A geodesic stops minimizing exactly at its first conjugate point

Construct the rectangular quotient with periods $L_i>0$ in its first $r$ coordinates. [construct, F5] For $1\le r\le n$, let $\mathrm{\Lambda }=\{(L_1{k}_1,\dots ,L_r{k}_r,0,\dots ,0):k_i\in \mathbb{Z}\}\subset {\mathbb{R}}^n,M={\mathbb{R}}^n\mathrm{/}\mathrm{\Lambda },$ and let $q:{\mathbb{R}}^n\to M$ be the quotient map. Balls of radius less than $\frac{1}{2}{\min }_i{L}_i$ have disjoint translates, so their quotient images give charts on which $q$ is a diffeomorphism; chart changes are translations. The Euclidean metric therefore defines a consistent quotient metric, and in each such chart its coefficients are constant. These charts show that $M$ is boundaryless; it is connected because it is the continuous image of ${\mathbb{R}}^n$. The same charts show that $q$ is a covering and that it preserves speed on every lifted curve piece.

Every geodesic lifts to an affine line, giving the exponential formula and an all-time extension. [step 1.1, F3, F6, F7] Every geodesic in this quotient lifts, by [F3], to a curve in ${\mathbb{R}}^n$. On each periodic chart, [F6] gives ${\mathrm{\Gamma }}^k{}_{ij}=0$, so [F7] says that the lift has zero ordinary second derivative. The chart transitions are translations, so these affine pieces combine into one affine line $\tilde{\gamma }(t)=x+tw$ on its whole parameter interval. Its projection $q(x+tw)$ exists for all real $t$ and extends the geodesic; thus $M$ is geodesically complete. For $p=q(x)$ and the tangent vector identified with $w$ by the local chart, this also gives ${\exp }_p(w)=q(x+w)\mathrm{.}$

Quotient distance is the minimum Euclidean length among all lattice-separated lifts. [step 1.1, F1, F3, F4, F5] For $x,y\in {\mathbb{R}}^n$, let $\alpha$ be any piecewise $C^1$ path from $q(x)$ to $q(y)$. Its lift from $x$ ends at $y+\lambda$ for some $\lambda \in \mathrm{\Lambda }$ by [F3]. Speed preservation from step 1.1 and the Euclidean endpoint length bound give $L_g(\alpha )=L_{{\mathbb{R}}^n}(\tilde{\alpha })\ge \mathrm{\mid }y+\lambda -x\mathrm{\mid }\mathrm{.}$ Refining the finitely many curve pieces into quotient-chart neighborhoods makes each lifted piece a smooth inverse-branch image, so the lift is piecewise $C^1$ and its length is computed by [F5]. For each smooth piece, the fundamental theorem of calculus writes its displacement as the integral of its velocity; the triangle inequality for the finite sum of those integrals gives the displayed endpoint bound. Conversely, the straight segment from $x$ to $y+\lambda$ projects to a path of length $\mathrm{\mid }y+\lambda -x\mathrm{\mid }$. Thus [F4] gives the infimum of these lengths. It is a minimum: $\lambda =0$ gives the bound $\mathrm{\mid }y-x\mathrm{\mid }$, and every candidate no longer than this satisfies $\mathrm{\mid }\lambda \mathrm{\mid }\le 2\mathrm{\mid }y-x\mathrm{\mid }$. Thus each integer coordinate satisfies $\mathrm{\mid }{k}_i\mathrm{\mid }\le 2\mathrm{\mid }y-x\mathrm{\mid }\mathrm{/}{L}_i$, leaving only finitely many tuples, so a least candidate exists. We have proved $d_g(q(x),q(y))={\min }_{\lambda \in \mathrm{\Lambda }}\mathrm{\mid }y+\lambda -x\mathrm{\mid }\mathrm{.}$ The quotient is the product of the $r$ periodic circles and ${\mathbb{R}}^{n-r}$; the formula splits into the sum of the squared circle distances and the squared Euclidean distance. Each circle is complete: for its coordinate quotient $q_i:\mathbb{R}\to \mathbb{R}\mathrm{/}{L}_i\mathbb{Z}$, the formula gives $d(q_i(u),q_i(v))\le \mathrm{\mid }u-v\mathrm{\mid }$, so $q_i{\mathrm{\mid }}_{[0,L_i]}$ is continuous and surjective with compact image. For any Cauchy sequence, the closures of its nonempty tails are nested nonempty closed subsets; compactness gives a common point, and the Cauchy property forces convergence to it. Thus each circle is complete. Coordinatewise convergence then makes the finite product with its sum-of-squares distance complete. Thus every model used below meets the Riemannian-completeness hypothesis of [F1].

The quotient is flat, and every Jacobi field lifts to an affine vector field. [step 2.1, F2, F6, F8, F9] All coordinate derivatives of the constant Christoffel symbols vanish, so [F8] gives $R=0$ in every quotient chart. Lift a Jacobi field to the affine geodesic in ${\mathbb{R}}^n$ using the derivative of the covering chart; chart changes are translations, so the lifted vector components agree on overlaps. Writing $J=J^k{\mathrm{\partial }}_k$, [F9] and the coordinate Christoffel symbols give $(D_{t}J{)}^k={\dot{J}}^k+{\mathrm{\Gamma }}^k{}_{ij}{\dot{x}}^i{J}^j$. The second term is zero by [F6], and [F2] therefore reduces to ${\tilde{J}}^{\mathrm{\prime }\mathrm{\prime }}(t)=0.$ Consequently $\tilde{J}(t)=A+tB$. If $\tilde{J}(s_0)=\tilde{J}(s_1)=0$ for $s_0<s_1$, then $(s_1-s_0)B=0$, hence $B=A=0$. The lift, and therefore $J$, is zero. By the stated two-endpoint criterion, this quotient has no conjugate points along any geodesic.

The rectangular torus Dirichlet cell is the product of half-period intervals. [step 2.2, algebra] For a full rectangular torus ($r=n$), step 2.2 gives $d_g(q(0),q(x){)}^2={\sum }_{i=1}^n{\min }_{k_i\in \mathbb{Z}}\mathrm{\mid }{x}_i+k_i{L}_i{\mathrm{\mid }}^2\mathrm{.}$ The coordinate minima are attained independently, so choosing a nearest integer in each of the finitely many coordinates attains their summed value. By definition, the Euclidean Dirichlet cell of the origin is $V_0=\{x\in {\mathbb{R}}^n:\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }x-\lambda \mathrm{\mid }\text{ for every }\lambda \in \mathrm{\Lambda }\}\mathrm{.}$ This cell is exactly ${\prod }_{i=1}^n[-L_i\mathrm{/}2,L_i\mathrm{/}2]\mathrm{.}$ Indeed comparison with the adjacent lattice points $\pm L_i{e}_i$ forces $-L_i\mathrm{/}2\le x_i\le L_i\mathrm{/}2$; inside that box each coordinate of $0$ is nearest among all its period translates, so summing the coordinate inequalities proves comparison with every lattice point. At a face, the origin and the adjacent translate tie as nearest lifts; intersections of faces have more ties. Thus the torus boundary consists of points with multiple minimizing lifts, as claimed by the scaffold calculation.

The length-$L$ circle has cut time $L\mathrm{/}2$, with two minimizing lifts at its cut point. [F1, step 2.1, step 2.2] Take $n=r=1$, so $M=\mathbb{R}\mathrm{/}L\mathbb{Z}$ is a circle, and choose $p=q(0)$ and the unit direction $v=1$. By step 2.1, $\gamma (t)=q(t)$; step 2.2 gives $d_g(p,\gamma (t))={\min }_{k\in \mathbb{Z}}\mathrm{\mid }t-kL\mathrm{\mid }\mathrm{.}$ For $0\le t\le L\mathrm{/}2$, the minimum is $t$: the $k=0$ lift has length $t$, every $k\le -1$ is farther, and every $k\ge 1$ has distance at least $L-t\ge t$. At $t=L\mathrm{/}2$, precisely the lifts $L\mathrm{/}2$ and $-L\mathrm{/}2$ tie, giving two distinct minimizing semicircles. For every $t>L\mathrm{/}2$, the lift ending at $t-L$ is shorter, since $\mathrm{\mid }t-L\mathrm{\mid }<t$. Hence the minimizing positive times are exactly $(0,L\mathrm{/}2]$, and [F1] gives $c_p(v)=L\mathrm{/}2.$ The endpoint still minimizes; every later point on this ray fails to minimize.

The complete cylinder has the same finite horizontal cut time and no conjugate points. [step 2.2, step 3.1, step 3.3] The cylinder $S^1(L)\times \mathbb{R}$ is the quotient with $n=2$, $r=1$; its two coordinates are the circular and axial directions. Step 3.1 proves that every cylinder geodesic has no conjugate instant, while the distance formula in step 2.2 and the calculation in step 3.3 give cut time $L\mathrm{/}2$ on the horizontal ray and two minimizing paths at its endpoint. This is the cylinder model cited by Lee.

A complete circle has a finite cut time but no conjugate instant, refuting the proposed claim. [step 1.1, step 2.2, step 3.1, step 3.3, step 4.1] The circle is connected and boundaryless by step 1.1, complete by step 2.2, and has a finite cut time $L\mathrm{/}2$ by step 3.3, but has no conjugate point by step 3.1. The complete cylinder supplies the same phenomenon by step 4.1. Therefore a geodesic can cease to minimize without reaching any conjugate point, so the cut time may precede every conjugate point.

Empty, zero-dimensional, one-dimensional, endpoint, degeneracy and choice cases are settled. [A1, F1, step 2.2, step 3.3, step 5.1] The model has $n\ge 1$ and $r\ge 1$; the one-dimensional circle already provides the witness, and the two-dimensional cylinder is Lee's cited model. An empty manifold has no base point, and in dimension zero there is no unit direction, so neither case is an instance of the universal cut-time setup. At $t=0$ the radial segment is constant and minimizing; the finite cut-time endpoint $L\mathrm{/}2$ is included, while every $t>L\mathrm{/}2$ fails strictly. Periods are assumed positive, and a constant geodesic is not used. The construction chooses no family of points, paths or lattice vectors: for each distance calculation the bounded candidate set is finite. The compact metric-space completeness argument in step 2.2 uses only nested closed tail closures; quotient path lifts are unique and all lattice searches are finite. The declared ${\mathrm{A}\mathrm{C}}_{\omega }$ enters only as the inherited hypothesis [F1] for cut time; no local calculation spends it and no full Axiom of Choice is used. This is an existential counterexample, not an if-and-only-if assertion.