LECTURE 22:THEOREMS ON CURVATURE V.S.TOPOLOGY1.COMPLETE RIEMANNIAN MANIFOLDS WITH NON-POSITIVE CURVATUREIA Hessian comparison fordLet (M,g)be a Cartan-Hadamard manifold,i.e.a complete simply-connectedRiemannian manifold with non-positive curvature.We have seen that there isno conjugate point for such manifolds, and by Cartan-Hadamard theorem, exp, :TM -→ M is diffeomorphism. In particular, between any pair of points there isa unique geodesic [which has to be minimizingl, and there is no cut point for Cartan-HadamardmanifoldsWe first prove that d, is strictly convex on Cartan-Hadamard manifolds:Proposition 1.1. For any p in a Cartan-Hadamard manifold (M,g), V?d ≥ 2g.Moreover, if the sectional curvature is negative, then ?d >2g at any qp.Proof. Let be a normal geodesic with (o) = p. For any E TqM, where q = (),(V2d))g(%(1), ) = (V0 Vdp, ) = (V(0)(2t)), ) = 2(%(0), )It remains to prove that for u I (t), one has (V?dp)g(u, u) ≥ 2(u, u).Let V be the normal Jacobi field along with V(0) = 0, V(l) = . Then(V?d)(t)(V,V)=(Vv(t)Vde, V)= (Vv(t)(2t)), V)=2t(Vv(t)(), V) =2t(Vs(t) V, V).Denote () - Vov). Then limt-0+ () = +0, since v() = Vo) + (t2).(V(t),V(t)By definition we haveF()- v)+(V-20 (Vr(t) V, V)?≥ -f2(t),(V,V)(V,V)2where we used the fact (Vs(t)V(t)V, V)=(R(%, V)%, V)=-K(, V)/^ V|P ≥ 0 andCauchy-Schwartz inequality (V(t)V,V() V)(V, v) ≥ (V(t)V,V)2. It follows111-f'(T)dt7dTinf2(T)f(t)T-0+f()-f(t)f(T)So we get tf(t) ≥1 for all t, and the first conclusion follows.口For the second conclusion, one just notice f'(t) >-f?(t) for t > 0
LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY 1. Complete Riemannian manifolds with non-positive curvature ¶ A Hessian comparison for d 2 p . Let (M, g) be a Cartan-Hadamard manifold, i.e. a complete simply-connected Riemannian manifold with non-positive curvature. We have seen that there is no conjugate point for such manifolds, and by Cartan-Hadamard theorem, expp : TpM → M is diffeomorphism. In particular, between any pair of points there is a unique geodesic [which has to be minimizing], and there is no cut point for CartanHadamard manifolds. We first prove that d 2 p is strictly convex on Cartan-Hadamard manifolds: Proposition 1.1. For any p in a Cartan-Hadamard manifold (M, g), ∇2d 2 p ≥ 2g. Moreover, if the sectional curvature is negative, then ∇2d 2 p > 2g at any q ̸= p. Proof. Let γ be a normal geodesic with γ(0) = p. For any v ∈ TqM, where q = γ(l), (∇2 d 2 p )q( ˙γ(l), v) = ⟨∇γ˙ (l)∇d 2 p , v⟩ = ⟨∇γ˙ (l)(2tγ˙), v⟩ = 2⟨γ˙(l), v⟩. It remains to prove that for v ⊥ γ˙(l), one has (∇2d 2 p )q(v, v) ≥ 2⟨v, v⟩. Let V be the normal Jacobi field along γ with V (0) = 0, V (l) = v. Then (∇2 d 2 p )γ(t)(V, V )=⟨∇V (t)∇d 2 p , V ⟩=⟨∇V (t)(2tγ˙), V ⟩=2t⟨∇V (t)( ˙γ), V ⟩=2t⟨∇γ˙ (t)V, V ⟩. Denote f(t) = ⟨∇γ˙ (t)V,V (t)⟩ ⟨V (t),V (t)⟩ . Then limt→0+ f(t) = +∞, since V (t) = t∇γ˙ (0)V + O(t 2 ). By definition we have f ′ (t) = ⟨∇γ˙ (t)∇γ˙ (t)V, V ⟩ + ⟨∇γ˙ (t)V, ∇γ˙ (t)V ⟩ ⟨V, V ⟩ − 2 ⟨∇γ˙ (t)V, V ⟩ 2 ⟨V, V ⟩ 2 ≥ −f 2 (t), where we used the fact ⟨∇γ˙ (t)∇γ˙ (t)V, V ⟩=⟨R( ˙γ, V ) ˙γ, V ⟩=−K( ˙γ, V )|γ˙ ∧ V | 2 ≥ 0 and Cauchy-Schwartz inequality ⟨∇γ˙ (t)V,∇γ˙ (t)V ⟩⟨V, V ⟩ ≥ ⟨∇γ˙ (t)V, V ⟩ 2 . It follows t ≥ Z t 0 −f ′ (τ ) f 2 (τ ) dτ = Z t 0 ( 1 f(τ ) ) ′ dτ = 1 f(t) − limτ→0+ 1 f(τ ) = 1 f(t) , So we get tf(t) ≥ 1 for all t, and the first conclusion follows. For the second conclusion, one just notice f ′ (t) > −f 2 (t) for t > 0. □ 1
2LECTURE 22:THEOREMSONCURVATUREV.S.TOPOLOGYIThefundamental group of Riemannian manifolds with K≤0.Asafirst application, weproveTheorem 1.2 (Cartan).Let (M,g) be a Cartan-Hadamard manifold, and :M -→M an isometry with on =Id for some n.Then admits a fired point.Proof. Fix p E M and consider the functiong : M -→R, q → g(q) = d(q,p) +d'(q,p(p)) +++d(q, on-1(p)Then g is strictly convex and g(g) → +oo as d(q,p) → +oo. So g admits a unique口minimum at some point p. Since g((q))= g(q) we conclude p(p)=p.As a corollarywegetCorollary 1.3. Let (M,g) be a complete Riemannian manifold with non-positivesectional curvature. Then Ti(M) is torsion free [ie. no nontrivial finite order element).Proof. If i(M) admits a finite order element r, then the corresponding Deck trans-formation ff:M -→M is of finite order, and thus by Cartan's theorem above, f-口admits a fixed point. This implies f. = Id and T = e.As an immediate consequence, we seeCorollary 1.4. For any compact manifold M, Rpm × M admits no metric of non-positive sectional curvature.IAweak cosinelaw for Cartan-Hadamard manifolds.As a second application of the convexity of de, we prove the following weakcosine lawfor Cartan-Hadamard manifolds.Proposition 1.5. Let (M, g) be Cartan-Hadamard manifold. Consider the geodesictriangle with vertices pi,P2,P E M. Let a,b,c be the lengths of sides and A, B,Cbethe corresponding opposite angles.Then(1)a2+b22abcosC≤c2.(2) A+B+C≤T.Further more, if the sectional curvature is negative, then these inequalities are strict.Proof. Let be a normal geodesic from p3 to pi, and let f(t) = d(p2, (t). Thenf(0) = d (p2, P3) = a2andf(0) =2d(p2,p3)(Vdp,(0))= -2a cosC.By Lemma 2.4 in Lecture 21, f"(+) = (V2d)(t)((t), (t). By Proposition 1.1,f"() ≥ 2 for all T. Thus we getc2 = f(b) ≥ f(0) + f'(0)6 + 62 = a2 + 62 - 2ab cos C
2 LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY ¶ The fundamental group of Riemannian manifolds with K ≤ 0. As a first application, we prove Theorem 1.2 (Cartan). Let (M, g) be a Cartan-Hadamard manifold, and φ : M → M an isometry with φ n = Id for some n. Then φ admits a fixed point. Proof. Fix p ∈ M and consider the function g : M → R, q 7→ g(q) = d 2 (q, p) + d 2 (q, φ(p)) + · · · + d 2 (q, φn−1 (p)). Then g is strictly convex and g(q) → +∞ as d(q, p) → +∞. So g admits a unique minimum at some point ˜p. Since g(φ(q)) = g(q) we conclude φ(˜p) = ˜p. □ As a corollary we get Corollary 1.3. Let (M, g) be a complete Riemannian manifold with non-positive sectional curvature. Then π1(M) is torsion free [i.e. no nontrivial finite order element]. Proof. If π1(M) admits a finite order element τ , then the corresponding Deck transformation fτ : Mf → Mf is of finite order, and thus by Cartan’s theorem above, fτ admits a fixed point. This implies fτ = Id and τ = e. □ As an immediate consequence, we see Corollary 1.4. For any compact manifold M, RPm × M admits no metric of nonpositive sectional curvature. ¶ A weak cosine law for Cartan-Hadamard manifolds. As a second application of the convexity of d 2 p , we prove the following weak cosine law for Cartan-Hadamard manifolds. Proposition 1.5. Let (M, g) be Cartan-Hadamard manifold. Consider the geodesic triangle with vertices p1, p2, p3 ∈ M. Let a, b, c be the lengths of sides and A, B, C be the corresponding opposite angles. Then (1) a 2 + b 2 − 2ab cos C ≤ c 2 . (2) A + B + C ≤ π. Further more, if the sectional curvature is negative, then these inequalities are strict. Proof. Let γ be a normal geodesic from p3 to p1, and let f(t) = d 2 (p2, γ(t)). Then f(0) = d 2 (p2, p3) = a 2 and f ′ (0) = 2d(p2, p3)⟨∇dp2 , γ˙(0)⟩ = −2a cos C. By Lemma 2.4 in Lecture 21, f ′′(τ ) = (∇2d 2 p )γ(t)( ˙γ(t), γ˙(t)). By Proposition 1.1, f ′′(τ ) ≥ 2 for all τ . Thus we get c 2 = f(b) ≥ f(0) + f ′ (0)b + b 2 = a 2 + b 2 − 2ab cos C
3LECTURE22:THEOREMSONCURVATUREV.S.TOPOLOGYTo prove (2), one may compare the triangle in the plane with sides a,b,c [whichsatisfies the triangle inequality since they are distances of three points in a Riemannian manifold].Denote the angles by A',B',C'.Then by the cosine law in IR?we getA 2口forT+Oandtheconclusionfollows.I Preissman's theorem.What if M is not simply connected? We have just seen that if M admits anon-positive sectional curvature metric, then i(M) is torsion free.It turns outthat if M admits a negative sectional curvature metric, then any nontrivial abeliansubgroup of i(M) is an infinite cyclic group generated by one element:Theorem 1.6 (Preissman). Let (M,g) be a compact Riemannian manifold withnegative sectional curvature, and let [1] + H C i(M) be a nontrivial abeliansubgroup of thefundamental group.Then H=Z.Remark.The theorem was strengthened byByers to: under the same assumption,anynontrivialsolvablesubgroupofi(M)isinfinitecyclic.Erample.For any closed surface M.of genus g≥2, there is Riemannian metric ofconstant negative sectional curvature. The fundamental group of M, is(ai,bi,...,ag,bgaibiai"b-l...aibiai"b-"=e),which is not abelian, but all its abelian subgroups are isomorphic to Z.Asan immediateconsequence,weseeCorollary 1.7. Suppose m ≥ 2. For any compact manifold M, Tm × M admits nometric ofnegativesectionalcurvature.Remark. It was first proved by Gao and Yau in 1986 that any compact manifold ofdimension 3admits a metric with negative Ricci curvature.Then in 1994, Lohkampproved that any manifold of dimension at least 3 admits a completeRiemannianmetric of negative Ricci curvature. So there is no topological constraint for a mani-fold of dimension≥3 to admit Riemannian metrics with negativeRicci curvature.The idea of proof is as follows: Realize Deck transformations associated with allQ E H as “a discrete family of translations along a fixed geodesic". As a result, anontrivial discrete subgroups of H corresponds to a nontrivial subgroup of R, whichis isomorphic to Z
LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY 3 To prove (2), one may compare the triangle in the plane with sides a, b, c [which satisfies the triangle inequality since they are distances of three points in a Riemannian manifold]. Denote the angles by A′ , B′ , C′ . Then by the cosine law in R 2 we get A ≤ A ′ , B ≤ B ′ , C ≤ C ′ , which implies A + B + C ≤ π. Finally if the sectional curvature is negative, then by Proposition 1.1, f ′′(τ ) > 2 for τ ̸= 0 and the conclusion follows. □ ¶ Preissman’s theorem. What if M is not simply connected? We have just seen that if M admits a non-positive sectional curvature metric, then π1(M) is torsion free. It turns out that if M admits a negative sectional curvature metric, then any nontrivial abelian subgroup of π1(M) is an infinite cyclic group generated by one element: Theorem 1.6 (Preissman). Let (M, g) be a compact Riemannian manifold with negative sectional curvature, and let {1} ̸= H ⊂ π1(M) be a nontrivial abelian subgroup of the fundamental group. Then H ∼= Z. Remark. The theorem was strengthened by Byers to: under the same assumption, any nontrivial solvable subgroup of π1(M) is infinite cyclic. Example. For any closed surface Mg of genus g ≥ 2, there is Riemannian metric of constant negative sectional curvature. The fundamental group of Mg is ⟨a1, b1, · · · , ag, bg | a1b1a −1 1 b −1 1 · · · a1b1a −1 1 b −1 1 = e⟩, which is not abelian, but all its abelian subgroups are isomorphic to Z. As an immediate consequence, we see Corollary 1.7. Suppose m ≥ 2. For any compact manifold M, T m × M admits no metric of negative sectional curvature. Remark. It was first proved by Gao and Yau in 1986 that any compact manifold of dimension 3 admits a metric with negative Ricci curvature. Then in 1994, Lohkamp proved that any manifold of dimension at least 3 admits a complete Riemannian metric of negative Ricci curvature. So there is no topological constraint for a manifold of dimension ≥ 3 to admit Riemannian metrics with negative Ricci curvature. The idea of proof is as follows: Realize Deck transformations associated with all α ∈ H as “a discrete family of translations along a fixed geodesic”. As a result, a nontrivial discrete subgroups of H corresponds to a nontrivial subgroup of R, which is isomorphic to Z
4LECTURE22:THEOREMSONCURVATUREV.S.TOPOLOGYTranslations in Cartan-Hadamard manifolds.Sowe need to introduce the concept of translation.Definition 1.8. Let (M,g) be a complete simply-connected Riemannian manifold.and : R → M a geodesic. An isometry f : (M,g) → (M,g) is called a translationalong if f has no fixed point, and f(Im() = Im().Let (M,g) be any complete Riemannian manifold and π : M → M be theuniversal covering. We endow with M the pull back metric g = *g. Recall that foreach element α E i(M),one can define a deck transformation fa: M M:for each p e M, there is a loop based at p = r(p) whose homotopyclass is α. Let be the lift of with starting point p. Define fa(p)betheendpointof.One can prove that. f is well-defined,· fa is an isometry,. fgo fa= fBa for all a,βE i(M),. fa has no fixed point if a + e.Now suppose e + α E πi(M), and let be a minimizing closed geodesic in thehomotopy class α. Let be a lift of to M. Then by definition is a geodesic in(M,9), and fa(Im()) =Im(). So we getLemma1.9.fa:(M,g)→(M,g)isatranslationalongforanye+aEi(M).Asaconsequenceof thislemma,weproveCorollary 1.1o. Suppose (M,g)has negative sectional curvature, then any trans-lation f:M-→M fires only one geodesicl.Proof. Suppose there are two geodesics % and %2 in M such that f(%) = %. Firstwe claimthatn2=0.Otherwiseeithern2=(p)forsomep,which impliesf(p) =p which is a contradiction (since f has no fixed point), or there are at leasttwo points in n2, which contradicts with Cartan-Hadamard theorem.Nowchoosep;E,andletandbetheminimizinggeodesicconnectingpi,p2 and connecting f(pi),p2 respectively.Let 4 = f(%)be the minimizing geodesicconnecting f(pi), f(p2). Since f is an isometry, the “corresponding angles"at piand at f(pi) are the same, similarly for p2. As a result, the angles of the two geodesictriangle pip2f(pi) and p2f(pi)f(p2) add up to at least 2, which contradicts with口Proposition 1.5.'Here different parametrizations will be viewed as the same
4 LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY ¶ Translations in Cartan-Hadamard manifolds. So we need to introduce the concept of translation. Definition 1.8. Let (M, g) be a complete simply-connected Riemannian manifold, and γ : R → M a geodesic. An isometry f : (M, g) → (M, g) is called a translation along γ if f has no fixed point, and f(Im(γ)) = Im(γ). Let (M, g) be any complete Riemannian manifold and π : Mf → M be the universal covering. We endow with Mf the pull back metric ˜g = π ∗ g. Recall that for each element α ∈ π1(M), one can define a deck transformation fα : Mf → Mf: for each ˜p ∈ Mf, there is a loop γ based at p = π(˜p) whose homotopy class is α. Let ˜γ be the lift of γ with starting point ˜p. Define fα(˜p) be the endpoint of ˜γ. One can prove that • fα is well-defined, • fα is an isometry, • fβ ◦ fα = fβα for all α, β ∈ π1(M), • fα has no fixed point if α ̸= e. Now suppose e ̸= α ∈ π1(M), and let γ be a minimizing closed geodesic in the homotopy class α. Let ˜γ be a lift of γ to Mf. Then by definition ˜γ is a geodesic in (M, f g˜), and fα(Im(˜γ)) = Im(˜γ). So we get Lemma 1.9. fα : (M, f g˜) → (M, f g˜) is a translation along γ˜ for any e ̸= α ∈ π1(M). As a consequence of this lemma, we prove Corollary 1.10. Suppose (M, g) has negative sectional curvature, then any translation f : Mf → Mf fixes only one geodesic1 . Proof. Suppose there are two geodesics ˜γ1 and ˜γ2 in Mf such that f(˜γi) = ˜γi . First we claim that ˜γ1 ∩ γ˜2 = ∅. Otherwise either ˜γ1 ∩ γ˜2 = {p˜} for some ˜p, which implies f(˜p) = ˜p which is a contradiction (since f has no fixed point), or there are at least two points in ˜γ1 ∩ γ˜2, which contradicts with Cartan-Hadamard theorem. Now choose ˜pi ∈ γ˜i , and let ˜γ3 and ˜γ5 be the minimizing geodesic connecting ˜p1, p˜2 and connecting f(˜p1), p˜2 respectively. Let ˜γ4 = f(˜γ3) be the minimizing geodesic connecting f(˜p1), f(˜p2). Since f is an isometry, the “corresponding angles” at ˜p1 and at f(˜p1) are the same, similarly for ˜p2. As a result, the angles of the two geodesic triangle ˜p1p˜2f(˜p1) and ˜p2f(˜p1)f(˜p2) add up to at least 2π, which contradicts with Proposition 1.5. □ 1Here different parametrizations will be viewed as the same
LECTURE 22:THEOREMSONCURVATUREV.S.TOPOLOGY5Proof of Preissman's theorem.As above we denote by M the universal covering of M, and f& the deck trans-formation described above associated to q E i(M).First fix Q E H and let be the geodesic that is invariant under fa.Then forany βeH, one has fBα=faβ sinceH is abelian.Sofs(Im()) = fe(fα(Im())) = fα(fs(Im()))By the corollary above, one must havefs(Im()) =Im(),Vβ EH.As a consequence, is invariant under all fa's for α EHNowwedenotepo=(o).Sinceis invariantunder fo,foreachβeH,thereis a unique tg ER so that(tp) = fs(po).Note that this implies(tp +t) = fe((t))for any t, since as t varies, both sides are geodesics with the same initial condition.Nowwedefineamap$: H→R,β(β) = tg.Claim l: is a group homomorphism:For any βi, β2 E H,(tB1+ tp2)= fB10fp2(po) = fB1β2(po)=(tB1B2)Sowehavep(Biβ2)=tpiB2 =tpi+tp2Claim 2: p is injective:Suppose p(β) = 0, then po =(O) = fe(po). So β =e E πi(M)Claim 3:The image of isnot dense in R.Pick a strongly convex geodesic ball U = Br(p) centered at p =π(po)so thatπ-1(U)=UsUs,where eachU.is diffeomorphic to Uunder the covering map : M → M and are disjoint. Denote Uobe the one so that po e Uo. Then for each β + e, fe(po) Uo. SoItpl = d(po, fe(po) ≥ rfor any β+e.As a consequence of the first two claims, H is an additive subgroup of R. Butwe know that any additive subgroup of R is either dense or infinite cyclic. So thetheorem is proved
LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY 5 ¶ Proof of Preissman’s theorem. As above we denote by Mf the universal covering of M, and fα the deck transformation described above associated to α ∈ π1(M). First fix α ∈ H and let ˜γ be the geodesic that is invariant under fα. Then for any β ∈ H, one has fβα = fαβ since H is abelian. So fβ(Im(˜γ)) = fβ(fα(Im(˜γ))) = fα(fβ(Im(˜γ))). By the corollary above, one must have fβ(Im(˜γ)) = Im(˜γ), ∀β ∈ H. As a consequence, ˜γ is invariant under all fα’s for α ∈ H. Now we denote ˜p0 = ˜γ(0). Since ˜γ is invariant under fβ, for each β ∈ H, there is a unique tβ ∈ R so that γ˜(tβ) = fβ(˜p0). Note that this implies γ˜(tβ + t) = fβ(˜γ(t)) for any t, since as t varies, both sides are geodesics with the same initial condition. Now we define a map φ : H → R, φ(β) = tβ. Claim 1: φ is a group homomorphism: For any β1, β2 ∈ H, γ˜(tβ1 + tβ2 ) = fβ1 ◦ fβ2 (˜p0) = fβ1β2 (˜p0) = ˜γ(tβ1β2 ). So we have φ(β1β2) = tβ1β2 = tβ1 + tβ2 . Claim 2: φ is injective: Suppose φ(β) = 0, then ˜p0 = ˜γ(0) = fβ(˜p0). So β = e ∈ π1(M). Claim 3: The image of φ is not dense in R. Pick a strongly convex geodesic ball U = Br(p) centered at p = π(˜p0) so that π −1 (U) = ∪δUδ, where each Uδ is diffeomorphic to U under the covering map π : Mf → M and are disjoint. Denote U0 be the one so that ˜p0 ∈ U0. Then for each β ̸= e, fβ(˜p0) ̸∈ U0. So |tβ| = d(˜p0, fβ(˜p0)) ≥ r for any β ̸= e. As a consequence of the first two claims, H is an additive subgroup of R. But we know that any additive subgroup of R is either dense or infinite cyclic. So the theorem is proved
6LECTURE 22:THEOREMSONCURVATUREV.S.TOPOLOGY2.COMPLETERIEMANNIAN MANIFOLDS WITHPOSITIVE CURVATURENow let's turn to Riemannian manifolds with positive curvature.I Synge's Theorem.Another application of the second variationformula to Riemannian manifoldswith positive curvature isTheorem 2.1 (Synge).Let (M,g) be a compact Riemannian manifold with positivesectional curvature.(1) If M is even dimensional and orientable, then M is simply connected.(2) If M is odd dimensional, then M is orientable.Since Rpm admits a positive sectional curvature metric, given the fact “i(RPm) Z, for m ≥ 2", we conclude that “Rpm is orientable if and only if m is odd"Corollary 2.2.If (M,g)is a compact even dimensional Riemannian manifold ofpositive sectional curvature, and M is not orientable, then Ti(M)= Z2Proof. Let M be the orientable double covering of M, endowed with the inducedpull-back metric.Then M is orientable and satisfies all the conditions in Synge口Theorem. It follows that M is simply connected and thus i(M) = Z2.As a consequence, Rp2 × Rp2 admits no metric of positive sectional curvaturesince i(RIp2× RIp2)= (Z)?.Weremark that it is still unknown whether S?×$2admits a positive sectional curvature metric: this is the well-known Hopf conjectureRemark. In the odd dimensional case we cannot say too much of its fundamentalgroup.For example,foreachk thereis alens space S3/Zthathas constant sectionalcurvature 1 and fundamental group Z.I Proof of Synge's Theorem.WefirstproveLemma2.3. Let (M,g) be an orientableRiemannian manifold, and: Ja,bl→Mbe a smooth loop, i.e. (a) = (b) := p. Then the parallel transport Pa,b : T,M -)T,M has determinant 1.Proof. In Lecture 6 we have seen P O(T,M). It remains to show det P, > 0.Forthispurposewetakeapositivem-formwonM,andletfe,lbeapositivebasisof T,M, i.e. w(e1,..,em) > 0. Let e;(t) = Pat(e;) be the parallel transport of [et)along . Thenw(ei(t),...,em(t)) +0for all t. It follows that w(ei(b), -. ,em(b)) > 0. Butw(ei(b),...,em(b)) = (det Pab)w(ei,...,em),口so we must have det Pa > 0
6 LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY 2. Complete Riemannian manifolds with positive curvature Now let’s turn to Riemannian manifolds with positive curvature. ¶ Synge’s Theorem. Another application of the second variation formula to Riemannian manifolds with positive curvature is Theorem 2.1 (Synge). Let (M, g) be a compact Riemannian manifold with positive sectional curvature. (1) If M is even dimensional and orientable, then M is simply connected. (2) If M is odd dimensional, then M is orientable. Since RP m admits a positive sectional curvature metric, given the fact “π1(RP m) ∼= Z2 for m ≥ 2”, we conclude that “RP m is orientable if and only if m is odd”. Corollary 2.2. If (M, g) is a compact even dimensional Riemannian manifold of positive sectional curvature, and M is not orientable, then π1(M) ∼= Z2. Proof. Let M be the orientable double covering of M, endowed with the induced pull-back metric. Then M is orientable and satisfies all the conditions in Synge Theorem. It follows that M is simply connected and thus π1(M) ∼= Z2. □ As a consequence, RP2 × RP2 admits no metric of positive sectional curvature since π1(RP2 × RP2 ) ∼= (Z2) 2 . We remark that it is still unknown whether S 2 × S 2 admits a positive sectional curvature metric: this is the well-known Hopf conjecture. Remark. In the odd dimensional case we cannot say too much of its fundamental group. For example, for each k there is a lens space S 3/Zk that has constant sectional curvature 1 and fundamental group Zk. ¶ Proof of Synge’s Theorem. We first prove Lemma 2.3. Let (M, g) be an orientable Riemannian manifold, and γ : [a, b] → M be a smooth loop, i.e. γ(a) = γ(b) := p. Then the parallel transport P γ a,b : TpM → TpM has determinant 1. Proof. In Lecture 6 we have seen P γ a,b ∈ O(TpM). It remains to show det P γ a,b > 0. For this purpose we take a positive m-form ω on M, and let {ei} be a positive basis of TpM, i.e. ω(e1, · · · , em) > 0. Let ej (t) = P γ a,t(ej ) be the parallel transport of {ei} along γ. Then ω(e1(t), · · · , em(t)) ̸= 0 for all t. It follows that ω(e1(b), · · · , em(b)) > 0. But ω(e1(b), · · · , em(b)) = (det P γ a,b)ω(e1, · · · , em), so we must have det P γ a,b > 0. □
7LECTURE22:THEOREMSONCURVATUREV.S.TOPOLOGYProof of Synge's Theorem. (1) Suppose M is not simply connected.Then thereexists a nontrivial closed geodesic :[o,1] M which is length minimizing in itsfree homotopy class.Sincethe parallel transport Po.iESO(T,M)and satisfiesP0.1(%(0)) = %(0),we can find Xp EEp= ((O))- such thatPo,i(Xp) = Xp.(Here, we used the condition that dim M is even, so that dim E is odd!)NowletX(t) betheparallel vectorfield along with X(O)=X,.ThenX(1) = Po1(Xp) = Xp.Thus for the variation s of whose variation field is X, we haved?E(%) :(R(,X)-V,V,X,X)dt =R(,X,,X)dt 0 such thatRic(Xp)≥(m-1)k,VX,ESM.Then M is compact, and its diameter is bounded byTdiam(M) := sup dist(p,q)≤VRP,qEM
LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY 7 Proof of Synge’s Theorem. (1) Suppose M is not simply connected. Then there exists a nontrivial closed geodesic γ : [0, 1] → M which is length minimizing in its free homotopy class. Since the parallel transport P γ 0,1 ∈ SO(TpM) and satisfies P γ 0,1 ( ˙γ(0)) = ˙γ(0), we can find Xp ∈ Ep = ( ˙γ(0))⊥ such that P γ 0,1 (Xp) = Xp. (Here, we used the condition that dim M is even, so that dim E is odd!) Now let X(t) be the parallel vector field along γ with X(0) = Xp. Then X(1) = P γ 0,1 (Xp) = Xp. Thus for the variation γs of γ whose variation field is X, we have d 2 ds2 s=0 E(γs) = Z 1 0 ⟨R( ˙γ, X) ˙γ − ∇γ˙ ∇γ˙ X, X⟩dt = Z 1 0 R( ˙γ, X, γ, X˙ ⟩dt 0 such that Ric(Xp) ≥ (m − 1)κ, ∀Xp ∈ SM. Then M is compact, and its diameter is bounded by diam(M) := sup p,q∈M dist(p, q) ≤ π √ κ
8LECTURE22:THEOREMSONCURVATUREV.S.TOPOLOGYRemarks. (1) One cannot weaken the condition on Ricci curvature to Ric > 0 orevenK>0.Forexample,considertheparaboloid[(r, ,2) e R3 / z = r? +y2]It is a surface of revolution with K > O, which is not compact.(2)The estimate is optimal in the following sense:Let M be the standard sphereof radius ， then it has Ricci curvature (m -1)k and diameter - (Note: thediameter here is not the standard diameter as a subset in Rm.)(3) We will prove the following result of S. Y. Cheng later: If (M,g) satisfiesthe conditions of the Bonnet-Myers theorem and diam(M)= , then (M,g) isisometricto the standard sphere of radiusCorollary 2.5. Let (M,g) be a complete Riemannian manifold whose Ricci curva-ture is bounded below by a positive number. Then i(M) is finite.Proof. Let M be the universal covering of M, endowed with the pull-back metricg =*g. Then (M,g) is also a complete Riemannian manifold whose Ricci curvatureis bounded below by a positive number. ByBonnet-Myers theorem,M is compact.口As a consequence, π : M -→ M has to be a finite covering. So πi(M) is finite.It particular, we see that if M,N are compact, i(M) is infinite, then M × Nadmits no Riemannian metric of positive Ricci curvature.I Proof of Bonnet-Myers Theorem.Proof. For any p,q e M, let : [0, 1] -→ M be a minimizing geodesic joining p to q.It's enough to show L()≤ (which implies compactness of M by Hopf-Rinow theorem).By contradiction, suppose that L() = l > . Let [ei(t)) be parallel vectorfields along which form an orthonormal basis at each point (t) and so thatei(t) - . For j = 2,..,m, we defineV,(t) = sin(πt)e;(t)Then V;(0) = Vi(1) = 0, and V,V,V, = -π? sin(t)e;(t). ThusI(V,V) =/(R(%,V)- VVV,V)dt =sin?(t)(?+[R(ei,ej,ei,ej)dtSummingover2≤j≤m-1,wegetI(Vi, V) =sin2(元t)(m - 1)元2 - [’Ric(e1)dt < 0=2So there exists some j ≥ 2 so that I(Vj, V) < 0. If follows that there exists q = (to)with 0 < to < 1 which is conjugate to p along . In particular, is not length口minimizing. A contradiction
8 LECTURE 22: THEOREMS ON CURVATURE V.S. TOPOLOGY Remarks. (1) One cannot weaken the condition on Ricci curvature to Ric > 0 or even K > 0. For example, consider the paraboloid {(x, y, z) ∈ R 3 | z = x 2 + y 2 }. It is a surface of revolution with K > 0, which is not compact. (2) The estimate is optimal in the following sense: Let M be the standard sphere of radius √ 1 κ , then it has Ricci curvature (m − 1)κ and diameter √π κ . (Note: the diameter here is not the standard diameter as a subset in R m.) (3) We will prove the following result of S. Y. Cheng later: If (M, g) satisfies the conditions of the Bonnet-Myers theorem and diam(M) = √π κ , then (M, g) is isometric to the standard sphere of radius √ 1 κ . Corollary 2.5. Let (M, g) be a complete Riemannian manifold whose Ricci curvature is bounded below by a positive number. Then π1(M) is finite. Proof. Let Mf be the universal covering of M, endowed with the pull-back metric g˜ = π ∗ g. Then (M, f g˜) is also a complete Riemannian manifold whose Ricci curvature is bounded below by a positive number. By Bonnet-Myers theorem, Mf is compact. As a consequence, π : Mf → M has to be a finite covering. So π1(M) is finite. □ It particular, we see that if M, N are compact, π1(M) is infinite, then M × N admits no Riemannian metric of positive Ricci curvature. ¶ Proof of Bonnet-Myers Theorem. Proof. For any p, q ∈ M, let γ : [0, 1] → M be a minimizing geodesic joining p to q. It’s enough to show L(γ) ≤ √π κ (which implies compactness of M by Hopf-Rinow theorem). By contradiction, suppose that L(γ) = l > √π κ . Let {ei(t)} be parallel vector fields along γ which form an orthonormal basis at each point γ(t) and so that e1(t) = γ˙ (t) l . For j = 2, · · · , m, we define Vj (t) = sin(πt)ej (t). Then Vj (0) = Vj (1) = 0, and ∇γ˙ ∇γ˙ Vj = −π 2 sin(πt)ej (t). Thus I(Vj , Vj ) = Z 1 0 ⟨R( ˙γ, Vj ) ˙γ − ∇γ˙ ∇γ˙ Vj , Vj ⟩dt = Z 1 0 sin2 (πt)(π 2 + l 2R(e1, ej , e1, ej ))dt. Summing over 2 ≤ j ≤ m − 1, we get Xm j=2 I(Vj , Vj ) = Z 1 0 sin2 (πt)((m − 1)π 2 − l 2Ric(e1))dt < 0. So there exists some j ≥ 2 so that I(Vj , Vj ) < 0. If follows that there exists ¯q = γ(t0) with 0 < t0 < 1 which is conjugate to p along γ. In particular, γ is not length minimizing. A contradiction. □