Section 2. Basic formulas and identities in Riemannian geometry

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1 Section 2. Basic formulas and identities in Riemannian geometry Weimin Sheng and 1. Bianchi identities The first and second Bianchi identities are R ijkl + R iklj + R iljk = 0 R ijkl,m + R ijlm,k + R ijmk,l = 0. We may prove the above formulas directly. But it would be easy to prove them by using the Cartan structure equations. The twice contracted second Bianchi identity is g ij i R jk = 1 2 kr. This is equivalent to the Einstein tensor R ij 1Rg 2 ij being divergence-free: div (Rc 12 Rg = 0. Theorem 1 (Schur Let M be connected and dim M 3. If the sectional curvature of M is constant at each point, i.e. K (P = f (x for all 2-plane P T x M, then f (x = const., and M is a space form. Likewise, if the Ricci curvature is constant at each point, i.e. R ik = c (x g ij, then c (x = const., and M is Einstein. 1

2 Proof. By the definition of the sectional curvature, we have R ijkl = f (x (g ik g jl g il g jk then by the second Bianchi identity, we have (take normal coordinates at x M 0 = R ijkl,m + R ijlm,k + R ijmk,l = f,m (δ ik δ jl δ il δ jk + f,k (δ il δ jm δ im δ jl + f,l (δ im δ jk δ ik δ jm Since we assume dim M 3, for each m, we can find m, i, j, k, l with i = k, j = l and i m, j m and i j. Then we have f,m = 0. Since M is connected, then f (x = const.. The second claim follows in the same manner. QED 2. Lie derivative. For fixed point q M, one seeks a curve through q whose tangent vector at each point coincides with the value of a given vector X at this point, i.e. a curve c q (t is always tangent to the vector field X. Now we want to fix t and vary q; we put ϕ t (q := c q (t. Theorem 2. We have ϕ t ϕ s (q = ϕ t+s (q if s, t, s + t I q (2.1 and if ϕ t is defined on U M, it maps U diffeomorphically onto its image. Proof. We have c q (t + s = X (c q (t + s, hence c q (t + s = c cq(s (t. Starting from q, at time s one reaches the point c q (s, and if one proceeds a time t further, one reaches c q (t + s. One therefore reaches the same point if one walks from q on the integral curve for a time t + s, or if one walks a time t from c q (s. This shows (2.1. Inserting t = s into (2.1 for s I q, we obtain ϕ s ϕ s (q = ϕ 0 (q = q. 2

3 Thus the map ϕ s is the inverse of ϕ s, and the diffeomorphism property follows. QED Definition 1 A family (ϕ t t I ( I open interval with 0 I of diffeomorphisms from M to M satisfying (2.1 is called a local 1-parameter group of diffeomorphisms. In general, a local 1-parameter group need not be extendable to a group, since the maximal interval of definition I q of c q need not be all of R. This is already seen by easy example, e.g. M = R, X (τ = τ 2 d, i.e. c dτ (t = c 2 (t as differential equation. However, on a compact differentiable manifold, any vector field generates a 1-parameter group of diffeomorphisms. We call a vector field X is complete if there is a 1-parameter group of diffeomorphisms {ϕ t } t R generated by X. If M n is closed, then any smooth vector field is complete. Let α be a tensor of type (r, s and let X be a complete vector field generating a global 1-parameter group of diffeomorphisms ϕ t. The Lie derivative of α with respect to X is defined by 1 L X α =: lim t 0 t (α (ϕ t α. (2.2 Here (ϕ t : T p M T ϕt(pm is the differential of ϕ t. It acts on the cotangent bundle by (ϕ t = ( ϕ 1 t : T p M Tϕ M n t(p. We can naturally extend the action of (ϕ t to the tensor bundles of M n, which is used in (2.2. The Lie derivative, which measures the infinitesimal lack of diffeomorphism invariance of a tensor with respect to a 1-parameter group of diffeomorphism generated by a vector field, has the following properties: (1 If f is a function, then L X f = X (f. (2 If Y is a vector field, then L X Y = [X, Y ]. (3 If α and β are tensors, then L X (α β = (L X α β + α (L X β. (4 If α is an (0, r tensor, then for any vector fields X, Y 1,..., Y r (L X α (Y 1,..., Y r = X (α (Y 1,..., Y r For example, if α is a 2-tensor, then Σ r i=1α (Y 1,..., Y i 1, [X, Y i ], Y i+1,..., Y r = ( X α (Y 1,..., Y r + Σ r i=1α (Y 1,..., Y i 1, Yi X, Y i+1,..., Y r (2.3 (L X α ij = ( X α ij + g kl ( i X k α lj + j X k α il. 3

4 Recall that the gradient of a function f with respect to the metric g is defined by g (grad g f, X := Xf = df (X. In other words, grad g f is the dual of the 1-form df. using (2.3, we may show that the Lie derivative of the metric is given by (L X g (Y 1, Y 2 = g ( Y1 X, Y 2 + g (Y 1, Y2 X. In local coordinates, this implies (L X g ij = i X j + j X i. In particular, if f is a function, then ( Lgradgfg ij = 2 i j f. We say that a vector field X on (M n, g is Killing if L X g = 0. If X is a complete Killing vector field, then a 1-parameter group of diffeomorphisms ϕ t that it generates is a 1-parameter group of isometries of (M n, g. Definition 2. We say that a diffeomorphism ψ : (M n, g (N n, h is an isometry if ψ h = g. Two Riemannian manifolds are said to be isometric if there is an isometry from one to the other. 3. Differential forms. The volume form dµ of an oriented riemannian manifold (M n, g is defined in terms of a positively oriented orthonormal coframe {ω i } n i=1 by dµ = ω 1... ω n. The volume satisfies (dµ (e 1,..., e n = 1, where {e i } n i=1 is the orthonormal frame dual to {ω i } n i=1 (i.e. ωi (e j = δj. i In a positively oriented local coordinate system {x i }, we have dµ = det (g ij dx 1... dx n. In general, the wedge product of a p-form α and a q-form β is defined by α β (X 1,..., X p+q (p + q! =: A (α β (X 1,..., X p+q p!q! = Σ (J,K sign (J, K α ( ( X j1,..., X jp β Xk1,..., X kq, 4

5 where J = (j 1,..., j p and K = (k 1,..., k q are multi-indices and sign (J, K is the sign of the permutation (1,..., p + q (j 1,..., j p, k 1,..., k q. A p is the operator: p T M p T M p T M, A p (α = 1 p! Σ J (signj Jα. Recall that the exterior derivative of a p-form β satisfies (dβ (Y 1,..., Y p+1 = Σ p+1 i=1 ( 1i 1 Y i (β (Y 1,..., Y i,..., Y p+1 + Σ i<j ( 1 i+j β ([Y i, Y j ], Y 1,..., Y i,..., Y j,..., Y p+1. Using the product rule and the fact that is torsion-free, we may express dβ in terms of covariant derivatives as (dβ (Y 1,..., Y p+1 = Σ ( p+1 j=1 ( 1j 1 Yj β ( Y 1,..., Y j,..., Y p+1. (2.4 In local coordinates, this is (dβ i1...i p+1 = Σ p+1 j=1 ( 1j 1 β i1... i j...i p+1,i j where β i1...i p = β (,..., x i 1 x. For example, if β is a 1-form, then ip If β is a 2-form, then (dβ ij = β j,i β i,j. (dβ ijk = β jk,i β ik,j + β ij,k = β jk,i + β ki,j + β ij,k. Now we define the divergence of a (0, p-tensor as In particular, if X is a 1-form, then (divα i1...i p 1 = g jk α ki1...i p 1,j. div (X = g ij X j,i. 5

6 Given a p-form β and a vector field X, we define the interior product by (ι X β (Y 1,..., Y p 1 =: β (X, Y 1,..., Y p 1 for all vector fields Y 1,..., Y p 1. Recall that for any vector field X and any differential form γ L X γ = (d ι X + ι X d γ. (2.5 The inner product on p T M is defined by For example, γ, η =: g i 1j 1...g ipjp γ i1...i p η j1...j p. ω i 1... ω ip, ω j 1... ω jp = 1 p! det ( δ i kj l. Recall that given p-forms γ and η, their L 2 -inner product is defined by (γ, η L 2 =: γ, η dµ. M n The Hodge star operator : p T M n p T M, p = 0, 1,..., n is defined by γ, η dµ = γ η for any γ, η p T M. For example ( ω 1... ω p = ω p+1... ω n for a positively oriented orthonormal coframe {ω i } n i=1. The adjoint operator δ of d acting on a p-form α is defined in terms of d and the Hodge star operator by the formula δα = ( 1 np+n+1 d α. In terms of covariant derivatives, the adjoint δ is given by (δα (X 1,..., X p 1 = Σ p i=1 ( e i α (e i, X 1,..., X p 1, (2.6 where {e i } n i=1 coordinates, is an orthonormal frame. That is, δα = divα, or in local (δα i1...i p 1 = g jk α ki1...i p 1,j. 6

7 One easily verifies that (dβ, α L 2 = (β, δα L 2 where α p T M and β p 1 T M. The Hodge Laplacian acting on differential p-forms is defined by d =: (dδ + δd. Note that d is a self-adjoint operator. Acting on functions, it is the same as the Laplace-Beltrami operator defined by ( =: div = g ij i j = g ij 2 x i x j Γk ij. x k There are other equivalent ways to define, such as f = Σ n a=1e a (e a f ( ea e a f, where {e a } n a=1 is an orthonormal frame. Or or f = T race (Hessf = T race f = Σ n a=1 2 f (e a, e a, f = 1 ( Σ n g g ij f i,j=1 g x i x j where g = det (g ij. More generally, the (rough Laplacian operator acting on tensors is given by = div = trace g 2 = g ij i j =: i i. More explicitly, given an (r, s-tensor β, β is an (r, s + 2-tensor, which we contract to get β ( ω 1,..., ω r, X 1,..., X s =, n β ( ω 1,..., ω r, X 1,..., X s ; e a, e a a=1 for all covectors ω 1,..., ω r and vectors X 1,..., X s. Lemma 1. For any function f, we have i f = i f + R ij f. 7

8 Proof. This follows from f i,jj = f j,ij = f j,ji + R k jijf k. QED Similarly, the Bochner formula for f 2 is f 2 = 2 f 2 + 2R ij i f j f + 2 i f i ( f. If X is a 1-form, then If β is a 2-form, then X i R ij X j = d X i. ( d β ij = β ij + 2R ikjl β kl R ik β kj R jk β ik. Now let α be a differential p-form. Using (2.4 and (2.6, in local coordinates we may write the Hodge Laplacian as ( d α i1...i r = ( 1 j+1 g kl ij k α li1... i j...i r + g kl k l α i1...i r + ( 1 j g kl k ij α li1... i j...i r = ( α i1...i r + ( 1 j g kl ( k ij ij k αli1... i j...i r. 4. Integration by parts. Recall that Stokes Theorem says that Theorem 3. If α is an (n 1-form on a compact oriented manifold M n with (possibly empty boundary M, then dα = α. M Let (M n, g be an oriented Riemannian manifold with boundary M. The orientation on M defines an orientation on M. Locally, on the boundary, choose a positively oriented frame field {e i } n i=1 such that e 1 = ν is the unit outward normal. Then the frame field {e i } n i=2 is positively oriented on M. Let {ω i } n i=1 denote the orthonormal coframe field dual to {e i} n i=1. The volume form of M is dµ = ω 1... ω n 8 M

9 and the volume form of M is It is not difficult to see that Indeed, dσ = ω 2... ω n. dσ = ι ν (dµ. ι ν (dµ (e 2,..., e n = (dµ (e 1, e 2,..., e n = 1. The divergence theorem says Theorem 4. Let (M n, g be a compact oriented Riemannian manifold. If X is a vector field, then div (X dµ = X, ν dσ, where div (X = X i, i. Proof. Define the (n 1-form α by Using d 2 = 0 and (2.5 we compute M M α = ι X (dµ. dα = d ι X (dµ = (d ι X + ι X d (dµ = L X (dµ = div (X dµ, where to obtain the last equality, we may compute in an orthonormal frame e 1,..., e n : L X (dµ (e 1,..., e n = n dµ (e 1,..., ei X,..., e n i=1 = div (X dµ (e 1,..., e n. Now Stokes Theorem implies that div (X dµ = dα = α M M M = ι X (dµ = M 9 M X, ν dσ.

10 To verify the last equality, we used (ι X (dµ (e 2,..., e n = (dµ (X, e 2,..., e n = X, ν (dµ (e 1,..., e n = X, ν. Hence the theorem is proved. QED Corollary 1. Let (M n, g be a compact oriented Riemannian manifold. If α is an (r, s-tensor and β is an (r, s 1-tensor, then α, β dv = div (α, β dv. Proof. Let X j = α k 1...k r M M ji 2...i s β k 1...k r i 2...i s. We compute that divx = div (α, β + α, β, and the result follows from the divergence theorem. QED 10