Vacuum Spacetimes with a Constant Weyl Eigenvalue

Transcription

1 Vacuum Spacetimes with a Constant Weyl Eigenvalue Alan Barnes School of Engineering and Applied Science Aston University, Birmingham UK Gent Meeting, Tuesday 8th April /28

2 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab 2/28

3 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab This eigenvalue (Weyl scalar) will be assumed non-zero. Otherwise, if λ = 0, we have: 2/28

4 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab This eigenvalue (Weyl scalar) will be assumed non-zero. Otherwise, if λ = 0, we have: EITHER a completely general Petrov type III or type N vacuum spacetime too hard! 2/28

5 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab This eigenvalue (Weyl scalar) will be assumed non-zero. Otherwise, if λ = 0, we have: EITHER a completely general Petrov type III or type N vacuum spacetime too hard! OR a special Petrov type I spacetime with zero Weyl scalar; but these do not exist (Brans, 1975) 2/28

6 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab This eigenvalue (Weyl scalar) will be assumed non-zero. Otherwise, if λ = 0, we have: EITHER a completely general Petrov type III or type N vacuum spacetime too hard! OR a special Petrov type I spacetime with zero Weyl scalar; but these do not exist (Brans, 1975) So the Petrov types to be considered are I, II & D only. 2/28

7 The Problems Vacuum spacetimes with a cosmological constant Λ are considered R ab = Λg ab in which there is an eigenbivector V ab of the (self-dual) Weyl tensor with a constant eigenvalue 2λ C ab cd V cd = 2λV ab This eigenvalue (Weyl scalar) will be assumed non-zero. Otherwise, if λ = 0, we have: EITHER a completely general Petrov type III or type N vacuum spacetime too hard! OR a special Petrov type I spacetime with zero Weyl scalar; but these do not exist (Brans, 1975) So the Petrov types to be considered are I, II & D only. The case where the Weyl scalars are constant non-zero multiples of a single (in general complex) field are also considered. 2/28

8 Motivation New exact solutions of Petrov type I and II? 3/28

9 Motivation New exact solutions of Petrov type I and II? Can Brans s theorem (ie. non-existence of Petrov I Einstien spaces with Ψ 2 = 0) be generalised? 3/28

10 Motivation New exact solutions of Petrov type I and II? Can Brans s theorem (ie. non-existence of Petrov I Einstien spaces with Ψ 2 = 0) be generalised? Coley et al. looked at spacetimes where all invariants constructed from the curvature and its derivatives are constant. 3/28

11 Motivation New exact solutions of Petrov type I and II? Can Brans s theorem (ie. non-existence of Petrov I Einstien spaces with Ψ 2 = 0) be generalised? Coley et al. looked at spacetimes where all invariants constructed from the curvature and its derivatives are constant. For Einstein spaces can we get by with just the assumption of constant Weyl eigenvalues (equivalently constant quadraic and cubic invariants of the Weyl tensor I and J)? 3/28

12 Motivation New exact solutions of Petrov type I and II? Can Brans s theorem (ie. non-existence of Petrov I Einstien spaces with Ψ 2 = 0) be generalised? Coley et al. looked at spacetimes where all invariants constructed from the curvature and its derivatives are constant. For Einstein spaces can we get by with just the assumption of constant Weyl eigenvalues (equivalently constant quadraic and cubic invariants of the Weyl tensor I and J)? Einstein spaces of embedding class two with torsion necessarily have a constant Weyl eigenvalue (= 2Λ/3) Yakupov (1969). Can we prove (or find a counterexample to) his claim that there are no such solutions? 3/28

13 Petrov Classification I Many approaches to the Petrov classification; all essentially equivalent in 4-D: eigenbivector/invariant subspace structure of the Weyl tensor (Petrov s approach) 4/28

14 Petrov Classification I Many approaches to the Petrov classification; all essentially equivalent in 4-D: eigenbivector/invariant subspace structure of the Weyl tensor (Petrov s approach) ditto for the self-dual Weyl tensor C ab cd = Cab cd + ic ab cd 4/28

15 Petrov Classification I Many approaches to the Petrov classification; all essentially equivalent in 4-D: eigenbivector/invariant subspace structure of the Weyl tensor (Petrov s approach) ditto for the self-dual Weyl tensor C ab cd = Cab cd + ic ab cd invariant subspace structure of Q ab = E ab + ib ab Q ab e b = λe a NB no factor 2 where E ab & B ab are the electric and magnetic parts of the Weyl tensor wrt a unit timelike vector u a E ab = C acbd u c u d H ij = Cacbd uc u d 4/28

16 Petrov Classification I Many approaches to the Petrov classification; all essentially equivalent in 4-D: eigenbivector/invariant subspace structure of the Weyl tensor (Petrov s approach) ditto for the self-dual Weyl tensor C ab cd = Cab cd + ic ab cd invariant subspace structure of Q ab = E ab + ib ab Q ab e b = λe a NB no factor 2 where E ab & B ab are the electric and magnetic parts of the Weyl tensor wrt a unit timelike vector u a E ab = C acbd u c u d H ij = Cacbd uc u d Number of distinct principal null directions l a satisfying l [e l b C a]bc[d l c l f ] = 0 4/28

17 Petrov Classification I Many approaches to the Petrov classification; all essentially equivalent in 4-D: eigenbivector/invariant subspace structure of the Weyl tensor (Petrov s approach) ditto for the self-dual Weyl tensor C ab cd = Cab cd + ic ab cd invariant subspace structure of Q ab = E ab + ib ab Q ab e b = λe a NB no factor 2 where E ab & B ab are the electric and magnetic parts of the Weyl tensor wrt a unit timelike vector u a E ab = C acbd u c u d H ij = Cacbd uc u d Number of distinct principal null directions l a satisfying l [e l b C a]bc[d l c l f ] = 0 Number of distinct 1-spinors in the totally symmetric Weyl spinor Ψ ABCD = α (A β B γ C δ D) 4/28

18 Petrov Classification II For Petrov types I and D there is an orthonormal eigen-tetrad (u a, e a 1, ea 2, ea 3 ) of Q AB such that its frame components (A, B = ) are λ Q AB = 0 λ 2 0 λ 1 + λ 2 + λ 3 = λ 3 where for type D: λ 1 = λ 2 = λ 3 /2. 5/28

19 Petrov Classification II For Petrov types I and D there is an orthonormal eigen-tetrad (u a, e a 1, ea 2, ea 3 ) of Q AB such that its frame components (A, B = ) are λ Q AB = 0 λ 2 0 λ 1 + λ 2 + λ 3 = λ 3 where for type D: λ 1 = λ 2 = λ 3 /2. For Petrov type II the corresponding frame components are 1 λ 3 /2 i 0 Q AB = i 1 λ 3 / λ 3 5/28

20 Petrov Classification III Using the associated null tetrad: l a = 1/ 2(u a +e3 a ), na = 1/ 2(u a e3 a ), ma = 1/ 2(e1 a +iea 2 ) the NP Weyl tensor components take the form: For Petrov Type I: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = (λ 2 λ 1 )/2 6/28

21 Petrov Classification III Using the associated null tetrad: l a = 1/ 2(u a +e3 a ), na = 1/ 2(u a e3 a ), ma = 1/ 2(e1 a +iea 2 ) the NP Weyl tensor components take the form: For Petrov Type I: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = (λ 2 λ 1 )/2 For Petrov Type II: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = 0 Ψ 4 0 6/28

22 Petrov Classification III Using the associated null tetrad: l a = 1/ 2(u a +e3 a ), na = 1/ 2(u a e3 a ), ma = 1/ 2(e1 a +iea 2 ) the NP Weyl tensor components take the form: For Petrov Type I: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = (λ 2 λ 1 )/2 For Petrov Type II: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = 0 Ψ 4 0 For Petrov Type D: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = 0 6/28

23 Petrov Classification III Using the associated null tetrad: l a = 1/ 2(u a +e3 a ), na = 1/ 2(u a e3 a ), ma = 1/ 2(e1 a +iea 2 ) the NP Weyl tensor components take the form: For Petrov Type I: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = (λ 2 λ 1 )/2 For Petrov Type II: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = 0 Ψ 4 0 For Petrov Type D: Ψ 1 = Ψ 3 = 0, Ψ 2 = λ 3 /2, Ψ 0 = Ψ 4 = 0 We will assume that λ 3 (and so Ψ 2 ) is constant. 6/28

24 Algebraically Special Cases (Type II & D) In the Newman-Penrose spin coefficient formalism the relevant Bianchi identites are (since Ψ 0 = Ψ 1 = Ψ 3 = 0): κψ 2 = 0 σψ 2 = 0 D(Ψ 2 + R/12) = 3ρΨ 2 δ(ψ 2 + R/12) = 3τΨ 2 δ(ψ 2 + R/12) = 3πΨ 2 κψ 4 (Ψ 2 + R/12) = 3µΨ 2 + σψ 4 7/28

25 Algebraically Special Cases (Type II & D) In the Newman-Penrose spin coefficient formalism the relevant Bianchi identites are (since Ψ 0 = Ψ 1 = Ψ 3 = 0): κψ 2 = 0 σψ 2 = 0 D(Ψ 2 + R/12) = 3ρΨ 2 δ(ψ 2 + R/12) = 3τΨ 2 δ(ψ 2 + R/12) = 3πΨ 2 κψ 4 (Ψ 2 + R/12) = 3µΨ 2 + σψ 4 Thus, as Ψ 2 ( 0) and R(= 4Λ) are constant: κ = σ = ρ = τ = 0 (and π = µ = 0) 7/28

26 Algebraically Special Cases (Type II & D) In the Newman-Penrose spin coefficient formalism the relevant Bianchi identites are (since Ψ 0 = Ψ 1 = Ψ 3 = 0): κψ 2 = 0 σψ 2 = 0 D(Ψ 2 + R/12) = 3ρΨ 2 δ(ψ 2 + R/12) = 3τΨ 2 δ(ψ 2 + R/12) = 3πΨ 2 κψ 4 (Ψ 2 + R/12) = 3µΨ 2 + σψ 4 Thus, as Ψ 2 ( 0) and R(= 4Λ) are constant: κ = σ = ρ = τ = 0 (and π = µ = 0) In addition, from the NP equation for ρ δτ or for Dµ δπ, it follows that Ψ 2 = Λ/3 0 or equivalently λ 3 = 2Λ/3. 7/28

27 Algebraically Special Cases (Type II & D) In the Newman-Penrose spin coefficient formalism the relevant Bianchi identites are (since Ψ 0 = Ψ 1 = Ψ 3 = 0): κψ 2 = 0 σψ 2 = 0 D(Ψ 2 + R/12) = 3ρΨ 2 δ(ψ 2 + R/12) = 3τΨ 2 δ(ψ 2 + R/12) = 3πΨ 2 κψ 4 (Ψ 2 + R/12) = 3µΨ 2 + σψ 4 Thus, as Ψ 2 ( 0) and R(= 4Λ) are constant: κ = σ = ρ = τ = 0 (and π = µ = 0) In addition, from the NP equation for ρ δτ or for Dµ δπ, it follows that Ψ 2 = Λ/3 0 or equivalently λ 3 = 2Λ/3. NB No algebraically special pure vacuum spacetimes with a constant Weyl scalar. 7/28

28 Algebraically Special Cases 2 Alg. sp. Einstein spaces with constant Weyl scalars have Ψ 2 = Λ/3 0. They include all possible alg. sp. proper Einstein spaces of embedding class 2 with torsion. 8/28

29 Algebraically Special Cases 2 Alg. sp. Einstein spaces with constant Weyl scalars have Ψ 2 = Λ/3 0. They include all possible alg. sp. proper Einstein spaces of embedding class 2 with torsion. The spacetime is a Kundt (1961) spacetime with metric: ds 2 = 2P 2 dzd z 2du(dv + Wdz + W d z + Hdu) where P = P(z, z, u) and H = H(z, z, u, v) are real and W = W (z, z, u, v) is complex. 8/28

30 Algebraically Special Cases 2 Alg. sp. Einstein spaces with constant Weyl scalars have Ψ 2 = Λ/3 0. They include all possible alg. sp. proper Einstein spaces of embedding class 2 with torsion. The spacetime is a Kundt (1961) spacetime with metric: ds 2 = 2P 2 dzd z 2du(dv + Wdz + W d z + Hdu) where P = P(z, z, u) and H = H(z, z, u, v) are real and W = W (z, z, u, v) is complex. W,v = 0 as a consequence of τ = 0. So W = W (z, z, u). 8/28

31 Algebraically Special Cases 2 Alg. sp. Einstein spaces with constant Weyl scalars have Ψ 2 = Λ/3 0. They include all possible alg. sp. proper Einstein spaces of embedding class 2 with torsion. The spacetime is a Kundt (1961) spacetime with metric: ds 2 = 2P 2 dzd z 2du(dv + Wdz + W d z + Hdu) where P = P(z, z, u) and H = H(z, z, u, v) are real and W = W (z, z, u, v) is complex. W,v = 0 as a consequence of τ = 0. So W = W (z, z, u). These spacetimes were studied by Lewandowski (1992) assuming Λ with reduced holonomy group 8/28

32 Algebraically Special Cases 2 Alg. sp. Einstein spaces with constant Weyl scalars have Ψ 2 = Λ/3 0. They include all possible alg. sp. proper Einstein spaces of embedding class 2 with torsion. The spacetime is a Kundt (1961) spacetime with metric: ds 2 = 2P 2 dzd z 2du(dv + Wdz + W d z + Hdu) where P = P(z, z, u) and H = H(z, z, u, v) are real and W = W (z, z, u, v) is complex. W,v = 0 as a consequence of τ = 0. So W = W (z, z, u). These spacetimes were studied by Lewandowski (1992) assuming Λ with reduced holonomy group He showed (using coordinate freedom in the metric): P = 1 + Λz z/2 H = Λv 2 /2 + H 0 (z, z, u) 8/28

33 Lewandowski Metrics Type II W is given by W = il,z where L is a real potential satisfying P 2 L,z z = ΛL. 9/28

34 Lewandowski Metrics Type II W is given by W = il,z where L is a real potential satisfying P 2 L,z z = ΛL. The general solution for the potential L is L = R(ΛP 1 zf (z, u) f,z (z, u)) where f (z, u) is an arbitrary function analytic in z. 9/28

35 Lewandowski Metrics Type II W is given by W = il,z where L is a real potential satisfying P 2 L,z z = ΛL. The general solution for the potential L is L = R(ΛP 1 zf (z, u) f,z (z, u)) where f (z, u) is an arbitrary function analytic in z. The remaining field equation implies H 0,z z = ΛL,z L, z Λ 2 P 2 L 2 Given L, this can be integrated to give H 0 up to addition of an arbitrary harmonic function Rh 0 (z, u) 9/28

36 Lewandowski Metrics Type II W is given by W = il,z where L is a real potential satisfying P 2 L,z z = ΛL. The general solution for the potential L is L = R(ΛP 1 zf (z, u) f,z (z, u)) where f (z, u) is an arbitrary function analytic in z. The remaining field equation implies H 0,z z = ΛL,z L, z Λ 2 P 2 L 2 Given L, this can be integrated to give H 0 up to addition of an arbitrary harmonic function Rh 0 (z, u) The solutions are quite general depending on two arbitrary functions: f (z, u) & h 0 (z, u). In general they admit no Killling vectors except... 9/28

37 Type D case In the general case Ψ 2 = Λ/3 and Ψ 4 is given by Ψ 4 = i( / u Λv)(P 2 L, z ), z + (P 2 H 0, z ), z Λ(PL, z ) 2 10/28

38 Type D case In the general case Ψ 2 = Λ/3 and Ψ 4 is given by Ψ 4 = i( / u Λv)(P 2 L, z ), z + (P 2 H 0, z ), z Λ(PL, z ) 2 For type D the condition Ψ 4 = 0 implies (P 2 L, z ), z = 0 and hence L, z = g(z, u)/p 2 for some function g(z, u). 10/28

39 Type D case In the general case Ψ 2 = Λ/3 and Ψ 4 is given by Ψ 4 = i( / u Λv)(P 2 L, z ), z + (P 2 H 0, z ), z Λ(PL, z ) 2 For type D the condition Ψ 4 = 0 implies (P 2 L, z ), z = 0 and hence L, z = g(z, u)/p 2 for some function g(z, u). This implies that f (z, u) = A(u)z 2 + B(u)z + C(u) where A, B & C are arbitrary functions of u and H 0 = h(u). 10/28

40 Type D case In the general case Ψ 2 = Λ/3 and Ψ 4 is given by Ψ 4 = i( / u Λv)(P 2 L, z ), z + (P 2 H 0, z ), z Λ(PL, z ) 2 For type D the condition Ψ 4 = 0 implies (P 2 L, z ), z = 0 and hence L, z = g(z, u)/p 2 for some function g(z, u). This implies that f (z, u) = A(u)z 2 + B(u)z + C(u) where A, B & C are arbitrary functions of u and H 0 = h(u). The remaining coordinate freedom preserving the metric can be used to set W = h(u) = 0. 10/28

41 Type D case In the general case Ψ 2 = Λ/3 and Ψ 4 is given by Ψ 4 = i( / u Λv)(P 2 L, z ), z + (P 2 H 0, z ), z Λ(PL, z ) 2 For type D the condition Ψ 4 = 0 implies (P 2 L, z ), z = 0 and hence L, z = g(z, u)/p 2 for some function g(z, u). This implies that f (z, u) = A(u)z 2 + B(u)z + C(u) where A, B & C are arbitrary functions of u and H 0 = h(u). The remaining coordinate freedom preserving the metric can be used to set W = h(u) = 0. The type D metric is decomposable into two 2-spaces of constant curvature Robinson-Bertotti metric ds 2 = 2(1 + Λz z/2) 2 dzd z 2dudv Λv 2 du 2 Homogeneous with 6-D isometry group 10/28

42 Algebraically General Case (Petrov type I) Consider first the case where all 3 Weyl scalars are constant. Thus Ψ 2 and Ψ 0 (= Ψ 4 ) are constant. 11/28

43 Algebraically General Case (Petrov type I) Consider first the case where all 3 Weyl scalars are constant. Thus Ψ 2 and Ψ 0 (= Ψ 4 ) are constant. The Bianchi identities reduce in this case to purely algebraic equations: (π 4α)Ψ 0 = 3κΨ 2 (µ 4γ)Ψ 0 = 3σΨ 2 (ρ 4ɛ)Ψ 0 = 3λΨ 2 (τ 4β)Ψ 0 = 3νΨ 2 λψ 0 = 3ρΨ 2 σψ 0 = 3µΨ 2 κψ 0 = 3πΨ 2 νψ 0 = 3τΨ 2 11/28

44 Algebraically General Case (Petrov type I) Consider first the case where all 3 Weyl scalars are constant. Thus Ψ 2 and Ψ 0 (= Ψ 4 ) are constant. The Bianchi identities reduce in this case to purely algebraic equations: (π 4α)Ψ 0 = 3κΨ 2 (µ 4γ)Ψ 0 = 3σΨ 2 (ρ 4ɛ)Ψ 0 = 3λΨ 2 (τ 4β)Ψ 0 = 3νΨ 2 λψ 0 = 3ρΨ 2 σψ 0 = 3µΨ 2 κψ 0 = 3πΨ 2 νψ 0 = 3τΨ 2 Thus, writing ψ = Ψ 0 /(3Ψ 2 ), we have ρ = ψλ τ = ψν µ = ψσ π = ψκ, α = (ψ + ψ 1 )κ/4, β = (ψ + ψ 1 )ν/4, ɛ = (ψ + ψ 1 )λ/4, γ = (ψ + ψ 1 )σ/4 11/28

45 Algebraically General Case (Petrov type I) Consider first the case where all 3 Weyl scalars are constant. Thus Ψ 2 and Ψ 0 (= Ψ 4 ) are constant. The Bianchi identities reduce in this case to purely algebraic equations: (π 4α)Ψ 0 = 3κΨ 2 (µ 4γ)Ψ 0 = 3σΨ 2 (ρ 4ɛ)Ψ 0 = 3λΨ 2 (τ 4β)Ψ 0 = 3νΨ 2 λψ 0 = 3ρΨ 2 σψ 0 = 3µΨ 2 κψ 0 = 3πΨ 2 νψ 0 = 3τΨ 2 Thus, writing ψ = Ψ 0 /(3Ψ 2 ), we have ρ = ψλ τ = ψν µ = ψσ π = ψκ, α = (ψ + ψ 1 )κ/4, β = (ψ + ψ 1 )ν/4, ɛ = (ψ + ψ 1 )λ/4, γ = (ψ + ψ 1 )σ/4 ψ 0, ±1 (type D) ψ ±1/3 (zero Weyl eigenvalue) 11/28

46 Algebraically General Case 2 Substituting these equations in the NP Ricci identities only the spin coefficients κ, σ, ν & λ remain. 12/28

47 Algebraically General Case 2 Substituting these equations in the NP Ricci identities only the spin coefficients κ, σ, ν & λ remain. Maple and Norbert Van den Bergh s package for the NP formalism was used extensively. 12/28

48 Algebraically General Case 2 Substituting these equations in the NP Ricci identities only the spin coefficients κ, σ, ν & λ remain. Maple and Norbert Van den Bergh s package for the NP formalism was used extensively. 1st & 7th NP equations are two linear simultaneous equations (non-singular for ψ ±1) for Dλ & δκ which may be solved for these two derivs. 12/28

49 Algebraically General Case 2 Substituting these equations in the NP Ricci identities only the spin coefficients κ, σ, ν & λ remain. Maple and Norbert Van den Bergh s package for the NP formalism was used extensively. 1st & 7th NP equations are two linear simultaneous equations (non-singular for ψ ±1) for Dλ & δκ which may be solved for these two derivs. Similarly equations for Dν & κ, δλ & δσ and δν & σ are obtained. 8 spin coeffient derivatives now known 12/28

50 Algebraically General Case 2 Substituting these equations in the NP Ricci identities only the spin coefficients κ, σ, ν & λ remain. Maple and Norbert Van den Bergh s package for the NP formalism was used extensively. 1st & 7th NP equations are two linear simultaneous equations (non-singular for ψ ±1) for Dλ & δκ which may be solved for these two derivs. Similarly equations for Dν & κ, δλ & δσ and δν & σ are obtained. 8 spin coeffient derivatives now known Commutator δd Dδ can now be applied to λ & ν and commutator δ δ to σ & κ to produce integrability conditions but first... 12/28

51 Some Algebraic Equalities The 2nd & 8th NP equations are both equalities for Dσ δκ and lead to a purely algebraic identity relating the curvature to a cross-ratio of spin coeffs: 24(κν σλ)(1 ψ 2 ) 12Ψ 2 (1 3ψ 2 ) R = 0. The 10th & 17th NP equations also produce this. 13/28

52 Some Algebraic Equalities The 2nd & 8th NP equations are both equalities for Dσ δκ and lead to a purely algebraic identity relating the curvature to a cross-ratio of spin coeffs: 24(κν σλ)(1 ψ 2 ) 12Ψ 2 (1 3ψ 2 ) R = 0. The 10th & 17th NP equations also produce this. A second algebraic relation for the cross-ratio is obtained from the 2nd, 6th, 10th and 12th NP eqs: 6(κν σλ)(1 ψ 2 )(1 5ψ 2 ) ψ 2 (18ψ 2 Ψ 2 42Ψ 2 + R) = 0. 13/28

53 Some Algebraic Equalities The 2nd & 8th NP equations are both equalities for Dσ δκ and lead to a purely algebraic identity relating the curvature to a cross-ratio of spin coeffs: 24(κν σλ)(1 ψ 2 ) 12Ψ 2 (1 3ψ 2 ) R = 0. The 10th & 17th NP equations also produce this. A second algebraic relation for the cross-ratio is obtained from the 2nd, 6th, 10th and 12th NP eqs: 6(κν σλ)(1 ψ 2 )(1 5ψ 2 ) ψ 2 (18ψ 2 Ψ 2 42Ψ 2 + R) = 0. Equivalently but simpler: Ψ 2 = (κν σλ)(9ψ2 1) 9ψ 2 R = 4(κν σλ)(3ψ 2 + 1) 2 /3 13/28

54 Some Algebraic Equalities The 2nd & 8th NP equations are both equalities for Dσ δκ and lead to a purely algebraic identity relating the curvature to a cross-ratio of spin coeffs: 24(κν σλ)(1 ψ 2 ) 12Ψ 2 (1 3ψ 2 ) R = 0. The 10th & 17th NP equations also produce this. A second algebraic relation for the cross-ratio is obtained from the 2nd, 6th, 10th and 12th NP eqs: 6(κν σλ)(1 ψ 2 )(1 5ψ 2 ) ψ 2 (18ψ 2 Ψ 2 42Ψ 2 + R) = 0. Equivalently but simpler: Ψ 2 = (κν σλ)(9ψ2 1) 9ψ 2 R = 4(κν σλ)(3ψ 2 + 1) 2 /3 Thus, if ψ = ±i/ 3 we have pure vacuum 13/28

55 Commutators 1 Applying the commutators, equations obtained for κ λ σ δλ κ ν σ δν νdσ λδσ νdκ λδκ 14/28

56 Commutators 1 Applying the commutators, equations obtained for κ λ σ δλ κ ν σ δν νdσ λδσ νdκ λδκ There remain 5 independent NP equations for Dσ δκ Dκ δλ Dσ λ λ δν λ δν 14/28

57 Commutators 1 Applying the commutators, equations obtained for κ λ σ δλ κ ν σ δν νdσ λδσ νdκ λδκ There remain 5 independent NP equations for Dσ δκ Dκ δλ Dσ λ λ δν λ δν The remaining NP equation for δκ δν is dependent 1st, 3rd & 4th of these. 14/28

58 Commutators 1 Applying the commutators, equations obtained for κ λ σ δλ κ ν σ δν νdσ λδσ νdκ λδκ There remain 5 independent NP equations for Dσ δκ Dκ δλ Dσ λ λ δν λ δν The remaining NP equation for δκ δν is dependent 1st, 3rd & 4th of these. If σ 0, the 5 NP and the first 3 commutator eqs can be solved for the remaining 8 derivatives of κ, ν, σ & λ. 14/28

59 Commutators 1 Applying the commutators, equations obtained for κ λ σ δλ κ ν σ δν νdσ λδσ νdκ λδκ There remain 5 independent NP equations for Dσ δκ Dκ δλ Dσ λ λ δν λ δν The remaining NP equation for δκ δν is dependent 1st, 3rd & 4th of these. If σ 0, the 5 NP and the first 3 commutator eqs can be solved for the remaining 8 derivatives of κ, ν, σ & λ. The fourth commutator then simplifies to ψ(1 + 3ψ 2 )(κν σλ) 2 = 0. Thus ψ = ±i/ 3 and we must have pure vacuum 14/28

60 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. 15/28

61 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. 15/28

62 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. Dλ & σ eqs. imply κ 2 + κν = 0 and ν 2 + κ ν = 0. Thus ν = ɛ 1 κ & κ 2 + ɛ 1 κ 2 = 0 where ɛ 1 = ±1. 15/28

63 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. Dλ & σ eqs. imply κ 2 + κν = 0 and ν 2 + κ ν = 0. Thus ν = ɛ 1 κ & κ 2 + ɛ 1 κ 2 = 0 where ɛ 1 = ±1. As a consequence Dκ = Dν = κ = ν = 0. 15/28

64 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. Dλ & σ eqs. imply κ 2 + κν = 0 and ν 2 + κ ν = 0. Thus ν = ɛ 1 κ & κ 2 + ɛ 1 κ 2 = 0 where ɛ 1 = ±1. As a consequence Dκ = Dν = κ = ν = 0. The Dσ λ eq. implies 6ψ 2 + ɛ 1 3ψ ψ + 1 = 0. This is only consistent only if ɛ 1 = 1 & ψ = ±i/ (3). Thus the spacetime is pure vacuum. 15/28

65 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. Dλ & σ eqs. imply κ 2 + κν = 0 and ν 2 + κ ν = 0. Thus ν = ɛ 1 κ & κ 2 + ɛ 1 κ 2 = 0 where ɛ 1 = ±1. As a consequence Dκ = Dν = κ = ν = 0. The Dσ λ eq. implies 6ψ 2 + ɛ 1 3ψ ψ + 1 = 0. This is only consistent only if ɛ 1 = 1 & ψ = ±i/ (3). Thus the spacetime is pure vacuum. WLOG set ψ = +i/ 3 (by the rotation m a = im a if necessary). Then κ = ±k(1 + ɛ 2 i)/ 2 where k = κ and ɛ 2 = ±1 15/28

66 The case σ = 0 If σ = 0, then κ 0 (Golberg-Sachs). The δσ eq. implies λν = 0. But ν = 0 leads to a contradiction since it implies κν σλ = 0 and hence Ψ 2 = 0. So λ = 0. Dλ & σ eqs. imply κ 2 + κν = 0 and ν 2 + κ ν = 0. Thus ν = ɛ 1 κ & κ 2 + ɛ 1 κ 2 = 0 where ɛ 1 = ±1. As a consequence Dκ = Dν = κ = ν = 0. The Dσ λ eq. implies 6ψ 2 + ɛ 1 3ψ ψ + 1 = 0. This is only consistent only if ɛ 1 = 1 & ψ = ±i/ (3). Thus the spacetime is pure vacuum. WLOG set ψ = +i/ 3 (by the rotation m a = im a if necessary). Then κ = ±k(1 + ɛ 2 i)/ 2 where k = κ and ɛ 2 = ±1 It now follows that δκ = δκ = 0 so all spin coefficients are constant. 15/28

67 The case σ 0 More Commutators Returning to the general case: we have Λ = 0 (pure vacuum only) and WLOG ψ = 1/ 3. 16/28

68 The case σ 0 More Commutators Returning to the general case: we have Λ = 0 (pure vacuum only) and WLOG ψ = 1/ 3. The last equation means that the 3 Weyl scalars are multiples of the three cube roots of 1. 16/28

69 The case σ 0 More Commutators Returning to the general case: we have Λ = 0 (pure vacuum only) and WLOG ψ = 1/ 3. The last equation means that the 3 Weyl scalars are multiples of the three cube roots of 1. Applying the commutation relations to the newly found spin coefficient derivatives, 8 complex 3rd order homogeneous algebraic relations are found for κ, σ, ν & λ. 16/28

70 The case σ 0 More Commutators Returning to the general case: we have Λ = 0 (pure vacuum only) and WLOG ψ = 1/ 3. The last equation means that the 3 Weyl scalars are multiples of the three cube roots of 1. Applying the commutation relations to the newly found spin coefficient derivatives, 8 complex 3rd order homogeneous algebraic relations are found for κ, σ, ν & λ. The symmetry of the system means it can be solved by hand whilst Maple s Groebner package fails. 16/28

71 The case σ 0 More Commutators Returning to the general case: we have Λ = 0 (pure vacuum only) and WLOG ψ = 1/ 3. The last equation means that the 3 Weyl scalars are multiples of the three cube roots of 1. Applying the commutation relations to the newly found spin coefficient derivatives, 8 complex 3rd order homogeneous algebraic relations are found for κ, σ, ν & λ. The symmetry of the system means it can be solved by hand whilst Maple s Groebner package fails. The 8 eqs. are only consistent if κ, σ, ν & λ are all non-zero constants (and for the case σ = λ = 0 above). 16/28

72 The Solutions Including the case σ = 0 there are 3 solutions (where k is an arbitrary positive constant and ɛ 2 = ±1): σ = λ = 0, ν = + κ, κ = ±k(1 + ɛ 2 i)/ 2, Ψ 2 = 8k 2 /3 (1) 17/28

73 The Solutions Including the case σ = 0 there are 3 solutions (where k is an arbitrary positive constant and ɛ 2 = ±1): σ = λ = 0, ν = + κ, κ = ±k(1 + ɛ 2 i)/ 2, Ψ 2 = 8k 2 /3 (1) λ = σ = ɛ 2 κ, ν = κ, κ = ±k( 3 + i)/2, Ψ 2 = 4k 2 ( 1 + i 3)/3 (2) 17/28

74 The Solutions Including the case σ = 0 there are 3 solutions (where k is an arbitrary positive constant and ɛ 2 = ±1): σ = λ = 0, ν = + κ, κ = ±k(1 + ɛ 2 i)/ 2, Ψ 2 = 8k 2 /3 (1) λ = σ = ɛ 2 κ, ν = κ, κ = ±k( 3 + i)/2, Ψ 2 = 4k 2 ( 1 + i 3)/3 (2) λ = σ = iɛ 2 κ, ν = +κ, κ = ±k(1 i 3)/2, Ψ 2 = 4k 2 (1 + i 3)/3 (3) 17/28

75 The Solutions Including the case σ = 0 there are 3 solutions (where k is an arbitrary positive constant and ɛ 2 = ±1): σ = λ = 0, ν = + κ, κ = ±k(1 + ɛ 2 i)/ 2, Ψ 2 = 8k 2 /3 (1) λ = σ = ɛ 2 κ, ν = κ, κ = ±k( 3 + i)/2, Ψ 2 = 4k 2 ( 1 + i 3)/3 (2) λ = σ = iɛ 2 κ, ν = +κ, κ = ±k(1 i 3)/2, Ψ 2 = 4k 2 (1 + i 3)/3 (3) In all three cases ρ = iλ/ 3, τ = iν/ 3, µ = iσ/ 3, π = iκ/ 3. α = iκ/ 3, β = iν/ 3, ɛ = iλ/ 3, γ = iσ/ 3. 17/28

76 The Metric The spacetime is homogeneous with metric: ds 2 = 1 ( 4k 2 dx 2 + e 4x/ 3 dy 2 + e 2x/ 3 cos(2x)(dz 2 dt 2 ) ) 2e 2x/ 3 sin(2x)dzdt 18/28

77 The Metric The spacetime is homogeneous with metric: ds 2 = 1 ( 4k 2 dx 2 + e 4x/ 3 dy 2 + e 2x/ 3 cos(2x)(dz 2 dt 2 ) ) 2e 2x/ 3 sin(2x)dzdt This pure vacuum metric is originally due to Petrov (1962) who derived it using group theoretic methods. It has a 4-D isometry group. 18/28

78 The Metric The spacetime is homogeneous with metric: ds 2 = 1 ( 4k 2 dx 2 + e 4x/ 3 dy 2 + e 2x/ 3 cos(2x)(dz 2 dt 2 ) ) 2e 2x/ 3 sin(2x)dzdt This pure vacuum metric is originally due to Petrov (1962) who derived it using group theoretic methods. It has a 4-D isometry group. It is the only Petrov type I Einstein space with constant Weyl eigenvalues. 18/28

79 The Metric The spacetime is homogeneous with metric: ds 2 = 1 ( 4k 2 dx 2 + e 4x/ 3 dy 2 + e 2x/ 3 cos(2x)(dz 2 dt 2 ) ) 2e 2x/ 3 sin(2x)dzdt This pure vacuum metric is originally due to Petrov (1962) who derived it using group theoretic methods. It has a 4-D isometry group. It is the only Petrov type I Einstein space with constant Weyl eigenvalues. So a fortiori it is the only Petrov type I homogeneous Einstein space. 18/28

80 The Metric The spacetime is homogeneous with metric: ds 2 = 1 ( 4k 2 dx 2 + e 4x/ 3 dy 2 + e 2x/ 3 cos(2x)(dz 2 dt 2 ) ) 2e 2x/ 3 sin(2x)dzdt This pure vacuum metric is originally due to Petrov (1962) who derived it using group theoretic methods. It has a 4-D isometry group. It is the only Petrov type I Einstein space with constant Weyl eigenvalues. So a fortiori it is the only Petrov type I homogeneous Einstein space. The canonical complex null tetrad is not unique; it depends on which of the 3 spacelike eigenvectors of Q AB is used to construct the real null vectors. 18/28

81 Only One Metric! Hence three solutions for the spin coeffs. are to be expected. Also the sign ambiguities in the solutions stem from those in the eigenvectors. 19/28

82 Only One Metric! Hence three solutions for the spin coeffs. are to be expected. Also the sign ambiguities in the solutions stem from those in the eigenvectors. The three solutions for the spin coefficients all correspond to the same underlying spacetime. 19/28

83 Only One Metric! Hence three solutions for the spin coeffs. are to be expected. Also the sign ambiguities in the solutions stem from those in the eigenvectors. The three solutions for the spin coefficients all correspond to the same underlying spacetime. A canonical orthonormal tetrad of Weyl eigen-one-forms corresponding to solution (1) is 3 u a dx a = ex/ ( ) cos x dt sin x dz 2k e 3a dx a = ex/ 3 ( ) sin x dt + cos x dz 2k e 1a dx a = 1 2 ( dx e 2x/ 3 dy ) 2k e 2a dx a = 1 2 ( dx + e 2x/ 3 dy ). 2k 19/28

84 One Constant Weyl Eigenvalue We consider Petrov type I Einstein spaces with only one constant Weyl eigenvalue (WLOG λ 3 ). 20/28

85 One Constant Weyl Eigenvalue We consider Petrov type I Einstein spaces with only one constant Weyl eigenvalue (WLOG λ 3 ). Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 and Ψ 2 is a non-zero constant. As before ψ = Ψ 0 /(3Ψ 2 ), but now ψ is not constant. 20/28

86 One Constant Weyl Eigenvalue We consider Petrov type I Einstein spaces with only one constant Weyl eigenvalue (WLOG λ 3 ). Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 and Ψ 2 is a non-zero constant. As before ψ = Ψ 0 /(3Ψ 2 ), but now ψ is not constant. Four Bianchi identities become purely algebraic: ρ = ψλ, τ = ψν, π = ψκ, µ = ψσ 20/28

87 One Constant Weyl Eigenvalue We consider Petrov type I Einstein spaces with only one constant Weyl eigenvalue (WLOG λ 3 ). Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 and Ψ 2 is a non-zero constant. As before ψ = Ψ 0 /(3Ψ 2 ), but now ψ is not constant. Four Bianchi identities become purely algebraic: ρ = ψλ, τ = ψν, π = ψκ, µ = ψσ The remaining four (non-trivial) ones reduce to Dψ = (ψ 2 1)λ 4ψɛ, ψ = (ψ 2 1)σ + 4ψγ δψ = (ψ 2 1)ν 4ψβ, δψ = (ψ 2 1)κ + 4ψα 20/28

88 One Constant Weyl Eigenvalue Subtraction of ψ (2nd Ricci identity) from the 8th, after substituting for ρ, τ, µ & π produces: κν σλ = (1 3ψ2 )Ψ 2 + R/12 2(1 ψ 2 ) as in case with 3 constant evs. 21/28

89 One Constant Weyl Eigenvalue Subtraction of ψ (2nd Ricci identity) from the 8th, after substituting for ρ, τ, µ & π produces: κν σλ = (1 3ψ2 )Ψ 2 + R/12 2(1 ψ 2 ) as in case with 3 constant evs. The 10th & 17th NP eqs. produce the same result 21/28

90 One Constant Weyl Eigenvalue Subtraction of ψ (2nd Ricci identity) from the 8th, after substituting for ρ, τ, µ & π produces: κν σλ = (1 3ψ2 )Ψ 2 + R/12 2(1 ψ 2 ) as in case with 3 constant evs. The 10th & 17th NP eqs. produce the same result This may be written in the form: κν σλ = 3 Ψ 2(3 Ψ 0 Ψ 4 /Ψ 2 2 ) + Λ 2(9 Ψ 0 Ψ 4 /Ψ 2 2 ) 21/28

91 One Constant Weyl Eigenvalue Subtraction of ψ (2nd Ricci identity) from the 8th, after substituting for ρ, τ, µ & π produces: κν σλ = (1 3ψ2 )Ψ 2 + R/12 2(1 ψ 2 ) as in case with 3 constant evs. The 10th & 17th NP eqs. produce the same result This may be written in the form: κν σλ = 3 Ψ 2(3 Ψ 0 Ψ 4 /Ψ 2 2 ) + Λ 2(9 Ψ 0 Ψ 4 /Ψ 2 2 ) which is invariant under boosts and rotations of the canonical null tetrad: l a = Al a, ñ a = A 1 n a, m a = e iθ m a. 21/28

92 One Constant Weyl Eigenvalue Pairs of simultaneous equations for Dλ & δκ, Dν & κ, δλ & δσ and δν & σ may be obtained as before by eliminating ρ, τ, µ, π from the Ricci identities and using the Bianchi identities to remove derivatives of ψ. These can be solved for the 8 single derivatives above. 22/28

93 One Constant Weyl Eigenvalue Pairs of simultaneous equations for Dλ & δκ, Dν & κ, δλ & δσ and δν & σ may be obtained as before by eliminating ρ, τ, µ, π from the Ricci identities and using the Bianchi identities to remove derivatives of ψ. These can be solved for the 8 single derivatives above. In addition there are 8 independent Ricci identities involving two coupled spin coeff. derivatives. 22/28

94 One Constant Weyl Eigenvalue Pairs of simultaneous equations for Dλ & δκ, Dν & κ, δλ & δσ and δν & σ may be obtained as before by eliminating ρ, τ, µ, π from the Ricci identities and using the Bianchi identities to remove derivatives of ψ. These can be solved for the 8 single derivatives above. In addition there are 8 independent Ricci identities involving two coupled spin coeff. derivatives. The 6 commutators applied to ψ produce only 4 independent relations. 22/28

95 One Constant Weyl Eigenvalue Pairs of simultaneous equations for Dλ & δκ, Dν & κ, δλ & δσ and δν & σ may be obtained as before by eliminating ρ, τ, µ, π from the Ricci identities and using the Bianchi identities to remove derivatives of ψ. These can be solved for the 8 single derivatives above. In addition there are 8 independent Ricci identities involving two coupled spin coeff. derivatives. The 6 commutators applied to ψ produce only 4 independent relations. Four more are obtained from the commutators (δd Dδ)λ, (δd Dδ)ν, ( δ δ)σ and ( δ δ)κ. 22/28

96 One Constant Weyl Eigenvalue Pairs of simultaneous equations for Dλ & δκ, Dν & κ, δλ & δσ and δν & σ may be obtained as before by eliminating ρ, τ, µ, π from the Ricci identities and using the Bianchi identities to remove derivatives of ψ. These can be solved for the 8 single derivatives above. In addition there are 8 independent Ricci identities involving two coupled spin coeff. derivatives. The 6 commutators applied to ψ produce only 4 independent relations. Four more are obtained from the commutators (δd Dδ)λ, (δd Dδ)ν, ( δ δ)σ and ( δ δ)κ. Finally 4 more equations are obtained by applying the differential operators D,, δ and δ to the cross-ratio expression. 28 equations in all for 32 derivs. 22/28

97 Constant Ratios of the Weyl Eigenvalues Consider Petrov type I spacetimes with the Weyl eigenvalues constant multiples of a single scalar field. 23/28

98 Constant Ratios of the Weyl Eigenvalues Consider Petrov type I spacetimes with the Weyl eigenvalues constant multiples of a single scalar field. This conditiion is equivalent to ψ = Ψ 4 /(3Ψ 2 ) being constant. As usual Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 23/28

99 Constant Ratios of the Weyl Eigenvalues Consider Petrov type I spacetimes with the Weyl eigenvalues constant multiples of a single scalar field. This conditiion is equivalent to ψ = Ψ 4 /(3Ψ 2 ) being constant. As usual Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 Four Bianchi identities simplify to DΨ 2 = 3(ρ ψλ)ψ 2 δψ 2 = 3(τ ψν)ψ 2 δψ 2 = 3(ψκ π)ψ 2 Ψ 2 = 3(ψσ µ)ψ 2 23/28

100 Constant Ratios of the Weyl Eigenvalues Consider Petrov type I spacetimes with the Weyl eigenvalues constant multiples of a single scalar field. This conditiion is equivalent to ψ = Ψ 4 /(3Ψ 2 ) being constant. As usual Ψ 1 = Ψ 3 = 0 and Ψ 0 = Ψ 4 Four Bianchi identities simplify to DΨ 2 = 3(ρ ψλ)ψ 2 δψ 2 = 3(τ ψν)ψ 2 δψ 2 = 3(ψκ π)ψ 2 Ψ 2 = 3(ψσ µ)ψ 2 The other 4 Bianchi identities lead to alternative equations for these 4 derivatives. Eliminating the derivatives of Ψ 2 : α = γ = 3ψ2 1 4ψ κ π 2 β = 3ψ2 1 4ψ ν τ 2 3ψ2 1 4ψ σ µ 2 ɛ = 3ψ2 1 4ψ λ ρ 2 23/28

101 Constant Ratios of the Weyl Eigenvalues The combination (q) (h) + 2 (l) 2 (f ) of the Ricci identities results in an expression for ( λ δν + δκ Dσ)(3ψ ψ 1)/(2 ψ) 24/28

102 Constant Ratios of the Weyl Eigenvalues The combination (q) (h) + 2 (l) 2 (f ) of the Ricci identities results in an expression for ( λ δν + δκ Dσ)(3ψ ψ 1)/(2 ψ) This combination of derivatives also appears in Ricci ids. (j) & (b) and so may be eliminated to produce: Ψ 2 = (κν σλ) 9ψ2 1 9ψ 2 24/28

103 Constant Ratios of the Weyl Eigenvalues The combination (q) (h) + 2 (l) 2 (f ) of the Ricci identities results in an expression for ( λ δν + δκ Dσ)(3ψ ψ 1)/(2 ψ) This combination of derivatives also appears in Ricci ids. (j) & (b) and so may be eliminated to produce: Ψ 2 = (κν σλ) 9ψ2 1 9ψ 2 Rearranging and eliminating ψ: κν σλ = Ψ 2 1 Ψ 2 2 /(Ψ 0Ψ 4 ). This equation is invariant under boosts and rotations of the canonical null tetrad. 24/28

104 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. 25/28

105 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. The Ricci identities provided 17 independent eqs. 25/28

106 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. The Ricci identities provided 17 independent eqs. The derivatives of the algebraic equation for Ψ 2 (after eliminating derivatives of Ψ 2 via the Bianchi ids.) provide 4 more. 25/28

107 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. The Ricci identities provided 17 independent eqs. The derivatives of the algebraic equation for Ψ 2 (after eliminating derivatives of Ψ 2 via the Bianchi ids.) provide 4 more. The 6 commutators applied to Ψ 2 provide 3 more eqs. 24 independent equations in all 25/28

108 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. The Ricci identities provided 17 independent eqs. The derivatives of the algebraic equation for Ψ 2 (after eliminating derivatives of Ψ 2 via the Bianchi ids.) provide 4 more. The 6 commutators applied to Ψ 2 provide 3 more eqs. 24 independent equations in all The difference of ( δδ δ δ)ψ 2 and ( D D )Ψ 2 is dependent on the Ricci ids. (δd Dδ)Ψ 2 and ( δ δ)ψ 2 are dependent on the Ricci ids. 25/28

109 Constant Ratios of the Weyl Eigenvalues There are 32 unknown derivatives of the spin coefficients κ, σ, ρ, τ, ν, λ, µ & π. The Ricci identities provided 17 independent eqs. The derivatives of the algebraic equation for Ψ 2 (after eliminating derivatives of Ψ 2 via the Bianchi ids.) provide 4 more. The 6 commutators applied to Ψ 2 provide 3 more eqs. 24 independent equations in all The difference of ( δδ δ δ)ψ 2 and ( D D )Ψ 2 is dependent on the Ricci ids. (δd Dδ)Ψ 2 and ( δ δ)ψ 2 are dependent on the Ricci ids. Both generalised systems are much richer than the case where all 3 Weyl eigenvalues are constant. They are currently being investigated. 25/28

110 Summary I All vacuum spacetimes (with Λ) which have non-zero constant Weyl scalars have been obtained in closed form; they are either homogeneous or the Petrov type II Kundt spacetimes studied by Lewandowski. 26/28

111 Summary I All vacuum spacetimes (with Λ) which have non-zero constant Weyl scalars have been obtained in closed form; they are either homogeneous or the Petrov type II Kundt spacetimes studied by Lewandowski. Coley et al. (2006) proved an analogous result: any spacetime for which all invariants constructed from the curvature tensor and its derivatives of all orders are constant is either homogeneous or Type II. 26/28

112 Summary I All vacuum spacetimes (with Λ) which have non-zero constant Weyl scalars have been obtained in closed form; they are either homogeneous or the Petrov type II Kundt spacetimes studied by Lewandowski. Coley et al. (2006) proved an analogous result: any spacetime for which all invariants constructed from the curvature tensor and its derivatives of all orders are constant is either homogeneous or Type II. In this work we assume the vacuum condition (with Λ) and the constancy of the algebraic invariants of the curvature tensor. 26/28

113 Summary I All vacuum spacetimes (with Λ) which have non-zero constant Weyl scalars have been obtained in closed form; they are either homogeneous or the Petrov type II Kundt spacetimes studied by Lewandowski. Coley et al. (2006) proved an analogous result: any spacetime for which all invariants constructed from the curvature tensor and its derivatives of all orders are constant is either homogeneous or Type II. In this work we assume the vacuum condition (with Λ) and the constancy of the algebraic invariants of the curvature tensor. MacCallum & Siklos (1992) showed there were no homogeneous proper Einstein spaces of Petrov types I or II. 26/28

114 Summary II Here for type I, homogeneity can be replaced by the weaker assumption of constant Weyl eigenvalues. Not true, however, for type II as there are many Lewandowski metrics. 27/28

115 Summary II Here for type I, homogeneity can be replaced by the weaker assumption of constant Weyl eigenvalues. Not true, however, for type II as there are many Lewandowski metrics. The only Petrov type I solution has Λ = 0 whereas the algebraically special solutions have Ψ 2 = Λ/ /28

116 Summary II Here for type I, homogeneity can be replaced by the weaker assumption of constant Weyl eigenvalues. Not true, however, for type II as there are many Lewandowski metrics. The only Petrov type I solution has Λ = 0 whereas the algebraically special solutions have Ψ 2 = Λ/3 0. Although the initial assumptions are rather weak, rather disappointingly all solutions have previously appeared in the literature. 27/28

117 Summary II Here for type I, homogeneity can be replaced by the weaker assumption of constant Weyl eigenvalues. Not true, however, for type II as there are many Lewandowski metrics. The only Petrov type I solution has Λ = 0 whereas the algebraically special solutions have Ψ 2 = Λ/3 0. Although the initial assumptions are rather weak, rather disappointingly all solutions have previously appeared in the literature. However, we have new characterisations of these solutions as the only Petrov type I, II & D solutions with constant Weyl eigenvalues whereas the original characterisations involve homogeneity or reduced holonomy. 27/28

118 Summary III The assumption for Petrov type I of three constant Weyl eigenvalues is relaxed in two distinct ways. 28/28

119 Summary III The assumption for Petrov type I of three constant Weyl eigenvalues is relaxed in two distinct ways. Firstly the case where only one eigenvalue is constant is considered. 28/28

120 Summary III The assumption for Petrov type I of three constant Weyl eigenvalues is relaxed in two distinct ways. Firstly the case where only one eigenvalue is constant is considered. Secondly the case where the eigenvalues are constant multiples of a single (in general complex) scalar field is considered. 28/28

121 Summary III The assumption for Petrov type I of three constant Weyl eigenvalues is relaxed in two distinct ways. Firstly the case where only one eigenvalue is constant is considered. Secondly the case where the eigenvalues are constant multiples of a single (in general complex) scalar field is considered. In both cases simple algebraic relations are found between the cross-ratio κν σλ of a canonical null tetrad and a quantity constructed from the curvature. 28/28

122 Summary III The assumption for Petrov type I of three constant Weyl eigenvalues is relaxed in two distinct ways. Firstly the case where only one eigenvalue is constant is considered. Secondly the case where the eigenvalues are constant multiples of a single (in general complex) scalar field is considered. In both cases simple algebraic relations are found between the cross-ratio κν σλ of a canonical null tetrad and a quantity constructed from the curvature. In both cases these relations are invariant under boosts and rotations of the canonical null tetrad leaving the directions of the real null vectors l a & n a invariant. Suggests the use of GHP to simplify the analysis? 28/28