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In unit-curvature Real projective space of dimension at least two, Riemannian distance is $d([v],[u])=\arccos |⟨v,u⟩|$ for unit lifts. A unit-speed geodesic has lift $\cos tv+\sin tw$, with $w$ unit and perpendicular to $v$, and therefore ceases to minimize after $\pi /2$. Its normal Jacobi fields with zero initial value are $\sin t$ times parallel vector fields along a curve, giving first conjugate point at time $\pi$. The two times differ in every direction.

On a complete connected Riemannian manifold, let $c(v)$ for a unit tangent vector $v$ be the supremum of times for which $t\mapsto {\exp }_p(tv)$ minimizes Riemannian distance from $p$. The cut locus consists of the endpoints ${\exp }_p(c(v)v)$ with $c(v)<\infty$. Radial minimization fails beyond this first cut time; distance spheres can cease to be smooth there.

The Morse lemma gives coordinates $f=f(z)-|u{|}^2+|v{|}^2$. Choose a Riemannian metric Euclidean in these coordinates. A downward gradient flow converging to $z$ has $u=0$ eventually, while $v(t)=e^{-2(t-t_0)}v(t_0)$. Thus its Riemannian distance from $z$ decays exponentially. A partition of unity makes this a global metric choice; changing the metric may change the trajectories.

A geodesic sphere is a level set of Riemannian distance from a point. For sufficiently small positive $r$, it is the exponential map image of the radius-$r$ tangent sphere, a smooth compact hypersurface. The Gauss lemma makes its tangent space orthogonal to the radial direction. Larger distance spheres need not be smooth at the cut locus.

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.

Hyperbolic distance is the Riemannian distance induced by the hyperbolic metric.

The length of a piecewise smooth curve is its arc length: $L(\alpha )=\int \Vert {\alpha }^{\prime }(t)\Vert dt$, using either the Euclidean space norm or the norm from a specified Riemannian metric. A straight segment from $p$ to $q$ in Euclidean space has length $\Vert q-p\Vert$, while a route that bends or retraces can be longer. On a Riemannian manifold, comparing these lengths over all routes defines Riemannian distance.

Assume the trace Ricci curvature satisfies $\mathrm{Ric}\ge (n-1)kg$, with $k>0$. If points $p,q$ have Riemannian distance $D=\pi /\sqrt{k}$, the balls $B(p,r)$ and $B(q,D-r)$ are disjoint. Bishop volume comparison with a positive-curvature model bounds their volumes below by $\mathrm{Vol}(M)v_k(r)/v_k(D)$ and $\mathrm{Vol}(M)v_k(D-r)/v_k(D)$. Since $v_k(r)+v_k(D-r)=v_k(D)$, both inequalities must be equalities. Taking $r\downarrow 0$ gives $\mathrm{Vol}M=v_k(D)$, and spherical rigidity of maximal total volume yields the round sphere.

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.

The Riemannian exponential map at $p$ is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v$ is the affinely parametrized geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. Its derivative at zero is the identity. The inverse function theorem therefore gives a normal neighbourhood $U={\exp }_p(V)$, where $V\subset T_{p}M$ is an open star-shaped neighbourhood of zero on which ${\exp }_p$ is a diffeomorphism. Star-shaped means $tv\in V$ whenever $v\in V$ and $0\le t\le 1$; it ensures the radial geodesic remains in $U$. This is distinct from a convex normal neighbourhood, which imposes a unique short geodesic between every pair of its points.

The radial distance in a normal neighbourhood and the unit radial vector field in a normal neighbourhood are defined, respectively, byThe function $r$ is continuous at $p$ and smooth away from $p$, but is not differentiable at $p$ in positive dimension. Some conventions use the scaled radial field $W_q=(d{\exp }_p{)}_{v}v$ and the smooth radial function $r^2/2$ instead; the associated gradient identities will distinguish them.

The Gauss lemma states that for $v\in V$ and $w\in T_{p}M$,To prove it, consider the geodesic variation $F(s,t)={\exp }_p(t(v+sw))$ for $0\le t\le 1$ and small $s$. Set $T={\partial }_{t}F$ and $J={\partial }_{s}F$. The Levi-Civita connection has zero torsion tensor, so $D_{t}J=D_{s}T$. Each $t$-curve is a geodesic, hence $D_{t}T=0$, and its squared speed is the constant $|v+sw{|}^2$. At $s=0$,Since $J(0)=0$, integration gives $g(T,J)=t{g}_p(v,w)$. At $t=1$, $T=(d{\exp }_p{)}_{v}v$ and $J=(d{\exp }_p{)}_{v}w$, proving the Gauss lemma. Taking $w=v$ shows $|E|=1$; taking $w\perp v$ shows that $E$ is orthogonal to the images of tangent vectors to the tangent-space sphere, hence to the geodesic spheres in $U$.

For $u\in T_{p}M$ of unit length, write the unit-speed radial geodesic as ${\gamma }_u(t)={\exp }_p(tu)$. Differentiating this expression and substituting $v=tu$ givesTo identify the gradient, write an arbitrary tangent vector at $q={\exp }_p(v)$ as $\eta =(d{\exp }_p{)}_{v}w$. Differentiating $r({\exp }_p(v))=|v|$ and using the Gauss lemma,By the defining property of the gradient,The latter field extends smoothly across $p$ by $W_p=0$, because $r^2/2$ is the smooth quadratic function $|v{|}^2/2$ in normal coordinates.

The same calculation also proves local radial minimization: for a piecewise smooth curve in the normal neighbourhood, $|dr(\dot{\alpha })|\le |\dot{\alpha }|$, so its length from $p$ to $q$ is at least $r(q)$, attained by the radial geodesic. Take a small tangent ball with closure inside a larger normal ball. Any path leaving the small ball before reaching a point inside it has already accumulated length at least its radius. Therefore, for points in the small ball, $r(q)=d(p,q)$ even when competitors may leave the neighbourhood. This local argument does not assert that $r$ equals global Riemannian distance throughout every arbitrarily large star-shaped normal neighbourhood.

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 $t_0$ be the finite cut time and put $q=\gamma (t_0)$. Continuity of Riemannian distance shows that the segment to $q$ still minimizes and $d(p,q)=t_0$. Choose $t_j\downarrow t_0$ with $t_j>t_0$. By the Hopf-Rinow theorem, there is a minimizing geodesic from $p$ to $q_j=\gamma (t_j)$ with length ${\ell }_j=d(p,q_j)<t_j$ and initial unit vector $v_j$. Continuity gives ${\ell }_j\to t_0$.

The unit tangent sphere is compact, so after a subsequence $v_j\to v$. Smooth dependence of the Riemannian exponential map gives ${\exp }_p(t_{0}v)=q$. The geodesic with initial vector $v$ has length $t_0=d(p,q)$ up to $q$, hence minimizes. If $v\ne \dot{\gamma }(0)$, this is the required second minimizing geodesic.

Suppose instead $v=\dot{\gamma }(0)$ and $q$ is not a conjugate point. Then the differential of ${\exp }_p$ at $t_0\dot{\gamma }(0)$ is nonsingular. The inverse function theorem makes ${\exp }_p$ injective on a neighborhood of this vector. For large $j$, both ${\ell }_j{v}_j$ and $t_j\dot{\gamma }(0)$ belong to that neighborhood and exponentiate to $q_j$. They must be equal, implying ${\ell }_j=t_j$, a contradiction. Thus either a distinct minimizing geodesic exists or $q$ is conjugate. In the latter case it is the first conjugate point, since part (a) excludes any earlier one on a minimizing segment. We have proved the cut-point dichotomy for complete Riemannian manifolds.

Here the relevant concept is a Riemannian cut point, not the topological cut point obtained by disconnecting a space through removal of a point.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.

Choose a finite symmetric generating set $S$ for $\Gamma$, a base point $p$, and put $L={\max }_{s\in S}d(p,sp)$. If the word length of $\gamma$ is at most $m$, the triangle inequality and invariance of Riemannian distance under isometries give $d(p,\gamma p)\le Lm$.

A properly discontinuous group action has finite point stabilizer $F={\Gamma }_p$ and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move $p$ into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is $\rho >0$ such that distinct orbit points have distance more than $2\rho$: choose it using the finitely many orbit points within a fixed small ball of $p$, and translate the resulting separation by isometries. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.

Let $N_m$ be the number of distinct orbit points represented by the word metric ball of radius $m$. Their radius-$\rho$ balls are disjoint, have equal positive volume $v_0=\mathrm{Vol}B(p,\rho )$, and lie inside $B(p,Lm+\rho )$. The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that $\mathrm{Vol}B(p,R)/R^n$ is nonincreasing for $R>0$. ThereforeEach orbit point has at most $|F|$ preimages inside the word ball, since its full fiber is a coset of $F$. For the growth function of a finitely generated group, this givesHence $\Gamma$ has polynomial growth of a group of degree at most $n$. The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is polynomial growth of properly discontinuous isometry groups.

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.