《黎曼几何》课程教学资源（英文讲义）Lec07 THE CURVATURE TENSOR

LECTURE7:THECURVATURE TENSOR1.THECURVATURE TENSOROFALINEARCONNECTIONI Derivations on the graded tensor algebra.Let M be a smooth manifold endowed with a linear connection V.As we haveseen last time, induces a linear connectionV : F(TM) × F(TM) -→F(TM)on each tensor bundle @k-TM. Moreover, all these linear connections are compatibleas a whole set of connections in the sense that they are compatible with the tensorproduct operation and thecontraction operationfortensors.Let's take a closer look of the “tensor product compatibility".Denote byFo(@**TM) the graded tensor algebra of all smooth tensor fields on M. Then thetensor product compatibility means that for any smooth vector field X EF(TM),themapVx :T(@**TM) → T(**TM)satisfies Vx(ST)=xST+ SVxT.In other words, Vx is a derivation onF(@*,*TM).Nowlet Dbe the set of all derivations on the tensor algebraT(**TM),whichare by definition linear maps D such that D(ST) = DST+SDT. A standardfact (which is easy to verify via definition) is that D is a Lie algebra (with respect tocommutator), namely if Di,D, are two derivations, so is their commutator[D1, D2] = Di o D2 - D2 0 D1.Ecample. For any smooth vector field X, the Lie derivative Cx is a derivation onF(**TM). Moreover, Lx satisfies (when acting on any tensor field)C(x,y) = Lx o Ly - Ly o Cx.In other words,the linear map“XCx"is a Liealgebrahomomorphismfrom“theLie algebra of all smooth vector fields on M" to “the Lie algebra of all derivationson (@**TM)".NowconsiderthelinearmapΦ: F(TM)→D, X-→Vx.One may ask: Is @ a Lie algebra homomorphism? In other words, do we haveV(x,y = Vx o Vy - Vy o Vx?1

LECTURE 7: THE CURVATURE TENSOR 1. The curvature tensor of a linear connection ¶ Derivations on the graded tensor algebra. Let M be a smooth manifold endowed with a linear connection ∇. As we have seen last time, ∇ induces a linear connection ∇ : Γ∞(TM) × Γ ∞(⊗ k,lTM) → Γ ∞(⊗ k,lTM) on each tensor bundle ⊗k,lTM. Moreover, all these linear connections are compatible as a whole set of connections in the sense that they are compatible with the tensor product operation and the contraction operation for tensors. Let’s take a closer look of the “tensor product compatibility”. Denote by Γ ∞(⊗∗,∗TM) the graded tensor algebra of all smooth tensor fields on M. Then the tensor product compatibility means that for any smooth vector field X ∈ Γ ∞(TM), the map ∇X : Γ∞(⊗ ∗,∗TM) → Γ ∞(⊗ ∗,∗TM) satisfies ∇X(S ⊗ T) = ∇XS ⊗ T + S ⊗ ∇XT. In other words, ∇X is a derivation on Γ ∞(⊗∗,∗TM). Now let D be the set of all derivations on the tensor algebra Γ∞(⊗∗,∗TM), which are by definition linear maps D such that D(S⊗T) = DS⊗T +S⊗DT. A standard fact (which is easy to verify via definition) is that D is a Lie algebra (with respect to commutator), namely if D1, D2 are two derivations, so is their commutator [D1, D2] = D1 ◦ D2 − D2 ◦ D1. Example. For any smooth vector field X, the Lie derivative LX is a derivation on Γ ∞(⊗∗,∗TM). Moreover, LX satisfies (when acting on any tensor field) L[X,Y ] = LX ◦ LY − LY ◦ LX. In other words, the linear map “X 7→ LX” is a Lie algebra homomorphism from “the Lie algebra of all smooth vector fields on M” to “the Lie algebra of all derivations on Γ∞(⊗∗,∗TM)”. Now consider the linear map Φ : Γ∞(TM) → D, X 7→ ∇X. One may ask: Is Φ a Lie algebra homomorphism? In other words, do we have ∇[X,Y ] = ∇X ◦ ∇Y − ∇Y ◦ ∇X? 1

2LECTURE 7:THE CURVATURETENSORUnfortunately the answer is no in general' (as we will see soon). So we are naturallyled to study the map R(X,Y) : Fo(@k,TM) → T(kTM) defined byR(X,Y)T=VxVyT_VyVxT-V(xY)TLet's start with two simple cases:. First for k = l = 0, i.e. T= f E C(M), the map R(X,Y) is zero, sinceR(X,Y)f = VxVyf - VyVxf -V(x,)f = XYf - YXf -[X,YIf = 0.. Next we study the case k = 1,l = 0, i.e. T = w is a smooth 1-form. It turnsout that one can convert R(X,Y) on 1-forms to R(X,Y) on vector fields:Lemma 1.1. For any 1-form w e 2'(M),(R(X,Y)w)(Z) = -w(R(X,Y)Z)口Proof. Compute by definition. Details left as an exercise.In view of the fact that the graded tensor algebra T(**TM) is generated bysmooth functions, vector fields and 1-forms, together with the fact that R(X,Y) isagain a derivation on Fo(@**TM), we conclude that to study R(X,Y) on all tensorfields, it is enough to study R(X,Y) on vector fields!I The curvature tensor of a linear connection.WedefineDefinition 1.2. Let M be a smooth manifold and a linear connection on M. Wecall the map R : T(TM) × F(TM) × T(TM) -→F(TM) defined by(1)R(X,Y)Z=VxVZ-VyVxZ-V(x,YZthe curvature tensor ofVAs we explained above, R measures to what extend the map @ fails to be a Liealgebra homomorphism.Remark. In many books, the definition of curvature tensor is different from theabove formula by a negative sign. Both definitions have their own advantages. Sowhen you open a new book on Riemannian geometry, you should first glance at itsdefinition of the curvature tensor.Erample.For the standard linear connection on Rm, we havexa (yia,) = X'd(Yi)o,.which impliesxz-xz=Xia(yia,zh)ox-ia,(xia,z)o=Vxmzand thus its curvature tensor R = 0.IMaybe we should say fortunately the answer is no, otherwise there will be no Riemanniangeometry, and the world will be boring

2 LECTURE 7: THE CURVATURE TENSOR Unfortunately the answer is no in general1 (as we will see soon). So we are naturally led to study the map R(X, Y ) : Γ∞(⊗k,lTM) → Γ ∞(⊗k,lTM) defined by R(X, Y )T = ∇X∇Y T − ∇Y ∇XT − ∇[X,Y ]T. Let’s start with two simple cases: • First for k = l = 0, i.e. T = f ∈ C ∞(M), the map R(X, Y ) is zero, since R(X, Y )f = ∇X∇Y f − ∇Y ∇Xf − ∇[X,Y ]f = XY f − Y Xf − [X, Y ]f = 0. • Next we study the case k = 1, l = 0, i.e. T = ω is a smooth 1-form. It turns out that one can convert R(X, Y ) on 1-forms to R(X, Y ) on vector fields: Lemma 1.1. For any 1-form ω ∈ Ω 1 (M), (R(X, Y )ω)(Z) = −ω(R(X, Y )Z). Proof. Compute by definition. Details left as an exercise. □ In view of the fact that the graded tensor algebra Γ∞(⊗∗,∗TM) is generated by smooth functions, vector fields and 1-forms, together with the fact that R(X, Y ) is again a derivation on Γ∞(⊗∗,∗TM), we conclude that to study R(X, Y ) on all tensor fields, it is enough to study R(X, Y ) on vector fields! ¶ The curvature tensor of a linear connection. We define Definition 1.2. Let M be a smooth manifold and ∇ a linear connection on M. We call the map R : Γ∞(TM) × Γ ∞(TM) × Γ ∞(TM) → Γ ∞(TM) defined by (1) R(X, Y )Z = ∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z the curvature tensor of ∇. As we explained above, R measures to what extend the map Φ fails to be a Lie algebra homomorphism. Remark. In many books, the definition of curvature tensor is different from the above formula by a negative sign. Both definitions have their own advantages. So when you open a new book on Riemannian geometry, you should first glance at its definition of the curvature tensor. Example. For the standard linear connection ∇ on R m, we have ∇Xi∂i (Y j ∂j ) = X i ∂i(Y j )∂j . which implies ∇X∇Y Z − ∇Y ∇XZ = X i ∂i(Y j ∂jZ k )∂k − Y j ∂j (X i ∂iZ k )∂k = ∇[X,Y ]Z and thus its curvature tensor R ≡ 0. 1Maybe we should say fortunately the answer is no, otherwise there will be no Riemannian geometry, and the world will be boring

3LECTURE 7:THECURVATURETENSORErample.Consider M=Sm.LasttimewehaveseenthatVxY =VxY +(X,Y)nis the Levi-Civita connection on Sm, where is the standard connection on IRm+1.ItfollowsVxVyZ=VxVyZ+(X,VyZ)n=Vx(VyZ+(Y,Z)n)+Y(X,Z)n- (VyX,Z)n= VxVZ+X((Y,Z))n+(Y,Z)X +Y((X,Z))n-(VyX,Z)nIn view of the fact R=O, wegetR(X,Y)Z =X((Y,Z))n+ (Y,Z)X +Y((X,Z))n - (VX,Z)n-Y((X,Z))n-(X,Z)Y -X((Y,Z))n+(VxY,Z)n-((X,Y),Z)n=(Y,Z)X -(X,Z)Y.I The curvature tensor is a tensor.Now we prove that R is a tensor of type (1,3), in the senseR(w,X,Y,Z) := w(R(X,Y)Z))is really an element in P(1,3TM):Proposition 1.3. The curvature tensor R is a (1,3)-tensor.Proof.We need to proveR(fX,Y)Z = R(X, fY)Z - R(X,Y)(fZ) = fR(X,Y)ZHereweonlycheckoneof them:R(fX,Y)Z= fVxVZ- V(fVxZ) -V(fX)Y-Y(fX)Z= f(VxVyZ-VyVxZ-V(x,YZ)-(Yf)VxZ+(Yf)VxZ= fR(X,Y)Z.口The others are similar and are left as happy exercises.Locally, we write the tensor R (or R) as?R-Rik'dr @ dr droi.e. if we denoteR(O,O,)On= Riika2Note that here we are using a "non-standard" order: for the local expression of the (1,3)-tensorR, we write the“1"-part (i.e. the vector t)after the "3"-parts (i.e. the co-vectors). In some bookspeople use different orders like R(O,,)Ok= R'kigOt. In other words, in writing local expressionsof R(w,X,Y,Z), we always want to put the index for w next to the index for Z.The reason willbe clear in next lecture

LECTURE 7: THE CURVATURE TENSOR 3 Example. Consider M = S m. Last time we have seen that ∇XY = ∇XY + ⟨X, Y ⟩⃗n is the Levi-Civita connection on S m, where ∇ is the standard connection on R m+1 . It follows ∇X∇Y Z = ∇X∇Y Z + ⟨X, ∇Y Z⟩⃗n = ∇X(∇Y Z + ⟨Y, Z⟩⃗n) + Y ⟨X, Z⟩⃗n − ⟨∇Y X, Z⟩⃗n = ∇X∇Y Z + X(⟨Y, Z⟩)⃗n + ⟨Y, Z⟩X + Y (⟨X, Z⟩)⃗n − ⟨∇Y X, Z⟩⃗n. In view of the fact R = 0, we get R(X, Y )Z =X(⟨Y, Z⟩)⃗n + ⟨Y, Z⟩X + Y (⟨X, Z⟩)⃗n − ⟨∇Y X, Z⟩⃗n − Y (⟨X, Z⟩)⃗n − ⟨X, Z⟩Y − X(⟨Y, Z⟩)⃗n + ⟨∇XY, Z⟩⃗n − ⟨[X, Y ], Z⟩⃗n =⟨Y, Z⟩X − ⟨X, Z⟩Y. ¶ The curvature tensor is a tensor. Now we prove that R is a tensor of type (1,3), in the sense Re(ω, X, Y, Z) := ω(R(X, Y )Z)) is really an element in Γ∞(⊗1,3TM): Proposition 1.3. The curvature tensor R is a (1, 3)-tensor. Proof. We need to prove R(fX, Y )Z = R(X, fY )Z = R(X, Y )(fZ) = fR(X, Y )Z. Here we only check one of them: R(fX, Y )Z = f∇X∇Y Z − ∇Y (f∇XZ) − ∇(fX)Y −Y (fX)Z = f(∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z) − (Y f)∇XZ + (Y f)∇XZ = fR(X, Y )Z. The others are similar and are left as happy exercises. □ Locally, we write the tensor R (or Re) as2 R = Rijk l dxi ⊗ dxj ⊗ dxk ⊗ ∂l , i.e. if we denote R(∂i , ∂j )∂k = Rijk l ∂l , 2Note that here we are using a “non-standard” order: for the local expression of the (1, 3)-tensor R, we write the “1”-part (i.e. the vector ∂l) after the “3”-parts (i.e. the co-vectors). In some books people use different orders like R(∂i , ∂j )∂k = Rl kij∂l . In other words, in writing local expressions of Re(ω, X, Y, Z), we always want to put the index for ω next to the index for Z. The reason will be clear in next lecture

4LECTURE 7:THE CURVATURE TENSORthen the coefficients Rik are related to the Christoffel symbols byRik=o,F'jk-0,F'ik +I'is-F' jsswhich is a consequence of the factRk=R(Oi,,)o=Va(To)-Va,(To).Note that this also implies that the curvature tensor for the standard connection onRm is identically zero, since its Christoffel symbols are all zero.I The commutator of two coordinate covariant derivatives of a surface.Consider an embedded 2-dimensional parametric surface in M,P: U CR? → S=p(U) C M.Denotetheparameters for U bys and t.We let sand Ot be thetwo coordinatevectorfieldsaaas := dp(andOt := dp(OsOton the 2-dimensional surface S. Note that by locality, for any smooth vector fieldZ on surface S, the expression Va.Va,Z make sense and will be thought of as theiterated second ordercovariant derivative of Z withrespectto Os,Ot.Itturnsoutthat R measures the non-commutativity of such iterated covariant derivatives:Proposition 1.4. For any smooth vector field Z (defined on the surface S)VaVaZ-VaVa.Z=R(Os,Ot)Z.Proof. Take a coordinate chart of M near S so that S is defined by r3 =rm = 0. Then (s,t,r3,..:,rm) is a local coordinate system on M in a tubularneighborhood of S. Now the conclusion follows from the fact as coordinate vector口fields, [Ot, o.] = 0.Remark.Geometrically, R is closely related to "the holonomy along an infinitesimalpath which is the boundary of p((O,e) x (O, e)". Details are left as a term project.TFlat connection.We are interested in linear connections with vanishing curvature tensor, i.e. withR=0.For example, the standard connection on IRm satisfies R=0.The nicepointfor Rm is that the coordinate vector fields are parallel along any vector fieldsDefinition 1.5. Let M be a smooth manifolds with a connection v. We say (M.v)admits a local fat frame everywhere if near any point p, there is a set of vector fieldsXi,...,Xm on a neighborhood U of p such that(1) [frame] [X;(q) [ 1 ≤i≤m) form a basis of TqM for every q E U,(2) [flatness] VyX, =0 for all i and for all vector field Y.It is easy to see that if (M, V) admits a local flat frame everywhere, thenR(X,Y)X; = 0 for all X,Y, and thus R = 0 since R is a tensor. Conversely

4 LECTURE 7: THE CURVATURE TENSOR then the coefficients Rijk l are related to the Christoffel symbols by Rijk l = ∂iΓ l jk − ∂jΓ l ik + Γs jkΓ l is − Γ s ikΓ l js, which is a consequence of the fact Rijk l ∂l = R(∂i , ∂j )∂k = ∇∂i (Γs jk∂s) − ∇∂j (Γs ik∂s). Note that this also implies that the curvature tensor for the standard connection on R m is identically zero, since its Christoffel symbols are all zero. ¶ The commutator of two coordinate covariant derivatives of a surface. Consider an embedded 2-dimensional parametric surface in M, φ : U ⊂ R 2 → S = φ(U) ⊂ M. Denote the parameters for U by s and t. We let ∂s and ∂t be the two coordinate vector fields ∂s := dφ( ∂ ∂s) and ∂t := dφ( ∂ ∂t) on the 2-dimensional surface S. Note that by locality, for any smooth vector field Z on surface S, the expression ∇∂s∇∂tZ make sense and will be thought of as the iterated second order covariant derivative of Z with respect to ∂s, ∂t . It turns out that R measures the non-commutativity of such iterated covariant derivatives: Proposition 1.4. For any smooth vector field Z (defined on the surface S), ∇∂s∇∂tZ − ∇∂t∇∂sZ = R(∂s, ∂t)Z. Proof. Take a coordinate chart of M near S so that S is defined by x 3 = · · · = x m = 0. Then (s, t, x3 , · · · , xm) is a local coordinate system on M in a tubular neighborhood of S. Now the conclusion follows from the fact as coordinate vector fields, [∂t , ∂s] = 0. □ Remark. Geometrically, R is closely related to “the holonomy along an infinitesimal path which is the boundary of φ((0, ε) × (0, ε))”. Details are left as a term project. ¶ Flat connection. We are interested in linear connections with vanishing curvature tensor, i.e. with R = 0. For example, the standard connection on R m satisfies R = 0. The nice point for R m is that the coordinate vector fields are parallel along any vector fields. Definition 1.5. Let M be a smooth manifolds with a connection ∇. We say (M, ∇) admits a local flat frame everywhere if near any point p, there is a set of vector fields X1, · · · , Xm on a neighborhood U of p such that (1) [frame] {Xi(q) | 1 ≤ i ≤ m} form a basis of TqM for every q ∈ U, (2) [flatness] ∇Y Xi = 0 for all i and for all vector field Y . It is easy to see that if (M, ∇) admits a local flat frame everywhere, then R(X, Y )Xi = 0 for all X, Y , and thus R ≡ 0 since R is a tensor. Conversely

LECTURE 7:THE CURVATURETENSOR5Proposition 1.6.Let M be a smooth manifold, be a linear connection. ThenR =o if and only if (M, ) admits a local fat frame everywhere.Proof. It remains to prove the“only if"part. Without loss of generality, we maytake U to be a coordinate neighborhood and p = (O,-.: ,O) the origin. We startwith any basis [ui,... , Um) of TpM and let Xi(p) = vi. We extend Xi to the "line"[(a,0, ..,O)) byparallel transporting thevector X;(p) along the curve %o(t) =(t,0,...,O). Then we extend further to the“plane"[(a,b,0,..,O)) by paralleltransporting each X;(o(a)) along %a(t) = (a,t, 0, **, 0). Repeating this procedure,we get a set of smooth (why?) vector fields [Xi,.-,Xm] on the whole of U. Byconstruction, they are a frame. It remains to prove the flatness.First by construction, we have.VaX, =0 at any point on the line (a, 0,.-,0). Va, X, = 0 at any point on the plane (a, b, 0, ..., O)Moreover, since R=0 and [i,] =0, we getVa VaX = Va Va,Xi= 0on the plane (a,b,O,...,O).As a consequence, Va,X, is parallel along each line(t) = (a,t,0, ..,0), with initial condition (VaX)(a, 0, ...,0) = 0. By unique-ness, one must have Va, X, = 0 along each a. In other words, we get. Va X,=0, Va,X, =0 at any point on the plane (a,b, 0, ...,O)Bythesame argument, weget.VaX, = 0, VaX, = 0, VaX, = 0 at any point of the form (a,b, c, 0, ...,0).Continuing this argument, one can see that Va,X, = o for all i,j, at all points in口U. As a consequence, Xi,... , Xm form a local flat frame.As aconsequence,Corollary 1.7. If (M,g) is a Riemannian manifold for which the curvature of theLevi-Civita connection vanishes, then (M,g) is locally isometric to (Rm,go).Proof. In the proof above, we take [ui, ... , Um] to be an orthonormal basis. Thenafter parallel transport, the vector fields Xi,...,Xm are orthonormal everywhereSince is Levi-Civita connection, it is torsion free. It follows that for any i,j0=Vx,X,-Vx,X,-[X,X] =-[Xi,X]By Frobenius theorem, there exists a local coordinate chart with X, = O, for all i口In this chart, we have gij = (O, O,) = dij, and thus locally g = dr @ da.Definition 1.8. A linear connection V on M is called flat if R = 0. A Riemannianmanifold is called flat if theLevi-Civita connection is flat

LECTURE 7: THE CURVATURE TENSOR 5 Proposition 1.6. Let M be a smooth manifold, ∇ be a linear connection. Then R = 0 if and only if (M, ∇) admits a local flat frame everywhere. Proof. It remains to prove the “only if” part. Without loss of generality, we may take U to be a coordinate neighborhood and p = (0, · · · , 0) the origin. We start with any basis {v1, · · · , vm} of TpM and let Xi(p) = vi . We extend Xi to the “line” {(a, 0, · · · , 0)} by parallel transporting the vector Xi(p) along the curve γ0(t) = (t, 0, · · · , 0). Then we extend further to the “plane” {(a, b, 0, · · · , 0)} by parallel transporting each Xi(γ0(a)) along γa(t) = (a, t, 0, · · · , 0). Repeating this procedure, we get a set of smooth (why?) vector fields {X1, · · · , Xm} on the whole of U. By construction, they are a frame. It remains to prove the flatness. First by construction, we have • ∇∂1Xi = 0 at any point on the line (a, 0, · · · , 0). • ∇∂2Xi = 0 at any point on the plane (a, b, 0, · · · , 0). Moreover, since R = 0 and [∂1, ∂2] = 0, we get ∇∂2∇∂1Xi = ∇∂1∇∂2Xi = 0 on the plane (a, b, 0, · · · , 0). As a consequence, ∇∂1Xi is parallel along each line γa(t) = (a, t, 0, · · · , 0), with initial condition (∇∂1Xi)(a, 0, · · · , 0) = 0. By unique￾ness, one must have ∇∂1Xi = 0 along each γa. In other words, we get • ∇∂1Xi = 0, ∇∂2Xi = 0 at any point on the plane (a, b, 0, · · · , 0). By the same argument, we get • ∇∂1Xi = 0, ∇∂2Xi = 0, ∇∂3Xi = 0 at any point of the form (a, b, c, 0, · · · , 0). Continuing this argument, one can see that ∇∂jXi = 0 for all i, j, at all points in U. As a consequence, X1, · · · , Xm form a local flat frame. □ As a consequence, Corollary 1.7. If (M, g) is a Riemannian manifold for which the curvature of the Levi-Civita connection vanishes, then (M, g) is locally isometric to (R m, g0). Proof. In the proof above, we take {v1, · · · , vm} to be an orthonormal basis. Then after parallel transport, the vector fields X1, · · · , Xm are orthonormal everywhere. Since ∇ is Levi-Civita connection, it is torsion free. It follows that for any i, j, 0 = ∇XiXj − ∇XjXi − [Xi , Xj ] = −[Xi , Xj ]. By Frobenius theorem, there exists a local coordinate chart with Xi = ∂i for all i. In this chart, we have gij = ⟨∂i , ∂j ⟩ = δij , and thus locally g = Pdxi ⊗ dxi . □ Definition 1.8. A linear connection ∇ on M is called flat if R = 0. A Riemannian manifold is called flat if the Levi-Civita connection is flat

6LECTURE 7:THE CURVATURE TENSOR2.SYMMETRIES OF THE CURVATURE TENSORFOR TORSION FREE CONNECTIONI Curvature tensor as commutator of 2 (for torsion free connection).Regard as a map of the form : T~(@*TM) →F(@*+1TM).We have studied the composition 2 : F(0,0TM) → F(0.2TM) which mapsany f C(M) to its HessianV?f : T(TM) ×F(TM) -→C(M),V?f(X,Y)=VVxf -VvyxfIn general, for any T e I(@kTM), the second covariant derivative map 2sends T to the (k,l + 2)-tensor 2T, which, by definition, equalsV?T(...,X,Y) = (VyVxT)(..) - (VvxxT)(...)ForsimplicitywilldenoteVXY : I(&*TM) →I(αTM), T -→V?T(. ,X,Y).One should be aware of the difference between the second covariant derivative andthe iterated covariant derivative.From now on suppose is torsion free, which as we have seen, is equivalent tothe fact that Vxyf is symmetric with respect to the entries X and Y. A naturalquestion is: if is torsion free, is xyT symmetric with respect to X and Y forall T? The answer is no. For example, if T = Z is a vector field,VxZ-VxyZ=VxVZ-VyVxZ-VvxyZ+VvyxZ= VxVyZ - VyVxZ- V(xnZ = R(X,Y)Zwhere in the last step we used the fact is torsion free, i.e. VxY -VyX -[X,Y].More generally,by exactlythe same computation one hasLemma2.1 (Ricciidentity).Letbe any torsion free connection,thenVx(T) - VXr(T) = R(X,Y)T, VT E T(TM).In conclusion, for a torsion free connection, the operator R(X,Y) (on any ten-sors) measures the non-commutativity of second covariant derivatives.Remark. One may ask: What about higher order covariant derivatives? It turnsout that under the torsion free assumption, the operator R(X,Y) also measures thenon-commutativity of higher covariant derivatives.For example,VY,zXT-V2,yxT=(VxV?T)(Y,Z)-(VxV?T)(Z,Y)=VxVyzT-VxyzT-VvxzT-VxV2yT+Vxz.T+V2xyT=-Vx(R(Y,Z)T)+R(VxY, Z)T+R(Y, VxZ)T

6 LECTURE 7: THE CURVATURE TENSOR 2. Symmetries of the curvature tensor for torsion free connection ¶ Curvature tensor as commutator of ∇2 (for torsion free connection). Regard ∇ as a map of the form ∇ : Γ∞(⊗ k,lTM) → Γ ∞(⊗ k,l+1TM). We have studied the composition ∇2 : Γ∞(⊗0,0TM) → Γ ∞(⊗0,2TM) which maps any f ∈ C ∞(M) to its Hessian ∇2 f : Γ∞(TM) × Γ ∞(TM) → C ∞(M), ∇2 f(X, Y ) = ∇Y ∇Xf − ∇∇Y Xf. In general, for any T ∈ Γ ∞(⊗k,lTM), the second covariant derivative map ∇2 sends T to the (k, l + 2)-tensor ∇2T, which, by definition, equals ∇2T(· · · , X, Y ) = (∇Y ∇XT)(· · ·) − (∇∇Y XT)(· · ·). For simplicity will denote ∇2 X,Y : Γ∞(⊗ k,lTM) → Γ ∞(⊗ k,lTM), T 7→ ∇2T(· · · , X, Y ). One should be aware of the difference between the second covariant derivative and the iterated covariant derivative. From now on suppose ∇ is torsion free, which as we have seen, is equivalent to the fact that ∇2 X,Y f is symmetric with respect to the entries X and Y . A natural question is: if ∇ is torsion free, is ∇2 X,Y T symmetric with respect to X and Y for all T? The answer is no. For example, if T = Z is a vector field, ∇2 Y,XZ − ∇2 X,Y Z = ∇X∇Y Z − ∇Y ∇XZ − ∇∇XY Z + ∇∇Y XZ = ∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z = R(X, Y )Z, where in the last step we used the fact ∇ is torsion free, i.e. ∇XY − ∇Y X = [X, Y ]. More generally, by exactly the same computation one has Lemma 2.1 (Ricci identity). Let ∇ be any torsion free connection, then ∇2 Y,X(T) − ∇2 X,Y (T) = R(X, Y )T, ∀T ∈ Γ ∞(⊗ k,lTM). In conclusion, for a torsion free connection, the operator R(X, Y ) (on any ten￾sors) measures the non-commutativity of second covariant derivatives. Remark. One may ask: What about higher order covariant derivatives? It turns out that under the torsion free assumption, the operator R(X, Y ) also measures the non-commutativity of higher covariant derivatives. For example, ∇3 Y,Z,XT −∇3 Z,Y,XT = (∇X∇2T)(Y, Z)−(∇X∇2T)(Z, Y ) =∇X∇2 Y,ZT −∇2 ∇XY,ZT −∇2 Y,∇XZT −∇X∇2 Z,Y T +∇2 ∇XZ,Y T +∇2 Z,∇XY T =−∇X(R(Y, Z)T)+R(∇XY, Z)T +R(Y, ∇XZ)T

LECTURE 7:THECURVATURETENSOR7IThe First/Algebraic Bianchi identity.Nowwe studysymmetries of thecurvaturetensorR.Bydefinitiononeimmedi-ately see that for any linear connection V, R admits the following anti-symmetry:(2)R(X,Y)Z = -R(Y,X)ZIn local coordinates it can be written asRijkl = -Rjik.It turns out that fortorsion freeconnections,Radmits more symmetries.Proposition 2.2 (The First Bianchi identity). If is a torsion-free, then(3)R(X,Y)Z + R(Y,Z)X + R(Z,X)Y = 0Proof.Sincefortorsion free connection,wehaveVxY-VyX-[X,Y] =O, soR(X,Y)Z + R(Y,Z)X + R(Z,X)Y=VxVyZ-VyVxZ-V(xmZ+VyzX-VzVyX-Vy,z)X+VzVxy-VxVzY-V(zx)Y= Vx[Y,Z] + Vy[Z, X] + Vz[X,Y] - V[x,y)z - V(y,z]X - V(z,x)Y= [X, [Y, Z]] +[Y, [Z, X]] + [Z, [X, Y]]= 0,口where in the last step weused the Jacobi identity forvectorfields.In local coordinates thefirst Bianchi identity canbewritten asRuk+Riki+Ruij=0.The Second/Differential Bianchi identity.Itturns outthat not onlyR has the above cvclicsvmmetry.butalsoits covariantderivative VRhas a similar cyclic symmetry.To understand VR, let's go back tothe (1,3)-tensor R defined byR(w, X,Y,Z) = w(R(X,Y)Z).It follows by definition that vR is the (1,4)-tensor given by(VR)(w, X,Y,Z.W)=(VwR)(w,X,Y,Z)=Vw(w(R(X,Y)Z)) - (Vww)(R(X,Y)Z) -w(R(VwX,Y)Z)-w(R(X, VwY)Z) -w(R(X,Y)VwZ)=w(Vw(R(X,Y)Z)-R(VwX, Y)Z-R(X, VwY)Z-R(X,Y)VwZ),So it is reasonable to define VwR as(VwR)(X,Y,Z) := Vw(R(X,Y)Z)-R(VwX,Y)Z-R(X, VwY)Z-R(X,Y)VwZ

LECTURE 7: THE CURVATURE TENSOR 7 ¶ The First/Algebraic Bianchi identity. Now we study symmetries of the curvature tensor R. By definition one immedi￾ately see that for any linear connection ∇, R admits the following anti-symmetry: (2) R(X, Y )Z = −R(Y, X)Z In local coordinates it can be written as R l ijk = −R l jik . It turns out that for torsion free connections, R admits more symmetries. Proposition 2.2 (The First Bianchi identity). If ∇ is a torsion-free, then (3) R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = 0. Proof. Since for torsion free connection, we have ∇XY − ∇Y X − [X, Y ] = 0, so R(X, Y )Z + R(Y, Z)X + R(Z, X)Y = ∇X∇Y Z − ∇Y ∇XZ − ∇[X,Y ]Z + ∇Y ∇ZX − ∇Z∇Y X − ∇[Y,Z]X + ∇Z∇XY − ∇X∇ZY − ∇[Z,X]Y = ∇X[Y, Z] + ∇Y [Z, X] + ∇Z[X, Y ] − ∇[X,Y ]Z − ∇[Y,Z]X − ∇[Z,X]Y = [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y ]] = 0, where in the last step we used the Jacobi identity for vector fields. □ In local coordinates the first Bianchi identity can be written as R l ijk + R l jki + R l kij = 0. ¶ The Second/Differential Bianchi identity. It turns out that not only R has the above cyclic symmetry, but also its covariant derivative ∇R has a similar cyclic symmetry. To understand ∇R, let’s go back to the (1, 3)-tensor Re defined by Re(ω, X, Y, Z) = ω(R(X, Y )Z). It follows by definition that ∇Re is the (1,4)-tensor given by (∇Re)(ω, X, Y, Z, W)=(∇W Re)(ω, X, Y, Z) =∇W (ω(R(X, Y )Z)) − (∇W ω)(R(X, Y )Z) − ω(R(∇W X, Y )Z) − ω(R(X, ∇W Y )Z) − ω(R(X, Y )∇W Z) =ω ￾ ∇W (R(X, Y )Z)−R(∇WX, Y )Z−R(X, ∇WY )Z−R(X, Y )∇WZ
. So it is reasonable to define ∇W R as (∇W R)(X, Y, Z) := ∇W (R(X, Y )Z)−R(∇W X, Y )Z−R(X, ∇W Y )Z−R(X, Y )∇W Z

8LECTURE 7:THE CURVATURE TENSORProposition 2.3 (The Second Bianchi Identity).Suppose is torsion free, then(VxR)(Y,Z, W) + (VyR)(Z,X, W) + (VzR)(X,Y,W) = 0.Proof. By definition,(VxR)(Y,Z,W) + (VyR)(Z,X, W) + (VzR)(X,Y,W)-Vx(R(Y.Z)W)-R(VxY.Z)W-R(Y.VxZ)W-R(Y.Z)VxW+Vy(R(Z,X)W) - R(VZ,X)W - R(Z,VX)W - R(Z,X)VyW+Vz(R(X,Y)W) - R(VzX,Y)W- R(X, VzY)W - R(X,Y)VzW.Using the torsion-freeness and (2), one can simplify the middle two columns toR([X, Z],Y)W + R([Y,X], Z)W + R([Y,X], Z)W.Now expand each R using its definition, the whole expression becomes a summationof27terms,the first 9termsbeing+VxVyVzW-VxVzVyW-VxVyzW+V(x,z)VyW- Vy(x,z)W- Vx,z).W-VyVzVxW+VzVVxW+VyzVxW,the second and third 9termsaresimilarto thefirst9 termsabove:onejustreplaceX,Y,Z by Y, Z,X and Z, X,Y respectively. It is not hard to check that all thoseexpressions containing three 's (12terms in total) cancel out trivially,all thoseexpressions containing two 's (also 12 terms in total) cancel out by using the fact[X, Y - -[Y, X], and the remaining three termsV[x,2],W + V[y,x),z)W + V[z,j,xW = 0口in view of the Jacobi identity.In local coordinates we can write Va, R = Riuk ndr dri dak O, Then thesecond Bianchi identity can be written asRyk in + Rjnk ; + Rnik y = 0.Remark. If we denoteto be the cyclic sum over X,Y, Z, then the first andOX.Y.Zsecond Bianchi identities can be written as R(X,Y)Z =0and (VxR)(Y,Z,W) = 0OX,Y,zOx,Y,zMore generally, if is not torsion free, then in terms of the torsion tensor T,Z R(X,Y)Z= Z (VxT)(Y,Z)+T(T(X,Y),Z)OX,Y,ZOx,Y,zand (VxR)(Y,z,W)+ R(T(X,Y),Z)W = 0.OX.Y,zOX,Y,z

8 LECTURE 7: THE CURVATURE TENSOR Proposition 2.3 (The Second Bianchi Identity). Suppose ∇ is torsion free, then (∇XR)(Y, Z, W) + (∇Y R)(Z, X, W) + (∇ZR)(X, Y, W) = 0. Proof. By definition, (∇XR)(Y, Z, W) + (∇Y R)(Z, X, W) + (∇ZR)(X, Y, W) =∇X(R(Y, Z)W) − R(∇XY, Z)W − R(Y, ∇XZ)W − R(Y, Z)∇XW+ ∇Y (R(Z, X)W) − R(∇Y Z, X)W − R(Z, ∇Y X)W − R(Z, X)∇Y W+ ∇Z(R(X, Y )W) − R(∇ZX, Y )W − R(X, ∇ZY )W − R(X, Y )∇ZW. Using the torsion-freeness and (2), one can simplify the middle two columns to R([X, Z], Y )W + R([Y, X], Z)W + R([Y, X], Z)W. Now expand each R using its definition, the whole expression becomes a summation of 27 terms, the first 9 terms being + ∇X∇Y ∇ZW − ∇X∇Z∇Y W − ∇X∇[Y,Z]W + ∇[X,Z]∇Y W − ∇Y ∇[X,Z]W − ∇[[X,Z],Y ]W − ∇Y ∇Z∇XW + ∇Z∇Y ∇XW + ∇[Y,Z]∇XW, the second and third 9 terms are similar to the first 9 terms above: one just replace X, Y, Z by Y, Z, X and Z, X, Y respectively. It is not hard to check that all those expressions containing three ∇’s (12 terms in total) cancel out trivially, all those expressions containing two ∇’s (also 12 terms in total) cancel out by using the fact [X, Y ] = −[Y, X], and the remaining three terms ∇[[X,Z],Y ]W + ∇[[Y,X],Z]W + ∇[[Z,Y ],X]W = 0 in view of the Jacobi identity. □ In local coordinates we can write ∇∂n R = R l ijk ;n dxi ⊗ dxj ⊗ dxk ⊗ ∂l , Then the second Bianchi identity can be written as R l ijk ;n + R l jnk ;i + R l nik ;j = 0. Remark. If we denote P ⃝X,Y,Z to be the cyclic sum over X, Y, Z, then the first and second Bianchi identities can be written as X ⃝X,Y,Z R(X, Y )Z = 0 and X ⃝X,Y,Z (∇XR)(Y, Z, W) = 0. More generally, if ∇ is not torsion free, then in terms of the torsion tensor T , X ⃝X,Y,Z R(X, Y )Z = X ⃝X,Y,Z ￾ (∇XT )(Y, Z) + T (T (X, Y ), Z)
and X ⃝X,Y,Z (∇XR)(Y, Z, W) + X ⃝X,Y,Z R(T (X, Y ), Z)W = 0