《黎曼几何》课程教学资源（英文讲义）Lec03 THE RIEMANNIAN DISTANCE.pdf_大学文库

2 LECTURE 3: THE RIEMANNIAN DISTANCE By the same change of variable argument one can prove Lemma 1.3. Let φ : (M, gM) → (N, gN ) be a local isometry, and γ be a regular parametric curve in M. Then LengthM(γ) = LengthN (φ(γ)). ¶ Arc length parametrization. Although there are lots of different ways to parametrize a curve, there is one parametrization, the arc-length parametrization defined below, that is best and thus will be extensively used in this course. To describe the arc-length parametrization, we start with any regular parametric curve γ : [a, b] → M. We will call s(t) = Z t a q ⟨γ˙(τ ), γ˙(τ )⟩γ(τ)dτ the arc-length function of γ. Obviously s = s(t) is a strictly increasing function mapping the interval [a, b] to the interval [0, Length(γ)]. We will denote by t = t(s) the inverse function of s = s(t). One can reparametrize γ via the parameter s, γ1(s) = γ(t(s)), 0 ≤ s ≤ Length(γ). This is called the arc-length parametrization. It has the following nice property Proposition 1.4. For the arc-length parametrization, ⟨γ˙ 1(s), γ˙ 1(s)⟩γ1(s) ≡ 1. Proof. At s = s(t), we have t ′ (s) = (s ′ (t))−1 = ⟨γ˙(t), γ˙(t)⟩ −1/2 . So ⟨γ˙ 1(s), γ˙ 1(s)⟩γ1(s) = ⟨γ˙(t)t ′ (s), γ˙(t)t ′ (s)⟩γ1(s(t)) = t ′ (s) 2 ⟨γ˙(t), γ˙(t)⟩γ(t) = 1. □ Conversely by the definition of the arc-length function s(t) above, we see that if a parametrization of γ satisfies ⟨γ, ˙ γ˙⟩ ≡ 1, then it is simply a translation of the arc-length parametrization. Usually a curve with ∥γ˙ ∥ ≡ 1 is called a normal curve, and a curve with ∥γ˙ ∥ ≡ c is called a curve of constant speed. Now suppose (M, g) is a 1-dimensional Riemannian manifold. Then locally near each point, a local coordinate neighborhood is a curve, and the arc-length parameter s is a local coordinate. By definition, g11 = g(∂s, ∂s) = 1, and thus locally g = ds ⊗ ds. As an immediate consequence, we get Corollary 1.5. Any two 1-dimensional Riemannian manifolds are locally isometric. Remark. Discussions above can be easily extended to piecewise C 1 curves in M: If γ : [a, b] → M is a continuous map, and there exists a partition a = a0 < a1 < a2 < · · · < aN = b

LECTURE3:THERIEMANNIANDISTANCE5Theorem (Myers-Steenrod). Let (M,gm) and (N,gn) be Riemann-ian manifolds,and dist,distn bethe corresponding distancefunc-tions.If:(M,distm)→(N,dist)is surjective and distance-preserving, then it is an isometry (and in particular it is smooth).In other words, the concept“isometry" in the world of Riemannian geometry coin-cides with the concept “isometry"in the world of metric spaces.Their proof wassimplified and the theorem was strengthened by Palais in 1959 toTheorem (Palais). The Riemannian distance of a Riemannian manifold determines its structure as a manifold (the smooth structure)and itsRiemannianmetric.Remark.Onemayfurtheraskthefollowingquestion:Given a metric d on a smooth manifold M, is it true that d can berealized as the Riemannian distance for some Riemannian metric g?The answer is no: Consider the taxicab metric on IR2. Then any two points can beconnected by infinitely many"shortest curves".On the other hand, as we will seelater, on a Riemannian manifold, near any point there is a neighborhood in whichany point can be connected to the given point by a unique shortest curve.I Continuity of the distance function.We may further study the topology generated by the metric dist. First we proveProposition 2.3.For any fired p, the functionf() = dist(p)is continuous on M (with respect to the manifold topology))Proof. Since manifolds are second countable, it is enough to prove sequential con-tinuity of f.As in the proof of the previous theorem we take a coordinate patch(, U, V) centered at q with p(U) = V = Bi(O) c Rm. Let qi be a sequence of pointsthat tends to g with respect to the manifold topology, i.e. for any k, there existsN(k) such that for all i ≥ N(k), p(qi) E Bi/k(O). We want to prove f(q) → f(q).By triangle inequality (see (c) above), we haveIf(qi) -f(q)/ ≤dist(q, qi).So itsuffices toprovedist(q,qi)→0as i→oo.Again we let h = (-1)*(glu) be the induced metric on V. DenoteA=sup(thegreatest eigenvalueof thematrix (hi)r|reBr/2()Then for any r e By/2(o) and any X e TV, we have(X, X)h ≤A(X,X)g0:So if we take%: [0, 1] -→ V, %(t) = tp(qi)

LECTURE 3: THE RIEMANNIAN DISTANCE 5 Theorem (Myers-Steenrod). Let (M, gM) and (N, gN ) be Riemannian manifolds, and distM, distN be the corresponding distance functions. If φ : (M, distM) → (N, distN ) is surjective and distancepreserving, then it is an isometry (and in particular it is smooth). In other words, the concept “isometry” in the world of Riemannian geometry coincides with the concept “isometry” in the world of metric spaces. Their proof was simplified and the theorem was strengthened by Palais in 1959 to Theorem (Palais). The Riemannian distance of a Riemannian manifold determines its structure as a manifold (the smooth structure) and its Riemannian metric. Remark. One may further ask the following question: Given a metric d on a smooth manifold M, is it true that d can be realized as the Riemannian distance for some Riemannian metric g? The answer is no: Consider the taxicab metric on R 2 . Then any two points can be connected by infinitely many “shortest curves”. On the other hand, as we will see later, on a Riemannian manifold, near any point there is a neighborhood in which any point can be connected to the given point by a unique shortest curve. ¶ Continuity of the distance function. We may further study the topology generated by the metric dist. First we prove Proposition 2.3. For any fixed p, the function f(·) = dist(·, p) is continuous on M (with respect to the manifold topology). Proof. Since manifolds are second countable, it is enough to prove sequential continuity of f. As in the proof of the previous theorem we take a coordinate patch (φ, U, V ) centered at q with φ(U) = V = B1(0) ⊂ R m. Let qi be a sequence of points that tends to q with respect to the manifold topology, i.e. for any k, there exists N(k) such that for all i ≥ N(k), φ(qi) ∈ B1/k(0). We want to prove f(qi) → f(q). By triangle inequality (see (c) above), we have |f(qi) − f(q)| ≤ dist(q, qi). So it suffices to prove dist(q, qi) → 0 as i → ∞. Again we let h = (φ −1 ) ∗ (g|U ) be the induced metric on V . Denote Λ = sup{the greatest eigenvalue of the matrix (hij )x | x ∈ B1/2(0)}. Then for any x ∈ B1/2(0) and any X ∈ TxV , we have ⟨X, X⟩h ≤ Λ⟨X, X⟩g0 . So if we take γ˜i : [0, 1] → V, γ˜i(t) = tφ(qi)

6LECTURE3:THERIEMANNIANDISTANCEbe the “straight line segment" from 0 = (q) to (qi), then for i ≥ N(k),Lengh() - i.i)au V/ /hi,i)ioat - A Lengha() bSince : (U,g)→ (V,h) is an isometry, we conclude thatdist(q, qi)≤ Length(-1 0%) = Lengthn(%) ≤ V/k,口for all i ≥ N(k). This completes the proof.Remark. The distance functions dist(,p) are among the most important functions(maybe the only natural geometric functions) on a Riemannian manifold. Note the triangleinequality implies that the“distance to p"function f(q)= dist(p,q) is not onlycontinuous,but also Lipschitz continuous (with respect to the distance). One mayfurther ask: are they smooth? Even in the Euclidean case, it is obvious that dist(,p)is not smooth at r =p(this singularity can be eliminated by considering f2).On the otherhand.as wewill seelater,thisfunction is smooth in a“punctured neighborhood'U-(p) near p, but might have singularities at other points.Erample.Consider the round sphere (S?,ground). Let p= (O,O,-1)be the southpole. It is easy to see that for q = (r, y, z),f(q) = d(p,q) = (1 - arccos z) = π(1 - arccos V1- r2 - y2)which is not smooth at both the south pole and the north pole. [In fact for any compactRiemannian manifold, the function d,() = dist(p, -) is not smooth at some point q + p. Since therestriction of Euclidean distance function dEuc(p, ) to S2 is smooth on S?- (p), we conclude thatthe restriction of the Euclidean distance function to S? is NOT the Riemannian distance for anyRiemannian metric on s2!I The metric topology.As a consequence of the continuity, we proveCorollary 2.4. The metric topology on M induced by the metric dist coincides withthe manifold topology on M.Proof.The continuity of f()= dist(p,)implies that any metric open ball is alsoopen in the manifold topology.Conversely, for any (manifold)open neighborhoodU of q in M, by shrinking U we may assume U is a coordinate neighborhood and(, U, V = Bi(0)) is a coordinate chart. Repeat the proof of Theorem 2.2 we see anypoint p g U must have distance d(p, @) ≥ V/2, which is equivalent to say that themetric open ball of radius V/2 sits in U. So the two topologies on M coincide. We will call the metric ballB(q,r) = (p E M / dist(p, q) <r)a geodesic ball of radius r centered at q. For r small it has the topology of anEuclidean ball, while for r large it may have very complicated topology

6 LECTURE 3: THE RIEMANNIAN DISTANCE be the “straight line segment” from 0 = φ(q) to φ(qi), then for i ≥ N(k), Lengthh (˜γi) = Z 1 0 q ⟨γ˜˙ i , γ˜˙ i⟩hdt ≤ √ Λ Z 1 0 q ⟨γ˜˙ i , γ˜˙ i⟩g0 dt = √ Λ Lengthg0 (˜γi) ≤ √ Λ k . Since φ : (U, g) → (V, h) is an isometry, we conclude that dist(q, qi) ≤ Lengthg (φ −1 ◦ γ˜i) = Lengthh (˜γi) ≤ √ Λ/k, for all i ≥ N(k). This completes the proof. □ Remark. The distance functions dist(·, p) are among the most important functions (maybe the only natural geometric functions) on a Riemannian manifold. Note the triangle inequality implies that the “distance to p” function f(q) = dist(p, q) is not only continuous, but also Lipschitz continuous (with respect to the distance). One may further ask: are they smooth? Even in the Euclidean case, it is obvious that dist(·, p) is not smooth at x = p(this singularity can be eliminated by considering f 2 ). On the other hand, as we will see later, this function is smooth in a “punctured neighborhood” U − {p} near p, but might have singularities at other points. Example. Consider the round sphere (S 2 , ground). Let p = (0, 0, −1) be the south pole. It is easy to see that for q = (x, y, z), f(q) = d(p, q) = π(1 − arccos z) = π(1 − arccos p 1 − x 2 − y 2 ) which is not smooth at both the south pole and the north pole. [In fact for any compact Riemannian manifold, the function dp(·) = dist(p, ·) is not smooth at some point q ̸= p. Since the restriction of Euclidean distance function dEuc(p, ·) to S 2 is smooth on S 2 − {p}, we conclude that the restriction of the Euclidean distance function to S 2 is NOT the Riemannian distance for any Riemannian metric on S 2 ! ¶ The metric topology. As a consequence of the continuity, we prove Corollary 2.4. The metric topology on M induced by the metric dist coincides with the manifold topology on M. Proof. The continuity of f(·) = dist(p, ·) implies that any metric open ball is also open in the manifold topology. Conversely, for any (manifold) open neighborhood U of q in M, by shrinking U we may assume U is a coordinate neighborhood and (φ, U, V = B1(0)) is a coordinate chart. Repeat the proof of Theorem 2.2 we see any point p ̸∈ U must have distance d(p, q) ≥ √ λ/2, which is equivalent to say that the metric open ball of radius √ λ/2 sits in U. So the two topologies on M coincide. □ We will call the metric ball B(q, r) = {p ∈ M | dist(p, q) < r} a geodesic ball of radius r centered at q. For r small it has the topology of an Euclidean ball, while for r large it may have very complicated topology