Math 241A. Instructor: Guofang Wei. Fall 2000

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1 Math 241A Instructor: Guofang Wei Fall 2 The basic idea of this course is that curvature bounds give information about manifolds, which in turn gives topological results. A typical example is the Bonnet-Myers Theorem. Intuitively, Bigger curvature Smaller manifold 1. 1 Volume Comparison Theorem 1.1 Volume of Riemannian Manifold Recall: For U R n, vol(u) = U 1 dv = U 1 dx 1 dx n. Note - dx 1 dx n is called the volume density element. Change of variable formula: Suppose ψ : V U is a diffeomorphism, with U, V R n. Suppose ψ(x) = y. Then dv = 1 dy 1 dy n = Jac(ψ) dx 1 dx n. U U On a Riemannian manifold M n, let ψ α : U α R n be a chart. Set E ip = (ψα 1 ) ( x i ). In general, the E ip s are not orthonormal. Let {e k } n be an orthonormal basis of T p M. Then E ip = a ik e k. The volume of V k=1 1 This quarter we use that bigger curvature smaller volume 1

2 the parallelepiped spanned by {E ip } is det(a ik ). Now g ij = det(g ij ) = det(a ij ) 2. Thus vol(u α ) = ψ(u α ) det(g ij ) (ψ 1 α ) dx 1 dx n n a ik a kj, so Note - dv = det(g ij ) (ψα 1 ) dx 1 dx n is called a volume density element, or volume form, on M. We have our first result, whose proof is left as an exercise. Lemma Volume is well defined. Definition Let M be a Riemannian manifold, and let {U α } be a covering of M by domains of coordinate charts. Let {f α } be a partition of unity subordinate to {U α }. The volume of M is vol(m) = 1 dv = f α dv. M α ψ(u α ) Lemma The volume of a Riemannian manifold is well defined. 1.2 Computing the volume of a Riemannian manifold Partitions of unity are not practically effective. Instead we look for charts that cover all but a measure zero set. Example For S 2, use stereographic projection. In general, we use the exponential map. We may choose normal coordinates or geodesic polar coordinates. Let p M n. Then exp p : T p M M is a local diffeomorphism. Let D p T p M be the segment disk. Then if C p is the cut locus of p, exp p : D p M C p is a diffeomorphism. Lemma C p has measure zero. Hence we may use exp p to compute the volume element dv = det(g ij ) dx 1 dx n. Now polar coordinates are not defined at p, but {p} has measure zero. We have 2 k=1

3 exp p : D p {} diffeo M C p {p}. Set E i = (exp p ) ( theta i ) and E n = (exp p ) ( ). To compute g r ij s, we want E i and E n explicitly. Since exp p is a radial isometry, g nn = 1 and g ni = for 1 i < n. Let J i (r, θ) be the Jacobi field with J i () = and J i () = θ i. Then E i (exp p (r, θ) = J i (r, θ). If we write J i and in terms of an orthonormal basis {e r k}, we have n J i = a ik e k. Thus k=1 det(g ij )(r, θ) = det(a ik ) = J 1 J n 1 r The volume density, or volume element, of M is dv = J 1 J n 1 r drdθ n 1 = A(r, θ) drdθ n 1 Example R n has Jacobi equation J = R(T, J)T. If J() = and J () = θ i then J(r) = r θ i. Thus the volume element is dv = r n 1 drdθ n 1. Example S n has J i (r) = sin(r) θ i. Hence dv = sin n 1 (r) drdθ n 1. Example H n has J i (r) = sinh(r) θ i. Hence dv = sinh n 1 (r) drdθ n 1. Example Volume of unit disk in R n ω n = S n 1 1 r n 1 drdθ n 1 = 1 n S n 1 dθ n 1 Note - S n 1 dθ n 1 = 2(π)n/2 Γ(n/2). 3

4 1.3 Comparison of Volume Elements Theorem Suppose M n has Ric M (n 1)H. Let dv = A(r, θ) drdθ n 1 be the volume element of M and let dv H = A H (r, θ) drdθ n 1 be the volume element of the model space (simply connected n-manifold with K H). Then is a nonincreasing function in r. A(r, θ) A H (r, θ) Proof 2. We show that Since r A(r, θ) ( A H (r, θ) )2. A(r, θ) 2 = J 1 J n 1 r, J 1 J n 1 r, we wish to show that J 1 J n 1 r, J 1 J n 1 r A H (r, θ) 2 A(r, θ) 2 J H 1 J H n 1 r, J H 1 J H n 1 r. Thus we wish to show that n 1 2 i=1 J 1 J i J n 1 r, J 1 J n 1 r J 1 J n 1 r, J 1 J n 1 r n 1 2 i=1 J H 1 (J H i ) J H n 1 r, J H 1 J H n 1 r J H 1 J H n 1 r, J H 1 J H n 1 r (1) At r = r, let J i (r ) be orthonormal such that J n (r ) = r r=r. Then for 1 i < n, n 1 J i (r ) = b ik J k (r ). k=1 2 Compare to the proof of the Rauch Comparison Theorem. 4

5 Define J n 1 i (r) = b ik J k (r), where the b ik s are fixed. Then each J i is a k=1 linear combination of Jacobi fields, and hence is a Jacobi field. The left hand side of (1), evaluated at r = r is 2 n 1 i=1 J 1 J i J n 1 r, J 1 J n 1 r J 1 J n 1 r, J 1 J n 1 r n 1 = 2 J i (r ), J n 1 i (r ) = 2 I( J i, J i ), i=1 i=1 r=r where I is the index form I(v, v) = r v, v + R(T, v)t, v dt. Note that for a Jacobi field J, I(J, J) = = r r = v, v r=r. J, J + R(T, J)T, J dt J, J J, J + R(T, J)T, J dt Let E i be a parallel field such that E i (r ) = J i (r ), and let w i = sin Hr sin E Hr i. By the Index Lemma, Jacobi fields minimize the index form provided there n 1 are no conjugate points. Thus we have 2 I( J i, J n 1 i ) 2 I(w i, w i ). By the n 1 curvature condition, 2 i=1 side of (1), evaluated at r = r. A(r, θ) Thus is nonincreasing in r. A H (r, θ) Remarks: i=1 n 1 i=1 I(w i, w i ) 2 I( J i H, J i H ), which is the right hand i=1 1. lim r A(r, θ) A H (r, θ) = 1, so A(r, θ) A H(r, θ). 2. (Rigidity) If A(r, θ) = A H (r, θ) for some r, then A(r, θ) = A H (r, θ) for all r r. But then B(p, r ) is isometric to B(r ) SH n, where SH n is the model space. But then the Jacobi fields in M correspond to the Jacobi fields in the model space, so that M is isometric to the model space. 5

6 3. We cannot use the Index Lemma to prove an analogous result for Ric M (n 1)H. In fact, there is no such result. For example, consider Einstein manifolds with Ric (n 1)H. 4. If K M H, we may use the Rauch Comparison Theorem to prove a similar result inside the injectivity radius. 5. (Lohkamp) Ric M (n 1)H has no topological implications. Any smooth manifold M n, with n 3, has a complete Riemannian metric with Ric M. 6. Ric M (n 1)H may still have geometric implications. For example, if M is compact with Ric M < then M has a finite isometry group. 1.4 Volume Comparison Theorem Theorem (Bishop-Gromov) If M n has Ric M (n 1)H then is nonincreasing in R. vol(b(p, R)) vol(b H (R)) Proof. We have volb(p, R) = = 1dv B(p,R) R S p (r) A(r, θ)dθ n 1 dr, where S p (r) = {θ S p : rθ D p }. Note that S p (r 1 ) S p (r 2 ) if r 1 r 2. The theorem now follows from two lemmas: Lemma If f(r)/g(r) is nonincreasing in r, with g(r) >, then is nonincreasing in R. R f(r)dr R g(r)dr 6

7 Proof of Lemma.- The numerator of the derivative is R R R R ( g(r)dr)( f(r)dr) ( f(r)dr)( g(r)dr) so Now Thus = f(r)( = g(r)( f(r) g(r) f(r) g(r) R so the derivative is nonpositive. R R g(r)f(r)dr f(r) g(r) g(r)dr) g(r)( [ f(r) g(r)dr) g(r) R g(r)f(r) f(r)g(r), R R f(r)dr R g(r)dr, f(r)g(r)dr. f(r)dr) R f(r)dr ] R g(r)dr Lemma (Comparison of Lower Area of Geodesic Sphere) Suppose r lies inside the injectivity radius of the model space SH n, so that if H >, r < π/ H. Then A(r, θ) dθ S p(r) n 1 A S H (r) dθ n 1 n 1 is nonincreasing in r. Proof of Lemma. In the model space, A H (r, θ) does not depend on θ, so we write A H (r). Note that if r R, S p(r) A(R, θ) dθ n 1 S n 1 A H (R) dθ n 1 = = S n dθ n 1 S n 1 dθ n 1 S p(r) S p(r) S p(r) A(r, θ)dθ n 1 S n 1 A H (r) dθ n 1, A(R, θ) A H (R) dθ n 1 A(r, θ) A H (r) dθ n 1 7

8 since S p (r) S p (R) and A(r,θ) is nonincreasing in r. The theorem now follows. A H (r) Note that if R is greater than the injectivity radius then volb(p, R) decreases. Thus the volume comparison theorem holds for all R. Corollaries: 1. (Bishop Absolute Volume Comparison) Under the same assumptions, volb(p, r) volb H (r). 2. (Relative Volume Comparison) If r R then volb(p, r) volb(p, R) volbh (r) volb H (R). If equality holds for some r then equality holds for all r r, and B(p, r ) is isometric to B H (r ). Proofs: (1) holds because lim r volb(p, r) volb H (r) = 1. (2) is a restatement of the the volume comparison theorem. Sometimes we let R = 2r in (2). Then (2) gives a lower bound on the ratio volb(p, r), called the doubling constant. If vol(m) V then we obtain a volb(p, R) lower bound on the volume of small balls. Generalizations: 1. The same proof shows that the result holds for vol Γ B(p, R), where Γ S p = S n 1 T p M. In particular, the result holds for annuli ( R r ) and for cones. 2. Integral Curvature 3. Stronger curvature conditions give submanifold results. 8

9 2 Applications of Volume Comparison 2.1 Cheng s Maximal Diameter Rigidity Theorem Theorem (Cheng) Suppose M n has Ric M (n 1)H >. By the Bonnet-Myers Theorem, diam M π/ H. If diam M = π/ H, Cheng s result states that M is isometric to the sphere S n H with radius 1/ H. Proof. (Shiohama) Let p, q M have d(p, q) = π/ H. Then vol B(p, π/(2 H)) vol M = vol B(p, π/(2 H)) vol B(p, π/ H) vol B H(π(/2 H)) vol B H (π/ H) = 1/2 Thus vol B(p, π/2 H) (vol M)/2. Similarly for q. Hence vol B(p, π/(2 H)) = (vol M)/2, so we have equality in the volume comparison. By rigidity, B(p, π/(2 H)) is isometric to the upper hemisphere of SH n. Similarly for B(q, π/2 H), so vol M = vol, SH n. Question: What about perturbation? Suppose Ric M (n 1)H and diam M π/ H ε. In general there is no result for ε >. There are spaces not homeomorphic to S n, provided n 4, with Ric (n 1)H and diam π/ H ε. Still, if Ric (n 1)H and vol M vol SH n ε(n, H) then M n diffeo SH n. 2.2 Growth of Fundamental Group Suppose Γ is a finitely generated group, say Γ = g 1,..., g k. Any g Γ can be written as a word g = g n i k i, where k i {1,..., k}. Define the length of i this word to be i n i, and let g be the mininimum of the lengths of all word representations of g. Note that depends on the choice of generators. Fix a set of generators for Γ. The growth function of Γ is Γ(s) = #{g Γ : g s}. Example If Γ is a finite group then Γ(s) Γ. 9

10 Example Γ = Z Z. Then Γ = g 1, g 2, where g 1 = (1, ) and g 2 = (, 1). Any g Γ can be written as g = s 1 g 1 + s 2 g 2. To find Γ(s), we want s 1 + s 2 s. Γ(s) = 2s = 2s s 2(2(s t) + 1) t=1 s (4s 4t + 2) t=1 = 2s s 2 + 2s 4 2 t t=1 = 4s 2 + 4s + 1 4(s(s + 1)/2) = 4s 2 + 4s + 1 2(s 2 + s) = 2s 2 + 2s + 1 In this case we say Γ has polynomial growth. Example Γ free abelian on k generators. Then Γ(s) = Γ has polynomial growth of degree k. k i= ( k i )( s i ). Definition Γ is said to have polynomial growth of degree n if for each set of generators the growth function Γ(s) as n for some a >. Γ is said to have exponential growth if for each set of generators the growth function Γ(s) e as for some a >. Lemma If for some set of generators, Γ(s) as n for some a >, then Γ has polynomial growth of degree n. If for some set of generators, Γ(s) e as for some a >, then Γ has exponential growth. Example Z k has polynomial growth of degree k. Example Z Z has exponential growth. Note that for each group Γ there always exists a > so that Γ(s) e as. Definition A group is called almost nilpotent if it has a nilpotent subgroup of finite index. 1

11 Theorem (Gromov) A finitely generated group Γ has polynomial growth iff Γ is almost nilpotent. Theorem (Milnor) If M n is complete with Ric M, then any finitely generated subgroup of π 1 (M) has polynomial growth of degree n. Proof. Let M have the induced metric. Then Ric M, and π 1 (M) acts isometrically on M. Suppose Γ = g 1,..., g k be a finitely generated subgroup of π 1 (M). Pick p M. Let l = max d(g i i p, p). Then if g π 1 (M) has g s, d(g p, p) sl. On the other hand, for any cover there exists ε > such that B(g p, ε) are pairwise disjoint for all g π 1 (M). Note that gb( p, ε) = B(g p, ε). Now B(g p, ε) B( p, sl + ε); g s since the B(g p, ε) s are disjoint and have the same volume, Thus Γ(s) volb( p, ε) volb( p, sl + ε). Γ(s) volb( p, sl + ε) volb( p, ε) volb Rn(, sl + ε). volb R n(, ε) Now vol(b R n(, s) = s S n 1 r n 1 drdθ n 1 = 1 n sn S n 1 dθ n 1 = s n ω n, sl + εn so Γ(s). ε n Since l and ε are fixed, we may choose a so that Γ(s) as n. 11

12 Example Let H be the Heisenberg group 1 x z 1 y : x, y, x R, 1 and let H Z = 1 n 1 n 2 1 n 3 1 : n i Z. Then H/H Z is a compact 3-manifold with π 1 (H/H Z ) = H Z. The growth of H Z is polynomial of degree 4, so H/H Z has no metric with Ric. Remarks: 1. If Ric M 1/k 2 >, then M is compact. Thus π 1 (M) is finitely generated. It is unknown whether π 1 (M) is finitely generated if M is noncompact. 2. Ricci curvature gives control on π(m), while sectional curvature gives control on the higher homology groups. For example, if K then the Betti numbers of M are bounded by dimension. 3. If M is compact then growth of π 1 (M) volume growth of M. Related Results: 1. (Gromov) If Ric M then any finitely generated subgroup of π 1 (M) is almost nilpotent. 2. (Cheeger-Gromoll, 1972) If M is compact with Ric then π 1 (M) is abelian up to finite index. 3. (Wei, 1988; Wilking 1999) Any finitely generated almost nilpotent group can be realized as π 1 (M) for some M with Ric. 4. Milnor s Conjecture (Open) If M n has Ric M then π 1 (M) is finitely generated. 12

13 In 1999, Wilking used algebraic methods to show that π 1 (M) is finitely generated iff any abelian subgroup of π 1 (M) is finitely generated (provided Ric M ). Sormani showed in 1998 that if M n has small linear diameter growth, i.e. if lim sup r diam B(p, r) r then π 1 (M) is finitely generated. < s n = n (n 1)3 n Basic Properties of Covering Space Suppose M M has the covering metric. ( ) n 1 n 1, n 2 1. M compact there is a compact set K M such that {γk} γ π1 (M) covers M. K is the closure of a fundamental domain. 2. {γk} γ π1 (M) is locally finite. Definition Suppose δ >. Set S = {γ : d(k, γk) δ}. Note that S is finite. Lemma If δ > D = diam M then S generates π 1 (M). In fact, for any a K, if d(a, γk) (δ D)s + D, then γ s. Proof. There exists y γk such that d(a, y) = d(a, γk). Connect a and y a minimal geodesic σ. Divide σ by y 1,..., y s+1 = y, where d(y i, y i+1 ) δ D and d(a, y i ) < D. Now {γk} γ π1 (M) covers M, so there exist γ i π 1 (M) and x i K such that γ i (x i ) = y i. Choose γ s+1 = γ and γ 1 = Id. Then γ = γ1 1 γ 2 γs 1 γ s+1. But γ 1 i γ i+1 S, since Thus γ s. d(x i, γ 1 i γ i+1 x i ) = d(γ i x i, γ i+1 x i ) = d(y i, γ i+1 x i ) d(y i, y i+1 ) + d(y i+1, γ i+1 x i ) = d(y i, y i+1 ) + d(x i+1, x i ) δ. 13

14 Theorem (Milnor 1968) Suppose M is compact with K M <. Then π 1 (M) has exponential growth. Note that K M H < since M is compact. The volume comparison holds for K H, but only for balls inside the injectivity radius. Since K M <, though, the injectivity radius is infinite. Proof of Theorem. By the lemma, γk B(a, (δ D)s + D), γ s so Γ(s)vol(K) volb(a, (δ d)s + D). Note that since K M H <. Now for r large. Thus volb(a, (δ D)s + D) volb H (a, (δ D)s + D), volb H (r) = Γ(s) S n 1 r = nω n r ( sinh Hr H ) n 1 drdθ n 1 ( sinh Hr H ) n 1 drdθ n 1 nω n 2(2 H) n 1 (n 1) H e Hr, volb(a, (δ D)s + D) vol(k) C(n, H)e (δ D) Hs, where C(n, H) is constant. Corollary The torus does not admit a metric with negative sectional curvature. 2.3 First Betti Number Estimate Suppose M is a manifold. The first Betti number of M is b 1 (M) = dim H 1 (M, R). 14

15 Now H 1 (M, Z) = π 1 (M)/[π 1 (M), π 1 (M)], which is the fundamental group of M made abelian. Let T be the group of torsion elements in H 1 (M, Z). Then T H 1 (M, Z) and Γ = H 1 (M, Z)/T is a free abelian group. Moreover, b 1 (M) = rank(γ) = rank(γ ), where Γ is any subgroup of Γ with finite index. Theorem (Gromov, Gallot) Suppose M n is a compact manifold with Ric M (n 1)H and diam M D. There is a function C(n, HD 2 ) such that b 1 (M) C(n, HD 2 ) and lim x C(n, x) = n and C(n, x) = for x >. In particular, if HD 2 is small, b 1 (M) n. Proof. First note that if M is compact and Ric M > then π 1 (M) is finite. In this case b 1 (M) =. Also, by Milnor s result, if M is compact with Ric M then b 1 (M) n. As above, b 1 (M) = rank(γ), where Γ = π 1 (M)/[π 1 (M), π 1 (M)]/T. Set M = M/[π 1 (M), π 1 (M)]/T be the covering space of M corresponding to Γ. Then Γ acts isometrically as deck transformations on M. Lemma For fixed x M there is a subgroup Γ Γ, [Γ : Γ ] finite, such that Γ = γ 1,..., γ 2, where: 1. d(x, γ i (x)) 2diam M and 2. For any γ Γ {e}, d(x, γ(x)) > diam M. Proof of Lemma. For each ε let Γ ε Γ be generated by {γ Γ : d(x, γ(x)) 2diam M + ε}. Then Γ ε has finite index. For if M/Γε is a covering space corresponding to Γ/Γ ε. Then [Γ : Γ ε ] is the number of copies of M in M/Γ ε. We show that diam( M/Γ ε ) 2diam M + 2ε so that M/Γ ε. Suppose not, so there is z M such that d(x, z) = diam M +ε. Then there is γ Γ that d(γ(x), z) diam M. Then if π ε is the covering M M/Γ ε, d(π ε x, π ε γ(x)) d(π ε x, π ε z) d(π ε z, π ε γ(x)) = ε. diam M + ε diam M 15

16 Thus γ / Γ ε. But d(x, γ(x)) d(x, z) + d(z, γ(x)) 2diam M + ε Thus M/Γ ε is compact, so Γ ε has finite index. Moreover, {γ Γ : d(x, γ(x)) 3diam M } is finite, Γ ε is finitely generated. Also, note that for ε small, {γ Γ : d(x, γ(x)) 2diam M } = {γ Γ : d(x, γ(x)) 2diam M + ε}. Pick such an ε >. Since Γ ε Γ has finite index, b 1 (M) = rank(γ ε ). Now Γ ε is finitely generated, say Γ ε = γ 1,..., γ m ; pick linearly independent generators γ 1,..., γ b1 so that Γ = γ 1,..., γ b1 has finite index in Γ ε. Let Γ = γ 1,..., γ b1, where γ k = l k1 γ l kk γ k and the coefficients l ki are chosen so that l kk is maximal with respect to the constraints: 1. γ k Γ {γ Γ : d(x, γ(x)) 2diam M } and 2. span{ γ 1,..., γ k } span{γ 1,..., γ k } with finite index. Then Γ Γ has finite index, and d(x, γ i (x)) 2diam M for each i. Finally, suppose there exists γ Γ {e} with d(x, γ(x)) diam M, write γ = m 1 γ γ k, with m k. Then d(x, γ 2 (x)) 2d(x, γ(x)) 2diam M, but γ 2 = 2m 1 γ m k γ k = (terms involving γ i, i < k) + 2m k l kk γ k, which contradicts the choice of the coefficients l ki. Proof of Theorem. Let Γ = γ 1,..., γ b1 be as in the lemma. Then d(γ i (x), γ j (x)) = d(x, γ 1 i γ j (x)) > D = diam M, where i j. Thus B(γ i (x), D/2) B(γ j (x), D/2) = for i j. Also B(γ i (x), D/2) B(x, 2D + D/2) 16

17 for all i, so that b 1 i=1 B(γ i (x), D/2) B(x, 2D + D/2). Hence volb(x, 2D + D/2) b 1 volbh (2D + D/2). volb(x, D/2) volb H (D/2) Since the result holds for H, assume H <. Then volb H (2D + D/2) volb H (D/2) = = = 5D/2 ( S sinh n 1 Ht H D/2 ( S sinh n 1 Ht H ) n 1 dtdθ ) n 1 dtdθ 5D/2 (sinh Ht) n 1 dt D/2 (sinh Ht) n 1 dt 5D H/2 (sinh r) n 1 dr D H/2 (sinh r) n 1 dr Let U(s) = {γ Γ : γ s}. Then B(γx, D/2) B(x, 2Ds + D/2), whence γ U(s) #U(s) = volb(x, 2Ds + D/2) volb(x, D/2) volbh (2Ds + D/2) volb H (D/2) (2s+ 1 2 )D H (sinh r) n 1 dr D H/2 (sinh r) n 1 dr 2(2s )n (D H) n ( 1 2 )n (D H) n = 2 n+1 (2s )n Thus b 1 (M) = rank(γ ) n, so that for HD 2 small, b 1 (M) n. Conjecture: For M n with Ric M (n 1)H and diam M D, the number of generators of π 1 (M) is uniformly bounded by C(n, H, D). 17

18 2.4 Finiteness of Fundamental Groups Lemma (Gromov, 198) For any compact M n and each x M there are generators γ 1,..., γ k of π 1 (M) such that d( x, γ i x) 2diam M and all relations of π 1 (M) are of the form γ i γ j = γ l. Proof. Let < ε < injectivty radius. Triangulate M so that the length of each adjacent edge is less than ε. Let x 1,..., x k be the vertices of the triangulation, and let e ij be minimal geodesics connecting x i and x j. Connect x to each x i by a minimal geodesic σ i, and set σ ij = σ 1 j e ij σ i. Then l(σ ij ) < 2diam M + ε, so d( x, σ ij x) < 2diam M + ε. We claim that {σ ij } generates π 1 (M). For any loop at x is homotopic to a 1-skeleton, while σ jk σ ij = σ ik as adjacent vertices span a 2-simplex. In addition, if 1 = σ π 1 (M), σ is trivial in some 2-simplex. Thus σ = 1 can be expressed as a product of the above relations. Theorem (Anderson, 199)) In the class of manifolds M with Ric M (n 1)H, vol M V and diam M D there are only finitely many isomorphism types of π 1 (M). Remark: The volume condition is necessary. For example, S 3 /Z n has K 1 and diam = π/2, but π 1 (S 3 /Z n ) = Z n. In this case, vol(s 3 /Z n ) as n. Proof of Theorem. Choose generators for π 1 (M) as in the lemma; it is sufficient to bound the number of generators. Let F be a fundamental domain in M that contains x. Then Also, vol(f ) = vol(m), so k k γ i (F ) B( x, 3D). i=1 vol B( x, 3D) volm volbh (3D). V This is a uniform bound depending on H, D and V. Theorem (Anderson, 199) For the class of manifolds M with Ric M (n 1)H, vol M V and diam M D there are L = L(n, H, V, D) and N = N(n, H, V, D) such that if Γ π 1 (M) is generated by {γ i } with each l(γ i ) L then the order of Γ is at most N. 18

19 Proof. Let Γ = γ 1,..., γ k π 1 (M), where each l(γ i ) L. Set U(s) = {γ Γ : γ s}, and let F M be a fundamental domain of M. Then γ i (F ) γ j (F ) has measure zero for i j. Now γ(f ) B( x, sl + D), so γ U(s) #U(s) volbh (sl + D). V Note that if U(s) = U(s + 1), then U(s) = Γ. Also, U(1) 1. Thus, if Γ has order greater than N, then U(N) N. Set L = D/N and s = N. Then Hence Γ N = volbh (2D) V N U(N) volbh (2D). V + 1, so Γ is finite. 3 Laplacian Comparison 3.1 What is the Laplacian? We restrict our attention to functions, so the Laplacian is a function : C (M) C (M) Invariant definition of the Laplacian Suppose f C (M). The gradient of f is defined by f, X = Xf. Note that the gradient depends on the metric. We may also define the Hessian of f to be the symmetric bilinear form Hess f : χ(m) χ(m) C (M) by Hess f(x, Y ) = 2 X,Y f = X(Y f) ( X Y )f = X f, Y. 19

20 The Laplacian of f is the trace of Hess f, f = tr(hess f). Note that if {e i } is an orthonormal basis, we have f = tr X f, Y n = ei f, e i i=1 = div f Laplacian in terms of geodesic polar coordinates Fix p M and use geodesic polar coordinates about p. For any x M C p, x p, connect p to x by a normalized minimal geodesic γ so γ() = p and γ(r) = x. Set N = γ (r), the outward pointing unit normal of the geodesic sphere. Let e 2,..., e n be an orthonormal basis tangent to the geodesic sphere, and extend N, e 2,..., e n to an orthonormal frame in a neighborhood of x. Then if e 1 = N, f = n ei f, e i = i=1 n (e i (e i f) ( ei e i )f). i=1 Note that ei e i = ei e i, N N + ( ei e i ) T = ei e i, N N + ( ei e i ), where is the induced connection on B(p, r). Thus f = N(Nf) ( N N)f + i=2 n (e i (e i f) ( ei e i )f) i=2 = 2 f n r + (e 2 i (e i f) ( n ei e i )f) ( ei e i, N N)f = f + m(r, θ) r f + 2 f r 2, where is the induced Laplacian on the sphere and m(r, θ) = n i=2 e i e i, N is the mean curvature of the geodesic sphere in the inner normal direction. 2 i=2

21 3.1.3 Laplacian in local coordinates Let ϕ : U M n R n be a chart, and let e i = (ϕ 1 ) ( x i ) be the corresponding coordinate frame on U. Then f = k,l 1 det gij k ( det g ij g kl l )f, where g ij = e i, e j and (g ij ) = (g ij ) 1. Notes: 1. f = 2 f + m(r, θ) + f. Let m r 2 r H (r) be the mean curvature in the inner normal direction of B H (x, r). Then m H (r) = (n 1) 1 if H = r H cot Hr if H >. H coth Hr if H < 2. We have m(r, θ) = A (r, θ) A(r, θ), where A(r, θ)drdθ is the volume element. 3. We also have In m(r, θ) = n ei e i, N. k= R n, g = dr 2 + r 2 dθ ( n 1 2 SH n, g = dr2 sin + ) 2 Hr H dθ 2 n 1 ( H n H, g = dr2 sinh + ) 2 Hr H dθ 2 n 1. By Koszul s formula, ei e i, N = e i, [e i, N]. In Euclidean space, N = r, 1 r e i are orthonormal. In S n H, N = r, 21 H sin Hr e i

22 are orthonormal, while are orthonormal in H n H. N = r, 3.2 Laplacian Comparison H sinh Hr e i On a Riemannian manifold M n, the most natural function to consider is the distance function r(x) = d(x, p) with p M fixed. Then r(x) is continuous, and is smooth on M ({p} C p ). We consider r where r is smooth. If x M ({p} C p ), connect p and x with a normalized, minimal geodesic γ. Then γ() = p, γ(r(x)) = x and r = γ (r). In polar coordinates, Thus r = m(r, θ). = 2 r 2 + m(r, θ) r +. Theorem (Laplacian Comparison, Mean Curvature Comparison) Suppose M n has Ric M (n 1)H. Let H be the Laplacian of S n H and m H(r) be the mean curvature of B H (r) M n H. Then: 1. r H r (Laplacian Comparison) 2. m(r, θ) m H (r) (Mean Curvature Comparison) Proof. We first derive an equation. Let N, e 2,..., e n be an orthonormal basis at p, and extend to an orthonormal frame N, e 2,..., e n by parallel translation along N. Then N e i =, so N ei N, e i = N ei N, e i. Also, ei N N =. Thus Ric(N, N) = = n R(e i, N)N, e i i=2 n ei N N N ei N [ei,n]n, e i i=2 = n N ei N, e i i=2 22 n [ei,n]n, e i. i=2

23 Now so In addition, n n ei N, e i = e i N, e i N, ei e i i=2 i=2 = n N, ei e i i=2 = m(r, θ), n Ric(N, N) = m (r, θ) [ei,n]n, e i. i=2 ei N = j ei N, e j e j + ei N, N N. But so 2 ei N, N = e i N, N =, ei N = j ei N, e j e j. Thus n [ei,n]n, e i = i=2 n i=2 n ei N, E j ej N, e i j=2 = Hess (r) 2, where A 2 = tr(aa t ). Hence Ric(N, N) = m (r, θ) Hess (r) 2. Now A 2 = λ λ 2 n, where the λ i s are the eigenvalues of A. Let λ 1,..., λ n be the eigenvalues of Hess r; since N N = we may assume λ 1 =. Then Hess (r) 2 = λ λ 2 n (λ λ n ) 2 /(n 1), since A, I 2 A 2 I 2 with A diagonal and A, B = tr(ab t ). But λ λ n = m(r, θ), so Hess (r) 2 m(r, θ)2 (n 1). 23

24 so Since Ric(N, N) (n 1)H, we have (n 1)H + m(r, θ) 2 /(n 1) m (r, θ). Set u = (n 1)/m(r, θ), so m(r, θ) = (n 1)/u. Then H + 1/u 2 (1/u 2 )u, Hu u, which is u Hu Thus r u r Hu = r. If H =, we have u r. In this case (n 1)/r m(r, θ), so m(r, θ) m H (r). If H >, (tan 1 ( Hu))/ h r. Now as r, m(r, θ) (n 1)/r. Thus u as r. Hence Hu tan( hr). Thus H(n 1)/m(r, θ) tan( Hr), so H(n 1) m(r, θ) tan( Hr) = m H(r). Note that inside the cut locus of M, the mean curvature is positive, so the inequality is unchanged when we may multiply by m(r, θ). If H <, similar arguments show that 3.3 Maximal Principle m(r, θ (n 1) H coth( Hr) = m H(r). We first define the Laplacian for continuous functions, and then relate the Laplacian to local extrema. Lemma Suppose f, h C 2 (M) and p M. Then if 24

25 then 1. f(p) = h(p) 2. f(x) h(x) for all x in some neighborhood of p 1. f(p) = h(p) 2. Hess (f)(p) Hess (h)(p) 3. f(p) h(p). Proof. Suppose v T p M. Pick γ : ( ε, ε) M so that γ() = p and γ () = v. Then (f h) γ : ( ε, ε) R, so the result follows from the real case. Definition Suppose f C (M). We say that f(p) a in the barrier sense if for any ε > there exists a function f ε, called a support function, such that 1. f ε C 2 (U) for some neighborhood U of p 2. f ε (p) = f(p) and f(x) f ε (x) for all x U 3. f ε (p) a ε. Note that f ε is also called support from below, or a lower barrier, for f at p. A similar definition holds for upper barrier. Theorem (Maximal Principle) If f C (M) and f then f is constant is a neighborhood of each local maximum. In particular, if f has a global maximum, then f is constant. Proof. If f > then f cannot have a local maximum. Suppose f, f has a local maximum at p, but f is not constant at p. We perturb f so that F >. Consider the geodesic sphere B(p, r). For r sufficiently small, there is z B(p, r) with f(z) < f(p). We define h in a neighborhood of p such that 1. h > 2. h < on V = {x : f(x) = p} B(p, r) 25

26 3. h(p) = To this end, set h = e αψ 1. Then We want ψ such that 1. ψ(p) = h = αe αψ ψ h = α 2 e αψ ψ, ψ + αe αψ ψ = αe αφ (α ψ 2 + ψ 2. ψ(x) < on some neighborhood containing V 3. ψ Choose coordinates so p and z (r,,..., ). Set ψ = x 1 β(x x 2 n), where β is chosen large enough that ψ < on some open set in S n 1 r z. Then ψ satisfies the above conditions. Since ψ 2 1 and ψ is continuous, we may choose α large enough that h >. Now consider f δ = f + δh on B(p, r). For δ small, f δ (p) = f(p) > max f δ (x). B(p,r) Thus, for δ small, f δ has a local maximum in the interior of B(p, r). Call this point q, and set N = h(q) >. Since f(q), there is a lower barrier function for f at q, say g, with g > δn/2. Then (g + δh)(q) = g + δ h > δn/2 and g + δh is a lower barrier function for f δ at q. Thus f δ (q) >, which is a contradiction. Theorem (Regularity) If f C (M) and f in the barrier sense, then f is C. If f, f is called harmonic. 26

27 3.4 Splitting Theorem Definition A normalized geodesic γ : [, ) M is called a ray if d(γ(), γ(t)) = t for all t. A normalized geodesic γ : (, ) is called a line if d(γ(t), γ(s)) = s t for all s t. Definition M is called connected at infinity if for all K M, K compact, there is a compact K K such that every two points in M K can be connected in M K. Lemma If M is noncompact then for each p M there is a ray γ with γ() = p. If M is disconnected at infinity then M has a line. Example A Paraboloid has rays but no lines. Example R 2 has lines. Example A cylinder has lines. Example A surface of revolution has lines. The theorem that we seek to prove is: Theorem (Splitting Theorem: Cheeger Gromoll 1971) Suppose that M n is noncompact, Ric M, and M contains a line. Then M is isometric to N R, with the product metric, where N is a smooth (n 1)- manifold with Ric N. Thus, if N contains a line we may apply the result to N. To prove this theorem, we introduce Busemann functions. Definition If γ : [, ) M is a ray, set b γ t (x) = t d(x, γ(t)). Lemma We have 1. b γ t (x) d(x, γ()). 2. For x fixed, b γ t (x) is nondecreasing in t. 3. b γ t (x) b γ t (y) d(x, y). 27

28 Proof. (1) and (3) are the triangle inequality. For (2), suppose s < t. Then b γ s(x) b γ t (x) = (s t) d(x, γ(s)) + d(x, γ(t)) = d(x, γ(t)) d(x, γ(s)) d(γ(s), γ(t)) Definition If γ : [, ) M is a ray, the Busemann function associated to γ is b γ (x) = lim b γ t (x) t = lim t d(x, γ(t)). t By the above, Busemann functions are well defined and Lipschitz continuous. Intuitively, b γ (x) is the distance from γ( ). Also, since b γ (x) is linear along γ(t). b γ (γ(s)) = lim t d(γ(s), γ(t)) t = lim t (t s) = s, t Example In R n, the rays are γ(t) = γ()+γ ()t. In this case, b γ (x) = x γ(), γ (). The level sets of b γ are hyperplanes. Lemma If M has Ric M and γ is a ray on M then (b γ ) in the barrier sense. Proof. For each p M, we construct a support function of b γ at p. We first construct asymptotic rays of γ at p. Pick t i. For each i, connect p and γ(t i ) by a minimal geodesic σ i. Then {σ i()} S n 1, so there is a subsequential limit γ (). The geodesic γ is called an asymptotic ray of γ at p; note that γ need not be unique. We claim that b γ (x)+b γ (p) is a support function of b γ at p. For b γ (p) =, so the functions agree at p. In addition, γ is a ray, so γ(t) is not a cut point of γ along p. Hence d( γ(t), ) is smooth at p, so b γ (x) is smooth in a neighborhood of p. 28

29 Now b γ (x) = lim t d(x, γ(t)) t lim t d(x, γ(s)) + d( γ(t), γ(s)) t = lim t + s d(x, γ(s)) s + d( γ(t), γ(s)) t = lim t + b γ t s(x) b γ s( γ(t)); letting s, we obtain We also have Thus b γ (x) lim t t + b γ (x) b γ ( γ(t)). b γ (p) = lim t i d(p, γ(t i )) i = lim t i d(p, σ i (t)) d(σ i (t), γ(t i )) i = d(p, γ(t)) + lim t i d(σ i (t), γ(t i )) i = d(p, γ(t)) + b γ ( γ(t)) = t + b γ ( γ(t)). b γ (x) + b γ (p) lim t + b γ (x) b γ ( γ(t)) t + b γ ( γ(t)) t = b γ (x), so b γ (x)+b γ (p) is a support function for b γ at p. By a similar argument, each b γ t (x) + b γ (p) is a support function for b γ at p. Finally, since Ric M, (b γ t (x) + b γ (p)) = (t d(x, γ(t))) = (x, γ(t)) n 1 d(x, γ(t)), which tends to as t. Thus (b γ ) in the barrier sense. The level sets of b γ t are geodesic spheres at γ(t). The level sets of b γ t are geodesic spheres at γ( ). 29

30 Lemma Suppose γ is a line in M, Ric M. Then γ defines two rays, γ + and γ. Let b + and b be the associated Busemann functions. Then: 1. b + + b on M. 2. b + and b are smooth. 3. Given any point p M there is a unique line passing through p that is perpendicular to v = {x : b + (x) = } and consists of asymptotic rays. Proof. For 1. Observe that b + (x) + b (x) = lim t (t d(x, γ + (t))) + lim t (t d(x, γ (t))) = lim t 2t (d(x, γ + (t)) d(x, γ (t))) 2t d(γ + (t), γ (t)) =. Since b + (γ()) + b (γ()) =, is a global maximum. But so b + + b. 2. We have b + = b. Thus (b + + b ) = b + + b, b + = b, so both b + and b are smooth by regularity. 3. At p there are asymptotic rays γ + and γ. We first show that γ + + γ is a line. Since d( γ + (s 1 ), γ (s 2 )) d( γ (s 2 ), γ + (t)) d( γ + (s 1 ), γ + (t)) holds for all t, we have = (t d( γ + (s 1 ), γ + (t))) (t d( γ (s 2 ), γ + (t))) d( γ + (s 1 ), γ (s 2 )) b + ( γ + (s 1 )) b + ( γ (s 2 )) = b + ( γ + (s 1 )) + b ( γ (s 2 )) b + ( γ + (s 1 )) + b + (p) + b ( γ (s 2 )) + b (p) = s 1 + s 2. 3

31 Thus γ + + γ is a line. But our argument shows that any two asymptotic rays form a line, so the line is unique. Set ṽ t = ( b + ) 1 (t ). Then if y ṽ t which shows γ ṽ t. we have d(y, γ + (t)) b + (y) b + ( γ + (t)) = t t Finally, since b + (x) + b + (p) b + (x), But b + = b and b + = b, so = d( γ + (t ), γ + (t)), ( b + (x) + b + (p)) b + (x). b (x) + b (p) b (x). Since b (x) + b (p) b (x) as well, b (x) + b (p) = b (x). Thus the level sets of b + are the level sets of b +, which proves the result. Note that b + : M R is smooth. Since b + is linear on γ with a Lipschitz constant 1, b + = 1. Thus v = (b + ) 1 () is a smooth (n-1) submanifold of M. Proof of Splitting Theorem. Let φ : R v M be given by (t, p) γ(t) = exp p tγ (), where γ is the unique line passing through p, perpendicular to v. By the existence and uniqueness of γ, φ is bijective. Since exp p is a local diffeomorphism and γ () = ( b + )(v) smooth, φ is a diffeomorphism. To show that φ is an isometry, set v t = (b + ) 1 (t) and let m(t) be the mean curvature of v t. Then m(t) = b + =. In the proof of the Laplacian comparison, we derived Ric(N, N) + Hess (r) 2 = m (r, θ), where N = γ. Note that γ is the integral curve of b + passing through p, so γ = m(t) and b + = N. In our case, Ric(N, N) and m (r, θ) =, so Hess (b + ) = Hess (r). 31

32 Thus Hess (b + ) =, so that b + is a parallel vector field. Now φ is an isometry in the t direction since exp p is a radial isometry. Suppose X is a vector field on v. Then R(N, X)N = N X N X N N [X,N] N. But N N =, and we may extend X in the coordinate direction so that [X, N] =. Since X N = X b + =, we have R(N, X)N =. Let J(t) = φ (x) = d ds (φ(c(s))) s=t, where c : ( ε, ε) v has c (t) = X. Then J(t) is a Jacobi field, J (t) = and J N. Thus J(t) is constant. Hence φ (X) = X, so φ is an isometry. Remark: Since Hess (b + ) =, we have X b + = for all vector fields X. By the de Rham decomposition, φ is a locally isometric splitting. Summary of Proof of Splitting Theorem. 1. Laplcaian Comparison in Barrier Sense 2. Maximal Principle 3. Bochner Formula: Generalizes Ric(N, N) + Hess (r) = m (r, θ) 4. de Rham Decomposition Also, the Regularity Theorem was used. 3.5 Applications of the Splitting Theorem Theorem (Cheeger-Gromoll 1971) If M n is compact with Ric M iso then the universal cover M N R k, where N is a compact (n k)- manifold. Thus π 1 (M) is almost π 1 (Flat Manifold), i.e. F π 1 (M) B k, where F is a finite group and B k is the fundamental group of some compact flat manifold. 32

33 B k is called a Bieberbach group. Proof. By the splitting theorem, M N R k, where N has no line. We show N is compact. Note that isometries map lines to lines. Thus, if ψ Iso( M), then ψ = (ψ 1, ψ 2 ), where ψ 1 : N N and ψ 2 : R k R k are isometries. Suppose N is not compact, so N contains a ray γ : [, ) N. Let F be a fundamental domain of M, so F is compact, and let p 1 be the projection M N. Pick t i. For each i there is g i π 1 (M) such that g i (γ(t i )) p 1 (F ). But p 1 ( F ) is compact, so we may assume g i (γ(t i )) p N. Set γ i (t) = g i (γ(t + t i )). Then γ i : [ t i, ) N is minimal, and {γ i } converges to a line σ in N. Thus N is compact. For the second statement, let p 2 : π 1 (M) Iso(R k ) be the map ψ = (ψ 1, ψ 2 ) ψ 2. Then Ker(p 2 ) π 1 (M) Im(p 2 ) is exact. Now Ker(p 2 ) = {(ψ 1, )}, while Im(ψ 2 ) = {(, ψ 2 )}. Since Ker(p 2 ) gives a properly discontinuous group action on a compact manifold, Ker(p 2 ) is finite. On the other hand, Im(p 2 ) is an isometry group on R k, so Im(p 2 ) is a Bieberbach group. Remark: The curvature condition is only used to obtain the splitting M N R k. Thus, if the conclusion of the splitting theorem holds, the curvature condition is unnecessary. Corollary If M n is compact with Ric M and Ric M > at one point, then π 1 (M) is finite. Remark: This corollary improves the theorem of Bonnet-Myers. The corollary can also be proven using the Bochner technique. In fact, Aubin s deformation gives another metric that has Ric M > everywhere. Corollary If M n has Ric M then b 1 (M) n, with equality if and only if M n iso T n, where T n is a flat torus. Definition Suppose M n is noncompact. Then M is said to have the geodesic loops to infinity property if for any ray γ in M, any g π 1 (M, γ()) and any compact K M there is a geodesic loop c at γ t in M K such that g = [c] = [(γ t ) 1 c γ t ]. Example M = N R If the ray γ is in the splitting direction, then any g π 1 (M, γ) is homotopic to a geodesic loop at infinity along γ. 33

34 Theorem (Sormani, 1999) If M n is complete and noncompact with Ric M > then M has the geodesic loops to infinity property. Theorem (Line Theorem) If M n does not have the geodesic loops to infinity property then there is a line in M. Application:(Shen-Sormani) If M n is noncompact with Ric M > then H n 1 (M, Z) =. 3.6 Excess Estimate Definition Given p, q M, the excess function associated to p and q is e p,q (x) = d(p, x) + d(q, x) d(p, q). For fixed p, q M, write e(x).if γ is a minimal geodesic connecting p and q with γ() = p and γ(1) = q, let h(x) = min d(x, γ(t)). Then t 1 e(x) 2h(x). Let y be the point along γ between p and q with d(x, y) = h(x). Set s 1 = d(p, x), t 1 = d(p, y) s 2 = d(q, x), t 2 = d(q, y). We consider triangles pqx for which h/t 1 is small; such triangles are called thin. Example In R n, s 1 = h 2 + t 2 1 = t (h/t1 ) 2. For a thin triangle, we may use a Taylor expansion to obtain s 1 t 1 (1 + (h/t 1 ) 2 ). Thus e(x) = s 1 + s 2 t 1 t 2 h 2 /t 1 + h 2 /t 2 = h(h/t 1 + h/t 2 ) 2h(h/t), where t = min{t 1, t 2 }. Thus e(x) is small is h 2 /t is small. 34

35 If M has K then the Toponogov comparison shows that s 1 h2 + t 2 1, so the same estimate holds. Lemma e(x) has the following basic properties: 1. e(x). 2. e γ =. 3. e(x) e(y) 2d(x, y). 4. If M has Ric, where s = min{s 1, s 2 }. (e(x)) (n 1)(1/s 1 + 1/s 2 ) (n 1)(2/s), Proof. (1), (2) and (3) are clear. (4) is a consequence of the following Laplacian comparison. Lemma Suppose M has Ric (n 1)H. Set r(x) = d(p, x), and let f : R R. Then, in the barrier sense, 1. If f then f(r(x)) H f r=r(x). 2. If f then f(r(x)) H f r=r(x). Proof. Recall that = 2 on the geodesic sphere. Hence r 2 + m(r, θ) r +, where is the Laplacian f(r(x)) = f + m(r, θ)f = f + rf, so we need only show r H r in the barrier sense. We have proved the result where r is smooth, so need only prove at cut points. Suppose q is a cut point of p. Let γ be a minimal geodesic with γ() = p and γ(l) = q. We claim that d(γ(ε), x) + ε is an upper barrier function of r(x) = d(p, x) at q, as 1. d(γ(ε), x) + ε d(p, x), 35

36 2. d(γ(ε), q) + ε = d(p, q) and 3. d(γ(ε), x) + ε is smooth near q, since q is not a cut point of γ(ε) for ε >. Since (d(γ(ε), x) + ε) H (d(γ(ε), x)) we have the result. = m H (d(γ(ε), x)) Definition The dilation of a function is dil(f) = min x,y m H (d(p, x)) + cε = H (r(x)) + cε, f(x) f(y). d(x, y) By property (2) of e(x), we have dil(e(x)) 2. Theorem Suppose U : B(y, R + η) R is a Lipschitz function on M, Ric M (n 1)H and 1. U, 2. dil(u) a, 3. u(y ) = for some y B(y, R) and 4. U b in the barrier sense. Then U(y) ac + G(c) for all < c < R, where G(r(x)) is the unique function on M H such that: 1. G(r) > for < r < R. 2. G (r) < for < r < R. 3. G(R) =. 4. H G b. 36

37 Proof. Suppose H =, n 3. We want H G = b. Since H = 2 r 2 + m H (r, θ) r +, we solve G + (n 1)G /r = b G r 2 + (n 1)G r = br 2, which is an Euler type O.D.E. The solutions are G = G p + G h, where G p = b/2nr 2 and G h = c 1 + c 2 r (n 2). Now G(R) = gives while G < gives for all < r < R. Thus c 2 Hence G(r) = b 2n (r2 + 2 G(R) = and G <. For general H <, b 2n R2 + c 1 + c 2 R (n 2) =, b n r (n 2)c 2r (n 1) > R G(r) = b r b n(n 2) Rn. n 2 r (n 2) n t r n 2 R2. Note that G > follows from ( ) n 1 sinh Ht sinh dsdt. Hs Note that H G b by the Laplacian comparison. To complete the proof, fix < c < R. If d(y, y ) c, U(y) = U(y) U(y ) ad(y, y ) ac ac + G(c). If d(y, y ) > c then consider G defined on B(y, R + ε), where < ε < η. Letting ε gives the result. Consider V = G U. Then V = G U, V B(R+ε) and V (y ) >. Now y is in the interior of B(y, R + ε) B(y, c), so V (y ) > for some y B(y, c). Since U(y) U(y ) ad(y, y ) = ac 37

38 and we have G(c) U(y ) = V (y ) >, U(y) ac + U(y ) < ac + G(c). We now apply this result to e(x). Here e(x), a = 2 and R = h(x). We assume s(x) 2h(x). On B(x, R), e 4(n 1), s(x) so b = 4(n 1)/s(x). Thus e(x) 2c + G(c) = 2(n 1) 2c + (c 2 + frac2n 2h n c (n 2) + n ns n 2 h2 ) for all < c < h. To find the minimal value for ar + G(r), < r < R, consider a + G (r) = a + b 2n (2r 2Rn r 1 n ) =. This gives r(r n /r n 1) = an/b. To get an estimate, choose r small. Then R n /r n is large, so R n /r n 1 an/b. Hence r = is close to a minimal point. For the excess function, choose ( 2h n c = s ) 1 n 1 ( ) R n 1 b n 1 an ( ) 1 h n 4(n 1) n 1 s. 2n Then G(c) = 2(n 1) ns ( (2h n s ) 2/(n 1) + 2 n 2 hn ( 2h n s ) n 2 ) n 1 n n 2 h2. 38

39 and so Now ( 2h n s ) 2 n 1 = h 2 ( 2h s 2h s 1 n n 2 > 1, ) 2 n 1, 2(n 1) 2 G(c) n n 2 s ( 2(n 1) 2h n n(n 2) s 2c. h n ) 1 n 1 ( 2h n s ) 1 n 1 1 Thus e(x) 2c + G(c) = 2c + 2c ( 2h n = s ( h n 8 s ) 1 n 1 ) 1 n 1. Remarks: 1. A more careful estimate is ( ) ( ) n 1 c3 h n 1 ( ) 1 n 1 h n 1 e(x) 2 = 8h, n 2 2 s ( ) where c 3 = n n s 1 h s 2 and h < min(s h 1, s 2 ). 2. In general, if Ric M (n 1)H then e(x) hf ( h/s ) for some continuous F satisfying F () =. F is given by an integral; consider the proof of the estimate in the case Ric M. 39

40 3.7 Applications of the Excess Estimate Theorem (Sormani, 1998) Suppose M n complete and noncompact with Ric M. If, for some p M, lim sup r diam( B(p, r)) r < 4s n, where s n = n 4, n 1 then π 1 (M) is finitely generated. Compare this result with: Theorem (Abresch and Gromoll) If M is noncompact with Ric M, K 1 and diameter growth o(r 1 n ), then M has finite topological type. Note: Diameter growth is the growth of diam B(p, r). When Ric, diam B(p, r) r. To say M has finite topological type is to say that each H i (M, Z) is finite. To prove Sormani s result we choose a desirable set of generators for π 1 (M). Lemma For M n complete we may choose a set of generators g 1,..., g n,... of π 1 (M) such that: 1. g i span{g 1,..., g i 1 }. 2. Each g i can be represented by a minimal geodesic loop γ i based at p such that if l(γ i ) = d i then d(γ(), γ(d i /2)) = d i /2, and the lift γ i based at p is a minimal geodesic. Proof. Fix p M. Let G = π 1 (M). Choose g 1 G such that d( p, g 1 ( p)) d( p, g( p)) for all g G {e}. Note that since G acts discretely on M, only finitely many elements of G satisfy a given distance restraint. Let G i = g 1,..., g i 1. Choose g i G G i such that d( p, g i ( p)) d( p, g( p)) for all g G G i. If π 1 (M) is finitely generated, we have a sequence g 1,..., g n,... ; otherwise we have a list. The g i s satisfy (1). Let γ i be the minimal geodesic connecting p to g i ( p). Set γ i = π( γ i ), where is the covering π : M M. We claim that if l(γi ) = d i then d(γ(), γ(d i /2)) = d i /2. 4

41 Otherwise, for some i and some T < d i /2, γ i (T ) is a cut point of p along γ i. Since M and M are locally isometric, and γ i (T ) is not conjugate to p along γ, γ i (T ) is not conjugate to p along γ. Hence we can connect p to γ i (T ) with a second minimal geodesic σ. Set h 1 = σ 1 γ i [,T ] and Now h 1 is not a geodesic, so h 2 = γ i [T,di ] σ. Similarly, d( p, h 1 ( p)) < 2T < d i. d( p, h 2 ( p)) < T + d i T = d i. Hence h 1, h 2 G i. But then γ i = h 2 h 1 G i, which is a contradiction. Lemma Suppose M n has Ric, n 3 and γ is a geodesic loop based at p. Set D = l(γ). Suppose 1. γ [,D/2], and γ [D/2,D] are minimal. 2. l(γ) l(σ) for all [σ] = [γ]. Then for x B(p, RD), R 1/2 + s n, we have d(x, γ(d/2)) (R 1/2)D + 2s n D. Remark: γ(d/2) is a cut point of p along γ. Since d(p, x) > D/2, any minimal geodesic connecting p and x cannot pass through γ(d/2). Thus d(γ(d/2), x) > d(p, x) d(p, γ(d/2)) = RD D/2 = (R 1/2)D. The lemma gives a bound on how much larger d(γ(d/2), x) is. Proof. It is enough to prove for R = 1/2 + s n. For if R > 1/2 + s n, we may choose y B(p, (1/2 + s n )) such that d(x, γ(d/2)) = d(x, y) + d(y, γ(d/2)). 41

42 Then d(x, γ(d/2)) d(x, y) + 3s n D (R (1/2 + s n ))D + 3s n D = (R 1/2)D + s n D. Suppose there exists x B(p, (1/2 + s n )D) such that d(x, γ(d/2)) = H < 3s n D. Let c be a minimal geodesic connecting x and γ(d/2). Let p be a lift of p, and lift γ to γ starting at p. If g = [γ], then γ connects p and g( p). Lift c to c starting at γ(d/2), and lift c γ [,D/2] to c γ [,D/2]. Then d( p, x) d(p, x) = (1/2 + s n )D, and d(g( p), x) (1/2 + s n )D. Thus e p,g( p) ( x) = d( p, x) + d(g( p), x) d( p, g( p)) (1/2 + s n )D + (1/2 + s n )D D = 2s n D. But, by the excess estimate, if s 2h, In this case, h H < 3s n D. Also, ( ) 1 h n n 1 e( x) 8. s s (1/2 + s n )D > D/2. Since s n < 1/12 for n 2, we have s 2h. Thus ( ) (3sn D) n 1 n 1 e( x) 8. D/2 But this gives 2s n D 8D (2(3s n ) n ) 1 n 1, 42

43 whence s n > n 4. n 1 We may now prove Sormani s result. Proof of Theorem. Pick a set of generators {g k } as in the lemma, where g k is represented by γ k. If x k B(p, (1/2 + s n )d k ), where d k = l(γ k ), we showed that d(x k, γ(d k /2)) 3s n d k. Let y k B(p, d k /2) be the point on a minimal geodesic connecting p and x k. Then lim sup r diam( B(p, r)) r d(y k, γ k (d k /2)) lim k d k /2 2s n d k lim k d k /2 = 4s n, so we have a contradiction if there are infinitely many generators. The excess estimate can also be used for compact manifolds. Lemma Suppose M n with Ric M (n 1). Then given δ > there is ε(n, δ) > such that if d(p, q) π ε then e p,q (x) δ. This lemma can be used to prove the following: Theorem There is ε(n, H) such that if M n has Ric M (n 1), diam M π ε and K M H then M is a twisted sphere. Proof of Lemma. Fix x and set e = e p,q (x). Then B(x, e/2), B(p, d(p, x) e/2) and B(q, d(x, q) e/2) are disjoint. Thus vol(m) vol(b(x, e/2)) + vol(b(p, d(p, x) e/2)) + vol(b(q, d(q, x) e/2)) ( ) vol(b(x, e/2)) vol(b(p, d(p, x) e/2)) vol(b(q, d(q, x) e/2)) = vol(m) + + vol(b(x, π)) vol(b(p, π)) vol(b(q, π)) ( ) v(n, 1, e/2) + v(n, 1, d(p, x) e/2) + v(n, 1, d(q, x) e/2) vol(m), v(n, 1, π) where v(n, H, r) = vol(b(r)), B(r) M n H. 43

44 Now in S n (1), vol(b(r)) = r sinn 1 t dt is a convex function of r. Thus we have v(n, 1, π) v(n, 1, e/2) + v(n, 1, d(p, x) e/2) + v(n, 1, d(q, x) e/2) ( ) d(p, x) + d(q, x) e v(n, 1, e/2) + 2v n, 1, 2 ( ) d(p, q) = v(n, 1, e/2) + 2v n, 1,. 2 Hence ( v(n, 1, e/2) v(n, 1, π) 2v n, 1, which tends to as ε. Thus e. ) d(p, q), 2 44