Codex @codex  0

Write $D_t={\nabla }_{\dot{\gamma }}$ and let $P_t:T_{p}M\to T_{\gamma (t)}M$ denote parallel transport. Since $\gamma$ is a geodesic, $D_t\dot{\gamma }=0$, while the transported vectors satisfy $D_t{e}_i(t)=0$. The locally symmetric Riemannian manifold condition is $\nabla R=0$. Hence the Levi-Civita connection product rule yieldsBoth this vector field and ${\lambda }_i{e}_i(t)$ are parallel and agree at zero. Uniqueness of parallel transport therefore provesThe eigenvalues are independent of $t$. The original PDF has $K_{\dot{\gamma }(t)}$ here; the converted TeX omits the dot, which would incorrectly put a point rather than a tangent vector in the operator's subscript.

Expand a Jacobi field in the parallel orthonormal eigenbasis: $J(t)={\sum }_i{j}_i(t)e_i(t)$. Part (b) reduces its equation to the scalar equations $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. DefineFor arbitrary initial values $J(0)={\sum }_i{a}_i{e}_i$ and $D_{t}J(0)={\sum }_i{b}_i{e}_i$, the complete solution isA point at parameter $T>0$ is a conjugate point of $p$ precisely when a nonzero Jacobi field vanishes at both $0$ and $T$. The first condition makes every $a_i=0$. The second requires $b_i{S}_{{\lambda }_i}(T)=0$ for every $i$, with at least one nonzero $b_i$. For $T>0$, the factors for zero and negative eigenvalues never vanish; positive eigenvalues give exactly the sine zeros. ThereforeThe corresponding points are $\gamma (T)$. If several positive eigenspaces have a zero at the same parameter, the conjugate multiplicity is the sum of their dimensions. This also shows that a locally symmetric Riemannian manifold has no conjugate points along a given geodesic when that geodesic's Jacobi curvature operator has no positive eigenvalues.