《黎曼几何》课程教学资源（英文讲义）Lec23 RAUCH COMPARISON THEOREM

LECTURE 23:RAUCHCOMPARISONTHEOREMNow webegin to study the so-called comparison theorems.As wehave seenlast time, a comparison on curvature tensor will induce a comparison on Jacobifields, which would further give a comparison of geometry (triangles for the Cartan-Hadamard manifolds) or analysis (Hessian of d, for the Cartan-Hadamard man-ifolds), or restrict the possible behavior of geodesics (as in the proof of Synge'stheorem and Bonnet-Myers theorem). In all these cases we finally arrive at some re-strictions on global geometry/topology of the manifold. In the next couple lectureswe develop such ideasmore systematically.1.THEINDEX COMPARISONI Basic index comparison lemma.In the proof of Synge's Theorem and Bonnet-Myers Theorem, we used parallelvector fields to construct variations of a given geodesic. The advantage of parallelvector fields is that inner products (and thus lengths, angles) are preserved along geodesic. Another class of vector fields that are widely used in constructing variationsare Jacobi fields, which are variation fields of geodesic variations. Usually one startwith two geodesics on two manifolds whose curvatures are pointwise comparable,then compare two Jacobi fields with same initial value on these geodesics.So our basic setting for comparison is the following:. Let (M,g) and (M,g) be Riemannian manifolds of dimension m..Let :[0,a] →M and : [0,a] →M be normal geodesics with (O) =pand (0)= p..For each te[0,al, letK-(t) = min[K(II(t) / II(t) C T(t)M, dim II(t) = 2 and (t) E I(t),K+(t) = max[K((t) / il(t) C T(t)M, dimi(t) = 2 and (t) e iig().. We say two vectors Xe T(c) M and X. Ta(c)M are roughly the same if[x| =[xl and《Xe, (c) = (xe,(c).Now let X, X be Jacobi fields along and respectively, with X(0) = 0 andX(0) = 0. To compare X and X, we usually assume either"X(a) and X(a) are roughly the same"or"V(o)X and V(o)X are roughly the same

LECTURE 23: RAUCH COMPARISON THEOREM Now we begin to study the so-called comparison theorems. As we have seen last time, a comparison on curvature tensor will induce a comparison on Jacobi fields, which would further give a comparison of geometry (triangles for the Cartan￾Hadamard manifolds) or analysis (Hessian of d 2 p for the Cartan-Hadamard man￾ifolds), or restrict the possible behavior of geodesics (as in the proof of Synge’s theorem and Bonnet-Myers theorem). In all these cases we finally arrive at some re￾strictions on global geometry/topology of the manifold. In the next couple lectures we develop such ideas more systematically. 1. The Index Comparison ¶ Basic index comparison lemma. In the proof of Synge’s Theorem and Bonnet-Myers Theorem, we used parallel vector fields to construct variations of a given geodesic. The advantage of parallel vector fields is that inner products (and thus lengths, angles) are preserved along ge￾odesic. Another class of vector fields that are widely used in constructing variations are Jacobi fields, which are variation fields of geodesic variations. Usually one start with two geodesics on two manifolds whose curvatures are pointwise comparable, then compare two Jacobi fields with same initial value on these geodesics. So our basic setting for comparison is the following: • Let (M, g) and (M, f g˜) be Riemannian manifolds of dimension m. • Let γ : [0, a] → M and γe : [0, a] → Mf be normal geodesics with γ(0) = p and γe(0) = ˜p. • For each t ∈ [0, a], let K−(t) = min{K(Πγ(t)) | Πγ(t) ⊂ Tγ(t)M, dim Πγ(t) = 2 and ˙γ(t) ∈ Πγ(t)}, Ke +(t) = max{Ke(Πeγe(t)) | Πeγ(t) ⊂ Tγ˜(t)M, f dim Πeγ˜(t) = 2 and ˙γe(t) ∈ Πeγe(t)}. • We say two vectors Xc ∈ Tγ(c)M and Xec ∈ Tγ˜(c)Mf are ✿✿✿✿✿✿✿✿ roughly✿✿✿✿ the✿✿✿✿✿✿ same if |Xc| = |Xec| and ⟨Xc, γ˙(c)⟩ = ⟨Xec, γ˜˙(c)⟩. Now let X, Xe be Jacobi fields along γ and γe respectively, with X(0) = 0 and Xe(0) = 0. To compare X and Xe, we usually assume either “X(a) and Xe(a) are roughly the same”, or “∇γ˙ (0)X and ∇e γ˜˙ (0)Xe are roughly the same”. 1

2LECTURE23:RAUCHCOMPARISONTHEOREMThe following lemma is quite obvious whose proof is left as an exercise:Lemma 1.1. Let X, X be Jacobi fields along and with X(0) = X(0) = 0, andwriteX=c+dt+x,X=+dt+×Suppose either "X(a) and X(a) are roughly the same", or “V(o)X and s(o)X areroughly the same", then c= c and d = d.As a result, to compare two Jacobi fields whose initial or boundary values areroughly the same, it is enough to compare their “normal components".Now we prove the basic index comparison theorem:Theorem 1.2. Let X, X be Jacobi fields along , such that X(0) = 0, X(0) = 0,and suppose X(a) and X(a) are roughly the same. Assume further that(1) has no conjugate points on [0, a],(2) K+(t) ≤ K-(t) holds for all t E [0,a].thenI(X, X)≤I(x,x).Moreover, if K+(t) < K-(t) for some t< a, then I(X, X) <I(X,X)Proof. By Lemma 1.1 we may assume X, X are normal. Let [ei(t), ..: ,em(t)) and[ei(t), . ,em(t)) be orthonormal frames that are parallel along and , such thatei(t) =(t), éi(t) =(t), and e2(a) =X(a)/α, é2(a) =X(a)/α,where Q=[x(a)|=x(a)| o since has no conjugate point.If we denoteX(t) = Xi(t)e;(t), X(t) = Xi(t)é;(t)respectively, then obviously we have. xi(0) = Xi(0) = 0 for all i,. X2(a) = X2(a) = α and Xi(a) = Xi(a) = 0 for all i 2,. X'(t) = X'(t) = o for all t e [o, al (since both X and X are normal).As in the proof of Bonnet-Myers theorem we transplant X to by definingY(t) = Xi(t)e;(t)Then Y(O) = 0, Y(a) = X(a). Since X is a Jacobi fieldI(X, X) ≤ I(Y,Y)

2 LECTURE 23: RAUCH COMPARISON THEOREM The following lemma is quite obvious whose proof is left as an exercise: Lemma 1.1. Let X, Xe be Jacobi fields along γ and γ˜ with X(0) = Xe(0) = 0, and write X = cγ˙ + dtγ˙ + X ⊥, Xe = ˜cγ˜˙ + ˜dtγ˜˙ + Xe⊥. Suppose either “X(a) and Xe(a) are roughly the same”, or “∇γ˙ (0)X and ∇e γ˜˙ (0)Xe are roughly the same”, then c = ˜c and d = ˜d. As a result, to compare two Jacobi fields whose initial or boundary values are roughly the same, it is enough to compare their “normal components”. Now we prove the basic index comparison theorem: Theorem 1.2. Let X, Xe be Jacobi fields along γ, γe such that X(0) = 0, Xe(0) = 0, and suppose X(a) and Xe(a) are roughly the same. Assume further that (1) γ has no conjugate points on [0, a], (2) Ke +(t) ≤ K−(t) holds for all t ∈ [0, a]. then I(X, X) ≤ I(X, e Xe). Moreover, if K+(t) < K−(t) for some t < a, then I(X, X) < I(X, e Xe). Proof. By Lemma 1.1 we may assume X, Xe are normal. Let {e1(t), · · · , em(t)} and {e˜1(t), · · · , e˜m(t)} be orthonormal frames that are parallel along γ and ˜γ, such that e1(t) = ˙γ(t), e˜1(t) = γ˜˙(t), and e2(a) = X(a)/α, e˜2(a) = Xe(a)/α, where α = |X(a)| = |Xe(a)| ̸= 0 since γ has no conjugate point. If we denote X(t) = X i (t)ei(t), Xe(t) = Xei (t)˜ei(t) respectively, then obviously we have • Xi (0) = Xei (0) = 0 for all i, • X2 (a) = Xe2 (a) = α and Xi (a) = Xei (a) = 0 for all i ̸= 2, • X1 (t) = X˜ 1 (t) = 0 for all t ∈ [0, a] (since both X and Xe are normal). As in the proof of Bonnet-Myers theorem we transplant Xe to γ by defining Y (t) = Xei (t)ei(t). Then Y (0) = 0, Y (a) = X(a). Since X is a Jacobi field, I(X, X) ≤ I(Y, Y )

3LECTURE23:RAUCHCOMPARISONTHEOREMOn the other handI(Y,Y) = / (IV,YP +(R(%,Y)%,Y)) dt/ (E(x(t)* -E(x(t)*k(%,) dt≤ / (E(x (t)2 -Z(x(t)K-(t) d≤ / (Z(x(t)? -Z(X(t)2K+(t) dt≤ / (I,XP+(R(%,X),x) dt= I(X,X).It follows that I(X, X) ≤ I(X, X).Finally if K+(t) < K-(t) for some t < a, then the second inequality above isstrict, and thus I(X, X) < I(X, X)口I Local Hessian comparison.As a consequence of the basic index comparison theorem, we proveTheorem 1.3 (Local Hessian Comparison). Let (M,g), (M,g) be complete Rie-mannian manifolds, : [0,a] → M and : [0,a] → M be minimizing normalgeodesics in M and M respectively, so thatK+(t) ≤ K-(t) holds for all t E [0, a].Fir 0< b< a and write q= (b), q=(b). Suppose Xg e TqM and X, e TqM areroughly the same. ThenV?dp(Xq, Xa) ≤ V2dp(Xg,Xa)Moreover, the equality holds if and only if K+(t) =K-(t) for all t e [o,b]l.Proof. Since and are length minimizing, and b < a, we see q Cut(p) andq g Cut(p). Let X be the Jacobi field with X(O) = 0, X(b) =Xg, then as we haveseen in Lecture 21,(V2dp)q(Xq,Xa) = (V(g)X,Xa) = I(X,X),口Now the conclusion follows from the index comparison theorem above.Since = Trv?, by taking trace we get, under the same assumptions,△dp(q) ≤△dp(g)

LECTURE 23: RAUCH COMPARISON THEOREM 3 On the other hand, I(Y, Y ) = Z a 0 ￾ |∇γ˙ Y | 2 + ⟨R( ˙γ, Y ) ˙γ, Y ⟩
dt = Z a 0 X( ˙Xe i (t))2 − X(X˜i (t))2K( ˙γ, Y )  dt ≤ Z a 0 X( ˙Xe i (t))2 − X(X˜i (t))2K−(t)  dt ≤ Z a 0 X( ˙Xe i (t))2 − X(X˜i (t))2Ke +(t)  dt ≤ Z a 0  |∇e γ˜˙ Xe| 2 + ⟨Re(γ, ˜˙ Xe)γ, ˜˙ Xe⟩  dt = I(X, e Xe). It follows that I(X, X) ≤ I(X, e Xe). Finally if K+(t) < K−(t) for some t < a, then the second inequality above is strict, and thus I(X, X) < I(X, e Xe). □ ¶ Local Hessian comparison. As a consequence of the basic index comparison theorem, we prove Theorem 1.3 (Local Hessian Comparison). Let (M, g),(M, f g˜) be complete Rie￾mannian manifolds, γ : [0, a] → M and γ˜ : [0, a] → Mf be minimizing normal geodesics in M and Mf respectively, so that Ke +(t) ≤ K−(t) holds for all t ∈ [0, a]. Fix 0 < b < a and write q = γ(b), q˜ = ˜γ(b). Suppose Xq ∈ TqM and Xeq˜ ∈ Tq˜Mf are roughly the same. Then ∇2 dp(Xq, Xq) ≤ ∇e 2 ˜dp˜(Xeq˜, Xeq˜). Moreover, the equality holds if and only if Ke +(t) = K−(t) for all t ∈ [0, b]. Proof. Since γ and γe are length minimizing, and b < a, we see q ̸∈ Cut(p) and qe ̸∈ Cut(˜p). Let X be the Jacobi field with X(0) = 0, X(b) = Xq, then as we have seen in Lecture 21, (∇2 dp)q(Xq, Xq) = ⟨∇γ˙ (q)X, Xq⟩ = I(X, X). Now the conclusion follows from the index comparison theorem above. □ Since ∆ = Tr∇2 , by taking trace we get, under the same assumptions, ∆dp(q) ≤ ∆e ˜dp˜(˜q)

4LECTURE23:RAUCHCOMPARISONTHEOREM2.RAUCH'SJACOBIFIELDCOMPARISONTHEOREMI Rauch comparison theorem.Now we state and prove Rauch's comparison theorem, which relates the sectionalcurvature of a Riemannian manifold to the length of Jacobi fields (and thus the ratesat which geodesics spread apart).Theorem 2.1 (Rauch comparison theorem). Let X, X be Jacobi fields along %,with X(0) = X(0) = 0, such that V(o)X and V(o)X are roughly the same. Assume(1) has no conjugate points on [0,a],(2) K+(t) o for all t e (oal. Let c 0 on (O,c). For any b e (0, c), we definex(t)x(t)、x(t)=X(t) =X(6)1X(b)Applying Theorem 1.2 to Xb, X, on [0, b] we get(V(6)Xb, Xb(b) = I(Xb, X6) ≤ I(Xb, Xb) = (V(6)Xb, X(b),i.e.((6)x,x(b)1(b)1 u(b) (V(b)X, X(b))2u(6)《X(6),X(b))= 2 (6)(X(6), X(6))<t)i(t)So for any t e (O, c), we have.This is exactlywhatwe need.++

4 LECTURE 23: RAUCH COMPARISON THEOREM 2. Rauch’s Jacobi field comparison theorem ¶ Rauch comparison theorem. Now we state and prove Rauch’s comparison theorem, which relates the sectional curvature of a Riemannian manifold to the length of Jacobi fields (and thus the rates at which geodesics spread apart). Theorem 2.1 (Rauch comparison theorem). Let X, Xe be Jacobi fields along γ, γ˜ with X(0) = Xe(0) = 0, such that ∇γ˙ (0)X and ∇e γ˜˙ (0)Xe are roughly the same. Assume (1) γ has no conjugate points on [0, a], (2) Ke +(t) ≤ K−(t) holds for all t ∈ [0, a]. Then γ˜ has no conjugate points on [0, a], and for all t ∈ [0, a], |X(t)| ≤ |Xe(t)|. Moreover, if there is 0 0 for all t ∈ (0, a]. Let c ≤ a be the greatest number so that ˜u(t) > 0 on (0, c). For any b ∈ (0, c), we define Xb(t) = X(t) |X(b)| , Xeb(t) = Xe(t) |Xe(b)| . Applying Theorem 1.2 to Xb, Xeb on [0, b] we get ⟨∇γ˙ (b)Xb, Xb(b)⟩ = I(Xb, Xb) ≤ I(Xeb, Xeb) = ⟨∇e γ˜˙ (b)Xeb, Xeb(b)⟩, i.e. 1 2 u˙(b) u(b) = ⟨∇γ˙ (b)X, X(b)⟩ ⟨X(b), X(b)⟩ ≤ ⟨∇e γ˜˙ (b)X, e Xe(b)⟩ ⟨Xe(b), Xe(b)⟩ = 1 2 u˜˙(b) u˜(b) . So for any t ∈ (0, c), we have u˙ (t) u(t) ≤ u˜˙ (t) u˜(t) . This is exactly what we need

LECTURE23:RAUCHCOMPARISONTHEOREM5To summary, we proved that [x(t)/ ≤|X(t)| for t e (0, c). If c 0,contradicting with the choice of c. So we must have c=a. In particular, has noconjugate points on [O, a].Finally if there is 0< to < t such that K+(to) < K-(to), thenI(Xto, Xto)< I(Xto, Xto),口and thus the inequality for t is strict. This completes the proofAlthough Rauch comparison theorem is stated for two general manifolds whosesectional curvatures are comparable, in most applications one of the two manifoldsis a model space that has constant sectional curvature, and the second one is themanifoldunderstudywhosecurvatureisboundedeitherfrombelow orfromaboveForexample,accordingtotheexplicitformulaforJacobifieldsonspheres,thedistance between any two consecutive conjugate points of the sphere with sectionalcurvaturekis/k.Asa resultwegetCorollary 2.2. Suppose the sectional curvature of (M,g) satisfies0<Ci≤K≤C2where Ci,C2 are constants.Letbe any geodesic in M.Then the distance Dbetween any two consecutive conjugate points of satisfies元元DOne should compare this with Sturm comparison theorem in ODE.I Rauch comparison theorem: second form.Since any Jacobi field X along with X(O) = 0 can be written explicitly asX(t) = t(d expp)t(o)(Vs(0)x),one can rewrite Rauch comparison theorem above asTheorem 2.3 (Rauch comparison theorem, Second form). Let , be geodesicswith p = (O), p = (O), and suppose Xp E T,M, Xp E TpM are roughly the same.Assumealso(1) has no conjugate points on [0,a],(2) K+(t) ≤ K-(t) holds for all t E [0,a].ThenI(d expp)tr(0)Xp/ ≤ I(d expp)t(0)XplMoreover, the equality is strict for t if there erists 0 < to < t with K+(to) < K-(to)

LECTURE 23: RAUCH COMPARISON THEOREM 5 To summary, we proved that |X(t)| ≤ |Xe(t)| for t ∈ (0, c). If c 0, contradicting with the choice of c. So we must have c = a. In particular, ˜γ has no conjugate points on [0, a]. Finally if there is 0 < t0 < t such that K+(t0) < K−(t0), then I(Xt0 , Xt0 ) < I(Xet0 , Xet0 ), and thus the inequality for t is strict. This completes the proof. □ Although Rauch comparison theorem is stated for two general manifolds whose sectional curvatures are comparable, in most applications one of the two manifolds is a model space that has constant sectional curvature, and the second one is the manifold under study whose curvature is bounded either from below or from above. For example, according to the explicit formula for Jacobi fields on spheres, the distance between any two consecutive conjugate points of the sphere with sectional curvature κ is π/√ κ. As a result we get Corollary 2.2. Suppose the sectional curvature of (M, g) satisfies 0 < C1 ≤ K ≤ C2, where C1, C2 are constants. Let γ be any geodesic in M. Then the distance D between any two consecutive conjugate points of γ satisfies π √ C2 ≤ D ≤ π √ C1 . One should compare this with Sturm comparison theorem in ODE. ¶ Rauch comparison theorem: second form. Since any Jacobi field X along γ with X(0) = 0 can be written explicitly as X(t) = t(d expp )tγ˙ (0)(∇γ˙ (0)X), one can rewrite Rauch comparison theorem above as Theorem 2.3 (Rauch comparison theorem, Second form). Let γ, γ˜ be geodesics with p = γ(0), p˜ = ˜γ(0), and suppose Xp ∈ TpM, Xep˜ ∈ Tp˜Mf are roughly the same. Assume also (1) γ has no conjugate points on [0, a], (2) Ke +(t) ≤ K−(t) holds for all t ∈ [0, a]. Then |(d expp )tγ˙ (0)Xp| ≤ |(d expp˜ )tγ˜˙ (0)Xep|. Moreover, the equality is strict for t if there exists 0 < t0 < t with K+(t0) < K−(t0)

6LECTURE23:RAUCHCOMPARISONTHEOREMNote that for any Riemannian manifold (M,g)and any point p E M, (T,M,gp)isa Riemannian manifold that has constant sectional curvature 0.Applying Theorem2.3 to (M,g) and (TpM,9p) we getCorollary 2.4. Let (M,g) be a complete Riemannian manifold with sectional cur-vatureK≤0.Then for anypEM,anyXpETpM andYpETpM =Tx,(TpM),I(d exPp)x,(Yp)/ ≥ /YplIn particular, for any curve in T,M, one hasL()≤ L(exPpo)Moreover, the equality is strict if K< 0.Note that by using this corollary one can give another proof of Proposition 1.5(the weak cosine law for Cartan-Hadamard manifolds) in Lecture 22

6 LECTURE 23: RAUCH COMPARISON THEOREM Note that for any Riemannian manifold (M, g) and any point p ∈ M, (TpM, gp) is a Riemannian manifold that has constant sectional curvature 0. Applying Theorem 2.3 to (M, g) and (TpM, gp) we get Corollary 2.4. Let (M, g) be a complete Riemannian manifold with sectional cur￾vature K ≤ 0. Then for any p ∈ M, any Xp ∈ TpM and Yp ∈ TpM = TXp (TpM), |(d expp )Xp (Yp)| ≥ |Yp|. In particular, for any curve γ in TpM, one has L(γ) ≤ L(expp ◦γ). Moreover, the equality is strict if K < 0. Note that by using this corollary one can give another proof of Proposition 1.5 (the weak cosine law for Cartan-Hadamard manifolds) in Lecture 22