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 with γ(0) = p, γ˜(0) =p. ˜ For each t ∈ [0, a], we denote − K (t) = min{K(Πγ(t)) | γ˙ (t) ∈ Πγ(t)}, + ˙ Ke (t) = max{Ke(Πe γ˜(t)) | γ˜(t) ∈ Πe γ˜(t)}.

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 |∇γ˙ (0)X| = |∇e γ˜˙ (0)Xe|, 3 hγ˙ (0), ∇γ˙ (0)Xi = hγ˜(0), ∇e γ˜˙ (0)Xei. Assume further that i γ has no conjugate points on [0, a], ii 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)|. 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 |X(a)| = |Xe(a)|. Assume further that i and ii above holds. Then I(X,X) ≤ I(X,e Xe).

Proof. Let {e1(t), ··· , em(t)} and {e˜1(t), ··· , e˜m(t)} 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 = |X(a)| = |Xe(a)|= 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 ∈ [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 ) = |∇γ˙ Y | − 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 ˙ ˙ ≤ |∇e γ˜˙ Xe| − 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) = |X(t)|2, u˜(t) = |Xe(t)|2. Thenu ˜(t)/u(t) is well-defined, and by L0Hospital’s rule, u˜(t) u˜¨(t) |∇ X|2 lim = lim = γ˙ (0) = 1. t→0 u(t) t→0 u¨(t) 2 |∇γ˜˙ (0)Xe|

d u˜(t) Therefore, to prove |X| ≤ |Xe|, 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 ∈ (0, a]. Let c ≤ a be the greatest number so thatu ˜(t) > 0 on (0, c). For any b ∈ (0, c), we define X(t) Xe(t) Xb(t) = , Xeb(t) = . |X(b)| |Xe(b)|

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 ) = − h∇γ˙ ∇γ˙ X + R(γ, ˙ X)γ, ˙ Y idt + h∇γ˙ X,Y i|0. 0 So

I(Xb,Xb) = h∇γ˙ (b)Xb,Xb(b)i and I(Xeb, Xeb) = h∇e γ˜˙ (b)Xeb, Xeb(b)i. It follows

h∇γ˙ (b)Xb,Xb(b)i ≤ h∇e γ˜˙ (b)Xeb, Xeb(b)i, i.e. 1 u˙(b) h∇ X,X(b)i h∇e ˙ 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 ∈ (0, c), we have u(t) ≤ u˜(t) . This is exactly what we need. To summary, we proved that |X(t)| ≤ |Xe(t)| for t ∈ (0, c). If c < a, then |Xe(c)| ≥ |X(c)| > 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 variation. In particular, if X is a Jacobi field along γ : [0, a] → M with

X(0) = 0, ∇γ˙ (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 ∈ TpM, Xep˜ ∈ Tp˜Mf satisfies ˙ hXp, γ˙ (0)i = hXep˜, γ˜(0)i, |Xp| = |Xep˜|.

Then |(d expp)tγ˙ (0)Xp| ≤ |(d expp˜)tγ˜˙ (0)Xep| 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 with with another “model” Riemannian manifold of constant . 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 . 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).

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 4OPQ, where O is the origin of TpM, so that |OP | = a, |OQ| = b and ∠O = C.

Let η be the pre-image of the geodesic c in TpM. Then |PQ| ≤ L(η) ≤ c, where the second inequality follows from corollary 2.1. Now apply the Euclidean cosine law to 4OPQ. (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 √ ≤ D ≤ √ . 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.