Great circles as round-sphere geodesics

Differentiating $⟨p,p⟩=1$ along sphere curves shows $T_p{S}^n\subseteq p^{\perp }$. Both spaces have dimension $n$ by [F1], so equality holds. For tangent fields $X,Y$, differentiating $⟨Y,p⟩=0$ gives $⟨D_{X}Y,p⟩=-⟨X,Y⟩$; hence the tangent projection of the ambient derivative is ${\tilde{\mathrm{\nabla }}}_{X}Y=D_{X}Y+⟨X,Y⟩p\mathrm{.}$ The ordinary componentwise product rule makes this an affine connection. Its normal correction is orthogonal to tangent vectors, so differentiating the Euclidean pairing proves metric compatibility. Also $D_{X}Y-D_{Y}X=[X,Y]$ componentwise, while the displayed normal correction is symmetric in $X,Y$; thus its torsion vanishes. By [F2], $\tilde{\mathrm{\nabla }}$ is the round sphere's Levi--Civita connection.

Applying the formula from step 1.1 along $\gamma$ to a tangent field $V$ gives $D_t^{\tilde{\mathrm{\nabla }}}V=V^{\mathrm{\prime }}+⟨{\gamma }^{\mathrm{\prime }},V⟩\gamma \mathrm{.}$ In particular, [F3] says that $\gamma$ is a geodesic exactly when ${\gamma }^{\mathrm{\prime }\mathrm{\prime }}+\mathrm{\mid }{\gamma }^{\mathrm{\prime }}{\mathrm{\mid }}^2\gamma =0.$

Suppose $\gamma$ is a geodesic. Its speed is a constant $c\ge 0$ by [F4]. If $c=0$, every ambient component of $\gamma$ has zero derivative and [L4] makes $\gamma$ constant. For a nonconstant geodesic, therefore, $c>0$. Put $p=\gamma (t_0)$ and $u={\gamma }^{\mathrm{\prime }}(t_0)\mathrm{/}c$. The sphere constraint gives $\mathrm{\mid }p\mathrm{\mid }=1$ and $⟨p,{\gamma }^{\mathrm{\prime }}(t_0)⟩=0$, so $p,u$ are orthonormal; step 2.1 gives ${\gamma }^{\mathrm{\prime }\mathrm{\prime }}=-c^2\gamma$.

Define $q(t)=\cos (c(t-t_0))p+\sin (c(t-t_0))u$. By [L1], [L2], and the orthonormality from step 3.1, $q(t)\in S^n$, $q(t_0)=p$, $q^{\mathrm{\prime }}(t_0)=cu={\gamma }^{\mathrm{\prime }}(t_0)$, and $q^{\mathrm{\prime }\mathrm{\prime }}=-c^{2}q$. For $h=\gamma -q$, the nonnegative function $E=\mathrm{\mid }{h}^{\mathrm{\prime }}{\mathrm{\mid }}^2+c^2\mathrm{\mid }h{\mathrm{\mid }}^2$ satisfies $E^{\mathrm{\prime }}=2⟨h^{\mathrm{\prime }},h^{\mathrm{\prime }\mathrm{\prime }}+c^{2}h⟩=0.$ By [L4], $E$ is constant; its value at $t_0$ is zero, so $h=0$ throughout $I$. This also covers an included endpoint $t_0$, using the one-sided derivatives and endpoint continuity in [L4].

The orthonormal vectors $p,u$ span a two-plane through the origin, and [L3] shows that the formula in step 4.1, defined for every real $t$, covers its unit circle with constant speed $c$. It is a geodesic by step 2.1, so it extends the original curve. Moreover, step 4.1 applies on the domain of any other extension with the same initial data at $t_0$ and identifies that extension with this formula; hence this all-real extension is unique and, since no interval properly contains $\mathbb{R}$, maximal. Its image is the whole great circle. Conversely, starting with orthonormal $p,u$ and $c>0$, [L1]--[L2] give $\mathrm{\mid }{q}^{\mathrm{\prime }}\mathrm{\mid }=c$ and $q^{\mathrm{\prime }\mathrm{\prime }}=-c^{2}q$, so step 2.1 gives $D_t^{\tilde{\mathrm{\nabla }}}{q}^{\mathrm{\prime }}=0$ and [F3] makes $q$ a geodesic; a constant curve is a geodesic by [F3]. The case $n=1$ is included: the two-plane is all of ${\mathbb{R}}^2$ and its unit circle is $S^1$. No point, direction, or plane is selected from a family: all are supplied or obtained uniquely from $\gamma ,t_0$, so the argument uses no choice principle.