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Lecture 20: Rauch Comparison Theorem and Its Applications

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Lecture 20: Rauch Comparison Theorem and Its Applications LECTURE 20: RAUCH COMPARISON THEOREM AND ITS APPLICATIONS 1. Rauch Comparison Theorem Let (M; g) and (M;f g~) be Riemannian manifolds of dimension m. Let γ : [0; a] ! M andγ ~ : [0; a] ! Mf be normal geodesics with γ(0) = p; γ~(0) =p: ~ For each t 2 [0; a], we denote − K (t) = minfK(Πγ(t)) j γ_ (t) 2 Πγ(t)g; + _ Ke (t) = maxfKe(Πe γ~(t)) j γ~(t) 2 Πe γ~(t)g: Theorem 1.1 (Rauch comparison theorem). Let X; Xe be Jacobi fields along γ; γ~ respectively, such that _ 1 X(0) = 0; Xe(0) = 0; 2 jrγ_ (0)Xj = jre γ~_ (0)Xej; 3 hγ_ (0); rγ_ (0)Xi = hγ~(0); re γ~_ (0)Xei: Assume further that i γ has no conjugate points on [0; a]; ii Ke +(t) ≤ K−(t) holds for all t 2 [0; a]: Then γ~ has no conjugate points on [0; a], and for all t 2 [0; a], jX(t)j ≤ jXe(t)j: We first prove Lemma 1.2. Let X; Xe be normal Jacobi fields along γ; γ~ respectively, such that a X(0) = 0; Xe(0) = 0; b jX(a)j = jXe(a)j: Assume further that i and ii above holds. Then I(X; X) ≤ I(X;e Xe): Proof. Let fe1(t); ··· ; em(t)g and fe~1(t); ··· ; e~m(t)g be orthonormal frames that are parallel along γ andγ ~ respectively, such that e1(t) =γ _ (t); e~1(t) = γ~_ (t) and e2(a) = X(a)/α; e~2(a) = Xe(a)/α; 1 2 LECTURE 20: RAUCH COMPARISON THEOREM AND ITS APPLICATIONS where α = jX(a)j = jXe(a)j= 6 0 since γ has no conjugate point. If we denote i i X(t) = X (t)ei(t); Xe(t) = Xe (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 6= 2, • X1(t) = X~ 1(t) = 0 for all t 2 [0; a]. We define a vector field Y along γ by i Y = Xe (t)ei(t): Then Y (0) = 0, Y (a) = X(a). So by corollary 2.4 in lecture 17, I(X; X) ≤ I(Y; Y ): On the other handside, Z a 2 I(Y; Y ) = jrγ_ Y j − Rm(_γ; Y; γ;_ Y ) dt 0 Z a X _ X = (Xe i(t))2 − (X~ i(t))2K(_γ; Y ) dt 0 Z a X _ X ≤ (Xe i(t))2 − (X~ i(t))2K−(t) dt 0 Z a X _ X ≤ (Xe i(t))2 − (X~ i(t))2Ke +(t) dt 0 Z a 2 _ _ ≤ jre γ~_ Xej − Rmg(γ;~ X;e γ;~ Xe) dt 0 = I(X;e Xe): It follows that I(X; X) ≤ I(X;e Xe). Proof of Rauch Comparison Theorem. The conditions 1 and 3 implies that the tangential components of X and Xe have the same length at correponding points. So WLOG, we may assume that both X and Xe are normal Jacobi fields. We denote u(t) = jX(t)j2; u~(t) = jXe(t)j2: Thenu ~(t)=u(t) is well-defined, and by L0Hospital's rule, u~(t) u~¨(t) jr Xj2 lim = lim = γ_ (0) = 1: t!0 u(t) t!0 u¨(t) 2 jrγ~_ (0)Xej d u~(t) Therefore, to prove jXj ≤ jXej, it is enough to prove dt u(t) ≥ 0, or equivalently, u~_(t)u(t) − u~(t)_u(t) ≥ 0: LECTURE 20: RAUCH COMPARISON THEOREM AND ITS APPLICATIONS 3 Since γ has no conjugate point, u(t) > 0 for all t 2 (0; a]. Let c ≤ a be the greatest number so thatu ~(t) > 0 on (0; c). For any b 2 (0; c), we define X(t) Xe(t) Xb(t) = ; Xeb(t) = : jX(b)j jXe(b)j Applying the previous lemma to Xb; Xeb on [0; b] , we get I(Xb;Xb) ≤ I(Xeb; Xeb): Recall that after integration by parts Z a a I(X; Y ) = − hrγ_ rγ_ X + R(_γ; X)_γ; Y idt + hrγ_ X; Y ij0: 0 So I(Xb;Xb) = hrγ_ (b)Xb;Xb(b)i and I(Xeb; Xeb) = hre γ~_ (b)Xeb; Xeb(b)i: It follows hrγ_ (b)Xb;Xb(b)i ≤ hre γ~_ (b)Xeb; Xeb(b)i; i.e. 1 u_(b) hr X; X(b)i hre _ X;e Xe(b)i 1 u~_(b) = γ_ (b) ≤ γ~(b) = : 2 u(b) hX(b);X(b)i hXe(b); Xe(b)i 2 u~(b) u_ (t) u~_ (t) So for any t 2 (0; c), we have u(t) ≤ u~(t) . This is exactly what we need. To summary, we proved that jX(t)j ≤ jXe(t)j for t 2 (0; c). If c < a, then jXe(c)j ≥ jX(c)j > 0, contradicting with the choice of c. So we must have c = a. In particular,γ ~ has no conjugate points on [0; a]. This completes the proof. Recall that any Jacobi field is a variation field of some geodesic variation. In particular, if X is a Jacobi field along γ : [0; a] ! M with X(0) = 0; rγ_ (0)X = Xp; then the corresponding variation can be chosen as γs(t) = expp(t(_γ(0) + sXp)); and as a consequence, the Jacobi field X is explicitly given by X(t) = t(d expp)tγ_ (0)Xp: One can rewrite Rauch comparison theorem above as Theorem 1.3 (Rauch comparison theorem, Second form). Suppose i and ii holds. Denote p = γ(0) and p~ =γ ~(0), and suppose Xp 2 TpM; Xep~ 2 Tp~Mf satisfies _ hXp; γ_ (0)i = hXep~; γ~(0)i; jXpj = jXep~j: Then j(d expp)tγ_ (0)Xpj ≤ j(d expp~)tγ~_ (0)Xepj 4 LECTURE 20: RAUCH COMPARISON THEOREM AND ITS APPLICATIONS Remark. If we replace the inequality K− ≥ Ke + in ii by the strict one K− > Ke +; then the inequality in lemma 1.2 is strict. As a consequence, the inequalities in theorems 1.1 and 1.3 are strict for t > 0. 2. Some Direct Applications In applications, we will mainly compare a given Riemannian manifold with with another \model" Riemannian manifold of constant curvature. Here we list several interesting applications, and leave the proofs as exercises. Corollary 2.1. Let (M; g) be a complete Riemannian manifold with non-positive sectional curvature. Then for any p 2 M, any Xp 2 TpM and Yp 2 TpM = TXp (TpM), j(d expp)Xp (Yp)j ≥ jYpj: In particular, for any curve γ in TpM, one has L(γ) ≤ L(expp ◦γ): Proof. Apply theorem 1.3 to (M; g) and (TpM; gp). Corollary 2.2. Let (M; g) be a complete simply connected Riemannian manifold with non-positive sectional curvature, and consider a geodesic triangle in M whose side lengths are a; b; c with opposite angles A; B; C respectively. Then (1) a2 + b2 − 2ab cos C ≤ c, (2) A + B + C ≤ π: Moreover, if M has negative sectional curvature, then the inequalities are strict. Proof. (1) Denote the vertex at the angle C by p. In the tangent space TpM, draw a triangle 4OP Q, where O is the origin of TpM, so that jOP j = a; jOQj = b and \O = C: Let η be the pre-image of the geodesic c in TpM. Then jPQj ≤ L(η) ≤ c; where the second inequality follows from corollary 2.1. Now apply the Euclidean cosine law to 4OP Q. (2) Construct a triangle in R2 whose side lengths are a; b; c. This is possible since a; b; c are lengths of sides of a geodesic triangle (meaning each side is a shortest geodesic connecting the corresponding vertices). Denote the corresponding angles of this new triangle by A0;B0;C0. Then by (1), we have C ≤ C0. Similarly A ≤ A0 and B ≤ B0. If K < 0, then according to the remark after theroem 1.3, the inequality in lemma 2.1 is also strict for Xp;Yp 6= 0. So the conclusion follows. LECTURE 20: RAUCH COMPARISON THEOREM AND ITS APPLICATIONS 5 Corollary 2.3. 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 π p ≤ D ≤ p : C2 C1 Proof. This follows from the explicit formula for Jacobi fields on spheres. Remark. One should compare this with the Sturm comparison theorem in ODE..
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