Differential Geometry I: Einstein Summation Notation

Differential Geometry I: Einstein Summation Notation Helen K. Lei September 6, 2010

Acknowledgement. H. K. Lei wishes to thank Monica Visan and Rowan Killip for helpful discussions, etc. 1 η.κ.λ

Contents 1 Notation 4 2 Connection: D X Y, Y 5 3 Torsion: T (X, Y ) = D X Y D Y X [X, Y ], T γ αβ = Γγ αβ Γγ βα 6 4 Curvature: R(X, Y ) = D X D Y D Y D X D [X,Y ], R λ γαβ = αγ λ βγ βγ λ αγ + Γ µ βγ Γλ αµ Γ µ αγγ λ βµ 7 5 Covariant Derivatives: (1, 0), (1, 1) Tensors 9 6 Covariant Derivatives: (2, 0) Tensors 11 7 Christoffel Symbols 12 8 Riemannian Connection 14 9 Contraction Against g: Duality and Raising and Lowering of Indices 16 10 Laplacian (For Functions) 18 10.1 Contravariant Derivative and Gradient..................... 18 10.2 Laplacian as Trace of Hessian.......................... 18 10.3 Laplacian as div( f)............................... 20 10.4 Other (Coordinate) Expressions......................... 21 11 Covariant Derivative for General Tensors 22 12 Laplacian for Higher Order Tensors 24 13 Change of Variables, Pullback Metric, and Normal Coordinates 25 2

14 (Riemannian) Curvature Tensors and Bianchi Identities 27 15 Harmonic/Wave Coordinate Condition and Consequences 29 3 η.κ.λ

Chapter 1 Notation We have some manifold M and fix some frame { α } and the corresponding dual frame {θ γ } (so that θ γ ( α ) = δ β α, where δ β α = 1 if α = β and 0 otherwise). The tangent space at the point p M is denoted T p M and Γ(M) denotes the space of vector fields. We will use Einstein summation notation, i.e., repeated indices (one upper and one lower) is summed. By convention, covariant indices (e.g., corresponding to tangent basis element or components of dual vectors) are below whereas contravariant indices (e.g., components of tangent vectors or dual basis elements) are above. We will use D α (or α ) to denote the covariant derivative in the α th coordinate direction whereas α denotes directional derivative. Given a (semi )Riemannian metric g, g αβ = g( α, β ), whereas g αβ denotes the corresponding entry in the inverse matrix to g, i.e., g αβ g βγ = δ α γ. In many places (especially in the beginning) we are loosely following parts of [1]. 4

Chapter 2 Connection: D X Y, Y We start with the definition of a connection, which is a mapping D : T P (M) Γ(M) T P (M) satisfying bilinearity and Leibnitz rule: D X (fy ) = X(f)Y + fd X Y. Here X(f) is the usual directional derivative: X(f) = df(x). Writing X = X α α, Y = Y β β and expanding, we see that D X Y = X α [( α Y β ) β + Y β (D α β )], which leads to the Christoffel symbols Γ γ αβ such that D α β = Γ γ αβ γ so that D X Y = X α [( α Y β ) β + Y β Γ γ αβ γ]. Picking out the γ component (D X Y ) γ, we have (D X Y ) γ = X α [ α Y γ + Y β Γ γ αβ ] Setting X = α and abbreviating D α = D α, we have (D α Y ) γ = α Y γ + Y β Γ γ αβ. We define Y to be the (1, 1) tensor given by the matrix coefficients (D α Y ) γ, i.e., Y = (D α Y ) γ (θ α γ ) where {θ α } is the dual frame. Notice that Y satisfies Y (X, f) = (D X Y )(f) 5

Chapter 3 Torsion: T (X, Y ) = D X Y D Y X [X, Y ], T γ αβ = Γγ αβ Γγ βα Given the definition of the connection, we can now define a host of objects, starting with the torsion, which is a mapping T : Γ(M) Γ(M) Γ(M) defined by T (X, Y ) = D X Y D Y X [X, Y ]. The commutator [X, Y ] is interpreted via its action on functions f C (M): Leibnitz rule, we see that we have e.g. (again with X = X α α Y = Y β β ), Using XY (f) = X α α (Y (f)) = X α α (Y β β f) = X α (( α Y β ) β f + Y β αβ f) (Here αβ = α β = βα.) We thus have [T (X, Y )] γ = [X α (D α Y ) γ Y β (D β X) γ ] [X α α Y γ Y β β X γ ]. (the first term on the right cancels the corre- Now we have e.g., (D α Y ) γ = α Y γ + Y β Γ γ αβ sponding term from XY ) we see that [T (X, Y )] γ = X α Y β Γ γ αβ Xα Y β Γ γ βα = X α Y β (Γ γ αβ Γγ βα ) := T γ αβ Xα Y β. Note that the torsion does not depend on the first derivatives of the components of X and Y (these terms canceled out). Therefore, as before, we can define a (2, 1) tensor by the matrix coefficients (T γ αβ ): T (X, Y )f := T γ αβ (θα θ β γ )(X, Y, f). 6

Chapter 4 Curvature: R(X, Y ) = D X D Y D Y D X D [X,Y ], R λ γαβ = α Γ λ βγ β Γ λ αγ+γ µ βγ Γλ αµ Γ µ αγγ λ βµ The curvature is a mapping defined by R : Γ(M) Γ(M) Hom(Γ, Γ) R(X, Y ) = D X D Y D Y D X D [X,Y ]. Let us write X = X α α, Y = Y β β, Z = Z γ γ. First, we have as before that (assuming αβ = βα ) [X, Y ] λ = X α ( α Y λ ) Y β ( β X λ ) So On the other hand, D [X,Y ] Z = [X, Y ] λ (( λ Z γ ) γ + Z γ D λ γ ) D X D Y Z = X α D α (Y β D β Z) = [X, Y ] λ (( λ Z γ ) γ + Z γ Γ µ λγ µ) = [X, Y ] λ (( λ Z µ ) + Z γ Γ µ λγ ) µ = X α [( α Y β )D β Z + Y β D α D β Z] = X α ( α Y β )( β Z λ + Z γ Γ λ βγ ) λ + X α Y β D α D β Z = X α ( α Y λ )( λ Z µ + Z γ Γ µ λγ ) µ + X α Y λ D α D λ Z, where in the last line we have performed a change of indices β λ and λ µ, from which it is clear that the first term matches the contribution from the XY part of D [X,Y ] Z. Similarly, the corresponding term from the Y X part of D [X,Y ] Z matches the corresponding term from D Y D X Z. I.e., we have observed that D X D Y Z = D [X,Y ] Z + X α Y β (D α D β D β D α )Z or R(X, Y )Z = X α Y β (D α D β D β D α )Z. 7

Next we have D α D β Z = D α [( β Z γ ) γ + Z γ D β γ ] = D α [(( β Z µ ) + Z γ Γ µ βγ ) µ] = [ αβ Z µ + α (Z γ Γ µ βγ )] µ + [( β Z µ ) + Z γ Γ µ βγ ]Γλ αµ λ = [ αβ Z λ + α (Z γ Γ λ βγ ) + (( βz µ ) + Z γ Γ µ βγ )Γλ αµ] λ = [ αβ Z λ + ( α Z γ )Γ λ βγ ) + Zγ ( α Γ λ βγ ) + (( βz µ ) + Z γ Γ µ βγ )Γλ αµ] λ Interchanging α and β, we find that D β D α Z = [ βα Z λ + ( β Z γ )Γ λ αγ + Z γ ( β Γ λ αγ) + (( α Z µ ) + Z γ Γ µ αγ)γ λ βµ ] λ. First we can cancel the αβ and βα terms. Next observe that 1) ( α Z γ )Γ λ βγ matches up with ( α Z µ )Γ λ βµ and 2) ( βz µ )Γ λ αµ matches up with ( β Z γ )Γ λ αγ if we rename γ µ. Thus all terms involving derivatives of Z γ will be gone, and we are left with [ ] [R(X, Y )Z] λ = X α Y β Z γ α Γ λ βγ βγ λ αγ + Γ µ βγ Γλ αµ Γ µ αγγ λ βµ := X α Y β Z γ Rγαβ λ. I.e., Rγαβ λ is the λ component of R( α, β ) γ. Finally, in this case we have a (3, 1) tensor [R(X, Y )Z]f := Rγαβ λ (θγ θ α θ β λ )(Z, X, Y, f). 8 η.κ.λ

Chapter 5 Covariant Derivatives: (1, 0), (1, 1) Tensors For a function, the covariant derivative corresponds to D X (f) = X(f), the usual directional derivative, and for vector fields D X Y is the connection. We can now generalize the notion of covariant derivatives to higher order tensors in the natural way, satisfying the following properties: D X preserves the type of tensor. D X commutes with contraction, in the sense that e.g., if ω is a 1 form and Y is a vector field, then D X [ω(y )] = (D X ω)(y ) + ω(d X (Y )). D X satisfies the Leibnitz rule: D X (u v) = (D X u) v + u (D X v). Let us now compute the covariant derivative of a 1 form ω = ω γ θ γ. Then the Leibnitz rule will allow us to compute the covariant derivative of general (r, s) tensors by induction. Let us first consider X = X α α and Y = Y β β to show that D X ω is indeed tensorial. It is trivial to contract ω with Y : We therefore have ω(y ) = ω γ θ γ (Y β β ) = ω γ Y β δ γ β = ω γy γ. D X (ω γ Y γ ) = D X (ω(y )) = (D X ω)(y ) + ω(d X (Y )). We will therefore obtain an expression for D X (ω)(y ) if we compute D X (ω γ Y γ ) and ω(d X (Y )). The resulting expression will show that D X (ω) is a tensor, and then the appropriate coordinate expression will emerge by considering X = α, Y = β. First, ω γ Y γ is a function, so D X (ω γ Y γ ) = X α α (ω γ Y γ ) = X α [( α ω γ )Y γ + ω γ ( α Y γ )]. On the other hand, D X (Y ) is the connection, so as before, we obtain D X (Y ) = X α [( α Y λ ) + Y β Γ λ αβ ] λ. 9

So we have ω(d X (Y )) = ω γ X α [( α Y γ ) + Y β Γ γ αβ ]. Therefore (D X ω)(y ) = D X (ω γ Y γ ) ω(d X (Y )) = X α [( α ω γ )Y γ + ω γ ( α Y γ )] ω γ X α [( α Y γ ) + Y β Γ γ αβ ] = X α [( α ω γ )Y γ ω γ Y β Γ γ αβ ] = X α Y β [ α ω β ω γ Γ γ αβ ] := X α Y β (D α ω) β. I.e., D X ω = X α D α ω = [X α (D α ω) β ]θ β. Thus D X ω(y ) is indeed tensorial, and we have the (2, 0) tensor ω: ω(x, Y ) := (D α ω) β (θ α θ β )(X, Y ). For comparison purposes, let us now write down the covariant derivatives of a (1, 0) tensor and (0, 1) tensor: (D α Y ) λ = α Y λ + Y β Γ λ αβ (D α ω) λ = α ω λ ω γ Γ γ αλ (5.1) So from the Leibnitz rule, we see that e.g., the covariant derivative of v u a (1, 1) tensor would be (with u = u µ µ, v = v µ θ µ ) D α (v u) = ( α u λ + u β Γ λ αβ )(v λ) + (θ λ u)( α v λ v γ Γ γ αλ ) = v µ ( α u λ + u β Γ λ αβ )(θµ λ ) + (θ λ µ )( α v λ v γ Γ γ αλ )uµ = v µ ( α u λ + u β Γ λ αβ )(θµ λ ) + (θ µ λ )( α v µ v γ Γ γ αµ)u λ = [v µ ( α u λ + u β Γ λ αβ ) + ( αv µ v γ Γ γ αµ)u λ ](θ µ λ ) = [D α (v u)] λ µ (θ µ λ ), and thus, we have a (2, 1) tensor (u v): [ (u v)](x, Y, f) := [D α (u v)] λ µ (θ α θ µ λ )(X, Y, f). Remark. Another way to compute the covariant derivative would be to first write e.g., v u = ((h µ λ θλ ) µ ) and then apply the Leibnitz rule. Indeed, it is with this way of computing that we can see that the formulas in (5.1) generalize to higher order tensors. We will shortly carry out the computation for (2, 0) tensors in this way. It is perhaps clear by now the reason for the name covariant derivative: Starting with a k times covariant tensor ω, we end up with a k + 1 times covariant tensor ω, by anti contracting with a vector field direction of differentiation. 10 η.κ.λ

Chapter 6 Covariant Derivatives: (2, 0) Tensors Let now g be a (2, 0) tensor, and let g λµ = g( λ, µ ) be its components, so that g = g λµ (θ λ θ µ ). By Leibnitz rule, we then have Proceeding as before, we find that D α g = D α (g λµ θ λ ) θ µ + g λµ θ λ D α (θ µ ). D α [θ µ ( γ )] = (D α θ µ )( γ ) + θ µ (D α γ ) = [D α θ µ ]( γ ) + Γ µ αγ. Since θ µ ( γ ) = δ µ γ, the left hand side is equal to zero, and hence Similarly, [D α θ µ ]( γ ) = Γ µ αγ. D α [g λµ θ λ ( β )] = D α (g λµ θ λ )( β ) + g λµ θ λ [D α β ] = D α (g λµ θ λ )( β ) + g λµ Γ λ αβ. Since D α [g λµ θ λ ( β )] = D α (g λµ δ λ β ) = αg βµ, we learn that Thus, we learn that D α (g λµ θ λ )( β ) = α g βµ g λµ Γ λ αβ. D α g = ( α g βµ g λµ Γ λ αβ )θβ θ µ g λµ Γ µ αγθ λ θ γ. Renaming indices µ γ, λ β, we see that D α g = [ α g βγ g λγ Γ λ αβ g βµγ µ αγ](θ β θ γ ). We note that this is a generalization of the second formula in (5.1). Remark. The coefficient in the last display is the (α, β, γ) th component of the (3, 0) tensor g, where if X, Y, Z are vector fields, then g(x, Y, Z) = (D X g)(y, Z). By the usual abuse of notation (or poetic license), we may denote such a component by a host of different things: (D α g) βγ = ( g) αβγ = α g βγ = g βγ;α. 11

Chapter 7 Christoffel Symbols Recall that the Christoffel symbols are defined by Equivalently, Γ γ αβ = (D α β ) γ = θ γ (D α β ). D α β = Γ γ αβ γ. Let us consider two different local charts, or (from the pointwise perspective) a change of bases: { α } and { k } such that α = A k α k. (Of course, from the coordinate perspective, it is clear that A k α = xk x α.) So if X is a vector field, then we can write X α A k α k = X α α = X = X k k, and if ω is a 1 form, then ω λ B λ k θ k = ω λ θ λ = ω = ω k θk, where (B λ k ) is the matrix inverse of (Ak λ ). Now let ω = ω λ θ λ = ω k θk be a 1 form and let us express the scalar quantity ω[d X (Y )] in both coordinate systems: ω[d X (Y )] = ω λ θ λ X α D α (Y β β ) = ω λ X α [( α Y λ ) + Y β Γ λ αβ ] = ω k X i [( i Y k ) + Y j Γ k ij]. On the other hand, changing basis explicitly, we also have ω[d X (Y )] = ω λ B λ k Xα A i α[ i (A k γy γ ) + Y β A j β Γk ij] = ω λ B λ k Xα A i α[y γ i A k γ + A k γ i Y γ + Y β A j β Γk ij] Choosing X and ω so that (for any particular indices λ and α), we have ω l = δ λ,l, X l = δ α,l, we get that α Y λ + Y β Γ λ αβ = Bλ k Ai α[y γ i A k γ + A k γ i Y γ + Y β A j β Γk ij]. (7.1) We can also rewrite α Y λ = A i α i Y λ 12

and note that B λ k Ak γ = δ λ γ, and therefore in (7.1), α Y λ cancels the corresponding term on the right hand side. Finally, choosing Y so that (for any particular index β), we have Y l = δ β,l, we see that Γ λ αβ = Bλ k Ai α[ i A k β + Aj β Γk ij] = B λ k Ai αa γ i γa k β + Bλ k Ai αa j β Γk ij = B λ k ( αa k β ) + Bλ k Ai αa j β Γk ij, where the last equality follows directly since A i αa γ i = δγ α. We see that Γ λ αβ is not the component of a tensor, since that would mean the Bλ k ( αa k β ) term should be absent. On the other hand, if Γ λ αβ represents another connection, then by the same computation (since Bk λ( αa k β ) does not depend on the connection, only the coordinates) Cαβ λ = Γ λ αβ Γλ αβ is indeed the component of a (2, 1) tensor. Conversely, we can add any (2, 1) tensor to a connection to form another connection (since tensoriality ensures the relevant properties of a connection). Thus connections are unique up to adding (2, 1) tensors. 13 η.κ.λ

Chapter 8 Riemannian Connection A (pseudo) Riemannian manifold is a manifold equipped with a symmetric, invertible 2 form g. Here symmetric means g(x, Y ) = g(y, X) and invertible means the matrix with entries g αβ = g( α, β ) is invertible. A Riemannian connection is a connection which is torsion free and commutes with the covariant derivative, i.e., it satisfies Γ γ αβ = Γγ βα g = 0 Given these conditions, we can explicitly solve for the Γ γ αβ s. From before, we have (cyclically permuting indices) 0 = (D α g) βγ = α g βγ g λγ Γ λ αβ g βλγ λ αγ 0 = (D β g) γα = β g γα g λα Γ λ βγ g γλγ λ βα (8.1) 0 = (D γ g) αβ = γ g αβ g λβ Γ λ γα g αλ Γ λ γβ Note that a priori there are 3! = 6 such equations, but since g αβ = g βα and Γ γ αβ = Γγ βα, we have also that (D α g) βγ = (D α g) γβ and hence we only have three equations. Solving these equations, we learn that Γ λ αβ = 1 2 [ αg βγ + β g γα γ g αβ ]g λγ. (8.2) We notice that equations (8.1) directly imply that the connection is metric (and vice versa), i.e., Z g(x, Y ) = g(d Z X, Y ) + g(x, D Z Y ). (8.3) By tensoriality, it is enough to check this relation for coordinate vector fields: γ g( α, β ) = γ g αβ = g λβ Γ λ γα + g αλ Γ λ γβ = g(γ λ γα λ, β ) + g( α, Γ λ γβ λ) = g(d γ α, β ) + g( α, D γ β ). 14

Remark. Conversely, it is possible to first define the covariant derivative implicitly via the equation (which holds in R d ) 2g(D Y X, Z) = (L X g)(y, Z) + (dθ X )(Y, Z) where L X denotes the Lie derivative, and θ X (Y ) = g(x, Y ), and then compute in coordinates to see that indeed the Christoffel symbols are as given in (8.2). This is done in [2]. Finally, we note that cyclically permuting indices in the last computation and adding two of the resulting equations and subtracting the third (and using that the connection is torsion free) we also learn that the Riemannian connection satisfies Finally, tautologically, we have α g βγ + β g γα γ g αβ = 2g(D α β, γ ) = 2Γ λ αβ g λγ. Γ λ αβ = g(d α β, γ )g λγ := Γ αβ,γ g λγ. 15 η.κ.λ

Chapter 9 Contraction Against g: Duality and Raising and Lowering of Indices We note that contraction against g αβ dualizes the tensor and hence lowers indices (since indices are placed oppositely of the actual object), whereas contraction against g αβ has the opposite effect. That is, contraction against g αβ sends (p, q) tensors to (p+1, q 1) tensors, whereas contraction against g αβ sends (p, q) tensors to (p 1, q + 1) tensors. First we have that X α g αβ = g(x, β ) := X β represents the 1 form dual to the vector field X = X α α : X β θ β : Y g(y, X), i.e., X β θ β (Y ) = Y T [g]x = Y β g αβ X α = Y β X β = g(x, Y ). In particular, X β g αβ X α = g(x, X). Thought another way, (in matrix form) [g][x β ] gives the column vector [g( α, X)] (here [X β ] denotes the column vector with entries being components of X) so [g] 1 [g][x β ] = [X], i.e., g γα g αβ X β = X γ, or X = g γα g( α, X) γ In the particular cases when { α } form an orthonormal basis with respect to g, we have [g] = Id, so that X = g( α, X) α. The dual situation is if ω α θ α is a 1 form, then X β = g αβ ω α are the components of a vector. What we have is nothing other than the usual duality provided by g: We have maps T p M T p M : X ω(x) : ω(x)( ) = g(x, ) (lowering of indices) and its inverse T p M T p M : ω X(ω) : ω( ) = g(, X(ω)) (raising of indices) (X(ω) is exactly the vector provided by the Riesz Representation Theorem.) 16

Indeed, we have that ω( α ) = ω α, and thus the vector X(ω) satisfies g( α, X(ω)) = ω α and thus as observed before, we have [g] 1 [ω α ] = [X(ω)] : g βα ω α = (X(ω)) β. 17 η.κ.λ

Chapter 10 Laplacian (For Functions) 10.1 Contravariant Derivative and Gradient By raising indices with respect to g αβ (dualizing) we obtain the contravariant derivative: D β = g αβ D α. Let us first understand this in the case of a function, which we view as a 0 form. In accord with the notation we have used to denote the covariant derivative, f is a 1 form: f = ( f) α θ α = (D α f)θ α := ( α f)θ α, so that α f denote components of a 1 form. Given a metric tensor g, the usual dualizing gives a vector field X( f) such that for all α, Therefore, α f = g(x( f), α ). X( f) = (g αγ α f) γ = (g αγ D α f) γ = (D γ f) γ := ( γ f) γ, so that γ f denote the components of a vector field. Finally, by the usual abuse of notation, we will denote X( f) = f = ( γ f) γ. 10.2 Laplacian as Trace of Hessian We define the Laplacian g f := tr( ( f)) = θ α ( α ( f)) i.e., it is the sum of the α component of the covariant derivatives of f in the direction α. A more general formulation is as follows: Let us consider the second order operator defined implicitly as g( X f, Y ) = D X g( f, Y ) g( f, D X Y ) = D X D Y f D DX Y f. 18

(Here we have used the fact that covariant differentiation is commensurate with the metric g.) This indeed implicitly defines the Hessian of f and further, the definition Hessf(X, Y ) = g( X f, Y ) (Hessf(X, Y ) =) 2 X,Y f = (D X D Y D DX Y )f can be easily generalized to (p, 0) tensors as the second covariant derivative, which is then a (p+2, 0) tensor (here we are viewing f as a 0 form). It is easy to check that for functions, the second covariant derivative is symmetric. Indeed, let h αβ denote components of the Hessian, then h αβ = g(d α f, β ) = α β f Γ γ αβ γf, and since the connection is torsion free and α β = β α, we can clearly interchange α and β in the above. Finally, we can retrieve g f as the trace of Hess(f), i.e., h αβ g αβ = g αβ g(d α f, β ) = g αβ (D α f) γ g γβ = δ α γ (D α f) γ = θ α (D α f). (In the particular case where we have an orthonormal frame, we have g f = α h αα.) Let us now compute g f in coordinates. We have Now using the relation that g f = θ α ( α ( f)) the last term in the above display becomes = θ α [ α (g βγ β f) γ + (g βγ β f)γ λ αγ λ ] = g αβ α β f + α g αβ β f + (g βγ β f)δ α λ Γλ αγ. Γ λ αγ = 1 2 gµλ ( α g γµ + γ g µα µ g αγ ), 1 2 ( βf)g βγ δ α λ gµλ ( α g γµ + γ g µα µ g αγ ) = 1 2 ( βf)g βγ g µλ ( λ g γµ + γ g µλ µ g λγ ) = 1 2 ( βf)g βγ g µλ ( γ g µλ ), where to obtain the first equality we have performed the trace δλ α and to obtain the second equality, we note that interchanging λ and µ and using that g is symmetric, we have g µλ λ g γµ = g λµ µ g γλ = g µλ µ g λγ and hence the only term that remains is g µλ ( γ g µλ ). We have thus obtained the expression (after γ α) [ g f = g αβ α + α g αβ + 1 ] 2 gαβ g µλ ( α g µλ ) β f 19 η.κ.λ

10.3 Laplacian as div( f) On the other hand, we may start with the volume form (see e.g., [3]) dv = g dx 1 dx n (Here g = det([g])) which induces global inner products X, Y = g(x, Y ) dv, f, g = M M fh dv. Here X, Y are vector fields and f, h are functions. Now via integration by parts we may implicitly define divergence as which by explicit computation gives divx, f = X, f divx = 1 g α (X α g ). Indeed, with X = X α α, X, f = g(x, f) dv = X α α f g dx 1 dx n = α (X α g )f dx 1 dx n = 1 α (X α g )f dv g = (divx) f dv Plugging in X α = α f = g αβ β f, we obtain that g f = 1 g α (g αβ g β f) Let us check that this agrees with our previous definition. We have [ 1 α (g αβ g β f) = g αβ α + α g αβ + ] α g g αβ β f g g Now we recall Jacobi s Formula Thus we have α g g = [ g αβ α + α g αβ + 1 2 gµλ g = g g µλ. = g µλ α g µλ g and the two definitions of the Laplacian are the same. = g µλ α g µλ ] α g g αβ β f. g 20 η.κ.λ

10.4 Other (Coordinate) Expressions Recall the expression g f = [ g αβ α + α g αβ + 1 ] 2 gαβ g µλ ( α g µλ ) β f. We will now rewrite this expression in terms of the Christoffel symbols. We note that Now we use the fact that g αβ Γ γ αβ γ = 1 2 gαβ g γλ ( α g βλ + β g λα λ g αβ ) γ. µ (g αβ g βγ ) = µ δ α γ = 0 to rewrite the first two terms in the previous expression as Γ γ γ := g αβ Γ γ αβ γ = 1 2 gαβ (g βλ α g γλ + g λα β g γλ + g γλ λ g αβ ) γ = 1 2 (δα λ αg γλ + δ β λ βg γλ + g αβ g γλ λ g αβ ) γ (Incidentally, by symmetry of g, = ( α g αγ + 1 2 gαβ g γλ λ g αβ ) γ = ( α g αγ + 1 2 gγλ g αβ λ g αβ ) γ. 1 2 gµλ ( α g µλ ) = 1 2 gµλ (Γ γ αµg λγ + Γ γ αλ g µγ) = g µλ Γ γ αµg λγ = Γ µ αµ.) Denoting by g = g αβ α β the reduced wave operator, we see that we have g = g Γ γ γ. Alternatively, g f = D α ( α f) (D α α )(f). 21 η.κ.λ

Chapter 11 Covariant Derivative for General Tensors Let us recall that the covariant derivative preserves the type of tensor, and starting with the connection D X Y we are able to find the covariant derivative of 1 forms since D X commutes with evaluation: If ω is a 1 form, then Given that D X also satisfies Leibnitz rule: D X [(ω(y )] = (D X ω)(y ) + ω(d X (Y )) D X (u v) = (D X u) v + u (D X v), we should be able to compute the covariant derivative of general (p, q) tensors by induction. We will do so here and along the way develop some notation to handle higher order tensors. First let us recall that if X = X α α is a vector field and ω = ω α θ α is a 1 form, then D β X = ( β X λ + X α Γ λ βα ) λ D β ω = ( β ω λ ω γ Γ γ βλ )θλ. We will shortly see that with appropriate notation the above form easily generalize to (p, 0) and (0, q) tensors, respectively. To start we will use multi index: [α] := α 1, α 2,..., α p and write e.g., X [α] [α] := X α 1,α 2,...,α p ( α1 α2 αp ). First we let T be a (0, q) tensor so that T = T [α] [α]. Applying the Leibnitz rule once, we get that D β T = ( β T [α] ) [α] + T [α] (D β [α] ). To understand the second term, let us for now suppose p = 2, so that we have D β ( α1 α2 ) = D β α1 α2 + α1 D β α2 = Γ γ βα 1 γ α2 + α1 Γ γ βα 2 γ. 22

This pattern clearly persists if p > 2: Namely, we will have sum of terms where Γ γ βα k γ will replace the γ th spot in the tensor product. Let us now introduce the notation (α k) [α] (Γ γ βα k γ ) to mean exactly this. The parenthesis around α k in the superscript denotes that the term following the should be inserted in the k th spot, and the α k s are summed as usual (ranging over all single indices contained in [α]). Thus we learn that D β (T [α] [α] ) = ( β T [α] ) [α] + T [α] (α k) [α] (Γ γ βα k γ ). We would now like to to re sum so that both terms isolate the index γ. For this purpose let us introduce the assignment operator: so that (using multi linearity) D β (T [α] [α] ) = Next let S be a (p, 0) tensor so that Proceeding as before we have T [α] k:γ := T [α] δ α k,γ [ β T [α] k:γ + T [α] Γ γ βαk ] ( (α k) [α] γ ). S = S [α] θ [α]. D β S = ( β S [α] )θ [α] + S [α] (D β θ [α] ) = ( β S [α] )θ [α] + S [α] θ [α] (α k ) ( Γα k βγ θγ ) [ ] = ( β S [α] k:γ) S [α] Γ α k βγ (θ [α] (α k ) θγ ). Finally, if R is now a general (p, q) tensor so that then we have R = R [µ] [α] (θ[α] [µ] ), D β R = ( β R [µ] [α] )(θ[α] [µ] ) + R [µ] [α] D β(θ [α] [µ] ) = ( β R [µ] [α] )(θ[α] [µ] ) + R [µ] [α] = ( β R [µ] [α] )(θ[α] [µ] ) + R [µ] [α] = ( β R [µ] [α] )(θ[α] [µ] ) + R [µ] [α] [ = β R [µ] k:γ [α] l:λ + R[µ] [α] k:γ Γλ βα l R [µ] l:λ [α] [ ] θ [α] (α k ) ( Γα k βγ θγ ) [µ] + θ [α] D β [µ] [ ] θ [α] (α k ) ( Γα k βγ θγ ) [µ] + θ [α] (µ l) [µ] Γ γ βµ l γ [ Γ λ βµ l θ [α] ( (µ l) Γ α k βγ [µ] λ ) Γ α k βγ (θ[α] ] [(θ [α] (α k ), (α l) [µ] ) (θ γ, λ )] (α k ) θγ ) [µ] ] 23 η.κ.λ

Chapter 12 Change of Variables, Pullback Metric, and Normal Coordinates In many cases it will be convenient to use the normal coordinate system (by tensoriality, it is sufficient to prove a tensor identity in any convenient coordinate system). Such a coordinate system at a point P is a local coordinate system such that g αβ (P ) = δ β α and γ g αβ (P ) = 0, for all α, β, γ, that is, we require orthonormality and vanishing of the Christoffel symbols at the point P. It is straightforward to construct such a coordinate system: Choose a local chart ϕ around P so that ϕ(p ) = 0 and let {x α } denote the local coordinate system. Via a linear transformation, choose a frame in R n so that g αβ (P ) = δ β α. Here g is the appropriate pullback (see below) of the Euclidean inner product. Change coordinates via a map F : {x α } {y β } by x λ y λ = Γ λ αβ (P )yα y β and note that at P, [DF ] = Id, where [DF ] denotes the Jacobian matrix, so that by the inverse function theorem {y β } are coordinates. We will now need to consider the pullback metric, which in a general context is defined as follows: If F : (M, g M, x) (N, g N, y), then g M (X, Y ) = g N (F (X), F (Y )). Here F : T p M T F (p) M is the pushforward of F, defined so that if h : N R, then By the chain rule, we learn that [X(h F )](p) = [F (X)(h)](F (p)). (h F ) x α (p) = h y β y β (F (p)). xα That is, α = (DF ) β α β 24

or inverting, β = ((DF ) 1 ) α β α = (DF 1 ) α β α, where the last equality follows by the inverse function theorem. Note that DF has one covariant and one contravaiant index: The vector fields transform covariantly (in the same direction as the function F ) whereas the coordinates themselves transform contravariantly. Notice also that strictly speaking we have identified [DF ] and [DF ] T, since it is clear that e.g., (DF ) β α means yβ x α. Thus if we change variables from x to y and F : (M, g, y) (N, g, x), then in the new coordinates the components of the metric tensor are given by In our case, we have g αβ = g((df 1 ) λ α λ, (DF 1 ) µ β µ). [F (y)] λ = x λ = y λ Γ λ αβ yα y β, so (DF 1 ) λ µ = δ λ µ Γ λ αµy α and (recall that g is the metric in the new coordinates {y β }). g( α, β ) = g((df 1 ) µ α µ, (DF 1 ) λ β λ) = (δ α µ Γ µ γαy γ )(δβ λ Γλ γβ yγ )g µλ = [δ αδ µ β λ δµ αγ λ γβ yγ δβ λ Γµ γαy γ + Γ µ γαγ λ γβ (yγ ) 2 ]g µλ = g αβ (g αλ Γ λ γβ + g µβγ µ γα)y γ + Γ µ γαγ λ γβ (yγ ) 2 g µλ. Now we recall that at the point P, DF = Id and so y = x = ϕ(p ) = 0 and therefore γ g αβ = γ g αβ (g αλ Γ λ γβ + g µβγ µ γα) = 0, where again we have used the fact that γ g( α, β ) = g(d γ α, β ) + g( α, D γ β ). 25 η.κ.λ

Chapter 13 (Riemannian) Curvature Tensors and Bianchi Identities Next let us consider the curvature R(X, Y )Z = (D X D Y D Y D X D [X,Y ] )Z and the associated (3, 1) tensor Rγαβ λ (θγ θ α θ β λ ) where Rγαβ λ = [R( α, β ) γ ] λ = α Γ λ βγ βγ λ αγ + Γ µ βγ Γλ αµ Γ µ αγγ λ βµ. In a normal coordinate system, the last two terms vanish, and we get that Rγαβ λ = αγ λ βγ βγ λ αγ. Cyclically permuting α, β, γ, we also get R λ αβγ = βγ λ γα γ Γ λ βα R λ βγα = γγ λ αβ αγ λ γβ. Adding these three equations and using the fact that the connection is torsion free, we get the first Bianchi identity R λ γαβ + Rλ αβγ + Rλ βγα = 0. Next we will differentiate the curvature, to obtain Cyclically permuting µ, α, β, we also get µ R λ γαβ = µαγ λ βγ µβγ λ αγ. α R λ γβµ = αβγ λ µγ αµ Γ λ βγ β R λ γµα = βµ Γ λ αγ βα Γ λ µγ. Adding these three equations and using the fact that αβ = α β = β α, we get the second Biachi identity µ R λ γαβ + αr λ γβµ + βr λ γµα = 0. If we lower index, then we get g µλ R λ γαβ = R µγαβ = g(r( α, β ) γ, µ ), 26

which are components of a (4, 0) Riemannian curvature tensor X, Y, Z, W g(r(x, Y )Z, W ) := R(X, Y, Z, W ). Note that from the explicit expression for the components, we see that R(X, Y, Z, W ) = R(Y, X, Z, W ) = R(Y, X, W, Z). Since the upper index in Rγαβ λ is unchanged in derivation of the Bianchi identities, we recover the Bianchi identities for these dualized tensors R µγαβ + R µαβγ + R µβγα = 0 σ R µγαβ + α R µγβσ + β R µγσα = 0. Using the first Bianchi identity together with the symmetry property derived, we also obtain that R is symmetric in the first two and last two entries R(X, Y, Z, W ) = R(Z, W, X, Y ). Taking trace of the α and µ components in R µγαβ, we obtain (components of) the Ricci tensor: Since R µγαβ = R αβµγ, we easily see that R βγ = g αµ R µγαβ = g αµ g(r( α, β ) γ, µ ) = [R( α, β ) γ )] α. R βγ = R γβ, i.e., the Ricci tensor is symmetric. Also, we have the explicit expression R γβ = g µα R µγαβ = g µα g µλ ( α Γ λ βγ βγ λ γα) = δ α λ ( αγ λ βγ βγ λ γα) = α Γ α βγ βγ α γα. Finally, taking the trace one last time, we obtain the scalar curvature R = g βγ R βγ. 27 η.κ.λ

Chapter 14 Harmonic/Wave Coordinate Condition and Consequences By (locally) solving the Dirichlet problem, we may choose harmonic coordinates x λ so that g x λ = 0. In these coordinates we see that g x λ = [g αβ α β + Γ γ γ ]x λ = Γ λ = 0. Therefore in harmonic coordinates we see that g = g = g = g αβ α β. Next we can see that the Ricci tensor also simplifies, up to O(( g) 2 ). We have R γβ = α Γ α βγ βγ α γα + O(( g) 2 ) = 1 2 gαλ [ α ( β g γλ + γ g λβ λ g βγ ) β ( α g γλ + γ g λα λ g αγ )] + O(( g) 2 ) = 1 2 gαλ [ α ( γ g λβ λ g βγ ) β ( γ g λα λ g αγ )] + O(( g) 2 ) = 1 2 gαλ α λ g βγ + 1 2 gαλ ( α γ g λβ β γ g λα + β λ g αγ ) + O(( g) 2 ) (Note that we are no longer in a normal coordinate system, but the O(Γ 2 ) terms are also O(( g) 2 ).) Now let us observe that given the wave coordinate condition that g αβ Γ λ αβ = 0, we have g αλ γ ( α g λβ β g λα + λ g αβ ) = g αλ γ (Γ µ αλ g βµ) Thus we may continue the expression for R βγ as = g αλ g βµ γ Γ µ αλ = O(( g) 2 ) R βγ = 1 2 gαλ α λ g βγ + 1 2 gαλ γ ( α g λβ β g λα + λ g αβ ) + 1 2 gαλ ( β λ g αγ γ λ g αβ ) + O(( g) 2 ) = 1 2 g g βγ + 1 2 gαλ ( β λ g αγ γ λ g αβ ) + O(( g) 2 ). 28

Finally we note that the remaining term other than g g βγ is symmetric in β and γ and hence in the expression 2R βγ = R βγ + R γβ it would be zero and so we conclude that R βγ = 1 2 g g βγ + O(( g) 2 ). 29 η.κ.λ

Bibliography [1] T. Aubin. A Course in Differential Geometry. Providence, RI: American Mathematical Society (2001). [2] P. Petersen. Riemannian Geometry. New York, NY: Springer (2006). [3] S. Rosenberg. The Laplacian on a Riemannian Manifold. Cambridge, UK: Cambridge University Press (1997). 30