An equator geodesic of the round-sphere factor: conjugate instant at distance pi with multiplicity one

(The geodesic and its parallel frame.) Along $\gamma$ one has $x^1\equiv \pi \mathrm{/}2$, so ${\mathrm{\Gamma }}^1{}_{22}=-\sin (\pi \mathrm{/}2)\cos (\pi \mathrm{/}2)=0$ and ${\mathrm{\Gamma }}^2{}_{12}=\cot (\pi \mathrm{/}2)=0$; by [L1] every Christoffel symbol vanishes along $\gamma$. By [L2], ${\mathrm{\nabla }}_{{\mathrm{\partial }}_2}{\mathrm{\partial }}_2=0$, so $D_{s}T=0$ and $\gamma$ is a geodesic with affine parameter $s$; likewise $D_s{\mathrm{\partial }}_a=0$ for $a=0,1,2,3$, so the coordinate frame is parallel along $\gamma$ and $D_s$ is the componentwise derivative for fields written in this frame. Finally $g(T,T)={\sin }^2(\pi \mathrm{/}2)=1$.

(The Jacobi field.) Let $J(s):=\sin (s){\mathrm{\partial }}_1{\mathrm{\mid }}_{\gamma (s)}$. By step 1.1, $D_{s}J=\cos (s){\mathrm{\partial }}_1$ and $D_s^{2}J=-\sin (s){\mathrm{\partial }}_1$. By [L1] and [L4], $R(T,J)T=\sin (s)R({\mathrm{\partial }}_2,{\mathrm{\partial }}_1){\mathrm{\partial }}_2=-\sin (s)R({\mathrm{\partial }}_1,{\mathrm{\partial }}_2){\mathrm{\partial }}_2=-\sin (s){\sin }^2(\pi \mathrm{/}2){\mathrm{\partial }}_1=-\sin (s){\mathrm{\partial }}_1$. Hence $D_s^{2}J=R(T,J)T$: $J$ is a Jacobi field by [L3]. It is nonzero (for instance $J(\pi \mathrm{/}2)={\mathrm{\partial }}_1\mathrm{\ne }0$) and $J(0)=J(\pi )=0$.

(The general Jacobi field and the first conjugate instant.) Let $K=a{\mathrm{\partial }}_0+b{\mathrm{\partial }}_1+c{\mathrm{\partial }}_2+d{\mathrm{\partial }}_3$ be a smooth field along $\gamma$. By step 1.1, $D_s^{2}K=a^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\partial }}_0+b^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\partial }}_1+c^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\partial }}_2+d^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\partial }}_3$, while $R(T,K)T=bR({\mathrm{\partial }}_2,{\mathrm{\partial }}_1){\mathrm{\partial }}_2=-b{\mathrm{\partial }}_1$ by [L1] and [L4]: the Jacobi equation is the system $a^{\mathrm{\prime }\mathrm{\prime }}=c^{\mathrm{\prime }\mathrm{\prime }}=d^{\mathrm{\prime }\mathrm{\prime }}=0$, $b^{\mathrm{\prime }\mathrm{\prime }}=-b$, with general solution $a=a_0+a_{1}s$, $b=b_0\cos s+b_1\sin s$, $c=c_0+c_{1}s$, $d=d_0+d_{1}s$. If $K(0)=0$ and $K(s_1)=0$ for some $s_1\in (0,\pi )$: the first condition gives $a_0=b_0=c_0=d_0=0$; the second gives $a_1=c_1=d_1=0$ and $b_1\sin s_1=0$, so $b_1=0$ because $\sin s_1\mathrm{\ne }0$. Hence $K=0$: no instant in $(0,\pi )$ is conjugate to $0$, and the first conjugate instant of $\gamma$ is $\pi$.

(Conjugacy at $\pi$.) By step 2.1 and [L3], $\pi$ is a conjugate instant of $\gamma$ relative to $0$ with $p=\gamma (0)$ and $\gamma (\pi )=(0,\pi \mathrm{/}2,3\pi \mathrm{/}2,0)\mathrm{\ne }p$: the two distinct points $p$ and $\gamma (\pi )$ of $M$ are conjugate along $\gamma$.

(Multiplicity one and the kernel of the exponential differential.) By step 2.2, a Jacobi field with $K(0)=K(\pi )=0$ has $a_0=b_0=c_0=d_0=a_1=c_1=d_1=0$, hence is $K=b_1\sin s{\mathrm{\partial }}_1$: the space of such fields is one-dimensional, so the multiplicity of the conjugate instant $\pi$ is $1$ by [L3]. For the kernel statement, let $v:=T(p)={\mathrm{\partial }}_2{\mathrm{\mid }}_p$ and note that the geodesic $\tau \mapsto {\exp }_p(\tau \pi v)=\gamma (\tau \pi )$ exists for $\tau \in [0,1]$, so $\pi v\in {\mathcal{E}}_p$. By [L5], $\ker d({\exp }_p{)}_{\pi v}=\pi \{u\in T_{p}M:J_u(\pi )=0\}$; solving the system of step 2.2 with initial data $J_u(0)=0$, $D_s{J}_u(0)=u$ gives $J_u(s)=u_{0}s{\mathrm{\partial }}_0+u_1\sin s{\mathrm{\partial }}_1+u_{2}s{\mathrm{\partial }}_2+u_{3}s{\mathrm{\partial }}_3$ for $u={\sum }_a{u}_a{\mathrm{\partial }}_a{\mathrm{\mid }}_p$, so $J_u(\pi )=0$ if and only if $u_0=u_2=u_3=0$, that is $u\in \mathbb{R}{\mathrm{\partial }}_1{\mathrm{\mid }}_p$. Hence $\ker d({\exp }_p{)}_{\pi v}=\pi \mathbb{R}{\mathrm{\partial }}_1{\mathrm{\mid }}_p$ is one-dimensional, in agreement with the multiplicity computed above.