A Mathematical Comparison of the Schwarzschild and Kerr Metrics
aa r X i v : . [ phy s i c s . g e n - ph ] S e p A MATHEMATICAL COMPARISON OF THESCHWARZSCHILD AND KERR METRICS
J.-F. PommaretCERMICS, Ecole des Ponts ParisTech, [email protected]://cermics.enpc.fr/ ∼ pommaret/home.html ABSTRACT
A few physicists have recently constructed the generating compatibility conditions (CC) of theKilling operator for the Minkowski (M) , Schwarzschild (S) and Kerr (K) metrics. They discoveredsecond order CC, well known for M, but also third order CC for S and K. In a recent paper, wehave studied the cases of M and S, without using specific technical tools such as Teukolski scalarsor Killing-Yano tensors. However, even if S( m ) and K( m, a ) are depending on constant parametersin such a way that S → M when m → → S when a →
0, the CC of S do not providethe CC of M when m → a →
0. In thispaper, using tricky motivating examples of operators with constant or variable parameters, weexplain why the CC are depending on the choice of the parameters. In particular, the only purelyintrinsic objects that can be defined, namely the extension modules, may change drastically. Asthe algebroid bracket is compatible with the prolongation/projection (PP) procedure, we providefor the first time all the CC for K in an intrinsic way, showing that they only depend on theunderlying Killing algebras and that the role played by the Spencer operator is crucial. We getK < S < M with 2 < <
10 for the Killing algebras and explain why the formal search of the CC forM, S or K are strikingly different, even though each Spencer sequence is isomorphic to the tensorproduct of the Poincar´e sequence for the exterior derivative by the corresponding Lie algebra.
KEY WORDS
Formal integrability; Involutivity; Compatibility condition; Janet sequence; Spencer sequence;Minkowski metric; Schwarzschild metric; Kerr metric.1 ) INTRODUCTION
In order to explain the type of problems we want to solve, let us start adding a constant pa-rameter to the example provided by Macaulay in 1916 that we have presented in a previous paperfor other reasons ([14]). We first recall the following key definition:
DEFINITION 1.1 : A system of order q on E is an open vector subbundle R q ⊆ J q ( E ) withprolongations ρ r ( R q ) = R q + r = J r ( R q ) ∩ J q + r ( E ) ⊆ J r ( J q ( E )) and symbols ρ r ( g q ) = g q + r = S q + r T ∗ ⊗ E ∩ R q + r ⊆ J q + r ( E ) only depending on g q ⊆ S q T ∗ ⊗ E . For r, s ≥
0, we denoteby R ( s ) q + r = π q + r + sq + r ( R q + r + s ) ⊆ R q + r the projection of R q + r + s on R q + r , which is thus definedby more equations in general. The system R q is said to be formally integrable (FI) if we have R ( s ) q + r = R q + r , ∀ r, s ≥
0, that is if all the equations of order q + r can be obtained by means of only r prolongations. The system R q is said to be involutive if it is FI with an involutive symbol g q .We shall simply denote by Θ = { f ∈ E | j q ( f ) ∈ R q } the ”set” of (formal) solutions. It is finallyeasy to prove that the Spencer operator D : J q +1 ( E ) → T ∗ ⊗ J q ( E ) restricts to D : R q +1 → T ∗ ⊗ R q .The most difficult but also the most important theorem has been discovered by M. Janet in1920 ([7]) and presented by H. Goldschmidt in a modern setting in 1968 ([5]). However, the firstproof with examples is not intrinsic while the second, using the Spencer operator, is very technicaland we have given a quite simpler different proof in 1978 ([7], also [9],[10]) that we shall use lateron for studying the Killing equations for the Schwarzschild and Kerr metrics: THEOREM 1.2 : If R q ⊂ J q ( E ) is a system of order q on E such that its first prolongation R q +1 ⊂ J q +1 ( E ) is a vector bundle while its symbol g q +1 is also a vector bundle, then, if g q is2-acyclic, we have ρ r ( R (1) q ) = R (1) q + r . COROLLARY 1.3 : (PP procedure ) If a system R q ⊂ J q ( E ) is defined over a differential fiel K ,then one can find integers r, s ≥ R ( s ) q + r is formally integrable or even involutive.Starting with an arbitrary system R q ⊂ J q ( E ), the main purpose of the next crucial exampleis to prove that the generating CC of the operator D = Φ ◦ j q : E j q −→ J q ( E ) Φ −→ J q ( E ) /R q = F ,though they are of course fully determined by the first order CC of the final involutive system R ( s ) q + r produced by the up/down PP procedure, are in general of order r + s + 1 like the Riemannor Weyl operators, but may be of strictly lower order. MOTIVATING EXAMPLE 1.4 : With m = 1 , n = 3 , q = 2, let us consider the second orderlinear system R ⊂ J ( E ) with dim ( R ) = 8 and parametric jets { y, y , y , y , y , y , y , y } ,defined by the two inhomogeneous PD equations where a is a constant parameter: P y ≡ y = u, Qy ≡ y + a y = v First of all we have to look for the symbol g defined by the two linear equations y = 0 , y = 0.The coordinate system is not δ -regular and exchanging x with x , we get the Janet board: (cid:26) y = 0 y = 0 1 2 31 2 • It follows that g is involutive, thus 2-acyclic and we obtain from the main theorem ρ r ( R (1)2 ) = R (1)2+ r . However, R (1)2 ⊂ R with a strict inclusion because R (1)2 with dim ( R (1)2 ) = 7 is now definedby the 3 equations: y = u, a y = v − u , y + a y = v We may start again with R (1)2 and study its symbol g (1)2 defined by the 3 linear equations withJanet tabular: 2 y = 0 a y = 0 y = 0 1 2 31 2 • • • Since that moment, we have to consider the two possibilities: • a = 0: The initial system becomes y = u, y = v and has an involutive symbol. It is thusinvolutive because it is trivially FI as the left members are homogeneous with only one generatingfirst order CC, namely u − v = 0. We have dim ( g r ) = 4 + r and the following commutativeand exact diagrams: 0 0 0 0 ↓ ↓ ↓ ↓ → g → S T ∗ ⊗ E → T ∗ ⊗ F → F → ↓ ↓ ↓ k → R → J ( E ) → J ( F ) → F → ↓ ↓ ↓ ↓ → R → J ( E ) → F → ↓ ↓ ↓ ↓ ↓ ↓ ↓ → → → → → ↓ ↓ ↓ ↓ → → → → → ↓ ↓ ↓ ↓ → → → → ↓ ↓ ↓ → Θ → E D −→ F D −→ F → D = Q [ d , d , d ] = Q [ d ]:0 → D −→ D −→ D p −→ M → p is the canonical projection onto the residual differential module. • a = 0: When the coefficients are in a differential field of constants, for example if a ∈ Q is invert-ible, we may choose a = 1 like Macaulay ([14]). It follows that g (1)2 is still involutive but we havethe strict inclusion g (1)2 ⊂ g and thus the strict inclusion R (1)2 ⊂ R because dim ( R (1)2 ) = 7 < R (2)2 ⊂ R (1)2 because dim ( R (22 ) = 6 as R (2)2 is defined by the 4 equations with Janet tabular: y = uy = v − u y = v − v + u y + y = v • • • • As R (2)2 is easily seen to be involutive, we achieve the PP procedure, obtaining the strict intrinsicinclusions and corresponding fiber dimensions: R (2)2 ⊂ R (1)2 ⊂ R ⇔ < < ρ r ( R (2)2 ) = ρ r (( R (1)2 ) (1) ) = ( ρ r ( R (1)2 )) (1) = ( R (1)2+ r ) (1) = R (2)2+ r .It remains to find out the CC for ( u, v ) in the initial inhomogeneous system. As we have used twoprolongations in order to exhibit R (2)2 , we have second order formal derivatives of u and v in theright members. Now, as we have an involutive system, we have first order CC for the new rightmembers and could hope therefore for third order generating CC. However, we have successivelythe 4 CC: y = d ( v − u ) = d u ⇒ v − u − u = 0 y = d ( v − v + u ) = d ( v − u ) ⇒ v − u − u = 0 y + y = d v = d u + ( v − u ) ⇒ y + y = d v = d ( v − u ) + ( v − v + u ) ⇒ only one second and one third order CC: v − u − u = 0 , v − u − u = 0but, surprisingly , we are left with the only generating second order CC v − u − u = 0 whichis coming from the fact that the operator P commutes with the operator Q .We let the reader prove as an exercise (See [14],[20] for details) that dim ( R r +2 ) = 4 r + 8 , ∀ r ≥ dim ( R ) = 12 , dim ( R ) = 16 in the following commutative and exact diagrams where E is the trivial vector bundle with dim ( E ) = 1 and dim ( g r +2 ) = r + 4 , ∀ r ≥ ↓ ↓ ↓ → g → S T ∗ ⊗ E → S T ∗ ⊗ F → h → ↓ ↓ ↓ ↓ → R → J ( E ) → J ( F ) → F → ↓ ↓ ↓ ↓ → R → J ( E ) → J ( F ) → ↓ ↓ ↓ ↓ ↓ → → → → → ↓ ↓ ↓ ↓ → → → → → ↓ ↓ ↓ ↓ → → → → ↓ ↓ → Θ → E D −→ F D −→ F → D as before:0 → D −→ D −→ D p −→ M → not a Janet sequence because R is not involutive. MOTIVATING EXAMPLE 1.5 : We now prove that the case of variable coefficients can leadto strikingly different results, even if we choose them in the differential field K = Q ( x , x , x ) ofrational functions in the coordinates that we shall meet in the study of the S and K metrics. Inorder to justify this comment, let us consider the simplest situation met with the second order4ystem R ⊂ J ( E ): R ⊂ J ( E ) (cid:8) y = u, y + x y = v We may consider successively the following systems of decreasing dimensions 8 > > > R ′ = R (1)2 ⊂ R (cid:8) y = u, x y + y = v − u , y + x y = vR ′′ = R (2)2 ⊂ R ′ (cid:26) y = u, y = v − u − x u , y + x y = v, y = − x v + x u + ( x ) u + 2 v − u R ′′′ = R (3)2 ⊂ R ”2 y = u, y = v − u − x u , y + x y = v, x y = − v + u + x u + 2 v y = − x v + x u + ( x ) u + 2 v − u y = − x v + x u + ( x ) u + 2 v − u The last system is involutive with the following Janet tabular: y = 0 y = 0 y = 0 y = 0 y = 0 y = 0 1 2 31 2 • • • • • •• • • The generic solution is of the form y = b ( x ) + cx and it is rather striking that such a system hasconstant coefficients (This will be exactly the case of the S and K metrics but similar examplescan be found in [10]). We could hope for 9 generating CC up to order 4 but tedious computations,left to the reader as a tricky exercise, prove that we have in fact, as before, only 2 generating thirdorder CC described by the following involutive system, namely: A ≡ v − u − x u − u = 0 B ≡ v − u − ( x ) v + ( x ) u + 2 x v + ( x ) u − x u − v = 0satisfying the only first order CC: d B − d A + ( x ) d A = 0Introducing the ring D = K [ d , d , d ] = K [ d ] of differential operators with coefficients in K , weobtain the sequence of D -modules:0 −→ D −→ D −→ D −→ D p −→ M −→ MOTIVATING EXAMPLE 1.6 : We comment a tricky example first provided by M. Janet in1920, that we have studied with details in ([9],[11]). Using jet notations with n = 3 , m = 1 , q =2 , K = Q ( x ), let us consider the inhomogeneous second order system: R ⊂ J ( E ) (cid:8) y − x y = u, y = v We let the reader prove that the space of solutions has dimension 12 over Q and that we have r = 0 , s = 5 in such a way that R (5)2 is involutive and even finite type with a zero symbol. Accord-ingly, we have dim ( R (5)2 ) = 12. Passing to the differential module point of view, it follows that dim K ( M ) = 12 and rk D ( M ) = 0. According to the general results presented, we have thus to use5 prolongations and could therefore wait for CC up to order ... 6 !!!. In fact, and we repeat that here is no hint at all for predicting this result in any intrinsic way , we have only two generatingCC, one of order 3 and ... one of order 6 indeed, namely: A ≡ v − x v − u − v = 0 B ≡ v − x v − u − x ( v − x v − u )+( x ) ( v − x v − u ) − u + 2 x u − u = 0satisfying the only fourth order CC: C ≡ A − x A + ( x ) A − B = 0It follows that we have the unexpected differential resolution:0 → D −→ D −→ D −→ D p −→ M → rk D ( M ) = 1 − − a by replacing the coefficient x by ax , we obtain 2 ay = v − ax v − u and obtain the same conclusions as before. We point out the fact that, when a = 0, the system y = u, y = v , which is trivially FI because it is homogeneous, has a symbol g which is neither involutive (otherwise it should admit a first order CC), nor even par = ( y , y , y , y ) , par = ( y , y , y , y ) , par = ( y , y , y , y )and the long δ -sequence:0 → g δ −→ T ∗ ⊗ g δ −→ ∧ T ∗ ⊗ g δ −→ ∧ T ∗ ⊗ T ∗ → → δ −→ δ −→ δ −→ → dim ( B ( g )) = 12 − , dim ( Z ( g )) = 12 − ⇒ dim ( H ( g ) = 9 − = 0.However, g is involutive with the following Janet tabular for the vertical jets ( v ijk ) ∈ S T ∗ : v = 0 v = 0 v = 0 v = 0 v = 0 v = 0 1 2 31 2 • • • • • • • Accordingly, R is thus involutive and the only CC v − u = 0 is of order 2 because we needone prolongation only to reach involution and thus 2-acyclicity. MOTIVATING EXAMPLE 1.7 : With m = 1 , n = 2 , q = 2 , K = Q , let us consider the inho-mogeneous second order system: y = u, y − y = v We obtain at once through crossed derivatives y = u − v − v and, by substituting, two fourthorder CC for ( u, v ), namely: A ≡ u − v − v − u = 0 , B ≡ u − u − v = 0satisfying B + B − A = 0 . However, we may also obtain a single CC for ( u, v ), namely C ≡ d u − u − d v = 0 and we check at once A = d C + C, B = d C while C = d B − d A + A . We let the reader prove that dim ( R r ) = 4 , ∀ r ≥
0. Hence, if (
A, B ) is a section of F C is a section of F ′ , the jet prolongation sequence:0 → R → J ( E ) → J ( F ) → F → → → → → → not formally exact because 4 −
28 + 30 − = 0, while the corresponding long sequence:0 → R r +4 → J r +4 ( E ) → J r +2 ( F ) → J r ( F ′ ) → → → ( r + 5)( r + 6) / → ( r + 3)( r + 4) → ( r + 1)( r + 2) / → − r +11 r +30)2 + ( r + 7 r + 12) − ( r +3 r +2)2 = 0 but not strictlyexact because R is quite far from being FI as we have even R (4)2 = 0.It follows from these examples and the many others presented in ([20]) that we cannot agreewith ([1-4]). Indeed, it is clear that one can use successive prolongations in order to look for CC oforder 1 , , , ... and so on, selecting each time the new generating ones and knowing that Noethe-rian arguments will stop such a procedure ... after a while !. However, it is clear that, as long asthe numbers r and s are not known, it is not effectively possible to decide in advance about themaximum order that must be reached. Therefore, it becomes clear that exactly the same procedure must be applied when looking for the CC of the various Killing operators we want to study, theproblem becoming a ” mathematical ” one, surely not a ” physical ” one. IMPORTANT REMARK 1.8 : Taking the adjoint operators, it is essential to notice that ad ( D )generates the CC of ad ( D ) when a = 0, a result leading to ext ( M ) = 0 but this is not true when a = 0, a result leading to ext ( M ) = 0 ([10],[18],[21],[22]). Hence we discover on such an examplethat the intrinsic properties of a system with constant or even variable coefficients may drastically depend on these coefficients, even though the correspondig systems do not appear to be quitedifferent at first sight. Accordingly, we have:WHEN A SYSTEM IS NOT FI, NOTHING CAN BE SAID ” A PRIORI ” ABOUT THE CC,THAT IS WITHOUT ANY EXPLICIT COMPUTATION OR COMPUTER ALGEBRA.Comparing the sequences obtained in the previous examples, we may state:
DEFINITION 1.9 : A differential sequence is said to be formally exact if it is exact on the jetlevel composition of the prolongations involved. A formally exact sequence is said to be strictlyexact if all the operators/systems involved are FI (See [14] for more details). A strictly exactsequence is called canonical if all the operators/systems are involutive. The only known canonicalsequences are the Janet and Spencer sequences that can be defined independently from each other.With canonical projection Φ = Φ : J q ( E ) ⇒ J q ( E ) /R q = F , the various prolongations aredescribed by the following commutative and exact introductory diagram :0 0 0 ↓ ↓ ↓ → g q + r +1 → S q + r +1 T ∗ ⊗ E → S r +1 T ∗ ⊗ F → h r +1 → ↓ ↓ ↓ ↓ → R q + r +1 → J q + r +1 ( E ) ρ r +1 (Φ) −→ J r +1 ( F ) → Q r +1 → ↓ ↓ ↓ ↓ → R q + r → J q + r ( E ) ρ r (Φ) −→ J r ( F ) → Q r → ↓ ↓ ↓ snake ” lemma, we obtainthe useful long exact connecting sequence :0 → g q +1 → R q +1 → R q → h → Q → lower left ) with CC ( upper right ).We finally recall the Fundamental Diagram I that we have presented in many books and papers,relating the (upper) canonical Spencer sequence to the (lower) canonical Janet sequence , that onlydepends on the left commutative square D = Φ ◦ j q with Φ = Φ when one has an involutive system R q ⊆ J q ( E ) over E with dim ( X ) = n and j q : E → J q ( E ) is the derivative operator up to order q : ↓ ↓ ↓ ↓ → Θ j q → C D → C D → C D → ... D n → C n → ↓ ↓ ↓ ↓ → E j q → C ( E ) D → C ( E ) D → C ( E ) D → ... D n → C n ( E ) → k ↓ Φ ↓ Φ ↓ Φ ↓ Φ n → Θ → E D → F D → F D → F D → ... D n → F n → ↓ ↓ ↓ ↓ We shall use this result, first found exactly 40 years ago ([7]) but never acknowledged, in order toprovide a critical study of the comparison between the S and K metrics.
EXAMPLE 1.10 : The Janet tabular in Example 1.4 with a = 1 provides the fiber dimensions:0 0 0 0 ↓ ↓ ↓ ↓ → Θ j → D → D → D → → ↓ ↓ ↓ k → j → D → D → D → → k ↓ Φ ↓ Φ ↓ Φ ↓ → Θ → D → D → D → → ↓ ↓ ↓ −
16 + 14 − −
10 + 20 −
15 + 4 = 0 and 1 − − a = 0 and a = x .
2) SCHWARZSCHILD VERSUS KERRa) SCHWARZSCHILD METRIC
In the Boyer-Lindquist (BL) coordinates ( t, r, θ, φ ) = ( x , x , x , x ), the Schwarzschild metricis ω = A ( r ) dt − (1 /A ( r )) dr − r dθ − r sin ( θ ) dφ and ξ = ξ i d i ∈ T , let us introduce ξ i = ω ri ξ r with the 4 formal derivatives ( d = d t , d = d r , d = d θ , d = d φ ). With speed of light c = 1 and A = 1 − mr where m is a constant, the metric can be written in the diagonal form: A − /A − r
00 0 0 − r sin ( θ ) with a surprisingly simple determinant det ( ω ) = − r sin ( θ ).Using the notations of differential modules or jet theory, we may consider the infinitesimal Killingequations:Ω ≡ L ( ξ ) ω = 0 ⇔ Ω ij ≡ d i ξ j + d j ξ i − γ rij ξ r = 0 ⇔ Ω ij ≡ ω rj ξ ri + ω ir ξ rj + ξ r ∂ r ω ij = 0where we have introduced the Christoffel symbols γ through he standard Levi-Civita isomorphism j ( ω ) ≃ ( ω, γ ) while setting A ′ = ∂ r A in the differential field K of coefficients ([19]). As in the8acaulay example just considered and in order to avoid any further confusion between sectionsand derivatives, we shall use the sectional point of view and rewrite the previous 10 equations inthe symbolic form Ω ≡ L ( ξ ) ω ∈ S T ∗ where L is the formal Lie derivative : R ⊂ J ( T ) Ω ≡ − r sin ( θ ) ξ − rsin ( θ ) ξ − r sin ( θ ) cos ( θ ) ξ = 0Ω ≡ − r ξ − r sin ( θ ) ξ = 0Ω ≡ − A ξ − r sin ( θ ) ξ = 0Ω ≡ A ξ − r sin ( θ ) ξ = 0Ω ≡ − r ξ − rξ = 0Ω ≡ − A ξ − r ξ = 0Ω ≡ A ξ − r ξ = 0Ω ≡ − A ξ + A ′ A ξ = 0Ω ≡ − A ξ + Aξ = 0Ω ≡ A ξ + A ′ ξ = 0Though this system R ⊂ J ( T ) has 4 equations of class 3, 3 equations of class 2, 2 equations ofclass 1 and 1 equation of class 0, it is far from being involutive because it is finite type with secondsymbol g = 0 defined by the 40 equations v kij = 0 in the initial coordinates. From the symetry, it isclear that such a system has at least ∂ t ↔ ξ = 1 ⇔ ξ = A and, using cartesian coordinates ( t, x, y, z ), the 3 space rotations y∂ z − z∂ y , z∂ x − x∂ z , x∂ y − y∂ x .We obtain in particular, modulo Ω: ξ = − A ′ A ξ , ξ = + A ′ A ξ , ξ = − r ξ , ξ = − r ξ − cot ( θ ) ξ ⇒ ξ + ξ = 0 , ξ + ξ = − cot ( θ ) ξ We may also write the Schwarzschild metric in cartesian coordinates as: ω = A ( r ) dt + (1 − A ( r ) ) dr − ( dx + dy + dz ) , rdr = xdx + ydy + zdz and notice that the 3 × one and only one zero orderKilling equation rξ r = xξ x + yξ y + zξ z = 0 ⇒ ξ = ξ r = 0. Such a result also amounts to say thatthe spatial projection of any Killing vector on the radial spatial unit vector ( x/r, y/r, z/r ) vanishesbeause r must stay invariant.However, as we are dealing with sections , ξ = 0 implies ξ = 0 , ξ = 0 , ξ = 0 ... but NOT ( care ) ξ = 0, these later condition being only brought by one additional prolongation and we havethe strict inclusions R (3)1 ⊂ R (2)1 ⊂ R (1)1 = R that we rename as R ”1 ⊂ R ′ ⊂ R . Hence, it remainsto determine the dimensions of these subsystems and their symbols, exactly like in the Macaulayexample. We shall prove in the next section that two prolongations bring the five new equations: ξ = 0 , ξ = 0 , ξ = 0 , ξ = 0 , ξ = 0and a new prolongation only brings the single equation ξ = 0 .Knowing that dim ( R ) = dim ( R ) = 10, dim ( R ) = 5, dim ( R ) = 4, we have thus obtainedthe 15 equations defining R ′ = R (2)1 with dim ( R ′ ) = 20 −
15 = 5 and let the reader draw thecorresponding Janet tabular for the 4 equations of class 3, the 4 equations of class 1, the 3 equa-tions of class 0 and the 3 equations of class 2. The symbol g ′ has the two parametric jets ( v , v )and is not 2-acyclic. Adding ξ = 0 ⇔ ξ = 0, we finally achieve the PP procedure with the 16equations defining the system R ”1 = R (3)1 with dim ( R ”1 ) = 20 −
16 = 4, namely:9 ”1 ⊂ R ′ ⊂ R ⊂ J ( T ) ξ + cot ( θ ) ξ = 0 ξ + sin ( θ ) ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 2 0 1 32 0 1 32 0 1 32 0 1 32 0 1 • • • • • • • • • • • • • • × • • • • • •• • • • and we have replaced by ” × ” the only ” dot ” (non-multiplicative variable) that cannot providevanishing crossed derivatives and thus involution of the symbol g ”1 with the only parametric jets( v , v ). It is easy to check that R ”1 , having minimum dimension equal to 4, is formally integrable,though not involutive as it is finite type with dim ( g ”1 ) = 16 −
15 = 1 ⇒ g ”1 = 0 with parametric jet v and to exhibit 4 solutions linearly independent over the constants. We let the reader prove asan exercise that the dimension of the Spencer δ -cohomology at ∧ T ∗ ⊗ g ”1 is dim (( H ( g ”1 )) = 3 = 0but we have proved in ([19]) that its restriction to ( x , x ) is of dimension 1 only. We obtain:THIS SYSTEM DOES NOT DEPEND ON m ANY LONGER !.Denoting by R ”2 ⊂ R ⊂ J ( T ) with dim ( R ”2 ) = 4 the prolongation of R ′ ⊂ J ( T ), it isthe involutive system provided by the prolongation/projection (PP) procedure, we are in positionto construct the corresponding canonical/involutive (lower) Janet and (upper) Spencer sequencesalong the following fundamental diagram I that we recalled in the Introduction. In the presentsituation, the Spencer sequence is isomorphic to the tensor product of the Poincar´e sequence bythe underlying 4-dimensional Lie algebra G , namely: ∧ T ∗ ⊗ G d −→ ∧ T ∗ ⊗ G d −→ ... d −→ ∧ T ∗ ⊗ G −→ not depending any longer on m , we have now C r = ∧ r T ∗ ⊗ R ”2 and D is of order2 like j while all the other operators are of order 1:0 0 0 0 0 ↓ ↓ ↓ ↓ ↓ → Θ j → D → D → D → D → → ↓ ↓ ↓ ↓ ↓ → j → D → D → D → D → → k ↓ ↓ ↓ ↓ ↓ → Θ → D → D → D → D → D → → ↓ ↓ ↓ ↓ ↓ −
16 + 24 −
16 + 4 = 0 , −
60 + 160 −
180 + 96 −
20 = 0 , −
56 + 144 −
156 + 80 −
16 = 0We point out that, whatever is the sequence used or the way to describe D , then ad ( D ) isparametrizing the Cauchy operator ad ( D ) for the S metric. However, such an approach does nottell us explicitly what are the second and third order CC involved in the initial situation.10n actual practice, all the preceding computations have been finally used to reduce the Poincar´egroup to its subgroup made with only one time translation and three space rotations !. On thecontrary, we have proved during almost fourty years that one must increase the Poincar´e group(10 parameters), first to the Weyl group (11 parameters by adding 1 dilatation) and finally tothe conformal group of space-time (15 parameters by adding 4 elations) while only dealing withhe Spencer sequence in order to increase the dimensions of the Spencer bundles, thus the number dim ( C ) of potentials and the number dim ( C ) of fields (Compare to [6]). b) KERR METRIC We now write the Kerr metric in Boyer-Lindquist coordinates: ds = ρ − mrρ dt − ρ ∆ dr − ρ dθ − amrsin ( θ ) ρ dtdφ − ( r + a + mra sin ( θ ) ρ ) sin ( θ ) dφ where we have set ∆ = r − mr + a , ρ = r + a cos ( θ ) as usual and we check that: a = 0 ⇒ ds = (1 − mr ) dt − − mr dr − r dθ − r sin ( θ ) dφ as a well known way to recover the Schwarschild metric. We notice that t or φ do not appear inthe coefficients of the metric and thus, as the maximum subgroup of invariance of the Kerr metric must be contained in the maximum subgroup of invariance of the Schwarzschild metric because ofthe above limit when a →
0, we obtain the only possible 2 infinitesimal generators { ∂ t , ∂ φ } and wehave the fundamental diagram I with fiber dimensions:0 0 0 0 0 ↓ ↓ ↓ ↓ ↓ → Θ j → D → D → D → D → → ↓ ↓ ↓ ↓ ↓ → j → D → D → D → D → → k ↓ ↓ ↓ ↓ ↓ → Θ → D → D → D → D → D → → ↓ ↓ ↓ ↓ ↓ −
58 + 152 −
168 + 88 −
18 = 0. Comparing the surprisinglyhigh dimensions of the Janet bundles with the surprisingly low dimensions of the Spencer bundlesneeds no comment on the physical usefulness of the Janet sequence, despite its purely mathemat-ical importance. In addition, using the same notations as in the preceding section, we shall provethat we have now the additional zero order equations ξ r = 0 , ξ θ = 0 produced by the non-zerocomponents of the Weyl tensor and thus, at best , dim ( R (3)0 ) = 2 ⇔ dim (( R (2)1 ) = 2 as these zeroorder equations will be obtained after only two prolongations. They depend on j (Ω) and we shouldobtain therefore eventually dim ( Q ) = 10 + dim ( R ) ≥
12 CC of order 2 without any way way toknow abut the desired third order CC.Using now cartesian space coordinates ( x, y, z ) with ξ z = 0 , xξ x + yξ y = 0, we have only tostudy the following first order involutive system for ξ x = ξ with coefficients no longer dependingon ( a, m ), providing the only generator x∂ y − y∂ x : Φ ≡ ξ z = 0Φ ≡ ξ y − y ξ = 0Φ ≡ ξ x = 0 1 2 31 2 • • • and the fundamental diagram 11 0 0 0 ↓ ↓ ↓ ↓ → Θ j → D → D → D → → ↓ ↓ ↓ k → j → D → D → D → → k ↓ ↓ ↓ ↓ → Θ → D → D → D → → ↓ ↓ ↓ m, a ) any longer.Accordingly, this final result definitively proves that, as far as differential sequences are concerned :THE ONLY IMPORTANT OBJECT IS THE GROUP, NOT THE METRIC
3) SCHWARZSCHILD METRIC REVISITED
Let us now introduce the Riemann tensor ( ρ kl,ij ) ∈ ∧ T ∗ ⊗ T ∗ ⊗ T and use the metric in order toraise or lower the indices in order to obtain the purely covariant tensor ( ρ kl,ij ) ∈ ∧ T ∗ ⊗ T ∗ ⊗ T ∗ .Then, using r as an implicit summation index, we may consider the formal Lie derivative on sec-tions: R kl,ij ≡ ρ rl,ij ξ rk + ρ kr,ij ξ rl + ρ kl,rj ξ ri + ρ kl,ir ξ rj + ξ r ∂ r ρ kl,ij = 0that can be considered as an infinitesimal variation. As for the Ricci tensor ( ρ ij ) ∈ S T ∗ , we noticethat ρ ij = ρ ri,rj = 0 ⇒ R ij ≡ ρ rj ξ ri + ρ ir ξ rj + ξ r ∂ r ρ ij = 0 though we have only: ω rs R ri,sj = R ij + ω rs ρ ti,sj Ω st ⇒ R ij = R ri,rj = ω rs R ri,sj mod (Ω)The 6 non-zero components of the Riemann tensor are known to be: ρ , = + mr , ρ , = − m A r , ρ , = − m A sin ( θ )2 r ρ , = + m r A , ρ , = + m sin ( θ )2 r A , ρ , = − m r sin ( θ )First of all, we notice that: ξ + A ′ A ξ = 0 , ξ − A ′ A ξ = 0 ⇒ ξ + ξ = 0Ω ≡ − A ξ − r ξ = 0 , ξ + 1 r ξ = 0We obtain therefore: R , ≡ ρ , ( ξ + ξ ) + ξ r ∂ r ( ρ , ) = ξ ∂ ρ , = − mr ξ = 0 ⇒ ξ = 0 R , ≡ ρ , ( ξ + ξ ) + ξ r ∂ r ( ρ , ) ≡ ( − mAr ( − A ′ A − r ) − ( mA r ) ′ ) ξ = 3 mA r ξ = 0Similarly, we also get: R , ≡ ρ , ξ + ρ , ξ + ξ r ∂ r ρ , = 0 ⇒ mr ξ − mA r ξ = 0 ⇒ ξ = 0 R , ≡ ρ , ξ + ρ , ξ + ξ r ∂ r ρ , = 0 ⇒ mr ξ − mAsin ( θ )2 r ξ = 0 ⇒ ξ = 0and so on. We obtain for example, among the second order CC: R , ≡ − mr ξ = 0 , R , ≡ mA r ξ = 0 ⇒ R , − r A R , = 012nd thus, among the first prolongations, the third order CC that cannot be obtained by prolon-gation of the various second order CC while taking into account the Bianchi identities ([MSK]).Using the Spencer operator and the fact that ξ ∈ j (Ω), we obtain indeed: d ξ − ξ = d ξ − A ′ A ξ = 0 , d ξ − ξ = 0 , d ξ − ξ = 0In addition, introducing ξ ∈ j (Ω) in the right member as in the motivating examples, we have3 PD equations for ( ξ , ξ ), namely: ξ + cot ( θ ) ξ = − r ξ , ξ + sin ( θ ) ξ = 0 , ξ = − r ξ Using two prolongations and eliminating the third order jets, we obtain successively: ξ + sin ( θ ) ξ + 2 sin ( θ ) cos ( θ ) ξ = 0 − ξ = 1 r ξ − sin ( θ ) ξ − sin ( θ ) cos ( θ ) ξ + 2 ξ − cot ( θ ) ξ = − sin ( θ ) r ξ − sin ( θ ) cos ( θ ) ξ − cos ( θ ) ξ + 2 cot ( θ ) ξ = 2 sin ( θ ) cos ( θ ) r ξ sin ( θ ) cos ( θ ) ξ = − sin( θ ) cos ( θ ) r ξ − sin ( θ ) ξ = 2 sin ( θ ) r ξ Summing, we see that all terms in ξ and ξ disappear and that we are only left with terms in ξ ,including in particular the second order jets ξ , ξ , namely: ξ − sin ( θ ) ξ − sin ( θ ) cos ( θ ) ξ + 2 sin ( θ ) ξ = 0Setting U = ξ , V = ξ , V = ξ , W = ξ , W = ξ with ( U, V , V , W , W ) ∈ j (Ω), we obtainthe new strikingly unusual third order CC for Ω: d V − sin ( θ ) d V − sin ( θ ) cos ( θ ) V + 2 sin ( θ ) U = 0However, in our opinion at least, we do not believe that such a purely ”technical ” relation couldhave any ”physical ” usefulness and let the reader compare it with the CC already found in ([19],Lemma 3.B.3). In addition and contrary to this situation, we have successively: R , ≡ ρ , ξ + ρ , ξ + ξ∂ρ , = − mr ξ − m rA ξ = − m r ξ = 0 ⇒ ξ = 0 R , ≡ ρ , ξ + ρ , ξ + ξ∂ρ , = − mr ξ − msin ( θ )2 rA ξ = − m r ξ = 0 ⇒ ξ = 0 ρ , = 0 ⇒ R , ≡ ρ , ( ξ + ξ + ξ + ξ ) + ξ∂ρ , = 0 d R , + d R , + d R , = 3 m r ( d ξ − d ξ ) = 0 mod (Ω , Γ , R )a result showing that certain third order CC may be differential consequrences of the Bianchiidentities (See [19] for details). Finally, we notice that: R , ≡ ρ , ( ξ + ξ ) + ξ∂ρ , = 3 msin ( θ ) ξ = 0and, comparing to the previous computation for ( ξ , ξ ), nothing can be said about the generatingCC as long as the PP procedure has not been totally achieved with a FI or involutive system .13 ) KERR METRIC REVISITED Though we shall provide explicitly all the details of the computations involved, we shall changethe coordinate system in order to confirm theses results by using computer algebra in a much fasterway. The idea is to use the so-called ” rational polynomial ” coefficients while setting anew :( x = t, x = r, x = c = cos ( θ ) , x = φ ) ⇒ dx = − sin ( θ ) dθ ⇒ ( dx ) = (1 − c ) dθ in order to obtain over the differential field K = Q ( a, m )( t, r, c, φ ) = Q ( a, m )( x ): ds = ρ − mx ρ ( dx ) − ρ ∆ ( dx ) − ρ − ( x ) ( dx ) − amx (1 − ( x ) ) ρ dx dx − (1 − ( x ) )(( x ) + a + ma x (1 − ( x ) ) ρ )( dx ) with now ∆ = ( x ) − mx + a = r − mr + a and ρ = ( x ) + a ( x ) = r + a c . For a lateruse, it is also possible to set ω = − (1 − c )(( r + a ) − a ((1 − c )( a − mr + r )) / ( r + a c ).As this result will be crucially used later on, we have: LEMMA 4.1 : det ( ω ) = − ( r + a c ) . Proof : As an elementary result on matrices, we have: det ( ω ) = det a e b c e d = bc det (cid:18) a ee d (cid:19) = bc ( ad − e )with e = ω = amx (1 − ( x ) ) ρ because ds = ... + 2 ω dx dx + ... and det ( ω ) is thus equal to: ρ ∆(1 − ( x ) ) [ − ( ρ − mx ) ρ (1 − ( x ) )(( x ) + a + ma x (1 − ( x ) ) ρ ) − ( amx (1 − ( x ) )) ρ ]that is, after division by (1 − ( x ) ) and ρ :1∆ [ − ( ρ − mx )( ρ ( x ) + ρ a + ma x (1 − ( x ) )) − a m ( x ) (1 − ( x ) ))Finally, after eliminating the last term, we get:1∆ [ − ρ (( x ) + a ) − ρ ma x (1 − ( x ) ) + ρ mx (( x ) + a )]that is (Compare to [ ] and [ ]):1∆ [ − ρ (∆+ mx )+ ρ ma x ( x ) + ρ m ( x ) ] = 1∆ [ − ρ (∆+ mx )+ ρ mx ( a ( x ) +( x ) )] = − ρ in a coherent way with the result A ( − A )( − r (1 − c ) ( − r (1 − c )) = − r obtained for the S metricwhen a →
0. For a later use, we have obtained ω ω − ( ω ) = − (1 − c )∆. Q.E.D.Contrary to the Schwarzschild metric, the main ”trick” for studying the Kerr metric is to takeinto account that the partition between the zero and nonzero terms will not change if we modifythe coordinates, even if, of course, the nonzero terms may change. Meanwhile, we notice that themost important property of the Kerr metric is the off-diagonal term ω tφ = ω φt = − amsin ( θ ) ρ , thatis the coefficient of dtdφ in the metric ds which is indeed 2 ω tφ dtdφ . We may obtain thereforesuccessively the Killing equations for the Kerr type metric, using sections of jet bundles and writingsimply ξ∂ω = ξ r ∂ r ω = ξ ∂ ω + ξ ∂ ω while framing the principal derivative ξ ji of Ω ij :14 ⊂ J ( T ) Ω ≡ ω ξ + ω ξ ) + ξ∂ω = 0Ω ≡ ω ξ + ω ξ + ω ξ = 0Ω ≡ ω ξ + ξ∂ω = 0Ω ≡ ω ξ + ω ξ + ω ξ = 0Ω ≡ ω ξ + ω ξ = 0Ω ≡ ω ξ + ξ∂ω = 0Ω ≡ ω ξ + ω ( ξ + ξ ) + ω ξ + ξ∂ω = 0Ω ≡ ω ξ + ω ξ + ω ξ = 0Ω ≡ ω ξ + ω ξ + ω ξ = 0Ω ≡ ω ξ + ω ξ ) + ξ∂ω = 0With mod ( ξ ) = mod ( ξ , ξ ), multiplying Ω by ω , Ω by ω and adding, we notice that:2 ω ω ( ξ + ξ ) + 2 ω ( ω ξ + ω ξ ) + ξ∂ ( ω ω ) = 0Similarly, multiplying Ω by 2 ω ( care to the factor ω ) ( ξ + ξ ) + 2 ω ( ω ξ + ω ξ ) + ξ∂ ( ω ) = 0Substracting, we obtain therefore the tricky formula (see the previous Lemma):2( ω ω − ( ω ) )( ξ + ξ ) + ξ∂ ( ω ω − ( ω ) ) = 0Substituting, we obtain: ω ξ + ω ξ = 0 mod ( ξ ) , ω ξ + ω ξ = 0 mod ( ξ ) , ω ξ − ω ξ = 0 mod ( ξ )a situation leading to modify Ω , Ω and Ω , similar to the one found in the Minkowski casewith ξ , = 0 , ξ , + ξ , = 0 , ξ , = 0 mod ( ξ ) when ω = 0. We also obtain with Ω and Ω :( ω ω − ( ω ) ) ξ + ω ( ω ξ − ω ξ ) = 0 mod ( ξ )( ω ω − ( ω ) ) ξ − ω ( ω ξ − ω ξ ) = 0 mod ( ξ )and with Ω and Ω :( ω ω − ( ω ) ) ξ + ω ( ω ξ − ω ξ ) = 0 mod ( ξ )( ω ω − ( ω ) ) ξ − ω ( ω ξ − ω ξ ) = 0 mod ( ξ )Finally, multiplying Ω by ω , Ω by ω and adding, we finally obtain (see the Lemma again)2( ω ω )( ξ + ξ ) + ξ∂ ( ω ω ) = 0Using the rational coefficients belonging to the differential field K = Q ( m, a )( x , x ), thenonzero components of the corresponding Riemann tensor can be found in textbooks.One has the classical orthonormal decomposition: ds = ∆ ρ ( dt − asin ( θ ) dφ ) − ρ ∆ ( dr ) − ρ ( dθ ) − ( r + a ) sin ( θ ) ρ ( dφ − ar + a dt ) and defining: dX = √ ∆ ρ ( dt − asin ( θ ) dφ ) dX = ρ √ ∆ dr = ρ √ ∆ dx dX = ρdθ = − ρsin ( θ ) dx dX = ( r + a ) sin ( θ ) ρ ( dφ − ar + a dt )15n which the coefficient of ( dt ) is ∆ ρ − a sin ( θ ) ρ = 1 − mrρ while the coefficient of ( dφ ) is − ( r + a + mra sin ( θ ) ρ ) sin ( θ ) indeed. We have ds = ( dX ) − ( dX ) − ( dX ) − ( dX ) and make thus the Minkowski metric appearing in a purely alebraic way. We now use the new co-ordinates ( x = t, x = r, x = cos ( θ ) , x = φ ) and it follows that the conditions ξ = 0 , ξ = 0 areinvariant under such a change of basis because dX and dX are respectively proportional to dx and dx . Indeed, as ω = ω ( r, θ ) and thus ξ∂ω = 0, the new symbol g ′ of R ′ = R (2)1 ⊂ R ⊂ T ∗ ⊗ T while ρ ∈ ∧ T ∗ ⊗ T ∗ ⊗ T as mixed tensors.We obtain simpler formulas in the corresponding basis, in particular the 6 components with onlytwo different indices are proportional to mr ( r − a c )( r + a c ) while the 3 components with four differentindices are proportional to amc (3 r − a c )( r + a c ) .In the original rational coordinate system, the main nonzero useful components of the Riemanntensor can only be obtained by means of computer algebra . For helping the reader with the lit-erature, in particular the book ” Computational in Riemann Geometry ” wrote by Kenneth R.Koehler that can be found on the net with a free access. We notice that ω → − ω , that is to saychanging the sign of the metric, does not change the Christoffel symbols ( γ kij ) and the Riemanntensor ( ρ rl,ij ) but changes the sign of ( ρ kl,ij = ω kr ρ rl,ij ). For this reason, we have adopted the signconvention of this reference for the explicit computation of these later components as the productsand quotients used in the sequel will not be changed.We have successively: ρ , = − mr (2( r − mr + a )+ a (1 − c ))( r − a c )2( r + a c ) ( r − mr + a ) ρ , = mr ( r − mr + a +2 a (1 − c ))( r − a c )2(1 − c )( r + a c ) ρ , = mr (1 − c )( r − mr + a )( r − a c )2( r + a c ) ρ , = − mr ( r − a c )2(1 − c )( r + a c )( r − mr + a ) ρ , = − (1 − c ) mr ( r − a c r +4 a r − a c +3 a − a mr (1 − c ))( r − a c )2( r + a c ) ( r − mr + a ) ρ , = mr (2 r − a c r +5 a r − a c +3 a − a mr (1 − c ))( r − a c )2( r + a c ) ρ , = amc (2 r − a c +3 a )(3 r − a c )2( r + a c ) ρ , = − amc ( r − a c +3 a )(3 r − a c )2( r + a c ) ρ , = − amc (3 r − a c )2( r + a c ) It must be noticed that we have been able to factorize the six components with only two differentindices by ( r − a c ) and the three components with four different indices by (3 r − a c ), aresult not evident at first sight but coherent with the orthogonal decomposition.After tedious computations, we obtain: − ω ω ρ , = − ( − ρ ∆ )( amr (1 − c ) ρ )( − − c )∆ ) ρ , = − am r (1 − c )( r − a c )2( r + a c ) ( r − mr + a )which is indeed vanishing when a = 0 for the S metric, both with: ρ , + ρ , = a mc (1 − c )(3 r − a c )2( r + a c ) ρ , + ρ , = amc ( r + a )(3 r − a c )2( r + a c ) ρ , = a mc (3 r − a c )2( r + a c ) ρ , = amr (3 r − mr +3 a )( r − a c )2( r + a c ) ρ , = − a mc (1 − c )( r + a )(3 r − a c )2( r + a c ) ρ , = amr (1 − c )(3 r +3 a − mr )( r − a c )2( r + a c ) ( r − mr + a ) = amr (1 − c )( r − a c )2( r + a c ) + am r (1 − c )( r − a c )2( r + a c ) ( r − mr + a ) Introducing the formal Lie derivative R = L ( ξ ) ρ and using the fact that ρ ∈ ∧ T ∗ ⊗ T ∗ ⊗ T ∗ is a tensor, the system R (2)1 contains the new equations: R kl,ij ≡ ρ rl,ij ξ rk + ρ kr,ij ξ rl + ρ kl,rj ξ ri + ρ kl,ir ξ rj + ξ r ∂ r ρ kl,ij = 0Taking into account the original first order Killing equations, we obtain successively: R , ≡ ρ , ( ξ + ξ ) + 2 ρ , ξ + 2 ρ , ξ + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ ) + 2 ρ , ξ + 2 ρ , ξ + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ ) + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ ) + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ ) + 2 ρ , ξ + 2 ρ , ξ + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ ) + 2 ρ , ξ + 2 ρ , ξ + ξ∂ρ , = 0and we must add: R , ≡ ρ , ( ξ + ξ + ξ + ξ ) + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ + ξ + ξ ) + ξ∂ρ , = 0 R , ≡ ρ , ( ξ + ξ + ξ + ξ ) + ξ∂ρ , = 0These linear equations are not linearly independent because: ρ , + ρ , + ρ , = 0 ⇒ R , + R , + R , = 0Also, linearizing while using the Kronecker symbol δ , we get: ω ir ω kr = δ ki ⇒ Ω kl = − ω kr ω ls Ω rs Thus, introducing the Ricci tensor and linearizing, we get: ρ ij = ω rs ρ ri,sj = ω rs ρ ir,js = 0 ⇒ R ij = ω rs R ri,sj + ρ ik,jl Ω kl = ω rs R ir,js − ρ ik,jl ω kr ω ls Ω rs = ρ rj ξ ri + ρ ir ξ rj + ξ r ∂ r ρ ij = 0It follows that − R ij ≡ ω rs R ir,sj = 0 mod (Ω) and we have in particular: R ≡ ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + 2 ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + 2 ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + ω R , + ω R , = 0 mod (Ω)The first row proves that R , is a linear combination of R , and R , . Then, if we wantto solve the three other equations with respect to R , , R , and R , , the correspondingdeterminant is, up to sign: det ω ω ω ω ω ω = − ω ω ω = 017ccordingly, we only need to take into account R , , R , , R , , R , .Similarly, we also obtain: R ≡ ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + ω R , + ω ( R , + R , ) = 0 mod (Ω) R ≡ ω R , + ω R , + ω R , = 0 mod (Ω) R ≡ ω R , + ω R , + ω R , = 0 mod (Ω)where we have to set R , = 0 , R , = 0 ⇒ R , = 0.Hence, taking into account R = 0, we just need to use R , , R , and R , .However, using the previous lemma, we obtain the formal Lie derivative:2 det ( ω )( ξ + ξ + ξ + ξ ) + ξ∂det ( ω ) = 0and thus ξ∂ ( ρ , / ( p | det ( ω ) | )) = 0 with p | det ( ω ) | ) = r + a cos ( θ ).In addition, we have 2( ω ω )( ξ + ξ ) + ξ∂ ( ω ω ) = 0 and thus ξ∂ ( ρ , / ( ω ω )) = 0.We have also:2( ρ , ρ , )( ξ + ξ + ξ + ξ ) + ξ∂ ( ρ , ρ , ) = 0 ⇒ ξ∂ ( ρ , ρ , /det ( ω )) = 0The following invariants are obtained successively in a coherent way: | ρ , ρ , | = m r ( r − a c ) r + a c ) ⇒ | ρ , ρ , | / | det ( ω ) | = ( mr ( r − a c )2( r + a c ) ) ω ω = ( r + a c ) (1 − c )( r − mr + a ) ⇒ | ρ , | / ( ω ω ) = mr ( r − a c )2( r + a c ) However, as a ∈ K , then ρ , and ρ , can be both divided by a and we get the new invariant: ρ , /ρ , = 2 r − a c + 3 a r + a c These results are leading to ξ = 0 , ξ = 0 , thus to ξ = 0 , ξ = 0 and ξ + ξ = 0 aftersubstitution. In the case of the S-metric, only the first invariant can be used in order to find ξ = 0.Taking into account the previous result, we obtain the two equations: (cid:26) ρ , ( ξ + ξ ) + ρ , ξ + ρ , ξ = 0 ρ , ( ξ + ξ ) + ρ , ξ + ρ , ξ = 0Using the fact that we have now: ω ξ + ω ξ = 0 ⇔ ω ξ + ω ξ = 0we may multiply the first equation by ω , the second by ω and sum in order to obtain:( ω ρ , + ω ρ , ) ξ + ( ω ρ , + ω ρ , ) ξ = 0Using the previous identity for R , we obtain therefore: ω ρ , ξ + ω ρ , ξ = 0 ⇒ ω ξ + ω ξ = 0 ⇔ ω ξ − ω ξ = 0Taking into account the fact that ξ = ω ω ξ , ξ = − ω ω ξ and substituting, we finally obtain:( ω ω − ( ω ) ) ξ = 0 ⇒ ξ = 0 , ξ = 0 ⇔ ξ = 0 , ξ = 0 , ξ = 0 , ξ = 0A similar procedure could have been followed by using R , = 0 , R , = 0 and ρ = 0.18ow, we must distinguish among the 20 components of the Riemann tensor along with thefollowing tabular where we have to take into account the identity ρ , + ρ , + ρ , = 0 : ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , ρ , In this tabular, the vanishing components obtained by computer algebra are put in a box, thenonzero components of the left column do not vanish when a = 0 and the other components vanishwhen a = 0. Also, the 11 ( care ) lower components can be known from the 10 upper ones.Keeping in mind the study of the S-metric and the fact that ρ , = 0 , ρ , = 0 , ρ , =0 , ρ , = 0 while framing the leading terms not vanishing when a = 0, we get: R , ≡ ρ , ξ + ρ , ξ + ( ρ , + ρ , ) ξ + ρ , ξ + ρ , ξ = 0Then, taking into account the fact that ρ , = 0 , ρ , = 0 , ρ , = 0, we obtain similarly: R , ≡ ( ρ , + ρ , ) ξ + ρ , ξ + ρ , ξ + + ρ , ξ + ρ , ξ = 0The leading determinant does not vanish when a = 0 because, in this case, all terms are van-ishing and we are left with the two linearly independent framed terms, a result amounting to ξ = 0 ⇔ ξ = 0 and ξ = 0 ⇔ ξ = 0 in the case of the S-metric in ([19]).In the case of the K-metric, we may use the relations already framed in order to keep only the fourparametric jets ( ξ , ξ , ξ , ξ ) on the right side . We may also rewrite them as follows: (cid:26) ω ξ + ω ξ + ω ξ = 0 , ω ξ + ω ξ + ω ξ = 0 ω ξ + ω ξ + ω ξ = 0 , ω ξ + ω ξ + ω ξ = 0if we use the fact that ω = − ω / ( ω ω − ( ω ) ) in the inverse metric.As a byproduct, we are now left with the two (complicate) equations ξ + a ( ... ) = 0 and ξ + a ( ... ) = 0 where the dots means linear combinations of ( ξ , ξ ) with coefficients in K and thestudy of the Killing operator is quite more difficult in the case of the K-metric. Of course, it be-comes clear that the use of the formal theory is absolutely necessary as an intrinsic approach couldnot be achieved if one uses solutions instead of sections . Indeed the strict inclusion R ′ = R (2)1 ⊂ R cannot be even imagined if one does believe that ξ = 0 , ξ = 0 brings ξ = 0 and ξ = 0. Theprevious computation could have also be done with R , = 0 and R , = 0 because R = 0and R = 0.The next hard step will be to prove that the other linearized components of the Riemann tensordo not produce any new different first order equation. The main idea will be to revisit the newlinearized tabular with: R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , R , + R , + R , = 0 ω R , + ω R , − ( ω /ω )( ω R , + ω R , ) = 0 mod (Ω) ω R , + ω R , + ω R , = 0 mod (Ω) ω R , + 2 ω R , + ω R , + ω R , = 0 mod (Ω)and so on, allowing to compute the 11 ( care ) lower terms from the 2 + 4 + 4 = 10 upper ones.We have thus the following successive eleven logical inter-relations:( R , , R , ) → R , ( R , , R , , R , ) ( R ,R ,R ,R ,R ) −→ ( R , , R , , R , , R , , R , )( R , , R , , R , ) R −→ R , ( R , , R , ) R −→ R , ( R , , R , ) R −→ R , ( R , , R , ) R −→ R , ( R , , R , ) R −→ R , Keeping in mind the four additional equations and their consequences that have been alreadyframed, both with the vanishing components of the Riemann tensor, namely: ρ , = 0 , ρ , = 0 , ρ , = 0 , ρ , = 0 , ρ , = 0 , ρ , = 0 , ρ , = 0 , ρ , = 0we get successively: R , = 0 , R , = 0 , R , = 0 , R , = 0 , R , = 0 , R , = 0As we have already exhibited an isomorphism ( ξ , ξ , ξ , ξ ) → ( ξ , ξ , ξ , ξ ), we may use only thelater right set of parametric jet components. Using the previous logical relations while framingthe leading terms not vanishing a priori when a = 0, there is only one possibility to choose fourcomponents of the linearized Riemann tensor , namely: R , ≡ ( ρ , ξ + ρ , ξ ) + ( ρ , + ρ , ) ξ + ρ , ξ + ρ , ξ = 0 R , ≡ ( ρ , + ρ , ) ξ + ( ρ , ξ + ρ , ξ ) + ρ , ξ + ρ , ξ = 0 R , ≡ ρ , ξ + ρ , ξ + ( ρ , ξ + ρ , ξ ) + ( ρ , + ρ , ) ξ = 0 R , ≡ ρ , ξ + ρ , ξ + ( ρ , + ρ , ) ξ + ( ρ , ξ + ρ , ξ ) = 0In order to understand the difficulty of the computations involved, we propose to the reader, as anexercise, to prove ” directly ” that the two following relations: R , ≡ ( ρ , ξ + ρ , ξ ) + ( ρ , + ρ , ) ξ + ρ , ξ + ρ , ξ = 0 R , ≡ ( ρ , ξ + ρ , ξ ) + ( ρ , + ρ , ) ξ + ρ , ξ + ρ , ξ = 0are only linear combinations of the previous ones mod (Ω).20e are facing two technical problems ” spoilting ”, in our opinion , the use of the K metric: • With ω − in place of ω , we have ω ξ = − ω ξ + ... and the leading term of R , becomesproportional to ( ω ρ , − ω ρ , ) ξ + ... with a wrong sign indeed that cannot allow to use R . A similar comment is valid for the four successive leading terms. • In addition, we also discover the summation ρ , + ρ , in R , with a wrong sign indeedthat cannot allow to introduce ρ , as one could hope. A similar comment is valid for the foursuccessive summations.Nevertheless, we obtain the following unexpected formal linearized result that will be used ina crucial intrinsic way for finding out the generating second order and third order CC: THEOREM 4.2 : The rank of the previous system with respect to the four jet coodinates( ξ , ξ , ξ , ξ ) is equal to 2, for both the S and K metrics. We obtain in particular the two strikingidentities: R , + a (1 − c ) R , = 0 , R , + a ( r + a ) R , Proof : In the case of he S metric with a = 0, only the framed terms may not vanish and, denotingby ” ∼ ” a linear proportionality, we have already obtained mod ( j (Ω)): R , ∼ ξ , R , ∼ ξ , R , = 0 , R , = 0Hence, the rank of the system with respect to the 4 parametric jets ( ξ , ξ , ξ , ξ ) just drops to 2and this fact confirms the existence of the 5 additional first order equations obtained, as we saw,after two prolongations.In the case of the K metric with a = 0, the study is much more delicate.With a = 1, the coefficients of the 4 × a a a a a aa a a a a a a a Indeed, we have successively for the common factor − a (1 − c ): ( Row ξ → ρ , − ω ω ρ , = − mr ( r − a c )2( r + a c ) Row ξ → ρ , = amr (1 − c )( r − a c )2( r + a c ) ( Row ξ → ρ , + ρ , = amc ( r + a )(3 r − a c )2( r + a c ) Row ξ → ρ , = − a mc (1 − c )( r + a )(3 r − a c )2( r + a c ) ( Row ξ → ρ , − ω ω ρ , = amr (1 − c )( r − a c )2( r + a c ) Row ξ → ρ , − ω ω ρ , = − a mr (1 − c ) ( r − a c )2( r + a c ) ( Row ξ → ρ , = − a mc (3 r − a c )2( r + a c ) Row ξ → ρ , + ρ , = a mc (1 − c )(3 r − a c )2( r + a c ) and similarly for the common factor − a ( r + a ) :21 Row ξ → ρ , + ρ , = − amc ( r + a )(3 r − a c )2( r + a c ) Row ξ → ρ , = − a mc (3 r − a c )2( r + a c ) ( Row ξ → ρ , − ω ω ρ , = mr ( r + a ) ( r − a c )2( r + a c ) Row ξ → ρ , − ω ω ρ , = − amr ( r + a )( r − a c )2( r + a c ) ( Row ξ → ρ , = − a mc (1 − c )( r + a )(3 r − a c )2( r + a c ) Row ξ → ρ , + ρ , = a mc (1 − c )(3 r − a c )2( r + a c ) ( Row ξ → ρ , − ω ω ρ , = − amr ( r + a )( r − a c )2( r + a c ) Row ξ → ρ , − ω ω ρ , = a mr ( r − a c )2( r + a c ) We do not believe that such a purely computational mathematical result, though striking itmay look like , could have any useful physical application and this comment will be strengthenedby the next theorem provided at the end of this section. Q.E.D.
COROLLARY 4.3 : The Killing operator for the K metric has 14 generating second order CC.
Proof : According to the previous theorem, we have dim ( R (2)1 ) = dim ( R ) = 4 as we can choosethe 4 parametric jets ( ξ , ξ , ξ , ξ ) and g = 0. Using the introductory diagram with n = 4 , q =1 , r = 2 , E = T and thus dim ( J ( F )) − dim ( J ( T )) = 150 −
140 = 10, we obtain at once dim ( Q ) = 10 + dim ( R (2)1 ) = 14 in a purely intrinsic way. We may thus start afresh with the newfirst order system R ′ = R (2)1 ⊂ R ⊂ J ( T ) obtained from R after 2 prolongations.Q.E.D.Finally, we know from ([7-9],[16],[18],[19]) that if R q ⊂ J q ( T ) is a system of infinitesimal Lieequations, then we have the algebroid bracket and its link with the prolongation/projection (PP)procedure: [ R q , R q ] ⊂ R q ⇒ [ R ( s ) q + r , R ( s ) q + r ] ⊂ R ( s ) q + r , ∀ q, r, s ≥ R ′ = R (2)1 = π ( R ) is such that [ R ′ , R ′ ] ⊂ R ′ with dim ( R ′ ) = 20 −
16 = 4 becausewe have obtained a total of 6 new different first order equations. Using the first general diagramof the Introduction, we discover that the operator defining R has 10 + 4 = 14 CC of order 2, aresult obtained totally independently of any specific GR technical object like the Teukolski scalars or the
Killing-Yano tensors introduced in ([1-6]).It remains to make one more prolongation in order to study R ”1 = R (3)1 = π ( R ) ⊂ R ′ ⊂ R with strict inclusions in order to sudy the third order CC for Ω already described for theSchwarzschild metric in ([19]).We have on sections ( care ) the 16 (linear) equations of R ′ as follows: R ′ = R (2)1 ⊂ J ( T ) ξ = 0 , ξ = 0 ⇒ ω ξ + ω ξ + ω ξ = 0 , ξ = 0 , ξ = 0 ξ = 0 ⇒ ξ = 0 ξ + ... = 0 ⇒ ω ξ + ω ξ + ω ξ = 0 ξ + ... = 0 ⇒ ω ξ + ω ξ + ω ξ = 0 ,ω ξ + ω ξ + ω ξ = 0 ξ = 0 ⇒ ξ = 0 , ξ = 0 , ξ = 0and we may choose only the 2 parametric jets ( ξ , ξ ) among ( ξ , ξ , ξ , ξ ) to which we must add( ξ , ξ ) in any case as they are not appearing in the Killing equations and their prolongations.The system is not involutive because it is finite type with g ′ = 0 and g ′ cannot be thus involutive.22aking therefore into account that the metric only depends on ( x = r, x = cos ( θ )) we obtain after three prolongations the first order system: R ”1 ⊂ R ′ ⊂ R ⊂ J ( T ) ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 ξ = 0 0 1 2 30 1 2 30 1 2 30 1 2 30 1 2 • • • • • • • • • • • • • • • • • • • • • • • •• • • •• • • • Surprisingly and contrary to the situation found for the S metric , we have now a trivially involutivefirst order system with only solutions ( ξ = cst, ξ = 0 , ξ = 0 , ξ = cst ). However, the difficultyis to know what second members must be used along the procedure met for all the motivatingexamples. In particular, we have again identities to zero like d ξ − ξ = 0 , d ξ − ξ = 0 or, equivalently , d ξ − ξ = 0 , d ξ − ξ = 0 and thus 4 third order CC coming from the 4 followingcomponents of the Spencer operator: d ξ − ξ = 0 , d ξ − ξ = 0 , d ξ − ξ = 0 , d ξ − ξ = 0a result that cannot be even imagined from ([1-6]). Of course, proceeding like in the motivatingexamples, we must substitute in the right members the values obtained from j (Ω) and set forexample ξ = − ω ξ∂ω while replacing ξ and ξ by the corresponding linear combinations ofthe Riemann tensor already obtained for the right members of the two zero order equations.Using one more prolongation, all the sections ( care again ) vanish but ξ and ξ , a result leadingto dim ( R ”1 ) = 2 in a coherent way with the only nonzero Killing vectors { ∂ t , ∂ φ } . We have indeed: ξ = 0 ⇒ ξ = 0 ⇒ ξ = 0 , ξ = 0 ⇒ ξ = 0 ⇒ ξ = 0Like in the case of the S metric, R is not involutive but R is involutive. However, contraryto the S metric with g ”1 = 0, now g ”1 = 0 for the K metric and R ”1 is trivially involutive with a fullJanet tabular having 16 rows of first order jets and 2 rows of zero order jets. REMARK 4.4 : We have in general ([7],[10] p 339, 345): R ( s ) q + r = π q + r + sq + r ( R q + r + s ) = π q + r + sq + r ( J r ( R q + s ) ∩ J q + r + s ( E )) ⊆ J r ( π q + sq )(( R q + s ) ∩ J q + r ( E ) = J r ( R ( s ) q ) ∩ J q + r ( E ) = ρ r ( R ( s ) q )that is, in our case R (2)2 ⊆ ρ ( R (2)1 ). However, we have indeed the equality R (2)2 = ρ ( R (2)1 ) evenif the conditions of Theorem 1.1 are not satisfied because g ′ is not 2-acyclic. Indeed, the Spencermap δ : ∧ T ∗ ⊗ g ′ → ∧ T ∗ ⊗ T is not injective and we let the reader check as an exercise that itskernel is generated by { v , , v , } and the Spencer δ -cohomology is such that dim ( H ( g ′ ) = 2 = 0because the cocycles are defined by the equations v ki,jr + v kj,ri + v kr,ij = 0. Hence, contrary to whatcould be imagined , the major difference between the S and K metrics is not at all the existence ofoff-diagonal terms but rather the fact that R ”1 is not involutive with g ”1 = 0 for the S metric while23 ”1 is involutive with g ”1 = 0 for the K metric. This is the reason for which one among the fourthird order CC must be added with two prolongations for the S metric while the four third order CCare obtained in the same way from the Spencer operator for the K metric. Of course no classicalapproach can explain this fact which is lacking in ([1-4]).The following result even questions the usefulness of the whole previous approach: THEOREM 4.5 : The operator
Cauchy = ad ( Killing ) admits a minimum parametrization bythe operator
Airy = ad ( Riemann ) with 1 potential when n = 2, found in 1863. It admits a canon-ical self-adjoint parametrization by the operator Beltrami = ad ( Riemann ) with 6 potentials when n = 3, found in 1892 and modified to a mimimum parametrization by the operator M axwell with3 potentials, found in 1870. More generally, it admits a canonical parametrization by the operator ad ( Riemann ) with n ( n − /
12 potentials that can be modified to a relative parametrization by ad ( Ricci ) with n ( n + 1) / n ≥
4, found in2007. In all these cases, the corresponding potentials have nothing to do with the perturbation ofthe metric . Such a result is also valid for any Lie group of transformations, in particular for theconformal group in arbitrary dimension.
Proof : We provide successively the explicit corresponding parametrizations: • n = 2 : Multiplying the linearized Riemann operator by a test function φ and integrating byparts, we obtain ( care to the factor 2 involved ): φ ( d Ω − d Ω + d Ω ) = ( d φ Ω − d φ Ω + d φ Ω ) + div ( ... ) σ ij = σ ji ⇒ σ ij Ω ij = σ Ω + 2 σ Ω + σ Ω Cauchy operator d σ + d σ = f , d σ + d σ = f Airy operator σ = d φ, σ = σ = − d φ, σ = d φξ −→ Ω −→ R −→ Killing −→ Riemann −→ −→ ←− Cauchy ←− Airy ←− ←− f ←− σ ←− φ • n=3 We now present the original Beltrami parametrization: σ σ σ σ σ σ = d − d d − d d d − d d − d − d d d − d d − d d d − d d − d d φ φ φ φ φ φ which does not seem to be self-adjoint but is such that d r σ ir = 0. Accordingly, the Beltrami parametrization of the
Cauchy operator for the stress is nothing else than the formal adjoint ofthe
Riemann operator. However, modifying slightly the rows, we get the new operator matrix:24 σ σ σ σ σ σ = d − d d − d d d − d d − d − d d d − d d − d d d − d d − d d φ φ φ φ φ φ which is indeed self-adjoint . Keeping ( φ = A, φ = B, Φ = C ) with ( φ = 0 , φ = 0 , φ = 0),we obtain the M axwell parametrization: σ σ σ σ σ σ = d d − d − d d d − d d d ABC which is minimum because n ( n − / it is not evident at all to prove that it is also involutive as we must look for δ -regular coordinates (See [15] for the technical details). • n ≥ we do believe that it is not possible to avoid using differ-ential homological algebra , in particular extension modules . As we found it already in many books([8],[9],[16],[18]) or papers ([12],[13],[21-23]), the linear Spencer sequence is (locally) isomorphicto the tensor product of a Poincar´e type sequence for the exterior derivative by a Lie algebra G with dim ( G ) ≤ n ( n + 1) / n = 4, this dimension is 10 for the M-metric, 4 for the S-metric and 2 for theK-metric. As a byproduct, the adjoint sequence roughly just exchanges the exterior derivatives upto sign and one has for example, when n = 3, the relations ad ( grad ) = − div, ad (( div ) = − grad .It follows that, if D generates the CC of D , then ad ( D ) is parametrizing ad ( D ), a fact notevident at all , even when n = 2 for the Cosserat couple-stress equations exactly described by ad ( D )([12]). Passing to the differential modules point of view with the ring (even an integral domain) D = K [ d , ..., d n ] = K [ d ] of differential operators with coefficients in a differential field K , thisresult amounts to say that ext D ( M, D ) = ext ( M ) = 0. As it is known that such a result does notdepend on the differential resolution used or, equivalently , on the differential sequence used, if D generates the CC of D in the Janet sequence, then ad ( D ) is parametrizing ad ( D ) and this result isstill true even if D is not involutive. In such a situation, which is the one considered in this paper,the Killing operators for the M-metric, the S-metric and the K-metric are such that, whatever arethe generating CC D (second order for the M-metric, a mixture of second and third order for theS-metric and K-metric), then ad ( D ) is, in any case , parametrizing the Cauchy operator ad ( D ) for any D : T → S T ∗ : ξ → L ( ξ ) ω . Once more, the central object is the group, not the metric . Thesame results are also valid for any Lie group of transformations, in particular for the conformalgroup in arbitrary dimension, even if the operator D is of order 3 when n = 3 as we shall seebelow ([16],[20-23]). Q.E.D. REMARK 4.6 : Accordingly, the situation met today in GR cannot evolve as long as peoplewill not acknowledge the fact that the components of the Weyl tensor are similarly playing thepart of torsion elements (the so-called
Lichnerowicz waves in [17]) for the equations
Ricci = 0, aresult only depending on the group structure of the conformal group of space-time that brings thecanonical splitting
Riemann = W eyl ⊕ Ricci without any reference to a backgroung metric as itis usually done ([8],[9],[15],[18],[21-23]). It is an open problem to know why one may sometimesfind a SELF-ADJOINT OPERATOR. It is such a confusion that led to introduce the so-called
Einstein parametrizing operator ([17]). 25
XAMPLE 4.7 : (
Weyl tensor for n=3 and euclidean metric ) We proved in ([16], p 156-158) andmore recently in ([15],[22],[23]) that, for n = 3, the natural ”geometric object ” corresponding tothe Weyl tensor is no longer described by a second order differential operator but by a third orderdifferential operator ˆ D with first order CC ˆ D in the differential sequence:0 −→ ˆΘ −→ ˆ D −→ ˆ D −→ ˆ D −→ −→ D -modules:0 −→ D D −→ D D −→ D D −→ D p −→ ˆ M −→ p is the canonical residual projection. The true reason is that the symbol ˆ g of ˆ D is finitetype with second prolongation ˆ g = 0 while its first prolongation ˆ g is not in any case the identities ˆ D ◦ ˆ D = 0and ˆ D ◦ ˆ D = 0.Of course, these operators can be obtained by using computer algebra like in ([16], Appendix 2)but one may check at once that ˆ D and ˆ D are completely different operators while the operator ˆ D is far from being self-adjoint even though it is described by a 5 × × curl operator in 3-dimensional classical geometry because ad ( grad ) = − div .It does not seem that these results are known today.The starting point is the 3 × D , namely: d − d − d d d d d − d d − d d d Substracting the fourth row from the first row and mutiplying the fourth row by , we obtain theoperator matrix: d − d d d d d − d d − d d d Adding the fourth row to the first, we obtain the operator matrix: d − d d d d d − d d − d d d Adding the first row to the fourth row and dividing by 2, we obtain the operator matrix: d − d d d d d d − d d d Multiplying the second, fourth and fifth row by −
1, then multilying the central column of thematrix thus obtained by −
1, we finally obtain the operator matrix D ′ :26 d − d − d d d d d d d − d We now care about transforming ˆ D given in ([16], p 158) by the 5 × − d d − d − d d − d d d − d d Dividing the first column by 2 and the fourth column by −
2, then using the central row as a newtop row while using the former top row as new bottom row, we obtain the operator matrix D ′ : d − d d d d d − d d d − d and check that ad ( ˆ D ′ ) = − ˆ D ′ like in the Poincar´e sequence for n = 3 where ad ( div ) = − grad .As the new corresponding operator ˆ D ′ is homogeneous and of order 3 ( care ), we obtain locally ad ( ˆ D ′ ) = ˆ D ′ , a result not evident at first sight (Compare to [16], p 157).The combination of this example with the results announced in ([22]) brings the need to revisitalmost entirely the whole conformal geometry in arbitrary dimension and we notice the essentialrole performed by the Spencer δ -cohomology in this new framework.
5) CONCLUSION
First of all, we may summarize the results previously obtained by saying that ”