《黎曼几何》课程教学资源（英文讲义）Lec26 APPLICATIONS OF THE VOLUME COMPARISON THEOREM.pdf

2LECTURE26:APPLICATIONSOFTHEVOLUMECOMPARISONTHEOREMSo B(p,)U B(q, ) = M. According to Bishop-Gromov comparison theorem,B(p)and B(q)are both isometric to hemisphere in Sp.It follows thatVol(M,g) = Vol(S) and thus M is isometric to Sm.口Remark.The maximal diameter theorem was firstproved by Toponogov under thestronger assumption K≥k.I Volume growth rate.Another immediate consequence of volume comparison theorem isCorollary 1.3. Let (M,g) be a complete Riemannian manifold with Ric≥ 0, thenVolB(p, r) ≤ Vol(Bo(r)) = wmrmwhere Wm is the volume of unit ball in RmNote that for any p, q, with I = d(p,q), one hasB(p,r) C B(q,r +I) C B(p,r + 21)It follows that the asymptotic volume ratioVol(B(p,r))am := limWmrmr-→is independent of p, and am ≤ 1, with equality if and only if (M,g) is isometricwith (Rm,go).Remark. We say (M,g) has large volume growth [or Euclidean volume growth] if aM > 0.It has beenproved by Li (1986)and Anderson (1990)thati(M)/is bounded by,and in particular, M is simply connected if am > 1/2. In 1994 Perelman provedthat there exists om>0 such that if am ≥1-om, then M is contractible.Remark. The asymptotic volume ratio QM appears in many problems. For example,S.Brendlerecently proved the following isoperimetric inequality:Let (M,g)beacompletenoncompactRiemannianmanifoldwithnonnegativeRiccicurvature,thenfor any compact domain2 inM with boundary 2,(Area(02)1/(m-1)1/m(m-1) (Area(Bo(1)1/(m-1)≥OM(Vol(2)1/m(Vol(Bo(1)1/mIt is easy to construct manifolds with Ric≥ 0 and am =O.For example, theRiemannianmanifoldM=SkxR(withtheproductmetric)hasnonnegativeRiccicurvature, and the volume Vol(B(p,r) is of order rl. In particular, sm-1 × R has"linear"volume growth.It turns out that foranycomplete non-compact Riemannianmanifold (M,g)with Ric ≥0, this is the least possible volume growth:Theorem 1.4(Calabi-Yau).Let (M,g)be a completenon-compactRiemannianmanifold with Ric ≥ O. Then there erists a positive constant c depending only on pandmsothatVol(B(p, r)) ≥ cr, Vr > 2

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)}

4LECTURE26:APPLICATIONSOFTHEVOLUMECOMPARISONTHEOREMFor simplicity we denote do = diam(M, g). It is not hard to check(a) K is compact and π(K) = M.(b)Foranye+gETi(M),g·KnKcoK.(c) M = Ugemi(M)g · K.(d) K c B(p, do).Our first geometric strengthened version of Theorem A.1 is a set of generators withan estimate of the"length" of any element via such generators:Theorem A.2. The setT = (gE Ti(M) : 3r, y EK such that d(,g y) ≤ 1)is a finite generator set of i(M).Moreover, for any g E Ti(M), if there ecistsr,y E K such that d(r,g-y) ≤ s, then there erists gi,.:,gs E f such that g91...9s.Proof. For any g e , if we pick r, y e K so that d(r,g y) ≤ 1, then for any q e K,d(g q,p) ≤ d(g q,g p) +d(gp,gy) +d(g y, r) + d(r,p) ≤3do + 1.So we get g - K C B(p, 3do + 1) and as a result, I is a finite set.To prove the second conclusion, weproceed by induction. The conclusion holdstrivially for s = 1. Suppose it holds for s = k, and there exists r,y e K such thatd(r,g y) ≤ k + 1. On the minimizing geodesic connecting r and g - y, one can finda point, which, in view of (c) above, can be written as g. z for some g e i(M)and z E K, such that d(r, g'-z) ≤1 and d(g'-z, g-y) ≤ k. By induction hypothesis,g eT and (g)-g = g1 ... gk for some gi,..., gkeT. This completes the proof. In a second geometric strengthened version of Theorem A.1, we give another setof generators with relations of given form:Theorem A.3 (Gromov).Let (M,g)be a compact Riemannian manifold.Thenone can find a finite generator set F = (gi,..,gn] so that.d(p,gi·p)≤2dofor1<i<n.. all relations among these generators are of the form gigigk=e.Proof. Since M is compact, one can find a triangulation of M so that the distancebetween any pair of adjacent vertices is less than < inj(M) and such that any loopis homotopic to a loop in the 1-skeleton of the triangulation. Let (ui, ., ua) be theset of vertices, and eij the set of edges (so ej is only defined for some i,j). Fix p = r(p)and realize i(M)with basepoint p.Let o,be the minimal geodesic from p to vi.Then for any edge eij, oij := oieijoj' is a loop based at p, withL(oi) ≤ 2do + E.Since any loops in M based at p is homotopic to a loop in the 1-skeleton of thetriangulation, and since oieijejkol ~ Qidij jko,l, the loops Qij's generates i(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)

5LECTURE26:APPLICATIONSOFTHEVOLUMECOMPARISONTHEOREMObserve that if three vertices are adjacent to each other, then they span a 2-simplexijk,and the loop ijo jku is null-homotopic, ie.Cijojkoil=e. So each2-simplex gives rise to a relation among i,'s of the given form. Conversely, if is null-homotopic loop in the 1-skeleton based at p, then is contractible in the2-skeleton, and thus there is set of 2-simplex △ijk so that the relation = e is aproduct of the elementary relations of the form oijo jkoi' = e.Finallyby discreteness of theactionofi(M)on M onehas,for esmall enough,(g E i(M) I d(p, g -p) < 2do +e) = (g E πi(M) / d(p, g - p) ≤ 2do).口This finishes the proof.IDetour:Growth ofafinitelygenerated groupWe give some abstract definitions in algebra.Let G be a finitely generated group,and F = (gi, .. , gn] a generator set. The growth function of G with respect to Iis defined to be the number of group elements that can be represented as a productof atmostk generators,i.e.N&(k) = #(g EGIEl ≤k and gi,..., gin E T s.t. g= gtl...gtl)We will use Igl to represent the smallest I such that g = gt1 . gtl, and call it thelength of g with respect to the given generator set.Definition 2.1. Let G be finitely generated, and T is a finite set that generates G.(1) We say that G is of polynomial growth (of order n) ifNE(k)≤ cknfor some constant c depending only on G,I.(2)We say G is of erponential growth if there is c> 0and a> 1 such thatNE(k) ≥ cak.Remark. Note that if F' is another finite set of generators, then there exists integersCi, C so that any element of F can be represented via at most ci elements of F', andany element of F' can be represented via at most c elements of T.It follows thatNE(K) ≥ N(cik),NE(K) ≥ NE(c2k)So the concept of polynomial/exponential growth is independent of the choice of thegenerator set.Erample.Herearetwo simpleexamples:. For G = Z Z, we may take I = (1, O), (0, 1)) and it is quite obvious thatN(k) ~ 2k2. It follows that Z @ Z has quadratic growth.. For G = Z * Z, we may take I = [(1, 0), (0, 1) and it is not hard to getN(k) ~ 4t=0 3' = 2 . 3k+1. It follows that Z * Z has exponential growth

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˜) < 2d0 + ε} = {g ∈ π1(M) | d(˜p, g · p˜) ≤ 2d0}. This finishes the proof. □ ¶ Detour: Growth of a finitely generated group. We give some abstract definitions in algebra. Let G be a finitely generated group, and Γ = {g1, · · · , gN } a generator set. The growth function of G with respect to Γ is defined to be the number of group elements that can be represented as a product of at most k generators, i.e. N Γ G(k) = #{g ∈ G | ∃l ≤ k and gi1 , · · · , gil ∈ Γ s.t. g = g ±1 i1 · · · g ±1 il }. We will use |g| to represent the smallest l such that g = g ±1 i1 · · · g ±1 il , and call it the length of g with respect to the given generator set. Definition 2.1. Let G be finitely generated, and Γ is a finite set that generates G. (1) We say that G is of polynomial growth (✿✿ of✿✿✿✿✿✿ order✿✿n) if N Γ G(k) ≤ ckn for some constant c depending only on G, Γ. (2) We say G is of exponential growth if there is c > 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