PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH 1. Differential Forms To start our discussion, we will define a special class of type (0,r) tensors: Definition 1.1. A differential form of order r or an r-form is a totally antisymmetric tensor of type (0, r). Before, we built higher rank tensors out of covariant and contravariant vectors using the tensor product. In a similar way, a p-form will be decomposed into 1-forms using the wedge product. Definition 1.. The wedge product of r 1-forms is the totally anti-symmetric tensor product dx µ1 dx µ... dx µr = 1 sgn(p )dx µp (1) dx µp ()... dx µp (r) r! P S r we can easily show that the wedge product satisfies the following conditions: dx µ1 dx µ... dx µr = 0 if some index µ appears at least twice. dx µ1 dx µ... dx µr = sgn(p )dx µ P (1) dx µ P ()... dx µ P (r) dx µ1 dx µ... dx µr is linear in each dx µ. 1 Denoting the vector space of r-forms at p M by Ω r p(m), the set of r-forms forms a basis of Ω r p(m) and an element ω Ω r p(m) may be expressed as ω = 1 r! w µ 1µ...µ r dx µ1 dx µ... dx µr The dimension of Ω r p(m) is dimω r p = ( ) m r where m is the dimension of the manifold M. Due to this, the vector spaces Ω r p(m) and Ωp m r (M) are isomorphic since they have the same dimension. We can generalize the wedge product to an exterior product of a q-form and an r-form : Ω q p(m) Ω r p(m) Ω q+r p (M) Denoting ω Ω q p(m) and ζ Ω r p(m), the action of the q + r form ω ζ on q + r arbitrary vectors, V i T p M is [ω ζ](v 1,..., V q+r ) = 1 sgn(p )ω(v P (1),..., V P (q) )ζ(v P (q+1),..., V P (q+r) ) q!r! P S q+r Notice that if q + r > m then ω ζ vanishes automatically. There is one more operation we must define before we can discuss non-coordinate bases and the Cartan Structure equations. We must define the exterior derivative of an r-form. Definition 1.3. The exterior derivative d r is a map Ω r (M) Ω r+1 (M) whose action on an r-form ω = 1 r! ω µ 1,...µ r dx µ1 dx µ... dx µr is defined as d r ω = 1 ( ) r! x ν ω µ 1...µ r dx ν dx µ1 dx µ... dx µr. 1
PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH Without proof, we have three useful statements about the exterior derivative: For ζ Ω 1 (M) and ω Ω r (M) we have d(ζ ω) = dζ ω + ( 1) q ζ dω For X, Y T p (M) and ω Ω 1 (M), we have dω(x, Y ) = X(ω(Y )) Y (ω(x)) ω([x, Y ]). For an r-form ω Ω r (M) this becomes r dω(x 1,..., X r+1 ) = ( 1) i+1 X i ω(x 1,..., ˆX i,...x r+1 ) i=1 + i<j( 1) i+j ω([x i, X j ], X 1,..., ˆX i,..., ˆX j,..., X r+1 ) where a hatted quantity indicates it is being omitted. Defining the inner product i X Ω r (M) Ω r 1 (M) where X T p M, and ω Ω r (M) as i X ω(x 1,..., X r 1 ) = ω(x, X 1,..., X r 1 ) where X i T p M, i = 1,...r are arbitrary vectors. The Lie derivative of ω Ω 1 (M) in the direction of X is L X ω = (di X + i X d)ω Amazingly, there is no change in this formula for a general r-form ω Ω r (M) 1. Non-Coordinate Bases In this class we have worked in a coordinate basis, so that T p M was spanned by {e µ } = { } x and T µ p M spanned by {θ µ } = {dx µ }. There are alternative choices for our bases of these spaces, in particular if the manifold M is equipped with a metric g then we can consider the linear combinations e i = e µ i x µ, {e µ i } GL(m, R) with det e µ i > 0. We have a new frame of basis vectors {e i } obtained by a GL(m, R)-transformation of the original coordinate basis which preserves the orientation. To differentiate between the original coordinate basis and the noncoordinate basis, we will use the indices i, j, k, l [1, 4] when we discuss vectors represented in the new basis. We require that {e i } must be orthonormal with respect to g g(e i, e k ) = e µ i ej ν g µν = η ij If the metric was Riemannian then η ij would be replaced with δ ij. Using the inverse of e µ i which we will denote as e i µ this identity yields g µν = e i µe j νη ij A vector is independent of the basis chosen, so we have V = V µ e µ = V i e i = V i e µ i e µ 1 See Nakahara, section 5.8 Thus, we have e i µ ei ν = δµ ν and ei µ e µ j = δ i j.
PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH 3 and so, V µ = V i e µ i, V i = e i µv µ. If we introduce the dual basis {θ i } such that the inner product < θ i, e j >= δ i j, then θ i = e i µdx µ. We can express the metric in terms of {θ i } as g µν dx µ dx ν = δ ij θ i θ j {e i } and {θ i } are called non-coordinate bases The coefficients e µ i are called the veirbeins if the space is four dimensional, and vielbeins if it is n-dimensional with n > 4. Unlike the coordinate basis, the non-coordinate basis has non-vanishing Lie bracket, where c k Example.1. The metric on S is [e i, e j ] p = c k ij (p)e k P ij (p) = e k ν[e µ µ ej ν e µ µe ν ](p) i g = dθ + sin θdφ = (θ 1 ) + (θ ) where θ 1 = dθ and θ = sin θdφ. The zweibeins are e 1 θ = 1, e1 φ = 0, e θ = 0, e φ = sin θ. The non-vanishing components of c k ij are c 1 = c 1 = cot θ. 3. Cartan s Structure Equations The most common definitions for the curvature tensor R and the torsion tensor T are R(X, Y )Z = X Y Z Y X Z [X,Y ] Z T (X, Y ) = X Y Y X [X, Y ] Let {e i } and {θ i } be a non-coordinate basis and dual basis respectively, then the vector fields {e i } satisfy [e i, e j ] = cij ke k. We define connection coefficients with respect to the basis {e i } as i e j = ei e j = Γ k ije k Denoting the coordinate basis as {ê α }, we have e i = e µ i ê µ then this expression becomes from which it follows e µ i ( µ ej ν + ej λ Γ ν µλ)ê ν = Γ k ijek ν ê ν Γ k ij = e k νe µ i ( µ ej ν + ej λ Γ ν µλ) = e k νe µ i µ e ν j i j.
4 PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH The components of T and R in this basis are given by T i jk = < θ i, T (e j, e k ) >=< θ i, j e k k e j [e j, e k ] > = Γ i jk Γ i kj c i jk R i jkl = < θ i, k l e j l k e j [ek,e l ]e j > = < θ i, k (Γ l0 lj e l 0 l Γ l0 kj e l 0 ) c l0 kl l 0 e j > = e k [Γ i lj] e l [Γ i kj] + Γ l0 lj Γi kl 0 Γ l0 kj Γi ll 0 c l0 kl Γi l 0j To simplify the calculations we introduce a matrix-valued one-form ω i j called the connection one-form by ω i j = Γ i kjθ k. it may be shown that ω i j plays an important role in computing the curvature and torsion tensors 3 Theorem 3.1. The connection one-form ω i j satisfies Cartan s Structure Equations dθ i + ω i k θ j = T i dω i j + ω j k ωk j = R i j where T i = 1 T i jk θj θ k is the torsion two-form and R i j = 1 Ri jkl θk θ l is the curvature two form. Furthermore, the Bianchi identities follow by taking the exterior derivative of these equations dt i + ω i jt j = R i j θ j dr i j + ω i k R k j R i k ω k j = 0 4. The Local Frame In an m-dimensional Riemannian manifold the metric tensor g µν degrees of freedom, while the vielbein e µ i has m degrees of freedom. For a given metric, there are uncountably many non-coordinate bases which yield the same metric. Each of these bases are related to the other by the local orthogonal rotations θ i θ i (p) = Λ i j(p)θ j (p) at each point p. The vielbein transform as e i µ(p) = e a µ (p) = Λ i j(p)e j µ(p) has m(m+1) Remark 4.1. Unlike the indices µ, ν, λ,... which transform under coordinate changes, the indices i, j, k, l,... transform under the local orthogonal rotation and are inert under coordinate changes. Since the metric tensor is invariant under the rotation, Λ i j, we have Λ i jη il Λ l k = η jk 3 See Nakahara, section 7.8. for proof of theorem.
PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH 5 for any Lorentzian spacetime. Therefore Λ i j (p) SO(m 1, 1), the subset of the Lorentzian transformations with determinant 1. Treating this as a Lie group, it has dimension m(m 1) = m m(m + 1) which is the difference between the degrees of freedom of e µ i and g µν. Under the local frame rotation Λ i j (p) the indices i, j, k, l are rotated while the indices ν, µ, λ,... are not affected. Under the rotation, the basis vector transforms as e i e i = e j (Λ 1 ) j i With this rule, we see that upper non-coordinate indices are rotated by Λ i j while the lower indices are rotated by (Λ 1 ) i j Example 4.. Given a tensor field of type (1,1) we have t = t i j e iθ j in the new frame {e i } = {e j(λ 1 ) j i } and θ i } = {Λ i j θj } then t i j t i j = Λ i kt k l(λ 1 ) l j It should be clear that the torsion two-form and curvature two-form transform homogeneously due to this, i.e. T j Λ i j and i R k = Λ i k Rk l (Λ 1 ) l j. Using this we can determine the transformation rule for the connection oneform ω i j. Applying a frame transformation to the torsion tensor, and applying the identity dλλ 1 + ΛdΛ 1 = 0 we find ω i j = Λ i kω k l(λ 1 ) l j + Λ i k(dλ 1 ) k j 5. Levi-Civita Connection in a Non-Coordinate Basis Supposing is the Levi-Civita connection on (M, g), i.e., X g = 0 and Γ λ [µnu] = 0, the components Γ λ µν and Γ i jk are related to each other by a transformation rule. If we define the Ricci rotation coefficients Γ ijk = η il Γ l jk the metric compatibility becomes Γ ijk = η il e l λe µ j µe λ k = η il ek λ e µ j µe l λ = η kl e l λe µ j µe λ i = Γ kji where µ g = 0 has been used. Lowering the first index on the connection one-form ω ij = η ik ω k j this is simply ω ij = ω ji This condition allows us to compute cij k relation c k ij e k = [e i, e j ] = i e j j e i of the basis {e i}, using the commutation where we have exploited the torsion-free condition. Therefore in terms of connection coefficients c k ij = Γ k ij Γ k ji Substituting this into the expression for the Riemann curvature tensor, this is R i jkl = e k [Γ i lj] e l [Γ i kj] + Γ l0 lj Γi kl 0 Γ l0 kj Γi ll 0 [Γ l0 kl Γl0 lk )Γi l 0j
6 PHYS 4390: GENERAL RELATIVITY NON-COORDINATE BASIS APPROACH Example 5.1. Recall that the sphere S the zweibein, e i µ, has non-zero components: e 1 θ = 1, e φ = sin θ The metric condition implies ω 11 = ω = 0 hence ω 1 1 = ω = 0 and so the one-forms obtained from the torsion free condition are d(dθ) + ω 1 (sin θdφ) = 0 d(sin θdφ) + ω 1 dθ = 0 From this we can solve to find ω 1 = cos θdφ and the metric condition ω 1 = ω 1 tells us that ω 1 = cos θdφ. The Riemann tensor may be found from the second structure equation ω 1 ω 1 = 1 R61 1ijθ i θ j dω 1 = 1 R1 ij θ1 θ j dω 1 = 1 R 1ij θ1 θ j ω 1 ω 1 = 1 R ij θ1 θ j The non-vanishing components are R 1 1 = R 1 1 = sin θ and R 11 = R 11 = sin θ. To return this to the coordinate basis, simply use e µ i and e i µ to transform the indices, i.e., R θ φθφ = ei θ e j φ ek θe l φr i jkl = 1 sin θ R1 1 = 1 sin θ Example 5.. The Schwarzschild metric is ( ds = 1 M ) ( dt 1 M ) 1 dr r (dθ + sin θdφ ) = (θ 0 ) (θ 1 ) (θ ) (θ 3 ) r r where θ 0 = ( 1 M r ) 1 ( dt, θ 1 = 1 M r ) 1 dr, θ = rdθ, θ 3 = r sin θdφ with parameters ranging 0 < m < r, 0 θπ and 0 φ < π. As before the metric condition implies ω i i = 0 for i = 0, 1,, 3.