At a conjugate endpoint the index form is degenerate

For every unit $u\in T_p{S}_R^n$ the curve $s_u(t)=\cos (t\mathrm{/}R)p+R\sin (t\mathrm{/}R)u$ is a unit-speed geodesic of $S_R^n$, and in particular $s_v(t)={\exp }_p(tv)$ for all real $t$; consequently $\gamma =s_v{\mathrm{\mid }}_{[0,\pi R]}$ and $\gamma (\pi R)=-p$. [F1, F2, F8, given] Since $u\perp p$ and $\mathrm{\mid }u\mathrm{\mid }=1$ with $\mathrm{\mid }p\mathrm{\mid }=R$, the Pythagorean identity gives $\mathrm{\mid }{s}_u(t){\mathrm{\mid }}^2=R^2{\cos }^2(t\mathrm{/}R)+R^2{\sin }^2(t\mathrm{/}R)=R^2$, so $s_u$ maps into $S_R^n$. The chain rule gives $s_u^{\mathrm{\prime }}(t)=-(1\mathrm{/}R)\sin (t\mathrm{/}R)p+\cos (t\mathrm{/}R)u$ and $s_u^{\mathrm{\prime }\mathrm{\prime }}(t)=-(1\mathrm{/}{R}^2)s_u(t)$; thus $\mathrm{\mid }{s}_u^{\mathrm{\prime }}{\mathrm{\mid }}^2={\sin }^2(t\mathrm{/}R)+{\cos }^2(t\mathrm{/}R)=1$, and $s_u^{\mathrm{\prime }\mathrm{\prime }}$ is a multiple of $s_u$, hence orthogonal to the sphere at $s_u(t)$. By [F1] the tangential projection of the ambient acceleration is the covariant derivative of $s_u^{\mathrm{\prime }}$ along $s_u$, so $s_u$ is a unit-speed geodesic. It satisfies $s_u(0)=p$ and $s_u^{\mathrm{\prime }}(0)=u$, so by [F2] it equals the maximal geodesic ${\exp }_p(tu)$; at $t=\pi R$ the values $\cos \pi =-1$ and $\sin \pi =0$ of [F8] give $s_u(\pi R)=-p$.

Part (a) holds. [F6, F9, given] If $\gamma (a)$ and $\gamma (b)$ are conjugate along $\gamma$, [F6] supplies a nonzero Jacobi field $J$ along $\gamma$ with $J(a)=J(b)=0$. A Jacobi field is smooth, hence a member of the index-form space, and its endpoint values place it in the fixed-endpoint subspace ${\mathcal{X}}_0(\gamma )$. For $W\in {\mathcal{X}}_0(\gamma )$ the integration-by-parts identity of [F9] applies with a smooth $J$ and no derivative jumps; the integral vanishes pointwise because $J$ is Jacobi, and the boundary term $[g(D_{t}J,W){]}_a^b$ vanishes because $W(a)=W(b)=0$. Hence $I_{\gamma }(J,W)=0$ for every $W\in {\mathcal{X}}_0(\gamma )$, and $I_{\gamma }$ is degenerate with the nonzero null vector $J$.

Let $\theta \in (0,\pi )$ and let $w\in T_p{S}_R^n$ be a unit vector with $w\perp v$; such a $w$ exists because $n\ge 2$ makes $p^{\perp }\cap v^{\perp }$ at least one-dimensional. Put $u_{\theta }=\cos \theta v+\sin \theta w$ and $s_{\theta }=s_{u_{\theta }}$ on $[0,\pi R]$. Then $s_{\theta }$ is a unit-speed curve on $S_R^n$ with $s_{\theta }(0)=p$ and $s_{\theta }(\pi R)=-p$, its length is $\pi R$, and $s_{\theta }\mathrm{\ne }\gamma$. [F1, F5, F8, step 1.1, given] The vector $u_{\theta }$ is a unit vector of $T_p{S}_R^n$ because $v,w$ are orthonormal tangent vectors at $p$; step 1.1 therefore applies to $s_{\theta }$. The endpoints are $s_{\theta }(0)=\cos 0p+0=p$ and $s_{\theta }(\pi R)=\cos \pi p+R\sin \pi u_{\theta }=-p$ by [F8]. Unit speed makes the length equal to the parameter length $\pi R$ by [F5]. At the midpoint, $s_{\theta }(\pi R\mathrm{/}2)=\cos (\pi \mathrm{/}2)p+R\sin (\pi \mathrm{/}2)u_{\theta }=R{u}_{\theta }$ differs from $s_v(\pi R\mathrm{/}2)=Rv$ for $0<\theta <\pi$, because $u_{\theta }=\cos \theta v+\sin \theta w$ with $\sin \theta \mathrm{\ne }0$ and $w\perp v$; hence $s_{\theta }\mathrm{\ne }\gamma$.

In the setting of (b), $d(p,-p)=\pi R$ and $\gamma$ is minimizing. [F1, F5, F7, F10, F11, step 1.1] For any piecewise $C^1$ joining path $\sigma$, put $q(t)=⟨p,\sigma (t)⟩\mathrm{/}{R}^2$ and $\vartheta (t)=\arccos q(t)\in [0,\pi ]$. On an interval with $0<\vartheta <\pi$, $⟨\sigma ,{\sigma }^{\mathrm{\prime }}⟩=0$ and differentiation gives $\mathrm{\mid }{\vartheta }^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }⟨p-q\sigma ,{\sigma }^{\mathrm{\prime }}⟩\mathrm{\mid }\mathrm{/}(R^2\sqrt{1-q^2})\le \mathrm{\mid }{\sigma }^{\mathrm{\prime }}\mathrm{\mid }\mathrm{/}R$ by [F7], because $\mathrm{\mid }p-q\sigma \mathrm{\mid }=R\sqrt{1-q^2}$. For $0<\delta <\pi \mathrm{/}2$, let $t_{+}$ be the first time $\vartheta =\pi -\delta$ and $t_{-}$ the last time before $t_{+}$ with $\vartheta =\delta$. Continuity, the endpoint values and compactness of the level sets supply these times, and $\delta \le \vartheta \le \pi -\delta$ on $[t_{-},t_{+}]$. Thus $\vartheta$ is piecewise $C^1$ on this subinterval by [F11]. Integrating the derivative bound piece by piece using [F10] gives $L(\sigma )\ge R(\pi -2\delta )$. Letting $\delta \downarrow 0$ yields $L(\sigma )\ge \pi R$. By step 1.1, $\gamma$ joins the poles with unit speed for time $\pi R$, so $L(\gamma )=\pi R$ and the infimum in [F5] is attained by $\gamma$.

The fixed-endpoint index form on $[0,\pi R]$ is degenerate, and its radical has dimension $n-1$. [F1, F3, F4, F6, F9, F10, F11, step 1.1, step 1.2] For each $w\in T_p{S}_R^n$ perpendicular to $v$, vary the initial direction through $u_s=\cos sv+\sin sw$ when $w$ is unit, scaling linearly for general $w$. Step 1.1 makes $s_{u_s}(t)$ a geodesic variation; its variation field is $J_w(t)=R\sin (t\mathrm{/}R)w$. By [F3] it is Jacobi, and it vanishes at $0$ and $\pi R$. Its initial covariant derivative is $w$, so the $n-1$ independent choices of $w\in v^{\perp }\cap T_p{S}_R^n$ give independent endpoint-vanishing Jacobi fields. The tangent field $tT(t)$ is also Jacobi by [F3], applied to the geodesic variation $(s,t)\mapsto \gamma ((1+s)t)$; it vanishes at $0$, has initial derivative $v$, and is nonzero at $\pi R$. By [F4], these $n$ fields span all Jacobi fields with $J(0)=0$, because their initial derivatives form a basis of $T_p{S}_R^n$. Their endpoint evaluation has one-dimensional image, generated by $(\pi R)T(\pi R)$, so its kernel has dimension $n-1$. It remains to justify the radical on the full piecewise-$C^1$ domain, rather than only the domain of [F6]. Choose an orthonormal basis $e_2,\dots ,e_n$ of $p^{\perp }\cap v^{\perp }$ by [F11]. Along $\gamma$, the frame $E_1=T$, $E_i=e_i$ is orthonormal and parallel by [F1]: the constant $e_i$ have zero ambient derivative and $T^{\mathrm{\prime }}$ is normal. Write a radical field $V$ in this frame as a continuous piecewise-$C^1$ vector function $x$. Put $B_{ij}(t)=g(R(E_j,T)T,E_i)$, a smooth matrix. The definition in [F6] gives $0=I(V,W)=\int (⟨x^{\mathrm{\prime }},y^{\mathrm{\prime }}⟩-⟨Bx,y⟩)dt$ for every endpoint-zero piecewise-$C^1$ coefficient vector $y$. On any closed piece $[c,d]$ of $x$, define $H(t)={\int }_c^{t}B(s)x(s)ds$, $h=x^{\mathrm{\prime }}+H$ and $\bar{h}=(d-c{)}^{-1}{\int }_c^{d}h$. Take $y(t)={\int }_c^t(h(s)-\bar{h})ds$ on $[c,d]$ and zero outside. It is an admissible continuous piecewise-$C^1$ test since $y(c)=y(d)=0$. By [F10], $H^{\mathrm{\prime }}=Bx$ and integration by parts gives $0={\int }_c^d⟨h,y^{\mathrm{\prime }}⟩={\int }_c^d\mathrm{\mid }h-\bar{h}{\mathrm{\mid }}^2\mathrm{.}$ Therefore $h=\bar{h}$ everywhere on $[c,d]$, so $x^{\mathrm{\prime }}=\bar{h}-H$ is $C^1$, including one-sided endpoints. Hence $V$ is piecewise $C^2$. Since it annihilates every piecewise-$C^2$ endpoint-zero test, [F6] makes it a smooth endpoint-vanishing Jacobi field. Conversely every such Jacobi field annihilates the entire piecewise-$C^1$ space by [F9], exactly as in step 1.2. Thus the radical is precisely the already computed $(n-1)$-dimensional space.

For every $t\in [0,\pi R]$ the distance between $\gamma (t)$ and $s_{\theta }(t)$ is at most $R\theta$. [F1, F5, F8, step 1.1, step 2.1] Fix $t$ and define $\rho (s)=\cos (t\mathrm{/}R)p+R\sin (t\mathrm{/}R)(\cos sv+\sin sw)$ for $s\in [0,\theta ]$. Then $\rho$ is smooth, $\rho (s)\in S_R^n$ by the Pythagorean identity, $\rho (0)=s_v(t)=\gamma (t)$ and $\rho (\theta )=s_{\theta }(t)$. Its velocity is ${\rho }^{\mathrm{\prime }}(s)=R\sin (t\mathrm{/}R)(-\sin sv+\cos sw)$, a tangent vector of ambient norm $R\sin (t\mathrm{/}R)\le R$, because $v,w$ are orthonormal and orthogonal to $p$; by [F1] the Riemannian speed of $\rho$ is that ambient norm. Hence $L(\rho )\le R\theta$ by [F5], and since $\rho$ joins $\gamma (t)$ to $s_{\theta }(t)$, the distance is at most $L(\rho )\le R\theta$. Taking the supremum over $t$ gives ${\sup }_{t\in [0,\pi R]}d(\gamma (t),s_{\theta }(t))\le R\theta$.

The geodesic $\gamma$ of (b) is not a strict local minimizer. [step 2.1, step 2.2, step 3.1] Given $\epsilon >0$, choose $\theta \in (0,\pi )$ with $R\theta <\epsilon$ and put $\sigma =s_{\theta }$. By step 2.1, $\sigma$ is a piecewise $C^1$ path from $p$ to $-p$ with $\sigma \mathrm{\ne }\gamma$ and $L(\sigma )=\pi R=L(\gamma )$, and by step 3.1 it satisfies ${\sup }_{t}d(\sigma (t),\gamma (t))<\epsilon$. Thus every uniform neighborhood of $\gamma$ contains a competitor of the same length, while step 2.2 shows that $\gamma$ is minimizing; strict local minimality fails.

Boundary, degeneracy and choice audit. [A1, F6, step 1.2, step 2.1, step 2.3, step 4.1] The segment $[a,b]$ of (a) is nondegenerate, and included endpoints use the one-sided conventions of the index-form and Jacobi-field definitions. In (a) the empty and zero-dimensional cases are vacuous because no conjugate pair exists, and in dimension one the vanishing space is zero by the multiplicity bound of [F6]. In (b), $R>0$ and $n\ge 2$; the case $n=1$ is excluded because the competitor construction needs a normal direction orthogonal to $v$, and on the circle the two semicircles are not arbitrarily close. The competitors $s_{\theta }$ are distinct from $\gamma$ and have exactly the same length, so no strict inequality of lengths is claimed at the conjugate endpoint; the open geodesic before $\pi R$ is not addressed here. The vector $w$ of step 2.1 is chosen once from the nonzero finite-dimensional space $p^{\perp }\cap v^{\perp }$; no choice function on a family is used, and exactly the declared ${\mathrm{A}\mathrm{C}}_{\omega }$ is inherited through the index-form and Jacobi-field interfaces. The proposition asserts degeneracy and non-strictness; it claims no converse and does not say that every conjugate endpoint fails to minimize. $\mathrm{\square }$