mapping satisfying $p = \mathbf{x}(0,0)$ for point $p$. Let $(u^{1}, u^{2})$ be the coordinates from $\mathbf{x} : U \to M$ to $U$. Let’s say $\mathbf{X} = \sum \limits_{i} X^{i}\mathbf{x}_{i}$ is the tangent vector and we have $\boldsymbol{\gamma}(s) = \mathbf{x}(\gamma^{1}(s), \gamma^{2}(s))$. Now, consider the following initial value problem of an ODE system.

$$\begin{align} (\gamma^{k})^{\prime \prime} =&\ -\sum_{i,j}{\Gamma_{ij}}^{k}(\gamma^{i})^{\prime} (\gamma^{j})^{\prime} \\ \gamma^{i}(s_{0}) =&\ 0 \nonumber \\ (\gamma^{i})^{\prime}(s_{0}) =&\ X^{i} \nonumber \end{align}$$

By Picard’s theorem, there exists a unique solution to this ODE system in some neighborhood of $s_{0}$. Since being a unit-speed curve $\boldsymbol{\gamma}$ which satisfies $(1)$ is the necessary and sufficient condition for a geodesic, we only need to check if such $\boldsymbol{\gamma}$ is unit-speed.

Let’s say we have $f(s) = \left| \boldsymbol{\gamma}^{\prime}(s) \right|^{2} = \sum\limits_{i,j} g_{ij}(\gamma^{i})^{\prime}(s)(\gamma^{j})^{\prime}(s)$. To finish the proof, we show that $f(s) = 1$ holds. By the chain rule, because we have $\dfrac{d g_{ij}}{d s} = \sum\limits_{k} \dfrac{\partial g_{ij}}{\partial u^{k}}(\gamma^{k})^{\prime}$,

$$f^{\prime}(s) = \sum\limits_{i,j,k}\dfrac{\partial g_{ij}}{\partial u^{k}}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime} + \sum\limits_{i,j} g_{ij}(\gamma^{i})^{\prime \prime}(\gamma^{j})^{\prime} + \sum\limits_{i,j} g_{ij}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime \prime}$$

At this point, considering that we have $\dfrac{\partial g_{ij}}{\partial u^{k}} = \dfrac{\partial \left\langle \mathbf{x}_{i}, \mathbf{x}_{j} \right\rangle}{\partial u^{k}} = \left\langle \mathbf{x}_{ik}, \mathbf{x}_{j} \right\rangle + \left\langle \mathbf{x}_{i}, \mathbf{x}_{jk} \right\rangle$ and $\left\langle \mathbf{x}_{ik}, \mathbf{x}_{j} \right\rangle = \sum\limits_{l}{\Gamma_{ik}}^{l}g_{lj}$ is valid, substituting the first term gives us the following.

$$\begin{align*} f^{\prime}(s) =&\ \sum\limits_{i,j,k,l}g_{li}{\Gamma_{jk}}^{l}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime} + \sum\limits_{i,j,k,l}g_{lj}{\Gamma_{ik}}^{l}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime}\\ &+ \sum\limits_{i,j} g_{ij}(\gamma^{i})^{\prime \prime}(\gamma^{j})^{\prime} + \sum\limits_{i,j} g_{ij}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime \prime} \end{align*}$$

Swapping the dummy indices of the third and fourth terms and rearranging gives us the following.

$$\begin{align*} f^{\prime}(s) =&\ \sum\limits_{i,j,k,l}g_{li}{\Gamma_{jk}}^{l}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime} + \sum\limits_{i,j,k,l}g_{lj}{\Gamma_{ik}}^{l}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime}(\gamma^{j})^{\prime}\\ &+ \sum\limits_{l,i} g_{li}(\gamma^{l})^{\prime \prime}(\gamma^{i})^{\prime} + \sum\limits_{j,l} g_{jl}(\gamma^{j})^{\prime}(\gamma^{l})^{\prime \prime} \\ =&\ \sum\limits_{i,j,k,l}g_{li}\left[ (\gamma^{l})^{\prime \prime} + {\Gamma_{jk}}^{l}(\gamma^{k})^{\prime}(\gamma^{j})^{\prime} \right] (\gamma^{i})^{\prime} + \sum\limits_{i,j,k,l}g_{lj}\left[ (\gamma^{l})^{\prime \prime} + {\Gamma_{ik}}^{l}(\gamma^{k})^{\prime}(\gamma^{i})^{\prime} \right] (\gamma^{j})^{\prime} \end{align*}$$

At this point, the expressions in each bracket, since $\boldsymbol{\gamma}$ satisfies $(1)$, are $0$. Therefore, we get $f^{\prime}(s) = 0$, and since $f$ is a constant and we assumed $\mathbf{X}$ is a unit vector,

$$f(s_{0}) = \left| \boldsymbol{\gamma}^{\prime}(s_{0}) \right|^{2} = \left| \mathbf{X} \right|^{2} = 1$$

Thus, $f = 1$ holds, and $\boldsymbol{\gamma}$ is a unit-speed curve.

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