[proofplan] We exploit the deck-transformation action of $\Gamma = \pi_1(M)$ on $\tilde M$ to manufacture longer and longer minimizing geodesic segments and then extract a line by an Arzelà–Ascoli argument. Compactness of $M$ together with non-compactness of $\tilde M$ forces $\Gamma$ to be infinite, so we can pick $\gamma_n \in \Gamma$ with $d_{\tilde g}(\tilde x, \gamma_n \tilde x) \to \infty$ and join $\tilde x$ to $\gamma_n \tilde x$ by a unit-speed minimizing geodesic. Translating each segment by an element of $\Gamma$ to push its midpoint into a fixed compact fundamental domain, we extract a subsequence of midpoint tangent vectors converging in the unit tangent bundle over a compact set; smooth dependence of geodesics on initial data then yields a globally minimizing geodesic $\sigma_\infty: \mathbb R \to \tilde M$, i.e., a line. [/proofplan]

[step:Set up the deck-transformation action and check that $\tilde M$ is geodesically complete]Since $M$ is connected and $\pi: \tilde M \to M$ is the universal Riemannian cover, the deck-transformation group \begin{align*} \Gamma := \operatorname{Deck}(\pi: \tilde M \to M) \end{align*} is canonically isomorphic to $\pi_1(M, x_0)$ for any basepoint $x_0 \in M$, and $\Gamma$ acts on $\tilde M$ properly discontinuously and freely by deck transformations. Because $\tilde g = \pi^* g$, the covering map $\pi: (\tilde M, \tilde g) \to (M, g)$ is a local isometry, so each deck transformation $\gamma: \tilde M \to \tilde M$ is a global isometry of $(\tilde M, \tilde g)$. The orbit space $\Gamma \backslash \tilde M$ is canonically identified with $M$ as a Riemannian manifold. We verify $(\tilde M, \tilde g)$ is geodesically complete. Since $M$ is compact, it is geodesically complete by the Hopf–Rinow theorem. A Riemannian covering of a complete manifold is complete: given a unit-speed geodesic $\tilde \sigma: [0, T) \to \tilde M$ with $T < \infty$, its projection $\sigma := \pi \circ \tilde \sigma$ is a unit-speed geodesic in $M$ (because $\pi$ is a local isometry and geodesics are local), hence extends to $[0, T + \varepsilon)$ in $M$ by completeness of $M$, and the extension lifts uniquely to $\tilde M$ by the path-lifting property of the covering map. So $\tilde \sigma$ extends to $[0, T + \varepsilon)$, contradicting maximality. Hence every geodesic of $\tilde M$ extends to all of $\mathbb R$.[/step]

[guided]The universal Riemannian cover is by definition the universal cover $\tilde M \to M$ equipped with the pullback metric $\tilde g = \pi^* g$. The pullback makes $\pi$ a local isometry: for each $\tilde p \in \tilde M$, the differential $d\pi_{\tilde p}: T_{\tilde p}\tilde M \to T_{\pi(\tilde p)} M$ is a linear isometry. From this we extract two pieces of structure that the rest of the proof depends on: an isometric action by deck transformations, and geodesic completeness of $\tilde M$. **The deck-transformation group acts by isometries.** Define \begin{align*} \Gamma := \operatorname{Deck}(\pi: \tilde M \to M), \end{align*} the group of automorphisms of the cover, canonically isomorphic to $\pi_1(M, x_0)$ for any basepoint. Why is each $\gamma \in \Gamma$ an isometry of $(\tilde M, \tilde g)$? A deck transformation is by definition a self-map of $\tilde M$ commuting with the projection: $\pi \circ \gamma = \pi$. Differentiating at any $\tilde p$ gives $d\pi_{\gamma(\tilde p)} \circ d\gamma_{\tilde p} = d\pi_{\tilde p}$. Since $d\pi$ is a linear isometry at every point of $\tilde M$, this equality forces $d\gamma_{\tilde p}: T_{\tilde p}\tilde M \to T_{\gamma(\tilde p)}\tilde M$ to be a linear isometry too. So $\gamma^* \tilde g = \tilde g$ pointwise, i.e., $\gamma$ is a Riemannian isometry. The action is also properly discontinuous and free (general covering-space theory), and the orbit space $\Gamma \backslash \tilde M$ is canonically identified with $M$ as a Riemannian manifold. **$(\tilde M, \tilde g)$ is geodesically complete.** This is what lets us join arbitrary pairs of points by minimizing geodesics, the engine of the entire argument. Why does completeness of $M$ imply completeness of $\tilde M$? Suppose $\tilde \sigma: [0, T) \to \tilde M$ is a unit-speed geodesic with $T < \infty$ that does not extend. We derive a contradiction. The projection $\sigma := \pi \circ \tilde \sigma$ is a unit-speed curve in $M$; it is locally a geodesic because $\pi$ is a local isometry and the geodesic equation is local. Since $M$ is compact, by the Hopf–Rinow theorem $M$ is geodesically complete, so $\sigma$ extends to a geodesic on $[0, T + \varepsilon)$ for some $\varepsilon > 0$. The path-lifting property of covering maps (any path in the base, given a lift of its initial point, has a unique lift in the cover) lifts the extended $\sigma$ uniquely to a curve in $\tilde M$ starting from $\tilde\sigma(0)$, which agrees with $\tilde\sigma$ on $[0, T)$ and extends it to $[0, T + \varepsilon)$. This contradicts maximality of $\tilde\sigma$. So every geodesic of $\tilde M$ extends to all of $\mathbb R$. By Hopf–Rinow on $\tilde M$, any two points are joined by a unit-speed minimizing geodesic — a fact we will use repeatedly to produce minimizing segments of arbitrary length.[/guided]

[step:Use non-compactness of $\tilde M$ to show $\Gamma$ is infinite and to produce far-away orbit points]We claim $\Gamma$ is infinite. The action of $\Gamma$ on $\tilde M$ by isometries has orbit space $M$, which is compact. Pick a closed fundamental domain \begin{align*} D \subset \tilde M \end{align*} for the action of $\Gamma$, meaning $D$ is closed, $\bigcup_{\gamma \in \Gamma} \gamma D = \tilde M$, and the projection $\pi: D \to M$ is surjective. Since $M$ is compact, we may choose $D$ to be compact: lift a finite cover of $M$ by evenly covered open sets, each of whose closures is compact and projects diffeomorphically to its image. If $\Gamma$ were finite with $|\Gamma| = N < \infty$, then \begin{align*} \tilde M = \bigcup_{\gamma \in \Gamma} \gamma D \end{align*} would be a finite union of compact sets (each $\gamma D$ is the isometric image of the compact set $D$ under the diffeomorphism $\gamma$, hence compact), hence $\tilde M$ would be compact. This contradicts the hypothesis that $\tilde M$ is non-compact. So $\Gamma$ is infinite. Fix any basepoint $\tilde x \in D$. We claim \begin{align*} \sup_{\gamma \in \Gamma} d_{\tilde g}(\tilde x, \gamma \tilde x) = \infty. \end{align*} Suppose not: then $d_{\tilde g}(\tilde x, \gamma \tilde x) \leq R$ for some $R < \infty$ and all $\gamma \in \Gamma$, so the orbit $\Gamma \cdot \tilde x$ is contained in the closed ball $\overline{B}_{\tilde g}(\tilde x, R)$. By Hopf–Rinow on $\tilde M$ (geodesically complete), this closed ball is compact. But the action is properly discontinuous, so the orbit $\Gamma \cdot \tilde x$ is a discrete subset of $\tilde M$; a discrete set inside a compact set is finite, contradicting $|\Gamma| = \infty$ (since the action is free, $|\Gamma \cdot \tilde x| = |\Gamma|$). Therefore there exists a sequence $\gamma_n \in \Gamma$ with \begin{align*} L_n := d_{\tilde g}(\tilde x, \gamma_n \tilde x) \longrightarrow \infty. \end{align*}[/step]

[guided]The goal of this step is two-fold: show that $\Gamma$ is infinite, and use this to produce a sequence of orbit points marching out to infinity in $\tilde M$. Both rely on a careful choice of *compact* fundamental domain. **Constructing a compact fundamental domain $D$.** A fundamental domain for the properly discontinuous free action $\Gamma \curvearrowright \tilde M$ is a closed set $D \subset \tilde M$ with $\bigcup_{\gamma \in \Gamma} \gamma D = \tilde M$ and the interiors of distinct translates $\gamma D, \gamma' D$ disjoint. We can choose $D$ to be compact when $M$ is. The construction: cover $M$ by finitely many open sets $V_1, \ldots, V_k$ each evenly covered by $\pi$ — i.e., $\pi^{-1}(V_i)$ is a disjoint union of open sets each mapping diffeomorphically to $V_i$. Pick closed compact $K_i \subset V_i$ with $\bigcup_i K_i = M$ (possible because $M$ is compact and locally compact Hausdorff). For each $i$ lift $K_i$ to a compact $\tilde K_i \subset \tilde M$ projecting diffeomorphically onto $K_i$. Set \begin{align*} D := \bigcup_{i=1}^k \tilde K_i \subset \tilde M. \end{align*} This $D$ is compact (finite union of compact sets), closed, and the $\Gamma$-translates of $D$ cover $\tilde M$ because $\Gamma \cdot \tilde K_i = \pi^{-1}(K_i)$ and the $K_i$ cover $M$. **$\Gamma$ is infinite.** Suppose for contradiction that $|\Gamma| = N < \infty$. Then \begin{align*} \tilde M = \bigcup_{\gamma \in \Gamma} \gamma D \end{align*} is a finite union of compact sets — each $\gamma D$ is the image of the compact $D$ under the diffeomorphism $\gamma$, hence compact — and therefore $\tilde M$ is compact. This contradicts the standing hypothesis that $\tilde M$ is non-compact. Hence $\Gamma$ must be infinite. **The orbit $\Gamma \cdot \tilde x$ is unbounded.** Fix any basepoint $\tilde x \in D$. We claim \begin{align*} \sup_{\gamma \in \Gamma} d_{\tilde g}(\tilde x, \gamma \tilde x) = \infty. \end{align*} Why must this hold? Suppose instead $d_{\tilde g}(\tilde x, \gamma \tilde x) \leq R$ for all $\gamma \in \Gamma$ and some $R < \infty$. Then $\Gamma \cdot \tilde x \subset \overline{B}_{\tilde g}(\tilde x, R)$. By the Hopf–Rinow theorem applied to the geodesically complete $\tilde M$, the closed ball $\overline{B}_{\tilde g}(\tilde x, R)$ is compact. Proper discontinuity of the action says: for every compact $K \subset \tilde M$, only finitely many $\gamma \in \Gamma$ satisfy $\gamma K \cap K \neq \varnothing$. Apply this with $K = \overline{B}_{\tilde g}(\tilde x, R)$: since $\tilde x \in K$ and $\gamma \tilde x \in K$, we have $\gamma \tilde x \in \gamma K \cap K$, so only finitely many $\gamma$ qualify. Freeness of the action means distinct $\gamma$ give distinct orbit points, so $|\Gamma| < \infty$, contradicting the previous paragraph. **Producing the sequence.** Since the orbit is unbounded, we can pick a sequence $\gamma_n \in \Gamma$ with \begin{align*} L_n := d_{\tilde g}(\tilde x, \gamma_n \tilde x) \longrightarrow \infty. \end{align*} This is the input to the next step: $\tilde x$ and $\gamma_n \tilde x$ are pairs of points whose distance grows without bound, and we will join them by minimizing geodesics whose lengths $L_n$ also grow without bound.[/guided]

[step:Connect $\tilde x$ to $\gamma_n \tilde x$ by a minimizing unit-speed geodesic and reparametrize so the midpoint is at $0$]Since $(\tilde M, \tilde g)$ is geodesically complete and connected, the Hopf–Rinow theorem provides for each $n$ a unit-speed minimizing geodesic \begin{align*} \tilde \sigma_n: [0, L_n] &\to \tilde M, \\ \tilde \sigma_n(0) &= \tilde x, \quad \tilde \sigma_n(L_n) = \gamma_n \tilde x, \end{align*} with $d_{\tilde g}(\tilde \sigma_n(s), \tilde \sigma_n(t)) = |s - t|$ for all $s, t \in [0, L_n]$. Reparametrize by shifting time so that the midpoint sits at parameter $0$: define \begin{align*} \sigma_n: [-a_n, b_n] &\to \tilde M, & \sigma_n(t) &:= \tilde \sigma_n(t + L_n/2), \end{align*} where $a_n := L_n / 2 = b_n$. Then $\sigma_n$ is a unit-speed minimizing geodesic on $[-a_n, b_n]$ with $a_n, b_n \to \infty$.[/step]

[guided]We have a sequence of pairs of points $(\tilde x, \gamma_n \tilde x)$ in $\tilde M$ with separation $L_n \to \infty$. To get a limit geodesic we need to (i) realize each pair by a unit-speed minimizing geodesic and (ii) parametrize so the segments share a natural reference point at parameter $0$ — namely the midpoint, which we will later push into the compact fundamental domain. **Producing the geodesic segment.** Since $(\tilde M, \tilde g)$ is geodesically complete and connected, the Hopf–Rinow theorem guarantees that any two points are joined by a unit-speed minimizing geodesic. Apply this to the pair $\tilde x, \gamma_n \tilde x$: there exists \begin{align*} \tilde \sigma_n: [0, L_n] &\to \tilde M, \\ \tilde \sigma_n(0) &= \tilde x, \quad \tilde \sigma_n(L_n) = \gamma_n \tilde x, \end{align*} with the minimizing identity $d_{\tilde g}(\tilde \sigma_n(s), \tilde \sigma_n(t)) = |s - t|$ for all $s, t \in [0, L_n]$. The unit-speed parametrization is a normalization choice — we want all the $\sigma_n$ to share a single time scale so a limit makes sense. **Centering at the midpoint.** The midpoint of the original parametrization is at $t = L_n/2$, but for the convergence argument ahead we want the midpoints to live at the *same* parameter value across all $n$, namely $t = 0$, so that we can extract a limit geodesic $\sigma_\infty: \mathbb R \to \tilde M$ with $\sigma_\infty(0) = \tilde p_\infty$. Define \begin{align*} \sigma_n: [-a_n, b_n] &\to \tilde M, & \sigma_n(t) &:= \tilde \sigma_n(t + L_n/2), \end{align*} where $a_n := L_n/2 = b_n$. The new parameter interval $[-a_n, b_n] = [-L_n/2, L_n/2]$ has $a_n, b_n \to \infty$, so eventually it covers any compact interval $[-T, T]$ — which is what we need for the Arzelà–Ascoli step. **The reparametrization preserves all metric properties.** A time translation is an isometry of the parameter interval, so unit-speed-ness and the minimizing identity transfer cleanly. For any $t_1, t_2 \in [-a_n, b_n]$, \begin{align*} d_{\tilde g}(\sigma_n(t_1), \sigma_n(t_2)) = d_{\tilde g}(\tilde \sigma_n(t_1 + a_n), \tilde \sigma_n(t_2 + a_n)) = |(t_1 + a_n) - (t_2 + a_n)| = |t_1 - t_2|, \end{align*} where the middle equality uses that $\tilde \sigma_n$ is minimizing on its full domain $[0, L_n]$ and both $t_1 + a_n, t_2 + a_n$ lie in $[0, L_n]$. So $\sigma_n: [-a_n, b_n] \to \tilde M$ is a unit-speed minimizing geodesic with parameter interval growing to all of $\mathbb R$ and midpoint at $\sigma_n(0) = \tilde\sigma_n(L_n/2)$.[/guided]

[step:Translate each $\sigma_n$ by an element of $\Gamma$ so the midpoint $\sigma_n(0)$ lies in a fixed compact set]The midpoint $\sigma_n(0) \in \tilde M$ projects to a point $\pi(\sigma_n(0)) \in M$. We want all midpoints to lie in a fixed compact set, which is achieved by the fundamental-domain trick. Recall the compact fundamental domain $D \subset \tilde M$ from the previous step. For each $n$, choose $\delta_n \in \Gamma$ such that \begin{align*} \delta_n \cdot \sigma_n(0) \in D. \end{align*} Such $\delta_n$ exists: $\bigcup_{\gamma \in \Gamma} \gamma D = \tilde M$, so $\sigma_n(0) \in \gamma_n^{-1} D$ for some $\gamma_n^{-1} \in \Gamma$, and we set $\delta_n := \gamma_n^{-1}$. Replace $\sigma_n$ by \begin{align*} \hat \sigma_n: [-a_n, b_n] &\to \tilde M, & \hat \sigma_n(t) &:= \delta_n \cdot \sigma_n(t). \end{align*} Since $\delta_n: \tilde M \to \tilde M$ is an isometry, $\hat \sigma_n$ is a unit-speed minimizing geodesic on $[-a_n, b_n]$ (isometries preserve geodesic-ness, parametrization speed, and the minimizing property). Moreover $\hat \sigma_n(0) = \delta_n \cdot \sigma_n(0) \in D$. Henceforth we drop the hat and write $\sigma_n$ for $\hat \sigma_n$. We have unit-speed minimizing geodesics $\sigma_n: [-a_n, b_n] \to \tilde M$ with $a_n, b_n \to \infty$ and $\sigma_n(0) \in D$, a fixed compact set.[/step]

[guided]This translation step is the geometric core of the argument. Why is it needed at all? Without it, the midpoints $\sigma_n(0)$ could march off to infinity in $\tilde M$ as $n \to \infty$ — they could be anywhere along the geodesic from $\tilde x$ out to $\gamma_n \tilde x$, with no a priori control. We could not then extract a convergent subsequence of midpoints. The fix is to use the $\Gamma$-action to pull each midpoint back into the compact fundamental domain $D$, exploiting that $\Gamma$ acts by isometries (so the geodesic structure is preserved) and that $\bigcup_\gamma \gamma D = \tilde M$ (so some translate always lands in $D$). **Choice of $\delta_n$.** Recall the compact fundamental domain $D \subset \tilde M$ from the previous step. Since $\bigcup_{\gamma \in \Gamma} \gamma D = \tilde M$, the midpoint $\sigma_n(0) \in \tilde M$ lies in some translate $\gamma D$, say $\sigma_n(0) \in \gamma'_n D$ for some $\gamma'_n \in \Gamma$. Setting $\delta_n := (\gamma'_n)^{-1}$, we have $\delta_n \cdot \sigma_n(0) \in D$: \begin{align*} \delta_n \cdot \sigma_n(0) \in D. \end{align*} **Translating the geodesic.** Replace $\sigma_n$ by its $\delta_n$-translate: \begin{align*} \hat \sigma_n: [-a_n, b_n] &\to \tilde M, & \hat \sigma_n(t) &:= \delta_n \cdot \sigma_n(t). \end{align*} Why does $\hat\sigma_n$ remain a unit-speed minimizing geodesic? Because $\delta_n: \tilde M \to \tilde M$ is a Riemannian isometry (Step 1), and isometries preserve every metric quantity that defines "unit-speed minimizing geodesic": they map geodesics to geodesics, preserve the parameter speed $|\dot \sigma_n(t)|_{\tilde g}$, and preserve distances (so the minimizing identity $d_{\tilde g}(\hat\sigma_n(s), \hat\sigma_n(t)) = d_{\tilde g}(\sigma_n(s), \sigma_n(t)) = |s-t|$ is automatic). In particular $\hat\sigma_n(0) = \delta_n \cdot \sigma_n(0) \in D$ as required. **A subtle point about the endpoints.** The deck transformation $\delta_n$ also moves the endpoints: the new endpoints are $\delta_n \tilde x$ and $\delta_n \gamma_n \tilde x$. Their distance is preserved by the isometry, so it is still $L_n \to \infty$. We do not need to track these endpoints explicitly in what follows — only the midpoint $\hat\sigma_n(0) \in D$ — but it is reassuring that the "growing length" property is intact. **Notation collapse.** Henceforth we drop the hat and write $\sigma_n$ for $\hat \sigma_n$. The summary of the situation: we have unit-speed minimizing geodesics $\sigma_n: [-a_n, b_n] \to \tilde M$ with $a_n, b_n \to \infty$ and $\sigma_n(0) \in D$, where $D$ is a fixed compact subset of $\tilde M$. This compactness of the midpoint locus is exactly what unlocks the Arzelà–Ascoli-style extraction in the next step.[/guided]

[step:Extract a subsequence of midpoint tangent vectors converging in the unit tangent bundle]The unit tangent bundle of $\tilde M$ is \begin{align*} T^1 \tilde M := \{(\tilde p, v) : \tilde p \in \tilde M,\; v \in T_{\tilde p} \tilde M,\; \tilde g_{\tilde p}(v, v) = 1\}. \end{align*} This is a smooth fiber bundle over $\tilde M$ with $(n-1)$-sphere fibers, where $n = \dim \tilde M$. Its restriction over the compact set $D$, \begin{align*} T^1 \tilde M \big|_D := \{(\tilde p, v) \in T^1 \tilde M : \tilde p \in D\}, \end{align*} is compact (a sphere bundle over a compact base, with the topology induced from $T \tilde M$). The sequence \begin{align*} (\sigma_n(0), \dot \sigma_n(0)) \in T^1 \tilde M \big|_D \end{align*} lies in this compact space (each $\sigma_n$ is unit speed, so $|\dot \sigma_n(0)|_{\tilde g} = 1$, and $\sigma_n(0) \in D$). By sequential compactness, there is a subsequence (re-index) and a limit \begin{align*} (\sigma_n(0), \dot \sigma_n(0)) \longrightarrow (\tilde p_\infty, v_\infty) \in T^1 \tilde M \big|_D \end{align*} with $|v_\infty|_{\tilde g} = 1$.[/step]

[guided]This is the Arzelà–Ascoli step, repackaged as sequential compactness of the unit tangent bundle. The strategy: a unit-speed geodesic on a complete manifold is determined by its initial position and initial velocity, so a sequence of unit-speed geodesics is encoded by a sequence in the unit tangent bundle. If those points sit in a compact subset of the unit tangent bundle, we can extract a limit, which will serve as the initial data of the limit geodesic in the next step. **The unit tangent bundle.** Define \begin{align*} T^1 \tilde M := \{(\tilde p, v) : \tilde p \in \tilde M,\; v \in T_{\tilde p} \tilde M,\; \tilde g_{\tilde p}(v, v) = 1\}, \end{align*} the bundle of unit tangent vectors. It is the level set $\{\tilde g(v,v) = 1\} \subset T\tilde M$ of a smooth submersion (the squared-norm function), hence a smooth submanifold of $T\tilde M$ of codimension $1$. Equivalently it is a smooth fiber bundle over $\tilde M$ with fiber the unit sphere $S^{n-1}$ in $T_{\tilde p}\tilde M$, where $n = \dim \tilde M$. **Why the restriction over $D$ is compact.** Restrict to the compact base: \begin{align*} T^1 \tilde M \big|_D := \{(\tilde p, v) \in T^1 \tilde M : \tilde p \in D\}. \end{align*} Why is this compact? The unit tangent bundle is closed in $T\tilde M$ because it is cut out by the continuous equation $\tilde g(v,v) = 1$. The full tangent bundle $T\tilde M|_D \to D$ has fiber $\mathbb R^n$ (non-compact), but the unit-norm subset has fiber $S^{n-1}$ (compact). Locally we can trivialize $T\tilde M|_D$ over a finite cover of $D$, and over each trivializing chart $T^1\tilde M$ looks like (chart) $\times S^{n-1}$, which is compact. A finite union of such pieces, glued along their overlaps inside the closed subset $T^1\tilde M$ of $T\tilde M$, is compact. **The sequence and its limit.** The sequence of midpoint tangent vectors lives in this compact space: \begin{align*} (\sigma_n(0), \dot \sigma_n(0)) \in T^1 \tilde M \big|_D. \end{align*} The first coordinate $\sigma_n(0)$ lies in $D$ by Step 4, and the second coordinate $\dot\sigma_n(0)$ has norm $1$ because each $\sigma_n$ is unit-speed. Why does sequential compactness apply? $T^1\tilde M|_D$ is a finite-dimensional smooth manifold, hence second-countable and Hausdorff, hence metrizable; compact metrizable spaces are sequentially compact. So after passing to a subsequence (re-indexed for convenience), \begin{align*} (\sigma_n(0), \dot \sigma_n(0)) \longrightarrow (\tilde p_\infty, v_\infty) \in T^1 \tilde M \big|_D, \end{align*} with $\tilde p_\infty \in D$ and $|v_\infty|_{\tilde g} = 1$ (since the unit-norm condition is closed under limits). The limit pair $(\tilde p_\infty, v_\infty)$ is the initial data of the line we will construct in the next step: $v_\infty$ is the unit direction in which a globally minimizing geodesic will run from the limit basepoint $\tilde p_\infty$.[/guided]

[step:Define the limit geodesic and prove it is a line]Define \begin{align*} \sigma_\infty: \mathbb R &\to \tilde M, & \sigma_\infty(t) &:= \exp_{\tilde p_\infty}(t v_\infty), \end{align*} the unit-speed geodesic with initial position $\tilde p_\infty$ and initial velocity $v_\infty$. It is defined on all of $\mathbb R$ because $\tilde M$ is geodesically complete. [claim:$\sigma_n$ converges to $\sigma_\infty$ uniformly on every compact interval $[-T, T] \subset \mathbb R$] [proof] Geodesics depend smoothly on initial conditions. More precisely, the geodesic flow \begin{align*} \phi: \mathbb R \times T \tilde M &\to \tilde M, & \phi(t, \tilde p, v) &:= \exp_{\tilde p}(tv), \end{align*} is smooth on the maximal domain of existence (which equals $\mathbb R \times T \tilde M$ here by completeness). In particular $\phi$ is continuous. Fix $T > 0$. For $n$ large enough, $a_n, b_n > T$, so $\sigma_n$ is defined on $[-T, T]$. We have \begin{align*} \sigma_n(t) = \exp_{\sigma_n(0)}(t \dot \sigma_n(0)) = \phi(t, \sigma_n(0), \dot \sigma_n(0)), \end{align*} and $(\sigma_n(0), \dot \sigma_n(0)) \to (\tilde p_\infty, v_\infty)$ in $T \tilde M$. Continuity of $\phi$ on $[-T, T] \times T \tilde M$ (a continuous function on a compact subset of its domain is uniformly continuous) gives $\phi(t, \sigma_n(0), \dot \sigma_n(0)) \to \phi(t, \tilde p_\infty, v_\infty) = \sigma_\infty(t)$ uniformly in $t \in [-T, T]$. [/proof] [/claim] [claim:$\sigma_\infty$ is a line, i.e., a unit-speed geodesic with $d_{\tilde g}(\sigma_\infty(s), \sigma_\infty(t)) = |s - t|$ for all $s, t \in \mathbb R$] [proof] $\sigma_\infty$ is unit-speed because $|v_\infty|_{\tilde g} = 1$. We must show globally minimizing. Fix $s, t \in \mathbb R$. Pick $T > \max(|s|, |t|)$. For $n$ large, $a_n, b_n > T$, so $\sigma_n$ is defined on $[-T, T] \supseteq \{s, t\}$ and is *minimizing* on $[-a_n, b_n] \supseteq [-T, T]$. In particular, \begin{align*} d_{\tilde g}(\sigma_n(s), \sigma_n(t)) = |s - t|. \end{align*} Letting $n \to \infty$, by uniform convergence on $[-T, T]$ we have $\sigma_n(s) \to \sigma_\infty(s)$ and $\sigma_n(t) \to \sigma_\infty(t)$. By continuity of the Riemannian distance function $d_{\tilde g}: \tilde M \times \tilde M \to [0, \infty)$, \begin{align*} d_{\tilde g}(\sigma_\infty(s), \sigma_\infty(t)) = \lim_{n \to \infty} d_{\tilde g}(\sigma_n(s), \sigma_n(t)) = |s - t|. \end{align*} Since $s, t \in \mathbb R$ were arbitrary, $\sigma_\infty$ is globally minimizing on $\mathbb R$. [/proof] [/claim] Thus $\sigma_\infty: \mathbb R \to \tilde M$ is a line.[/step]

[guided]We have the limit initial data $(\tilde p_\infty, v_\infty) \in T^1 \tilde M|_D$ from Step 5. The remaining work is to integrate this data into a globally defined geodesic on $\mathbb R$ and show it is *minimizing* across the whole real line — that is, a line. The argument has two ingredients: smooth dependence of geodesics on initial conditions (which gives uniform-on-compacts convergence of $\sigma_n$ to a limit geodesic) and continuity of the Riemannian distance (which transports the minimizing identity to the limit). **Defining the limit geodesic.** Use the exponential map at $\tilde p_\infty$ to integrate the initial direction $v_\infty$. Set \begin{align*} \sigma_\infty: \mathbb R &\to \tilde M, & \sigma_\infty(t) &:= \exp_{\tilde p_\infty}(t v_\infty). \end{align*} This is the unit-speed geodesic with $\sigma_\infty(0) = \tilde p_\infty$ and $\dot\sigma_\infty(0) = v_\infty$. Why is it defined on all of $\mathbb R$? Because $(\tilde M, \tilde g)$ is geodesically complete (Step 1), so $\exp_{\tilde p_\infty}$ is defined on all of $T_{\tilde p_\infty}\tilde M$, hence on all real multiples of $v_\infty$. It is unit-speed because $|v_\infty|_{\tilde g} = 1$ and geodesic speed is constant. **$\sigma_n$ converges to $\sigma_\infty$ uniformly on every compact interval $[-T, T] \subset \mathbb R$.** This is where smooth dependence on initial conditions enters. Define the geodesic flow \begin{align*} \phi: \mathbb R \times T \tilde M &\to \tilde M, & \phi(t, \tilde p, v) &:= \exp_{\tilde p}(tv). \end{align*} By completeness, $\phi$ is defined on the full domain $\mathbb R \times T\tilde M$, and the standard ODE-theoretic regularity result for the geodesic equation gives that $\phi$ is smooth, in particular continuous. Fix $T > 0$. For $n$ sufficiently large, $a_n, b_n > T$ (since $a_n, b_n \to \infty$), so $\sigma_n$ is defined on $[-T, T]$ and we may write \begin{align*} \sigma_n(t) = \exp_{\sigma_n(0)}(t \dot \sigma_n(0)) = \phi(t, \sigma_n(0), \dot \sigma_n(0)). \end{align*} By Step 5, $(\sigma_n(0), \dot \sigma_n(0)) \to (\tilde p_\infty, v_\infty)$ in $T\tilde M$. Pick a compact neighborhood $K' \subset T\tilde M$ of $(\tilde p_\infty, v_\infty)$ containing $(\sigma_n(0), \dot \sigma_n(0))$ for all $n$ large. Restricted to the compact set $[-T, T] \times K'$, the continuous map $\phi$ is uniformly continuous. Hence \begin{align*} \sup_{t \in [-T, T]} d_{\tilde g}\big(\phi(t, \sigma_n(0), \dot\sigma_n(0)),\, \phi(t, \tilde p_\infty, v_\infty)\big) \longrightarrow 0, \end{align*} i.e., $\sigma_n \to \sigma_\infty$ uniformly on $[-T, T]$. **$\sigma_\infty$ is minimizing on all of $\mathbb R$.** Fix any $s, t \in \mathbb R$. We must show $d_{\tilde g}(\sigma_\infty(s), \sigma_\infty(t)) = |s - t|$. Pick $T > \max(|s|, |t|)$ so that $s, t$ are interior to $[-T, T]$. For $n$ large enough that $a_n, b_n > T$, the segment $\sigma_n$ is defined on $[-T, T]$ and is *minimizing* on its full domain $[-a_n, b_n] \supseteq [-T, T]$ (Step 3). In particular, \begin{align*} d_{\tilde g}(\sigma_n(s), \sigma_n(t)) = |s - t|. \end{align*} Now pass to the limit. Uniform convergence on $[-T, T]$ implies pointwise convergence, so $\sigma_n(s) \to \sigma_\infty(s)$ and $\sigma_n(t) \to \sigma_\infty(t)$. The Riemannian distance $d_{\tilde g}: \tilde M \times \tilde M \to [0, \infty)$ is continuous (it is a metric inducing the manifold topology), so \begin{align*} d_{\tilde g}(\sigma_\infty(s), \sigma_\infty(t)) = \lim_{n \to \infty} d_{\tilde g}(\sigma_n(s), \sigma_n(t)) = |s - t|. \end{align*} Since $s, t \in \mathbb R$ were arbitrary, $\sigma_\infty$ is globally minimizing. **Conclusion.** $\sigma_\infty: \mathbb R \to \tilde M$ is a unit-speed geodesic with $d_{\tilde g}(\sigma_\infty(s), \sigma_\infty(t)) = |s-t|$ for all $s, t \in \mathbb R$ — i.e., a line. **A remark on the strategy.** The key general principle is: a uniform-on-compacts limit of minimizing geodesic segments is itself minimizing, *provided* the segments grow long enough to engulf any prescribed pair of parameters. Without the growth $a_n, b_n \to \infty$ — which traces back to the unbounded orbit $L_n \to \infty$ in Step 2 — we would only get a minimizing segment of finite length, not a line. The non-compactness hypothesis on $\tilde M$ is precisely what produces this unbounded growth.[/guided]