LECTURE24:THE GLOBALHESSIANANDTOPONOGOVCOMPARISONLast time we proved various local comparison theorems that holds away fromcut locus. Today weturn to global comparisonthatholds onM.1.THEHESSIANCOMPARISONTHEOREM:GLOBALFORMI Hessian of distance for Mm.Let Mm be a space form, i.e. a complete connected Riemannian manifold withconstant sectional curvature k. Let : [o,ll → Mm be a normal geodesic in Mmfrom p to q Cut(p) U (p), then for any X, e ((i)+, the Jacobi field V along with V(O) = 0 and V(U) = X issnk(t)X,(t)V(t) =snk(1)where Xq(t) is the parallel vector field along with X(l) = Xq, and we used( sin(Vkt)k>0cos(Vkt),k>0,Vksnk(t) =k=0and cnk(t) = sns(t) =k = 0,1N1sinh(V-kt)cosh(V-kt), k < 0.k<0V-kAs a result, foranyY eT,M,cnk(l)(V2dp)a(Xq, Ya) = (Vs()V,Ya) =(Xq, Yq)snk(l)So the Hessian of dp at the point g, with respect to an orthonormal basis ei() =(),e2(U), ..-,em(), is00/0...cnx()00sna()(V2dp)g::.-cnk()00snx(0)The Hessian matrix is almost a constant matrix, with the only exception thatthere is a zero for the top-left entry. We will carefully choose a function f so thatV2(f odp) is a constant matrix. For this purpose we calculate, for any Xq, Yq e TqM,V?(fodp)(Xg,Ya)=(VxV(fodp),Yq)=(Vx,(f(dp)Vdp),Ya)=f"(dp)(VxVdp,Ya)+f"(dp)(Vdp,Xa)(Vdp,Ya)=f'(dp)2dp(Xg,Y)+f"(dp)<(0),Xa)((l),Yg)
LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON Last time we proved various local comparison theorems that holds away from cut locus. Today we turn to global comparison that holds on M. 1. The Hessian comparison theorem: global form ¶ Hessian of distance for Mm κ . Let Mm κ be a space form, i.e. a complete connected Riemannian manifold with constant sectional curvature κ. Let γ : [0, l] → Mm κ be a normal geodesic in Mm κ from p to q ̸∈ Cut(p) ∪ {p}, then for any Xq ∈ ( ˙γ(l))⊥, the Jacobi field V along γ with V (0) = 0 and V (l) = Xq is V (t) = snk(t) snk(l) Xq(t), where Xq(t) is the parallel vector field along γ with Xq(l) = Xq, and we used snk(t) = sin(√ kt) √ k , k > 0 t, k = 0 sinh(√ −kt) √ −k , k 0, 1, k = 0, cosh(√ −kt), k < 0. As a result, for any Yq ∈ TqM, (∇2 dp)q(Xq, Yq) = ⟨∇γ˙ (l)V, Yq⟩ = cnk(l) snk(l) ⟨Xq, Yq⟩. So the Hessian of dp at the point q, with respect to an orthonormal basis e1(l) = γ˙(l), e2(l), · · · , em(l), is (∇2 dp)q = 0 0 · · · 0 0 cnκ(l) snκ(l) · · · 0 . . . . . . . . . . . . 0 0 · · · cnκ(l) snκ(l) . The Hessian matrix is almost a constant matrix, with the only exception that there is a zero for the top-left entry. We will carefully choose a function f so that ∇2 (f ◦dp) is a constant matrix. For this purpose we calculate, for any Xq, Yq ∈ TqM, ∇2 (f ◦dp)(Xq,Yq)=⟨∇Xq∇(f ◦dp), Yq⟩=⟨∇Xq (f ′ (dp)∇dp), Yq⟩ =f ′ (dp)⟨∇Xq∇dp, Yq⟩+f ′′(dp)⟨∇dp,Xq⟩⟨∇dp,Yq⟩ =f ′ (dp)∇2 dp(Xq,Yq)+f ′′(dp)⟨γ˙(l),Xq⟩⟨γ˙(l),Yq⟩, 1
2LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISONwhere is the minimizing geodesic from p to q. As a result, for ?(f o dp) to be aconstant matrix under the given basis, we should choose f so that'(t) cnx(t)= f"(t).snr(t)So the simplest solution is to take f to satisfy f'(t) = snk(t), i.e. take f to be=cos(V/kr)if k> 0,Amd,(r) =if k= 0,snr(t)dt =-cosh(V-kr)if k 0we assume a,b,c < /Vk. Let 1 : [0,a] → Mm be the normal geodesic with(0) = B, (a) = C and let 2 : [0, b] -→ Mm be the normal geodesic with 2(0) =C, 2(b) = A.Now take p = B and = 2, i.e. consider (t) = md, o dg o 2(t). Thenp(0) = md,(a),p(b) =md,(c)and(0) = sn(a)(%1(a), 2(0)) = -sn(a) cos C.So if k = 0, we getα212p(t)=号-acos Ct+22and thus p(b) = md (c) becomes c2 = a? + b? - 2ab cos CFor K≠o,weget(t) = I+csn:()+ Ccn(t),The initial conditions (0) = mds(a) and '(O) = -sns(a)cosC implies ci Icnk(a). So the equation p(b) = mdx(c) becomes the fol--snk(a)cosC,c2lowingcosinelawinMm,cn(c)=cn(a)cnk(b)+ksnk(a)snk(b)cosC.Asadirectcorollary,wegetProposition 1.1. In Mm, take an angle α with side lengths l and l2 fired. Letf(a) be the distance between the end points. Then f(a) is increasing in α
2 LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON where γ is the minimizing geodesic from p to q. As a result, for ∇2 (f ◦ dp) to be a constant matrix under the given basis, we should choose f so that f ′ (t) cnκ(t) snκ(t) = f ′′(t). So the simplest solution is to take f to satisfy f ′ (t) = snk(t), i.e. take f to be mdκ(r) = Z r 0 snκ(t)dt = 1−cos(√ κr) κ , if κ > 0, r 2 2 , if κ = 0, 1−cosh(√ −κr) κ , if k 0 we assume a, b, c < π/√ κ. Let γ1 : [0, a] → Mm κ be the normal geodesic with γ1(0) = B, γ1(a) = C and let γ2 : [0, b] → Mm κ be the normal geodesic with γ2(0) = C, γ2(b) = A. Now take p = B and γ = γ2, i.e. consider φ(t) = mdκ ◦ dB ◦ γ2(t). Then φ(0) = mdκ(a), φ(b) = mdκ(c) and φ ′ (0) = snκ(a)⟨γ˙ 1(a), γ˙ 2(0)⟩ = −snκ(a) cos C. So if κ = 0, we get φ(t) = a 2 2 − a cos Ct + 1 2 t 2 and thus φ(b) = mdκ(c) becomes c 2 = a 2 + b 2 − 2ab cos C. For κ ̸= 0, we get φ(t) = 1 κ + c1snκ(t) + c2cnκ(t). The initial conditions φ(0) = mdκ(a) and φ ′ (0) = −snκ(a) cos C implies c1 = −snκ(a) cos C, c2 = − 1 κ cnκ(a). So the equation φ(b) = mdκ(c) becomes the following cosine law in Mm κ , cnκ(c) = cnκ(a)cnκ(b) + κsnκ(a)snκ(b) cos C. As a direct corollary, we get Proposition 1.1. In Mm κ , take an angle α with side lengths l1 and l2 fixed. Let f(α) be the distance between the end points. Then f(α) is increasing in α
LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISON3I Compare in the barrier sense.It turns out that the Hessian and Laplacian comparison theorems holds globallyon the whole of M in several weak sense: in thebarrier sense,in theviscosity senseand in the distribution sense'. Here we only discuss the first one, since the conditionis often easier to check.The notion of barrier sense wasfirst introduced by Calabiin1958Definition 1.2. Let f be a continuous function defined on (M,g).(1) If g e C2(U) is defined in a neighborhood U of p, andf(p) = g(p),and f(q)≤g(q), Vq EUthen we call g an upper barrier function of f at p.(2) If for any e > o, there is an upper barrier function ge of f at p, such thatAge(p) ≤c+e, then we say△f(p)≤c in the barrier sense.(3) If for any normal geodesic with o(0) = q, one has (f o o)"(0) ≤ c in thebarrier sense, then we say(2f)(q)≤c.Idinthebarriersense.Erample. Note that by taking M = (a, b), we get a definition of “f"(to) ≤ c inthe barrier sense" for continuous function f : (a,b) → R. For example, considerf(r) = -rl. Then g = 0 is a upper barrier function of f at = 0. As a result,f"(O)≤0 in the barrier sense.As observed by Calabi, one can easily construct upper barrier functions for thedistance function:Erample. If is a minimizing geodesic from p to q, then for 0 < n < d(p, q), thefunction rn(r)=n +d(r,(n)) is an upper barrier function for d, at qThe proof of the following lemma will be left as an exercise.Lemma 1.3. Let f:(a,b)-→R be continuous.(1) If f is C2, then f"(to) ≤c in the barrier sense if and only if f"(to) ≤c inthe usual sense.(2) If f takes its minimum at to, and f"(to)< c in the barrier sense, then c≥0.(3) If f"<O in the barrier sense, then f is concave.We also mention thefollowingHopf strong maximal principle withoutproof.Theorem 1.4 (Hopf-Calabi strong maximum principle). Let 2 M be a con-nected open set. Suppose f 0 in M in the barrier sense, and f has an interiorminimum, then f is constant on 2.'It can be proven that if f ≤ g holds in the barrier sense, then it also holds in the viscositysense and in the distribution sense
LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON 3 ¶ Compare in the barrier sense. It turns out that the Hessian and Laplacian comparison theorems holds globally on the whole of M in several weak sense: in the barrier sense, in the viscosity sense and in the distribution sense1 . Here we only discuss the first one, since the condition is often easier to check. The notion of barrier sense was first introduced by Calabi in 1958: Definition 1.2. Let f be a continuous function defined on (M, g). (1) If g ∈ C 2 (U) is defined in a neighborhood U of p, and f(p) = g(p), and f(q) ≤ g(q), ∀q ∈ U, then we call g an upper barrier function of f at p. (2) If for any ε > 0, there is an upper barrier function gε of f at p, such that ∆gε(p) ≤ c + ε, then we say ∆f(p) ≤ c in the barrier sense. (3) If for any normal geodesic σ with σ(0) = q, one has (f ◦ σ) ′′(0) ≤ c in the barrier sense, then we say (∇2 f)(q) ≤ c · Id in the barrier sense. Example. Note that by taking M = (a, b), we get a definition of “f ′′(t0) ≤ c in the barrier sense” for continuous function f : (a, b) → R. For example, consider f(x) = −|x|. Then g = 0 is a upper barrier function of f at x = 0. As a result, f ′′(0) ≤ 0 in the barrier sense. As observed by Calabi, one can easily construct upper barrier functions for the distance function: Example. If γ is a minimizing geodesic from p to q, then for 0 < η < d(p, q), the function rη(x) = η + d(x, γ(η)) is an upper barrier function for dp at q. The proof of the following lemma will be left as an exercise. Lemma 1.3. Let f : (a, b) → R be continuous. (1) If f is C 2 , then f ′′(t0) ≤ c in the barrier sense if and only if f ′′(t0) ≤ c in the usual sense. (2) If f takes its minimum at t0, and f ′′(t0) ≤ c in the barrier sense, then c ≥ 0. (3) If f ′′ ≤ 0 in the barrier sense, then f is concave. We also mention the following Hopf strong maximal principle without proof. Theorem 1.4 (Hopf-Calabi strong maximum principle). Let Ω ⊂ M be a connected open set. Suppose ∆f ≤ 0 in M in the barrier sense, and f has an interior minimum, then f is constant on Ω. 1 It can be proven that if ∆f ≤ g holds in the barrier sense, then it also holds in the viscosity sense and in the distribution sense
4LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISONI The global Hessian comparison theorem.Now weare ready to state and proveTheorem 1.5 (The global Hessian comparison theorem). Let (M, g) be a Riemann-ian manifold with sectional curvature K ≥ k. Then for any p E M,V(mds o dp) ≤cnk o dp Id in the barrier sense.Proof.According to the(local)Hessiancomparison theorem that weproved lasttime, the theorem holds at smooth points q of dp.We first prove the conclusion atthe point p:Forany normal geodesic with o(O)=p, we havemdko doo(t) =md(It)) =mds(t)Thus (mdo d,o )"(0)= md(0) = cn(0).It remains to prove the conclusion for a non-smooth point q + p of dp. So we let : [o,i] -→ M be a minimizing normal geodesic from p to q, and let be a normalgeodesic with o(0) = q. Note that by Bonnet-Myers theorem, I ≤ if k > 0.For 0 0, given any > 0, for n is small enough, we have(md 0 rn o)"(0) ≤ cnk(l) +e,whichimplies?(md o d) ≤ cnk o d,in the barrier sense.For k > 0 and I = , we will prove (M,g) is isomorphic to Sm, in which case we口may take α such that (O) = (l), and the desired conclusion follows
4 LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON ¶ The global Hessian comparison theorem. Now we are ready to state and prove Theorem 1.5 (The global Hessian comparison theorem). Let (M, g) be a Riemannian manifold with sectional curvature K ≥ κ. Then for any p ∈ M, ∇2 (mdκ ◦ dp) ≤ cnk ◦ dp · Id in the barrier sense. Proof. According to the (local) Hessian comparison theorem that we proved last time, the theorem holds at smooth points q of dp. We first prove the conclusion at the point p: For any normal geodesic σ with σ(0) = p, we have mdk ◦ dp ◦ σ(t) = mdk(|t|) = mdk(t). Thus (mdk ◦ dp ◦ σ) ′′(0) = md′′ k (0) = cnk(0). It remains to prove the conclusion for a non-smooth point q ̸= p of dp. So we let γ : [0, l] → M be a minimizing normal geodesic from p to q, and let σ be a normal geodesic with σ(0) = q. Note that by Bonnet-Myers theorem, l ≤ √π κ if κ > 0. For 0 0, given any ε > 0, for η is small enough, we have (mdκ ◦ rη ◦ σ) ′′(0) ≤ cnκ(l) + ε, which implies ∇2 (mdκ ◦ dp) ≤ cnk ◦ dp in the barrier sense. For κ > 0 and l = √π κ , we will prove (M, g) is isomorphic to S m κ , in which case we may take σ such that ˙σ(0) = ˙γ(l), and the desired conclusion follows. □
LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISON52.THE TOPONOGOV COMPARISON THEOREMThe purpose of this section is to prove a very useful global comparison theorem,due to Toponogov in 1959. It quantifies the assertion (c.f. PSet 4) that a pair ofgeodesics emanating from a point p spread apart more slowly in a region of highcurvature than they would in a region of low curvature.I Geodesic triangles and hinges.Definition 2.1. Let (M, g) be complete.(1) A geodesic triangle △ABC consists of three points A, B,C in M (which arecalled the ertices) and three minimizing normal geodesics (which are calledthe sides)AB,Bc,CA joining each two of them.Ifonly two sides, say AB and Ac, are minimizing, whilethethird is a normalgeodesic [which need not be minimizing] that satisfies the triangle inequalityL(BC) ≤ L(AB) + L(AC),then we will call ABC a generalized geodesic triangle(2) A geodesic hinge /BAC consists of a point A in M (which is again calledthe verter) and two minimizing normal geodesics AB, Ac (called the sides)emanating from A, with end points B and C in M.If one side is minimizing, while the other side is a normal geodesic[which neednot be minimizingl, we call ZBAC a generalized geodesic hinge.In what follows when we say hinge or triangle, we always mean generalizedgeodesic hinge or generalized geodesic triangle.Remark. In the definition of generalized geodesic hinge, we required that at leastone curve is minimizing.Otherwise the Toponogov comparison theorem below mayfail: If one take two geodesics of length in Mk+e, then the other endpoints of thecomparing hinge in M,will meet.Lemma2.2. Let (M,g)be a completeRiemannian manifold of dimension m whosesectional curvature K≥ k. Then(1)Let/BAC be a generalized geodesic hinge inM.If k>O we furtherassumethat all the sides of LABC have lengths no more than Then there isa generalized geodesic hinge BAC in Mm with same angle and the samecorresponding side lengths. [We will call it a comparing hinge.](2)Let ABC be a generalized geodesic triangle in M. If k > 0 we furtherassume that all the sides of ABC have lengths no more than k-Thenthere is a triangle ABC in Mm whose corresponding sides have the samelength asABC.[We will call it a comparing triangle.]
LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON 5 2. The Toponogov Comparison Theorem The purpose of this section is to prove a very useful global comparison theorem, due to Toponogov in 1959. It quantifies the assertion (c.f. PSet 4) that a pair of geodesics emanating from a point p spread apart more slowly in a region of high curvature than they would in a region of low curvature. ¶ Geodesic triangles and hinges. Definition 2.1. Let (M, g) be complete. (1) A geodesic triangle △ABC consists of three points A, B, C in M (which are called the vertices) and three minimizing normal geodesics (which are called the sides) γAB, γBC, γCA joining each two of them. If only two sides, say γAB and γAC, are minimizing, while the third is a normal geodesic [which need not be minimizing] that satisfies the triangle inequality L(γBC) ≤ L(γAB) + L(γAC), then we will call △ABC a generalized geodesic triangle. (2) A geodesic hinge ∠BAC consists of a point A in M (which is again called the vertex ) and two minimizing normal geodesics γAB, γAC (called the sides) emanating from A, with end points B and C in M. If one side is minimizing, while the other side is a normal geodesic[which need not be minimizing], we call ∠BAC a generalized geodesic hinge. In what follows when we say hinge or triangle, we always mean generalized geodesic hinge or generalized geodesic triangle. Remark. In the definition of generalized geodesic hinge, we required that at least one curve is minimizing. Otherwise the Toponogov comparison theorem below may fail: If one take two geodesics of length √π κ in Mκ+ε, then the other endpoints of the comparing hinge in Mκ will meet. Lemma 2.2. Let (M, g) be a complete Riemannian manifold of dimension m whose sectional curvature K ≥ k. Then (1) Let ∠BAC be a generalized geodesic hinge in M. If κ > 0 we further assume that all the sides of ∠ABC have lengths no more than √π κ . Then there is a generalized geodesic hinge ∠BeAeCe in Mm k with same angle and the same corresponding side lengths. [We will call it a comparing hinge.] (2) Let △ABC be a generalized geodesic triangle in M. If κ > 0 we further assume that all the sides of △ABC have lengths no more than √π κ . Then there is a triangle △AeBeCe in Mm k whose corresponding sides have the same length as △ABC. [We will call it a comparing triangle.]
6LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISONI Toponogov Comparison Theorem.Now we state and prove Toponogov Comparison Theorem,in which we canactually compare the distance functions instead of only comparing their Hessian.Theorem 2.3 (Toponogov Comparison Theorem). Let (M,g) be a complete Rie-mannian manifold with sectional curvature K > k. Then(1) (Hinge Version) Let ZBAC be a generalized geodesic hinge in M andZBAC a comparing hinge in Mm2. If k > 0 we further assume that the sidesof ZBAC have lengths no more than . Then dist(B,C)≤ dist(B,C).(2) (Triangle Version) Let ABC be a generalized geodesic triangle in M andABC a comparing triangle in Mm.Ifk>O wefurtherassume that all thesides of ABC have lengths no more than .Then the angles in ABCopposites to the minimizing geodesics are greater than the corresponding an-gles inABC.Proof.We first observe that according to the cosine law in M?, for any hinge withsides 1, 2 and angle Q, the functionf(α) = d(%(lo), 1(l1)is increasing for a E (O, ). So the Hinge version implies the triangle version.To prove the Hinge version, we needLemma 2.4. Let f : [o,I] -→ R be a continuous function that isdifferentiable at t =0, with f(0)=0 and f(0)≤0, wherel≤ ifk > 0. Moreover.assumef"(t)+kf(t)≤0in the barrier sense,then f(t)0 small, letpe(t) = d(%(lo -e), (t)), t E [0,l]-Then pe is smooth for t > 0 small enough, pe(0) = lo - andp(0) = (-%(0), 1(0)) = - cos α.Bytheglobal Hessian comparisontheorem,(md, o pe)"(t) ≤ cnko p(t) = 1 -K md, o p(t), in the barrier sense.We may perform the same computation in M? to conclude that for pe(t) = do(lo-s)((t),one has pe(O) = lo -e, p,(O) =- cosα and(mdro pe)"(t) = 1 -k mdkop(t)2obviously can replace Mm by M?
6 LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON ¶ Toponogov Comparison Theorem. Now we state and prove Toponogov Comparison Theorem, in which we can actually compare the distance functions instead of only comparing their Hessian. Theorem 2.3 (Toponogov Comparison Theorem). Let (M, g) be a complete Riemannian manifold with sectional curvature K ≥ k. Then (1) (Hinge Version) Let ∠BAC be a generalized geodesic hinge in M and ∠BeAeCe a comparing hinge in Mm k 2 . If κ > 0 we further assume that the sides of ∠BAC have lengths no more than √π κ . Then dist(B, C) ≤ dist(B, e Ce). (2) (Triangle Version) Let △ABC be a generalized geodesic triangle in M and △AeBeCe a comparing triangle in Mm k . If κ > 0 we further assume that all the sides of △ABC have lengths no more than √π κ . Then the angles in △ABC opposites to the minimizing geodesics are greater than the corresponding angles in △AeBeCe. Proof. We first observe that according to the cosine law in M2 κ , for any hinge with sides γ1, γ2 and angle α, the function f(α) = d(γ0(l0), γ1(l1)) is increasing for α ∈ (0, π). So the Hinge version implies the triangle version. To prove the Hinge version, we need Lemma 2.4. Let f : [0, l] → R be a continuous function that is differentiable at t = 0, with f(0) = 0 and f ′ (0) ≤ 0, where l ≤ √π κ if κ > 0. Moreover, assume f ′′(t) + κf(t) ≤ 0 in the barrier sense, then f(t) ≤ 0 for all t ∈ [0, l]. We first assume this lemma and proceed. For simplicity we denote γ0 = γAB, γ1 = γAC and denote l0 = L(γ0), l1 = L(γ1), α = ∠BAC. Assume γ0 is minimizing. For ε > 0 small, let ρε(t) = d(γ0(l0 − ε), γ1(t)), t ∈ [0, l1]. Then ρε is smooth for t > 0 small enough, ρε(0) = l0 − ε and ρ ′ ε (0) = ⟨−γ˙ 0(0), γ˙ 1(0)⟩ = − cos α. By the global Hessian comparison theorem, (mdκ ◦ ρε) ′′(t) ≤ cnk ◦ ρ(t) = 1 − κ mdκ ◦ ρ(t), in the barrier sense. We may perform the same computation in M2 κ to conclude that for ˜ρε(t) = dγ˜0(l0−ε)(˜γ1(t)), one has ˜ρε(0) = l0 − ε, ˜ρ ′ ε (0) = − cos α and (mdκ ◦ ρ˜ε) ′′(t) = 1 − κ mdκ ◦ ρ˜(t). 2Obviously can replace Mm k by M2 k
LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISON7So if we let f(t)=mdoPe-md,ope,thenf"(t) +kf(t)≤0in the barrier sense,and f(0) = 0, f'(0) = snr(pe(0)p(0) - snr(p(0)p:(0) = 0. By Lemma 2.4, wehave f(t) ≤0 for all t e [0, i]. It follows that pe(t) ≤ pe(t) for all t e [0, li]. Letting口→ O weget the desired conclusion.I Proof of Lemma 2.4.Let fe(t) = f(t) - Esnk(t), then fe(0) = 0 and f'(0) ≤ -e 0 small enough. In what follows we prove fe(t) 0Wemayassumeto0small so that [-5, - 0] [0, o]. Let o(t) = -sin(Vk+6(t + 0), so that$"(t) + (K + ) = 0. Suppose lo,to) takes its maximum at ti. Let ge,e' be anupperbarrier function of fat ti,i.e.ge,e (ti) = fe(ti), ge,er(t) ≥ fe(t) near ti,and such thatge,e,(ti)<≤(-k)fe(ti) +e*.Then ti is a maximum for 9 since < 0. It follows= 9e,e(t)o(t) - ge,e(t)p'(t)9esp2(t)equals 0 at ti, and thusge,e(ti)0(ti) - ge,e (ti)p"(t1)q01tp2(t1)ge,e(ti) +(K+)gee (t) E' +ge,e (ti)d(ti)d(ti)Letting el → 0 we get fe(ti) = ge,er(ti) ≥ 0, a contradiction
LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON 7 So if we let f(t) = mdκ ◦ ρε − mdκ ◦ ρ˜ε, then f ′′(t) + κf(t) ≤ 0 in the barrier sense, and f(0) = 0, f′ (0) = snκ(ρε(0))ρ ′ ε (0) − snκ(˜ρ ′ ε (0))˜ρ ′ ε (0) = 0. By Lemma 2.4, we have f(t) ≤ 0 for all t ∈ [0, l1]. It follows that ρε(t) ≤ ρ˜ε(t) for all t ∈ [0, l1]. Letting ε → 0 we get the desired conclusion. □ ¶ Proof of Lemma 2.4. Let fε(t) = f(t) − εsnκ(t), then fε(0) = 0 and f ′ ε (0) ≤ −ε 0 small enough. In what follows we prove fε(t) ≤ 0 for all t ∈ [0, l]. Letting ε → 0 we get the desired conclusion. By contradiction we let t0 be the smallest positive root of fε. Case 1: κ ≤ 0 . Suppose fε|[0,t0] takes its minimum at t1. Then we get, f ′′ ε (t1) + κfε(t1) = f ′′(t1) + κf(t1) − ε(sn′′ κ (t1) + κsnκ(t1)) ≤ 0 in the barrier sense. By Lemma 1.3(2), we get −κfε(t1) ≥ 0, i.e. fε(t1) ≥ 0, a contradiction. Case 2: κ > 0 . We may assume t0 0 small so that [−δ, √ π κ+δ − δ] ⊃ [0, t0]. Let ϕ(t) = − sin(√ κ + δ(t + δ)), so that ϕ ′′(t) + (K + δ)ϕ = 0. Suppose fε ϕ |[0,t0] takes its maximum at t1. Let gε,ε′ be an upper barrier function of fε at t1, i.e. gε,ε′(t1) = fε(t1), gε,ε′(t) ≥ fε(t) near t1, and such that g ′′ ε,ε′(t1) ≤ (−κ)fε(t1) + ε ′ . Then t1 is a maximum for gε,ε′ ϕ since ϕ < 0. It follows gε,ε′ ϕ ′ (t) = g ′ ε,ε′(t)ϕ(t) − gε,ε′(t)ϕ ′ (t) ϕ2 (t) equals 0 at t1, and thus 0 ≥ gε,ε′ ϕ ′′ (t1) = g ′′ ε,ε′(t1)ϕ(t1) − gε,ε′(t1)ϕ ′′(t1) ϕ2 (t1) = g ′′ ε,ε′(t1) + (K + δ)gε,ε′(t1) ϕ(t1) ≥ ε ′ + δgε,ε′(t1) ϕ(t1) . Letting ε ′ → 0 we get fε(t1) = gε,ε′(t1) ≥ 0, a contradiction
8LECTURE24:THEGLOBALHESSIANANDTOPONOGOV COMPARISONI Application to fundamental group.Asan application,weproveTheorem 2.5 (Gromov). Let (M,g) be a complete Riemannian manifold with sec-tional curvature K ≥O. Then Ti(M) is generated by no more thanVol(Sm-1)C(m) = Vol(Sm-1(元/6)generators, where Sm-1(π/6) is a geodesic ball of radius /6 in Sm-1.Proof.We will consider i(M)as thegroup of Deck transformations on the universalcovering M. Fix p e M and choose inductively a generating set of πi(M) as follows. We first choose e + gi E Ti(M) so thatd(p,g1-p) ≤d(p,g p),Vg Ei(M) / (e). Suppose gi,.. , gk-1 are chosen. We then choose gk & (g1,*+*, gk-1) so thatd(p,gkp)≤d(p,g-p),Vg E i(M) /<g1,...,gk-1).Let k be a minimizing geodesic in (M,g) from p to gk p. We claim that theangle between any two such geodesics is at least . Then the conclusion follows.We prove the claim by contradiction. Suppose the angle between and k+lis less than . For simplicity we denote ik = d(p, gk - p). Then according to theToponogov comparison theorem,d(gk+1·,gk-)?<++-luk+≤+tThis impliesd(p,gk+gk-p)=d(gk+t-p,gkp)<lk+1=d(p,gk+1 p),口which contradicts with the choice of gk+1.Remark. By the same way one can prove the following theorem of Gromov:Theorem 2.6 (Gromov). For k negative, there is a constant C =C(m,k,D) so that for any complete Riemannian manifold (M,g)withK≥k and diam(M,g)≤D,thefundamental groupi(M)is gener-ated by no more than C(m, k, D) generators.Note that a bound on diameter is needed. To see this, one can look at the exampleof surface of genus g
8 LECTURE 24: THE GLOBAL HESSIAN AND TOPONOGOV COMPARISON ¶ Application to fundamental group. As an application, we prove Theorem 2.5 (Gromov). Let (M, g) be a complete Riemannian manifold with sectional curvature K ≥ 0. Then π1(M) is generated by no more than C(m) = Vol(S m−1 ) Vol(Sm−1 (π/6)) generators, where S m−1 (π/6) is a geodesic ball of radius π/6 in S m−1 . Proof. We will consider π1(M) as the group of Deck transformations on the universal covering Mf. Fix ˜p ∈ Mf and choose inductively a generating set of π1(M) as follows: • We first choose e ̸= g1 ∈ π1(M) so that d(˜p, g1 · p˜) ≤ d(˜p, g · p˜), ∀g ∈ π1(M) \ {e}. • Suppose g1, · · · , gk−1 are chosen. We then choose gk ̸∈ ⟨g1, · · · , gk−1⟩ so that d(˜p, gk · p˜) ≤ d(˜p, g · p˜), ∀g ∈ π1(M) \ ⟨g1, · · · , gk−1⟩. Let ˜γk be a minimizing geodesic in (M, f g˜) from ˜p to gk · p˜. We claim that the angle between any two such geodesics is at least π 3 . Then the conclusion follows. We prove the claim by contradiction. Suppose the angle between ˜γk and ˜γk+l is less than π 3 . For simplicity we denote ˜lk = d(˜p, gk · p˜). Then according to the Toponogov comparison theorem, d(gk+l · p, g ˜ k · p˜) 2 < l2 k + l 2 k+l − lklk+l ≤ l 2 k+l . This implies d(˜p, g−1 k+l gk · p˜) = d(gk+l · p, g ˜ k · p˜) < lk+l = d(˜p, gk+l · p˜), which contradicts with the choice of gk+l . □ Remark. By the same way one can prove the following theorem of Gromov: Theorem 2.6 (Gromov). For k negative, there is a constant C = C(m, k, D) so that for any complete Riemannian manifold (M, g) with K ≥ k and diam(M, g) ≤ D, the fundamental group π1(M) is generated by no more than C(m, k, D) generators. Note that a bound on diameter is needed. To see this, one can look at the example of surface of genus g