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Along the downward gradient flow, the gradient-flow dissipation identity, the chain rule and the defining property of the Riemannian gradient giveA smooth vector field on a compact manifold is complete, so the trajectory exists for every $t\ge 0$. In particular, its function value never increases.

If the trajectory is still in $C_{[a,b]}$ at time $t$, its monotone function value has stayed in $[a,b]$ throughout $[0,t]$. ConsequentlyThus a uniform exit bound isThe trajectory is outside the closed band for every $t>T$. Once it leaves below level $a$, monotonicity prevents its return. The argument also covers the zero-width band $a=b$.

For the total occupation-time bound for a gradient flow, let $m={\min }_{M}f$ and $F={\max }_{M}f$, which exist by compactness. For every $R>0$, the same gradient flow identity givesTaking $R\to \infty$, by monotone convergence theorem, gives the total occupation-time boundThis allows arbitrarily many visits to $C_{\epsilon }$; it bounds their total duration. Neither of these two estimates needs the Morse function hypothesis.

For exponential convergence of a Morse gradient flow, first observe that a limiting point $z=x_{\infty }$ is a critical point. Indeed, writing ${\varphi }_s$ for the flow and using continuous dependence on initial data,Differentiation at $s=0$ gives $\nabla f(z)=0$.

We use the Morse lemma: near a nondegenerate critical point of Morse index $\lambda$, there are coordinates $(u,v)\in {\mathbb{R}}^{\lambda }\times {\mathbb{R}}^{n-\lambda }$ in whichChoose a Riemannian metric equal to the Euclidean metric on a smaller ball in these coordinates. Such a global Riemannian metric exists by a partition of unity: patch this metric to any background metric outside a slightly larger ball. One can do this simultaneously at every critical point, since a Morse function on a compact manifold has only finitely many critical points. The metric is chosen before forming its gradient flow; changing the metric can change trajectories.

For this metric, any trajectory converging to $z$ eventually remains in the smaller coordinate ball. From some time $t_0$ onward its equations are exactlyConvergence forces $u(t_0)=0$, and thenThe straight radial segment stays in the coordinate ball, so its length bounds the Riemannian distance from above:Increase $C$ beyond both $e^{2t_0}|v(t_0)|$ and ${\max }_{0\le t\le t_0}{e}^{2t}d({\gamma }_x(t),z)$. Then the requested strict estimate holds for all $t\ge 0$:This proof uses the Morse lemma, existence and uniqueness for smooth ordinary differential equations, continuous dependence on initial data, and the existence of a partition of unity. The expanding coordinates explain why convergence to a saddle occurs only along its stable manifold.

For a counterexample, use an arc coordinate $x$ around a point of the circle, its flat local metric, and a globally smooth function that equals $x^4$ on this arc. Such a function can be made by a smooth cutoff, agreeing with a positive constant outside a larger arc. For $x_0>0$ sufficiently small the downward gradient flow remains in the arc and isIt converges to the degenerate critical point $0$, and its Riemannian distance from $0$ is $x(t)$ for sufficiently large $t$. For every $k>0$, $e^{kt}x(t)\to \infty$, so no exponential bound is possible. This counterexample is on a compact manifold as well.

Use the subextremal four-dimensional Reissner-Nordstrom spacetime, with $0<|Q|<M$. Its static radial function isTake the time-symmetric two-ended bridge through the outer bifurcation surface. Its initial data have $K_{ij}=0$ andon each $r\ge r_{+}$ end. The bridge is smooth: in proper radial distance $s$, $r-r_{+}=f^{\prime }(r_{+})s^2/4+O(s^4)$. Each asymptotically flat end is an infinite Riemannian distance away. Hence the spatial Riemannian manifold is complete and admits no proper same-dimensional smooth isometric extension as a connected spatial manifold.

Its maximal Cauchy development includes the two exteriors and the adjacent future and past regions between $r_{+}$ and $r_{-}$. It ends at inner Cauchy horizons, not at a curvature singularity. Since the simple root at $r_{-}$ is removable in horizon-penetrating coordinates and all curvature invariants are finite there, the exact solution extends across these Cauchy horizons. The extension is no longer globally determined by the given initial data.

The shaded region is the maximal Cauchy development; dashed upper and lower null edges are inner Cauchy horizons. The displayed neighboring diamonds illustrate smooth continuation, rather than the entire infinite extension.

The strong cosmic censorship conjecture concerns generic admissible initial data, in a specified extension regularity. Exact charged spherical data are exceptional. Perturbations can produce mass inflation at the inner Cauchy horizon, obstructing suitably regular extensions; the precise conjecture depends on the matter model and whether extensions are required to be $C^0$, $C^2$, or another regularity. Thus this exact extendible example does not refute a generic strong cosmic censorship conjecture.

For an Einstein-Maxwell example the gravitational triple $(\Sigma ,h,K)$ must be accompanied by electromagnetic initial data. One may take zero magnetic field and the smooth radial electric flux $Q/r^2$ through the bridge. The charges at the two ends have opposite signs when measured with outward normals. This is an example in the electrovacuum theory, not vacuum initial data.

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.

For a connected Riemannian manifold $(M,g)$, its Riemannian distance iswhere the infimum is over the piecewise smooth curves from $p$ to $q$. Connectedness of a smooth manifold implies path connectedness, so this set of curves is nonempty.

The Gauss lemma says that the differential of ${\exp }_p$ preserves the radial inner product: for $v,w\in T_{p}M$,Consequently radial geodesics from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_{p}M$. In a sufficiently small normal neighbourhood of $p$, this impliesevery competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.

The axioms $d_g(p,q)\ge 0$, symmetry, and the triangle inequality follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\ge r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.

Choosewhere $r_p$ is a normal radius at $p$. Take piecewise smooth curves $c_j$ from $p$ to $q$ such that $L_g(c_j)\to d_g(p,q)$. Each $c_j$ first leaves the normal ball $B_g(p,\delta )$ at a point $p_j$. The Gauss lemma gives $d_g(p,p_j)=\delta$. The geodesic sphereis compact, because the tangent-space sphere is compact and ${\exp }_p$ is defined on it. After taking a convergent subsequence, let $p_j\to p_0$.

The part of $c_j$ after $p_j$ has length at least $d_g(p_j,q)$, so continuity of the Riemannian distance givesThe triangle inequality gives the reverse inequality. Therefore

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 $p$.

In a normal neighbourhood $U={\exp }_p(V)$ with $V$ star-shaped, define $r(q)=|{\exp }_p^{-1}(q)|$. Away from $p$, the Gauss lemma gives $\nabla r=(d{\exp }_p{)}_v(v/|v|)$ for $v={\exp }_p^{-1}(q)$. Thus $|\nabla r|=1$ and this unit vector field is tangent to unit-speed radial geodesics. The function $r$ is not smooth at $p$, but $r^2/2$ is smooth and $\nabla (r^2/2)=(d{\exp }_p{)}_{v}v$ extends by zero. On a sufficiently small normal ball, $r$ equals the Riemannian distance from $p$.

// Target: riemannian-geometry.bigb

For a unit-speed geodesic $\gamma (t)={\exp }_p(tv)$, a finite cut time $c(v)$ is the last time to which its segment from $p$ minimizes Riemannian distance. Its endpoint is the Riemannian cut point along that geodesic. It lies in the cut locus; it need not be a cut point in the topological sense. The segment up to the cut time still minimizes by continuity of distance.

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.

In a relatively compact coordinate ball the Riemannian metric norm is bounded above and below by positive multiples of the Euclidean norm. A coordinate segment gives the upper distance bound. A path leaving the ball has a positive lower length bound before its first exit; paths staying in it have the Euclidean displacement lower bound. Consequently distinct points have positive Riemannian distance, and the resulting metric topology equals the manifold topology. Connectedness makes the distance finite because smooth manifolds are locally path connected.

Riemannian geometry studies smooth manifolds equipped with a Riemannian metric: an inner product at each tangent space, varying smoothly with position. This local measuring rule determines angles, arc length, Riemannian distance and geodesics. An inhabitant of a surface can discover its geometry by measuring within it, without knowing how it sits in a larger Euclidean space.

A circular cylinder is a particularly useful example. Its surface bends in space, yet unrolling it preserves the induced metric: a small patch has the geometry of a flat sheet. By contrast, a round sphere has positive Gaussian curvature. The distinction between intrinsic flatness of a circular cylinder and extrinsic curvature separates the geometry an inhabitant measures from the way an outside observer sees the surface bend.

Geodesics supply the analogue of straight lines. On a round sphere they follow great circles; on a circular cylinder they become straight lines when lifted to an unrolled sheet. Their arc length lets us compare candidate routes, as in great-circle distance and Riemannian distance on a circular cylinder. The same measuring rule also defines differential operators such as the Laplace-Beltrami operator, connecting geometry with diffusion and spectral geometry.

Bernhard Riemann's 1854 inaugural lecture proposed studying spaces of arbitrary dimension through their infinitesimal measuring rules, including a positive quadratic expression for the squared line element. This moved the subject beyond a particular surrounding space: the measuring rule itself became part of the geometry. ^[ref].

A Riemannian metric is a smoothly varying positive-definite inner product $g_p$ on each tangent space $T_{p}M$ of a smooth manifold. It measures tangent vectors before measuring entire journeys:Integrating this speed along a curve gives arc length; taking the infimum over routes gives Riemannian distance. A smooth manifold together with this measuring rule is a Riemannian manifold.

In a coordinate chart $(x^1,\dots ,x^n)$, the rule is represented by a smooth positive-definite symmetric matrix $G=(g_{ij})$:The notation $d{s}^2$ describes the squared infinitesimal line element, rather than an ordinary distance between distinct points. Changing coordinates changes the matrix entries; it preserves the measured norm of the same tangent vector.

For the plane with $d{s}^2=4d{x}^2+d{y}^2$, the vector $(1,0)$ has norm $2$ and $(0,1)$ has norm $1$. The coordinate change $X=2x,Y=y$ turns this into $d{s}^2=d{X}^2+d{Y}^2$: coefficients that look different can describe the same geometry. On a circular cylinder the induced metric $R^{2}d{\theta }^2+d{z}^2$ similarly becomes Euclidean when unrolled. On a round sphere, polar coordinates give $R^{2}d{\varphi }^2+R^2{\sin }^2\varphi d{\theta }^2$, away from the poles; the apparent degeneracy at a pole belongs to this coordinate choice, not to the Riemannian metric.

The point of the construction is that measurements make sense intrinsically, even without a chosen surrounding Euclidean space. It gives a common starting point for geodesics, curvature encoded by the Riemann curvature tensor, and the Laplace-Beltrami operator. ^[ref].

A connected Riemannian manifold is two-point homogeneous if its isometry group is transitive on ordered pairs at each fixed Riemannian distance. The equal-distance condition is necessary because isometries preserve distance. Round spheres and Euclidean spaces are examples. The unit tangent transitivity characterizes two-point homogeneity lemma shows that this pair condition is equivalent to transitivity on the unit tangent bundle.