[proofplan] The argument is a convexity calculation for $f(t) := |J(t)|_g^2$ along a non-trivial Jacobi field $J$ vanishing at $t = 0$. Two covariant differentiations together with the Jacobi equation $J'' = -R(J, \dot\gamma)\dot\gamma$ yield $\tfrac{1}{2} f''(t) = |J'(t)|_g^2 - K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t)$, where $\sigma(t)$ is the $2$-plane $\operatorname{span}(\dot\gamma(t), J(t))$ at points of linear independence (the right-hand side extends by the value $|J'|^2 \geq 0$ at points of dependence, where the wedge vanishes). Under $K \leq 0$ this is a sum of two non-negative quantities, so $f''(t) \geq 0$. Combined with $f(0) = 0$, $f'(0) = 0$, and $f''(0) = 2|J'(0)|_g^2 > 0$ (the latter from $J'(0) \neq 0$, since otherwise the Jacobi-equation initial-value problem forces $J \equiv 0$), Taylor expansion gives $f(t) > 0$ for small $t > 0$; non-decreasingness of $f'$ then propagates strict positivity to all later $t$, contradicting any putative second zero $J(t_1) = 0$. [/proofplan] [step:Set up the squared-norm function $f(t) := |J(t)|_g^2$ and assume a conjugate pair] Let $\gamma : I \to M$ be a geodesic on $(M, g)$ with $K(\sigma) \leq 0$ for every two-plane $\sigma \subseteq T_{\gamma(t)}M$ along $\gamma$. Suppose for contradiction that $\gamma$ has a pair of conjugate points: there exist $t_0 < t_1$ in $I$ and a non-trivial Jacobi field $J$ along $\gamma$ with $J(t_0) = J(t_1) = 0$. After translating the parameter we may assume $t_0 = 0$, so $J(0) = 0$, $J(t_1) = 0$ for some $t_1 > 0$, and $J \not\equiv 0$. Define \begin{align*} f : I &\to \mathbb{R}, \\ t &\mapsto g_{\gamma(t)}(J(t), J(t)) = |J(t)|_g^2. \end{align*} The map $t \mapsto J(t)$ is a smooth section of $\gamma^* TM$ by [Structure of Jacobi Fields](/theorems/2715), and $g_{\gamma(t)}$ is smooth in $t$, so $f \in C^\infty(I)$. [/step] [step:Compute the first two derivatives of $f$ via metric compatibility] Write $J'(t) := \nabla_t J(t)$ and $J''(t) := \nabla_t \nabla_t J(t)$ for the first and second covariant derivatives of $J$ along $\gamma$. By metric compatibility of the Levi-Civita connection, \begin{align*} f'(t) &= \tfrac{d}{dt} g(J, J)(t) = 2\, g(J'(t), J(t)), \\ f''(t) &= 2\, g(J''(t), J(t)) + 2\, g(J'(t), J'(t)) = 2 \big[ g(J''(t), J(t)) + |J'(t)|_g^2 \big]. \end{align*} The Jacobi equation along $\gamma$, with the chapter sign convention $R = -\nabla \circ \nabla$ for the curvature operator (which produces the form below — see [Structure of Jacobi Fields](/theorems/2715)), reads \begin{align*} J''(t) = -R(J(t), \dot\gamma(t)) \dot\gamma(t). \end{align*} Pairing with $J(t)$ and using $R(X, Y, Z, W) := g(R(X, Y) Z, W)$: \begin{align*} g(J''(t), J(t)) = -g(R(J, \dot\gamma)\dot\gamma, J)(t) = -R(J, \dot\gamma, \dot\gamma, J)(t). \end{align*} By the [Symmetries of the Riemann Curvature Tensor](/theorems/2704), the $(0,4)$-tensor $R$ is skew in the first pair and skew in the second pair, so $R(J, \dot\gamma, \dot\gamma, J) = -R(J, \dot\gamma, J, \dot\gamma)$. Therefore \begin{align*} g(J''(t), J(t)) = R(J, \dot\gamma, J, \dot\gamma)(t), \end{align*} and \begin{align*} \tfrac{1}{2} f''(t) = |J'(t)|_g^2 + R(J, \dot\gamma, J, \dot\gamma)(t). \end{align*} [/step] [step:Express the curvature term via sectional curvature and conclude $f''(t) \geq 0$] For each $t \in I$ at which $\dot\gamma(t)$ and $J(t)$ are linearly independent, let $\sigma(t) := \operatorname{span}(\dot\gamma(t), J(t)) \subseteq T_{\gamma(t)}M$. By the definition of sectional curvature applied to the pair $(J(t), \dot\gamma(t))$ (consistent with the chapter convention $R = -\nabla \circ \nabla$, which yields the standard sign for $K$ via $K(\sigma) = -R(X, Y, X, Y)/(|X|_g^2 |Y|_g^2 - g(X, Y)^2)$), \begin{align*} K(\sigma(t)) = -\frac{R(J, \dot\gamma, J, \dot\gamma)(t)}{|J|_g^2(t) \cdot |\dot\gamma|_g^2(t) - g(J, \dot\gamma)^2(t)} = -\frac{R(J, \dot\gamma, J, \dot\gamma)(t)}{|\dot\gamma \wedge J|_g^2(t)}, \end{align*} where $|X \wedge Y|_g^2 := |X|_g^2 \cdot |Y|_g^2 - g(X, Y)^2 \geq 0$ (Cauchy–Schwarz). Equivalently, \begin{align*} R(J, \dot\gamma, J, \dot\gamma)(t) = -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t). \end{align*} At points of linear dependence the right-hand side is interpreted as $0$, since $|\dot\gamma \wedge J|_g^2 = 0$ and the curvature pairing $R(J, \dot\gamma, J, \dot\gamma)$ also vanishes there (write $J = \lambda \dot\gamma$ and use skew-symmetry of $R$ in the first pair). Therefore the identity \begin{align*} R(J, \dot\gamma, J, \dot\gamma)(t) = -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t), \end{align*} extended by $0$ at points where $\sigma(t)$ is undefined, holds for every $t \in I$. Substituting into the formula from the previous step: \begin{align*} \tfrac{1}{2} f''(t) = |J'(t)|_g^2 - K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t). \end{align*} By hypothesis $K(\sigma(t)) \leq 0$ along $\gamma$, so $-K(\sigma(t)) \geq 0$. Combined with $|\dot\gamma \wedge J|_g^2(t) \geq 0$: \begin{align*} -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t) \geq 0, \qquad |J'(t)|_g^2 \geq 0, \end{align*} hence \begin{align*} \tfrac{1}{2} f''(t) \geq 0, \quad \text{i.e.,} \quad f''(t) \geq 0 \text{ for all } t \in I. \end{align*} [guided] We have the curvature pairing $R(J, \dot\gamma, J, \dot\gamma)(t)$ on the right-hand side of $\tfrac{1}{2} f''(t) = |J'(t)|_g^2 + R(J, \dot\gamma, J, \dot\gamma)(t)$ from the previous step. To turn this into something we can sign-control, we want to express it via the sectional curvature $K$, since the hypothesis is stated in terms of $K$. How does the chapter's sign convention play out? The chapter fixes $R = -\nabla \circ \nabla$, which produces the Jacobi equation $J'' = -R(J, \dot\gamma)\dot\gamma$ and pairs cleanly with the sectional-curvature normalisation \begin{align*} K(\sigma) = -\frac{R(X, Y, X, Y)}{|X|_g^2 \cdot |Y|_g^2 - g(X, Y)^2} \end{align*} on a $2$-plane $\sigma = \operatorname{span}(X, Y)$ with $X, Y$ linearly independent. Apply this to $X = J(t)$, $Y = \dot\gamma(t)$ at any $t$ where they are linearly independent, and write $\sigma(t) := \operatorname{span}(\dot\gamma(t), J(t))$. Then \begin{align*} K(\sigma(t)) = -\frac{R(J, \dot\gamma, J, \dot\gamma)(t)}{|J|_g^2(t) \cdot |\dot\gamma|_g^2(t) - g(J, \dot\gamma)^2(t)} = -\frac{R(J, \dot\gamma, J, \dot\gamma)(t)}{|\dot\gamma \wedge J|_g^2(t)}, \end{align*} where the wedge norm $|X \wedge Y|_g^2 := |X|_g^2 \cdot |Y|_g^2 - g(X, Y)^2 \geq 0$ is non-negative by Cauchy–Schwarz, and is the squared area of the parallelogram spanned by $X$ and $Y$. Solving for the curvature pairing, \begin{align*} R(J, \dot\gamma, J, \dot\gamma)(t) = -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t). \end{align*} What about points $t$ where $\dot\gamma(t)$ and $J(t)$ are linearly dependent? There the $2$-plane $\sigma(t)$ is undefined and the formula above seems to break down. But notice both sides vanish: writing $J = \lambda \dot\gamma$ for some scalar $\lambda$ at such a point, skew-symmetry of $R$ in its first pair gives $R(\lambda \dot\gamma, \dot\gamma, J, \dot\gamma) = \lambda R(\dot\gamma, \dot\gamma, J, \dot\gamma) = 0$, while $|\dot\gamma \wedge J|_g^2 = 0$ from the same dependence. Hence we extend the identity by $0$ at such points and the equation \begin{align*} R(J, \dot\gamma, J, \dot\gamma)(t) = -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t) \end{align*} holds for every $t \in I$. Substituting into the formula $\tfrac{1}{2} f''(t) = |J'(t)|_g^2 + R(J, \dot\gamma, J, \dot\gamma)(t)$ from the previous step, \begin{align*} \tfrac{1}{2} f''(t) = |J'(t)|_g^2 - K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t). \end{align*} Now we can read off the sign. This is the Riccati-type identity at the heart of the convexity argument. Its right-hand side decomposes into two contributions, both controllable in sign: - $|J'(t)|_g^2 \geq 0$ is the Pythagoras-style term from the squared-norm derivative $f'' = 2(g(J'', J) + |J'|_g^2)$. It is always non-negative, regardless of curvature. - $-K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t)$ is the curvature contribution. This is where the hypothesis $K \leq 0$ enters: it forces $-K(\sigma(t)) \geq 0$, and combined with $|\dot\gamma \wedge J|_g^2(t) \geq 0$ we get \begin{align*} -K(\sigma(t)) \cdot |\dot\gamma \wedge J|_g^2(t) \geq 0, \qquad |J'(t)|_g^2 \geq 0. \end{align*} Both terms on the right of the Riccati identity are non-negative, so \begin{align*} \tfrac{1}{2} f''(t) \geq 0, \quad \text{i.e.,} \quad f''(t) \geq 0 \text{ for all } t \in I. \end{align*} The hypothesis $K \leq 0$ is consumed exactly here. Without it, $-K \cdot |\dot\gamma \wedge J|_g^2$ could be negative and $f''$ could change sign, allowing $f$ to oscillate and produce conjugate points. Geometrically: in non-positive curvature, geodesics emanating from a common point spread apart, and the spread $f(t) = |J(t)|_g^2$ is convex. The Bonnet–Myers theorem (positive curvature forcing conjugate points) and the Hadamard–Cartan theorem (non-positive curvature forbidding them) sit at opposite ends of this dichotomy. [/guided] [/step] [step:Use $J'(0) \neq 0$ to obtain $f(t) > 0$ for $t$ in a right neighbourhood of $0$] We claim $J'(0) \neq 0$. If $J'(0) = 0$, then $J$ satisfies the Jacobi equation — a homogeneous second-order linear ODE on $T_{\gamma(t)} M \cong \mathbb{R}^n$ in any parallel orthonormal frame — with vanishing initial data $J(0) = 0$ and $\nabla_t J(0) = 0$. By the uniqueness statement of the [Structure of Jacobi Fields](/theorems/2715) (Part 1), the initial-value problem has a unique solution, which in this case is the zero solution. Hence $J \equiv 0$, contradicting non-triviality. Therefore $J'(0) \neq 0$. Since $J(0) = 0$: \begin{align*} f(0) = 0, \qquad f'(0) = 2\, g(J'(0), J(0)) = 0, \qquad f''(0) = 2|J'(0)|_g^2 - 2 K(\sigma(0)) \cdot 0 = 2|J'(0)|_g^2 > 0, \end{align*} where we used $|\dot\gamma \wedge J|_g^2(0) = |\dot\gamma(0)|_g^2 \cdot 0 - 0 = 0$ from $J(0) = 0$. Taylor's theorem at $t = 0$ gives, for $t \to 0$, \begin{align*} f(t) = \tfrac{1}{2} f''(0) t^2 + o(t^2) = |J'(0)|_g^2 \cdot t^2 + o(t^2). \end{align*} Choose $\delta > 0$ small enough that $f(t) \geq \tfrac{1}{2} |J'(0)|_g^2 \cdot t^2$ for all $t \in (0, \delta]$. Then \begin{align*} f(t) > 0 \quad \text{for all } t \in (0, \delta], \qquad f(\delta) \geq \tfrac{1}{2} |J'(0)|_g^2 \cdot \delta^2 > 0. \end{align*} [/step] [step:Propagate strict positivity to all $t > 0$ via non-decreasingness of $f'$] We have $f''(t) \geq 0$ for all $t \in I$, so $f'$ is non-decreasing on $I$. By the mean value theorem applied to $f$ on $[0, \delta]$, there exists $\tau \in (0, \delta)$ with \begin{align*} f'(\tau) = \frac{f(\delta) - f(0)}{\delta} = \frac{f(\delta)}{\delta} > 0, \end{align*} using $f(\delta) > 0$ from the previous step. Since $f'$ is non-decreasing, $f'(t) \geq f'(\tau) > 0$ for all $t \in [\tau, \infty) \cap I$, so $f$ is strictly increasing on $[\tau, \infty) \cap I$. In particular, for every $t \geq \tau$ with $t \in I$, \begin{align*} f(t) \geq f(\tau) \geq f(\delta) - 0 \cdot (\delta - \tau) > 0 \end{align*} (the simpler bound $f(t) \geq f(\tau) > 0$ follows from monotonicity). For $t \in (0, \tau)$, the Taylor estimate $f(t) \geq \tfrac{1}{2} |J'(0)|_g^2 \cdot t^2$ holds since $\tau < \delta$, giving $f(t) > 0$. Combining: \begin{align*} f(t) > 0 \quad \text{for all } t \in (0, \infty) \cap I. \end{align*} [/step] [step:Conclude $J$ has no zero past $0$, contradicting the assumption] By assumption $J(t_1) = 0$, i.e., $f(t_1) = 0$, for some $t_1 > 0$ in $I$. This contradicts the previous step, which established $f(t) > 0$ for every $t \in (0, \infty) \cap I$. Hence $\gamma$ admits no pair of conjugate points: for every non-trivial Jacobi field $J$ along $\gamma$ with $J(t_0) = 0$ at some $t_0$, the field $J$ vanishes nowhere else. This completes the proof. [guided] The argument has the structure: convexity ($f'' \geq 0$ from $K \leq 0$) + initial data ($f(0) = f'(0) = 0$, $f''(0) > 0$) + an open initial growth window ($f$ is strictly positive on $(0, \delta]$) — these combine to force $f$ strictly positive on all of $(0, \infty) \cap I$, ruling out a second zero. Why does convexity rule out a second zero? Suppose $f(t_1) = 0$ for some $t_1 > 0$. Since $f \geq 0$ and $f''$ is non-negative, $f$ is convex. A non-negative convex function vanishing at $0$ and at $t_1 > 0$ must vanish on the entire interval $[0, t_1]$ (any value in between would force the chord from $(0, 0)$ to $(t_1, 0)$ to lie above the graph, giving $f \leq 0$ on $[0, t_1]$, hence $f \equiv 0$ there). But $f(t) \geq \tfrac{1}{2}|J'(0)|^2 t^2 > 0$ for small $t > 0$, contradiction. The argument we wrote above is slightly more efficient: we use $f'(t) \geq f'(\tau) > 0$ for $t \geq \tau$ to make $f$ strictly increasing past $\tau$, which directly forbids $f(t_1) = 0$ for $t_1 \geq \tau$. For $t_1 \in (0, \tau)$, the Taylor estimate gives $f(t_1) > 0$ directly. The role of $J'(0) \neq 0$: this is the engine of the strict-initial-growth $f(t) \geq \tfrac{1}{2}|J'(0)|^2 t^2$. Without it, $f$ could begin flat and the argument would not get off the ground. Non-triviality of $J$ together with the Jacobi-equation uniqueness forces $J'(0) \neq 0$, closing this gap. [/guided] [/step]