No conjugate points under nonpositive sectional curvature

The energy identity. [F1, F6, given] Put $\phi (t):=g_{\gamma (t)}(J(t),J(t))=\mathrm{\mid }J(t){\mathrm{\mid }}^2$, a smooth nonnegative function on $I$. Differentiating twice along $\gamma$ and using [F1] gives ${\phi }^{\mathrm{\prime }}=2g(D_{t}J,J),{\phi }^{\mathrm{\prime }\mathrm{\prime }}=2g(D_t^{2}J,J)+2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2=2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2-2g(R(J,T)T,J)\mathrm{.}$ At an included endpoint the identity is read with one-sided derivatives, which is exactly the convention fixed by [F1].

The curvature term is nonpositive. [A1, F2, F5, F6, step 1.1] Write $f:=g(J,T)$ and split $J=fT+J^{\perp }$ with $g(J^{\perp },T)=0$; this is an orthogonal decomposition of the tangent space at each point because $\mathrm{\mid }T\mathrm{\mid }=1$ by [F6]. Linearity of $R$ in its first slot and [F5] give $R(J,T)T=R(J^{\perp },T)T+fR(T,T)T=R(J^{\perp },T)T$. Moreover [F5] gives $g(R(J^{\perp },T)T,T)=0$, so $g(R(J,T)T,J)=g(R(J^{\perp },T)T,J^{\perp })+fg(R(J^{\perp },T)T,T)=g(R(J^{\perp },T)T,J^{\perp })\mathrm{.}$ At a time $t$ with $J^{\perp }(t)=0$ both sides vanish. At a time with $J^{\perp }(t)\mathrm{\ne }0$, the vectors $J^{\perp }(t)$ and $T(t)$ are orthogonal and nonzero, so ${\sigma }_t:=\mathrm{span}(J^{\perp }(t),T(t))$ is a tangent two-plane, and [F2] applied with $X=J^{\perp }(t)$ gives $g(R(J^{\perp },T)T,J^{\perp })=\mathrm{Rm}(J^{\perp },T,T,J^{\perp })=K({\sigma }_t)\mathrm{\mid }{J}^{\perp }(t){\mathrm{\mid }}^2\le 0.$ Thus $g(R(J,T)T,J)\le 0$ at every $t\in I$, and step 1.1 yields ${\phi }^{\mathrm{\prime }\mathrm{\prime }}\ge 2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2\ge 0$.

$\phi$ is convex. [F7, step 2.1] By step 2.1 the second derivative of the $C^2$ function $\phi$ is nonnegative on the interior of $I$; at included endpoints the one-sided second derivatives exist and are nonnegative as well. Hence by [F7] the function $\phi$ is convex on $I$: for $s<t<u$ in $I$, $\phi (t)\le \frac{u-t}{u-s}\phi (s)+\frac{t-s}{u-s}\phi (u)\mathrm{.}$

Vanishing at two times forces $J\equiv 0$. [F4, F7, step 3.1, given] Assume $J(0)=0$ and $J(t_1)=0$ with $t_1>0$. Then $\phi (0)=\phi (t_1)=0$, and $\phi \ge 0$ because it is a squared norm. Applying the convexity inequality of step 3.1 with $s=0$, $u=t_1$ gives $\phi (t)\le 0$, hence $\phi (t)=0$ and $J(t)=0$, for every $t\in [0,t_1]$; so $J$ vanishes identically on $[0,t_1]$. Choose $t_0\in (0,t_1)$. Then $J(t_0)=0$, and since $J$ is identically zero on an interval around $t_0$, also $D_{t}J(t_0)=0$. By the uniqueness statement of [F4] the zero field is the only Jacobi field with these initial data, so $J\equiv 0$ on all of $I$.

Consequences and boundary cases. [F3, F4, step 4.1] A nonzero Jacobi field with $J(0)=0$ therefore has no positive zero in $I$. For a general conjugate pair $t_1<t_2$ in $I$, the shifted curve $\tilde{\gamma }(s):=\gamma (t_1+s)$ is again a unit-speed geodesic, the shifted field $\tilde{J}(s):=J(t_1+s)$ is again Jacobi, and $\tilde{J}(0)=0$; if a nonzero Jacobi field vanished at both $t_1$ and $t_2$ then step 4.1 would force it to vanish identically, contradicting [F3]. So no two distinct points of $I$ are conjugate along $\gamma$, and by the same shift argument with $\gamma$ reversed the conclusion is symmetric in the two endpoints. If $\gamma$ is constant, [F3] already gives the conclusion and $J$ is affine in the fixed tangent space, so the claim holds. If $M$ has dimension $0$ the only field is $0$. If $M$ has dimension $1$, then $J^{\perp }\equiv 0$ and step 2.1 reduces to ${\phi }^{\mathrm{\prime }\mathrm{\prime }}=2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2\ge 0$, so the same convexity argument applies without any two-plane. If $0$ or $t_1$ is an included endpoint of $I$, the second derivative in step 1.1 is one-sided and is nonnegative by step 2.1, and the convexity inequality of [F7] is applied on the closed interval $[0,t_1]$; nothing changes.