LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM Last time we proved the Bishop-Gromov comparison theorem, namely for any complete Riemannian manifold satisfying Ric ≥ (m − 1)k, the functions Area(S(p, r)) Area(Sk(r)) and Vol(B(p, r)) Vol(Bk(r)) are non-increasing in r, and both tends to 1 as r → 0+. Today we shall give some applications. 1. Applications to geometric quantities ¶ Cheng’s maximal diameter theorem. As an application of the volume comparison theorem, we get a one-sentence proof of Bonnet-Myers theorem: If d(p,q)>√π k , then 0 0, then Vol(M, g) ≤ Vol(S m k ), and the equality holds if and only if (M, g) is isometric to S m k . In fact, with a bit more work, one can prove Theorem 1.2 (S.Y. Cheng). Let (M, g) be a complete Riemannian manifold with Ric ≥ (m − 1)k > 0, and diam(M, g) = √π k , then (M, g) is isometric to S m k . Proof. By Bishop-Gromov volume comparison theorem, for any p ∈ M, Vol(B(p, π 2 √ k )) Vol(M) = Vol(B(p, π 2 √ k )) Vol(Bπ/√ k (p)) ≥ Vol(Bk( π 2 √ k )) Vol(Bk( √π k )) = 1 2 . Now let p, q ∈ M so that dist(p, q) = √π k . The the above inequality implies Vol(B(p, π 2 √ k )) ≥ 1 2 Vol(M), Vol(B(q, π 2 √ k )) ≥ 1 2 Vol(M). Since B(p, π 2 √ k ) ∩ B(q, π 2 √ k ) = ∅, we must have Vol(B(p, π 2 √ k )) Vol(M) = Vol(Bk( π 2 √ k )) Vol(Bk( √π k )) = 1 2 , Vol(B(q, π 2 √ k )) Vol(M) = Vol(Bk( π 2 √ k )) Vol(Bk( √π k )) = 1 2 . 1
2 LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM So B(p, π 2 √ k ) ∪ B(q, π 2 √ k ) = M. According to Bishop-Gromov comparison theorem, B(p, π 2 √ k ) and B(q, π 2 √ k ) are both isometric to hemisphere in S m k . It follows that Vol(M, g) = Vol(S m k ) and thus M is isometric to S m k . □ Remark. The maximal diameter theorem was first proved by Toponogov under the stronger assumption K ≥ k. ¶ Volume growth rate. Another immediate consequence of volume comparison theorem is Corollary 1.3. Let (M, g) be a complete Riemannian manifold with Ric ≥ 0, then VolB(p, r) ≤ Vol(B0(r)) = ωmr m, where ωm is the volume of unit ball in R m. Note that for any p, q, with l = d(p, q), one has B(p, r) ⊂ B(q, r + l) ⊂ B(p, r + 2l). It follows that the asymptotic volume ratio αM := limr→∞ Vol(B(p, r)) ωmrm is independent of p, and αM ≤ 1, with equality if and only if (M, g) is isometric with (R m, g0). Remark. We say (M, g) has large volume growth [or Euclidean volume growth] if αM > 0. It has been proved by Li (1986) and Anderson (1990) that |π1(M)| is bounded by 1 αM , and in particular, M is simply connected if αM > 1/2. In 1994 Perelman proved that there exists δm > 0 such that if αM ≥ 1 − δm, then M is contractible. Remark. The asymptotic volume ratio αM appears in many problems. For example, S. Brendle recently proved the following isoperimetric inequality: Let (M, g) be a complete noncompact Riemannian manifold with nonnegative Ricci curvature, then for any compact domain Ω in M with boundary ∂Ω, (Area(∂Ω))1/(m−1) (Vol(Ω))1/m ≥ α 1/m(m−1) M (Area(∂B0(1)))1/(m−1) (Vol(B0(1)))1/m . It is easy to construct manifolds with Ric ≥ 0 and αM = 0. For example, the Riemannian manifold M = S k ×R l (with the product metric) has nonnegative Ricci curvature, and the volume Vol(B(p, r)) is of order r l . In particular, S m−1 × R has “linear” volume growth. It turns out that for any complete non-compact Riemannian manifold (M, g) with Ric ≥ 0, this is the least possible volume growth: Theorem 1.4 (Calabi-Yau). Let (M, g) be a complete non-compact Riemannian manifold with Ric ≥ 0. Then there exists a positive constant c depending only on p and m so that Vol(B(p, r)) ≥ cr, ∀r > 2
LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM 3 Proof. (Following Gromov.) Since M is complete and non-compact, for any p ∈ M there exists a ray, i.e. a geodesic γ : [0,∞) → M with γ(0) = p such that dist(p, γ(t)) = t for all t > 0. (Exercise: Prove the existence of a ray.) For any t > 3 2 , using the Bishop-Gromov volume comparison theorem, we get Vol(B(γ(t), t + 1)) Vol(B(γ(t), t − 1)) ≤ ωm(t + 1)m ωm(t − 1)m = (t + 1)m (t − 1)m . On the other hand, by triangle inequality, B(p, 1) ⊂ B(γ(t), t+ 1)\B(γ(t), t−1). So Vol(B(p, 1)) Vol(B(γ(t), t − 1)) ≤ Vol (B(γ(t), t + 1)\B(γ(t), t − 1)) Vol(B(γ(t), t − 1)) ≤ (t + 1)m − (t − 1)m (t − 1)m , i.e. Vol(B(γ(t), t − 1)) ≥ Vol(B(p, 1)) (t − 1)m (t + 1)m − (t − 1)m ≥ C(m)Vol(B(p, 1))t, where C(m) is the infimum of the function 1 t (t−1)m (t+1)m−(t−1)m on [ 3 2 , ∞), which is positive. Now the theorem follows from the fact B(p, r) ⊃ B(γ( r + 1 2 ), r + 1 2 − 1). □ Remark. A complete non-compact Riemannian manifolds with lim inf r→∞ Vol(B(p, r)) r = C > 0 are called manifold with linear volume growth. Unlike the maximal volume growth case, the constant c = c(p, m) depends on p and may tends to 0 as p → ∞. 2. Applications to the fundamental group ¶ The fundamental group of compact manifolds. We start with a theorem concerning the topology of manifolds: Theorem A.1. For any compact manifold M, π1(M) is finitely generated. For a topological proof, we refer to Hatcher’s book ✿✿✿✿✿✿✿✿✿✿ Algebraic ✿✿✿✿✿✿✿✿✿✿ Topology (Cor. A.8 and A.9). In what follows we will give a couple geometric proofs. As we can see, by introducing a Riemannian metric on M, we are able to find a nice generator set that satisfies additional requirements and thus is better in applications. To state the geometric versions of Theorem A.1, we first endow a Riemannian metric g on M. As usual we let π : Mf → M be the universal covering, endowed with the pull-back metric ˜g = π ∗ g. Fix any ˜p ∈ Mf and consider the fundamental domain centered at ˜p, K = {q˜ ∈ Mf | d(˜q, p˜) ≤ d(˜q, g · p˜), ∀e ̸= g ∈ π1(M)}
4 LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM For simplicity we denote d0 = diam(M, g). It is not hard to check (a) K is compact and π(K) = M. (b) For any e ̸= g ∈ π1(M), g · K ∩ K ⊂ ∂K. (c) Mf = ∪g∈π1(M)g · K. (d) K ⊂ B(p, d e 0). Our first geometric strengthened version of Theorem A.1 is a set of generators with an estimate of the “length” of any element via such generators: Theorem A.2. The set Γ = {g ∈ π1(M) : ∃x, y ∈ K such that d(x, g · y) ≤ 1} is a finite generator set of π1(M). Moreover, for any g ∈ π1(M), if there exists x, y ∈ K such that d(x, g · y) ≤ s, then there exists g1, · · · , gs ∈ Γ such that g = g1 · · · gs. Proof. For any g ∈ Γ, if we pick x, y ∈ K so that d(x, g · y) ≤ 1, then for any ˜q ∈ K, d(g · q, ˜ p˜) ≤ d(g · q, g ˜ · p˜) + d(g · p, g ˜ · y) + d(g · y, x) + d(x, p˜) ≤ 3d0 + 1. So we get g · K ⊂ B(p, e 3d0 + 1) and as a result, Γ is a finite set. To prove the second conclusion, we proceed by induction. The conclusion holds trivially for s = 1. Suppose it holds for s = k, and there exists x, y ∈ K such that d(x, g · y) ≤ k + 1. On the minimizing geodesic connecting x and g · y, one can find a point, which, in view of (c) above, can be written as g ′ · z for some g ′ ∈ π1(M) and z ∈ K, such that d(x, g′ · z) ≤ 1 and d(g ′ · z, g · y) ≤ k. By induction hypothesis, g ′ ∈ Γ and (g ′ ) −1 g = g1 · · · gk for some g1, · · · , gk ∈ Γ. This completes the proof. □ In a second geometric strengthened version of Theorem A.1, we give another set of generators with relations of given form: Theorem A.3 (Gromov). Let (M, g) be a compact Riemannian manifold. Then one can find a finite generator set Γ = {g1, · · · , gn} so that • d(˜p, gi · p˜) ≤ 2d0 for 1 ≤ i ≤ n. • all relations among these generators are of the form gigjg −1 k = e. Proof. Since M is compact, one can find a triangulation of M so that the distance between any pair of adjacent vertices is less than ε < inj(M) and such that any loop is homotopic to a loop in the 1-skeleton of the triangulation. Let {v1, · · · , vq} be the set of vertices, and eij the set of edges (so eij is only defined for some i, j). Fix p = π(˜p) and realize π1(M) with basepoint p. Let σi be the minimal geodesic from p to vi . Then for any edge eij , σij := σieijσ −1 j is a loop based at p, with L(σij ) ≤ 2d0 + ε. Since any loops in M based at p is homotopic to a loop in the 1-skeleton of the triangulation, and since σieijejkσ −1 k ∼ σiσijσjkσ −1 k , the loops σij ’s generates π1(M)
LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM 5 Observe that if three vertices are adjacent to each other, then they span a 2- simplex ∆ijk, and the loop σijσjkσki is null-homotopic, i.e. σijσjkσ −1 ik = e. So each 2-simplex gives rise to a relation among σij ’s of the given form. Conversely, if σ is null-homotopic loop in the 1-skeleton based at p, then σ is contractible in the 2-skeleton, and thus there is set of 2-simplex ∆ijk so that the relation σ = e is a product of the elementary relations of the form σijσjkσ −1 ik = e. Finally by discreteness of the action of π1(M) on Mf one has, for ε small enough, {g ∈ π1(M) | d(˜p, g · p˜) 0 and a > 1 such that N Γ G(k) ≥ cak . Remark. Note that if Γ′ is another finite set of generators, then there exists integers c1, c2 so that any element of Γ can be represented via at most c1 elements of Γ′ , and any element of Γ′ can be represented via at most c2 elements of Γ. It follows that N Γ G(k) ≥ N Γ ′ G (c1k), NΓ ′ G (k) ≥ N Γ G(c2k). So the concept of polynomial/exponential growth is independent of the choice of the generator set. Example. Here are two simple examples: • For G = Z ⊕ Z, we may take Γ = {(1, 0),(0, 1)} and it is quite obvious that NΓ G(k) ≈ 2k 2 . It follows that Z ⊕ Z has quadratic growth. • For G = Z ∗ Z, we may take Γ = {(1, 0),(0, 1)} and it is not hard to get NΓ G(k) ≈ 4 Pk l=0 3 l = 2 · 3 k+1. It follows that Z ∗ Z has exponential growth
6 LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM ¶ Growth of fundamental group: negative curvature case. Back to Riemannian manifolds. Let (M, g) be a compact Riemannian manifold. According to theorem A.1, π1(M) is finitely generated. It is natural to study its growth rate. As one can imagine, the growth rate of π1(M) is related to the volume growth rate of the Riemannian universal covering (M, f g˜) of (M, g), since the fundamental group π1(M) acts isometrically on Mf as the group of deck transformations. In view of Lohkamp’s result that any smooth manifold M of dimension m ≥ 3 admits a Riemannian metric with Ric 0 so that K ≤ −k. Note that in this case the injectivity radius is +∞. So by Theorem 2.2, Vol(B(˜p, d0 + s)) ≥ Vol(B−k(d0 + s)). Finally use the fact Vol(B−k(r)) = Z r 0 Z Sm−1 ( sinh(√ kt) √ k ) m−1 dΘdt ≥ ce √ kr we get the desired estimate N Γ G(s) ≥ Vol(B(˜p, d0 + s)) Vol(K) ≥ Vol(B−k(d0 + s)) Vol(K) ≥ ce √ k(d0+s) . □
LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM 7 ¶ Growth of fundamental group: non-compact. From the proof above we see that the reason for “the fundamental group of ‘a compact Riemannian manifold with negative sectional curvature’ to be of exponential grown” is that the geodesic balls in the negative curvature space Mm k have exponential volume growth. On the other hand, the geodesic balls in the Euclidean space Mm 0 has volume growth of order m. So one may naturally anticipate Proposition 2.4. Let (M, g) be a compact Riemannian manifold with Ric ≥ 0, then π1(M) has polynomial growth of order ≤ m. This is true, and it is a direct consequence of the following more general theorem: Theorem 2.5 (Milnor). Let M be a complete Riemannian manifold with Ric ≥ 0 and let G ⊂ π1(M) be any finitely generated subgroup. Then G has polynomial growth of order no more than m. Proof. Let Γ be a finite set of generators of G. Fix a point ˜p ∈ Mf and let l = max{dist(˜p, gip˜) | gi ∈ Γ}. Then by triangle inequality, for any g = gi1 · · · gik ∈ Γ k ⊂ G, dist(˜p, gp˜) ≤ kl. On the other hand side, we can pick δ = 1 3 min{dist(˜p, gp˜) | e ̸= g ∈ G} > 0 so that the balls B(gp, δ ˜ ) are all disjoint for g ∈ G. So B(˜p, kl + δ) ⊃ ∪g∈ΓkB(gp, δ ˜ ) and thus Vol(B(˜p, kl + δ)) ≥ N Γ GVol(B(p, δ)). Since the Riemannian universal covering (M, f g˜) has nonnegative Ricci curvature, applying the Bishop-Gromov’s volume comparison theorem we get N Γ G ≤ Vol(B(˜p, kl + δ)) Vol(B(p, δ)) ≤ (kl + δ) m δm ≤ ckm. □ As consequences one gets Corollary 2.6. For g ≥ 2, Σg × R k admits no complete metric of Ric ≥ 0. Example. Let H be the Heisenberg group H = 1 x z 0 1 y 0 0 1 : x, y, z ∈ R and let HZ be the subgroup that contains elements with x, y, z ∈ Z. Then one can show that the growth rate of HZ is of order 4. So H/HZ admits no metric with Ric ≥ 0
8 LECTURE 26: APPLICATIONS OF THE VOLUME COMPARISON THEOREM Remark. According to a celebrated theorem of Gromov, if G is finitely generated and has polynomial growth, then G is virtually nilpotent [i.e. G contains a nilpotent subgroup of finite index]. As we have seen in Lecture 24, if (M, g) has sectional curvature K ≥ 0, then π1(M) is finitely generated. It was conjectured by Milnor in 1968 that Conjecture. For any complete Riemannian manifold (M, g) with Ric ≥ 0, the fundamental group π1(M, g) is finitely generated. The conjecture was proved for m = 3 by Liu in 2013, and was disproved for m ≥ 7 by Brue-Naber-Semola in 2023. It is still open for m = 4, 5, 6. ¶ Finiteness of fundamental group. Finally we apply Theorem A.3 and the volume comparison to prove Theorem 2.7 (Anderson). In the set of all Riemannian manifolds M with Ric ≥ (m − 1)k, Vol(M, g) ≥ V, diam(M) ≤ D, there are only finitely many isomorphism types of the fundamental groups π1(M). Proof. Let Γ = {g1, · · · , gn} be a set of generators as in Theorem A.3. Then there are at most 2n 3 isomorphism types of abstract groups G generated by Γ. It remains to show that n is bounded, which is a consequence of volume comparison. In fact, consider the fundamental domain K again. Then g · K’s are disjoint. Moreover, for each g ∈ Γ, g · K is contained in B(˜p, 4d0). So by volume comparison, nV ≤ nVol(M, g) = nVol(K) ≤ Vol(B(˜p, 4d0)) ≤ Vol(Bk(4d0)) ≤ Vol(Bk(4D)) and thus the conclusion follows. □ Remark. This is one of the many finiteness theorems that people have proved for Riemannian manifolds satisfying suitable bounds on curvature/volume/diameter etc. Note that this theorem fails if we drop the volume lower bound or diameter upper bound condition: • Consider the set of all lens space L(p, q) = S 3/Zp. Then they have constant sectional curvature 1 (and thus constant Ricci curvature m−1), and constant diameter1 π/2, but volume tends to 0 as p tends to infinity. Obviously this set has infinitely many different fundamental groups. • Consider the set of surfaces Σg endowed with hyperbolic metrics, i.e. with constant sectional curvature −1. Then they have constant Ricci curvature −(m − 1), volume bounded below (tends to infinity by Gauss-Bonnet), but the diameters are not bounded above. 1To see this one first note that for the lens space L(p, q), viewed as S 3 ⊂ C 2 quotient by the action k · (z1, z2) = (e 2πik/pz1, e2πikq/pz2), any smooth curve connecting π(1, 0) and π(0, 1) is the quotient of a curve connecting (1, 0) and a point of the form (0, e2πil/p). So diam(L(p, q)) ≥ π/2. On the other hand, by the Grove-Shiohama theorem that we will prove next time, we can’t have diam(L(p, q)) > π/2