Cut locus of a point on a round sphere

The defining function $F(x)=⟨x,x⟩$ has derivative $d{F}_x(w)=2⟨x,w⟩$. Since $x\mathrm{\ne }0$ on $S_R^n$, this derivative is surjective onto $\mathbb{R}$ at every level-set point. By [F1], $S_R^n$ is a nonempty, connected, boundaryless Riemannian $n$-manifold and $T_x{S}_R^n=x^{\perp }$.

Fix $q\in S_R^n$ and any piecewise-$C^1$ path $\alpha :[0,1]\to S_R^n$ from $p$ to $q$. On each smooth piece put $u(s)=⟨p,\alpha (s)⟩\mathrm{/}{R}^2$. Cauchy--Schwarz in [F7] gives $\mathrm{\mid }u\mathrm{\mid }\le 1$. Differentiating $\mathrm{\mid }\alpha {\mathrm{\mid }}^2=R^2$ gives $⟨\alpha ,\dot{\alpha }⟩=0$, and hence $u^{\mathrm{\prime }}=\frac{⟨p-u\alpha ,\dot{\alpha }⟩}{R^2},\mathrm{\mid }p-u\alpha \mathrm{\mid }=R\sqrt{1-u^2},\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \frac{\sqrt{1-u^2}}{R}\mathrm{\mid }\dot{\alpha }\mathrm{\mid }\mathrm{.}$ For $0<\epsilon <1$, define ${\theta }_{\epsilon }=\arccos ((1-\epsilon )u)$. Its argument lies strictly between $-1$ and $1$, so [F7] and the chain rule give $\mathrm{\mid }{\theta }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }=\frac{(1-\epsilon )\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }}{\sqrt{1-(1-\epsilon {)}^2{u}^2}}\le \frac{(1-\epsilon )\sqrt{1-u^2}}{R\sqrt{1-(1-\epsilon {)}^2{u}^2}}\mathrm{\mid }\dot{\alpha }\mathrm{\mid }\le \frac{\mathrm{\mid }\dot{\alpha }\mathrm{\mid }}{R}\mathrm{.}$ The last inequality follows because $1-(1-\epsilon {)}^2{u}^2-(1-\epsilon {)}^2(1-u^2)=1-(1-\epsilon {)}^2\ge 0$.

Fix $x\in S_R^n$ and $w\in T_x{S}_R^n$. If $w=0$, the constant curve is a geodesic on all of $\mathbb{R}$. Otherwise set $a=\mathrm{\mid }w\mathrm{\mid }$ and define ${\gamma }_{x,w}(t)=\cos (at\mathrm{/}R)x+\frac{R}{a}\sin (at\mathrm{/}R)w,t\in \mathbb{R}\mathrm{.}$ Since $⟨x,w⟩=0$ and $\mathrm{\mid }x\mathrm{\mid }=R$, $\mathrm{\mid }w\mathrm{\mid }=a$, the trigonometric identity in [F7] gives $\mathrm{\mid }{\gamma }_{x,w}(t)\mathrm{\mid }=R$; differentiating gives ${\gamma }_{x,w}(0)=x$, ${\dot{\gamma }}_{x,w}(0)=w$, $\mathrm{\mid }{\dot{\gamma }}_{x,w}(t)\mathrm{\mid }=a$, and ${\ddot{\gamma }}_{x,w}(t)=-(a^2\mathrm{/}{R}^2){\gamma }_{x,w}(t)$. The acceleration is normal to the sphere, so its tangential projection vanishes. By [F3] this curve is an affinely parametrized geodesic. It is defined for every real $t$; uniqueness in [F4] identifies it with the maximal geodesic for $(x,w)$. Therefore the sphere is geodesically complete.

On each smooth piece, [F8] yields $\mathrm{\mid }{\theta }_{\epsilon }(b)-{\theta }_{\epsilon }(a)\mathrm{\mid }\le {\int }_a^b\mathrm{\mid }{\theta }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le R^{-1}{\int }_a^b\mathrm{\mid }\dot{\alpha }\mathrm{\mid }$. Summing over the pieces and using the triangle inequality gives $\mathrm{\mid }{\theta }_{\epsilon }(1)-{\theta }_{\epsilon }(0)\mathrm{\mid }\le L_g(\alpha )\mathrm{/}R$. As $\epsilon \downarrow 0$, continuity of the principal inverse cosine in [F7] gives $R\arccos \text{\hspace{0.17em}⁣}\left(\frac{⟨p,q⟩}{R^2}\right)\le L_g(\alpha )\mathrm{.}$ This lower bound holds for every competitor in [F6].

The nonempty, connected, boundaryless hypotheses follow from step 1.1. Hopf--Rinow [F5] and geodesic completeness from step 2.1 therefore make $(S_R^n,d_g)$ metrically complete, as required by [F9].

Put $\theta =\arccos (⟨p,q⟩\mathrm{/}{R}^2)\in [0,\pi ]$. Since $\mathrm{\mid }p\mathrm{\mid }=\mathrm{\mid }q\mathrm{\mid }=R$ and $⟨p,q⟩=R^2\cos \theta$, $\mathrm{\mid }q-\cos \theta p{\mathrm{\mid }}^2=R^2{\sin }^2\theta$, so $\theta =0$ implies $q=p$ and $\theta =\pi$ implies $q=-p$. If $0<\theta <\pi$, define $v=(q-\cos \theta p)\mathrm{/}(R\sin \theta )$. Direct inner-product calculation gives $v\in p^{\perp }$ and $\mathrm{\mid }v\mathrm{\mid }=1$. The geodesic in step 2.1 with initial data $(p,v)$ reaches $q$ at time $R\theta$ and has unit speed. If $\theta =0$ then $q=p$ and the constant path has length zero. If $\theta =\pi$ then $q=-p$; because $n\ge 2$, $p^{\perp }$ contains a unit vector $v$, and the same formula reaches $-p$ at time $\pi R$ with unit speed. Thus a path of length $R\theta$ exists in every case. Combining this upper bound with step 2.2 proves $d_g(p,q)=R\arccos \text{\hspace{0.17em}⁣}\left(\frac{⟨p,q⟩}{R^2}\right)\mathrm{.}$

Let $v\in T_p{S}_R^n$ be any unit vector. Step 2.1 gives its radial geodesic ${\gamma }_v(t)=\cos (t\mathrm{/}R)p+R\sin (t\mathrm{/}R)v$. Hence $⟨p,{\gamma }_v(t)⟩\mathrm{/}{R}^2=\cos (t\mathrm{/}R)$, and step 3.2 gives $d_g(p,{\gamma }_v(t))=R\arccos (\cos (t\mathrm{/}R))\mathrm{.}$ For $0<t\le \pi R$, the inverse-cosine definition in [F7] makes this equal to $t$. For $t>\pi R$, its range $[0,\pi ]$ gives $d_g(p,{\gamma }_v(t))\le \pi R<t$. Thus the positive minimizing-time set is exactly $(0,\pi R]$, including the endpoint, and [F9] yields $c_p(v)=\pi R$.

At the finite endpoint, every unit direction has ${\gamma }_v(\pi R)=-p$. Since $n\ge 2$, the tangent space has unit vectors, so the cut-point set in [F9] is nonempty and equals $\{-p\}$. The empty case cannot occur because $p$ is supplied and $S_R^n$ contains $R{e}_{n+1}$; the zero- and one-dimensional spheres are outside the stated $n\ge 2$ claim. The zero initial vector in the completeness check was treated in step 2.1, and the degenerate angular case $q=p$ in step 3.2. The radius is strictly positive; the cut-time definition excludes $t=0$, includes $t=\pi R$, and every later time fails strictly by step 4.1. Choosing a unit vector for the single antipodal path in step 3.2 uses only the nonzero finite-dimensional tangent space for that supplied $p$, not a choice function on a family. Exactly the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ is propagated through [F3]--[F5] and [F9]; no full AC is used. This example makes no iff claim.