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
2LECTURE10:RIEMANNIANMANIFOLDSWITHCONSTANTCURVATUREScontract the second and fourth entries:0=(VRm)=→0 = c2,4C1,3 (VRm).O3,4,5O3,4,5Note thatCv,yCu,x(VRm)(U, V, X, Y,Z) = V(cv,ycU,x Rm)(U, V,X, Y, Z) = (VS)(Z),whilecv,ycu,x(VRm)(U, V,Y, Z, X) = -cu,x(Vcvy Rm)(U, V,Z, Y, X)= -Cu,x(VRc)(U, Z, X)andCvyCu,x(VRm)(U, V,Z, X, Y) = -cvyV(cux Rm)(U, V,X, z,Y)= -cv(VRc)(V, Z, Y)So we arrived atVS = 2c1,3VRc,口which completes the proof.In local coordinates, the contracted Bianchi identity can be written asOS = 2g" Rcik:j.I Schur's theorem.NowwearereadytoproveTheorem 1.2 (Schur). Let (M,g) be a connected Riemannian manifold of dimen-sion m > 3.(1) If Ric(X,)=f(p) depends only on p,then (M,g) has constant Ricci curvature(2)If K(II,)=f(p) depends only on p,then (M,g)has constant sectional curvature.Proof. (1) Under the assumption we have Rcp = f(p)gp. It followsS(p) = Tr(Rcp) = f(p)Tr(gp) = mf(p).So by the contracted Bianchi identity and the fact Vg = 0 (which implies gij;k = 0),mOkf =OkS=2gRcik:j=2g(fg)ikj=2g(0,f)gik =20xf.It follows that Ouf = O for any k and thus f is a constant.(2) If K(II,) = f(p), thenRic(Xp) = (m - 1)f(p)口So by (1), f is constant.2Remark.Obviouslythetheoremfailsindimension2,inwhichcasethesectional/Riccicurvature is always a function on M but need not be a constant.2we will give another direct proof of this fact using moving frames next time
2 LECTURE 10: RIEMANNIAN MANIFOLDS WITH CONSTANT CURVATURES contract the second and fourth entries: 0 = X ⃝3,4,5 (∇Rm) =⇒ 0 = c2,4c1,3 X ⃝3,4,5 (∇Rm). Note that cV,Y cU,X(∇Rm)(U, V, X, Y, Z) = ∇(cV,Y cU,XRm)(U, V, X, Y, Z) = (∇S)(Z), while cV,Y cU,X(∇Rm)(U, V, Y, Z, X) = −cU,X(∇cV,Y Rm)(U, V, Z, Y, X) = −cU,X(∇Rc)(U, Z, X) and cV,Y cU,X(∇Rm)(U, V, Z, X, Y ) = −cV,Y ∇(cU,XRm)(U, V, X, Z, Y ) = −cV,Y (∇Rc)(V, Z, Y ) So we arrived at ∇S = 2c1,3∇Rc, which completes the proof. □ In local coordinates, the contracted Bianchi identity can be written as ∂kS = 2g ijRcik;j . ¶ Schur’s theorem. Now we are ready to prove Theorem 1.2 (Schur). Let (M, g) be a connected Riemannian manifold of dimension m ≥ 3. (1) If Ric(Xp) = f(p) depends only on p, then (M, g) has constant Ricci curvature. (2) If K(Πp) = f(p) depends only on p, then (M, g) has constant sectional curvature. Proof. (1) Under the assumption we have Rcp = f(p)gp. It follows S(p) = Tr(Rcp) = f(p)Tr(gp) = mf(p). So by the contracted Bianchi identity and the fact ∇g = 0 (which implies gij;k = 0), m∂kf = ∂kS = 2g ijRcik;j = 2g ij (fg)ik;j = 2g ij (∂jf)gik = 2∂kf. It follows that ∂kf = 0 for any k and thus f is a constant. (2) If K(Πp) = f(p), then Ric(Xp) = (m − 1)f(p). So by (1), f is constant.2 □ Remark. Obviously the theorem fails in dimension 2, in which case the sectional/Ricci curvature is always a function on M but need not be a constant. 2We will give another direct proof of this fact using moving frames next time
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
LECTURE10:RIEMANNIANMANIFOLDSWITHCONSTANTCURVATURES5I Spaces with constant Ricci Curvatures: Einstein manifolds.Now let's turn to Riemannian manifolds with constant Ricci curvature, i.e. sat-isfying Ric(X,) =k for any p E M and any X, E SpM. It turns out that suchmanifolds play important roles in Einstein's general theory of relativity(in a slightlydifferent 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 mathematicalformulation of general relativity) has the form1RcSg+Ag=kT,2whereS is the scalar curvature function,Aisknown as the cosmological constantk is the Einstein gravitational constant, and T is the so-called stress-energy tensor.In the case of yacuum where T = O, the equation becomesSRc=- A)g2According to Schur's theorem, the function 号-A 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 ifthere exists a constant such thatRc= Λg.Einstein manifolds with = O are known as Ricci-flat manifolds.Obviously,if (M,g) has constant sectional curvature k, then (M,g)is an Einsteinmanifold sinceRc = c(Rm) = (m -1)kg.Since Tr(Rc) = S (the scalar curvature) and Tr(g) = m (the dimension of M), weconclude that the constant ^ for an Einstein manifold must bes入二mSince the tracelessRicci tensor E=Rc-g,we concludeCorollary 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 = O, then (M,g) hasconstant sectional curvature.Since W =0 in dimension 3,it followsProposition2.6.Form=2 or3,(M,g)isEinstein if and onlyif (M,g)hasconstant sectional curvature.4According to Gamow, Einstein regard the introduction of the cosmological term as“the biggestblunder of his life". So we name these manifolds as Einsteins manifolds as a punishment (joke)
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)