Characterization of disks of constant curvature and whose boundaries have constant geodesic curvature

Question

Is it true that if $D$ is a Riemannian $2$-disk having constant Gaussian curvature equal to $1$ and whose boundary has constant geodesic curvature, then $D$ is isometric to some geodesic ball of the unit sphere $\mathbb{S}^2 \subset \mathbb{R}^3$? I strongly suspect so, but I couldn't find a complete argument.

Answer 1

A Riemannian $2$-manifold $D$ with constant Gaussian curvature $1$ is locally isometric to the standard unit sphere $\mathbb{S}^2$. This known as Minding's Theorem, and a proof can be found in section 4-6 of Do Carmo's book, Differential geometry of curves and surfaces. It follows that there is an isometric immersion $D \rightarrow \mathbb{S}^2$.

Its boundary $C$ is therefore an immersed curve in $\mathbb{S}^2 \subset \mathbb{R}^3$. We want to show that if $C$ has constant geodesic curvature, then it is the intersection of $\mathbb{S}^2$ with a plane. Let $p \in C$ and $v$ be a unit vector tangent to $C$ at $p$.

Here is one way: On one hand, the existence and uniqueness theorem of ODEs, given $\kappa \ge 0$, there exists a unique unit speed immersed curve with constant geodesic curvature $\kappa$ passing through $p$ with velocity $v$. Therefore, this curve must be $C$. On the other hand, it is easy to show that there exists a unique plane $P$ containing $p$ and parallel to $v$ such that the curve $P\cap\mathbb{S}^2$ has constant geodesic curvature $\kappa$. It follows that $C = P\cap\mathbb{S}^2$.

Here is a second way: Parameterize $C$ by arclength. Let $(e_1,e_2,e_3)$ be an orthonormal frame along $C$ such that $e_1$ is tangent to $C$ and $e_3$ is normal to $\mathbb{S}^2$. It's straightforward to check that \begin{equation} \begin{bmatrix} e_1' \\ e_2' \\ e_3' \end{bmatrix} = \begin{bmatrix} 0 & \kappa & -1 \\ -\kappa & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} e_1 \\ e_2 \\ e_3 \end{bmatrix}, \end{equation} where $\kappa$ is the geodesic curvature of $C$. If $\kappa$ is constant, then it is easy to verify that $(e_2 + \kappa e_3)' = 0$. It follows that $C$ lies in a plane normal to $e_2 + \kappa e_3$ and has curvature equal to $\sqrt{1+\kappa^2}$.

Either way, $C = P \cap \mathbb{S}^2$ and $D$ is one of the two spherical caps with boundary $C$.