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Home , General relativity, Gravity, Riemann curvature tensor, Weyl tensor

Notes on the Weyl , decomposition of tensor, Ruse-Lanczos identity and duality of the tensor

Wytler Cordeiro dos Santos Universidade de Bras´ılia,CEP 70910-900, DF, Brasil

Abstract These notes have the pedagogical purpose of exposing mathematical accounts about of the . The deve- lopment of calculations is presented to obtain the curvature tensor from the conformal transformation, consequently resulting in the Weyl conformal tensor and the Riemann tensor reveals itself decomposed into its parts. The Ruse- Lanczos identity is a useful mathematical tool to build duality identities in the two pairs of antisymmetric indices of the Riemann tensor components. Keywords: Weyl tensor, decomposition of the Riemann tensor, dual tensor

1. Introduction

In accordance with tensorial approach of the Electromagnetic , the electromagnetic field is given in terms of the four-vector Fµ = Eµ + iBµ, where Eµ = (0, E1, E2, E3) and Bµ = (0, B1, B2, B3) are electric and magnetic field observated by an observer in an inertial reference frame, where the four- vector of the observer is a timelike µ ν ν vector u = (1, 0, 0, 0). Thus, the four-electromagnetico vector is given by Fµ = Fµνu + iF˜µνu , where the Fµν is the ˜ 1 κλ field-strength tensor of and the Fµν = 2 κλµνF is its dual field-strength tensor. In this formalism, ν ν we separete the eletric vector given by Eµ = Fµνu and magnetic vector given by Bµ = F˜µνu with aid of dual operation. In fact, in the inertial reference frame in which it is at rest, the vectors Eµ and Bµ, are spatial three-vectors, then we can put F = E + iB, where the six components of the field-strength tensor of electromagnetism Fµν are into complex three-vector F [1]. The duality of an like field-strength tensor of electromagnetism can be used for split vectorial components. With this in mind we can apply the dual operation to the pair of antisymmetric indices of . In the literature that deals with duality of the Riemann curvature tensor, decomposed into its irreducible components, the Weyl tensor, the Ricci tensor and the curvature , seldom makes explicit details of calculations involving the decomposition of the the Riemann curvature tensor. In the basic courses of , presents Weyl tensor as a definition without exposing the details of the calculations that lead to it. Some books, for example [2, 3], present the conformal transformation and indicate the path to be followed to obtain the Weyl curvature tensor. In this paper we solved this exercise to obtain the Weyl tensor from conformal transformation and explained the decomposition of Riemann curvature tensor. In order to obtain the duality properties of Riemann curvature tensor, it is necessary to study the Ruse-Lanczos identity [4]. Consequently, in the same way as for the duality of electromagnetismo we present the formalism of duality and complexification of the parts of Riemann curvature tensor. κ According to General Relativity, the Riemann curvature tensor, R µλν, is related to the distribution of , κ specifically by Ricci tensor, Rµν = R µκν, directly related to the local - tensor by Einstein’s field equations, 1 R − g R = 8πGT , (1) µν 2 µν µν

Email address: [email protected] (Wytler Cordeiro dos Santos)

1 where G is the gravitational ’s constant and we are overlooking the value of the . In free gravitational field, we have no matter, Tµν = 0 and consequently we have null Ricci tensor, Rµν = 0. Greek labels refer to four dimensional and we use signature (− + ++).

2. Conformal transformation

Hermann Weyl tried to unify electromagnetism and , this attempt was carried out in 1919 [5], however this attempt failed, but Weyl introduced a new principle of invariance in his theory, a scale change of the given by g¯ = e2ω g. This principle later evolved through Quantum Theory being known as ‘gauge invariance’ [6, 7, 8]. A scale change of the metric tensor is classified as conformal transformation in Riemannian leading to the Weyl conformal tensor. Let (M, g) be a spacetime where M is pseudo-Riemannian and g is a Lorentz metric tensor. A con- formal transformation [2, 3, 9] is not a change of coordinates but an actual change in the geometry under a ‘gauge transformation’ [10] as already mentioned, g¯ = e2ω g, (2) where g e g¯ are metric on the manifold M and ω = ω(x) is a real function of the spacetime coordinates. The ‘gauge transformation’ in equation (2) is a rescaling of the metric tensor and it is named Weyl rescaling. In the Weyl rescaling we can state that g¯ is conformally related to g, because the gauge transformation (2) pre- serves the angle between two vectors U and V of the tangent Tp(M). The inner product between the two vectors U and V, denoted by U · V = g(U, V) and the measure of the angle between two vectors follows from the definition, U · V g(U, V) cos θ = = √ √ . (3) |U||V| g(U, U) g(V, V) The measure of the angle between the same two vectors under Weyl rescaling (2) is given by,

g¯(U, V) e2ω g(U, V) g(U, V) cos θ¯ = √ √ = p p = √ √ = cos θ, (4) g¯(U, U) g¯(V, V) e2ω g(U, U) e2ω g(V, V) g(U, U) g(V, V) therefore comparing the equation (3) and (4), the Weyl rescaling is a conformal transformation that preserves the angle but changes the scale. In accordance with this statement, a conformal transformation on spacetime (M, g) preserves the local cone structure, where, timelike vectors remain timelike vectors, spacelike vectors remain spacelike vectors and null vectors remain null vectors. A very useful applications are the Penrose conformal diagrams that enable the whole of an infinite spacetime to be represented as a finite diagram, by applying a conformal transformation to the metric structure [9, 11].

2.1. Calculation of the connections or Christoffel symbols and Riemann curvature tensor in a conformal space Let us compute the Riemann curvature tensor and for the first step, we calculate the Christoffel symbols in a conformal space starting with 1 Γ¯ ρ = g¯ρσ(¯g + g¯ − g¯ ). (5) µν 2 µσ,ν νσ,µ µν,σ with a comma denoting a partial derivative. To facilitate some calculations let us admit that eω(x) → Ω(x) is some non-vanishing scalar funtion of in the Weyl local rescaling metric of the metric tensor (2),

2 µν −2 µν g¯µν = Ω gµν andg ¯ = Ω g , (6)

ρ replacing above equation (6) into Christoffel symbols Γ¯ µν and calculating the derivatives, we have, 1 Γ¯ ρ = Ω−2gρσ[2Ω(Ω, )g + Ω2g + 2Ω(Ω, )g + Ω2g − 2Ω(Ω, )g − Ω2g ] µν 2 ν µσ µσ,ν µ νσ νσ,µ σ µν µν,σ 1 1 = Ω−2gρσ[Ω2(g + g − g )] + Ω−2gρσ[2Ω(Ω, g + Ω, g − Ω, g )], 2 µσ,ν νσ,µ µν,σ 2 ν µσ µ νσ σ µν 2 that is ρ ρ −1 ρ ρ ρσ Γ¯ µν = Γ µν + Ω (δ µΩ,ν +δ νΩ,µ −g gµνΩ,σ ). (7) Now we can calculate the Riemann curvature tensor in a conformal space with metric tensor g¯,

ρ ρ ρ κ ρ κ ρ R¯ µσν = Γ¯ µν,σ − Γ¯ µσ,ν + Γ¯ µνΓ¯ σκ − Γ¯ σµΓ¯ νκ (8) where we need to calculate the derivative of Christoffel symbol (7), that it will result in the first term of Riemann curvature tensor (8),

ρ ρ ρ −1 ρ −1 ρκ −1 ρκ −1 Γ¯ µν,σ = Γ µν,σ + δ µ(Ω Ω,ν ),σ +δ ν(Ω Ω,µ ),σ −gµν,σ(g Ω Ω,κ ) − gµν(g Ω Ω,κ ),σ .

The term of Riemann curvature tensor (8) is similar to the above first term, for just changing the indexes ν ↔ σ, ρ ρ ρ −1 ρ −1 ρκ −1 ρκ −1 Γ¯ µσ,ν = Γ µσ,ν + δ µ(Ω Ω,σ ),ν +δ σ(Ω Ω,µ ),ν −gµσ,ν(g Ω Ω,κ ) − gµσ(g Ω Ω,κ ),ν . By subtracting the two terms above, we have that,

ρ −1 ρ −1 ρ −2 −1 −2 −1 δ µ(Ω Ω,ν ),σ −δ µ(Ω Ω,σ ),ν = δ µ[(−Ω Ω,σ Ω,ν +Ω ,νσ ) − (−Ω Ω,ν Ω,σ +Ω ,σν )] = 0 and then we have

ρ ρ ρ ρ ρ −1 ρ −1 ρκ −1 ρκ −1 Γ¯ µν,σ − Γ¯ µσ,ν = Γ µν,σ − Γ µσ,ν + δ ν(Ω Ω,µ ),σ −δ σ(Ω Ω,µ ),ν −gµν,σ(g Ω Ω,κ ) + gµσ,ν(g Ω Ω,κ ) ρκ −1 ρκ −1 +gµσ(g Ω Ω,κ ),ν −gµν(g Ω Ω,κ ),σ . (9)

Let us calculate the third term of Riemann curvature tensor of equation (8),

κ ρ κ −1 κ κ κλ ρ −1 ρ ρ ρλ Γ¯ µνΓ¯ σκ = [Γ µν + Ω (δ µΩ,ν +δ νΩ,µ −g gµνΩ,λ )][Γ σκ + Ω (δ σΩ,κ +δ κΩ,σ −g gσκΩ,λ )] κ ρ −1 κ ρ κ ρ ρ κλ = Γ µνΓ σκ + Ω (δ µΓ σκΩ,ν +δ νΓ σκΩ,µ −Γ σκg gµνΩ,λ ) −1 ρ κ ρ κ κ ρλ +Ω (δ σΓ µνΩ,κ +δ κΓ µνΩ,σ −Γ µνg gσκΩ,λ ) −2 κ ρ κ ρ κ ρλ +Ω (δ µδ σΩ,ν Ω,κ +δ µδ κΩ,ν Ω,σ −δ µg gσκΩ,ν Ω,λ κ ρ κ ρ κ ρλ +δ νδ σΩ,µ Ω,κ +δ νδ κΩ,µ Ω,σ −δ νg gσκΩ,µ Ω,λ κλ ρ κλ ρ κλ ρτ −g gµνδ σΩ,κ Ω,λ −g gµνδ κΩ,σ Ω,λ +g gµνg gσκΩ,λ Ω,τ ), the last term in above equation can be simplified,

κλ ρτ λ ρτ ρτ g gµνg gσκΩ,λ Ω,τ = δ σgµνg Ω,λ Ω,τ = gµνg Ω,σ Ω,τ . so that

κ ρ κ ρ −1 ρ ρ ρ κ ρ κλ ρ ρλ κ  Γ¯ µνΓ¯ σκ = Γ µνΓ σκ + Ω Γ σµΩ,ν +Γ σνΩ,µ +δ σΓ µνΩ,κ +Γ µνΩ,σ −g gµνΓ σκΩ,λ −g gσκΓ µνΩ,λ −2 ρ ρ ρ ρ ρλ ρλ +Ω δ σΩ,ν Ω,µ +δ µΩ,ν Ω,σ +δ σΩ,ν Ω,µ +δ νΩ,µ Ω,σ −g gµσΩ,ν Ω,λ −g gσνΩ,µ Ω,λ κλ ρ ρλ ρλ  −g gµνδ σΩ,κ Ω,λ −g gµνΩ,σ Ω,λ +gµνg Ω,σ Ω,λ , we can see that the last two terms cancel each other resulting in

κ ρ κ ρ −1 ρ ρ ρ ρ κ κλ ρ ρλ κ  Γ¯ µνΓ¯ σκ = Γ µνΓ σκ + Ω Γ σµΩ,ν +Γ σνΩ,µ +Γ µνΩ,σ +δ σΓ µνΩ,κ −g gµνΓ σκΩ,λ −g gσκΓ µνΩ,λ −2 ρ ρ ρ ρλ ρ κλ  +Ω 2δ σΩ,µ Ω,ν +δ µΩ,ν Ω,σ +δ νΩ,µ Ω,σ −g (gµσΩ,ν Ω,λ −gνσΩ,µ Ω,λ ) − gµνδ σg Ω,κ Ω,λ

The fourth and final term of Riemann curvature tensor of equation (8), is similar to the above third term, for just changing the indexes ν ↔ σ,

κ ρ κ ρ −1 ρ ρ ρ ρ κ κλ ρ ρλ κ  Γ¯ σµΓ¯ νκ = Γ σµΓ νκ + Ω Γ νµΩ,σ +Γ νσΩ,µ +Γ µσΩ,ν +δ νΓ µσΩ,κ −g gµσΓ νκΩ,λ −g gνκΓ µσΩ,λ −2 ρ ρ ρ ρλ ρ κλ  +Ω 2δ νΩ,µ Ω,σ +δ µΩ,σ Ω,ν +δ σΩ,µ Ω,ν −g (gµνΩ,σ Ω,λ −gσνΩ,µ Ω,λ ) − gµσδ νg Ω,κ Ω,λ .

The subtraction between the third and the fourth terms obtained above results in,

κ ρ κ ρ κ ρ κ ρ −1 ρ κ ρ κ κλ ρ ρ Γ¯ µνΓ¯ σκ − Γ¯ σµΓ¯ νκ = Γ µνΓ σκ − Γ σµΓ νκ + Ω [(δ σΓ µν − δ νΓ µσ)Ω,κ −g (gµνΓ σκ − gµσΓ νκ)Ω,λ 3 ρλ κ κ −2 ρ ρ ρκ −g (gσκΓ µν − gνκΓ µσ)Ω,λ ] + Ω [(δ σΩ,ν −δ νΩ,σ )Ω,µ −g (gµσΩ,ν Ω,κ −gµνΩ,σ Ω,κ ) κλ ρ ρ −g (gµνδ σ − gµσδ ν)Ω,κ Ω,λ ]. (10)

So, composing the Riemann curvature tensor (8),

ρ ρ ρ κ ρ κ ρ R¯ µσν = Γ¯ µν,σ − Γ¯ µσ,ν + Γ¯ µνΓ¯ σκ − Γ¯ σµΓ¯ νκ, by adding the terms obtained with the equations (9) and (10) that are part of the the Riemann tensor,

ρ ρ ρ ρ −1 ρ −1 ρκ −1 ρκ −1 R¯ µσν = Γ µν,σ − Γ µσ,ν + δ ν(Ω Ω,µ ),σ −δ σ(Ω Ω,µ ),ν −gµν,σ(g Ω Ω,κ ) + gµσ,ν(g Ω Ω,κ ) ρκ −1 ρκ −1 κ ρ κ ρ +gµσ(g Ω Ω,κ ),ν −gµν(g Ω Ω,κ ),σ + Γ µνΓ σκ − Γ σµΓ νκ −1 ρ κ ρ κ κλ ρ ρ ρλ κ κ +Ω [(δ σΓ µν − δ νΓ µσ)Ω,κ −g (gµνΓ σκ − gµσΓ νκ)Ω,λ −g (gσκΓ µν − gνκΓ µσ)Ω,λ ] −2 ρ ρ ρκ κλ ρ ρ +Ω [(δ σΩ,ν −δ νΩ,σ )Ω,µ −g (gµσΩ,ν Ω,κ −gµνΩ,σ Ω,κ ) − g (gµνδ σ − gµσδ ν)Ω,κ Ω,λ ]. or

ρ ρ ρ −1 ρ −1 ρκ −1 R¯ µσν = R µσν +δ ν(Ω Ω,µ ),σ −δ σ(Ω Ω,µ ),ν − g Ω Ω,κ (gµν,σ − gµσ,ν) | {z } | {z } (i) (ii) ρκ −1 ρκ −1 +gµσ(g Ω Ω,κ ),ν −gµν(g Ω Ω,κ ),σ | {z } (iii) −1 ρ κ ρ κ ρλ κ κ +Ω [(δ σΓ µν − δ νΓ µσ)Ω,κ −g (gσκΓ µν − gνκΓ µσ)Ω,λ | {z } | {z } (i) (ii) κλ ρ ρ −2 ρ ρ −g (gµνΓ σκ − gµσΓ νκ)Ω,λ] + Ω [(δ σΩ,ν −δ νΩ,σ )Ω,µ | {z } (iii) ρκ κλ ρ ρ −g (gµσΩ,ν Ω,κ −gµνΩ,σ Ω,κ ) − g (gµνδ σ − gµσδ ν)Ω,κ Ω,λ ]. (11)

We must note the underbraced terms, where the terms underbraced with (i) result in,

ρ −1 ρ −1 −1 ρ κ ρ κ ρ −1 ρ −1 δ ν(Ω Ω,µ ),σ −δ σ(Ω Ω,µ ),ν +Ω (δ σΓ µν − δ νΓ µσ)Ω,κ = δ ν∇σ(Ω Ω,µ ) − δ σ∇ν(Ω Ω,µ ).

The terms underbraced with (ii) result in,

ρκ −1 −1 ρλ κ κ −g Ω Ω,κ (gµν,σ − gµσ,ν) − Ω g (gσκΓ µν − gνκΓ µσ)Ω,λ = ρκ −1 λ λ λ λ g Ω Ω,κ (gµσ,ν − gσλΓ µν − gµλΓ σν + gµλΓ σν −gµν,σ + gνλΓ µσ) = | {z } zero ρκ −1 λ λ λ λ g Ω Ω,κ [(gµσ,ν − gσλΓ µν − gµλΓ σν) − (gµν,σ − gνλΓ µσ − gµλΓ σν)] = ρκ −1 g Ω Ω ,κ [∇νgµσ − ∇σgµν] = 0. The terms underbraced with (iii) result in,

ρκ −1 ρκ −1 κλ ρ ρ gµσ(g Ω Ω,κ ),ν −gµν(g Ω Ω,κ ),σ −g (gµνΓ σκ − gµσΓ νκ)Ω,λ = ρκ −1 ρ −1 κλ ρκ −1 ρ −1 κλ gµσ(g Ω Ω,κ ),ν +gµσΓ νκ(Ω g Ω,λ ) − [gµν(g Ω Ω,κ ),σ +gµνΓ σκ(Ω g Ω,λ )] = ρκ −1 ρκ −1 ρκ −1 ρκ −1 gµσ∇ν[g (Ω Ω,κ )] − gµν∇σ[g (Ω Ω,κ )] = gµσg ∇ν(Ω Ω,κ ) − gµνg ∇σ(Ω Ω,κ ).

With the three results above and going back into the equation (11), we can rewright it as,

ρ ρ ρ −1 ρ −1 ρκ −1 R¯ µσν = R µσν + δ ν∇σ(Ω Ω,µ ) − δ σ∇ν(Ω Ω,µ ) + gµσg ∇ν(Ω Ω,κ ) ρκ −1 ρ −1 −1 ρ −1 −1 −gµνg ∇σ(Ω Ω,κ ) + δ σ(Ω Ω,ν )(Ω Ω,µ ) − δ ν(Ω Ω,σ )(Ω Ω,µ ) ρκ −1 −1 ρκ −1 −1 ρ ρ κλ −1 −1 −g gµσ(Ω Ω,ν )(Ω Ω,κ ) + g gµν(Ω Ω,σ )(Ω Ω,κ ) + (gµσδ ν − gµνδ σ)g (Ω Ω,κ )(Ω Ω,λ ).

We can separate the above terms into groups with four brackets

ρ ρ R¯ µσν = R µσν 4 1 + g [gρκ∇ (Ω−1Ω, ) − gρκ(Ω−1Ω, )(Ω−1Ω, ) + δρ gκλ(Ω−1Ω, )(Ω−1Ω, )] µσ ν κ ν κ 2 ν κ λ 1 − g [gρκ∇ (Ω−1Ω, ) − gρκ(Ω−1Ω, )(Ω−1Ω, ) + δρ gκλ(Ω−1Ω, )(Ω−1Ω, )] µν σ κ σ κ 2 σ κ λ 1 + δρ g [gτκ∇ (Ω−1Ω, ) − gτκ(Ω−1Ω, )(Ω−1Ω, ) + δτ gκλ(Ω−1Ω, )(Ω−1Ω, )] ν µτ σ κ σ κ 2 σ κ λ 1 − δρ g [gτκ∇ (Ω−1Ω, ) − gτκ(Ω−1Ω, )(Ω−1Ω, ) + δτ gκλ(Ω−1Ω, )(Ω−1Ω, )]. σ µτ ν κ ν κ 2 ν κ λ Bracketed terms form tensors that can simplify the Riemann tensor, ρ ρ ρ ρ ρ τ ρ τ R¯ µσν = R µσν + gµσBν − gµνBσ + δ νgµτBσ − δ σgµτBν , (12) where we can define 1 B λ = gλµ∇ (Ω−1Ω, ) − gλµ(Ω−1Ω, )(Ω−1Ω, ) + δλ gµν(Ω−1Ω, )(Ω−1Ω, ). κ κ µ κ µ 2 κ µ ν We can calculate the in the above equation, −1 −2 −1 ∇κ(Ω Ω,µ ) = −Ω Ω,κ Ω,µ +Ω Ω;κµ , where we use the shorthand semicolon notation Ω;κµ = ∇κΩ,µ. So, we have, 1 B λ = Ω−1gλµΩ; −2Ω−2gλµΩ, Ω, + Ω−2δλ gµνΩ, Ω, , κ κµ κ µ 2 κ µ ν or with the two covariant indices, 1 B = Ω−1Ω; −2Ω−2Ω, Ω, + g gµνΩ, Ω, , κλ κλ κ λ 2 κλ µ ν The of the above tensor is obtained by 1 B κ = Ω−1gκµΩ; −2Ω−2gκµΩ, Ω, + Ω−2δκ gµνΩ, Ω, . κ κµ κ µ 2 κ µ ν κ being that δ κ = m where m is the of manifold. The trace is ! m − 4 B κ = Ω−1gµνΩ; + Ω−2gµνΩ, Ω, . (13) κ µν 2 µ ν When calculating the Ricci tensor from the equation (12) we can obtain σ σ σ σ σ σ R¯µν = R¯ µσν = R µσν + gµσBν − gµνBσ + δ νBσµ − δ σBνµ, or σ R¯µν = Rµν − gµνBσ − (m − 2)Bµν. (14) To obtain the curvature scalar it passes through the , κµ −2  κµ κ σ κµ  g¯ R¯µν = Ω g Rµν − δ µBσ − (m − 2)g Bµν , that results in κ −2  κµ κ σ κ R¯ ν = Ω g Rµν − δ νBσ − (m − 2)Bν , then the curvature scalar results in κ −2  κµ κ σ κ R¯ = R¯ κ = Ω g Rµκ − δ κ Bσ − (m − 2)Bκ or −2  κ R¯ = Ω R − 2(m − 1)Bκ . (15) κ Placing the result of tensor Bκ of the equation (13) into above equation (15) we can obtain, ( " ! #) m − 4 R¯ = Ω−2 R − 2(m − 1) Ω−1gµνΩ; + Ω−2gµνΩ, Ω, , µν 2 µ ν which can be simplified to −2 −3 µν −4 µν R¯ = Ω R − 2(m − 1)Ω g Ω;µν −(m − 1)(m − 4)Ω g Ω,µ Ω,ν ). (16) 5 3. The Weyl curvature tensor

The goal now is to calculate the Weyl curvature tensor. For that let us go back to the equation (12),

ρ ρ ρ ρ ρ ρ R¯ µσν = R µσν + gµσBν − gµνBσ + δ νBσµ − δ σBνµ. From equation (15) we have −2  κ R¯ = Ω R − 2(m − 1)Bκ such that

2 −2  κ g¯µνR¯ = (Ω gµν)Ω R − 2(m − 1)Bκ κ g¯µνR¯ = gµνR − 2(m − 1)gµνBκ , so putting in evidence the term of trace of tensor Bµν, we have that,

gµνR − g¯µνR¯ g B κ = . µν κ 2(m − 1) Now we must replace the result above in the equation (14),

κ R¯µν = Rµν − gµνBκ − (m − 2)Bµν, to get the following result g¯µνR¯ − gµνR R¯ − R = − (m − 2)B , µν µν 2(m − 1) µν thus we have that the tensor Bµν is g¯µνR¯ − gµνR Rµν − R¯µν B = + . (17) µν 2(m − 2)(m − 1) m − 2 The next step is to write the Riemann tensor (12) with all indexes down as follows way,

ρ 2 ρ ρ ρ ρ ρ g¯κρR¯ µσν = R¯κµσν = Ω gκρ(R µσν + gµσBν − gµνBσ + δ νBσµ − δ σBνµ) 2 R¯κµσν = Ω (Rκµσν + gµσBνκ − gµνBσκ + gκνBσµ − gκσBνµ), (18) and replace the equation (17) into equation above (18), such that, " ! ! g¯ R¯ − g R R − R¯ g¯ R¯ − g R R − R¯ R¯ = Ω2 R + g κν κν + κν κν − g κσ κσ + κσ κσ κµσν κµσν µσ 2(m − 2)(m − 1) m − 2 µν 2(m − 2)(m − 1) m − 2 ! !# g¯µσR¯ − gµσR Rµσ − R¯µσ g¯µνR¯ − gµνR Rµν − R¯µν + g + − g + . (19) κν 2(m − 2)(m − 1) m − 2 κσ 2(m − 2)(m − 1) m − 2 Separating the Ricci tensors terms from the curvature scalar and also grouping the barred terms of the non-barred, we obtain 1   Ω−2R¯ = R + g R − g R + g R − g R κµσν κµσν m − 2 µσ κν µν κσ κν µσ κσ µν R   + − g g + g g − g g + g g 2(m − 2)(m − 1) µσ κν µν κσ κν µσ κσ µν 1   + g R¯ − g R¯ + g R¯ − g R¯ m − 2 µσ κν µν κσ κν µσ κσ µν R¯   + − g g¯ + g g¯ − g g¯ + g g¯ . 2(m − 2)(m − 1) µσ κν µν κσ κν µσ κσ µν Now we must separate the barred and non-barred terms, 1   R¯   Ω−2R¯ + g R¯ − g R¯ + g R¯ − g R¯ + − g g¯ + g g¯ − g g¯ + g g¯ κµσν m − 2 µσ κν µν κσ κν µσ κσ µν 2(m − 2)(m − 1) µσ κν µν κσ κν µσ κσ µν 6 1   R   = R + g R − g R + g R − g R + g g − g g . κµσν m − 2 µσ κν µν κσ κν µσ κσ µν (m − 2)(m − 1) µν κσ µσ κν −2 Recall the local rescaling metric of the metric tensor from equation (6), where we have gµν = Ω g¯µν, so that, Ω−2   R¯ Ω−2   Ω−2R¯ + g¯ R¯ − g¯ R¯ + g¯ R¯ − g¯ R¯ + − g¯ g¯ + g¯ g¯ − g¯ g¯ + g¯ g¯ κµσν m − 2 µσ κν µν κσ κν µσ κσ µν 2(m − 2)(m − 1) µσ κν µν κσ κν µσ κσ µν 1   R   = R + g R − g R + g R − g R + g g − g g , κµσν m − 2 µσ κν µν κσ κν µσ κσ µν (m − 2)(m − 1) µν κσ µσ κν or " # 1   R¯   Ω−2 R¯ + g¯ R¯ − g¯ R¯ + g¯ R¯ − g¯ R¯ + g¯ g¯ − g¯ g¯ κµσν m − 2 µσ κν µν κσ κν µσ κσ µν (m − 2)(m − 1) µν κσ µσ κν 1   R   = R + g R − g R + g R − g R + g g − g g . κµσν m − 2 µσ κν µν κσ κν µσ κσ µν (m − 2)(m − 1) µν κσ µσ κν unless of the factor Ω−2, algebraically the left side is equal to the right side. Thus we can write the equation above in a compact format, −2 Ω C¯κµσν = Cκµσν, (20) where the tensor Cκλµν is called Weyl curvature tensor given by, 1   R   C = R + g R − g R + g R − g R + g g − g g , (21) κλµν κλµν m − 2 λµ κν λν κµ κν λµ κµ λν (m − 2)(m − 1) λν κµ λµ κν or in terms of antisymmetrization of index pair by square brackets we have,

2   2R gκ[µgν]λ C = R + g R + g R + . (22) κλµν κλµν m − 2 κ[ν µ]λ λ[µ ν]κ (m − 2)(m − 1) In the equation (20) we can apply gκρ so that we have

κρ −2 κρ g Ω C¯κµσν = g Cκµσν, κρ ρ g¯ C¯κµσν = C µσν, therefore, we have κ κ C¯ λµν = C λµν, (23) consequently the Weyl tensor has the special property that it is under conformal changes to the metric (2), and for this reason the Weyl tensor is also called the conformal tensor. κ Let us calculate the trace of Weyl curvature tensor given by, Cµν = C µκν. For that, let us go back to the equation (21) as follows, 1   R   Cκ = Rκ + g R κ − g R κ + δκ R − δκ R + g δκ − δµ g , µκν µκν m − 2 µκ ν µν κ ν κµ κ µν (m − 2)(m − 1) µν κ κ νκ this equation above results in 1   R   C = R + R − g R + R − mR + mg − g µν µν m − 2 µν µν µν µν (m − 2)(m − 1) µν µν 1 h i R   C = R + R (−m + 2) − g R + (m − 1)g µν µν m − 2 µν µν (m − 2)(m − 1) µν gµνR gµνR C = R − R − + = 0. (24) µν µν µν m − 2 m − 2 The Weyl curvature tensor is completely traceless. 2 If every point of a spacetime (M, g¯), the metric tensor isg ¯µν = Ω ηµν, where ηµν is the metric tensor of flat Minkowski spacetime, then (M, g¯) is said to be conformally flat. Since the Wey tensor for a flat Minkowski metric is null, it is also null for a conformally flat metric. 7 4. Decomposition of the Riemann curvature tensor

It is possible express the curvature Riemann tensor in terms of Weyl curvature tensor, Ricci tensor and scalar of curvature, when we isolate the Riemann curvature tensor in the equation we have (21), 1   R   R = C + g R + g R − g R − g R + g g − g g . (25) ρσµν ρσµν m − 2 σν ρµ ρµ σν σµ ρν ρν σµ (m − 2)(m − 1) ρν σµ ρµ σν

Let us consider the 4-dimensional space V4, where we must have m = 4 in above equation, such as, 1  R  R = C + g R + g R − g R − g R + g g − g g . (26) ρσµν ρσµν 2 σν ρµ ρµ σν σµ ρν ρν σµ 6 ρν σµ ρµ σν It is common to use the definition below 1 E = (g S + g S − g S − g S ), (27) ρσµν 2 ρµ σν σν ρµ ρν σµ σµ ρν where S ρσ is a traceless part of the Ricci tensor, the reduced Ricci tensor given by 1 S = R − g R (28) ρσ ρσ 4 ρσ so that ρσ g S ρσ = 0. (29) We can also identify the reduced Ricci tensor with the the , 1 G = R − g R, (30) µν µν 2 µν µ with the trace Gµ = G = −R, we have that 1 S = G − g G. (31) µν µν 4 µν

Notice that the tensor Eρσµν is expressed as 1 1 E = [(g R + g R − g R − g R ) − (g g + g g − g g − g g )R], (32) ρσµν 2 ρµ σν σν ρµ ρν σµ σµ ρν 8 ρµ σν σν ρµ ρν σµ σµ ρν or then, 1 R E = (g R + g R − g R − g R ) + (g g − g g ). (33) ρσµν 2 ρµ σν σν ρµ ρν σµ σµ ρν 4 ρν σµ ρµ σν Note that the first term of the equation above is equal to the second term of the equation (26), but that the second term R R of the expression above has a factor and the equation (26) has a factor . In this way we express the Riemann 4 6 curvature tensor as follows R   R = C + E − g g − g g , ρσµν ρσµν ρσµν 12 ρν σµ ρµ σν or then, R   R = C + E + g g − g g , ρσµν ρσµν ρσµν 12 ρµ σν ρν σµ where we can define the new tensor, R   G = g g − g g . (34) ρσµν 12 ρµ σν ρν σµ Thus we come to the Riemann curvature tensor in a 4-dimensional Riemann space V4,

Rρσµν = Cρσµν + Eρσµν + Gρσµν. (35) 8 We must calculate the trace of the tensor Gρσµν, as follows as R R R Gρ = (δρ g − δρ g ) = (4g − g ) = g , (36) σρν 12 ρ σν ν σρ 12 σν σν 4 σν and from equation (27), the calculation of the trace of the tensor Eρσµν is given by, 1 1 Eρ = (δρ S + g S ρ − δρ S − δρ S ) = (4S − 0 − S − S ) = S . (37) σρν 2 ρ σν σν ρ ν σρ σ ρν 2 σν σν σν σν With this it can be confirmed that the trace of the Riemann curvature tensor, given by:

ρ ρ ρ ρ R σρν = C σρν + E σρν + G σρν, results in the Ricci tensor, R R R Rρ = 0 + S + g = R − g + g = R . σρν σν 4 σν σν 4 σν 4 σν σν We know that the full Riemann curvature tensor has 20 independent components [2, 3, 12, 13, 14, 15] whereas the Ricci tensor has 10 independent components because it is , Rµν = Rνµ, then we can take a look again at the equation (26) and to observe that the Weyl curvature tensor must have the other 10 independent components which at any point are completely independent of the Ricci tensor components. The Riemann curvature tensor is a sum of the Weyl curvature tensor plus terms built out of the Ricci tensor and , then for solutions such as Schwarzschild and Kerr , we have Rµν = 0 and R = 0, and consequently the Weyl curvature tensor is exactly the same Riemann curvature tensors. We say, therefore, that the curvature Weyl tensor corresponds to the free gravitational field [16]. In basic courses of General Relativity we have studied the equation of variation [12, 13, 14, 15] that measures convergence or of two nearby initially parallel in a curved spacetime, depending on µ the local curvature R νρσ. If we consider an initial spherical distribution of non-interacting freely falling towards the , each moves on a straight line through the centre of the Earth. The particles more near the Earth fall faster because the gravitational attraction is stronger. Then the spherical distribution of particles no longer remains a shape but is distorted into an ellipsoid of the same volume. Thus the has produced a tidal in the sphere of particles that results in an elongation of the distribution in the direction of and a compression of the distribution in the transverse direction. Let us consider any pair of non- particles in a sphere of particles, each one is in and so they must move along the timelike geodesics xµ(τ) and yµ(τ) respectively, where τ is the proper experienced by the first particle. We can define a small separation vector between the two particle worldlines by ξµ(τ) = yµ(τ) − xµ(τ), then we have that the tidal of a gravitational field is given by, d2ξµ = Rµ uρuσξν, (38) dτ2 ρσν dxρ where uρ = . The tidal forces can be represented by curvature of a spacetime in which particles follow geodesics. dτ Therefore if we considere a spherical distrubution of particles in vacuum space nearby to gravitating astrophysical spherical boby, where the spacetime is Schwarzschild or Kerr spacetime, we have that Rµν = 0 and R = 0 and from µ µ equation (26), we have that the Riemann curvature tensor becomes R νρσ = C νρσ and the equation (38) for tidal forces of a free gravitational field reads, d2ξµ = Cµ uρuσξν. (39) dτ2 ρσν The Weyl tensor expresses the that a body feels when moving along a geodesic. When we consider a general gravitational hµν satisfying the empty-space linearised field equation and the Lorentz gauge condition, among some solutions there are plane- solutions in vacuum space [12, 13, 14, 15]. If we consider the famous circular ring of test particles in the transverse plane to propagation of the plane-wave in vacuum space, surrounding a central particle, we can verify that the wave deforms what was a ring as measured in the proper reference frame of the central particle into an , causing the ring to oscillate in various modes of

9 . Because the linear plane-wave is in the vacuum the Ricci tensor and curvature scalar are nulls, the oscillating curvature tensor is the Weyl curvature tensor that produce the tidal accelaration between various particles in accordance with equation (39). Therefore the Weyl tensor is locally measurable by simply watching the proper changes between nearby geodesics. The Weyl curvature is the only part of the curvature that exists in free gravitational field and it governs the propagation of gravitational through regions of space devoid of matter.

5. Ruse-Lanczos Identity

Let us consider at a point p in the manifold M those tensors W of type (0, 4) satisfying the algebraic similar to the Riemann curvature tensor,

Wλκµν = −Wκλµν = −Wλκνµ, Wλκµν = Wµνλκ (40) and Wκλµν + Wκµµλ + Wκνλµ = 0. (41) Tensors with the properties defined above, such as the curvature tensor, with two pairs of antisymmetric indices, there is a definition of two duality operations [4, 17, 18], the left dual ∼W and right dual W∼ defined by, 1 1 ∼W = W gρτgσφ  =  Wρσ (42) κλµν τφµν 2 κλρσ 2 κλρσ µν and 1 1 W∼ = W gρτgσφ  = W ρσ  . (43) κλµν κλτφ 2 µνρσ κλ 2 µνρσ As a consequence of double application of the left duality operation one have, 1 1 1 1 1 ∼∼W =  ∼Wρσ =  ρστφW = (−2)(δ τδ φ − δ φδ τ)W = − (W − W ), κλµν 2 κλρσ µν 2 κλρσ 2 τφµν 4 κ λ κ λ κλµν 2 κλµν λκµν with aid of equation (40), where one have Wλκµν = −Wκλµν, it follows as,

∼∼ Wκλµν = −Wκλµν. (44)

The double application of the right duality operation is similar to double left duality operation, where one have,

∼∼ Wκλµν = −Wκλµν. (45) In order to obtain the Ruse-Lanczos identity [4], we need to calculate left and right duality simultaneously as we will perform, ! 1 1 ∼W∼ =  Wρστφ  , κλµν 2 κλρσ 2 µντφ it can be rearranged as follows, 1 ∼W∼κλ = κλρσ W τφ. (46) µν 4 µντφ ρσ Let us pause here to calculate

κ κ κ κ δ µ δ ν δ τ δ φ λ λ λ λ κλρσ κ λ ρ σ δ µ δ ν δ τ δ φ  µντφ = −δ µ ν τ φ = − ρ ρ ρ ρ δ µ δ ν δ τ δ φ σ σ σ σ δ µ δ ν δ τ δ φ or λ λ λ λ λ λ λ λ λ λ λ λ δ ν δ τ δ φ δ µ δ τ δ φ δ µ δ ν δ φ δ µ δ ν δ τ κλρσ κ ρ ρ ρ κ ρ ρ ρ κ ρ ρ ρ κ ρ ρ ρ − µντφ = δ µ δ ν δ τ δ φ − δ ν δ µ δ τ δ φ + δ τ δ µ δ ν δ φ − δ φ δ µ δ ν δ τ σ σ σ σ σ σ σ σ σ σ σ σ δ ν δ τ δ φ δ µ δ τ δ φ δ µ δ ν δ φ δ µ δ ν δ τ 10 with more details we have

κλρσ κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  − µντφ = δ µ δ νδ τδ φ + δ τδ φδ ν + δ φδ νδ τ − δ φδ τδ ν − δ τδ νδ φ − δ νδ φδ τ κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  −δ ν δ µδ τδ φ + δ τδ φδ µ + δ φδ µδ τ − δ φδ τδ µ − δ τδ µδ φ − δ µδ φδ τ κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  +δ τ δ µδ νδ φ + δ νδ φδ µ + δ φδ µδ ν − δ φδ νδ µ − δ νδ µδ φ − δ µδ φδ ν κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  −δ φ δ µδ νδ τ + δ νδ τδ µ + δ τδ µδ ν − δ τδ νδ µ − δ νδ µδ τ − δ µδ τδ ν (47)

∼ ∼κλ 1 κλρσ τφ Now we must look at the equation (46), where we have W µν = 4  µντφWρσ , we are going to multiply the τφ tensor Wρσ with each above parentheses. The the first calculation is

κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  τφ δ µ δ νδ τδ φ + δ τδ φδ ν + δ φδ νδ τ − δ φδ τδ ν − δ τδ νδ φ − δ νδ φδ τ Wρσ κ  λ τφ λτ τλ = 2δ µ δ νWτφ + Wτν + Wντ Recalling that the tensor W is similar to algebraic symmetries of Riemann curvature tensor, we have from equation τλ λτ τ τφ τφ φ (40) that Wντ = Wτν and also W ντλ = Wνλ and besides that Wτφ = W τφ = W φ = W. Thus we have that,

κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  τφ κ λ κ λ δ µ δ νδ τδ φ + δ τδ φδ ν + δ φδ νδ τ − δ φδ τδ ν − δ τδ νδ φ − δ νδ φδ τ Wρσ = 2δ µδ νW − 4δ µWν . (48)

τφ The second parentheses of equation (47) multiplied by Wρσ results in,

κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  τφ κ λ κ λ δ ν δ µδ τδ φ + δ τδ φδ µ + δ φδ µδ τ − δ φδ τδ µ − δ τδ µδ φ − δ µδ φδ τ Wρσ = 2δ νδ µW − 4δ νWµ . (49)

τφ The third parentheses of equation (47) multiplied by Wρσ results in,

κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  τφ δ τ δ µδ νδ φ + δ νδ φδ µ + δ φδ µδ ν − δ φδ νδ µ − δ νδ µδ φ − δ µδ φδ ν Wρσ = λ κ λ κ κλ 2δ µWν − 2δ νWµ + 2Wµν . (50)

τφ And finally the fourth parentheses of equation (47) multiplied by Wρσ results in,

κ  λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ λ ρ σ  τφ δ φ δ µδ νδ τ + δ νδ τδ µ + δ τδ µδ ν − δ τδ νδ µ − δ νδ µδ τ − δ µδ τδ ν Wρσ = λ κ λ κ κλ −2δ µWν + 2δ νWµ − 2Wµν . (51) Turning these results (48), (49), (50) and (51) back into the equation (47) we have,

κλρσ κ λ κ λ κ λ κ λ λ κ λ κ κλ − µντφ = 2(δ µδ ν − δ νδ µ)W − 4(δ µWν − δ νWµ − δ µWν + δ νWµ ) + 4Wµν (52) Thus, we put the above equation (52) into equation (46) we have, 1 h i ∼W∼ = − 2(g g − g g )W − 4(g W − g W − g W + g W ) + 4W (53) κλµν 4 κµ λν κν λµ κµ λν κν λµ λµ κν λν κµ κλµν or ! ! ! ! 1 1 1 1 ∼W∼ + W = g W − g W − g W − g W + g W − g W − g W − g W . (54) κλµν κλµν λν κµ 4 κµ λµ κν 4 κν κµ λν 4 λν κν λµ 4 λµ The above equation is Ruse-Lanczos identity. We can simplify the above equation by definition, 1 P = W − g W, (55) κµ κµ 4 κµ so that we have ∼ ∼ Wκλµν + Wκλµν = gκµPλν − gκνPλµ + gλνPκµ − gλµPκν, (56) or ∼ ∼ Wκλµν + Wκλµν = gκ[µPν]λ + gλ[νPµ]κ. (57) The Ruse-Lanczos identity can be applied to Riemann curvature tensor and its components, conformal Weyl tensor Cκλµν as well as the others tensors Eκλµν and Gκλµν seen in equation (35). 11 6. Duality and complex formalism of the components of curvature tensor

The Riemann curvature tensor obeys the Ruse-Lanczos identity as follows in accordance with the equation (56), 1 where for Riemann curvature tensor we can recall the definition (28), S µν = Rµν − 4 gµν R, such as Pµν −→ S µν, we have, ∼ ∼ Rκλµν + Rκλµν = gκµS λν − gκνS λµ + gλνS κµ − gλµS κν. 1 Then we can recall the equation (27), where we have Eκλµν = 2 (gκµS λν + gλνS κµ − gκνS λµ − gλµS κν), so that the above equation becomes, ∼ ∼ Rκλµν + Rκλµν = 2Eκλµν. (58) µ Let us calculate the trace of the above equation with aid of the equation (37), where E λµν = S λν and denoting gκµ ∼R∼ = ∼R∼ we have that, κλµν λν ! 1 ∼R∼ + R = 2S = 2 R − g R , λν λν λν λν 4 λν that results in, ∼ ∼ Rµν = Gµν, (59) where Gµν is the Einstein tensor (30). The first component of Riemann curvature tensor that we apply the Ruse-Lanczos identity is the Weyl curvature tensor, recalling the equation (24) where Weyl curvature tensor is completely traceless, such as Pµν −→ Cµν = 0 into equation (56) for Weyl tensor and we have, ∼ ∼ Cκλµν + Cκλµν = 0. (60) The double application of the right duality operation into above equation we have,

∼ ∼∼ ∼ Cκλµν + Cκλµν = 0, with aid of equation (45) we have ∼ ∼ Cκλµν = Cκλµν. (61)

The second component of Riemann curvature tensor that we apply the Ruse-Lanczos identity is the tensor Gκλµν R   defined by equation (34), where it reads G = g g − g g . For this tensor we have from equation (36), the κλµν 12 κµ λν κν λµ R trace of this tensor is Gκ = g , such as the tensor P from definition (55) is null for tensor G , that it results µκν 4 µν µν κλµν into Ruse-Lanczos identity (56) in the below equation,

∼ ∼ Gκλµν + Gκλµν = 0. (62) and also the calculation performed for the above Weyl tensor, when we perform another right duality operation into above equation we have, ∼ ∼ Gκλµν = Gκλµν. (63)

The third component of Riemann curvature tensor that we apply the Ruse-Lanczos identity is the tensor Eκλµν defined by equation (27) to obtain the relation between the left dual and right dual. For this purpose we can return to equation (58) with the decomposition of Riemann curvature tensor obtained in the equation (35), where we have, Rκλµν = Cκλµν + Eκλµν + Gκλµν, so that,

∼ ∼ ∼ ∼ ∼ ∼ Cκλµν + Eκλµν + Gκλµν + Cκλµν + Eκλµν + Gκλµν = 2Eκλµν, associating similar terms we have,

∼ ∼ ∼ ∼ ∼ ∼ ( Cκλµν + Cκλµν) + ( Gκλµν + Gκλµν) + ( Eκλµν + Eκλµν) = 2Eκλµν, with aid of the equation (60) and (62) the above equation is reduced to

∼ ∼ Eκλµν = Eκλµν, 12 where another right duality operation into above equation results in,

∼ ∼ Eκλµν = −Eκλµν. (64)

For algebraic classification it is necessary to introduce the complex tensors

∗ ∼ Cκλµν = Cκλµν + iCκλµν, (65)

∗ ∼ Eκλµν = Eκλµν + iEκλµν, (66) and ∗ ∼ Gκλµν = Gκλµν + iGκλµν. (67) Applying left duality operation in the complex Weyl tensor in the equation (65) we obtain,

∼ ∗ ∼ ∼ ∼ Cκλµν = Cκλµν + i Cκλµν, here we can use the identities (61) and (60) so that the above equation produces

∼ ∗ ∼ ∼ Cκλµν = Cκλµν − iCκλµν = −i(Cκλµν + iCκλµν), or ∼ ∗ ∗ Cκλµν = −iCκλµν. (68) The same procedure for the other two tensors (66) and (67) yield,

∼ ∗ ∗ Eκλµν = iEκλµν. (69) and ∼ ∗ ∗ Gκλµν = −iGκλµν. (70) The Weyl tensor in complex formalism seen in equation (65) was classified first by Petrov for characterization of the exact solutions of General Relativity [18, 19]. Just like the electromagnetic case, with aid of a timelike vector, κ κ four-velocity vector u , where uκu = −1, we can separete the ‘electric’ and ‘magnetic’ parts of the Weyl tensor,

∗ λ ν λ ν ∼ λ ν Cκλµνu u = Cκλµνu u + iCκλµνu u = Eκµ + iBκµ. (71)

In the bivector language the Weyl tensor is decomposed in five complex components and it is shown by means of the Petrov classification how the local free gravitational field is composed of the linear sum of the distinct components, a transverse wave, a longitudinal wave and a Coulomb component [16, 18, 19, 20].

7. Conclusion

The standard elements of the theory of General Relativity discussed in the basic literature [12, 13, 14, 15], it is superficially approached the details and the importance of the Weyl tensor. The purpose of this paper was to indicate details of calculations about obtaining the Weyl tensor from the conformal transformations. The duality and complexity properties of the Weyl tensor explored in books and articles dealing with advanced topics of General Relativity assume that students know the decomposition properties of the Riemann tensor and their duality properties, but often do not bridge the existing gap between the introductory and advanced levels. Then, from that point of knowledge of the duality and complexity properties of the Weyl tensor, it is possible to understand subsequent themes such as the Petrov classification and the Newman-Penrose equations for General Relativity.

13 References

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