LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS We recall from last lecture that 1. The Hessian Comparison Theorem K t) = min{kπ γt) ) γt) Π γt) }, K + t) = max{k Π γt) ) γt) Π γt) }. and we have shown lemma 1.2 in lecture 2) that if X and X are normal Jacobi fields along normal geodesics γ : [, a] M and γ : [, a] M respectively, such that a X) =, X) =, b Xa) = Xa). and assume further that i γ has no conjugate points on [, a], ii K+ t) K t) holds for all t [, a]. then IX, X) I X, X). Using this we showed that the Jacobi fields with the same initial conditions are comparable. Today we would lie to compare the distance functions on different manifolds. We let M, g) be a complete Riemannian manifold. We fix p M and consider the distance function d p : M R, d p q) = distp, q). Then we have c.f. lecture 16 and PSet 3) d p is a Lipschitz continuous function. d p is smooth on M \ Cutp) {p}. For each q M \ Cutp) {p}, the gradient vector of d p at q is d p )q) = γ q d p q)), where γ q is the unique normal minimizing geodesic from p to q. For any q M \Cutp) {p} and Y q T q M, the Hessian of d p is given by 2 d p Y q, Y q ) = d p q) γ q d pq))x, Xd p q)). where X is the Jacobi field along γ q such that X) =, Xd p q)) = Y q. 1
2 LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS Since d p has length 1 almost everywhere, to compare the distance functions one must compare their second order derivatives, i.e. compare the Hessians. We recall from lecture 5 that the Hessian 2 f of a smooth function f on M is 2 fx, Y ) = X f, Y. We have shown that it is a symmetric, 2) tensor, and We are ready to state f = Tr 2 f). Theorem 1.1 The Hessian Comparison Theorem). Let M, g), M, g) be complete Riemannian manifolds, γ : [, b] M and γ : [, b] M be minimizing normal geodesics in M and M respectively, so that K + t) K t) holds for all t [, b]. Denote q = γa) and q = γa) a < b). Suppose X q T q M and X q T q M satisfy Then X q, γa) = X q, γa), X q = X q. 2 d p X q, X q ) 2 d p X q, X q ). Proof. For any Y q T q M, we have 2 d p γa), Y q ) = γa) d p, Y q = γa) γ, Y q =. The same relation also wors for M. So WLOG, we can assume X q γa) and X q γa). Let X, X be normal Jacobi fields along γ, γ respectively, so that X) =, Xa) = X q and X) =, Xa) = X q. Since Xa) = Xa), we get use the lemma 1.2 in lecture 2) IX, X) I X, X). Now the theorem follows from 2 d p X q, X q ) = a γa) X, Xa) = aix, X). Since = Tr 2, by taing trace we immediately get Corollary 1.2. Under the same assumptions as in theorem 1.1, we have d p q) d p q).
LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS 3 2. The Laplace comparison theorem In many applications, one would lie to weaen the conditions on the sectional curvatures in corollary 1.2 to conditions on the Ricci curvatures. This is not always true. However, we will prove below that this is possible if we assume that M has constant sectional curvature. Let s fix some notions. For any R, we will denote by M m the m-dimensional space form of constant curvature. In other words, M m = Sm ) or R m or H m ), depending on the sign of. Theorem 2.1 The Laplace Comparison Theorem). Let M, g) be a Riemannian manifold of dimension m whose Ricci curvature is bounded below, RicM, g) m 1). Let γ : [, b] M, γ : [, b] M m be minimizing normal geodesics in M and M m starting from γ) = p and γ) = p respectively. Then d p γt)) d p γt)). Remar. One can replace M m by a Riemannian manifold M which has constant sectional curvature along the geodesic γ, and assume that for any t [, b], Ric γt), γt)) m 1) = Ric γt), γt)). The proof is identically the same as below. Proof. Fix t. Let {e 1,, e m } be an orthonormal frame that is parallel along γ, such that e 1 = γ. For any i 2 let X i τ) be the normal Jacobi field along γ [,t] with X i ) = and X i t) = e i t). Then d p γt)) = 2 d p e i t), e i t)) = t IX i, X i ) Similarly d p γt)) = t I X i, X i ), where X i is the normal Jacobi field along γ [,t] such that X i ) =, X i t) = ẽ i t). It remains to prove IX i, X i ) I X i, X i ). We shall apply the same tric that we played in the proof of lemma 1.2 in lecture 2. In other words, we write for each τ t, X i τ) = a j i τ)e jτ).
4 LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS Then by assumptions X i ) = and X i t) = e i t), we see a j i ) = and aj i t) = δj i. On the other hand, since M m has constant sectional curvature, our computations in lecture 12 together with the conditions X i ) =, X i t) = ẽ i t) implies X i τ) = δ j i sn τ) sn t) ẽjτ) = sn τ) sn t) ẽiτ), where sn τ) = sin τ) or τ or sinh τ). Now for each 2 i m we define on γ [,t] a vector field X iτ) = sn τ) sn t) e iτ) Then X i has the same boundary condition as X i. Since X i s are Jacobi fields, we have IX i, X i ) IX i, X i). Finally we the conclusion follows from IX i, X i) = t γ X i 2 Rm γ, X i, γ, X i) ) dτ = t sn τ) i sn t) )2 sn ) τ) sn t) )2 K γ, e i ) dτ t = m 1) sn τ) sn t) )2 sn ) τ) sn t) )2 Ric γ) dt t m 1) sn τ) sn t) )2 sn ) τ) sn t) )2 m 1) dt t = m 1) sn τ) sn t) )2 sn ) τ) sn t) )2 Ric γ) dt = t X γ i 2 Rm γ, X i, γ, X ) i ) dτ = I X i, X i ). 3. The Toporogov Comparison Theorem Let M, g) be complete. Recall that a geodesic triangle ABC consists of three points A, B, C in M which are called the vertices) and three minimizing geodesics which are called the sides) γ AB, γ BC, γ CA joining each two of them. If only two
LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS 5 sides, say γ AB and γ AC, are minimal, while the third side is a geodesic, which may be not minimal, but still satisfies the triangle inequality Lγ BC ) Lγ AB ) + Lγ AC ), then we will call ABC a generalized geodesic triangle. Similarly we can define a geodesic hinge BAC, which consists of a point A in M which is again called the vertex) and two minimal geodesics γ AB, γ AC called the sides) emanating from A, with end points B and C in M. If one side, say γ AB, is minimal, while the other side γ AC is not, we will 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. Lemma 3.1. Let M, g) be a complete Riemannian manifold of dimension m whose sectional curvature K. Then 1) For each generalized geodesic hinge BAC in M, there is a hinge Bà C in M m with same angle and the corresponding sides are of same length as BAC. 2) For each generalized geodesic triangle ABC in M, there is a triangle à B C in M m whose corresponding sides have the same length as ABC. Proof. For = and <, there is a unique geodesic between any two points, and the lemma is clear. So in what follows we assume >. According to the theorem of Bonnet-Myers, one has diamm π. More over, if diamm = π, then according to the maximal diameter theorem of Cheng that we will later, M, g) is isometric to the standard sphere of radius 1 and thus the conclusion holds. So WLOG, we may assume that any minimizing geodesic has length less than π. In the hinge case, we can fix any à M m, and choose any B so that distã, B) = Lγ AB ) < π, and join them by the unique minimizing geodesic. Then we choose a direction at à so that the angle with the geodesic γã B is A. Use this direction one can generate a normal geodeisc emanating from Ã. Tae the point C to be the point on this geodesic with parameter Lγ AC ). Note that γã C is minimizing if γ AC is minimizing.) In the triangle case, we pic B and C as above, and then consider the two distance spheres B BdistA, B)) and B CdistA, C)). One can easily chec that these two spheres are non-empty and must intersect. We then tae any point à from the intersection and connect it to B and C by minimizing geodesics.
6 LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS One can thin of the following Toporogov comparison theorem as a global version of Hessian comparison theorem, where we can actually compare the distance function instead of only comparing their Hessian. Theorem 3.2 Toporogov Comparison Theorem). Let M, g) be a complete Riemannian manifold with sectional curvature K. Then 1) Hinge Version) Let BAC be a hinge in M and Bà C a comparing hinge in M m. Then distb, C) dist B, C). 2) Triangle Version) Let ABC be a triangle in M and à B C a comparing triangle in M m. Then the three angles in ABC are greater than the corresponding angles in à B C. The proof will be left as a possible final topic. Remars. 1) There is no analogous theorem for the case K. 2) One can prove that the hinge version and the triangle version of Toporogov comparison theorem are equivalent. 3) One can replace M m by M 2. 4) In the case >, one can show that the perimeter of the generalized geodesic triangle ABC is no more than 2π. As an immediate application, we will prove Theorem 3.3 Gromov). There is a constant C = Cm) so that for any complete Riemannian manifold M, g) with K, the fundamental group π 1 M) is generated by no more than Cm) generators. Proof. We will consider π 1 M) as the group of Dec transformations on the universal covering M. Fix p M and choose inductively a generating set of π 1 M) as follows: We first choose e g 1 π 1 M) so that for all g π 1 M) \ {e}, d p, g 1 p) d p, g p). Suppose g 1,, g 1 are chosen. We then choose g g 1,, g 1 so that d p, g p) d p, g p) holds for all g π 1 M) \ g 1,, g 1. Now let γ be a minimal geodesic in M, g) from p to g p. We claim that the angle between any two such minimal geodesics is at least π. Then the conclusion 3 follows. We prove the claim by contradiction. Suppose the angle between γ and γ +l is less than π. For simplicity we denote l 3 = d p, g p). Then according to the Toporogov comparison theorem, dg +l p, g p) 2 < l 2 + l 2 +l l l +l l 2 +l.
LECTURE 21: THE HESSIAN, LAPLACE AND TOPOROGOV COMPARISON THEOREMS 7 This implies d p, g 1 +l g p) = dg +l p, g p) < l +l = d p, g +l p), which contradicts with the choice of g +l. Remar. By the same way one can prove the following theorem of Gromov: Theorem 3.4 Gromov). For negative, there is a constant C = Cm,, D) so that for any complete Riemannian manifold M, g) with K and diamm, g) D, the fundamental group π 1 M) is generated by no more than Cm,, D) generators. Note that a bound on diameter is needed. To see this, one can loo at the example of surface of genus g. One of the most beautiful theorem in Riemannian geometry is Theorem 3.5 Gromov). There is a constant C = Cm) so that for any complete Riemannian manifold M, g) with K, the total betti number bm) = b i M) Cm). [Similarly one can state a version with K, but with the additional condition diamm, g) D. ] Gromov conjectured Cm) = 2 m. [This is the total Betti number for T m.]