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In a homogeneous Riemannian manifold, choose $r>0$ so that one closed metric ball $\overline{B}(o,r)$ is compact. Such a ball exists from local compactness and the metric topology. Every radius-$r$ ball is isometric to it. A Cauchy sequence eventually lies in one such compact ball, has a convergent subsequence, and therefore converges. This proves metric completeness; the Hopf-Rinow theorem gives geodesic completeness and minimizing geodesics between points.

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.

This metric pulls back the round spherical metric by $(u,v)\mapsto (\cos u\cos v,\cos u\sin v,\sin u)$. It has unit Gaussian curvature and Ricci curvature equal to its metric, yet the meridian reaches a missing boundary in finite time. Every curve between $(0,0)$ and $(0,L)$ has length at least $|L|/\sqrt{2}$, so its diameter is infinite. It shows that the positive Ricci bound alone does not replace geodesic completeness in the Bonnet-Myers theorem.

For a connected Riemannian manifold of dimension $n\ge 2$, assume geodesic completeness and $\mathrm{Ric}\ge (n-1)\kappa g$ for a constant $\kappa >0$. The Riemannian distance then has diameter at most $\pi /\sqrt{\kappa }$; the manifold is compact and its fundamental group is finite. The dimension restriction excludes the vacuous one-dimensional Ricci bound.

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.

Let $\pi :\tilde{M}\to M$ be the universal cover and equip it with the pullback Riemannian metric $\tilde{g}={\pi }^{\ast }g$. The covering map is a local isometry, so $\tilde{M}$ has constant sectional curvature one.

We must check completeness of the lifted Riemannian metric. A geodesic on $\tilde{M}$ projects to a geodesic on $M$. By geodesic completeness of $M$, the projected geodesic extends for all real time. Its unique lift with a specified initial point extends the original lifted geodesic, because $\pi$ is a local isometry. Hence $\tilde{M}$ is geodesically complete, and the Hopf-Rinow theorem also gives metric completeness.

The allowed constant-curvature form of Cartan-Ambrose-Hicks theorem says that a linear tangent-space isometry between complete simply connected Riemannian manifolds of the same constant sectional curvature extends to a global isometry. The unit round sphere $S^n$ is complete, simply connected for $n\ge 2$, and has sectional curvature one. Comparing it with $\tilde{M}$ therefore yieldsThis is the covering step in the classification of complete positive constant-curvature manifolds. As usual for this assertion, $n\ge 2$ is understood: in dimension one there are no sectional two-planes, and the universal cover of the circle is the line, not the circle.

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.

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.

The Riemannian distance on a connected manifold iswhere the infimum is over piecewise smooth curves joining $p$ to $q$. A connected manifold is path connected, so this infimum is finite. The metric is positive definite and induces the manifold topology; locally this follows from comparison with the Euclidean metric in a coordinate chart.

Geodesic completeness means that every geodesic with prescribed initial point and velocity extends to all parameter values in $\mathbb{R}$. The Hopf-Rinow theorem says that, for a connected finite-dimensional Riemannian manifold without boundary, this is equivalent to completeness of the distance metric and equivalent to compactness of every closed bounded set. It is also equivalent to the exponential map at one, and hence every, point being defined on the whole tangent space. Under these equivalent conditions, any two points are joined by a distance-minimizing geodesic.

A homogeneous Riemannian manifold has an isometry group acting transitively on its points. Fix $o\in M$. Local compactness and the local metric-topology comparison give an $r>0$ such that $\overline{B}(o,r)$ is compact: choose a relatively compact coordinate neighborhood of $o$, then a small closed metric ball contained in it. Homogeneity makes $\overline{B}(p,r)$ isometric to this fixed compact ball for every $p\in M$.

Let $(p_j)$ be a Cauchy sequence. Its tail lies inside $\overline{B}(p_N,r)$ for some $N$. Compactness gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge to the same limit. Thus the distance metric is complete. Hopf-Rinow now gives every homogeneous Riemannian manifold is geodesically complete. This is the completeness of homogeneous Riemannian manifolds; the compact ball used here is local and does not presuppose global completeness.

A two-point homogeneous Riemannian manifold has isometries acting transitively on ordered pairs at each fixed distance. We prove the unit tangent transitivity characterizes two-point homogeneity equivalence. First suppose two-point homogeneity holds. Applying it to pairs with identical endpoints gives ordinary homogeneity. Given $p,q$ and unit tangent vectors $u,v$, choose $a>0$ sufficiently small that $au$ and $av$ lie in normal neighbourhoods at $p$ and $q$. The Gauss lemma and local minimizing property giveThere is therefore an isometry $f$ taking these ordered pairs to one another. Isometries preserve the Levi-Civita connection and geodesics, soBoth vectors on the right lie in the injectivity neighborhood of ${\exp }_q$, because $d{f}_p$ preserves length. It follows that $d{f}_{p}u=v$. Thus isometries act transitively on the unit tangent bundle.

Conversely, suppose the stated transitivity on unit tangent vectors holds. In positive dimension it implies point homogeneity, and hence completeness by the preceding argument. Consider two ordered pairs at the same distance $a>0$. Hopf-Rinow supplies unit-speed minimizing geodesics ${\gamma }_1,{\gamma }_2:[0,a]\to M$ joining their respective endpoints. Choose an isometry taking ${\gamma }_1(0)$ to ${\gamma }_2(0)$ and ${\dot{\gamma }}_1(0)$ to ${\dot{\gamma }}_2(0)$. Then $f\circ {\gamma }_1$ and ${\gamma }_2$ solve the same geodesic initial-value problem, so they agree throughout $[0,a]$. In particular $f$ takes both endpoints of the first pair to those of the second. For $a=0$, point homogeneity suffices. If the connected manifold has dimension zero, it is a single point and both properties are immediate. Hence two-point homogeneity is equivalent to transitivity on unit tangent vectors. Completeness is the step allowing the local initial-direction condition to control pairs at arbitrary distance.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.

Let $\pi :T^{\ast }M\to M$ be the cotangent bundle projection. Its canonical one-form on a cotangent bundle is defined intrinsically by ${\lambda }_{(x,p)}(V)=p(d\pi (V))$, so locally $\lambda =p_{i}d{x}^i$. Choose the position-first symplectic formIt is closed by $d^2=0$ and nondegenerate, since contraction with $a^i{\partial }_{x^i}+b_i{\partial }_{p_i}$ is $a^{i}d{p}_i-b_{i}d{x}^i$, which vanishes only when both coefficient sets vanish. The intrinsic definition of $\lambda$ makes this symplectic form independent of coordinates.

Use the convention ${\iota }_{X_H}\omega =dH$. Then the Hamiltonian vector field and the Poisson bracket areso $X_H(F)=\{F,H\}$. Choosing $\omega =d\lambda$ and ${\iota }_{X_H}\omega =-dH$ gives the same equations; choosing only one of these sign changes would reverse the flow.

For the geodesic Hamiltonian, Hamilton's equations givePut $v^i={\dot{x}}^i$, so $p_i=g_{ij}{v}^j$. Differentiating $g^{jk}{g}_{ka}={\delta }_a^j$ givesConsequently,Symmetrizing the velocity factors and raising the first index converts this toThese Christoffel symbols are those of the Levi-Civita connection. Thus a Hamiltonian integral curve of a vector field projects to an affinely parametrized geodesic. Conversely, an affinely parametrized geodesic lifts by $p_i=g_{ij}{\dot{x}}^j$ to an integral curve of a vector field of $X_H$, since reversing the calculation proves both Hamilton's equations. This is the geodesic flow on the cotangent bundle. The conserved Hamiltonian is half the squared speed, and the zero-energy case gives the constant geodesics.

A quadratic homogeneous polynomial depends only on the symmetric part of its coefficient matrix. Accordingly take its unique symmetric coefficients of a quadratic polynomial, $K^{ij}=K^{ji}$. This is the standard implicit convention in identifying such polynomials with symmetric tensors. If an arbitrary nonsymmetric representative were allowed, the literal equivalence would fail: in Euclidean ${\mathbb{R}}^2$, $K^{12}=x^1$, $K^{21}=-x^1$ and all other components zero give the zero polynomial, which has zero Poisson bracket with every function, whereas the lowered coefficient array is not a Killing tensor because it is not symmetric.

Lower the indices of the symmetric coefficient tensor using the Riemannian metric. Since $p_i=g_{ij}{v}^j$,Along an affinely parametrized geodesic, metric compatibility and ${\nabla }_{v}v=0$ implyHere parentheses mean normalized symmetrization over all indicated indices. The left side is $\{K,H\}$ by the Hamiltonian vector field convention. Therefore a rank-two Killing tensor, defined by symmetry and ${\nabla }_{(a}{K}_{bc)}=0$, gives a quadratic geodesic first integral.

Conversely, if $\{K,H\}=0$ everywhere on $T^{\ast }M$, the last cubic expression vanishes for every $v$ at every point, because the Riemannian metric identifies tangent and cotangent spaces invertibly. A symmetric trilinear form is determined by its diagonal cubic polynomial: equivalently, compare its coefficients, or polarize the cubic. Hence ${\nabla }_{(a}{K}_{bc)}=0$. This proves both directions:with symmetry understood on the coefficient representative from the outset. The equivalence is local and does not require geodesic completeness.

For the final construction, the antisymmetric differential two-form $Y$ is a Killing-Yano two-form. Its defining equation says ${\nabla }_a{Y}_{bc}=-{\nabla }_b{Y}_{ac}$. Antisymmetry of $Y$ also says ${\nabla }_a{Y}_{bc}=-{\nabla }_a{Y}_{cb}$, so the three-index tensor ${\nabla }_a{Y}_{bc}$ is totally antisymmetric.

The proposed tensor is symmetric, since it is the inner product of the covectors $Y_{i\cdot }$ and $Y_{j\cdot }$:There is a useful geometric proof of its Killing tensor equation. Along any affinely parametrized geodesic, define $w_k=Y_{ik}{v}^i$. ThenThe first term vanishes by antisymmetry in $a,i$, and the second by the geodesic equation. Thus $w$ is carried by parallel transport. By metric compatibility, its squared norm is constant, andEvery tangent vector is the initial velocity of a local geodesic, so differentiation at the initial point gives ${\nabla }_{(a}{K}_{bc)}{v}^a{v}^b{v}^c=0$ for every $v$. The same cubic-coefficient argument provesThis proves that the square of a Killing-Yano two-form is a rank-two Killing tensor and supplies a nonnegative quadratic geodesic first integral. The argument also explains the conserved quantity: it is the squared norm of a covector that the Killing-Yano two-form makes parallel along every geodesic.

Let ${\gamma }_X(s)$ be the affinely parametrized geodesic of the affine connection with ${\gamma }_X(0)=p$ and ${\dot{\gamma }}_X(0)=X$. For fixed $t$, the curve $\eta (u)={\gamma }_X(tu)$ has initial velocity $tX$. The geodesic equation is homogeneous of degree two in the velocity, so its left-hand side for $\eta$ is $t^2$ times the left-hand side for ${\gamma }_X$, and is zero. Uniqueness of the geodesic initial value problem gives $\eta ={\gamma }_{tX}$.

By the definition of the affine exponential map, rescaling the initial velocity rescales affine parameter:At $t=0$ both sides equal $p$. The statement is made where the geodesics exist; without geodesic completeness, the affine exponential map is defined only on a suitable neighbourhood of zero in $T_{p}M$, not necessarily all of $T_{p}M$. An affine parameter distance need not be a metric length.

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,

The metric $g$ is geodesically complete when every maximal affinely parametrized geodesic is defined on all of $\mathbb{R}$; equivalently, ${\exp }_p$ is defined on every $T_{p}M$ for every $p\in M$.

The Hopf-Rinow theorem says that for a connected Riemannian manifold, the following are equivalent: geodesic completeness; completeness of the Riemannian distance $d_g$; compactness of every closed bounded subset; and the existence, between every two points, of a length-minimizing geodesic. It is enough in the exponential-map formulation that ${\exp }_p$ be defined on all of $T_{p}M$ for one point $p$.

The pointwise inequality $\hat{g}(X,X)\ge g(X,X)$ impliesEvery $d_{\hat{g}}$-Cauchy sequence $(x_j)$ is therefore $d_g$-Cauchy. Since $g$ is geodesically complete, the Hopf-Rinow theorem makes $(M,d_g)$ a complete metric space, so $x_j\to x$ in $d_g$ for some $x\in M$.

On a coordinate neighbourhood with compact closure around $x$, smooth positive-definite Riemannian metrics are uniformly equivalent. Thus there is $C>0$ such thatthere. For all sufficiently large $j$, a short $g$-geodesic from $x$ to $x_j$ stays in this neighbourhood, and henceThus $(M,d_{\hat{g}})$ is complete. Another application of Hopf-Rinow shows that $\hat{g}$ is geodesically complete.

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.