《黎曼几何》课程教学资源（英文讲义）Lec10 RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES.pdf_大学文库

LECTURE 1O: RIEMANNIAN MANIFOLDS WITH CONSTANTCURVATURESOn any smooth manifold there are numerous different Riemannian metrics, mostof which arenot interestingtous.Today wewill brieflydiscuss someresults onveryspecial Riemannian metrics, namely Riemannian metrics with constant curvatures(including sectional, Ricci, scalar and Einstein curvature).1. SCHUR'S THEOREM: FROM FIBER CONSTANT TO CONSTANTAs we have seen, the sectional curvature and the Ricci curvature are functionsnot defined on M itself, but defined on some fiber bundles over M, namely theGrassmannian 2-plane bundle Gr2(TM) and the sphere bundle SM.Before westudyRiemannianmanifolds with constant sectional or Ricci curvatures,let'sfirststudy an“intermediate"case, namely Riemannian manifolds whose sectional or Riccicurvatures arefiber-wise constant.It turns out that for connected Riemannianmanifolds of dimension m≥ 3, fiber-wise constant sectional/Ricci curvature willforcetheRiemannianmanifoldtohavegloballyconstantsectional/Riccicurvature.This result was first established by German mathematician F.Schur in 1886 (forthe sectional curvature case)IThe contracted Bianchi identity.The main tool in the proof of Schur's theorem is the second Bianchi identity(VRm)(U, V,X, Y,Z) + (VRm)(U, V,Y, Z, X) + (VRm)(U, V, Z, X,Y) = 0.We will first prove the stronger Ricci curvature version of Schur's theorem, for whichwhat we need isProposition1.1 (The contracted Bianchiidentity).For any Riemannian manifold,VS=2ci3VRc,where Cis is the metric contraction in the first and third entry.Proof.Since themetric contractions commutewith'V,wemay applymetric con-tractions to the Bianchi identity. Contracting the first and the third entries, thenwe note that the metric compatibility implies Vg* = 0, where g* = g*i,, is the “dual ofg".This in turn implies that the musical isomorphisms commute with the covariant derivativeV. Now consider the metric contraction ci,j that contracts the ith entry with the jth entry of a(O, k) tensor T, where 1 ≤ i + j< k. Since ci,j can be written as a composition of the standardcontraction Cj with a musical isomorphism, and since commutes with Cj, one can prove that also commutes with the metric contraction ci.j.1

LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES On any smooth manifold there are numerous different Riemannian metrics, most of which are not interesting to us. Today we will briefly discuss some results on very special Riemannian metrics, namely Riemannian metrics with constant curvatures (including sectional, Ricci, scalar and Einstein curvature). 1. Schur’s theorem: From fiber constant to constant As we have seen, the sectional curvature and the Ricci curvature are functions not defined on M itself, but defined on some fiber bundles over M, namely the Grassmannian 2-plane bundle Gr2(TM) and the sphere bundle SM. Before we study Riemannian manifolds with constant sectional or Ricci curvatures, let’s first study an “intermediate” case, namely Riemannian manifolds whose sectional or Ricci curvatures are fiber-wise constant. It turns out that for connected Riemannian manifolds of dimension m ≥ 3, fiber-wise constant sectional/Ricci curvature will force the Riemannian manifold to have globally constant sectional/Ricci curvature. This result was first established by German mathematician F. Schur in 1886 (for the sectional curvature case). ¶ The contracted Bianchi identity. The main tool in the proof of Schur’s theorem is the second Bianchi identity (∇Rm)(U, V, X, Y, Z) + (∇Rm)(U, V, Y, Z, X) + (∇Rm)(U, V, Z, X, Y ) = 0. We will first prove the stronger Ricci curvature version of Schur’s theorem, for which what we need is Proposition 1.1 (The contracted Bianchi identity). For any Riemannian manifold, ∇S = 2c1,3∇Rc, where c1,3 is the metric contraction in the first and third entry. Proof. Since the metric contractions commute with1 ∇, we may apply metric contractions to the Bianchi identity. Contracting the first and the third entries, then 1We note that the metric compatibility implies ∇g ∗ = 0, where g ∗ = g ij∂i∂j is the “dual of g”. This in turn implies that the musical isomorphisms commute with the covariant derivative ∇. Now consider the metric contraction ci,j that contracts the ith entry with the jth entry of a (0, k) tensor T, where 1 ≤ i ̸= j ≤ k. Since ci,j can be written as a composition of the standard contraction C i j with a musical isomorphism, and since ∇ commutes with C i j , one can prove that ∇ also commutes with the metric contraction ci,j . 1

LECTURE10:RIEMANNIANMANIFOLDSWITHCONSTANTCURVATURES32.RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURESI Manifolds with constant sectional curvatures.Now we study Riemannian manifolds with constant sectional curvatures,i.e.K(I)=k for all pEM and all I,ET,M.According to what weproved lasttime, (M,g) has constant curvature k if and only ifkRm=ggwhich is also equivalent to the fact “(M, g) has Weyl curvature tensor W = 0 andRiccicurvaturetensorRc=(m-1)kg"We have constructed,for any constant k, a simple Riemannian manifold whichhas constant sectional curvaturek,namely(a) (Sm, grouna) if k >0,(b) (Rm, go) if k = 0,(c) (Hm, -ghyperbolic) if k< 0.Of course there are many other constant sectional curvature manifolds, e.g.Any open subset in a Riemannian manifold of constant sectional curvatureis again a Riemannian manifold of constant sectional curvature. To makeour lives easier, we will exclude such examples3 by studying only connectedcomplete[i.e.when endowed with theRiemannian distance d,(M,d)is completeas:metric spacel Riemannian manifolds of constant sectional curvature, which areknownasspaceforms.. If (M,g) has constant sectional curvature, π : M → N is a smooth normal[i.e. the Deck transformation group acts freely on each fiber] covering map and gis invariant under all its Deck transformations, then (N, +g) has constantsectional curvature. Since universal cover is always normal, by this way wecan easily constructa constant positive sectional curvature metric on the real projectivespace Rpm = sm /Z2 and on the Lens space L(p, q),-aflat metric (constant curvature zerometric)on the torus Tm=Rm/Zm[in local coordinates dl, ..., gm on Tm = S' × ... × s1, the fat metric has the formg=doldol+...+dom@domja constant negative sectional curvature metric on any closed orientablesurface Zg of genus g ≥ 2.3Unfortunately it is not true that any constant sectional curvature Riemannian manifold is anopen submanifold of a complete constant sectional curvatureRiemannian manifold.Forexample,onemay start with 2(N,S] bethe standard sphere with the north/south poles removed, andconsider its universal covering (which is topologically R2 with pull-back Riemannian metric).Ac-cording to theKilling-Hopf theorembelow,themetric can'tbe complete.IncompleteRiemannianmanifolds arefarfrom well-understood

LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES 3 2. Riemannian manifolds with constant curvatures ¶ Manifolds with constant sectional curvatures. Now we study Riemannian manifolds with constant sectional curvatures, i.e. K(Πp) = k for all p ∈ M and all Πp ∈ TpM. According to what we proved last time, (M, g) has constant curvature k if and only if Rm = k 2 g○∧ g, which is also equivalent to the fact “(M, g) has Weyl curvature tensor W = 0 and Ricci curvature tensor Rc = (m − 1)kg”. We have constructed, for any constant k, a simple Riemannian manifold which has constant sectional curvature k, namely (a) (S m, 1 k ground) if k > 0, (b) (R m, g0) if k = 0, (c) (Hm, − 1 k ghyperbolic) if k < 0. Of course there are many other constant sectional curvature manifolds, e.g. • Any open subset in a Riemannian manifold of constant sectional curvature is again a Riemannian manifold of constant sectional curvature. To make our lives easier, we will exclude such examples3 by studying only ✿✿✿✿✿✿✿✿✿✿ connected ✿✿✿✿✿✿✿✿✿ complete [i.e. when endowed with the Riemannian distance d, (M, d) is complete as a metric space] Riemannian manifolds of ✿✿✿✿✿✿✿✿✿ constant ✿✿✿✿✿✿✿✿✿ sectional✿✿✿✿✿✿✿✿✿✿✿ curvature, which are known as space forms. • If (M, g) has constant sectional curvature, π : M → N is a smooth ✿✿✿✿✿✿✿ normal [i.e. the Deck transformation group acts freely on each fiber] covering map and g is invariant under all its Deck transformations, then (N, π∗g) has constant sectional curvature. Since universal cover is always normal, by this way we can easily construct – a constant positive sectional curvature metric on the real projective space RPm = S m/Z2 and on the Lens space L(p, q), – a flat metric (constant curvature zero metric) on the torus T m = R m/Z m, [in local coordinates θ 1 , · · · , θm on T m = S 1 × · · · × S 1 , the flat metric has the form g = dθ1 ⊗ dθ1 + · · · + dθm ⊗ dθm.] – a constant negative sectional curvature metric on any closed orientable surface Σg of genus g ≥ 2. 3Unfortunately it is not true that any constant sectional curvature Riemannian manifold is an open submanifold of a complete constant sectional curvature Riemannian manifold. For example, one may start with S 2 \ {N, S} be the standard sphere with the north/south poles removed, and consider its universal covering (which is topologically R 2 with pull-back Riemannian metric). According to the Killing-Hopf theorem below, the metric can’t be complete. Incomplete Riemannian manifolds are far from well-understood

ALECTURE 10:RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES. Conversely if (M, g) has constant sectional curvature, and : M → M is asmooth covering, then (M, r*g) has constant sectional curvature.SoitisreasonabletofocusfirstonsimplyconnectedcompleteRiemannianmanifoldsof constant sectional curvature, and the examples (a), (b), (c) above are all simplyconnected. Itturns out thatthey are the only ones,both locally (without completenessassumption)and globally (undertheassumption of completeness):Theorem 2.1 (Riemann). Let (M,g) be a Riemannian manifold with constant sec-tional curvature k, then any point p E M has a neighborhood that is isometric to anopen subset of (a) or (b) or (c).Theorem 2.2 (Killing-Hopf).Let (M,g) be a complete Riemannian manifold ofconstant sectional curvature k, then the Riemannian universal cover of (M,g) iseither (a) or (b) or (c)above (depending on the sign of k).We will postpone the proofs of both theorems to later.According to Killing-Hopf theorem, any complete Riemannian manifold of constant sectional curvature is the quotient of one of the three canonical examplesabove by a group (which is a subgroup of the corresponding isometry group) thatacts freely and properly discontinuously.As a result, only very few smooth man-ifolds can admit a constant sectional curvature metric. For example, there is noconstant sectional curvature metric on s? x s since its universal cover is s? × R,which is not one of the abovethree.In fact, we have seen all complete even dimensional Riemannian manifolds whichhas constant positive sectional curvature:Corollary 2.3. If (M,g) is a compact Riemannian manifolds of even dimensionm=2k, and ghas constantsectional curvature 1,then (M,g)is isometricto either(Sm, ground) or its quotient (RPm, g).Proof.Since M is compact, (M,g)is complete.ByKilling-Hopf theorem,(M,g)isthe quotient of (Sm,ground)bya subgroupT c Iso(Sm, ground) = O(m + 1)which acts on Sm freely and properly discontinuously.Now let I. If has an eigenvalue 1, then fixes a point in sm (which isthe unit eigenvector of ) and thus by freeness of the action, = Id.As a consequence, for Id e I, 1 is not an eigenvalue of We may considerits squarematrix, which is again an element inT.Sincem iseven,e SO(m+1)must has eigenvalue 1 and thus ? - Id. It follows that all eigenvalues of are -1,and thus = -Id.口So we must have T =[Id] or T = [Id], and the conclusion follows.Remark. The result fails in odd dimension, since as we have mentioned, all lensspaces admit constant sectional curvature Riemannian metric

4 LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES • Conversely if (M, g) has constant sectional curvature, and π : Mf → M is a smooth covering, then (M, π f ∗ g) has constant sectional curvature. So it is reasonable to focus first on simply connected complete Riemannian manifolds of constant sectional curvature, and the examples (a), (b), (c) above are all simply connected. It turns out that they are the only ones, both locally (without completeness assumption) and globally (under the assumption of completeness): Theorem 2.1 (Riemann). Let (M, g) be a Riemannian manifold with constant sectional curvature k, then any point p ∈ M has a neighborhood that is isometric to an open subset of (a) or (b) or (c). Theorem 2.2 (Killing-Hopf). Let (M, g) be a complete Riemannian manifold of constant sectional curvature k, then the Riemannian universal cover of (M, g) is either (a) or (b) or (c) above (depending on the sign of k). We will postpone the proofs of both theorems to later. According to Killing-Hopf theorem, any complete Riemannian manifold of constant sectional curvature is the quotient of one of the three canonical examples above by a group (which is a subgroup of the corresponding isometry group) that acts freely and properly discontinuously. As a result, only very few smooth manifolds can admit a constant sectional curvature metric. For example, there is no constant sectional curvature metric on S 2 × S 1 since its universal cover is S 2 × R, which is not one of the above three. In fact, we have seen all complete even dimensional Riemannian manifolds which has constant positive sectional curvature: Corollary 2.3. If (M, g) is a compact Riemannian manifolds of even dimension m = 2k, and g has constant sectional curvature 1, then (M, g) is isometric to either (S m, ground) or its quotient (RPm, g). Proof. Since M is compact, (M, g) is complete. By Killing-Hopf theorem, (M, g) is the quotient of (S m, ground) by a subgroup Γ ⊂ Iso(S m, ground) = O(m + 1) which acts on S m freely and properly discontinuously. Now let γ ∈ Γ. If γ has an eigenvalue 1, then γ fixes a point in S m (which is the unit eigenvector of γ) and thus by freeness of the action, γ = Id. As a consequence, for Id ̸= γ ∈ Γ, 1 is not an eigenvalue of γ. We may consider its square matrix γ 2 , which is again an element in Γ. Since m is even, γ 2 ∈ SO(m+1) must has eigenvalue 1 and thus γ 2 = Id. It follows that all eigenvalues of γ are −1, and thus γ = −Id. So we must have Γ = {Id} or Γ = {±Id}, and the conclusion follows. □ Remark. The result fails in odd dimension, since as we have mentioned, all lens spaces admit constant sectional curvature Riemannian metric

LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES 5 ¶ Spaces with constant Ricci Curvatures: Einstein manifolds. Now let’s turn to Riemannian manifolds with constant Ricci curvature, i.e. satisfying Ric(Xp) = k for any p ∈ M and any Xp ∈ SpM. It turns out that such manifolds play important roles in Einstein’s general theory of relativity(in a slightly different framework, i.e. pseudo-Riemannian geometry): The Einstein field equation (which, together with the geodesic equation that we will discuss later, form the core of the mathematical formulation of general relativity) has the form Rc − 1 2 Sg + Λg = κT, where S is the scalar curvature function, Λ is known as the cosmological constant4 , κ is the Einstein gravitational constant, and T is the so-called stress-energy tensor. In the case of ✿✿✿✿✿✿✿✿ vacuum where T = 0, the equation becomes Rc = (S 2 − Λ)g. According to Schur’s theorem, the function S 2 − Λ must be a constant, and thus (M, g) has constant Ricci curvature. Definition 2.4. We say a Riemannian manifold (M, g) is an Einstein manifold if there exists a constant λ such that Rc = λg. Einstein manifolds with λ = 0 are known as Ricci-flat manifolds. Obviously, if (M, g) has constant sectional curvature k, then (M, g) is an Einstein manifold since Rc = c(Rm) = (m − 1)kg. Since Tr(Rc) = S (the scalar curvature) and Tr(g) = m (the dimension of M), we conclude that the constant λ for an Einstein manifold must be λ = S m . Since the traceless Ricci tensor E = Rc − S m g, we conclude Corollary 2.5. (M, g) is an Einstein manifold if and only if E = 0. In particular if (M, g) is an Einstein manifold and W = 0, then (M, g) has constant sectional curvature. Since W = 0 in dimension 3, it follows Proposition 2.6. For m = 2 or 3, (M, g) is Einstein if and only if (M, g) has constant sectional curvature. 4According to Gamow, Einstein regard the introduction of the cosmological term as “the biggest blunder of his life”. So we name these manifolds as Einsteins manifolds as a punishment (joke)

6LECTURE 10:RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURESSo to find an Einstein manifold that is not of constant sectional curvature, onemustlook atmanifolds of dimension atleast 4.Todiscover“which manifold admitsan Einstein metric and which does not"is still a very active research topic today.Here is an example:Erample.Let M =Sm × Sm [ormoregenerallythe product of two m-dimensional Riemannianmanifolds that have the same constant sectional curvaturel,equipped with the product metricg=元igsm+元2gsm.Note that Sm has constant curvature 1, so that it is Einstein andRc(gsm)=(m-1)gsmItfollowsRc(g) = πRc(gsm) + 元2Rc(gsm) = (m - 1)元igsm +(m - 1)2gsm = (m - 1)gInotherwords, (M,g)is an Einsteinmanifold.On the other hand, (M,g) is not of constant sectional curvature for m > 1.This can be proved by using Killing-Hopf theorem, or by direct computation: for(p,q) e Sm × Sm, if we let ei,e2 be linearly independent vectors in T,sm and lete3ET.m,thenK(dt(ei), dt(e2)) = 1, K(dte(ei), dt)(e3)) = 0,where tq : Sm → Sm × Sm is the embedding that maps p to (p, q), while tp : Sm →Sm × Sm is the embedding that maps q to (p,q).Remark. Again there are many topological restrictions for a smooth manifold toadmit an Einstein metric. For example, by using the famous Chern-Gauss-Bonnettheorem.whichhasthefollowingformfororientableclosed4-manifolds.1(|Rm/2-4|Rc/2+$2)dv,x(M)= 32元2 Jwhere x(M) is the Euler characteristic of M, one can easily prove:Theorem (Berger). If M is an orientable closed 4-manifold and M admits an Ein-stein metric g, then x(M) ≥ O, and the equality holds if and only if g is flat.In particular, since x(s3× si)= 0 and s3× Si admits no flatmetric (byKilling-Hopf), we conclude that s3 × si admits no Einstein metric.The above result was further strengthened by Thorpe and Hitchin as follows:Theorem (Hitchin-Thorpe inequality). If M is an orientable closed 4-manifold andM admits anEinsteinmetricg,thenI-(M)I,x(M) ≥where T(M) is the signature of M. Moreover, if the equality holds, then the Einsteinmetric is a Ricci flat metric

6 LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES So to find an Einstein manifold that is not of constant sectional curvature, one must look at manifolds of dimension at least 4. To discover “which manifold admits an Einstein metric and which does not” is still a very active research topic today. Here is an example: Example. Let M = S m ×S m [or more generally the product of two m-dimensional Riemannian manifolds that have the same constant sectional curvature], equipped with the product metric g = π ∗ 1 gSm + π ∗ 2 gSm. Note that S m has constant curvature 1, so that it is Einstein and Rc(gSm) = (m − 1)gSm. It follows Rc(g) = π ∗ 1Rc(gSm) + π ∗ 2Rc(gSm) = (m − 1)π ∗ 1 gSm + (m − 1)π ∗ 2 gSm = (m − 1)g. In other words, (M, g) is an Einstein manifold. On the other hand, (M, g) is not of constant sectional curvature for m > 1. This can be proved by using Killing-Hopf theorem, or by direct computation: for (p, q) ∈ S m × S m, if we let e1, e2 be linearly independent vectors in TpS m and let e3 ∈ TqS m, then K(dι1 q (e1), dι1 q (e2)) = 1, K(dι1 q (e1), dι1 p (e3)) = 0, where ιq : S m → S m × S m is the embedding that maps p to (p, q), while ιp : S m → S m × S m is the embedding that maps q to (p, q). Remark. Again there are many topological restrictions for a smooth manifold to admit an Einstein metric. For example, by using the famous Chern-Gauss-Bonnet theorem, which has the following form for orientable closed 4-manifolds, χ(M) = 1 32π 2 Z M (|Rm| 2 − 4|Rc| 2 + S 2 )dv, where χ(M) is the Euler characteristic of M, one can easily prove: Theorem (Berger). If M is an orientable closed 4-manifold and M admits an Einstein metric g, then χ(M) ≥ 0, and the equality holds if and only if g is flat. In particular, since χ(S 3 × S 1 ) = 0 and S 3 × S 1 admits no flat metric (by Killing-Hopf), we conclude that S 3 × S 1 admits no Einstein metric. The above result was further strengthened by Thorpe and Hitchin as follows: Theorem (Hitchin–Thorpe inequality). If M is an orientable closed 4-manifold and M admits an Einstein metric g, then χ(M) ≥ 3 2 |τ (M)|, where τ (M) is the signature of M. Moreover, if the equality holds, then the Einstein metric is a Ricci flat metric

LECTURE 10:RIEMANNIANMANIFOLDS WITHCONSTANT CURVATURES>I Riemannian manifolds with constant scalar curvature.Now weturn toRiemannian manifoldswith constant scalar curvature.Thereare many such examples, e.g.. any Einstein manifold has constant scalar curvature,.the product of two Riemannian manifoldswith constant scalar curvaturesisagain a Riemannian manifold with constant scalar curvature since[exercise]Proposition 2.7. Let Si (i = 1,2) be the scalar curvature of (Mi, gi), and Sthe scalar curvature of (Mi × M2, πig1 +πg2). Then S(p,9) = Si(p)+ S2(g).For simplicity we only consider compact manifolds.It turns out that therealways exist lots of constant scalar curvature metrics on any compact manifold M.(1) Let's start with the case of dimension 2, i.e. S is a compact surface. Notethatin thiscase all curvatures arethesame.As wehaveseen,sincetheuniversal covering of S is either S? or R2 or D(the unit disc), M admits aconstant curvature metric. However, even in this case, one can say a lotmore: Suppose M is orientable (so that it has a Riemann surface structure), thenby lifting to the universal covering and using the famous uniformizationtheorem in complex analysis, one can proveTheorem. For any Riemannian metric g on an compact orientable surfaceS, there is u E Coo(S) so that (M,e2ug) has constant curvature.[Note that according to the Gauss-Bonnet theorem, in this case the sign of the constantcurvature depends on the topology of M.](2)In dimension m≥3,unlike thesectional curvature or the Ricci curvature, thescalar curvature encodes relativelyfew information.For example,Kazdanand Warner solved the prescribed scalar curvatureproblem and getTheorem (Kazdan-Warner). For any compact smooth manifold M of di-mension m≥3, eractly one of the following will happen:(a) For any f e Co(M),there erists a Riemannian metric g on M whosescalar curvature function is f.(b) Afunction f ECoo(M) is thescalar curvature of some Riemannianmetric on M if and only if either f = O, or f is negative somewhere.(c) A function f E Co(M) is the scalar curvature of some Riemannianmetric on M if and only if f is negative somewhere.As consequences we immediately get.any compact manifold of dimension m≥> 3 admits a Riemannian metricwhose scalar curvature is any given negative constant..ForanycompactmanifoldM,byProposition2.7themanifoldMxSm(where m ≥ 2) admits a positive scalar curvature metric, and thusany function on M x Sm can be realized as its scalar curvature

LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES 7 ¶ Riemannian manifolds with constant scalar curvature. Now we turn to Riemannian manifolds with constant scalar curvature. There are many such examples, e.g. • any Einstein manifold has constant scalar curvature, • the product of two Riemannian manifolds with constant scalar curvatures is again a Riemannian manifold with constant scalar curvature since[exercise] Proposition 2.7. Let Si (i = 1, 2) be the scalar curvature of (Mi , gi), and S the scalar curvature of (M1×M2, π∗ 1 g1+π ∗ 2 g2). Then S(p, q) = S1(p)+S2(q). For simplicity we only consider compact manifolds. It turns out that there always exist lots of constant scalar curvature metrics on any compact manifold M. (1) Let’s start with the case of dimension 2, i.e. S is a compact surface. Note that in this case all curvatures are the same. As we have seen, since the universal covering of S is either S 2 or R 2 or D(the unit disc), M admits a constant curvature metric. However, even in this case, one can say a lot more: Suppose M is orientable (so that it has a Riemann surface structure), then by lifting to the universal covering and using the famous ✿✿✿✿✿✿✿✿✿✿✿✿✿✿✿ uniformization ✿✿✿✿✿✿✿✿ theorem in complex analysis, one can prove Theorem. For any Riemannian metric g on an compact orientable surface S, there is u ∈ C ∞(S) so that (M, e2u g) has constant curvature. [Note that according to the Gauss-Bonnet theorem, in this case the sign of the constant curvature depends on the topology of M.] (2) In dimension m ≥ 3, unlike the sectional curvature or the Ricci curvature, the scalar curvature encodes relatively few information. For example, Kazdan and Warner solved the ✿✿✿✿✿✿✿✿✿✿✿ prescribed ✿✿✿✿✿✿ scalar✿✿✿✿✿✿✿✿✿✿✿ curvature ✿✿✿✿✿✿✿✿✿ problem and get Theorem (Kazdan-Warner). For any compact smooth manifold M of dimension m ≥ 3, exactly one of the following will happen: (a) For any f ∈ C ∞(M), there exists a Riemannian metric g on M whose scalar curvature function is f. (b) A function f ∈ C ∞(M) is the scalar curvature of some Riemannian metric on M if and only if either f ≡ 0, or f is negative somewhere. (c) A function f ∈ C ∞(M) is the scalar curvature of some Riemannian metric on M if and only if f is negative somewhere. As consequences we immediately get • any compact manifold of dimension m ≥ 3 admits a Riemannian metric whose scalar curvature is any given negative constant. • For any compact manifold M, by Proposition 2.7 the manifold M × S m(where m ≥ 2) admits a positive scalar curvature metric, and thus any function on M × S m can be realized as its scalar curvature

8LECTURE 10:RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURESOn the other hand, Atiyah-Singer index theorem gives topological obstruc-tions for the existence of positive scalar curvature metric.It is alsoknownthat any torus Tm admits no positive scalar curvature metric [Schoen-Yau for33,isthereuECoo(M)sothat(M,e2ug)hasconstantscalarcurvature?By computing the scalar curvature under conformal change, the problem isreduced to finding a positive solution to the partial differential equation4(m- 1)p=m-2Although there is a gap in Yamabe's origin solution (as pointed out by Trudinger),by combining the works of Yamabe(1960), Trudinger(1968), Aubin(1976) andSchoen(1984), the answer is YES. In particular, for any compact manifoldof dimension m ≥ 3, there exist lots of constant scalar curvature metrics.RiemannianmanifoldswithW=O(locallyconformallyflatmanifolds)Finally wereturn to constantsectional curvaturemetrics,butintheframeworkof conformal geometry. Of course globally there are topological restrictions for theexistence (by Killing-Hopf theorem), so we restrict ourselves to the local setting,i.e. study Riemannian manifolds (M, g) so that near each point p, there is an openneighborhood U and a smooth function u E C(U) so that the metric (U,e2"g) hasconstant sectional curvature. According to Theorem 2.1, if (U,e2"g) has constantsectional curvature,then by shrinking U if needed, themanifold (U,e2"g)is isometricto an open subset of one of the three canonical constant sectional curvature spaces,namely the Euclidean space or the sphere or the hyperbolic space (with scaledmetric). Let's take a closer look of there three metrics. For Rm and Hm we haveforRmgo=drldel+...+darmdamforHm(drdrl+...+dam?dem)ghyperbolic(rm)2while for Sm, we may use stereographic coordinates (t',..: ,tm) [so that any pointwhich is not the north pole on Sm can be written as (】toget[exercise]4forSm(dtl?dtl+...+dtm?dtm)ground(1 + |t/2)25One may pose the same problem for complete noncompact manifolds: The answer is no ingeneral, and counterexamples were constructed by Jin (1988).The problem of finding conditionsunder which the problem has a solution is still a topic of research today

8 LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES On the other hand, Atiyah-Singer index theorem gives topological obstructions for the existence of positive scalar curvature metric. It is also known that any torus T m admits no positive scalar curvature metric [Schoen-Yau for 3 ≤ m ≤ 7(related to the positive mass theorem), Gromov-Lawson for general m]. (3) Since for surfaces one can always find a constant scalar curvature metric in any given conformal class, it is natural to ask whether the same result holds in higher dimension [a classical problem in geometric analysis]: The Yamabe prolbem. Given a compact5 Riemannian manifold (M, g) of dimension m ≥ 3, is there u ∈ C ∞(M) so that (M, e2u g) has constant scalar curvature? By computing the scalar curvature under conformal change, the problem is reduced to finding a positive solution to the partial differential equation 4(m − 1) m − 2 ∆φ + Sφ = λφ m+2 m−2 . Although there is a gap in Yamabe’s origin solution (as pointed out by Trudinger), by combining the works of Yamabe(1960), Trudinger(1968), Aubin(1976) and Schoen(1984), the answer is YES. In particular, for any compact manifold of dimension m ≥ 3, there exist lots of constant scalar curvature metrics. ¶ Riemannian manifolds with W = 0 (locally conformally flat manifolds). Finally we return to constant sectional curvature metrics, but in the framework of conformal geometry. Of course globally there are topological restrictions for the existence (by Killing-Hopf theorem), so we restrict ourselves to the local setting, i.e. study Riemannian manifolds (M, g) so that near each point p, there is an open neighborhood U and a smooth function u ∈ C ∞(U) so that the metric (U, e2u g) has constant sectional curvature. According to Theorem 2.1, if (U, e2u g) has constant sectional curvature, then by shrinking U if needed, the manifold (U, e2u g) is isometric to an open subset of one of the three canonical constant sectional curvature spaces, namely the Euclidean space or the sphere or the hyperbolic space (with scaled metric). Let’s take a closer look of there three metrics. For R m and Hm we have g0 = dx1 ⊗ dx1 + · · · + dxm ⊗ dxm for R m ghyperbolic = 1 (xm) 2 dx1 ⊗ dx1 + · · · + dxm ⊗ dxm for H m while for S m, we may use stereographic coordinates (t 1 , · · · , tm) [so that any point which is not the north pole on S m can be written as ( 2t 1 1+|t| 2 , · · · , 2tm 1+|t| 2 , 1−|t| 2 1+|t| 2 )] to get [exercise] ground = 4 (1 + |t| 2 ) 2 dt1 ⊗ dt1 + · · · + dtm ⊗ dtm for S m. 5One may pose the same problem for complete noncompact manifolds. The answer is no in general, and counterexamples were constructed by Jin (1988). The problem of finding conditions under which the problem has a solution is still a topic of research today

LECTURE1O:RIEMANNIANMANIFOLDSWITHCONSTANTCURVATURES9So in all these three cases, after multiplying a conformal factor one gets a flatmetric.In other words,"locally conformally constant sectional curvatureis thesame as“"locally conformally flat":Definition 2.8. We say a Riemannian manifold (M,g) is locally conformally fat iffor any p E M, there is a neighborhood U of p and a smooth function u E Co(U)so that the metric g = e2ug is flat in U.Erample. We give a couple examples:. Any constant sectional curvature space is locally conformally flat(as just explained)..Any surface (with any real analytic Riemannian metric)is locally conformally flat:Theorem (Gauss, 1822). On any surface with real analytic Riemannian metricg, there erists coordinates r,y and smooth function u = u(r,y) so thatg =e2u(drdr +dy@dy)[Such coordinates are known as isothermal coordinates.]To find out necessary conditions for a metric g to be locally conformally fat, weneed tofind local quantitiesthat are invariant under a conformal change of metric.Lemma 2.9. If g =e2g, then W =e2uW.Proof.InPSet2wewill seeRm=e2u(Rm-gT)for somesymmetric2-tensorT.Decomposing both Rm and Rm intwo Weyl and non-Weyl parts wegetW-e2uW=gOT,for some symmetric 2-tensor T. So W - e2uW E ker(c) n Im(). By Corollary 2.10口in Lecture 8, we must have W -e2uW = 0.Since any flat metric has Weyl curvature tensor W - O, we concludeCorollary 2.10. If (M,g) is locally conformally flat, then W = 0.It turns out that the condition is alsonecessary form≥4:Theorem (Weyl-Schouten).Let (M,g) be a Riemannian manifold(1) For m ≥ 4, (M,g) is locally conformally flat if and only if W - 0.(2) For m = 3, (M, g) is locally conformally flat if and only if [A is the Schouten tenser](VxA)(Y,Z) - (VA)(X,Z) = 0, VX,Y,Z ET(TM).[The proof will be left as part of one possible final project.]For example, by calculating the Weyl curvature tensor one can easily prove thatwhen endowed with the standard product metric, both $? × $? and $? × R? are notlocally conformally flat, while Smi × Hm2 is locally conformally flat

LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES 9 So in all these three cases, after multiplying a conformal factor one gets a flat metric. In other words, “locally conformally constant sectional curvature” is the same as “locally conformally flat”: Definition 2.8. We say a Riemannian manifold (M, g) is locally conformally flat if for any p ∈ M, there is a neighborhood U of p and a smooth function u ∈ C ∞(U) so that the metric ¯g = e 2u g is flat in U. Example. We give a couple examples: • Any constant sectional curvature space is locally conformally flat(as just explained). • Any surface (with any real analytic Riemannian metric) is locally conformally flat: Theorem (Gauss, 1822). On any surface with real analytic Riemannian metric g, there exists coordinates x, y and smooth function u = u(x, y) so that g = e 2u (dx ⊗ dx + dy ⊗ dy). [Such coordinates are known as isothermal coordinates.] To find out necessary conditions for a metric g to be locally conformally flat, we need to find local quantities that are invariant under a conformal change of metric. Lemma 2.9. If g¯ = e 2u g, then W = e 2uW. Proof. In PSet 2 we will see Rm = e 2u (Rm − g○∧ T) for some symmetric 2-tensor T. Decomposing both Rm and Rm intwo Weyl and non-Weyl parts we get W − e 2uW = g○∧ T , e for some symmetric 2-tensor Te. So W − e 2uW ∈ ker(c) ∩ Im(Ψ). By Corollary 2.10 in Lecture 8, we must have W − e 2uW = 0. □ Since any flat metric has Weyl curvature tensor W = 0, we conclude Corollary 2.10. If (M, g) is locally conformally flat, then W = 0. It turns out that the condition is also necessary for m ≥ 4: Theorem (Weyl-Schouten). Let (M, g) be a Riemannian manifold. (1) For m ≥ 4, (M, g) is locally conformally flat if and only if W = 0. (2) For m = 3, (M, g) is locally conformally flat if and only if [A is the Schouten tenser] (∇XA)(Y, Z) − (∇Y A)(X, Z) = 0, ∀X, Y, Z ∈ Γ ∞(TM). [The proof will be left as part of one possible final project.] For example, by calculating the Weyl curvature tensor one can easily prove that when endowed with the standard product metric, both S 2 × S 2 and S 2 × R 2 are not locally conformally flat, while S m1 × Hm2 is locally conformally flat