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A connected complete Riemannian manifold that is disconnected at infinity contains a line in a Riemannian manifold. Choose points escaping in two components outside a fixed compact set and join them by minimizing geodesics. Each segment crosses the compact set. Parametrize from a crossing point; both parameter endpoints tend to infinity in opposite directions. Compactness of the unit tangent vectors over the crossing set gives a subsequence converging to an entire geodesic. Every finite subsegment is minimizing by continuity of distance, so the limit is a line.

If the Riemannian exponential map at one point $p$ of a connected finite-dimensional Riemannian manifold is defined on all of $T_{p}M$, every point can be joined to $p$ by a minimizing geodesic. The distance splitting through a small geodesic sphere selects a radial direction attaining $d(p,q)=r+d(\gamma (r),q)$. If this equality stopped before the target, repeat the splitting at its last point. A broken minimizing path cannot have a corner, so the new segment continues the same geodesic, contradicting maximality. This is the minimizing-geodesic lemma used in the radial continuation proof of Hopf-Rinow.

// Target: riemannian-geometry.bigb

For a connected finite-dimensional Riemannian manifold, completeness of the Riemannian distance, geodesic completeness, and compactness of every closed bounded subset are equivalent. These conditions imply that any two points are joined by a minimizing geodesic. Pairwise existence of minimizing geodesics alone does not imply completeness: an open Euclidean ball is incomplete, yet the line segment between any two of its points stays in the ball and minimizes length.

A line is a unit-speed geodesic $\gamma :\mathbb{R}\to M$ that minimizes between every two of its points: $d_g(\gamma (s),\gamma (t))=|s-t|$.

This is a globally minimizing geodesic, a Riemannian specialization of the general geometric line notion.

Let a nonzero Jacobi field $J$ vanish at times zero and $a$. On a longer geodesic segment ending at $b>a$, extend it by zero to a continuous piecewise smooth field $Z$. Then $I(Z,Z)=0$. Choose an endpoint-vanishing field $P$ with $P(a)=-D_{t}J(a)$; integration by parts gives $I(Z,P)=-|D_{t}J(a){|}^2<0$. Thus $I(Z+\epsilon P,Z+\epsilon P)<0$ for small positive $\epsilon$. Smooth approximation preserves negativity. The second variation of geodesic energy supplies a shorter path, so a minimizing geodesic cannot extend beyond its first conjugate point.

// Target: differential-geometry.bigb

Work on a connected, finite-dimensional Riemannian manifold without boundary. It is geodesically complete if every maximal affinely parametrized geodesic has parameter domain $\mathbb{R}$. Equivalently, for every $p$ the Riemannian exponential map ${\exp }_p$ is defined on all of $T_{p}M$: any finite affine time can be rescaled to time one, and negative times correspond to reversing the initial velocity.

The Hopf-Rinow lemma in its minimizing-geodesic form says: if ${\exp }_p$ is defined on all of $T_{p}M$ for one point $p$, then every $q$ can be joined to $p$ by a geodesic of length $d(p,q)$. We first establish the local ingredient, the distance splitting through a small geodesic sphere. If $D=d(p,q)>0$, choose $0<\epsilon <D$ sufficiently small that the closed normal ball about $p$ is compact and radial lengths give Riemannian distance, as proved using the Gauss lemma. Take paths from $p$ to $q$ of lengths tending to $D$. Their first intersections $x_j$ with its boundary sphere have a subsequence converging to some $x$ on the sphere. Their lengths are at least $\epsilon +d(x_j,q)$, giving $D\ge \epsilon +d(x,q)$. The triangle inequality gives the reverse inequality. ThereforeThis splitting uses only a compact local geodesic sphere, not completeness or the existence of a globally minimizing path.

We also need that a minimizing broken geodesic has no corner. To see this directly, let the incoming and outgoing unit tangents at a junction be $u$ and $v$. Choose one nearby point on each segment so that they and the junction lie in a common convex normal neighbourhood. Move the junction with velocity $z$, joining it to those two fixed points by the unique short geodesics. The Gauss lemma, or the first variation formula, gives the derivative of the sum of their lengths as $⟨u-v,z⟩$. If $u\ne v$, setting $z=v-u$ makes this derivative $-|u-v{|}^2<0$, contradicting minimality. Thus the tangent vectors agree, and uniqueness for the geodesic equation makes the broken path one smooth geodesic.

Now choose the unit-speed radial geodesic $\gamma$ through the point $x$ obtained by the distance splitting, with $\gamma (0)=p$ and $\gamma (\epsilon )=x$. By the hypothesis on ${\exp }_p$, it is defined at least up to time $D$. DefineThis is a nonempty closed set. If $t\in A$, the triangle inequality and the length of $\gamma {|}_{[0,t]}$ giveso $d(p,\gamma (t))=t$: that geodesic segment minimizes. Let $t_{\ast }=\max A$. If $t_{\ast }<D$, apply the distance splitting through a small geodesic sphere at $\gamma (t_{\ast })$ with radius $0<\delta <D-t_{\ast }$. There is a point $y$, joined to $\gamma (t_{\ast })$ by a short minimizing radial geodesic $\sigma$, for whichThe broken path $\gamma {|}_{[0,t_{\ast }]}\ast \sigma$ has length $t_{\ast }+\delta$. Moreover,Consequently that broken path minimizes. It has no corner, so $\sigma$ is the continuation of $\gamma$ and $y=\gamma (t_{\ast }+\delta )$. The last distance equality puts $t_{\ast }+\delta$ in $A$, a contradiction. Hence $D\in A$ and $\gamma (D)=q$. This proves the Hopf-Rinow lemma, with the case $p=q$ supplied by the constant geodesic.

The Hopf-Rinow theorem asserts the equivalence of

These conditions imply the existence of a minimizing geodesic between any two points. Here is the deduction, including the implication back to geodesic completeness. If the manifold is geodesically complete, the Hopf-Rinow lemma applies at each point. For every $R\ge 0$,One inclusion follows from the length of the radial geodesic; the other follows by taking an initial velocity of a minimizing one. The tangent-space ball is compact, so the metric ball is compact as its continuous image. Every closed bounded set is a closed subset of such a ball and is compact. A Cauchy sequence is bounded and therefore has a convergent subsequence in a compact ball; the Cauchy property makes the whole sequence converge. This proves completeness of the Riemannian distance.

Conversely, assume metric completeness and let $\gamma :[0,b)\to M$ be a geodesic with $b<\infty$. Its speed $c$ is constant, so $d(\gamma (s),\gamma (t))\le c|s-t|$. Completeness therefore gives a limit $q$ as $t\uparrow b$. Eventually $\gamma$ lies in a coordinate neighbourhood with compact closure on which the Riemannian metric is uniformly comparable to the Euclidean metric. Its coordinate velocity is bounded by the constant-speed estimate. The Christoffel symbols are bounded there, so the geodesic equationmakes its coordinate acceleration bounded. Thus the coordinate velocity also has a limit $v$ at time $b$. The local existence and uniqueness theorem for smooth ordinary differential equations extends the solution from initial data $(q,v)$ past $b$, contradicting maximality. Reversing time treats a finite left endpoint. Hence metric completeness implies geodesic completeness, and the preceding argument supplies compactness of closed bounded sets. Compactness of closed bounded sets already implies metric completeness by the Cauchy-sequence argument, completing all equivalences.

In fact, the one-point hypothesis of the Hopf-Rinow lemma suffices for the theorem: the same closed-ball image argument at that point puts every Cauchy sequence in a compact ball, giving metric completeness and then geodesic completeness at every point. By contrast, the existence of a minimizing geodesic for every pair alone does not imply completeness: an open Euclidean ball has minimizing straight segments between all its points but has Cauchy sequences converging to its missing boundary.

Use the curvature convention $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ of the earlier curvature calculation. A geodesic variation of $\gamma :[a,b]\to M$ is a smooth map $F:(-\epsilon ,\epsilon )\times [a,b]\to M$ with $F(0,t)=\gamma (t)$ such that every $t\mapsto F(s,t)$ is an affinely parametrized geodesic. Its variation vector field $J={\partial }_{s}F{|}_{s=0}$ satisfies the Jacobi equationIndeed $D_{s}T=D_{t}J$ by vanishing torsion tensor; differentiating $D_{t}T=0$ in $s$ and commuting the covariant derivatives gives this equation. A Jacobi field is a field satisfying this equation. Conversely, any prescribed pair $J(a),D_{t}J(a)$ is obtained by smoothly varying the initial point and velocity of a geodesic; smooth dependence of the geodesic equation then constructs a geodesic variation on the compact parameter interval, whose variation vector field is the prescribed Jacobi field by uniqueness of this linear equation.

For continuous piecewise smooth vector fields along a curve $V,W$ along $\gamma$, the Riemannian index form is the symmetric bilinear formContinuity at the breakpoints is part of the admissibility condition; without it an arbitrary jump could destroy the claimed index inequality. The curvature symmetries make $V\mapsto R(V,T)T$ self-adjoint. For fields with zero endpoints, this form is the second variation of geodesic energy $E=\frac{1}{2}{\int }_a^b|\dot{\gamma }{|}^{2}dt$. More explicitly, twice differentiating energy, commuting $D_s,D_t$ and integrating the term $⟨D_t(D_s{\partial }_{s}F),T⟩$ by parts leaves the displayed integral; $D_{t}T=0$ and fixed endpoints remove the remaining terms.

The endpoints $p=\gamma (a)$ and $q=\gamma (b)$ are conjugate points along $\gamma$ when there is a nonzero Jacobi field $J$ with $J(a)=J(b)=0$. Equivalently, the Riemannian exponential map at the initial velocity for this segment has singular differential: variations of initial velocity give precisely the Jacobi fields with zero initial value. For a unit-speed geodesic starting at $p$, its cut time isIf finite and the geodesic is defined there, its endpoint $\gamma (c(v))$ is the Riemannian cut point along that geodesic. Before the cut time the segment minimizes, at that time it still minimizes by continuity, and every positive extension ceases to minimize. This is the Riemannian notion, independent of whether deleting the endpoint disconnects the space.

Let ${\mathcal{H}}_0$ denote the continuous piecewise smooth vector fields along a curve with zero endpoints. The precise characterization at conjugate endpoints isThus Jacobi fields form the radical of the fixed-endpoint index form. For a smooth field $J$, integration by parts givesA Jacobi field makes both terms vanish. Conversely, if $I(J,W)=0$ for all tests, first take tests supported inside each smooth subinterval. The fundamental lemma gives the Jacobi equation on each piece. Integrating by parts piecewise then leaves, at an interior breakpoint $t_i$, the term $⟨D_{t}J(t_i^{-})-D_{t}J(t_i^{+}),W(t_i)⟩$. The values of the tests there are arbitrary, so all derivative jumps vanish. The continuous field and its continuous first covariant derivative solve the smooth Jacobi equation globally, making $J$ smooth by uniqueness of its initial-value problem. In particular, conjugacy means that this radical is nonzero. Merely imposing $I(J,J)=0$ is insufficient when the Riemannian index form is indefinite; a zero value of an indefinite quadratic form need not mean membership in its radical.

Now suppose there are no conjugate points on $(0,1]$. In a parallel orthonormal frame along $\gamma$, write the Jacobi equation as $j^{\prime \prime }+K(t)j=0$, where the symmetric matrix $K$ represents $Z\mapsto R(Z,T)T$. LetEvery Jacobi field vanishing at zero has components $j(t)=A(t)c$. The nonconjugacy assumption means $A(t)$ is invertible for every $0<t\le 1$, and hence the unique Jacobi field with the requested endpoint values isPut $U=W-V$, so $U(0)=U(1)=0$. Since $V$ solves the Jacobi equation, integration by parts gives $I(V,U)=0$ and thereforeTo establish the needed strict positivity, use the Riccati factorization of the Riemannian index form. Differentiation givesand the initial data make this conserved quantity zero. Consequently $S=A^{\prime }{A}^{-1}$ is symmetric and obeys the matrix Riccati equation $S^{\prime }+S^2+K=0$. For the components $u$ of $U$,Near zero, the differential equation gives $A(t)=t{I}_n+O(t^3)$ and $S(t)=t^{-1}{I}_n+O(t)$. Since $U$ is piecewise smooth and $U(0)=0$, $u(t)=O(t)$, so $u^{T}Su\to 0$. At time one the same boundary term vanishes because $u(1)=0$. Integration, including cancellation at all interior breakpoints by continuity, therefore givesIf equality holds, $u^{\prime }=Su$ on each piece, so $(A^{-1}u{)}^{\prime }=0$ on $(0,1]$. Continuity across the breakpoints gives $u=Ac$ with one constant vector $c$; $u(1)=0$ and invertibility of $A(1)$ force $c=0$. HenceThe nonzero endpoint condition is compatible with this proof, but the minimizing property itself holds for arbitrary specified final value.

For the relation to the Riemannian cut point, let $a>0$ be the first conjugate time along a unit-speed geodesic, and let $J$ be a nonzero Jacobi field on $[0,a]$ vanishing at both endpoints. Necessarily $D_{t}J(a)\ne 0$, otherwise uniqueness of the Jacobi equation would give $J=0$. Fix any later time $b>a$ for which the geodesic exists. Extend $J$ by zero to a continuous piecewise smooth field $Z$ on $[0,b]$. Then $I(Z,Z)=0$, while for any endpoint-vanishing smooth field $P$,Choose $P(a)=-D_{t}J(a)$, using a smooth bump times parallel fields, so this equals $-|D_{t}J(a){|}^2<0$. For small $\epsilon >0$,If a smooth variation is desired, smooth the single corner of this field while preserving the endpoints. The field and its derivative converge in the integral defining the Riemannian index form, so strict negativity persists. The second variation of geodesic energy then gives a fixed-endpoint path of smaller energy. By the Cauchy-Schwarz inequality, $L^2\le 2bE$ for paths on $[0,b]$, with equality for the original unit-speed geodesic; the new path is therefore shorter. This proves loss of geodesic minimality beyond a conjugate point, and henceA missing conjugate time is interpreted as infinity. If a finite conjugate time exists and the geodesic continues past it, every later segment fails to minimize, so its first loss of minimality occurs no later than that conjugate time.

Finally, the cut-point dichotomy for complete Riemannian manifolds is the following characterization, stated without proof. On a complete connected Riemannian manifold, the cut time is the first time at which either the endpoint is conjugate to $p$ along the given geodesic, or a distinct minimizing geodesic from $p$ reaches the same endpoint with the same length. Thus a finite Riemannian cut point is either the first conjugate point or the meeting point of at least two distinct minimizing geodesics from the initial point; both alternatives may occur together. Completeness is the hypothesis guaranteeing that competing minimizing geodesics exist at a nonconjugate cut endpoint. It is not needed for the preceding local index-form argument on an existing segment.

A connected Riemannian manifold is complete when its Riemannian distance is a complete metric space, meaning that every Cauchy sequence converges in the manifold. The Hopf-Rinow theorem says that for a connected finite-dimensional Riemannian manifold, this is equivalent to geodesic completeness, and also to compactness of every closed bounded subset. Under these equivalent conditions any two points are joined by a minimizing geodesic, and the Riemannian exponential map at every point is defined on its entire tangent space.

We use the usual connected, boundaryless-manifold convention for the global statements in Q2–Q5. Connectedness is essential, including for the diameter and group-growth conclusions. Without connectedness, the completeness and covering assertions apply to the target components met by the map; an unrelated incomplete component cannot be controlled by a local isometry into another component.

A connected complete Riemannian manifold is disconnected at infinity when the complement of some compact set has at least two unbounded connected components. Equivalently it has at least two ends. A line in a Riemannian manifold is a unit-speed geodesic $\ell :\mathbb{R}\to M$ satisfying $d(\ell (s),\ell (t))=|s-t|$ for every $s,t$.

Let $K$ be the compact separating set, enlarged inside a fixed compact metric ball if needed for bounds. Choose $x_j,y_j$ in two different unbounded components of $M\setminus K$, with both distances from $K$ tending to infinity. By the Hopf-Rinow theorem, join them by unit-speed minimizing geodesics. Every such segment meets $K$, since a path avoiding it cannot join its different components. Reparametrize a segment so that one crossing point is at time zero. Its parameter interval is $[-a_j,b_j]$, where $a_j,b_j\to \infty$ by the distances from the endpoints to $K$.

The unit tangent bundle over $K$ is compact, so a subsequence of the initial position/velocity pairs converges. Continuous dependence of the geodesic equation on its initial data, and completeness, give a limiting geodesic defined on all of $\mathbb{R}$. For any fixed $s<t$, the approximating segments are defined on $[s,t]$ for large $j$ and satisfy $d({\gamma }_j(s),{\gamma }_j(t))=t-s$. Continuity of the distance passes this equality to the limit. Thus the limiting geodesic is a line, proving a line from disconnection at infinity.

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.

For $p,q\in M$, define the Riemannian distance byA connected smooth manifold is locally path connected, so any two points are joined by a finite concatenation of coordinate paths. Thus the infimum is finite. It is nonnegative, $d_g(p,p)=0$, and reversing a curve proves symmetry. Concatenating two curves whose lengths approach their respective infima proves the triangle inequality.

For positivity and the topology, choose a coordinate ball about $p$ whose closure is contained in a chart. On its closure there are constants $0<c<C$ withA curve staying in the ball has length at least $c$ times its coordinate displacement. A curve leaving it must first reach its boundary, costing at least $c$ times the positive coordinate distance from $p$ to that boundary. Therefore a distinct $q$ has $d_g(p,q)>0$, whether it lies inside or outside the ball. For $q$ sufficiently close to $p$, the coordinate line segment gives $d_g(p,q)\le C|x(q)-x(p)|$, while the preceding lower bounds prevent a short curve from taking a shortcut outside the chart. These inequalities prove that the Riemannian distance induces the manifold topology. In particular it defines a genuine metric space.

A Riemannian manifold is geodesically complete if every maximal affinely parametrized geodesic exists for all real parameter values. We prove both implications of the Hopf-Rinow theorem rather than assuming global distance minimizers.

First assume completeness as a metric space. Let a constant-speed geodesic have a finite maximal right endpoint $b$. The bound $d_g(\gamma (t),\gamma (s))\le c|t-s|$ makes $\gamma (t)$ a Cauchy sequence as $t\uparrow b$, hence gives a limit $p\in M$. The distance topology just established puts the final portion inside a relatively compact coordinate ball about $p$. The positive lower bound on the Riemannian metric bounds all coordinate velocity components. The Christoffel symbols are bounded on the closure of that ball, and the geodesic equationthen bounds the coordinate acceleration. The velocity has a limit $v$ as $t\uparrow b$. Local existence and uniqueness for the geodesic equation with data $(p,v)$ continue $\gamma$ through $b$, contradicting maximality. The same argument applies at a finite left endpoint. Hence metric completeness implies geodesic completeness.

Conversely assume geodesic completeness. Fix $p,q$, put $r=d_g(p,q)>0$, and choose $0<ϵ<r$ so small that a closed radius-$2ϵ$ normal ball about $p$ lies inside a convex normal neighborhood. By the Gauss lemma, its radial geodesics minimize; a curve leaving the larger ball cannot improve on a radius-$ϵ$ radial segment. Thus the radius-$ϵ$ geodesic sphere $S_{ϵ}(p)$ is compact and consists of points at distance exactly $ϵ$ from $p$.

The distance splitting through a small geodesic sphere has an elementary proof here. Every path from $p$ to $q$ must cross $S_{ϵ}(p)$, and its initial portion has length at least $ϵ$. HenceThe triangle inequality proves the reverse bound. Continuity of $d_g(\cdot ,q)$ and compactness give a point $x$ achieving the minimum, so $d_g(x,q)=r-ϵ$. Let $\gamma (t)={\exp }_p(tv)$ be the complete unit-speed radial geodesic through $x$ at time $ϵ$.

Consider the closed subsetIt is nonempty since $ϵ\in A$, and has a largest member $t_{\ast }$. Suppose $t_{\ast }<r$. At $x_{\ast }=\gamma (t_{\ast })$, choose a small normal sphere of radius $\eta <r-t_{\ast }$ and repeat the splitting argument. It supplies a radial endpoint $y$ withThe triangle inequality gives $d_g(p,y)\ge t_{\ast }+\eta$, while the broken path following $\gamma$ to $x_{\ast }$ and then the short radial segment to $y$ has exactly that length. It is therefore minimizing. Its two velocities at the joining point must agree: if unit incoming and outgoing velocities are $u,w$, moving the join in direction $w-u$ changes the sum of lengths by $-|u-w{|}^2$ to first order, using the first variation of geodesic energy or the equivalent unit-speed length formula. A corner would strictly shorten it. Uniqueness of the geodesic equation consequently gives $y=\gamma (t_{\ast }+\eta )$, contradicting the maximality of $t_{\ast }$. Thus $t_{\ast }=r$, $\gamma (r)=q$, and $\gamma {|}_{[0,r]}$ is a minimizing geodesic.

This radial continuation proof of Hopf-Rinow also proves compactness of closed balls. For every $R\ge 0$,The inclusion from right to left follows from the length of a radial geodesic; the reverse inclusion uses the minimizing geodesic just constructed. The Riemannian exponential map is defined on all of $T_{p}M$ by geodesic completeness, and the closed tangent ball is compact in a finite-dimensional vector space. Its continuous image is compact. Every Cauchy sequence is bounded, hence has a convergent subsequence in such a ball, and the Cauchy property makes the whole sequence converge. Therefore geodesic completeness implies metric completeness, completing both directions.

For the hyperbolic plane, use the upper half-plane model with $g=(d{x}^2+d{y}^2)/y^2$, $y>0$. Its geodesic equation isEvery nonconstant unit-speed solution is either a vertical curve $x=a$, $y=b{e}^{\pm t}$ or, after translating and possibly reversing $t$,Substitution verifies both equations and $g(\dot{\gamma },\dot{\gamma })=1$. These curves cover every unit initial tangent: a nonvertical tangent at $(x,y)$ determines the circle center $a=x+y{y}^{\prime }/x^{\prime }$ and radius $R=\sqrt{(x-a{)}^2+y^2}$, and its orientation determines the sign of the parameter. Constant solutions and constant-speed rescalings cover arbitrary initial velocities. In every case $y(t)>0$ for all finite $t$, and the solution is defined for every real $t$. Thus the real hyperbolic plane is geodesically complete, directly, and the proved Hopf-Rinow theorem also makes its distance complete.

Use the curvature convention $\mathcal{R}$ of Solution 1, and put $T=\dot{\gamma }$. On continuous piecewise smooth perpendicular vector fields with zero endpoint values, the Riemannian index form is the symmetric bilinear formA Jacobi field is a smooth field $J$ along $\gamma$ satisfying $D_t^{2}J+\mathcal{R}(J,T)T=0$. The point $\gamma (t_0)$ is a conjugate point to $\gamma (0)$ along this geodesic when there is a nonzero Jacobi field with $J(0)=J(t_0)=0$. Such a field is perpendicular to $T$, because $⟨J,T⟩$ is affine and vanishes at both endpoints.

The conjugate-point criterion for the Riemannian index form, stated without proof, isAt a first conjugate endpoint the form is nonnegative but has a nonzero nullspace, consisting of endpoint-vanishing Jacobi fields. A conjugate point strictly inside gives a negative direction. This explicitly distinguishes the non-strict and strict senses of positivity in the question; in terminology where “positive definite” already means strict positivity, the second line is the positive-definiteness condition.

Choose $n-1$ perpendicular orthonormal fields $E_1,\dots ,E_{n-1}$ by parallel transport along $\gamma$, and put $V_i(t)=\sin (\pi t/L)E_i(t)$. They are admissible andThe tangent has unit length. The assumed Ricci curvature bound and the two elementary sine-square and cosine-square integrals give the sine index-form bound for positive Ricci curvature:Since $n\ge 2$, at least one of the admissible $V_i$ therefore has $I(V_i,V_i)<0$.

If the Riemannian manifold is complete, the Hopf-Rinow theorem supplies a minimizing geodesic between any two points. Its Riemannian index form is nonnegative: any smooth fixed-endpoint variation has nonnegative second variation of geodesic energy, since a length minimizer of constant speed also minimizes energy on a fixed parameter interval by the Cauchy-Schwarz inequality. The same conclusion for piecewise smooth fields follows by smooth approximation, or by piecewise variations. The negative direction above excludes a minimizing segment longer than $\pi /\sqrt{\kappa }$. Taking the supremum over pairs yields the diameter conclusion of the Bonnet-Myers theorem:

The completeness assumption is essential. An explicit incomplete positively curved strip with infinite diameter isThe map $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the unit sphere, so $K=1$ and, in dimension two, $\mathrm{Ric}=g$. Thus the required Ricci curvature bound holds with $n=2$, $\kappa =1$. The unit-speed meridian $t\mapsto (t,0)$ reaches the missing boundary $u=\pi /4$ in finite time, so the strip is geodesically incomplete. On the other hand every curve joining $(0,0)$ to $(0,A)$ hassince $\cos u\ge 1/\sqrt{2}$ throughout the strip. Therefore its intrinsic diameter is infinite, in particular larger than $\pi$. This is the unwrapped strip with $v\in \mathbb{R}$, not a strip with longitude identified modulo $2\pi$; that distinction prevents spherical shortcuts from invalidating the example.

The Bonnet-Myers theorem says that a connected geodesically complete $n$-dimensional Riemannian manifold, with $n\ge 2$ and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$, hasThe dimension and positive lower bound are essential: in dimension one the Ricci condition is vacuous, and a zero lower bound does not imply bounded diameter.

By Hopf-Rinow, any two distinct points have a unit-speed minimizing geodesic $\gamma :[0,L]\to M$. Choose parallel orthonormal fields $E_1,\dots ,E_{n-1}$ perpendicular to its tangent $T$. For the endpoint-vanishing fields $V_i(t)=\sin (\pi t/L)E_i(t)$, the second variation of geodesic energy gives nonnegative index formsSumming and using the Ricci lower bound produces the sine index-form bound for positive Ricci curvature:If $L>\pi /\sqrt{\kappa }$, the last expression is negative, a contradiction. This proves the diameter bound. Hopf-Rinow makes closed bounded sets compact, so the entire manifold is compact.

Give the universal cover the pullback metric. Local isometry preserves its Ricci bound, and lifting complete base geodesics proves completeness of the cover. The same diameter and compactness argument applies there. A fiber of the covering is closed and discrete, hence finite in this compact cover; its cardinality is that of the fundamental group. This proves the final assertion.

For a counterexample that also breaks the diameter conclusion, use the incomplete positively curved strip with infinite diameterThe map $\Phi (u,v)=(\cos u\cos v,\cos u\sin v,\sin u)$ is a local isometry to the round unit sphere: its coordinate derivatives are orthogonal, with squared lengths one and ${\cos }^{2}u$. Thus $K=1$ and $\mathrm{Ric}=g$, satisfying the required lower bound with $n=2$, $\kappa =1$.

The meridian $(u,v)=(t,0)$ is a unit-speed geodesic and reaches the excluded boundary at $t=\pi /4$, so this metric is incomplete. Every curve joining $(0,0)$ to $(0,L)$ has length at least $\int \cos u|v^{\prime }|dt\ge |L|/\sqrt{2}$, because $\cos u\ge 1/\sqrt{2}$ on the strip. Therefore $\boxed{\mathrm{diam}M=\infty }$ despite the positive Ricci bound. Completeness cannot be omitted.

A local isometry here is a local diffeomorphism whose pullback metric is the source metric. Use the usual convention that a covering map is surjective. Assume the target is connected and the source is nonempty; this is the connected-manifold setting of the intended assertion. For a disconnected target the precise result is a covering of the union of target components met by the map. For example, inclusion $\mathbb{R}\hookrightarrow \mathbb{R}\bigsqcup \mathbb{R}$ is a local isometry from a complete manifold but is not a surjective covering of the whole target.

The correct Hopf-Rinow theorem says that on a connected finite-dimensional Riemannian manifold, completeness for the induced distance is equivalent to geodesic completeness, and to compactness of every closed bounded subset. These conditions imply that every two points are joined by a minimizing geodesic. The last property by itself does not imply completeness: an open Euclidean ball has minimizing segments between every pair but is incomplete.

A local isometry preserves the Levi-Civita connection, by uniqueness of the torsion-free metric connection on each local isometry chart. Hence it preserves affinely parametrized geodesics, andwhenever the source geodesic is defined. This is geodesic preservation by a local isometry.

Suppose $M$ is complete. Given $y\in N$, join $f(x)$ to $y$ by a minimizing target geodesic using target completeness. The source geodesic whose initial velocity is its inverse image under $d{f}_x$ exists for its full parameter interval by source completeness; its image is the target geodesic by uniqueness. Thus $f$ is onto.

Choose a normal ball $U={\exp }_y(B_r(0))$ with ${\exp }_y$ a diffeomorphism on $B_r(0)\subset T_{y}N$. For every $x\in f^{-1}(y)$ put $U_x={\exp }_x(B_r(0))$. Source completeness defines these exponential maps throughout the tangent balls. The displayed exponential identity makes $f:U_x\to U$ a diffeomorphism: injectivity follows from injectivity of the target exponential map, and differentiating the identity shows the source exponential has invertible derivative on the ball, so its image is open.

These sheets are disjoint. If $z$ lay in two, lift the reversed radial target geodesic from $f(z)$ to $y$, starting at $z$. Its initial velocity is uniquely fixed by $d{f}_z$; uniqueness of the source geodesic forces the two sheet centres to be equal. They exhaust $f^{-1}(U)$: for any point over $U$, lift that reversed radial geodesic using source completeness, obtaining a centre over $y$, and then reverse it. Thus $U$ is evenly covered. This proves the complete local isometry is a covering assertion.

Conversely suppose $f$ is a covering. Given any initial tangent vector in $M$, the corresponding target geodesic extends over all $\mathbb{R}$ by target completeness. The path lifting theorem for coverings lifts it over every compact time interval, with uniqueness gluing the lifts to a curve on $\mathbb{R}$. In local covering charts the lift is a geodesic, because the charts are local isometries. It extends the source initial-value geodesic. Therefore $M$ is geodesically complete, and Hopf-Rinow yields metric completeness on each source component. In the stated connected-target setting,

For $v$ in the domain of the exponential map at $p$, let ${\gamma }_v$ be the unique geodesic with initial data $(p,v)$ and define ${\exp }_p(v)={\gamma }_v(1)$. The domain consists of initial velocities whose geodesics exist on $[0,1]$; it is open, contains zero, and need not be the whole tangent space. The ordinary differential equation theorem used here says that a smooth vector field has unique maximal integral curves, with an open flow domain and smooth dependence on initial point and time. Applied to the geodesic equation on $TM$, it gives smoothness of the exponential map. Rescaling the parameter gives ${\exp }_p(tv)={\gamma }_v(t)$ and hence $(d{\exp }_p{)}_0(v)=v$. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighbourhood of zero to one of $p$. Coordinates in an orthonormal tangent basis transported by this map are geodesic normal coordinates.

The geodesic sphere of radius $r$ is $S_r(p)=\{x:d(p,x)=r\}$. For small $r>0$, it is ${\exp }_p(\{v:\Vert v\Vert =r\})$, a smooth compact hypersurface; larger distance spheres need not be smooth. The Gauss lemma statesIn particular radial and angular directions are orthogonal and the radial coordinate measures arc length.

For a piecewise smooth path $c$, put $L(c)=\int \Vert \dot{c}(t)\Vert dt$. The Riemannian distance is $d(x,y)=\inf L(c)$ over such paths from $x$ to $y$. Connectedness makes it finite. Reversal and concatenation give symmetry and the triangle inequality; positivity and compatibility with the manifold topology follow locally from geodesic normal coordinates and the Gauss lemma.

Choose $r_0>0$ so that ${\exp }_p$ is a diffeomorphism on a neighbourhood of the closed tangent ball of radius $2r_0$. For $r<r_0$, radial geodesics realize the distance from $p$ on that ball. Indeed, Gauss lemma bounds any path remaining in the normal ball below by its radial change, while a path leaving the radius-$2r_0$ ball already has length at least $2r_0$. Thus the radius-$r$ sphere is the compact exponential image described above.

Given $q\ne p$, take $0<\delta <\min \{r_0,d(p,q)\}$. The continuous function $z\mapsto d(z,q)$ attains its minimum at some $p_0\in S_{\delta }(p)$. Every path from $p$ to $q$ crosses that sphere, since the distance from $p$ is continuous. Splitting at a crossing gives length at least $\delta +{\min }_{S_{\delta }(p)}d(\cdot ,q)$. Taking the infimum over paths, and using the triangle inequality in the other direction, proves the distance splitting through a small geodesic sphere equalityThis compact local argument does not assume completeness or the existence of a globally minimizing geodesic to $q$.

For the Lie-group clause, a bi-invariant Riemannian metric has adjoint-invariant identity inner product. Differentiating this invariance gives $⟨[X,Y],Z⟩+⟨Y,[X,Z]⟩=0$. The Koszul formula for left-invariant vector fields has no metric-derivative terms, and this skew-adjointness reduces it toThis is the Levi-Civita connection of a bi-invariant metric.

Let $\sigma$ be the integral curve through the identity of the left-invariant field with identity value $v$. Uniqueness and left translation show $\sigma (s+t)=\sigma (s)\sigma (t)$ wherever the local curves are defined. The field is complete: a fixed local existence interval about the identity translates to an equally long interval at every point of the group. Restarting the curve before any proposed finite endpoint extends it beyond that endpoint. Thus $\sigma$ exists on $\mathbb{R}$, and uniqueness now gives the group law for every $s,t$.

Since ${\nabla }_{X}X=0$, this integral curve is a geodesic. Conversely any geodesic starting at the identity has the same initial data as one of these curves and agrees with it by uniqueness. Therefore the geodesics of a bi-invariant metric are one-parameter subgroups conclusion isThe zero velocity gives the constant, trivial subgroup. Here ${\exp }_G$ is the Exponential map of a Lie group; for this metric it agrees at the identity with the Riemannian exponential.

Assume geodesic completeness. Minimize the distance to a target over a small compact geodesic sphere about the initial point. The distance splitting through a small geodesic sphere gives an initial direction satisfying $d(p,\gamma (t))+d(\gamma (t),q)=d(p,q)$. If this equality stops before the target, repeat the splitting at the last equality point. The resulting broken path minimizes, so its corner must smooth into the same continuing geodesic, a contradiction. This supplies a minimizing geodesic to every target. Every closed metric ball is then the image under the Riemannian exponential map of a closed tangent ball, hence compact. A Cauchy sequence has a convergent subsequence in such a ball and therefore converges, proving metric completeness without presupposing a global minimizing curve.

The distance on a connected Riemannian manifold is the infimum of the arc lengths of piecewise smooth curves joining the endpoints:It answers “how far must I travel while staying in this geometry?” The Riemannian metric measures an instantaneous tangent vector, whereas Riemannian distance is the global result of comparing complete routes.

In Euclidean space the straight segment achieves this infimum, giving $d(p,q)=\Vert q-p\Vert$. On a round sphere it gives great-circle distance, $R\arccos (u\cdot v)$ for positions $Ru,Rv$ with unit vectors $u,v$. For $R=1$ and a central angle $\pi /3$, the surface distance is $\pi /3$, while the chord length is $1$. On a circular cylinder, routes with different winding lead to the minimum in Riemannian distance on a circular cylinder.

The word infimum matters: a shortest route need not exist. In the Euclidean plane with the origin removed, the points $(-1,0)$ and $(1,0)$ still have Riemannian distance $2$, approached by paths detouring around an arbitrarily small circle. Every allowed route has length strictly greater than $2$, since the only Euclidean straight segment passes through the missing point. The Hopf-Rinow theorem guarantees a minimizing geodesic between any two points when a connected Riemannian manifold is complete.