A group-theoretic characterisation of Taub-Nut spacetime
aa r X i v : . [ m a t h - ph ] F e b A group-theoretic characterisation ofTaub-Nut spacetime
Schiden Yohannes and Domenico GiuliniInstitute for Theoretical PhysicsLeibniz University of HannoverAppelstrasse 2, 30167 HannoverGermany18th February 2021
We prove that any G = SU (2) × U (1) symmetric spacetime that is Ricciflat (i.e. solves the matter-free Λ = 0 Einstein equations) with non-null G -orbits is locally isometric to some maximally extended generalised Taub-NUTspacetime. This paper deals with Taub-NUT spacetime [19, 15] that is an exact solution to thematter-free (i.e. vacuum) equations of General Relativity without cosmological constant.This particular solution is known for its many surprising properties [12, 11, 13]. Also, itcan be generalised in various ways and embedded into the wider Pleba´nski–Demia´nskiclass of solutions; see, e.g., [3, chapter 12,16]. Here we are not interested in these gener-alisations and restrict attention to strict Taub-NUT only.We prove a result that gives rise to a new group-theoretic characterisation of Taub-NUT spacetime, or rather some obvious topological generalisation of it. The maintheorem will be presented and proved in Section 3. It states that any SU (2) × U (1)symmetric vacuum solution to Einstein’s equations with non-null group orbits is locallyisometric to some maximally extended generalised Taub-NUT geometry, where the “gen-eralisation” here consists in replacing the 3-sphere in the global R × S topology withthat of a lens space L ( n, SO (3) with spacelike S or R P orbits. A modernproof can be found in [18, Chapter 4.10.1-2].We call our result an incomplete analogue to the Jebsen-Birkhoff theorem because inthe generalised Taub-NUT case we have several inequivalent (i.e. non globally isometric)maximal extensions, the precise classification of which we currently investigate.There is another, different notion of generalised Taub-NUT introduced in [14], whichrelaxes the global isometry to be merely U (1) (i.e. dropping the SU (2) factor altogether)and requires the spacetime to contain a compact Cauchy horizon diffeomorphic to S towhich the U (1) action restricts to a free action with lightlike orbits. Hence the S Cauchyhorizon is the total space of a U (1) principal bundle with base S (Hopf bundle) andlightlike fibres. The set of such “generalised Taub-NUT” spacetimes forms an infinite-dimensional proper submanifold within the set of all U (1)-symmetric vacuum spacetimesof “roughly half the dimensions” [14, p. 108]. In this case there are uncountably manyinequivalent maximal extensions.In order to make our paper self contained, we will review the essential geometryand topology of Taub-NUT spacetime in Section 2, also providing a characterisation ofthe NUT-charge as “dual” to mass. Our main theorem concerning the group-theoreticcharacterisation will be stated and proved in Section 3. We end with a brief outlook inSection 4. In the following we will present some of the features of the Taub-NUT space-time withrespect to the interpretation given by Misner [12]. In this interpretation the topology ofspace-time is R × S and, using Euler coordinates, the metric is given by g = − l f ( r )( dψ + cos θdϕ ) + 1 f ( r ) dr + ( r + l )( dθ + sin θdϕ ) , (2.1)with f ( r ) = r − mr − l r + l . (2.2)The constant m ∈ R is interpreted as the mass and the constant l ∈ R \ { } is referredto as the NUT parameter. The four-dimensional isometry group of this space-time is SU (2) L × U (1) R induced by the left-invariant vector field ξ = ∂ ψ (generating right-translations) and the right-invariant vector fields ξ = − sin ϕ∂ θ − cot θ cos ϕ∂ ϕ + csc θ cos ϕ∂ ψ (2.3a) ξ = cos ϕ∂ θ − cot θ sin ϕ∂ ϕ + csc θ sin ϕ∂ ψ (2.3b) ξ = ∂ ϕ . (2.3c)2generating left-translations) on the 3-sphere, which we identify with the group manifoldof SU (2). Note that the subscripts L and R on SU (2) and U (1), respectively, are meantto indicate that these groups act via left- and right-multiplication on SU (2). The vectorfields ξ , ξ , ξ , ξ satisfy the commutation relations:[ ξ i , ξ j ] = − ε kij ξ k (2.4a)[ ξ , ξ i ] = 0 i, j, k = 1 , , . (2.4b)In terms of the left-invariant one-forms σ x = sin ψdθ − sin θ cos ψdϕ (2.5a) σ y = cos ψdθ + sin θ sin ψdϕ (2.5b) σ z = dψ + cos θdϕ (2.5c)the metric can be written as g = − l f ( r ) σ z + 1 f ( r ) dr + ( r + l )( σ x + σ y ) . (2.6)The orbit generated by ξ , ξ , ξ is three-dimensional, namely S , with the orbits gener-ated by ξ being subsets of it.In the given coordinates the analytical expressions become singular at r ± = m ±√ m + l , whereas all components of the Riemann tensor in an orthonormal tetrad,and hence in particular the Kretschmann scalar, are regular. This indicates that thesesingularities are, in fact, coordinate artefacts. They correspond to the Killing horizonsof the Killing vector field ∂ ψ . A possible coordinate transformation removing thesesingularities is given by ψ ′ = ψ + Z lf ( r ) dr (2.7)such that the metric in these new coordinates is given by g = − l f ( r )( dψ ′ + cos θdϕ ) + 2(2 l )( dψ ′ + cos θdϕ ) dr + ( r + l )( dθ + sin θdϕ ) . (2.8)Another coordinate transformation would be ψ ′′ = ψ − Z lf ( r ) dr, (2.9)giving g = − l f ( r )( dψ ′′ + cos θdϕ ) − l )( dψ ′′ + cos θdϕ ) dr + ( r + l )( dθ + sin θdϕ ) . (2.10)3ritten in terms of the left-invariant one-forms of S , it is immediate that the metricsare regular on the whole manifold R × S . Furthermore, it can be shown that bothspace-times are maximal [19]. In these coordinates both the stationary regions r < r − and r > r + , the so-called NUT-regions, and the region r − < r < r + , called the Taub-region, are included. In particular, the hypersurfaces of r = const. are 3-spheres beingspacelike in the Taub-region, timelike in the NUT-regions, and lightlike at r = r ± .Furthermore, with respect to the U (1) right multiplication the space-time can beconsidered to be a principal fibre bundle analogous to the Hopf bundle. Since the r = const. hypersurfaces are 3-spheres, there exist no equal-time hypersurfaces intersectingthese 3-spheres in two-spheres along which we could evaluate the Komar integral formass in the usual form . However, we can use the structure of S as U (1) principle fibrebundle over the base S , which has a natural connection given by the distribution oforthogonal complements to the fibre in each tangent space where the generating vectorfield of U (1) is non-null. It is then possible to uniquely identify horizontal and right-invariant k -forms with k -forms on the base S . Thus, considering the NUT regions,admitting the timelike Killing vector field ∂ ψ (generating the right- U (1) translation) theKomar mass of the space-time can be calculated. We will be using the orthonormaltetrad ϑ = 2 lf / ( r )( dψ + cos θdϕ ) (2.11a) ϑ = f − / ( r ) dr (2.11b) ϑ = ( r + l ) / dθ (2.11c) ϑ = ( r + l ) / sin θdϕ. (2.11d)For lim r →∞ f ( r ) = 1, we will calculate the Komar mass with respect to the Killing vectorfield k := − l ∂ ψ which is normalised at infinity r → ∞ . The metric-dual one-form ofthe timelike Killing vector field is then given by k ♭ = 2 lf ( r )( dψ + cos θdϕ ) (2.12)and hence dk ♭ = 2 lf ′ ( r ) dr ∧ ( dψ + cos θdϕ ) − lf ( r ) sin θdθ ∧ dϕ (2.13a)= − f ′ ( r ) ϑ ∧ ϑ − l f ( r ) r + l ϑ ∧ ϑ , (2.13b)where we used the standard notation that denotes the one-form image of the vector k under the metric isomorphism by k ♭ =: g ( k, · ). Now, picking the orientation defined by ω = ϑ ∧ ϑ ∧ ϑ ∧ ϑ , we get ∗ dk ♭ = f ′ ( r ) ϑ ∧ ϑ − l f ( r ) r + l ϑ ∧ ϑ (2.14a)= f ′ ( r )( r + l ) sin θdθ ∧ dϕ + 4 l f ( r ) r + l dr ∧ ( dψ + cos θdϕ ) . (2.14b)4hen with respect to an arbitrary hypersurface r = const. , the two-form is given by ∗ dk ♭ = − f ′ ( r )( r + l ) dσ z . (2.15)Since i σ z is a connection one-form for the principal fibre bundle, ∗ dk ♭ can be consideredas a multiple of the curvature form and hence is horizontal and right-invariant, as wellas closed. Thus, we can identify it with a closed two-form on the base space S , suchthat using the formula for the Komar mass, we have − π Z S ∞ ∗ dk ♭ = lim r →∞ − π f ′ ( r )( r + l ) Z S sin θdθ ∧ dϕ = − m. (2.16)Therefore m can be interpreted as the Komar mass of the space-time. Moreover, con-sidering dk ♭ instead of ∗ dk ♭ , the same line of argument can be applied to give − π Z S ∞ dk ♭ = lim r →∞ π lf ( r ) Z S sin θdθ ∧ dϕ = l. (2.17)So the constants m and l are related by Hodge duality.Duality also arises in the description of the dual-Bondi-mass of space-times whichare asymptotically empty and flat at null infinity and with vanishing Bondi news. Itcan be shown that in this case null infinity, for space-times having a non-vanishingdual-Bondi-mass, is topologically a Lens space L ( n,
1) and a principal fibre bundle( L ( n, , π, S ; S ), with the dual-Bondi-mass being proportional to the number of twists, n , in the bundle. Conversely, if null infinity is a non-trivial S principal fibre bundleover S , the news tensor field vanishes and there exists an infinitesimal translation suchthat the dual-Bondi-mass with respect to it is non-zero. In particular, the Taub-NUTspace-time can be shown to be asymptotically empty and flat at null infinity, with nullinfinity being a 3-sphere. The dual-Bondi-mass with respect to the infinitesimal transla-tion induced by the Killing vector field − l ∂ ψ can be computed to be the NUT parameter l [16]. In this section we intend to give a unique characterisation of the Taub-NUT space-timein terms of the isometry group and its orbits. In particular, the Taub-NUT space-timecan be seen to be the universal cover of a family of space-times admitting SU (2) × U (1)as an isometry group such that the group orbits of SU (2) × U (1) and SU (2) are three-dimensional and non-null.Since the metric of the Taub-NUT space-time induces a SU (2) L × U (1) R invariantmetric on the hypersurfaces r = const. , being diffeomorphic to SU (2), we will begin bystudying special metrics on SU (2). For Lorentz metrics on SU (2) we have5 emma 3.1. Let G = SU (2) and g a Lorentz metric on G such that it is SU (2) left-invariant and U (1) right-invariant, whereby U (1) R is considered as a subgroup of the SU (2) R . Then the orbits of the U (1) right-multiplication are timelike curves.Proof. The left-action SU (2) L × U (1) R ⊂ S ( U ) L × SU (2) R is simply obtained by re-stricting the standard left-action of S ( U ) L × SU (2) R on SU (2):( SU (2) L × U (1) R ) × SU (2) → SU (2) (3.1a)(( h, h ′ ) , g ) hgh ′− . (3.1b)Now let e ∈ G be the identity, then we have for h ∈ SU (2) L and h ′ ∈ U (1) R (( h, h ′ ) , e ) heh ′− = hh ′− . (3.2)Hence the isotropy group at e is the diagonal U (1) subgroup in SU (2) L × U (1) R , denotedby C h . Then ( dC h ) e : T e G → T e G (3.3)induces the adjoint representation Ad : U (1) → GL ( T e G ) (3.4a) Ad ( h ) = ( dC h ) e . (3.4b)Since the tangent space is three-dimensional, the action induced by this U (1) on thetangent space is given by a U (1) subgroup of the three-dimensional Lorentz group. Fur-thermore, because the U (1) subgroups of the three-dimensional Lorentz group consist ofrotations acting by orthogonal transformations in a spacelike plane, such that the corres-ponding orthogonal timelike direction is invariant, the three-dimensional tangent space The reader should be aware of the conceptual difference between left/right-multiplication on groupsand left/right-action of groups on sets: A left-action of a group G on a set S is simply a homomorphismΦ : G → Bij( S ), from the group G into the group of bijections of S , with group multiplication ofthe latter just being composition of maps. This means that Φ : g Φ g is such that Φ g ◦ Φ h = Φ gh .In contrast, a right-action is an anti-homomorphism ˜Φ : G → Bij( S ) which satisfies ˜Φ g ◦ ˜Φ h = ˜Φ hg .A left-action can be turned into a right-action (and vice versa) if we compose it with the groupinversion I : G → G , g I ( g ) := g − which is an anti-homomorphims, i.e. I ( gh ) = I ( h ) I ( g ). Then˜Φ := Φ ◦ I is a right-action if Φ is a left-action. Now, If S = G , there are two natural actions of G on itself, called L and R and given by left- and right-multiplication respectively: L g ( p ) := gp and R g ( p ) := pg . Associativity of group multiplication implies that these two actions commute (asmaps): L g ◦ R h = R h ◦ L g for all g, h ∈ G . Written in this way L is a left- and R is a right-actionwhich as such do not combine to any action, left or right. However, the right-multiplication can beturned into a left-action by composing R with I . In this way we get two different and commutingleft-actions of G on itself, one by left-multiplication with g ∈ G and one by right-multiplicationwith g − . Together they define a left-action of G × G on G , given by Φ ( g,h ) := L g ◦ R h − , that isΦ ( g,h ) ( p ) = gph − . In order to distinguish the group G that acts by left-multiplication from the onethat acts by right-multiplication (with the inverse) we distinguish them notationally and call them G L and G R , respectively. Restricting this to the diagonal subgroup G ∆ := { ( g, g ) : G ∈ G } ⊂ G L × G R gives the left-action of G on itself that is usually referred to as “conjugation”. v in the one-dimensionaltimelike subspace, we define the left-invariant vector field X ∈ Lie ( G ) by X ( g ) = dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( g exp ( tv )) . (3.5)The right action on this left-invariant vector field is determined by the adjoint of v ∈ T e G with respect to h − ∈ U (1),( dR h ) g ( X g ) = dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( g exp ( tv ) h ) (3.6a)= dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( gh C h − ( exp ( tv ))) (3.6b)= dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( gh exp ( tAd ( h − )( v ))) . (3.6c)Since the timelike direction is invariant with respect to the adjoint representation, weobtain ( dR h ) g ( X g ) = dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( gh exp ( tAd ( h − )( v ))) (3.7a)= dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( gh exp ( tv )) (3.7b)= X ( gh ) . (3.7c)Therefore the left-invariant vector field is also U (1) R –invariant. Furthermore, becauseit is a left-invariant vector field, it generates a U (1) R -action, considered as U (1) ′ R ⊂ SU (2) R . By being also invariant under U (1) R ⊂ SU (2) R , the two U (1) right actionshave to commute, hence U (1) R = U (1) ′ R . Therefore, the orbits of the U (1) right actioncoincide with the orbits of X and thus are timelike.Next we will prove that a SU (2) L × U (1) R –invariant metric on SU (2) can be put intoa canonical form: Lemma 3.2.
Let G = SU (2) and g a non-degenerate, symmetric bilinear form on G which is SU (2) L × U (1) R –invariant. Then g can be written as g = Aσ z + B ( σ x + σ y ) , (3.8) where σ x , σ y , σ z are left–invariant one-forms on G .Proof. Let Z be a fundamental vector field associated to the U (1) R -action and anyelement ix ∈ Lie ( U (1)) = i R , Z ( g ) = dd t (cid:12)(cid:12)(cid:12)(cid:12) t =0 ( g exp ( tix )) , g ∈ G. (3.9)7hen the vector field Z is left-invariant. We will complete it to a basis for Lie ( G ) bychoosing two linearly independent left-invariant vector fields in the orthogonal com-plement of Z , so X, Y ∈ Lie ( G ) such that X, Y ⊥ Z . Then, denoting the basis as e = Z, e = X, e = Y and their dual one forms by ω , ω , ω , g can be written as g = λ ( ω ) + µ ( ω ) + ν ( ω ) + κω ω . (3.10)Now since g is U (1) R –invariant, we have L e g = 0 . (3.11)If the structure constants are given by[ e i , e j ] = c kij e k (3.12)we have for their dual one-forms dω k = − X i 1) [5].As already proven a SU (2) L × U (1) R –invariant metric on SU (2) ∼ = S can be put intoa canonical form. Now we want to study the case for L ( n, S ⊂ C theleft action of SU (2) on S is the natural action of SU (2) on C and the Γ = Z n -actionon S for L ( n, 1) is given by( z , z ) ( e πi/n z , e πi/n z ) , ( z , z ) ∈ S . (3.33)Now we can define a left action of SU (2) on L ( n, 1) by SU (2) × L ( n, → L ( n, 1) (3.34a)( A, π ( p )) π ( Ap ) , (3.34b)where π : S → L ( n, 1) is the projection map, so the covering map. This induces awell-defined SU (2) left action on L ( n, U (1)right action.Moreover, given a SU (2) L × U (1) R –invariant metric it is also invariant with respectto the Z n -action (3.33) and hence the following construction is well-defined:Let g be a Z n –invariant metric on S . We define a metric on L ( n, 1) pointwise by g ′ q : T q L ( n, × T q L ( n, → R (3.35a) g ′ q ( X ′ , Y ′ ) := g p ( X, Y ) , X ′ , Y ′ ∈ T q L ( n, , X, Y ∈ T p S (3.35b)where π ( p ) = q and dπ p ( X ) = X ′ , dπ p ( Y ) = Y ′ . The Z n -invariance of g implies theindependence of the choice of a representative p and since the covering map is a localdiffeomorphism its differential at any point is a linear isomorphism. Therefore thisdefinition makes sense. The metric g ′ is in fact smooth, since given any smooth localsection s : U ⊂ L ( n, → S and smooth local vector fields X ′ , Y ′ on U we have on Ug ′ ( X ′ , Y ′ ) = g ( ds ( X ′ ) , ds ( Y ′ )) . (3.36)This linear map is a bijection, because conversely given any metric g ′ on L ( n, 1) wecan define a metric g = π ∗ g ′ on S , being just the preimage of the preceding construc-tion. Thus, we see that the Z n –invariant metrics on S are in bijection to metrics on11 ( n, SU (2) L × U (1) R –invariant metric on S is also invariant with respect tothe Z n -action (3.33), by definition of the induced action and the constructed bijectionabove, it is immediate that the SU (2) L × U (1) R –invariant metrics on S are in one-to-one correspondence to the metrics on L ( n, 1) which are invariant under the induced SU (2) L × U (1) R -action.Using Euler coordinates and Lemma 3.2, a Lorentz or Riemannian metric on L ( n, SU (2) L × U (1) R can always be written as g = εA ( dψ + cos θdϕ ) + B ( dθ + sin θdϕ ) , ε = ± θ and ϕ ranging from 0 to π and 0 to 2 π respectively and ψ being 4 π/n -periodic.In particular, based on the preceding results, we see that the metric (2.1) is invariantwith respect to the Z n -action (3.33) and hence there exists a well-defined metric on anyLens space L ( n, R × L ( n, , g ) with g = − l f ( r )( dψ + cos θdϕ ) + 1 f ( r ) dr + ( r + l )( dθ + sin θdϕ ) , (3.38)which will be called the generalized Taub-NUT space-time . The space-time ( R × L ( n, , g ′ ),with g ′ = − l f ( r )( dψ ′ + cos θdϕ ) + 2(2 l )( dψ ′ + cos θdϕ ) dr + ( r + l )( dθ + sin θdϕ )(3.39)will be called a maximal extension of the generalized Taub-NUT space-time .Now we can prove the following statement: Theorem 3.3. Every ( C -) solution to the vacuum Einstein field equations admitting SU (2) × U (1) as an isometry group, such that SU (2) × U (1) and SU (2) both havethree-dimensional non-null orbits in an open subset U , is locally isometric to a maximalextension of the generalized Taub-NUT space-time.Proof. Let p ∈ M be an arbitrary point of the space-time ( M, g ) and O ( p ) be the three-dimensional orbit of p with respect to the action of SU (2). We define the orthogonalcomplement of the tangent space of the orbit to be: N p := T p O ( p ) ⊥ . Then the induceddistributions N := ∪ p ∈ M N p , O := ∪ p ∈ M T p O ( p ) (3.40)are both integrable, for N is a one-dimensional distribution, which is always integrable,and O is by construction integrable, with its integral manifolds being the orbits. So wehave a involutive three-dimensional distribution, spanned by Killing vector fields and theinvolutive one-dimensional normal bundle with N ∩ O = 0, since the orbits are non-null.Thus, it is possible to introduce local coordinates { x µ } = { r, x , x , x } such that g = g rr dr + g ab ( x µ ) dx a dx b (3.41)12here r = const. are the integral manifolds of O , the orbits, which are homogeneousspaces. Since the three-dimensional orbits of a SU (2)-action admitting SU (2) × U (1) asisometry group have to be topologically the Lens spaces L ( n, 1) and the SU (2) L × U (1) R –invariant Lorentz or Riemannian metrics on L ( n, 1) can always be put into a canonicalform, the metric can be written as g = − εA ( r ) dr + εB ( r )( dψ + cos θdϕ ) + R ( r )( dθ + sin θdϕ ) , (3.42)where the case ε = 1 represents spacelike orbits and ε = − ε = 1, so spacelike orbits. The orthonormal tetrad wewill be using is ϑ = A ( r ) dr (3.43a) ϑ = B ( r )( dψ + cos θdϕ ) (3.43b) ϑ = R ( r ) dθ (3.43c) ϑ = R ( r ) sin θdϕ. (3.43d)In the following the argument of the functions A, B, R will be omitted and a primeindicates the derivative with respect to r . Then exterior differentiation and expressingthe results in terms of the tetrad leads to dϑ = 0 (3.44a) dϑ = B ′ AB ϑ ∧ ϑ − BR ϑ ∧ ϑ (3.44b) dϑ = R ′ AR ϑ ∧ ϑ (3.44c) dϑ = R ′ AR ϑ ∧ ϑ + cot θR ϑ ∧ ϑ . (3.44d)Since the tetrad is orthonormal we have ω µν + ω νµ = 0. Now using the first structureequation with an ansatz for every connection one-form of the form a µ ϑ µ , the unique13olution is given by ω = ω = B ′ AB ϑ (3.45a) ω = ω = R ′ AR ϑ (3.45b) ω = ω = R ′ AR ϑ (3.45c) ω = − ω = − B R ϑ (3.45d) ω = − ω = B R ϑ (3.45e) ω = − ω = B R ϑ − cot θR ϑ . (3.45f)Then using the second structure equations, the curvature 2-form can be calculated:Ω = dω + ω ∧ ω + ω ∧ ω = (cid:18) B ′′ AB − B ′ A ′ A B − B ′ AB (cid:19) dr ∧ ϑ + B ′ AB dϑ + BR ′ AR ϑ ∧ ϑ − BR ′ AR ϑ ∧ ϑ = (cid:18) B ′′ A B − B ′ A ′ A B − B ′ A B (cid:19) ϑ ∧ ϑ + B ′ A B ϑ ∧ ϑ − B ′ AR ϑ ∧ ϑ + BR ′ AR ϑ ∧ ϑ = (cid:18) B ′′ A B − B ′ A ′ A B (cid:19) ϑ ∧ ϑ + (cid:18) BR ′ AR − B ′ AR (cid:19) ϑ ∧ ϑ (3.46a)Ω = dω + ω ∧ ω + ω ∧ ω = (cid:18) R ′′ A R − R ′ A ′ A R − R ′ A R (cid:19) ϑ ∧ ϑ + R ′ A R ϑ ∧ ϑ − B ′ AR ϑ ∧ ϑ − BR ′ AR ϑ ∧ ϑ = (cid:18) R ′′ A R − R ′ A ′ A R (cid:19) ϑ ∧ ϑ − (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ (3.46b)Ω = dω + ω ∧ ω + ω ∧ ω = (cid:18) R ′′ A R − R ′ A ′ A R − R ′ A R (cid:19) ϑ ∧ ϑ + R ′ A R ϑ ∧ ϑ + R ′ cot θAR ϑ ∧ ϑ + B ′ AR ϑ ∧ ϑ + BR ′ AR ϑ ∧ ϑ − R ′ cot θAR ϑ ∧ ϑ = (cid:18) R ′′ A R − R ′ A ′ A R (cid:19) ϑ ∧ ϑ + 12 (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ (3.46c)14 = dω + ω ∧ ω + ω ∧ ω = (cid:18) − B ′ AR + BR ′ AR (cid:19) ϑ ∧ ϑ − BR ′ AR ϑ ∧ ϑ − B cot θ R ϑ ∧ ϑ + B ′ R ′ A BR ϑ ∧ ϑ − B R ϑ ∧ ϑ + B cot θ R ϑ ∧ ϑ = 12 (cid:18) BR ′ AR − B ′ AR (cid:19) ϑ ∧ ϑ + (cid:18) B ′ R ′ A BR + B R (cid:19) ϑ ∧ ϑ (3.47a)Ω = dω + ω ∧ ω + ω ∧ ω = (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ + BR ′ AR ϑ ∧ ϑ + B ′ R ′ A BR ϑ ∧ ϑ − B R ϑ ∧ ϑ = 12 (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ + (cid:18) B ′ R ′ A BR + B R (cid:19) ϑ ∧ ϑ (3.47b)Ω = dω + ω ∧ ω + ω ∧ ω = (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ + B ′ AR ϑ ∧ ϑ − B R ϑ ∧ ϑ + csc θR ϑ ∧ ϑ + R ′ cot θAR ϑ ∧ ϑ − R ′ cot θAR ϑ ∧ ϑ − cot θR ϑ ∧ ϑ + R ′ A R ϑ ∧ ϑ + B R ϑ ∧ ϑ = (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ + (cid:18) R + R ′ A R − B R (cid:19) ϑ ∧ ϑ . (3.47c)Using Ω µν = R µναβ θ α ∧ θ β , the non-vanishing components of the Riemann tensor, upto symmetry, are R = B ′′ A B − B ′ A ′ A B (3.48a) R = 2 R = − R = BR ′ AR − B ′ AR (3.48b) R = R = R ′′ A R − R ′ A ′ A R (3.48c) R = R = B ′ R ′ A BR + B R (3.48d) R = 1 R + R ′ A R − B R (3.48e)The components of the Ricci tensor are R = − R − R (3.49a) R = R + 2 R (3.49b) R = R = R + R + R , (3.49c)15ith all other components being zero. The Ricci scalar is then R = − R + R + R + R (3.49d)= 2 (cid:0) R + 2 R + 2 R + R (cid:1) . (3.49e)Thus, we obtain the following non-vanishing components of the Einstein tensor G = 2 R + R (3.49f) G = − R − R (3.49g) G = G = − R − R − R . (3.49h)Now we will solve the vacuum Einstein field equations. Suppose h dR, dR i = 0 in U .Then, if R is constant in U , R = R > 0, we have0 = G = B R + 1 R − B R (3.50a) ⇐⇒ B = 4 R = const . (3.50b)Inserting this condition in 0 = G we arrive at the following contradiction:0 = G = − R + 34 B R = 2 R . (3.51)On the other hand, dR can not be non-zero and lightlike, since the group orbits arenon-null. Therefore, considering h dR, dR i 6 = 0 in U , we will choose coordinates suchthat R ( r ) = r . Next, to solve the field equations, we will consider0 = G + G = 2 (cid:0) R − R (cid:1) (3.52a) ⇐⇒ r ( AB ′ + A ′ B ) + 14 A B (3.52b)= r ( AB ) ′ + 14 ( AB ) . (3.52c)Defining D ( r ) = A ( r ) B ( r ), we get the first order ordinary differential equation for Dr D ′ + 14 D = 0 , (3.53)which can be integrated to give D ( r ) = 4 r c r − c , (3.54)with c > 0. Thus we need to have r > c . Now to solve0 = G = 2 B ′ A Br − B r + 1 r + 1 A r (3.55)16e use D = A B and multiply the equation by 4 c r to get0 = r ( r − c )2 BB ′ − c B + 4 r c + ( r − c ) B (3.56a)= r ( r − c )( B ) ′ + ( r − c ) B + 4 r c . (3.56b)Then, by introducing F ( r ) := B ( r ), we obtain an inhomogeneous first order lineardifferential equation 0 = F ′ + r − c r ( r − c ) F + 4 rc r − c . (3.57)To solve this differential equation, we will first solve the corresponding homogeneousequation: 0 = F ′ h + r − c r ( r − c ) F h (3.58a)= F ′ h + 2( r − c ) − r r ( r − c ) F h (3.58b)= F ′ h + (cid:18) r − rr − c (cid:19) F h . (3.58c)Thus, by integrating, the solution is given by ln ( F h ) = − (cid:16) ln ( r ) − ln (cid:16)p r − c (cid:17)(cid:17) + c (3.59a) ⇐⇒ F h = c ′ √ r − c r , c ′ = e c . (3.59b)Now to obtain the general solution, we will multiply the inhomogeneous equation by c ′ /F h : 0 = r √ r − c F ′ + r − rc ( r − c ) / F + 4 r c ( r − c ) / (3.60a)= (cid:18) r √ r − c F (cid:19) ′ + 4 r c ( r − c ) / . (3.60b)Hence, the solution is given by F = c √ r − c r − √ r − c r Z r c ( r − c ) / dr. (3.61a)To solve the integral we will use the substitution u = √ r − c with dr = u √ u + c du suchthat Z u + c ) / c u u √ u + c du (3.62a)= Z c + 4 c u du (3.62b)=4 c u − c u . (3.62c)17herefore, the solution of the inhomogeneous differential equation is F = B = − c r + c √ r − c + 8 c r . (3.63)Thus, we have solved G = G = 0 and obtained functions B ( r ) and A ( r ) = D ( r ) B ( r ) implying R = R (3.64a)2 R = − R . (3.64b)One can check that these functions satisfy R = R , implying G = G = 0.Inserting the functions in the metric, we see that there is a singularity at r = c . Tocheck if the space-time has a curvature singularity, we will compute the Kretschmannscalar K = R αβµν R αβµν . (3.65)Using the symmetry of the Riemann tensor and the implications of the field equations,the Kretschmann scalar is given by K = 4( R ) − R ) − R ) − R ) + 4( R ) + 4( R ) + 4( R ) + 4( R ) + 4( R ) = 12 (cid:0) ( R ) − ( R ) (cid:1) = 34 c − r − (cid:18) c − c r − c c r + c r + 128 c (cid:16) r + 4 c p r − c (cid:17) (3.66)+ 48 c c r (cid:16) c + 2 r p r − c (cid:17) − c (cid:16) c + 2 r + 16 c r p r − c (cid:17)(cid:19) , (3.67)observing that it is regular at r = c , thus indicating that the singularity is due to apoor choice of coordinates. To solve the coordinate singularity, we will use the followingcoordinate transformation: r ′ = 12 √ c Z D ( r ) dr = p r − c , dr ′ = 12 √ c D ( r ) dr. (3.68)With respect to r ′ we have B ( r ′ ) = − c r ′ − c c r ′ − c r ′ + c . (3.69)18efining l := c > , m := c c , the metric takes the form g = − l B ( r ′ ) dr ′ + B ( r ′ )( dψ + cos θdϕ ) + ( r ′ + l ) ( dθ + sin θdϕ ) (3.70a)= r ′ + l r ′ − mr − l dr ′ − l r ′ − mr ′ − l r ′ + l ( dψ + cos θdϕ ) + ( r ′ + l ) ( dθ + sin θdϕ ) . (3.70b)So we see that we obtain the generalized Taub-NUT space-time and since we assumedthat the orbits are spacelike, we get in fact the Taub-region. Now if we consider the case ε = − 1, so timelike orbits, we have g = A ( r ) dr − B ( r )( dψ + cos θdϕ ) + R ( r )( dθ + sin θdϕ ) . (3.71)Using the orthonormal tetrad ϑ = B ( r )( dψ + cos θdϕ ) (3.72a) ϑ = A ( r ) dr (3.72b) ϑ = R ( r ) dθ (3.72c) ϑ = R ( r ) sin θdϕ. (3.72d)we obtain dϑ = B ′ AB ϑ ∧ ϑ − BR ϑ ∧ ϑ (3.73a) dϑ = 0 (3.73b) dϑ = R ′ AR ϑ ∧ ϑ (3.73c) dϑ = R ′ AR ϑ ∧ ϑ + cot θR ϑ ∧ ϑ . (3.73d)The connection one-forms are then given by ω = ω = B ′ AB ϑ (3.74a) ω = ω = − B R ϑ (3.74b) ω = ω = B R ϑ (3.74c) ω = − ω = − R ′ AR ϑ (3.74d) ω = − ω = − R ′ AR ϑ (3.74e) ω = − ω = − B R ϑ − cot θR ϑ . (3.74f)19ue to their strong resemblance to the case of spacelike orbits we will see that solvingthe vacuum Einstein equations is analogous. Using the second structure equations thecurvature 2-form can be calculatedΩ = − (cid:18) B ′′ A B − B ′ A ′ A B (cid:19) ϑ ∧ ϑ + (cid:18) BR ′ AR − B ′ AR (cid:19) ϑ ∧ ϑ (3.75a)Ω = 12 (cid:18) BR ′ AR − B ′ AR (cid:19) ϑ ∧ ϑ − (cid:18) B ′ R ′ A BR + B R (cid:19) ϑ ∧ ϑ (3.75b)Ω = 12 (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ − (cid:18) B ′ R ′ A BR + B R (cid:19) ϑ ∧ ϑ (3.75c)Ω = − (cid:18) R ′′ A R − R ′ A ′ A R (cid:19) ϑ ∧ ϑ − (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ (3.75d)Ω = − (cid:18) R ′′ A R − R ′ A ′ A R (cid:19) ϑ ∧ ϑ − (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ (3.75e)Ω = (cid:18) B ′ AR − BR ′ AR (cid:19) ϑ ∧ ϑ + (cid:18) R − R ′ A R + 34 B R (cid:19) ϑ ∧ ϑ . (3.75f)Then using Ω µν = R µναβ θ α ∧ θ β , the non-vanishing components of the Riemann tensor,up to symmetry, are R = − B ′′ A B + B ′ A ′ A B (3.76a) R = 2 R = − R = BR ′ AR − B ′ AR (3.76b) R = R = − B ′ R ′ A BR − B R (3.76c) R = R = − R ′′ A R + R ′ A ′ A R (3.76d) R = 1 R − R ′ A R + 34 B R . (3.76e)Analogously, we have the following non-vanishing components of the Einstein tensor G = 2 R + R (3.77a) G = − R − R (3.77b) G = G = − R − R − R . (3.77c)Now we will solve the vacuum Einstein field equations. Supposing h dR, dR i = 0 in U leads analogously to a contradiction. Therefore, we have h dR, dR i 6 = 0 in U and wecan choose coordinates such that R ( r ) = r . Then to solve the field equations we willconsider again 0 = G + G = 2 (cid:0) R − R (cid:1) , but since R and R correspondto curvature components − R and − R , respectively, for the case ε = 1 we seethat they satisfy the same differential equation. Hence defining D ( r ) = A ( r ) B ( r ), we20ave D ( r ) = 4 r c r − c , (3.78)with c > r > c . Now we will solve0 = G = 2 B ′ A Br − B r − r + 1 A r . (3.79)Using D = A B and multiplying the equation by 4 c r we obtain0 = r ( r − c )2 BB ′ − c B − r c + ( r − c ) B (3.80a)= r ( r − c )( B ) ′ + ( r − c ) B − r c . (3.80b)Then again introducing F ( r ) := B ( r ) we have the inhomogeneous first order lineardifferential equation 0 = F ′ + r − c r ( r − c ) F − rc r − c . (3.81)Hence, we see that the corresponding homogeneous equation coincides with the one forspacelike orbits. Thus, we have F h = c ′ √ r − c r . (3.82)Now to obtain the general solution we will multiply the inhomogeneous equation by c ′ /F h : 0 = r √ r − c F ′ + r − rc ( r − c ) / F − r c ( r − c ) / (3.83a)= (cid:18) r √ r − c F (cid:19) ′ − r c ( r − c ) / . (3.83b)Hence, the solution is given by F = B = c √ r − c r + √ r − c r Z r c ( r − c ) / dr (3.84a)= 4 c r + c √ r − c − c r (3.84b)Now with the same line of argument we use the coordinate transformation r ′ = 12 √ c Z D ( r ) dr = p r − c , dr ′ = 12 √ c D ( r ) dr. (3.85)21uch that B ( r ′ ) = 4 c r ′ + c c r ′ − c r ′ + c . (3.86)Then defining l := c > , m := − c c we again obtain the generalized Taub-NUT metric g = 4 l B ( r ′ ) dr ′ − B ( r ′ )( dψ + cos θdϕ ) + ( r ′ + l )( dθ + sin θdϕ ) (3.87a)= r ′ + l r ′ − mr − l dr ′ − l r ′ − mr ′ − l r ′ + l ( dψ + cos θdϕ ) + ( r ′ + l ) ( dθ + sin θdϕ ) , (3.87b)however describing the NUT-regions. If the orbits are not everywhere space- or timelike,we can join them smoothly along the null hypersurfaces by an extension of the formdescribed in the last chapter.Thus, with respect to the constants, m , l and n , we have a three parameter family ofvacuum space-times admitting SU (2) × U (1) as an isometry group, such that SU (2) × U (1) and SU (2) both have three-dimensional non-null orbits. The Taub-NUT space-time is then the unique universal cover. As in the case of the Taub-NUT space-time thegeneralized space-time can be considered to be a principal fibre bundle with respect to the U (1) right action with its first chern class being the constant n . Furthermore, recallingthe remarks in the last section, the constant m can be considered to be the Komarmass of the space-time. In particular, in this case null infinity is the Lens space L ( n, l , being the dual-Bondi-mass with respect to the infinitesimaltranslation induced by the Killing vector field − l ∂ ψ , is proportional to n . Moreover,being a non-trivial S principal fibre bundle over S implies that the NUT parameter isnon-zero. Taub-NUT is a very peculiar spacetime in many respects, not only mathematically, butalso concerning its possible physical interpretation. Yet it is frequently regarded forpossible applications in astrophysics and cosmology, thereby suggesting that it may betaken as an adequate model for some astrophysical object. Geodesic motions, shadows,and lensing in NUT-spacetime have been investigated in detail; see, e.g., [10, 8, 4]. Thequestion of whether and how NUT-spacetime could be regarded as the exterior geometryproduced by some star made of ordinary matter, like, e.g., a perfect fluid, is an old onewith partially controversial claims, in particular regarding the physical interpretationof the NUT charge. So far no compelling physical insight seems to exists as to whatknown properties of ordinary matter could source a non-zero NUT charge. Perfect-fluidsolutions with radially pointing vorticity fields have been constructed for that end, butthe solutions established in [2] are singular, as has been discussed in [17].22n view of this mismatch between hypothetical physical applications eventually leadingto measurements of the NUT parameter on one hand, and the lacking of a proper physicalunderstanding of what might possibly be a matter source (if any) of it on the other, itseems a viable strategy to first characterise the solution as uniquely as possible by itssymmetry properties. 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