I am trying to understand Riemannian geometry from the book "Riemannian Manifolds: An Introduction to Curvature" by John Lee. I have the following doubt.
A Riemannian geodesic on a Riemannian manifold $(M,g)$ is a smooth curve $\gamma:I\to M$ such that $D_t\dot{\gamma}\equiv 0.$ Lee shows that the "geodesics of the unit sphere $S^n$ are precisely the great circles, with constant speed parametrization". Lee's argument is the following. Consider the north pole $N\in S^n$ and the tangent vector $\partial/\partial x^1$. Then it is enough to find out geodesic with initial point $N$ and initial velocity $\partial/\partial x^1$. All other geodesics can be obtained by the isotropic and transitive action on $S^n$ by the orthogonal group. First claim is that $\gamma(I)\subseteq\{(x^i)_{i=1}^n:\sum_{i=1}^{n+1}{x^i}^2=1,x^2=x^3\dots=x^n=0\}.$ Since the map $\phi(x^1,\dots,x^{n+1})\mapsto (x^1,\dots,-x^{i},\dots,x^{n+1})$ is a Riemannian isometry, and $\frac{d}{dt}|_{t=0}\phi\circ\gamma=\partial/\partial x^1$ and $\phi((\gamma(0)))=N$, we must have that $\phi\circ\gamma$ s again a geodesic and by uniqueness, we must have that $\gamma=\phi\circ\gamma.$ This shows that $\gamma(I)\subseteq\{(x^i)_{i=1}^n:\sum_{i=1}^{n+1}{x^i}^2=1,x^2=x^3\dots=x^{n+1}=0\}.$ Now Let me take a constant speed parametrization of $\{(x^i)_{i=1}^n:\sum_{i=1}^{n+1}{x^i}^2=1,x^2=x^3\dots=x^n=0\}$. How can I show that it has to be geodesic? The same type of reasoning from hyperboloid model tells that if we consider the hyperboloid $\{(\zeta^1,\dots,\zeta^n,\tau):\tau>0,\tau^2-\sum_{i=1}^n{\zeta^i}^2=R^2\}$ then consider the north pole $N$ again and consider the vector $\partial/\partial\zeta^1$ at $N.$ One can show again the corresponding geodesic in the hyperbolic space will stay on $\zeta^2=\zeta^3=\dots=\zeta^n=0$. Now again we can parametrize by unit speed curve. But how can we show that it has to be the geodesic?