LECTURE8:THERIEMANNIANCURVATURE1.THERIEMANNCURVATURETENSORI The Riemann curvature tensor of type (O, 4).Given any linear connection on M, one gets a type (1,3) curvature tensor RR(X,Y)Z = VxVyZ-VyVxZ-V(x,Mzwhich measures the non-commutativity of “second order/iterated covariant deriva-tives". Locally one may write R asR=Rukdr dridrk oNow suppose (M, g) is a Riemannian manifold and is the Levi-Civita con-nection. By using the Riemannian metric g (via the musical isomorphism) one canconvert the (1,3)-tensor R to a (0,4)-tensor Rm EF(0,4TM) defined byRm(X,Y,Z, W) := -g(R(X,Y)Z,W).Definition 1.1. We call Rm the Riemann curvature tensor of (M,g)LocallyifwewriteRm=Rikdr'drjdrkdr!thenRijkl = Rm(Oi,Oj,Ok,O) =-g(RikmOm,O) =-miRikmIn other words, the Riemannian metric lower one of the the index".Erample.For Sm (equipped with thestandard round metric),wehaveseenR(X,Y)Z = (Y,Z)X - (X,Z)Y.Thus the Riemann curvature tensor isRm(X,Y, Z,W) = -(Y,Z)(X,W) + (X,Z)(Y,W)Now we introduce the Kulkarni-Nomizu product that converts 2 symmetric (0,2)tensors Ti and T2 into one (0,4)-tensorTiT2defined by(Ti@T2)(X,Y, Z, W) :=Ti(X,Z)T2(Y, W) + T(Y,W)T2(X, Z)- Ti(X, W)T2(Y,Z) - Ti(Y, Z)T2(X, W)As a result, we get a very brief expression for the Riemann curvature tensor of Sm,Rm=9g?
LECTURE 8: THE RIEMANNIAN CURVATURE 1. The Riemann curvature tensor ¶ The Riemann curvature tensor of type (0, 4). Given any linear connection ∇ on M, one gets a type (1, 3) curvature tensor R R(X, Y )Z = ∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z which measures the non-commutativity of “second order/iterated covariant derivatives”. Locally one may write R as R = Rijk l dxi ⊗ dxj ⊗ dxk ⊗ ∂l . Now suppose (M, g) is a Riemannian manifold and ∇ is the Levi-Civita connection. By using the Riemannian metric g (via the musical isomorphism) one can convert the (1, 3)-tensor R to a (0, 4)-tensor Rm ∈ Γ ∞(⊗0,4TM) defined by Rm(X, Y, Z, W) := −g(R(X, Y )Z, W). Definition 1.1. We call Rm the Riemann curvature tensor of (M, g). Locally if we write Rm = Rijkldxi ⊗ dxj ⊗ dxk ⊗ dxl , then Rijkl = Rm(∂i , ∂j , ∂k, ∂l) = −g(R m ijk ∂m, ∂l) = −gmlR m ijk . In other words, the Riemannian metric “lower one of the the index”. Example. For S m (equipped with the standard round metric), we have seen R(X, Y )Z = ⟨Y, Z⟩X − ⟨X, Z⟩Y. Thus the Riemann curvature tensor is Rm(X, Y, Z, W) = −⟨Y, Z⟩⟨X, W⟩ + ⟨X, Z⟩⟨Y, W⟩. Now we introduce the Kulkarni-Nomizu product ○∧ that converts 2 symmetric (0, 2)- tensors T1 and T2 into one (0, 4)-tensor T1○∧ T2 defined by (T1○∧ T2)(X, Y, Z, W) :=T1(X, Z)T2(Y, W) + T1(Y, W)T2(X, Z) − T1(X, W)T2(Y, Z) − T1(Y, Z)T2(X, W). As a result, we get a very brief expression for the Riemann curvature tensor of S m, Rm = 1 2 g○∧ g. 1
2LECTURE8:THERIEMANNIANCURVATUREI Symmetries of Rm.By definition the (1,3)-tensor R admits the anti-symmetryR(X,Y)Z = -R(Y,X)Z.Moreover, if is torsion free, then the curvature tensor R admits two more cyclicsymmetry, namely the first Bianchi identityR(X,Y)Z + R(Y, Z)X + R(Z, X)Y = 0and the second Bianchi identity(VxR)(Y, Z, W) + (VR)(Z, X,W) + (VzR)(X,Y,W) = 0Obviously one can convert the symmetries of R to symmetries of Rm,namely(1)Rm(X,Y,Z, W) + Rm(Y,X, Z, W) = 0,the first Bianchi identity(2)Rm(X,Y,Z, W) + Rm(Y, Z,X, W) + Rm(Z, X,Y,W) = 0,and the second Bianchi identity(3)(VxRm)(Y,Z, W, V) + (VyRm)(Z, X, W,V) + (VzRm)(X,Y,W,V) = 0,or in local coordinates asRijkl +Rjikl =0,Rijkl +Rikil +Rkijl=0Rijklin +Rinkli+Rnikl:j=0wherewedenoteRijkl:n=(Va,R)(Oi,O,Ok,O).By staring at the Riemann curvature tensor Rm of the standard Sm, we mayfind more (anti-)symmetries than the ones we have seen, e.g. one can exchange zwith W to get a negative sign, or even exchange X,Y with Zz, W. In fact these two(anti-)symmetries are consequences of metric compatibility, and thus hold for anyRiemannian manifold:Proposition 1.2.The Riemann curvature tensor Rm satisfies(4)Rm(X,Y,Z,W) = -Rm(X,Y,W, Z),and(5)Rm(X,Y,Z,W) = Rm(Z, W,X,Y).Proof. For simplicity we denote f =(Z, Z), then by metric compatibility,(VxZ,Z) =Xf -(Z, VxZ)in other words(VxZ, Z) =Xf
2 LECTURE 8: THE RIEMANNIAN CURVATURE ¶ Symmetries of Rm. By definition the (1, 3)-tensor R admits the anti-symmetry R(X, Y )Z = −R(Y, X)Z. Moreover, if ∇ is torsion free, then the curvature tensor R admits two more cyclic symmetry, namely the first Bianchi identity R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0. and the second Bianchi identity (∇XR)(Y, Z, W) + (∇Y R)(Z, X, W) + (∇ZR)(X, Y, W) = 0. Obviously one can convert the symmetries of R to symmetries of Rm, namely (1) Rm(X, Y, Z, W) + Rm(Y, X, Z, W) = 0, the first Bianchi identity (2) Rm(X, Y, Z, W) + Rm(Y, Z, X, W) + Rm(Z, X, Y, W) = 0, and the second Bianchi identity (3) (∇XRm)(Y, Z, W, V ) + (∇Y Rm)(Z, X, W, V ) + (∇ZRm)(X, Y, W, V ) = 0, or in local coordinates as Rijkl + Rjikl = 0, Rijkl + Rjkil + Rkijl = 0, Rijkl;n + Rjnkl;i + Rnikl;j = 0. where we denote Rijkl;n = (∇∂n R)(∂i , ∂j , ∂k, ∂l). By staring at the Riemann curvature tensor Rm of the standard S m, we may find more (anti-)symmetries than the ones we have seen, e.g. one can exchange Z with W to get a negative sign, or even exchange X, Y with Z, W. In fact these two (anti-)symmetries are consequences of metric compatibility, and thus hold for any Riemannian manifold: Proposition 1.2. The Riemann curvature tensor Rm satisfies (4) Rm(X, Y, Z, W) = −Rm(X, Y, W, Z), and (5) Rm(X, Y, Z, W) = Rm(Z, W, X, Y ). Proof. For simplicity we denote f = ⟨Z, Z⟩, then by metric compatibility, ⟨∇XZ, Z⟩ = Xf − ⟨Z, ∇XZ⟩, in other words, ⟨∇XZ, Z⟩ = 1 2 Xf
3LECTURE8:THERIEMANNIANCURVATUREItfollowsX(Yf)-(Vyz,Vxz)(VxVyZ,Z)=X(VyZ,Z)-(VyZ,VxZ)=So-Rm(X,Y,Z,Z) = (R(X,Y)Z,Z) = (VxVyZ - VyVxZ - V(x,yZ,Z)X(Yf) -Y(Xf)-[X,Y]f = 0.2As a consequence, we getRm(X,Y,Z,W) + Rm(X,Y,W,Z)=Rm(X,Y,Z + W,Z + W) - Rm(X,Y,Z,Z)- Rm(X,Y,W,W) = 0.which implies (4).The equation (5) is a consequence of (4) first one together with (1) and (2). Infact, by the first Bianchi identity (2)one hasRm(X,Y.Z.W)+ Rm(Y.Z.X.W)+ Rm(Z.X,Y.W) =0.Rm(Y,Z, W,X) + Rm(Z, W,Y,X) + Rm(W,Y,Z,X) = 0.Rm(Z,W,X,Y) + Rm(W,X,Z,Y) + Rm(X,Z, W,Y) = 0,Rm(W, X,Y, Z) + Rm(X,Y, W,Z) + Rm(Y, W, X,Z) = 0.Adding these equations and using (1) and (4), we getRm(Z,X,Y,W) + Rm(W,Y,Z,X) = 0,口which is equivalent to (5).By using (5)we may rewrite the second Bianchi identity (3)as(3')(VuRm)(Y,Z,V,W) + (VvRm)(Y,Z, W,U) + (VwRm)(Y,Z,U,V) = 0,Inlocal coordinates, the identities (4),(5)and (3')becomeRijkl =-Rijlk, Rijkl = Rlij, and Rijkl;n +Rijlnl + Rijnk;l =0.I The curvature operator R.According to (1)and (4),theRiemann curvature tensor Rm canbe consideredas acting on two bi-vectors X AY and z ^ W instead of acting on four vectorsX,Y,Z, W. In other words, we may write Rm asRm : A2(TM) × A(TM) →C(M).Since the Riemannian metric on M induces an inner product on each A2(T,M), onemay convert the tensor Rm into an operator R: A?(TM) → A?(TM) such that(R(X^Y),Z^W) = Rm(X ^Y,Z ^W) := Rm(X,Y,Z,W)Moreover, the symmetry equation (5) implies that R is a self-adjoint operator oneach A2(T,M). The operator R is called the curvature operator
LECTURE 8: THE RIEMANNIAN CURVATURE 3 It follows ⟨∇X∇Y Z, Z⟩ = X⟨∇Y Z, Z⟩ − ⟨∇Y Z, ∇XZ⟩ = 1 2 X(Y f) − ⟨∇Y Z, ∇XZ⟩. So −Rm(X, Y, Z, Z) = ⟨R(X, Y )Z, Z⟩ = ⟨∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z, Z⟩ = 1 2 X(Y f) − 1 2 Y (Xf) − 1 2 [X, Y ]f = 0. As a consequence, we get Rm(X, Y, Z, W) + Rm(X, Y, W, Z) =Rm(X, Y, Z + W, Z + W) − Rm(X, Y, Z, Z) − Rm(X, Y, W, W) = 0, which implies (4). The equation (5) is a consequence of (4) first one together with (1) and (2). In fact, by the first Bianchi identity (2) one has Rm(X, Y, Z, W) + Rm(Y, Z, X, W) + Rm(Z, X, Y, W) = 0, Rm(Y, Z, W, X) + Rm(Z, W, Y, X) + Rm(W, Y, Z, X) = 0, Rm(Z, W, X, Y ) + Rm(W, X, Z, Y ) + Rm(X, Z, W, Y ) = 0, Rm(W, X, Y, Z) + Rm(X, Y, W, Z) + Rm(Y, W, X, Z) = 0, Adding these equations and using (1) and (4), we get Rm(Z, X, Y, W) + Rm(W, Y, Z, X) = 0, which is equivalent to (5). □ By using (5) we may rewrite the second Bianchi identity (3) as (3′ ) (∇URm)(Y, Z, V, W) + (∇V Rm)(Y, Z, W, U) + (∇W Rm)(Y, Z, U, V ) = 0, In local coordinates, the identities (4), (5) and (3′ ) become Rijkl = −Rijlk, Rijkl = Rklij , and Rijkl;n + Rijln;l + Rijnk;l = 0. ¶ The curvature operator R. According to (1) and (4), the Riemann curvature tensor Rm can be considered as acting on two bi-vectors X ∧ Y and Z ∧ W instead of acting on four vectors X, Y, Z, W. In other words, we may write Rm as Rmg : Λ2 (TM) × Λ 2 (TM) → C ∞(M). Since the Riemannian metric on M induces an inner product on each Λ2 (TpM), one may convert the tensor Rm into an operator R : Λ2 (TM) → Λ 2 (TM) such that ⟨R(X ∧ Y ), Z ∧ W⟩ = Rmg(X ∧ Y, Z ∧ W) := Rm(X, Y, Z, W). Moreover, the symmetry equation (5) implies that R is a self-adjoint operator on each Λ2 (TpM). The operator R is called the curvature operator
4LECTURE8:THERIEMANNIANCURVATURE2.DECOMPOSITIONOFTHERIEMANNCURVATURETENSORI Some tensor algebra: symmetric tensors.Let V be any vector space.Recall that ^?V* C?v* represents the space ofanti-symmetric 2-tensors on V,while s?v*C*represents the space of sym-metric 2-tensors on V. Any 2-tensor T on V can be decomposed uniquely as thesummation of a symmetric 2-tensor and an anti-symmetric 2-tensor asT(u,u) +T(,u), T(u,v) -T(v,u)T(u,v)=22If dim V = m, then we havedim^2V*= m(m- 1)and dim S2v*= m(m+ 1)22Note that bydefinition,S(^2v*)contains (0,4)-tensors that aresymmetric withrespectto(1.2)(3.4)and anti-symmetricwithrespectto12and34,i.eT(X,Y,Z,W) = -T(Y,X,Z,W) = -T(X,Y,W,Z) = T(Z,W,X,Y)For example, one can easily check that for any two symmetric (0,2)-tensor S,T S?(V*), their Kulkarni-Nomizu product S@T E S?(^V*). The set S?(?V*) is avector space of dimensiondim S2(^2v) = m(m - 1)(m2 - m +2)(6)8Moreover, the space of 4-forms, 4V*, is a subspace of $2(^2V*) with dimensiondim ^*V*(7)Let α.β E 2y*be any two linear2-forms,both viewed as skew-symmetric2-tensors on V. Define the symmetric product of Q and β to be the (O,4)-tensor(8)(αO β)(X,Y,Z, W) := α(X,Y)β(Z, W) + α(Z, W)β(X,Y),Inlocal coordinates onecanwrite(aoB)ijkl=aiBu+QuBiObviously each αβ is in S2(^2*).It turns out that these (0,4)-tensors generatesthe whole space S2(^V*):Lemma 2.1. Any element in S?(^2V*) can be written as a linear combination ofelements oftheformαoβ.Proof. Let el, .,em be a basis of V*, thenEl=el ^e?,E?=el Ne3,..,Em(m-1)/2=em-1 Nemis a basis of A2V*, and E"Ei (i ≤ j) are linearly independent in S(^2V*) (check).口By dimension counting, we see these elements form a basis of s?(^?v*)
4 LECTURE 8: THE RIEMANNIAN CURVATURE 2. Decomposition of the Riemann curvature tensor ¶ Some tensor algebra: symmetric tensors. Let V be any vector space. Recall that ∧ 2V ∗ ⊂ ⊗2V ∗ represents the space of anti-symmetric 2-tensors on V , while S 2V ∗ ⊂ ⊗2V ∗ represents the space of symmetric 2-tensors on V . Any 2-tensor T on V can be decomposed uniquely as the summation of a symmetric 2-tensor and an anti-symmetric 2-tensor as T(u, v) = T(u, v) + T(v, u) 2 + T(u, v) − T(v, u) 2 . If dim V = m, then we have dim ∧ 2V ∗ = m(m − 1) 2 , and dim S 2V ∗ = m(m + 1) 2 . Note that by definition, S 2 (∧ 2V ∗ ) contains (0,4)-tensors that are symmetric with respect to (1, 2) ↔ (3, 4) and anti-symmetric with respect to 1 ↔ 2 and 3 ↔ 4, i.e. T(X, Y, Z, W) = −T(Y, X, Z, W) = −T(X, Y, W, Z) = T(Z, W, X, Y ). For example, one can easily check that for any two symmetric (0,2)-tensor S, T ∈ S 2 (V ∗ ), their Kulkarni-Nomizu product S○∧ T ∈ S 2 (∧ 2V ∗ ). The set S 2 (∧ 2V ∗ ) is a vector space of dimension (6) dim S 2 (∧ 2V ∗ ) = m(m − 1)(m2 − m + 2) 8 . Moreover, the space of 4-forms, ∧ 4V ∗ , is a subspace of S 2 (∧ 2V ∗ ) with dimension (7) dim ∧ 4V ∗ = m 4 . Let α, β ∈ ∧2V ∗ be any two linear 2-forms, both viewed as skew-symmetric 2-tensors on V . Define the symmetric product of α and β to be the (0, 4)-tensor (8) (α ⊙ β)(X, Y, Z, W) := α(X, Y )β(Z, W) + α(Z, W)β(X, Y ). In local coordinates one can write (α ⊙ β)ijkl = αijβkl + αklβij . Obviously each α⊙β is in S 2 (∧ 2V ∗ ). It turns out that these (0,4)-tensors generates the whole space S 2 (∧ 2V ∗ ): Lemma 2.1. Any element in S 2 (∧ 2V ∗ ) can be written as a linear combination of elements of the form α ⊙ β. Proof. Let e 1 , · · · , em be a basis of V ∗ , then E 1 = e 1 ∧ e 2 , E2 = e 1 ∧ e 3 , · · · , Em(m−1)/2 = e m−1 ∧ e m is a basis of Λ2V ∗ , and E i⊙E j (i ≤ j) are linearly independent in S 2 (∧ 2V ∗ ) (check). By dimension counting, we see these elements form a basis of S 2 (∧ 2V ∗ ). □
LECTURE8:THERIEMANNIANCURVATURE5TSome tensor algebra:Curvature-liketensorsTo explore the cyclic symmetry that arise in the first Bianchi identity,we defineDefinition 2.2. The Bianchi symmetrization of any T e S?(^2V*) is the 4-tensor(9)bT(X,Y,Z,W) =(T(X,Y,Z,W) + T(Y,Z,X,W) +T(Z,X,Y,W))So the first Bianchi identity forRm now becomes the simple equation b(Rm)=0.Definition 2.3.If T e S2(^2v*) and b(T) =0, we call T a curvature-like tensor.The set of all curvature-like tensors is denoted by 6.Erample. For any S,T E S?V*, we have b(S@T) =0 since3b(ST)ik=SiTi+SiTk-SuTik-SkTu+SiTu+SuTk-SiTk-SuTi+SuT+SuTu-SuTik-SiTu=0.As a result, S@T is a curvature-like tensor.Note that by definition, curvature-like tensors are exactly those tensors satisfyingall the algebraic symmetries that Rm admit, namely (1), (4), (5) and (2)To study the set of all curvature-like tensors, we first study the image of theBianchi symmetrization b.Note that for any a,β eA?(V*),3(b(QB))ijk=QiBu+QuB+akB+QaBik+akj+aBk=aiBu-QiBjt+QaBik+ajkBi-ajBi+auBi(aamTESAIn other words, we haveLemma 2.4. For any a, β A?(V*), b(ao β) =a ^βInviewof Lemma2.1,weconclude that themapbhas imageIm(b) = ^*V* C A?(V*),In particular, for any T E $?(^?V*) we have bT e S(?V*). By definition it isstraightforwardtocheckLemma 2.5. For any T E S(^2V*), one has b(b(T)) = T.So the Bianchi symmetrization map b, as a linear map b : $?(^2V*) → S2(^?V*),is a projection.It followsfrom the standard linearalgebra thatS2(^2V*) = Ker(b) ④ Im(b) = ④ ^4V*.As a consequence, is a vector space of dimensiondim = m(m - 1)(m2 - m + 2)m1(10)12m (m2 - 1).8
LECTURE 8: THE RIEMANNIAN CURVATURE 5 ¶ Some tensor algebra: Curvature-like tensors. To explore the cyclic symmetry that arise in the first Bianchi identity, we define Definition 2.2. The Bianchi symmetrization of any T ∈ S 2 (∧ 2V ∗ ) is the 4-tensor (9) bT(X, Y, Z, W) = 1 3 (T(X, Y, Z, W) + T(Y, Z, X, W) + T(Z, X, Y, W)). So the first Bianchi identity for Rm now becomes the simple equation b(Rm) = 0. Definition 2.3. If T ∈ S 2 (∧ 2V ∗ ) and b(T) = 0, we call T a curvature-like tensor. The set of all curvature-like tensors is denoted by C . Example. For any S, T ∈ S 2V ∗ , we have b(S○∧ T) = 0 since 3b(S○∧ T)ijkl =SikTjl + SjlTik − SilTjk − SjkTil + SjkTli + SliTjk − SjiTlk − SlkTji + SlkTij + SijTlk − SljTik − SikTlj =0. As a result, S○∧ T is a curvature-like tensor. Note that by definition, curvature-like tensors are exactly those tensors satisfying all the algebraic symmetries that Rm admit, namely (1), (4), (5) and (2). To study the set C of all curvature-like tensors, we first study the image of the Bianchi symmetrization b. Note that for any α, β ∈ Λ 2 (V ∗ ), 3(b(α ⊙ β))ijkl = αijβkl + αklβij + αjkβil + αilβjk + αkiβjl + αjlβki = αijβkl − αikβjl + αilβjk + αjkβil − αjlβik + αklβij = 4! 2!2! 1 4! X π∈S4 (α ⊗ β) π ijkl = (α ∧ β)ijkl. In other words, we have Lemma 2.4. For any α, β ∈ Λ 2 (V ∗ ), b(α ⊙ β) = 1 3 α ∧ β. In view of Lemma 2.1, we conclude that the map b has image Im(b) = ∧ 4V ∗ ⊂ Λ 2 (V ∗ ). In particular, for any T ∈ S 2 (∧ 2V ∗ ) we have bT ∈ S 2 (∧ 2V ∗ ). By definition it is straightforward to check Lemma 2.5. For any T ∈ S 2 (∧ 2V ∗ ), one has b(b(T)) = T. So the Bianchi symmetrization map b, as a linear map b : S 2 (∧ 2V ∗ ) → S 2 (∧ 2V ∗ ), is a projection. It follows from the standard linear algebra that S 2 (∧ 2V ∗ ) = Ker(b) ⊕ Im(b) = C ⊕ ∧4V ∗ . As a consequence, C is a vector space of dimension (10) dim C = m(m − 1)(m2 − m + 2) 8 − m 4 = 1 12 m2 (m2 − 1)