[proofplan] A conjugate point at time $T$ is detected by a nonzero Jacobi field vanishing at the two endpoints of the geodesic segment. We first reduce such a field to a normal Jacobi field, since the tangential component must vanish when both endpoint values are zero. Then we introduce the constant-curvature model sine function and use Rauch comparison only on intervals lying strictly before the first possible zero of the Jacobi field. Passing to that first zero gives a contradiction, because the model function is positive before $\pi/\sqrt{k}$ when $k>0$ and positive for every positive time when $k \leq 0$. [/proofplan]

[step:Reduce a conjugate point to a nonzero normal Jacobi field] Let $\gamma: [0,T] \to M$ be a unit-speed geodesic, where $T>0$. Suppose that $\gamma(T)$ is conjugate to $\gamma(0)$ along $\gamma$. By the definition of conjugate points, there exists a nonzero Jacobi field \begin{align*} J: [0,T] &\to TM \\ s &\mapsto J(s) \in T_{\gamma(s)}M \end{align*} along $\gamma$ such that \begin{align*} J(0)=0, \qquad J(T)=0. \end{align*} Define the scalar function \begin{align*} a: [0,T] &\to \mathbb{R} \\ s &\mapsto g_{\gamma(s)}(J(s),\dot{\gamma}(s)). \end{align*} Since $\gamma$ is geodesic and $J$ is a Jacobi field, the tangential component of $J$ satisfies $a''(s)=0$. Hence $a(s)=\alpha s+\beta$ for constants $\alpha,\beta \in \mathbb{R}$. The endpoint conditions give $a(0)=a(T)=0$, so $\alpha=\beta=0$. Thus $J(s)$ is orthogonal to $\dot{\gamma}(s)$ for every $s \in [0,T]$. Therefore every conjugate point along a unit-speed geodesic is detected by a nonzero normal Jacobi field vanishing at both endpoints. [/step]

[step:Compare the Jacobi field before its first possible zero]Assume first that $k>0$. Let \begin{align*} \operatorname{sn}_k: \left[0,\frac{\pi}{\sqrt{k}}\right] &\to \mathbb{R} \\ s &\mapsto \frac{1}{\sqrt{k}}\sin(\sqrt{k}s) \end{align*} be the constant-curvature model sine function. It satisfies \begin{align*} \operatorname{sn}_k(0)=0, \qquad \operatorname{sn}_k'(0)=1, \end{align*} and \begin{align*} \operatorname{sn}_k(s)>0 \end{align*} for every $s \in (0,\pi/\sqrt{k})$. Let $J$ be the nonzero normal Jacobi field from the previous step. Since $J(0)=0$ and $J$ is not identically zero, uniqueness for the Jacobi equation gives \begin{align*} \nabla_{\dot{\gamma}}J(0) \neq 0. \end{align*} Suppose $\tau \in (0,\pi/\sqrt{k})$ is the first positive zero of $J$, meaning that $J(\tau)=0$ and $J(s)\neq 0$ for every $s\in(0,\tau)$. For each $b\in(0,\tau)$, the Jacobi field $J$ is nonzero on $(0,b]$. By the Rauch Comparison Theorem for an upper sectional curvature bound, applied on $[0,b]$ to the unit-speed geodesic $\gamma$, the normal Jacobi field $J$, and the constant-curvature $k$ model Jacobi field with initial derivative length $|\nabla_{\dot{\gamma}}J(0)|$, we obtain \begin{align*} |J(s)| \geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(s) \end{align*} for every $s\in(0,b]$. Since $b<\tau$ was arbitrary, the same estimate holds for every $s\in(0,\tau)$. Taking the limit as $s\to \tau^-$ and using continuity of $J$ and $\operatorname{sn}_k$ gives \begin{align*} 0=|J(\tau)|\geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(\tau)>0, \end{align*} which is impossible. Hence $J$ has no positive zero in $(0,\pi/\sqrt{k})$.[/step]

[guided]The comparison theorem must be used before the first zero of $J$, not after assuming the conclusion we want. The relevant model function is \begin{align*} \operatorname{sn}_k: \left[0,\frac{\pi}{\sqrt{k}}\right] &\to \mathbb{R} \\ s &\mapsto \frac{1}{\sqrt{k}}\sin(\sqrt{k}s). \end{align*} This function is the length of a model Jacobi field in the simply [connected space](/page/Connected%20Space) form of constant sectional curvature $k$. It vanishes at $s=0$ and at $s=\pi/\sqrt{k}$, and it is strictly positive between those two times: \begin{align*} \operatorname{sn}_k(s)>0 \qquad \text{for } 0<s<\frac{\pi}{\sqrt{k}}. \end{align*} Now let \begin{align*} J: [0,T] &\to TM \\ s &\mapsto J(s) \end{align*} be a nonzero normal Jacobi field along the unit-speed geodesic $\gamma$ with $J(0)=0$. Since the Jacobi equation is a second-order linear ordinary differential equation along $\gamma$, a Jacobi field is uniquely determined by its value and covariant derivative at one time. Therefore the conditions $J(0)=0$ and $J\not\equiv 0$ imply \begin{align*} \nabla_{\dot{\gamma}}J(0) \neq 0. \end{align*} Assume, in order to test whether a zero can occur before the model zero, that $\tau\in(0,\pi/\sqrt{k})$ is the first positive zero of $J$. Thus \begin{align*} J(\tau)=0, \qquad J(s)\neq 0 \quad \text{for every } s\in(0,\tau). \end{align*} Fix an arbitrary number $b\in(0,\tau)$. On the interval $(0,b]$, the field $J$ is nonzero, so the nonvanishing hypothesis in the Rauch Comparison Theorem is satisfied. The remaining hypotheses are also satisfied: $\gamma$ is unit speed, $J$ is normal with $J(0)=0$, and every sectional curvature of a plane containing $\dot{\gamma}(s)$ is at most $k$ because the global assumption is $K_M\leq k$. Applying Rauch comparison on $[0,b]$ to $J$ and to the constant-curvature $k$ model Jacobi field whose initial derivative has length $|\nabla_{\dot{\gamma}}J(0)|$ gives \begin{align*} |J(s)| \geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(s) \end{align*} for every $s\in(0,b]$. Since $b<\tau$ was arbitrary, this inequality holds for every $s\in(0,\tau)$. Now pass to the alleged first zero. Both sides are continuous in $s$: the Jacobi field $J$ is smooth along $\gamma$, and $\operatorname{sn}_k$ is smooth on its domain. Taking $s\to\tau^-$ gives \begin{align*} 0=|J(\tau)|\geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(\tau)>0. \end{align*} The final strict inequality uses $|\nabla_{\dot{\gamma}}J(0)|>0$ and $\operatorname{sn}_k(\tau)>0$ because $0<\tau<\pi/\sqrt{k}$. This contradiction proves that $J$ cannot vanish at any positive time smaller than $\pi/\sqrt{k}$.[/guided]

[step:Exclude conjugate points before the model zero when $k>0$] Suppose, toward a contradiction, that there is a conjugate point at time $T$ with \begin{align*} 0<T<\frac{\pi}{\sqrt{k}}. \end{align*} By the first step, there exists a nonzero normal Jacobi field $J$ along $\gamma$ such that \begin{align*} J(0)=0, \qquad J(T)=0. \end{align*} Since $J$ is continuous and $J(T)=0$, the set of positive zeros of $J$ in $[0,T]$ is nonempty. Define \begin{align*} \tau:=\inf\{s\in(0,T]: J(s)=0\}. \end{align*} Continuity gives $J(\tau)=0$, and the definition of $\tau$ gives $J(s)\neq 0$ for every $s\in(0,\tau)$. Hence $\tau$ is the first positive zero of $J$, with $0<\tau\leq T<\pi/\sqrt{k}$. This contradicts the previous step, which proved that no such first positive zero can lie in $(0,\pi/\sqrt{k})$. Hence no conjugate point can occur at any time $T<\pi/\sqrt{k}$. Taking the infimum over all conjugate times gives \begin{align*} \operatorname{conj}(M) \geq \frac{\pi}{\sqrt{k}}. \end{align*} [/step]

[step:Use the nonpositive-curvature model to rule out all conjugate points] Assume now that $k \leq 0$. Define \begin{align*} \operatorname{sn}_k: [0,\infty) &\to \mathbb{R} \end{align*} by \begin{align*} \operatorname{sn}_0(s)&=s,\\ \operatorname{sn}_k(s)&=\frac{1}{\sqrt{-k}}\sinh(\sqrt{-k}s) \qquad \text{if } k<0. \end{align*} In both cases, \begin{align*} \operatorname{sn}_k(s)>0 \end{align*} for every $s>0$. Let $\gamma: [0,T]\to M$ be any unit-speed geodesic segment with $T>0$, and let $J$ be a nonzero normal Jacobi field along $\gamma$ satisfying $J(0)=0$. As before, uniqueness for the Jacobi equation gives \begin{align*} \nabla_{\dot{\gamma}}J(0)\neq 0. \end{align*} Suppose that $J$ has a positive zero, and let $\tau>0$ be its first positive zero. For every $b\in(0,\tau)$, the field $J$ is nonzero on $(0,b]$, so the nonvanishing hypothesis in Rauch comparison is satisfied on $[0,b]$. Applying the upper-curvature Rauch comparison theorem with the constant-curvature $k$ model gives \begin{align*} |J(s)| \geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(s) \end{align*} for every $s\in(0,b]$. Since $b<\tau$ is arbitrary, this holds for every $s\in(0,\tau)$. Passing to the limit as $s\to\tau^-$ yields \begin{align*} 0=|J(\tau)|\geq |\nabla_{\dot{\gamma}}J(0)|\,\operatorname{sn}_k(\tau)>0, \end{align*} which is impossible. Thus $J$ cannot vanish at any positive time. Therefore no nonzero Jacobi field can vanish at both $0$ and a positive time $T$. By the Jacobi-field characterization of conjugate points, $M$ has no conjugate points along any geodesic when $k\leq 0$. [/step]