No conjugate points in nonpositive constant curvature

For every Jacobi field $J$ along $\gamma$ the function $\phi =g(J,T)$ is affine on $[a,b]$. [F2, F3, F4, F7, F8] By [F3] and [F4], ${\phi }^{\mathrm{\prime }}=g(D_{t}J,T)+g(J,D_{t}T)=g(D_{t}J,T)$, and then ${\phi }^{\mathrm{\prime }\mathrm{\prime }}=g(D_t^{2}J,T)+g(D_{t}J,D_{t}T)=g(D_t^{2}J,T)$. By the Jacobi equation [F2] and last-pair skewness [F7], ${\phi }^{\mathrm{\prime }\mathrm{\prime }}=-g(R(J,T)T,T)=-\mathrm{Rm}(J,T,T,T)=0$ at every interior point. Since ${\phi }^{\mathrm{\prime }}$ is continuous on $[a,b]$ with vanishing derivative inside, [F8] makes ${\phi }^{\mathrm{\prime }}$ constant, and a further application of [F8] makes $\phi$ affine: $\phi (t)=\alpha t+\beta$.

Let $J$ be a normal Jacobi field along $\gamma$, so $g(J,T)=0$ on $[a,b]$, and put $u(t)=\mathrm{\mid }J(t){\mathrm{\mid }}^2$. Then $D_t^{2}J=-K{c}^{2}J$ and $u^{\mathrm{\prime }\mathrm{\prime }}=-2K{c}^{2}u+2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2$ on $(a,b)$, where $c^2=g(T,T)$ is the constant squared speed. [F2, F3, F4, F5, F6] The curvature term of the Jacobi equation is $R(J,T)T=K(g(T,T)J-g(J,T)T)=K{c}^{2}J$ by [F6] and normality, so [F2] gives $D_t^{2}J=-K{c}^{2}J$. The product rule [F4] gives $u^{\mathrm{\prime }}=2g(D_{t}J,J)$ and then $u^{\mathrm{\prime }\mathrm{\prime }}=2g(D_t^{2}J,J)+2g(D_{t}J,D_{t}J)=-2K{c}^{2}u+2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2$, where $c^2=g(T,T)$ is constant by [F3] and both $u$ and $\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2$ are continuous up to the endpoints by [F5].

Every Jacobi field $J$ with $J(a)=J(b)=0$ is normal on $[a,b]$. [given, step 1.1] By step 1.1, $\phi =g(J,T)$ is affine; the endpoint hypotheses give $\phi (a)=g(0,T(a))=0$ and $\phi (b)=g(0,T(b))=0$. An affine function vanishing at the two distinct points $a<b$ is identically zero: writing $\phi (t)=\alpha t+\beta$, the two equations give $\alpha (b-a)=0$, hence $\alpha =\beta =0$. Therefore $g(J,T)=0$ throughout $[a,b]$.

For a normal Jacobi field $J$ along $\gamma$ with $J(a)=J(b)=0$, the function $u=\mathrm{\mid }J{\mathrm{\mid }}^2$ satisfies $u^{\mathrm{\prime }\mathrm{\prime }}\ge 0$ on $(a,b)$ because $K\le 0$; hence $u$ is convex on $(a,b)$ by [F9], and $u\ge 0$ on $[a,b]$ with $u(a)=u(b)=0$ and $u$ continuous on $[a,b]$. [F5, F9, given, step 1.2] By step 1.2, $u^{\mathrm{\prime }\mathrm{\prime }}=-2K{c}^{2}u+2\mathrm{\mid }{D}_{t}J{\mathrm{\mid }}^2$; since $K\le 0$ and $u\ge 0$ the first term is nonnegative, and the second is nonnegative as well, so $u^{\mathrm{\prime }\mathrm{\prime }}\ge 0$ on $(a,b)$. [F9] therefore makes $u$ convex on the open interval. Nonnegativity and the endpoint values come from [F5] and the hypotheses, and continuity up to the endpoints is [F5].

If $J$ is a Jacobi field along $\gamma$ with $J(a)=J(b)=0$, then $J$ vanishes identically on $[a,b]$. [F5, step 2.1, step 2.2] By step 2.1 such a $J$ is normal, so step 2.2 applies to $u=\mathrm{\mid }J{\mathrm{\mid }}^2$. Fix $t\in (a,b)$. For $a<t_1<t<t_2<b$ convexity gives $u(t)\le \lambda u(t_1)+(1-\lambda )u(t_2)$ with $\lambda =(t_2-t)\mathrm{/}(t_2-t_1)\in (0,1)$. Letting $t_1\downarrow a$ and $t_2\uparrow b$ and using the endpoint values and continuity gives $u(t)\le 0$; with $u(t)\ge 0$ this forces $u(t)=0$, and positive definiteness of $g$ gives $J(t)=0$. As $t\in (a,b)$ was arbitrary, $J\equiv 0$ on $[a,b]$, the endpoint values being given.

Consequently $\gamma (a)$ and $\gamma (b)$ are not conjugate along $\gamma {\mathrm{\mid }}_{[a,b]}$ and no positive multiplicity occurs. [F1, given, step 3.1] By step 3.1 the only Jacobi field along the segment vanishing at both endpoints is the zero field, so the vanishing space is $\{0\}$; by [F1] the endpoints are not conjugate along the segment, and the multiplicity, being defined only for a conjugate pair, does not arise. Since $a<b$ were arbitrary in $I$, no pair of distinct times of $\gamma$ is conjugate along the corresponding subsegment.

Boundary, degeneracy and choice audit. [A1, F1, F2, F5, step 1.1, step 2.1, step 2.2, step 3.1] The subinterval $[a,b]$ is nondegenerate by hypothesis, and included endpoints carry the one-sided conventions of [F2] and [F5]. If $M$ is empty there is no geodesic; in dimension zero the only field is zero, so the vanishing space is $\{0\}$ by [F1]; in dimension one a field normal to $T$ vanishes, so step 2.1 already forces $J=0$. The constant-geodesic case is excluded by the hypothesis that $\gamma$ is nonconstant and would in any case be a non-conjugate case by the explicit clause of [F1]. The zero field is the only endpoint-vanishing Jacobi field, which is exactly the claim, and the case $K=0$ is included in the estimate $u^{\mathrm{\prime }\mathrm{\prime }}\ge 0$; the argument never divides by $K$. The function $u$ is a squared length and is allowed to vanish on all of $[a,b]$; no strict convexity is asserted. Assumption [A1] is inherited from the curvature and Riemann-tensor suppliers, and no selection or countable family is used. The example proves the stated one-way non-conjugacy claim and asserts no converse. $\mathrm{\square }$