A geodesic does not minimize past its first conjugate point

By [F1], choose a nonzero Jacobi field $J$ on $[a,c]$ with $J(a)=J(c)=0$. Put $q=D_{t}J(c^{-})$; if $q=0$, then the data $J(c)=D_{t}J(c^{-})=0$ and uniqueness [F3] force $J=0$, so $q\mathrm{\ne }0$. Define the continuous piecewise smooth field $V$ on $[a,b]$ by $V=J$ on $[a,c]$ and $V=0$ on $[c,b]$. Then $V(a)=V(c)=V(b)=0$, its only derivative jump is ${\mathrm{\Delta }}_c{D}_{t}V=-q$, and its Jacobi residual vanishes on both pieces by [F2].

By [F6], let $E$ be the unique parallel section along $\gamma$ with $E(c)=-q$. Set $\chi (t)=\frac{(t-a)(b-t)}{(c-a)(b-c)}$ and $W(t)=\chi (t)E(t)$. The denominator is positive because $a<c<b$; hence $W$ is smooth, $W(a)=W(b)=0$, and $W(c)=-q$.

Apply [F5] to $(V,Z)$ for any continuous piecewise $C^1$ fixed-endpoint field $Z$. Its endpoint term vanishes, its Jacobi-residual integral is zero, and the only jump is $-q$, so $I_{\gamma }(V,Z)=-g(-q,Z(c))=g(q,Z(c))$. Taking $Z=V$ gives $I_{\gamma }(V,V)=0$, while taking $Z=W$ gives $I_{\gamma }(V,W)=-g(q,q)<0$.

By bilinearity and symmetry [F4], for $X_{\delta }=V+\delta W$ one has $I_{\gamma }(X_{\delta },X_{\delta })=-2\delta g(q,q)+{\delta }^2{I}_{\gamma }(W,W)$. Put $A=g(q,q)>0$ and choose $0<\delta <A\mathrm{/}(\mathrm{\mid }{I}_{\gamma }(W,W)\mathrm{\mid }+1)$. Then $\delta \mathrm{\mid }{I}_{\gamma }(W,W)\mathrm{\mid }<A$, so $I_{\gamma }(X_{\delta },X_{\delta })<-\delta A<0$. The field $X_{\delta }$ is continuous, piecewise smooth, and zero at both endpoints.

Fix this $X=X_{\delta }$ and define $\beta (s,t)={\exp }_{\gamma (t)}(sX(t))$. The map $(s,t)\mapsto (\gamma (t),sX(t))$ is continuous and the open exponential domain in [F7] contains its zero section at $s=0$; compactness in [F9] supplies one $\epsilon >0$ for which $\beta$ is defined for all $\mathrm{\mid }s\mathrm{\mid }<\epsilon$ and $t\in [a,b]$. It is continuous and smooth on the strips $[a,c]$ and $[c,b]$; at $c$, both strip formulas and all their pure parameter derivatives agree because $X$ is continuous there. Since $X(a)=X(b)=0$, the variation fixes both endpoints. Its variation field is $X$ by [F8]. Apply the second-variation formula to $B(s,r,t)=\beta (s+r,t)$ on a smaller parameter square: its two fields are both $X$, so $e^{\mathrm{\prime }\mathrm{\prime }}(0)=I_{\gamma }(X,X)<0$ for $e(s)=E(\beta (s,\cdot ))$. Smoothness on the compact strips and [F10] make $e$ twice continuously differentiable.

The first-variation formula [F10] and the fixed endpoints give $e^{\mathrm{\prime }}(0)=0$. Thus [F11] makes $s=0$ a strict local maximum of $e$, so for a sufficiently small nonzero $s$ the fixed-endpoint competitor $\beta (s,\cdot )$ has $E(\beta (s,\cdot ))<E(\gamma )$.

Since $\gamma$ is a nonconstant affine Levi-Civita geodesic, [F12] gives constant positive speed and therefore $L(\gamma {)}^2=2(b-a)E(\gamma )$. Applying [F12] to the same-interval competitor from step 5.1 yields $L(\beta (s,\cdot ){)}^2\le 2(b-a)E(\beta (s,\cdot ))<L(\gamma {)}^2$, hence its length is strictly smaller too.

The supplied curve rules out an empty manifold. If $\dim M=0$, [F13] gives $\dot{\gamma }=0$, so the required nonconstant geodesic does not exist; dimension one needs no separate construction because no step divides by dimension or requires a normal direction. The hypothesis $a<c<b$ excludes a degenerate interval and makes both pieces nondegenerate; $J$ and $V$ use one-sided derivatives at $a,c,b$, and the perturbation fixes the two outer endpoints. A zero Jacobi field cannot witness conjugacy by [F1]; constant geodesics are excluded here and, in any case, have no conjugate endpoints by the stated definition. The exact axiom is [A1]; compactness gives only a finite subcover, the parallel field is unique, and the proof assumes no full AC. Both iff cases are inapplicable because this theorem is a one-way implication from an interior conjugate point to a strict competitor, not an equivalence. [A1, F1, F6, F9, F13, step 1.1, step 4.1, step 6.1] $\mathrm{\square }$