Upper sectional curvature bounds delay conjugate points

A conjugate field may be taken normal and radial. [F1, F2, F5, given] Suppose $\gamma (t_0)$ is conjugate to $p=\gamma (0)$ for some $t_0>0$. By [F1] there is a nonzero Jacobi field $J$ along $\gamma$ with $J(0)=J(t_0)=0$. Its tangential scalar $a(t):=g(J(t),\dot{\gamma }(t))$ satisfies, by the Jacobi equation and the skew-symmetry of the curvature tensor in the last two slots, $a^{\mathrm{\prime }\mathrm{\prime }}(t)=g(D_t^{2}J,\dot{\gamma }(t))=-g(R(J,\dot{\gamma })\dot{\gamma },\dot{\gamma })=-Rm(J,\dot{\gamma },\dot{\gamma },\dot{\gamma })=0,$ the term $g(D_{t}J,D_t\dot{\gamma })$ being absent because $\dot{\gamma }$ is parallel. Hence $a$ is affine, and $a(0)=g(J(0),\dot{\gamma }(0))=0$ and $a(t_0)=g(J(t_0),\dot{\gamma }(t_0))=0$ force $a\equiv 0$: the field $J$ is normal [F2]. Since $J\mathrm{\ne }0$ and $J(0)=0$, its initial derivative $u:=D_{t}J(0)$ is nonzero; otherwise $J$ would vanish identically [F5]. Thus $\mathrm{\mid }{D}_{t}J(0)\mathrm{\mid }=a_0>0$ and, by [F5], $J$ is the radial normal Jacobi field generated by $u$.

Comparison with the constant-curvature model. [F3, F4, F5, F6, step 1.1, given] Keep the hypothetical conjugate time $t_0$ of step 1.1 and put $T:=t_0$; when $k>0$ we are treating only $t_0<\pi \mathrm{/}\sqrt{k}$, and when $k\le 0$ there is no restriction on $T$. Let $M_k^n$ be the space form of constant curvature $k$, let $\tilde{\gamma }$ be a unit-speed geodesic in it and let $J_k(t):={\mathrm{sn}}_k(t){\mathrm{\Phi }}_{t}w,\mathrm{\mid }w\mathrm{\mid }=a_0,$ be the model radial field with the same initial-derivative norm as $J$ [F4]. In the model, the normal radial Jacobi tensor is ${\mathrm{sn}}_k(t)$ times the parallel identification: it solves $Y^{\mathrm{\prime }\mathrm{\prime }}=-kY$, $Y(0)=0$, $Y^{\mathrm{\prime }}(0)=\mathrm{id}$ in a parallel orthonormal normal frame, and so does ${\mathrm{sn}}_k\mathrm{id}$ by [F4], whence the two agree by uniqueness of the linear initial-value problem; therefore the model has no conjugate point along $\tilde{\gamma }$ in $(0,T]$, because for $k>0$ one has $T=t_0<\pi \mathrm{/}\sqrt{k}$ with ${\mathrm{sn}}_k>0$ on $(0,\pi \mathrm{/}\sqrt{k})$, and for $k\le 0$ one has ${\mathrm{sn}}_k>0$ on all of $(0,T]$ [F4, F5]. Apply Rauch's first form [F3] with $M_1:=M_k^n$, ${\gamma }_1:=\tilde{\gamma }$, $J_1:=J_k$ and $M_2:=M$, ${\gamma }_2:=\gamma$, $J_2:=J$: both fields vanish at $0$ and have initial-derivative norm $a_0>0$, the pointwise curvature hypothesis holds because every radial curvature of the model is $k$ while every radial curvature of $M$ is at most $k$ [F4, F6], and the model has no conjugate point in $(0,T]$ as just noted. The conclusion gives $a_0{\mathrm{sn}}_k(t)=\mathrm{\mid }{J}_k(t)\mathrm{\mid }\le \mathrm{\mid }J(t)\mathrm{\mid }(0\le t\le T)\mathrm{.}$ At $t=T=t_0$ the right-hand side vanishes, while the left-hand side is $a_0{\mathrm{sn}}_k(t_0)>0$: for $k>0$ this uses $t_0<\pi \mathrm{/}\sqrt{k}$ and for $k\le 0$ the positivity of ${\mathrm{sn}}_k$ at every positive time [F4]. This is a contradiction. Hence no conjugate instant lies in $(0,\pi \mathrm{/}\sqrt{k})$ when $k>0$, and none lies at any positive time when $k\le 0$. All data are those of the hypothetical conjugate pair and of one model geodesic, so the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ of [A1] is not drawn on beyond its declaration.