4. ApplicationsConsidering the classical Killing operator D : ξ → L ( ξ ) ω = Ω ∈ S 2 T * = F 0 where L ( ξ ) is the Lie derivative with respect to ξ and ω ∈ S 2 T * is a nondegenerate metric with d e t ( ω ) ≠ 0 . Accordingly, it is a lie operator with D ξ = 0 , D η = 0 ⇒ D [ ξ , η ] = 0 and we denote simply by Θ ⊂ T the set of solutions with [ Θ , Θ ] ⊂ Θ . Now, as we have explained many times, the main problem is to describe the CC of D ξ = Ω ∈ F 0 in the form D 1 Ω = 0 by introducing the so-called Riemann operator D 1 : F 0 → F 1 . We advise the reader to follow closely the next lines and to imagine why it will not be possible to repeat them for studying the conformal Killing operator. Introducing the well known Levi-Civita isomorphism j 1 ( ω ) = ( ω , ∂ x ω ) ≃ ( ω , γ ) by defining the Christoffel symbols γ i j k = 1 2 ω k r ( ∂ i ω r j + ∂ j ω i r − ∂ r ω i j ) where ( ω r s ) is the inverse matrix of ( ω i j ) and the formal Lie derivative, we get the second order system R 2 ⊂ J 2 ( T ) :
{ Ω i j ≡ ( L ( ξ 1 ) ω ) i j = ω r j ( x ) ξ i r + ω i r ( x ) ξ j r + ξ r ∂ r ω i j ( x ) = 0 Γ i j k ≡ ( L ( ξ 2 ) γ ) i j k = ξ i j k + γ r j k ( x ) ξ i r + γ i r k ( x ) ξ j r + γ i r k ( x ) ξ j r − γ i j r ( x ) ξ r k + ξ r ∂ r γ i j k ( x ) = 0
with sections ξ 2 : x → ( ξ k ( x ) , ξ i k ( x ) , ξ i j k ( x ) ) transforming like j 2 ( ξ ) : x → ( ξ k ( x ) , ∂ i ξ k ( x ) , ∂ i j ξ k ( x ) ) . The system R 1 ⊂ J 1 ( T ) has a symbol g 1 ≃ ∧ 2 T * ⊂ T * ⊗ T depending only on ω with d i m ( g 1 ) = n ( n − 1 ) / 2 and is finite type because its first prolongation is g 2 = 0 . It cannot be thus involutive and we need to use one additional prolongation. Indeed, using one of the main results to be found in ( [1] [5] [9] [10] [14] ), we know that, when R 1 is FI, then the CC of D are of order s + 1 where s is the number of prolongations needed in order to get a 2-acyclic symbol, that is s = 1 in the present situation, a result that should lead to CC of order 2 if R 1 were FI. However, it is known that R 2 is FI, thus involutive, if and only if ω has constant Riemannian curvature, a result first found by L.P. Eisenhart in 1926 which is only a particular example of the Vessiot structure equations discovered b E. Vessiot in 1903 ( [27] ), though in a quite different setting (See [1] [5] [9] [14] for an explicit modern proof and compare to the references ( [22] [23] ) of ( [6] ).
We may introduce the (formal) linearization Γ ∈ S 2 T * ⊗ T of the Christoffel symbols by linearizing the relations ω k r γ i j k = 1 2 ( ∂ i ω r j + ∂ j ω i r − ∂ r ω i j ) in such a way that Ω = 0 ⇒ Γ = 0 with:
ω k r Γ i j k = 1 2 ( d i Ω r j + d j Ω i r − d r Ω i j ) − γ i j k Ω k r
We may also introduce the Riemann tensor ρ l , i j k and its (formal) linearization:
R l , i j k ≡ ( L ( ξ 1 ) ρ ) l , i j k = − ρ l , i j s ξ s k + ρ s , i j k ξ l s + ρ l , s j k ξ i s + ρ l , i s k ξ j s + ξ r ∂ r ρ l , i j k
in order to obtain the Ricci tensor ρ i j = ρ i , r j r = ρ j i and its linearization:
R i j = R i , r j r = ρ r j ξ i r + ρ i r ξ j r + ξ r ∂ r ρ i j = R j i
allowing to introduce the Einstein tensor ϵ i j = ρ i j − 1 2 ω i j ω r s ρ r s = ρ i j − 1 2 ω i j ρ with linearization:
E i j ≡ ( L ( ξ 1 ) ϵ ) i j = R i j − 1 2 ω i j ω r s R r s − 1 2 ρ Ω i j + 1 2 ω i j ω r u ω s v ρ r s Ω u v
and we must notice (care) that the linearization of ρ = ω r s ρ r s is R = ω r s R r s − ρ r s ω r u ω s v Ω u v .
These formulas become particularly simple when ω is a solution of Einstein equations in vacuum, that is when ε i j = 0 ⇔ ρ i j = 0 ⇒ ρ = 0 .
LEMMA 4.1: When n = 4 and the fixed euclidean metric for simplicity, we have the useful formula:
E 00 = − ( R 12,12 + R 13,13 + R 23,23 )
Proof: We have 2 E 00 = 2 R 00 − ( R 00 + R 11 + R 22 + R 33 ) = R 00 − ( R 11 + R 22 + R 33 )
and R 00 = ( R 10,10 + R 20,20 + R 30,30 ) . However, we have also:
R 11 = R 01,01 + R 21,21 + R 31,31 R 22 = R 02,02 + R 12,12 + R 32,32 R 33 = R 03,03 + R 13,13 + R 23,23
Summing, we obtain ( R 11 + R 22 + R 33 ) = ( R 01,01 + R 02,02 + R 03,03 ) + 2 ( R 12,12 + R 13,13 + R 23,23 ) . It follows that E 00 = − ( R 12,12 + R 13,13 + R 23,23 ) and the three other E i i are obtained by circular permutations of ( 0,1,2,3 ) . We let the reader treat the general situation as an exercise.
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A) MINKOWSKI METRIC:
We have considered this situation in many books or papers and refer the reader to our arXiv page or to the recent references ( [22] [28] ). All the operators are first order between the vector bundles E = T , F 0 = T * ⊗ T / g 1 ≃ S 2 T * , F 1 = H 2 ( g 1 ) , F 2 = H 3 ( g 1 ) , F 3 = H 4 ( g 1 ) that are only depending on g 1 with dimensions 4, 10, 20, 20, 6 when n = 4 and Euler-Poincaré characteristic r k D ( M ) = 4 − 10 + 20 − 20 + 6 = 0 . The case of an arbitrary n, provided in ( [20] ), depends on various chases in commutative diagrams that will be exhibited later on for comparing the respective dimensions. This is not a Janet sequence because R 1 is FI but g 1 is not involutive.
B) SCHWARZSCHILD METRIC:
With the standard Boyer-Lindquist local coordinates ( x 0 = t , x 1 = r , x 2 = θ , x 3 = ϕ ) and a constant parameter m, we may introduce the field of constants k = ℚ ( m ) and all the systems or differential modules considered in the sequel will be defined over the ground differential field K = k ( t , r , sin ( θ ) , cos ( θ ) ) with differential structure obtained by setting ∂ 2 sin ( θ ) = cos ( θ ) , ∂ 2 cos ( θ ) = − sin ( θ ) together with sin 2 ( θ ) + cos 2 ( θ ) = 1 instead of using the so-called “rational coordinates” ( [23] ). With speed of light c = 1 and A = 1 − m r , we shall introduce the diagonal Schwarzschild metric ω = ( A ( r ) , − 1 / A ( r ) , − r 2 , − r 2 sin 2 ( θ ) ) with d e t ( ω ) = − r 4 sin 2 ( θ ) . Following closely the motivating examples already presented, our challenge is to prove that the purely mathematical formal study of the corresponding Killing system R 1 ⊂ J 1 ( T ) can be achieved as a simple exercise of formal integrability, with no extra physical technical tool, contrary to ( [6] [7] [8] ). As the computations will be explicitly done, the numbers of CC obtained will bring serious doubts about the validity of the results obtained in the above references, later confirmed with the K metric. First of all we obtain easily the following 10 first order Killing equations m o d ( Ω ) :
R 1 ⊂ J 1 ( T ) { Ω 33 → ξ 3 3 + 1 r ξ 1 + cot ( θ ) ξ 2 = 0 Ω 23 → ξ 3 2 + sin 2 ( θ ) ξ 2 3 = 0 Ω 13 → ξ 3 1 + A r 2 sin 2 ( θ ) ξ 1 3 = 0 Ω 03 → ξ 3 0 − r 2 A sin 2 ( θ ) ξ 0 3 = 0 Ω 22 → ξ 2 2 + 1 r ξ 1 = 0 Ω 12 → ξ 2 1 + A r 2 ξ 1 2 = 0 Ω 02 → ξ 2 0 − r 2 A ξ 0 2 = 0 Ω 11 → ξ 1 1 − m 2 A r 2 ξ 1 = 0 Ω 01 → ξ 0 1 − A 2 ξ 1 0 = 0 Ω 00 → ξ 0 0 + m 2 A r 2 ξ 1 = 0
where we have framed the leading jets.
This is a finite type system because we get Γ i j k ≡ ξ i j k + ⋯ = 0 with only one prolongation!
The only 9 non-zero Christoffel symbols on 40 are:
γ 00 1 = m A 2 r 2 , γ 01 0 = m 2 A r 2 , γ 11 1 = − m 2 A r 2 γ 12 2 = 1 r , γ 13 3 = 1 r , γ 22 1 = − A r , γ 23 3 = cot ( θ ) , γ 33 1 = − A r sin 2 ( θ ) , γ 33 2 = − sin ( θ ) cos ( θ )
We obtain for example:
Γ 22 1 ≡ ξ 22 1 + ( 1 − 3 m 2 r ) ξ 1 = 0 , Γ 33 1 ≡ ξ 33 1 + sin ( θ ) cos ( θ ) ξ 2 1 + ( 1 − 3 m 2 r ) sin 2 ( θ ) ξ 1 = 0
⇒ Γ 33 1 − sin 2 ( θ ) Γ 22 1 ≡ ξ 33 1 + sin ( θ ) cos ( θ ) ξ 2 1 − sin 2 ( θ ) ξ 22 1 = 0
after only one prolongation (care).
Then, using r as a summation index, we shall see that we have in general for the linearization of the Riemann and Ricci tensors:
R k l , i j ≡ ρ r l , i j ξ k r + ρ k r , i j ξ l r + ρ k l , r j ξ i r + ρ k l , i r ξ j r + ξ r ∂ r ρ k l , i j ≠ 0
R i j ≡ ρ r j ξ i r + ρ i r ξ j r + ξ r ∂ r ρ i j ≠ 0
The only 6 non-zero components of the Riemann tensor are:
ρ 01 , 01 = + m r 3 , ρ 02 , 02 = − m A 2 r , ρ 03 , 03 = − m A sin 2 ( θ ) 2 r ρ 12 , 12 = + m 2 A r , ρ 13 , 13 = + m sin 2 ( θ ) 2 A r , ρ 23 , 23 = − m r sin 2 ( θ )
but we must not forget hat we have indeed ρ i j = 0 for the 10 components of the Ricci tensor, in particular ρ i i = 0 for the diagonal components with i = 0 , 1 , 2 , 3 . We have in particular:
ρ 22 = ρ 2 , 02 0 + ρ 2 , 12 1 + ρ 2 , 32 3 = 1 A ρ 02 , 02 − A ρ 12 , 12 − 1 r 2 sin 2 ( θ ) ρ 23 , 23 = − m 2 r − m 2 r + m r = 0
We also obtain m o d ( Ω ) :
R 01 , 01 ≡ 2 ρ 01 , 01 ( ξ 0 0 + ξ 1 1 ) + ξ r ∂ r ( ρ 01 , 01 ) = ξ 1 ∂ 1 ρ 01 , 01 ⇒ R 01 , 01 ≡ − 3 m r 4 ξ 1 = 0 ⇒ ξ 1 = 0
and similarly:
R 01 , 02 ≡ ρ 01 , 01 ξ 2 1 + ρ 02 , 02 ξ 1 2 + ξ r ∂ r ρ 01 , 02 = 3 m 2 r 3 ξ 2 1 = 0 ⇒ ξ 2 1 = 0
R 02 , 12 ≡ ρ 12 , 12 ξ 0 1 + ρ 02 , 02 ξ 1 0 + ξ r ∂ r ρ 02 , 12 = m 2 r A ( ξ 0 1 − A 2 ξ 1 0 ) = 0
R 23 , 12 ≡ ρ 21 , 12 ξ 3 1 + ρ 23 , 32 ξ 1 3 + ξ r ∂ r ρ 23 , 12 = − m 2 r A ξ 3 1 + m r sin 2 ( θ ) ξ 1 3 = − 3 m 2 r A ξ 3 1 = 0
⇒ R 01 , 03 = 3 m 2 r 3 ξ 3 1 = − A r 2 R 23 , 12
ρ 01 , 23 = 0 ⇒ R 01 , 23 = ρ 01 , 23 ( ξ 0 0 + ξ 1 1 + ξ 2 2 + ξ 3 3 ) + ξ r ∂ r ρ 01 , 23 = 0
and so on, in order to avoid using computer algebra. However, the main consequence of this remark is to explain the existence of the 15 second order CC. Indeed, denoting by “~” a linear proportional dependence m o d ( Ω ) , we have the successive three cases:
( R 00 , R 11 , R 22 , R 33 ) R 01,01 ∼ R 02,02 ∼ R 03,03 ∼ R 12,12 ∼ R 13,13 ∼ R 23,23 → ξ 1 = 0 ( R 12 ) R 01,02 ∼ R 13,23 → ξ 2 1 = 0 ( R 13 ) R 01,03 ∼ R 12,23 → ξ 3 1 = 0 ( R 02 ) R 01,12 ∼ R 03,23 → ξ 2 0 = 0 ( R 03 ) R 01,13 ∼ R 02,23 → ξ 3 0 = 0 ( R 01 , R 23 ) R 01,23 → 0, R 02,03 → 0, R 02,12 → 0, R 02,13 → 0, R 03,13 → 0, R 12,13 → 0
as a way to obtain the 5 equalities to zero on the right and thus a total of 20 − 5 = 15 second order CC obtained by elimination. However, the present partition 15 = 5 + 4 + 6 is quite different from the partition 15 = 10 + 5 used by the authors quoted in the Introduction which is obtained by taking into account the vanishing assumption of the 10 components of the Ricci tensor. As such a result questions once more the mathematical foundations of general relativity, in particular the existence of gravitational waves, we provide a few additional technical comments.
The main point is a tricky formula which is not evident at all. Indeed, using the well known properties of the Lie derivative, we have the following geometric objects (not necessarily tensors) and their linearizations (generally tensors):
ω i j → Ω i j ∈ S 2 T * , γ i j k → Γ i j k ∈ S 2 T * ⊗ T ,
ρ k l , i j → R k l , i j ∈ ∧ 2 T * ⊗ T * ⊗ T , β k l , i j r → B k l , i j r ∈ ∧ 3 T * ⊗ T * ⊗ T
Then, using r as a summation index, we shall see that we have in general:
R k l , i j ≡ ρ r l , i j ξ k r + ρ k r , i j ξ l r + ρ k l , r j ξ i r + ρ k l , i r ξ j r + ξ r ∂ r ρ k l , i j ≠ 0
ρ k l , i j = ω k r ρ l , i j r ⇒ R k l , i j = ω k r R l , i j r + ρ l , i j r Ω k r ⇒ ω r s R r i , s j = R i j + ω r s ρ i , r j t Ω s t
ρ i j = ρ i , r j r ⇒ R i j = R i , r j r ≠ ω r s R r i , s j
We prove these results using local coordinates and the formal Lie derivative obtained while replacing j 1 ( ξ ) by ξ 1 (See [1] [5] [9] [14] for details). First of all, from the tensorial property of the Riemann tensor and the Killing equations Ω u s = ω k u ξ s k + ω k s ξ u k + ξ r ∂ r ω u s , we have:
R l , i j k ≡ ( L ( ξ 1 ) ρ ) l , i j k = − ρ l , i j s ξ s k + ρ s , i j k ξ l s + ρ l , s j k ξ i s + ρ l , i s k ξ j s + ξ r ∂ r ρ l , i j k
ω k u ( − ρ v , i j s ξ s k + ξ r ∂ r ρ v , i j s ) = ρ v , i j s ω k s ξ u k + ( ξ r ∂ r ω u s ) ρ v , i j s + ω k u ξ r ∂ r ρ v , i j k − ρ v , i j s Ω s u = ρ s v , i j ξ u s + ξ r ∂ r ρ u v , i j − ρ v , i j s Ω s u
and thus ω k u R v , i j k = R u v , i j − ρ v , i j s Ω s u .
ρ 1 , 11 1 = 0 ⇒ ρ 11 = ρ 1 , 01 0 + ρ 1 , 21 2 + ρ 1 , 31 3 = 1 A ρ 01 , 01 − 1 r 2 ρ 12 , 12 − 1 r 2 sin 2 ( θ ) ρ 13 , 13 = 0
We have for example, in this particular case:
ρ 1 , 01 0 = 1 A ρ 01 , 01 = m A r 3 ⇒ R 1 , 01 0 = ( L ( ξ ) ρ ) 1 , 01 0 = 2 ρ 1 , 01 0 ξ 1 1 + ξ 1 ∂ 1 ρ 1 , 01 0 = 2 m A r 3 ξ 1 1 + ξ 1 ∂ 1 ( m A r 3 )
The only use of R 01,01 is allowing to get ξ 1 = 0 in the previous list, but we have also exactly:
ω r s R r 1 , s 2 = 1 A R 01 , 02 − 1 r 2 sin 2 ( θ ) R 31 , 32 = − m 2 r 3 Ω 12 = R 12 + ω 11 ρ 1 , 12 2 Ω 12 ⇒ R 12 = 0
The use of R 01,02 or R 13,23 is allowing to get ξ 2 1 = 0 in the previous list with:
R 1 , 02 0 = − 3 m 2 r 3 ξ 1 , 2 + m 2 r 3 Ω 12 , R 1 , 32 3 = + 3 m 2 r 3 ξ 1 , 2 − m r 3 Ω 12
and thus also exactly:
ω r s ρ 1 , r 2 t Ω s t = − ω 11 ω 22 ρ 12 , 12 Ω 12 = − m 2 r 3 Ω 12 ⇒ R 12 = R 1 , 02 0 + R 1 , 32 3 + m 2 r 3 Ω 12 = 0
It follows that the 4 central second order CC of the list successively amounts to R 12 = 0 , R 13 = 0 , R 02 = 0 , R 03 = 0 , a result breaking the intrinsic/coordinate-free interpretation of the 10 Einstein equations and the situation is even worst for the other components of the Ricci tensor. Indeed, R 01 and R 23 only depend on the vanishing of R 02,12 , R 03,13 and R 02,03 , R 12,13 among the bottom CC of the list, while the diagonal terms R 00 , R 11 , R 22 , R 33 only depend, as we just saw, on the 6 non zero components of the Riemann tensor. We have thus obtained the totally unusual partition 10 = 4 + 4 + 2 along the successive blocks of the former list with:
{ R i j } = { R 00 , R 11 , R 22 , R 33 } + { R 12 , R 13 , R 02 , R 03 } + { R 01 , R 23 }
Finally, we notice that R 01 , 23 = 0 , R 02 , 31 = 0 ⇒ R 03 , 12 = 0 from the identity in ∧ 3 T * ⊗ T * :
R 01 , 23 + R 02 , 31 + R 03 , 12 = 0
and there is no way to have two identical indices in the first jets appearing through the (formal) Lie derivative just described. As for the third order CC, setting ξ 1 1 = A ′ 2 A ξ 1 ∈ j 2 ( Ω ) , we have at least the first prolongations of the previous second order CC to which we have to add the three new generating ones:
d 1 ξ 1 − ξ 1 1 = 0 , d 2 ξ 1 − ξ 2 1 = 0 , d 3 ξ 1 − ξ 3 1 = 0
provided by the Spencer operator, leading to the crossed terms d i ξ j 1 − d j ξ i 1 = 0 for i , j = 1 , 2 , 3 because the Spencer operator is not FI.
Setting now ξ 1 = U , ξ 2 1 = V 2 , ξ 3 1 = V 3 , ξ 2 0 = W 2 , ξ 3 0 = W 3 with ( U , V , W ) ∈ j 2 ( Ω ) , we have to look for the CC of the system R 1 ( 1 ) = R 1 already presented, then the system R 1 ( 2 ) with d i m ( R 1 ( 2 ) ) = 5 and finally R 1 ( 3 ) with d i m ( R 1 ( 3 ) ) = 4 which is formally integrable but not involutive because it is of finite type. Beside the only zero order equation ξ 1 = U , we have the following 15 first order ones:
ξ 0 0 = − A ′ 2 A U , ξ 1 0 = 1 A 2 d 0 U , ξ 2 0 = W 2 , ξ 3 0 = W 3 , ξ 0 1 = d 0 U , ξ 1 1 = A ′ 2 A U , ξ 2 1 = V 2 , ξ 3 1 = V 3 , ξ 0 2 = A r 2 W 2 , ξ 1 2 = − 1 A r 2 V 2 , ξ 0 3 = A r 2 sin 2 ( θ ) W 3 , ξ 1 3 = − 1 A r 2 sin 2 ( θ ) V 3 , ξ 2 2 = − 1 r U , ξ 3 2 + sin 2 ( θ ) ξ 2 3 = 0 , ξ 3 3 + cot ( θ ) ξ 2 = − 1 r U
Among the CC we must have d 2 V 3 − d 3 V 2 = 0 which is among the differential consequences of the Spencer operator as we saw but we must also have d 2 W 3 − d 3 W 2 = 0 and both seem to be new third order CC, together with the CC obtained by eliminating ξ 2 and ξ 3 from the three last equations after two prolongations as in ( [23] ):
d 3 V 3 + sin ( θ ) cos ( θ ) V 2 + sin 2 ( θ ) d 2 V 2 + 2 sin 2 ( θ ) U = 0
However, things are not so simple, even if we have in mind that ( V , W ) ∈ j 2 ( Ω ) , because the central sign in the previous formula is opposite to the sign found after one prolongation in the formula:
ξ 33 1 + sin ( θ ) cos ( θ ) ξ 2 1 − sin 2 ( θ ) ξ 22 1 = 0
and it is at this moment that we need introduce new differential geometric methods!
First of all, we have:
ρ l , i j k = ∂ i γ l j k − ∂ j γ l i k + γ l j r γ r i k − γ l i r γ r j k
and thus, because Γ ∈ S 2 T * ⊗ T is a tensor::
R l , i j k = d i Γ l j k − d j Γ l i k + γ l j r Γ r i k − γ l i r Γ r j k + γ r i k Γ l j r − γ r j k Γ l i r = ( d i Γ l j k − γ l i r Γ r j k + γ r i k Γ l j r ) − ( d j Γ l i k − γ l j r Γ r i k + γ r j k Γ l i r ) = ( d i Γ l j k − γ l i r Γ r j k − γ j i r Γ l r k + γ r i k Γ l j r ) − ( d j Γ l i k − γ l j r Γ r i k − γ i j r Γ l r k + γ r j k Γ l i r ) = ∇ i Γ l j k − ∇ j Γ l i k
by introducing the covariant derivative ∇ . We recall that ∇ r ω i j = 0 , ∀ r , i , j or, equivalently, that ( i d , − γ ) : ξ ∈ T → ( ξ k , ξ i k = − γ i r k ξ r ) ∈ R 1 is a R 1 -connection with ω s j γ i r s + ω i s γ j r s = ∂ r ω i j , a result allowing to move down the index k in the previous formulas (See [9] for more details).
We may thus take into account the Bianchi identities implied by the cyclic sums on ( i j r )
β k l , i j r ≡ ∇ r ρ k l , i j + ∇ i ρ k l , j r + ∇ j ρ k l , r i = 0 ⇔ β ≡ Σ c y c l ( ∂ ρ − γ ρ ) = 0
and their respective linearizations B k l , i j r = 0 as described below. We shall see later on that β and B are sections of the vector bundle F 2 defined by the short exact sequence:
0 → F 2 → ∧ 3 T * ⊗ g 1 → δ ∧ 4 T * ⊗ T → 0
with d i m ( F 2 ) = ( n ( n − 1 ) ( n − 2 ) / 6 ) ( n ( n − 1 ) / 2 ) − ( n ( n − 1 ) ( n − 2 ) ( n − 3 ) / 24 ) n = n 2 ( n 2 − 1 ) ( n − 2 ) / 24 because d i m ( g 1 ) = n ( n − 1 ) / 2 for any nondegenerate metric, that is 24 − 4 = 20 when n = 4 .
Such results cannot be even imagined by somebody not aware of the δ-acyclicity ( [1] [9] [13] ).
We have the linearized cyclic sums of covariant derivatives both with their respective symbolic descriptions, not to be confused with the non-linear corresponding ones:
B k l , r i j ≡ ∇ r R k l , i j + ∇ i R k l , j r + ∇ j R k l , r i = 0 m o d ( Γ ) ⇔ Σ c y c l ( d R − γ R − ρ Γ ) = 0 ⇔ B ≡ Σ c y c l ( ∇ R ) = Σ c y c l ( ρ Γ )
In order to recapitulate these new concepts obtained after one, two or three prolongations, we have successively ω → γ → ρ → β and the respective linearizations Ω → Γ → R → B .
The 24 Bianchi identities are related by the 4 linear relations like B 01 , 023 − B 02 , 013 + B 03 , 012 = 0 when n = 4 because B 00 , 123 = 0 . These relations are existinging between the 24 components of the Lanczos tensor because B ∈ F 2 ⊂ k e r ( δ ) in the previous short exact sequence ( [20] ).
With more details, we number the 20 linearly independent Bianchi identities as follows:
1 B 01 , 012 , 2 B 01 , 013 , 3 B 02 , 123 , 4 B 02 , 012 , 5 B 02 , 013 , 6 B 02 , 023 , 7 B 03 , 123 , 8 B 03 , 012 , 9 B 03 , 013 , 10 B 03 , 023 , 11 B 12 , 123 , 12 B 12 , 012 , 13 B 12 , 013 , 14 B 12 , 023 , 15 B 13 , 123 , 16 B 13 , 012 , 14 B 13 , 013 , 18 B 13 , 023 , 19 B 23 , 123 , 20 B 23 , 023
to which we add the 4 linearly dependent:
21 B 01,123 , 22 B 01,023 , 23 B 23,012 , 24 B 23,013
We successively study a few situations without any, with one or with two vanishing linearized Riemann components, taking into account that the four Einstein equations are described by:
12 , 17 , 20 for the index 0, 4 , 9 , 19 for the index 1, 1 , 10 , 15 for the index 2, 2 , 6 , 11 for the index 3.
BIANCHI 1 : B 01,012 ≡ ∇ 0 R 01,12 + ∇ 1 R 01,20 + ∇ 2 R 01,01 = − 3 m 2 r 3 ( Γ 02 0 + Γ 12 1 ) .
First of all, we have R 01 , 12 = − 3 m 2 r 3 ξ 2 0 , R 01 , 02 = 3 m 2 r 3 ξ 2 1 , R 01 , 01 = − 3 m r 4 ξ 1 and obtain:
∇ 0 R 01 , 12 = − 3 m 2 r 3 ξ 02 0 − γ 01 0 R 01 , 02 = − 3 m 2 r 3 ( ξ 02 0 + m 2 A r 2 ξ 2 1 ) = − 3 m 2 r 3 Γ 02 0
∇ 1 R 01 , 20 = ( − 3 m 2 r 3 ξ 12 1 + 9 m 2 r 4 ξ 2 1 ) − ( 2 γ 01 0 + γ 11 1 + γ 12 2 ) R 01 , 20 = − 3 m 2 r 3 ( ξ 12 1 − m 2 A r 2 ξ 2 1 ) + 6 m r 4 ξ 2 1 = − 3 m 2 r 3 Γ 12 1 + 6 m r 4 ξ 2 1
∇ 2 R 01 , 01 = − 3 m r 4 ξ 2 1 − 2 γ 12 2 R 02 , 01 = − 3 m r 4 ξ 2 1 − 2 r 3 m 2 r 3 ξ 2 1 = − 6 m r 4 ξ 2 1
and we notice that Γ 02 0 + Γ 12 1 = ξ 02 0 + ξ 12 1 = d 2 ( ξ 0 0 + ξ 1 1 ) = 0 ⇒ B 01 , 012 = 0 . As a byproduct, we have d 0 W 2 + m 2 A r 2 V 2 = 0 , d 1 V 2 − m 2 A r 2 V 2 = 0 ⇒ d 1 V 2 + d 0 W 2 = 0 .
BIANCHI 3 : B 02 , 123 ≡ ∇ 1 R 02 , 23 + ∇ 2 R 02 , 31 + ∇ 3 R 02 , 12 = 3 m A 2 r Γ 13 0 .
First of all, we have R 02 , 23 = 3 m A 2 r ξ 3 0 , R 02 , 31 = 0 , R 02 , 12 = 0 , R 01 , 31 = 3 m 2 r 3 ξ 3 0 and obtain:
∇ 1 R 02 , 23 = d 1 R 02 , 23 − γ 01 r R r 2 , 23 − γ 12 r R 0 r , 23 − γ 12 r R 02 , r 3 − γ 13 r R 02 , 2 r = 3 m A 2 r ξ 13 0 + ( 3 m 2 2 r 3 − 3 m A 2 r 2 ) ξ 3 0 − ( γ 01 0 − 3 r ) R 02 , 23 = 3 m A 2 r ξ 13 0 − ( 12 m 2 r 2 − 27 m 2 4 r 3 ) ξ 3 0 = 3 m A 2 r Γ 13 0 − 9 m A 2 r 2 ξ 3 0
∇ 2 R 02 , 31 = 0 − γ 02 r R r 2 , 31 − γ 22 r R 0 r , 31 − γ 23 r R 02 , r 1 − γ 12 r R 02 , 3 r = A r R 01 , 31 − cot ( θ ) R 02 , 31 − 1 r R 02 , 32 = 3 m A 2 r 2 ξ 3 0 + 3 m A 2 r 2 ξ 3 0 = 3 m A r 2 ξ 3 0
∇ 3 R 02 , 12 = 0 − γ 03 r R r 2 , 12 − γ 23 r R 0 r , 12 − γ 13 r R 02 , r 2 − γ 23 r R 02 , 1 r = − γ 23 3 R 03 , 12 − γ 13 3 R 02 , 32 − γ 23 3 R 02 , 13 = 3 m A 2 r 2 ξ 3 0
and we may use the fact that Γ 13 0 ≡ ξ 13 0 − 1 A r ( 1 − 3 m 2 r ) ξ 3 0 = 0 ⇒ d 1 W 3 − 1 A r ( 1 − 3 m 2 r ) W 3 = 0 .
BIANCHI 4 : B 02 , 012 ≡ ∇ 0 R 02 , 12 + ∇ 1 R 02 , 20 + ∇ 2 R 02 , 01 = 3 m 2 r 3 Γ 22 1 .
First of all we have R 02 , 12 = 0 , R 02 , 02 = 3 m A 2 r 2 ξ 1 , R 02 , 01 = 3 m 2 r 3 ξ 2 1 , R 12 , 12 = − 3 m 2 A r 2 ξ 1 and obtain:
∇ 0 R 02 , 12 = d 0 R 02 , 12 − γ 00 r R r 2 , 12 − γ 02 r R 0 r , 12 − γ 01 r R 02 , r 2 − γ 02 r R 02 , 1 r = 0 − γ 00 1 R 12 , 12 − γ 01 0 R 02 , 02 = ( − m A 2 r 2 ) ( − 3 m 2 A r 2 ) ξ 1 − ( m 2 A r 2 ) ( 3 m A 2 r 2 ) ξ 1 = 0
∇ 1 R 02 , 20 = d 1 ( − 3 m A 2 r 2 ξ 1 ) − 2 γ 01 r R r 2 , 20 − 2 γ 12 r R 0 r , 20 = − 3 m A 2 r 2 ξ 1 1 − ∂ 1 ( 3 m A 2 r 2 ) ξ 1 + 2 γ 01 0 R 02 , 02 + 2 γ 12 2 R 02 , 02 = ( − 3 m A 2 r 2 ) ( m 2 A r 2 ) ξ 1 − ∂ 1 ( 3 m A 2 r 2 ) ξ 1 + 3 m 2 2 r 4 ξ 1 + 3 m A r 3 ξ 1 = − 3 m 2 4 r 4 ξ 1 + ( 3 m r 3 − 3 m 2 r 4 + 3 m r 3 − 3 m 2 r 4 ) ξ 1 = 6 m r 3 ξ 1 − 27 m 2 4 r 4 ξ 1
∇ 2 R 02 , 01 = 3 m 2 r 3 ξ 22 1 − γ 02 r R r 2 , 01 − γ 22 r R 0 r , 01 − γ 02 r R 02 , r 1 − γ 12 r R 02 , 0 r = 3 m 2 r 3 ξ 22 1 − γ 22 1 R 01 , 01 − γ 12 2 R 02 , 02 = 3 m 2 r 3 ξ 22 1 − 9 m A 2 r 3 ξ 1 = 3 m 2 r 3 Γ 22 1 − 6 m r 3 ξ 1 + 27 m 2 4 r 4 ξ 1
This delicate checking proves that B 02 , 012 ≡ 3 m 2 r 3 Γ 22 1 = 0 ⇒ d 2 V 2 + ( 1 − 3 m 2 r ) U = 0 is a differential consequence of R 02 , 12 = 0 . We let the reader prove as an exercise that B 03 , 013 ≡ 3 m 2 r 3 Γ 33 1 = 0 ⇒ d 3 V 3 + sin ( θ ) cos ( θ ) V 2 + ( 1 − 3 m 2 r ) sin 2 ( θ ) U = 0 in order to recover 19 by eliminating U.
BIANCHI 6 : B 02 , 023 ≡ ∇ 0 R 02 , 23 + ∇ 2 R 02 , 30 + ∇ 3 R 02 , 02 = 3 m A 2 r Γ 03 0 .
First of all, we have R 02 , 23 = 3 m A 2 r ξ 3 0 , R 02 , 30 = 0 , R 02 , 02 = 3 m A 2 r 2 ξ 1 and obtain:
∇ 0 R 02 , 23 = d 0 R 02 , 23 − γ 00 r R r 2 , 23 − γ 02 r R 0 r , 23 − γ 02 r R 02 , r 3 − γ 03 r R 02 , 2 r = d 0 R 02 , 23 − γ 00 1 R 12 , 23 = 3 m A 2 r ξ 03 0 + 3 m 2 4 r 3 ξ 3 1 = 3 m A 2 r Γ 03 0
∇ 2 R 02 , 30 = 0 − γ 22 1 R 01 , 30 = A r R 01 , 30 = − 3 m A 2 r 2 ξ 3 1
∇ 3 R 02 , 02 = 3 m A 2 r 2 ξ 3 1
where Γ 03 0 ≡ ξ 03 0 + m 2 A r 2 ξ 3 1 = d 3 ( ξ 0 0 + m 2 A r 2 ξ 1 ) = 0 and R 12 , 23 = − 3 m 2 A r ξ 3 1 , R 01 , 03 = 3 m 2 r 3 ξ 3 1 .
Again, only this final result proves that B 02 , 023 ≡ 3 m A 2 r Γ 03 0 ⇒ d 0 W 3 + m 2 A r 2 V 3 = 0 is a differential consequence of R 02 , 03 = 0 .
BIANCHI 12 : B 12 , 012 ≡ ∇ 0 R 12 , 12 + ∇ 1 R 12 , 20 + ∇ 2 R 12 , 01 = − 3 m 2 r 3 Γ 22 0 .
First of all, Γ 22 0 ≡ ξ 22 0 + r A ξ 0 1 , R 12 , 12 = − 3 m 2 A r 2 ξ 1 , R 12 , 20 = 0 , R 12 , 01 = − 3 m 2 r 3 ξ 2 0 and we obtain:
∇ 0 R 12 , 12 = d 0 R 12 , 12 − 2 γ 01 r R r 2 , 12 − 2 γ 02 r R 1 r , 12 = d 0 R 12 , 12 − 2 γ 01 0 R 02 , 12 = − 3 m 2 A r 2 ξ 0 1
∇ 1 R 12 , 20 = 0 − ( γ 11 1 + 2 γ 12 2 + γ 01 0 ) R 12 , 20 = − 2 γ 12 2 R 12 , 20 = 0
∇ 2 R 12 , 01 = − 3 m 2 r 3 ξ 22 0 − γ 12 2 R 12 , 02 = − 3 m 2 r 3 ξ 22 0 = − 3 m 2 r 3 Γ 22 0 + 3 m 2 A r 2 ξ 0 1
a result leading to d 2 W 2 + r A d 0 U = 0 . Again, only the final sum has an intrinsic mathematical meaning with B 12 , 012 = − 3 m 2 r 3 Γ 22 0 .
BIANCHI 22 : B 01 , 023 ≡ ∇ 0 R 01 , 23 + ∇ 2 R 01 , 30 + ∇ 3 R 01 , 02 = 0
First of all, we have R 01 , 23 = 0 , R 01 , 03 = 3 m 2 r 3 ξ 3 1 , R 01 , 02 = 3 m 2 r 3 ξ 2 1 , R 02 , 03 = 0 and, as Ω 23 = 0 :
Γ 23 1 ≡ ξ 23 1 + γ 33 1 ξ 2 3 + γ 22 1 ξ 3 2 − γ 23 3 ξ 3 1 = ξ 23 1 − A r ( ξ 3 2 + sin 2 ( θ ) ξ 2 3 ) − cot ( θ ) ξ 3 1 = ξ 23 1 − cot ( θ ) ξ 3 1
Γ 13 2 ≡ ξ 13 2 + γ r 3 2 ξ 1 r + γ 1 r 2 ξ 3 r − γ 13 r = ξ 13 2 + γ 33 2 ξ 1 3 + ( γ 12 2 − γ 13 3 ) ξ 3 2 = ξ 13 2 − sin ( θ ) cos ( θ ) ξ 1 3
Γ 12 3 ≡ ξ 12 3 + γ r 2 3 ξ 1 r + γ 1 r 3 ξ 2 r − γ 12 r ξ r 3 = ξ 12 3 + γ 23 3 ξ 1 3 + ( γ 13 3 − γ 12 2 ) ξ 2 3 = ξ 12 3 + cot ( θ ) ξ 1 3
∇ 0 R 01 , 23 = d 0 R 01 , 23 − γ 00 r R r 1 , 23 − γ 01 r R 0 r , 23 − γ 02 r R 01 , r 3 − γ 03 r R 01 , 2 r = d 0 R 01 , 23 − γ 00 1 R 01 , 23 − γ 01 0 R 00 , 13 = 0
∇ 2 R 01 , 30 = d 2 R 01 , 30 − γ 12 2 R 02 , 30 − γ 23 3 R 01 , 30 = − 3 m 2 r 3 ξ 23 1 + 3 m 2 r 3 cot ( θ ) ξ 3 1 = − 3 m 2 r 3 ( ξ 23 1 − cot ( θ ) ξ 3 1 ) = − 3 m 2 r 3 Γ 23 1
∇ 3 R 01 , 02 = d 3 R 01 , 02 − γ 13 3 R 03 , 02 − γ 23 3 R 01 , 03 = 3 m 2 r 3 ξ 23 1 − 3 m 2 r 3 cot ( θ ) ξ 3 1 = 3 m 2 r 3 ( ξ 23 1 − cot ( θ ) ξ 3 1 ) = 3 m 2 r 3 Γ 23 1
We could also say that d 2 R 01 , 03 = 3 m 2 r 3 ( d 2 V 3 − ξ 23 1 ) + 3 m 2 r 3 ξ 23 1 and obtain therefore finally the formula B 01 , 023 = 0 ⇒ d 2 V 3 − d 3 V 2 = 0 is a differential consequence of R 01 , 23 = 0 .
We also check in particular:
∇ 0 R 01 , 23 ⇒ ρ r 1 , 23 Γ 00 r + ρ 0 r , 23 Γ 01 r + ρ 01 , r 3 Γ 02 r + ρ 01 , 2 r Γ 03 r ⇒ 0
∇ 2 R 01 , 30 ⇒ ρ r 1 , 30 Γ 02 r + ρ 0 r , 30 Γ 12 r + ρ 01 , r 0 Γ 23 r + ρ 01 , 3 r Γ 02 r ⇒ ρ 03 , 30 Γ 12 3 + ρ 01 , 10 Γ 23 1 ⇒ m A 2 r sin 2 ( θ ) Γ 12 3 − m r 3 Γ 23 1 ⇒ m A 2 r sin 2 ( θ ) ξ 12 3 − m 2 r 3 cot ( θ ) ξ 3 1 − m r 3 Γ 23 1
⇒ − d 2 ( m 2 r 3 ξ 3 1 ) − m A r sin ( θ ) cos ( θ ) ξ 1 3 + m r 3 cot ( θ ) ξ 3 1 − m 2 r 3 cot ( θ ) ξ 3 1 − m r 3 Γ 23 1 ⇒ − m 2 r 3 ( ξ 23 1 − cot ( θ ) ξ 3 1 ) − m 2 r 3 Γ 23 1 ⇒ − 3 m 2 r 3 Γ 23 1
∇ 3 R 01 , 02 ⇒ ρ r 1 , 02 Γ 03 r + ρ 0 r , 02 Γ 13 r + ρ 01 , r 2 Γ 03 r + ρ 01 , 0 r Γ 23 r ⇒ ρ 02 , 02 Γ 13 2 + ρ 01 , 01 Γ 23 1 ⇒ − m A 2 r Γ 13 2 + m r 3 Γ 23 1
⇒ − m A 2 r ( ξ 13 2 − sin 2 ( θ ) cot ( θ ) ξ 1 3 ) + m r 3 Γ 23 1 ⇒ m 2 r 3 ( ξ 23 1 − cot ( θ ) ξ 3 1 ) + m r 3 Γ 23 1 ⇒ 3 m 2 r 3 Γ 23 1
As a tricky exercise too, we advise the reader to treat similarly the case of 13 + 16 or 21 in order to obtain B 01 , 123 = 0 ⇒ d 2 W 3 − d 3 W 2 = 0 because R 01 , 23 = 0 and R 03 , 12 = 0 (care).
REMARK 4.B.1: Though a few conditions like d 2 V 3 − d 3 V 2 = 0 22 or d 2 W 3 − d 3 W 2 = 0 21 look like to be third order CC for Ω , we have thus proved that they come indeed from the first prolongations of the second order CC. The same comment is also valid for a few other striking CC. Using previous results, we have successively 6 other relations:
ξ 12 1 + ξ 02 0 = d 2 ( ξ 1 1 + ξ 0 0 ) = 0 ⇒ d 1 V 2 + d 0 W 2 = 0 1
ξ 13 1 + ξ 03 0 = d 3 ( ξ 1 1 + ξ 0 0 ) = 0 ⇒ d 1 V 3 + d 0 W 3 = 0 2
ξ 33 1 + sin ( θ ) cos ( θ ) ξ 2 1 − sin 2 ( θ ) ξ 22 1 = 0 ⇒ d 3 V 3 + sin ( θ ) cos ( θ ) V 2 − sin 2 ( θ ) d 2 V 2 = 0 19
ξ 33 0 + sin ( θ ) cos ( θ ) ξ 2 0 − sin 2 ( θ ) ξ 22 0 = 0 ⇒ d 3 W 3 + sin ( θ ) cos ( θ ) W 2 − sin 2 ( θ ) d 2 W 2 = 0 20
because d 2 W 2 + r A d 0 U = 0 12 and
d 3 W 3 + sin ( θ ) cos ( θ ) W 2 + r A sin 2 ( θ ) d 0 U = 0 17
ξ 02 1 − A 2 ξ 12 0 = d 2 ( ξ 0 1 − A 2 ξ 1 0 ) = 0 ⇒ d 0 V 2 − A 2 d 1 W 2 = 0 24
ξ 03 1 − A 2 ξ 13 0 = d 3 ( ξ 0 1 − A 2 ξ 1 0 ) = 0 ⇒ d 0 V 3 − A 2 d 1 W 3 = 0 23
From the 24 B, we have thus used 8 of them and are left with 24 − 8 = 16 expressions involving the 4 × 4 = 16 different first derivatives of the 4 functions ( V , W ) , namely B 3 to B 18 . Now, we notice that, among these 24 B, only 4 of them do contain three components R k l , i j that are not vanishing for the S-metric, namely 1 , 2 , 19 and 20 . They are providing the terms d r E r r for r = 0 , 1 , 2 , 3 in the divergence type condition for the linearized Einstein equations implied by the linearized Bianchi identities over the Schwarzschild metric. Accordingly, it does not seem possible to obtain any other third order CC apart from these 4 divergence conditions.
It remains to apply these results to the successive prolongations of the Killing equations, as we know from the intrinsic study achieved in ( [22] [23] ) that we have the successive Lie algebroids:
R 1 ( 4 ) = R 1 ( 3 ) ⊂ R 1 ( 2 ) ⊂ R 1 ( 1 ) = R 1 ⊂ J 1 ( T )
with respective dimensions 4 = 4 < 5 < 10 = 10 < 20 and R 1 ( 3 ) does not depend any longer on the S-parameter m.
The challenge will be to prove that… the only knowledge of these numbers is sufficient!
In an equivalent way as g 2 = 0 ⇒ g r + 2 = 0 , ∀ r ≥ 0 , we obtain successively:
d i m ( R 1 ) = d i m ( J 1 ( T ) ) − d i m ( S 2 T * ) = 20 − 10 = 10 ,
d i m ( R 2 ) = d i m ( J 2 ( T ) ) − d i m ( J 1 ( S 2 T * ) ) = 60 − 50 = 10,
d i m ( R 3 ) = d i m ( R 2 ) − 5 = 5 ⇒ d i m ( R r + 4 ) = d i m ( R 4 ) = d i m ( R 3 ) − 1 = 4
and shall use these results from now on.
First of all, using the introductory diagram when q = 1 , r = 1 , we may apply the Spencer δ-map to the symbol top row in order to obtain the diagram:
0 0 ↓ ↓ 0 → S 3 T * ⊗ T → S 2 T * ⊗ F 0 → h 2 → 0 ↓ ↓ 0 → T * ⊗ S 2 T * ⊗ T → T * ⊗ T * ⊗ F 0 → 0 ↓ ↓ 0 → ∧ 2 T * ⊗ g 1 → ∧ 2 T * ⊗ T * ⊗ T → ∧ 2 T * ⊗ F 0 → 0 ↓ ↓ ↓ 0 → ∧ 3 T * ⊗ T = ∧ 3 T * ⊗ T → 0 ↓ ↓ 0 0
Using the Spencer δ-cohomology H r ( g 1 ) = Z r ( g 1 ) / B r ( g 1 ) at ⋯ → ∧ r T * ⊗ g 1 → ⋯ , we obtain:
PROPOSITION 4.B.2: h 2 ≃ H 2 ( g 1 ) ⇒ n ( n + 1 ) / 2 ≤ d i m ( Q 2 ) ≤ n 2 ( n 2 − 1 ) / 12 whenever n ≥ 3 .
Proof: As there cannot be any CC of order one and thus Q 1 = 0 , we have the long exact connecting sequence 0 → R 3 → R 2 → h 2 → Q 2 → 0 and counting the dimensions with F 0 = S 2 T * , we have:
d i m ( Q 2 ) ≤ d i m ( h 2 ) = d i m ( S 2 T * ⊗ S 2 T * ) − d i m ( S 3 T * ⊗ T ) = n 2 ( n 2 − 1 ) / 12
This result is confirmed by a circular chase proving that the left bottom δ-map is an epimorphism and a snake chase in the last diagram providing the short exact sequence:
0 → h 2 → ∧ 2 T * ⊗ g 1 → δ ∧ 3 T * ⊗ T → 0 , 0 → 20 → 36 → 16 → 0
Indeed, as d e t ( ω ) ≠ 0 we may use the metric for providing an isomorphism T ≃ T * : ( ξ r ) → ( ξ i = ω r i ξ r ) in such a way that g 1 ≃ ∧ 2 T * is defined by ξ i , j + ξ j , i = 0 for both the M, S and K metrics.
However, introducing the conformal Killing system of infinitesimal Lie equations with symbol g ^ 1 defined by the ( n ( n + 1 ) / 2 ) − 1 linear equations ω r j ξ i r + ω i r ξ j r − 2 n ω i j ξ r r = 0 that do not depend on any conformal factor, we have the fundamental diagram II ( [1] [12] ):
0 ↓ 0 S 2 T * ↓ ↓ 0 → Z 2 ( g 1 ) → H 2 ( g 1 ) → 0 ↓ ↓ ↓ 0 → T * ⊗ g ^ 2 → δ Z 2 ( g ^ 1 ) → H 2 ( g ^ 1 ) → 0 ↓ ↓ ↓ 0 → S 2 T * → δ T * ⊗ T * → δ ∧ 2 T * → 0 ↓ ↓ 0 0
showing that we have the splitting sequence 0 → S 2 T * → H 2 ( g 1 ) → H 2 ( g ^ 1 ) → 0 providing a totally unusual interpretation of the successive Ricci, Riemann and Weyl tensors and the corresponding splitting. However, it must be noticed that the Weyl-type operator is of order 3 when n = 3 because n 2 ( n 2 − 1 ) / 12 − n ( n + 1 ) / 2 = n ( n + 1 ) ( n + 2 ) ( n − 3 ) / 12 but of order 2 for n ≥ 4 ( [4] [9] ). Similar results could be obtained for the Bianchi-type operator as we shall see.
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Using now the same procedure for the introductory diagram with r = 2 , we get the diagram:
0 0 0 ↓ ↓ ↓ 0 → S 4 T * ⊗ T → S 3 T * ⊗ F 0 → h 3 → 0 ↓ ↓ ↘ ↓ 0 → T * ⊗ S 3 T * ⊗ T → T * ⊗ S 2 T * ⊗ F 0 → T * ⊗ h 2 → 0 ↓ ↓ 0 → ∧ 2 T * ⊗ S 2 T * ⊗ T → ∧ 2 T * ⊗ T * ⊗ F 0 → 0 ↓ ↓ 0 → ∧ 3 T * ⊗ g 1 → ∧ 3 T * ⊗ T * ⊗ T → ∧ 3 T * ⊗ F 0 → 0 ↓ ↓ ↓ 0 → ∧ 4 T * ⊗ T = ∧ 4 T * ⊗ T → 0 ↓ ↓ 0 0
Using a snake chase and Theorem 3.2.3, we obtain the short exact sequence:
0 → h 3 → T * ⊗ h 2 → H 3 ( g 1 ) → 0, 0 → 60 → 80 → 20 → 0
A chase around the upper south-east arrow on the right is leading to the following corollary where g ′ 2 ⊂ S 2 T * ⊗ F 0 is the symbol of the system R ′ 2 ⊂ J 2 ( F 0 ) which is the image of J 3 ( T ) and Q ′ 1 is the cokernel of the central bottom map:
COROLLARY 4.B.3: There is a long exact connecting diagram:
0 0 ↓ ↓ 0 → S 4 T * ⊗ T → S 3 T * ⊗ F 0 → T * ⊗ h 2 → H 3 ( g 1 ) → 0 ↓ ∥ ↓ ↓ 0 → ρ 1 ( g ′ 2 ) → S 3 T * ⊗ F 0 → T * ⊗ Q 2 → Q ′ 1 → 0 ↓ ↓ ↓ 0 0 0
allowing to use the Bianchi identities as B ∈ F 2 ≃ H 3 ( g 1 ) and we have d i m ( Q ′ 1 ) ≤ d i m ( H 3 ( g 1 ) ) .
Proof: Using the notations of the introductory diagram and the fact that Q 1 = 0 , we have the following two commutative and exact diagrams obtained by choosing F 1 = Q 2 for the first, then F 1 = Q 3 for the second and so on, in a systematic manner as in the motivating examples:
0 0 0 0 ↓ ↓ ↓ ↓ 0 → ρ 1 ( g ′ 2 ) → S 3 T * ⊗ F 0 → T * ⊗ Q 2 → Q ′ 1 → 0 ↓ ↓ ↓ ↘ ∥ 0 → ρ 1 ( R ′ 2 ) → J 3 ( F 0 ) → J 1 ( Q 2 ) → Q ′ 1 → 0 ↓ ↓ ↓ ↓ 0 → R ′ 2 → J 2 ( F 0 ) → Q 2 → 0 ↓ ↓ ↓ 0 0 0
First, we have the short exact sequence 0 → R 2 → J 2 ( T ) → J 1 ( F 0 ) → 0 with 10 − 60 + 50 = 0 and get R ′ 0 = F 0 , R ′ 1 = J 1 ( F 0 ) and d i m ( ρ 1 ( R ′ 0 ) ) − d i m ( R ′ 1 ) = 0 , that is no CC of order 1.
Now, using the long exact sequence:
0 → R 3 → J 3 ( T ) → J 2 ( F 0 ) → Q 2 → 0
R ′ 2 ⊂ ρ 1 ( R ′ 1 ) = ρ 1 ( J 1 ( F 0 ) ) = J 2 ( F 0 ) ⇒ d i m ( ρ 1 ( R ′ 1 ) ) − d i m ( R ′ 2 ) = 150 − 135 = d i m ( Q 2 ) = 15
because d i m ( R ′ 2 ) = d i m ( J 3 ( T ) ) − d i m ( R 3 ) = 140 − 5 = 135 and there are 15 second order CC.
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Then, with d i m ( Q ′ 1 ) = x , we obtain by counting the dimensions:
d i m ( ρ 1 ( R ′ 2 ) ) = d i m ( J 3 ( F 0 ) ) + x − d i m ( J 1 ( Q 2 ) ) = 350 + x − 75 = x + 275,
d i m ( R ′ 3 ) = d i m ( J 4 ( T ) ) − d i m ( R 4 ) = d i m ( J 3 ( F 0 ) ) − d i m ( Q 3 ) = 276
⇒ y = d i m ( ρ 1 ( R ′ 2 ) ) − d i m ( R ′ 3 ) = x + 275 − 276 = x − 1
that is y ≥ 3 because x ≥ 4 and thus x = 4 ⇒ y = 3 if we only take into account the 4 divergence condition of the Einstein equations. The situation will be worst for the Kerr metric with y = 6 .
After one prolongation, we get:
0 0 0 0 ↓ ↓ ↓ ↓ 0 → ρ 1 ( g ′ 3 ) → S 4 T * ⊗ F 0 → T * ⊗ Q 3 → Q ′ 1 → 0 ↓ ↓ ↓ ∥ 0 → ρ 1 ( R ′ 3 ) → J 4 ( F 0 ) → J 1 ( Q 3 ) → Q ′ 1 → 0 ↓ ↓ ↓ ↓ 0 → R ′ 3 → J 3 ( F 0 ) → Q 3 → 0 ↓ ↓ ↓ 0 0 0
From this second diagram we obtain the commutative and exact diagram:
0 0 ↓ ↓ 0 → R ′ 4 → J 4 ( F 0 ) → Q 4 → 0 ↓ ∥ ↓ 0 → ρ 1 ( R ′ 3 ) → J 4 ( F 0 ) → J 1 ( Q 3 ) → Q ′ 1 → 0 ↓ 0
Indeed, setting again d i m ( Q ′ 1 ) = x , we obtain now similarly:
d i m ( ρ 1 ( R ′ 3 ) ) = d i m ( J 4 ( F 0 ) ) + x − d i m ( J 1 ( Q 3 ) ) = 700 + x − 370 = x + 330,
d i m ( R ′ 4 ) = d i m ( J 5 ( T ) ) − d i m ( R 5 ) = d i m ( J 4 ( F 0 ) ) − d i m ( Q 4 ) = 500
⇒ y = d i m ( ρ 1 ( R ′ 3 ) ) − d i m ( R ′ 4 ) = x + 330 − 500 = x − 170
that is y = 0 ⇔ x = 170 . We find exactly d i m ( F 2 ) = 170 like in ( [22], p. 1996) and the condition y = 0 just means that the CC of order 4 are generated by the CC of order 3, but we have only R ′ 4 ⊆ ρ 1 ( R ′ 3 ) in general.
With one more prolongation, applying again the δ-map to the top symbol sequence, we get the following commutative diagram:
0 0 0 ↓ ↓ ↓ 0 → S 6 T * ⊗ T → S 5 T * ⊗ F 0 → h 5 → 0 ↓ δ ↓ δ ↓ 0 → T * ⊗ S 5 T * ⊗ T → T * ⊗ S 4 T * ⊗ F 0 → T * ⊗ h 4 → 0 ↓ δ ↓ δ ↓ 0 → ∧ 2 T * ⊗ S 4 T * ⊗ T → ∧ 2 T * ⊗ S 3 T * ⊗ F 0 → ∧ 2 T * ⊗ h 3 → 0 ↓ δ ↓ δ ↓ 0 → ∧ 3 T * ⊗ S 3 T * ⊗ T → ∧ 3 T * ⊗ S 2 T * ⊗ F 0 → ∧ 3 T * ⊗ h 2 → 0 ↓ δ ↓ δ ↓ 0 → ∧ 4 T * ⊗ S 2 T * ⊗ T → ∧ 4 T * ⊗ T * ⊗ F 0 → 0 ↓ ↓ 0 0
where the right exact vertical column is 0 → 224 → 504 → 360 → 80 → 0 . It just remains to replace in the two upper right epimorphisms h 5 by T * ⊗ Q 4 and h 4 by Q 4 along with the following commutative diagram where we have chosen F 1 = Q 4 :
0 0 ↘ ↓ h 5 ↓ ↘ 0 → T * ⊗ h 4 → T * ⊗ Q 4
in order to obtain the long exact sequence 0 → S 6 T * ⊗ T → S 5 T * ⊗ F 0 → T * ⊗ Q 4 .
Finally, chasing in the following commutative and exact introductory diagram:
0 0 0 0 ↓ ↓ ↓ ↓ 0 → S 6 T * ⊗ T → σ 5 ( Φ ) S 5 T * ⊗ F 0 → T * ⊗ Q 4 → Q ′ 1 → 0 ↓ ↓ ↓ ↓ ∥ 0 → R 6 → J 6 ( T ) → ρ 5 ( Φ ) J 5 ( F 0 ) → J 1 ( Q 4 ) → Q ′ 1 → 0 ↓ ↓ ↓ ↓ ↓ 0 → R 5 → J 5 ( T ) → ρ 4 ( Φ ) J 4 ( F 0 ) → Q 4 → 0 ↓ ↓ ↓ ↓ 0 0 0 0
we deduce that R ′ 5 = ρ 1 ( R ′ 4 ) is involutive with d i m ( R ′ 5 ) = 840 − 4 = 836 and symbol g ′ 5 ≃ S 6 T * ⊗ T .
Unhappily, the reader will check at once that a similar procedure cannot be applied in order to prove that R ′ 4 = ρ 1 ( R ′ 3 ) . Indeed, if we still have a monomorphism 0 → h 4 → Q 4 we do not have a monomorphism h 3 → Q 3 because now this map has a kernel of dimension equal to d i m ( R 3 / R 3 ( 1 ) ) = 5 − 4 = 1 according to the corresponding long exact connecting sequence.
IT IS THUS NOT POSSIBLE TO PROVE THAT THERE ARE ONLY SECOND AND THIRD ORDER GENERATING CC IN A SIMPLE INTRINSIC WAY.
However, like in the first motivating example in which we should be waiting for third order CC but a direct computation was proving that only second order ones could be used, we have:
THEOREM 4.B.4: The CC of the first order operator D : T → F 0 are generated by a third order operator D 1 : F 0 → F 1 = Q 3 and we have thus R ′ 4 = ρ 1 ( R ′ 3 ) .
Proof: With F 0 = S 2 T * and F 1 = Q 3 while applying the Spencer operator, we obtain the following commutative diagram in which the two central vertical columns are locally exact ( [1] [5] ):
0 0 0 ↓ ↓ , ↓ 0 → Θ → T → D F 0 → D 1 F 1 ↓ j 4 ↓ j 4 ↓ j 3 ↘ ∥ 0 → R 4 → J 4 ( T ) → J 3 ( F 0 ) → F 1 → 0 ↓ d ↓ d ↓ d 0 → T * ⊗ R 3 → T * ⊗ J 3 ( T ) → T * ⊗ J 2 ( F 0 ) ↓ d ↓ d 0 → ∧ 2 T * ⊗ R 2 → ∧ 2 T * ⊗ J 2 ( T )
Chasing in this diagram by using the Snake lemma of the second section, we discover that the local exactness at F 0 of the top row is equivalent to the local exactness at T * ⊗ R 3 of the left column.
Now, we have the commutative diagram:
0 0 0 ↓ ↓ ↓ 0 → Θ → j 5 R 5 → d T * ⊗ R 4 → d ∧ 2 T * ⊗ R 3 ∥ ↓ ↓ ↓ 0 → Θ → j 4 R 4 → d T * ⊗ R 3 → d ∧ 2 T * ⊗ R 2 ↓ ↓ ↓ 0 → T * ⊗ ( R 3 / R 3 ( 1 ) ) → d ∧ 2 T * ⊗ ( R 2 / R 2 ( 1 ) ) ↓ ↓ 0 0
The top row is known to be locally exact as it is isomorphic to a part of the Poincaré sequence according to the commutative diagram with R 4 ≃ R 5 ≃ R 6 :
0 0 0 ↓ ↓ ↓ 0 → Θ → j 6 R 6 → d T * ⊗ R 5 → d ∧ 2 T * ⊗ R 4 ∥ ↓ ↓ ↓ 0 → Θ → j 5 R 5 → d T * ⊗ R 4 → d ∧ 2 T * ⊗ R 3 ↓ ↓ 0 0
The bottom row is purely algebraic as it is induced by the exact sequence obtained by applying the Spencer operator to the long exact connecting sequence and chasing along the south west diagonal:
0 → h 4 → δ T * ⊗ h 3 → δ ∧ 2 T * ⊗ h 2 → ∧ 4 T * ⊗ g 1 → 0
0 → 126 → 240 → 120 → 6 → 0
Changing the confusing notations used in ( [24] ), we prove that the bottom Spencer operator is injective. Indeed, we have the following representative parametric jets for the various Lie equations:
d i m ( R 2 ) = 10 ⇒ { ξ 0 , ξ 1 , ξ 2 , ξ 3 , ξ 0 1 , ξ 2 0 , ξ 3 0 , ξ 2 1 , ξ 3 1 , ξ 3 2 }
d i m ( R 3 ) = 5 ⇒ { ξ 0 , ξ 2 , ξ 3 , ξ 0 1 , ξ 3 2 } , d i m ( R 4 ) = 4 ⇒ { ξ 0 , ξ 2 , ξ 3 , ξ 3 2 }
d i m ( R 3 / R 3 ( 1 ) ) = 1 ⇒ { ξ 0 1 } , d i m ( R 2 / R 2 ( 1 ) ) = 5 ⇒ { ξ 1 , ξ 2 1 , ξ 3 1 , ξ 2 0 , ξ 3 0 }
We also recall the definition of the Spencer operator d : T * ⊗ J q + 1 ( T ) → ∧ 2 T * ⊗ J q ( T ) :
( ξ ν , i k ) → ξ μ , i j k = ( ∂ i ξ μ , j k − ∂ j ξ μ , i k + ξ μ + 1 j , i k − ξ μ + 1 i , j k )
Accordingly, we may choose local coordinates ( ξ 0, i 1 ) for a representative and a representative of the image by d is for example ( ξ ,0 i 1 = ξ 0, i 1 − ξ i ,0 1 ) . Now, as d i m ( R 3 / R 3 ( 1 ) ) = 4 − 3 = 1 , we may introduce the four local coordinates ξ 0, i 1 and ξ 1, i 0 such that ξ 0 , i 1 − A 2 ξ 1 , i 0 = 0 , ∀ i = 0 , 1 , 2 , 3 . We may also use the 6 × 5 = 30 local coordinates ( ξ , i j 1 , ξ 2, i j 1 , ξ 3, i j 1 , ξ 2, i j 0 , ξ 3, i j 0 ) in order to describe ∧ 2 T * ⊗ ( R 2 / R 2 ( 1 ) ) . In the kernel of d, we have in particular
ξ , 0 i 1 = ξ 0 , i 1 − ξ i , 0 1 = 0 ⇒ ξ 0 , i 1 = ξ i , 0 1 = 0 , ∀ i = 1 , 2 , 3 because ξ 1 1 = m 2 A r 2 ξ 1 in R 1 but also ξ , 01 0 = ξ 0 , 1 0 − ξ 1 , 0 0 = 0 ⇒ ξ 1 , 0 0 = 0 ⇒ ξ 0 , 0 1 = 0 because { ξ 0 } is among the parametric jets of R 3 and thus ξ 0 , i 1 = 0 , ∀ i = 0 , 1 , 2 , 3 . The bottom Spencer operator is thus injective and the bottom sequence is thus exact. A circular chase ends the proof: If b ∈ T * ⊗ R 3 is killed by d, then its projection c ∈ T * ⊗ ( R 3 / R 3 ( 1 ) ) is also killed by d and is such that c = 0 . Accordingly, ∃ a ∈ T * ⊗ R 4 with image b under the monomorphism T * ⊗ R 4 → T * ⊗ R 3 and such that d a = 0 . We may thus find e ∈ R 5 and f ∈ R 4 because R 5 ≃ R 4 with a = d e ⇒ b = d f .
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Like in the second motivating example, the sequence constructed in the previous theorem may have “jumps” in the order of the successive operators and we have therefore (Compare to [1] ):
COROLLARY 4.B.5: The symbol of D 1 is not 2-acyclic and the CC operator D 2 is thus of order 2. Accordingly, if one does want a formally exact canonical Janet sequence, the only possibility is to use the involutive operator D 1 of order 4 defined by R ′ 4 = ρ 1 ( R ′ 3 ) .
Proof: Recapitulating the results so far obtained, we have successively R ′ r + 1 ⊆ ρ 1 ( R ′ r ) with:
R ′ 0 = F 0 , d i m ( R ′ 1 ) = d i m ( J 2 ( T ) ) − d i m ( R 2 ) = 60 − 10 = 50 ⇒ R ′ 1 = J 1 ( F 0 ) , ρ 1 ( R ′ 1 ) = J 2 ( F 0 )
d i m ( R ′ 2 ) = d i m ( J 3 ( T ) ) − d i m ( R 3 ) = 140 − 5 = 135, d i m ( ρ 1 ( R ′ 1 ) ) − d i m ( R ′ 2 ) = 150 − 135 = 15
d i m ( R ′ 3 ) = d i m ( J 4 ( T ) ) − d i m ( R 4 ) = 280 − 4 = 276, d i m ( ρ 1 ( R ′ 2 ) ) − d i m ( R ′ 3 ) ≥ 3,
d i m ( R ′ 4 ) = d i m ( J 5 ( T ) ) − d i m ( R 5 ) = 504 − 4 = 500, R ′ 4 = ρ 1 ( R ′ 3 ) ,
d i m ( R ′ 5 ) = d i m ( J 6 ( T ) ) − d i m ( R 6 ) = 840 − 4 = 836, R ′ 5 = ρ 1 ( R ′ 4 ) .
the long exact sequence: 0 → ρ 1 ( R ′ 2 ) → J 3 ( F 0 ) → J 1 ( Q 2 ) → Q ′ 1 → 0 with d i m ( Q ′ 1 ) = x ≥ 4 because of the divergence CC condition for Einstein equations implied by the Bianchi identities.
It also follows that:
g ′ 1 ≃ T * ⊗ F 0 ⇒ d i m ( g ′ 1 ) = 40, S 3 T * ⊗ T ⊂ g ′ 2 ⇔ 80 < d i m ( g ′ 2 ) = 135 − 50 = 85 ,
S 4 T * ⊗ T ⊂ g ′ 3 ⇔ 140 < 141 = 276 − 135, S 5 + r T * ⊗ T ≃ g ′ 4 + r , ∀ r ≥ 0
and we have the basic commutative and exact “defining diagram” of the system R 2 ⊂ J 2 ( T ) :
0 0 ↓ ↓ 0 → S 2 T * ⊗ T → T * ⊗ F 0 → 0 ↓ ↓ ↓ 0 → R 2 → J 2 ( T ) → J 1 ( F 0 ) → 0 ↓ ↓ ↓ 0 → R 1 → J 1 ( T ) → F 0 → 0 ↓ ↓ ↓ 0 0 0
allowing to obtain the central vertical short exact sequence 0 → g ′ 1 → R ′ 1 → F 9 → 0 .
Now, it is known that a symbol g q of finite type is involutive if an only if it is vanishing ( [1] [5] [10] ). Using a similar proof, let us consider the commutative diagram of δ-sequences:
0 0 0 ↓ ↓ ↓ ⋯ → T * ⊗ S 6 T * ⊗ T → ∧ 3 T * ⊗ S 5 T * ⊗ T → ∧ 4 T * ⊗ S 4 T * ⊗ T → 0 ↓ ↓ ↓ ⋯ → T * ⊗ g ′ 5 → ∧ 3 T * ⊗ g ′ 4 → ∧ 4 T * ⊗ g ′ 3 → 0 ↓ ↓ 0 0
Using the fact that the upper sequence is known to be exact as a δ-sequence and that we have d i m ( S 4 T * ⊗ T ) = 140 < 141 = d i m ( g ′ 3 ) , an easy chase proves that the lower sequence cannot be exact and thus g ′ 3 cannot be involutive after counting the dimensions. The corollary follows from the fact that g ′ 4 = ρ 1 ( g ′ 3 ) ≃ S 5 T * ⊗ T is indeed 3-acyclic one step ahead by chasing and even involutive.
Finally, with vector bundles A , B such that d i m ( A ) = 1 , d i m ( B ) = 5 , we have the commutative diagram of δ-sequences in which we recall that g ′ 1 ≃ T * ⊗ F 0 :
0 0 0 0 ↓ ↓ ↓ ↓ T * ⊗ S 5 T * ⊗ T → ∧ 2 T * ⊗ S 4 T * ⊗ T → ∧ 3 T * ⊗ S 3 T * ⊗ T → ∧ 4 T * ⊗ S 2 T * ⊗ T → 0 ↓ ↓ ↓ ↓ T * ⊗ g ′ 4 → ∧ 2 T * ⊗ g ′ 3 → ∧ 3 T * ⊗ g ′ 2 → ∧ 4 T * ⊗ g ′ 1 → 0 ↓ ↓ ↓ ↓ 0 → ∧ 2 T * ⊗ A → ∧ 3 T * ⊗ B → 0 ↓ ↓ 0 0
Taking into account that the top row is exact and proceeding as in the last theorem with similar local coordinates, we get:
ξ , 123 1 = ξ 1 , 23 1 + ξ 2 , 31 1 + ξ 3 , 12 1 = 0 + 0 + 0 = 0
always.
ξ , 012 1 = ξ 0 , 12 1 + ξ 1 , 20 1 + ξ 2 , 01 1 = ξ 0 , 12 1 + 0 + 0 = 0 ⇒ ξ 0 , 12 1 = 0 ⇒ ξ 0 , i j 1 = 0 , ∀ i , j = 1 , 2 , 3
ξ , 10 i 0 = ξ 1 , 0 i 0 + ξ 0 , i 1 0 + ξ i , 01 0 = ξ 1 , 0 i 0 + 0 + 0 = 0 ⇒ ξ 1 , 0 i 0 = 0 ⇒ ξ 0 , 0 i 1 = 0 , ∀ i = 2 , 3 .
We are thus only left with ξ 0,01 1 that may not vanish though ξ 0 , 01 1 + ξ 0 , 10 1 + ξ 1 , 00 1 = 0 in any case and the bottom map δ is not injective.
Let us prove that g ′ 3 is not 2-acyclic because the central δ-sequence cannot be exact at ∧ 2 T * ⊗ g ′ 3 . Indeed, if it were, let c ∈ ∧ 2 T * ⊗ A be killed by δ. Then, we may lift c to b ∈ ∧ 2 T * ⊗ g ′ 3 such that δ b = f ∈ ∧ 3 T * ⊗ S 3 T * ⊗ T and obtain by commutativity δ f = 0 because the last vertical downarrow on the right is an isomorphism, thus a monomorphism. As the upper row is an exact sequence, we may thus find a ∈ ∧ 2 T * ⊗ S 4 T * ⊗ T such that f = δ a . Chasing circularly, it follows from the exactness assumption at ∧ 2 T * ⊗ g ′ 3 that we can find e ∈ T * ⊗ g ′ 4 ≃ T * ⊗ S 5 T * ⊗ T such that b = a + δ e = a ′ ∈ ∧ 2 T * ⊗ S 4 T * ⊗ T . It should follow that c = 0 and a contradiction, that is g ′ 3 cannot be 2-acyclic.
As we know from ( [1] [5] [10] ) that the order of the generating CC for D 1 is equal to s + 1 if one needs s prolongations in such a way that ρ s ( g ′ 3 ) = g ′ 3 + s becomes 2-acyclic. As we already know that g ′ 4 = ρ 1 ( g ′ 3 ) ≃ T * ⊗ S 5 T * ⊗ T is involutive, we get s = 1 and the generating CC D 2 of D 1 are of order 2. We have just a “jump” in the order and, for the details, refer the reader to the quite delicate Example 3.14 of ( [9], p 119-125) in which it is already difficult to discover how many new second order CC should be introduced though the initial system is trivially FI with coefficients in ℚ . Such a result could not even be imagined while using the methods of ( [6] [7] [8] ).
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There are “natural” reason for which we do not believe that these results could be useful in physics. Indeed, considering like in the previous reference the long exact sequence of jet bundles allowing to define F 2 when F 1 = Q 3 , namely:
0 → R 6 → J 6 ( T ) → 1 J 5 ( F 0 ) → 3 J 2 ( F 1 ) → 2 F 2 → 0
0 → 4 → 840 → 1 1260 → 3 1110 → 2 686 → 0
and the large values of these dimensions need no comment for any application.
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C) KERR METRIC:
We now write the Kerr metric in Boyer-Lindquist coordinates:
d s 2 = ρ 2 − m r ρ 2 d t 2 − ρ 2 Δ d r 2 − ρ 2 d θ 2 − 2 a m r sin 2 ( θ ) ρ 2 d t d ϕ − ( r 2 + a 2 + m r a 2 sin 2 ( θ ) ρ 2 ) sin 2 ( θ ) d ϕ 2
where we have set Δ = r 2 − m r + a 2 , ρ 2 = r 2 + a 2 cos 2 ( θ ) as usual and we check that we recover the Schwarschild metric when a = 0 . We notice that t or ϕ do not appear in the coefficients of the metric. We shall change the coordinate system in order to confirm theses results by using computer algebra and the idea is to use the so-called “rational polynomial” coefficients as follows:
( x 0 = t , x 1 = r , x 2 = c = cos ( θ ) , x 3 = ϕ ) ⇒ d x 2 = − sin ( θ ) d θ ⇒ ( d x 2 ) 2 = ( 1 − c 2 ) d θ 2
We obtain over the differential field K = ℚ ( a , m ) ( t , r , c , ϕ ) = ℚ ( a , m ) ( x ) :
d s 2 = ρ 2 − m x 1 ρ 2 ( d x 0 ) 2 − ρ 2 Δ ( d x 1 ) 2 − ρ 2 1 − ( x 2 ) 2 ( d x 2 ) 2 − 2 a m x 1 ( 1 − ( x 2 ) 2 ) ρ 2 d x 0 d x 3 − ( 1 − ( x 2 ) 2 ) ( ( x 1 ) 2 + a 2 + m a 2 x 1 ( 1 − ( x 2 ) 2 ) ρ 2 ) ( d x 3 ) 2
with now Δ = ( x 1 ) 2 − m x 1 + a 2 = r 2 − m r + a 2 and ρ 2 = ( x 1 ) 2 + a 2 ( x 2 ) 2 = r 2 + a 2 c 2 . For a later use, it is also possible to set ω 33 = − ( 1 − c 2 ) ( ( r 2 + a 2 ) 2 − a 2 ( 1 − c 2 ) ( a 2 − m r + r 2 ) ) / ( r 2 + a 2 c 2 ) and we have d e t ( ω ) = − ( r 2 + a 2 c 2 ) 2 . Framing the leading derivatives, we obtain:
R 1 ⊂ J 1 ( T ) { Ω 33 ≡ 2 ( ω 33 ξ 3 3 + ω 03 ξ 3 0 ) + ξ ∂ ω 33 = 0 Ω 23 ≡ ω 33 ξ 2 3 + ω 03 ξ 2 0 + ω 22 ξ 3 2 = 0 Ω 22 ≡ 2 ω 22 ξ 2 2 + ξ ∂ ω 22 = 0 Ω 13 ≡ ω 33 ξ 1 3 + ω 03 ξ 1 0 + ω 11 ξ 3 1 = 0 Ω 12 ≡ ω 22 ξ 1 2 + ω 11 ξ 2 1 = 0 Ω 11 ≡ 2 ω 11 ξ 1 1 + ξ ∂ ω 11 = 0 Ω 03 ≡ ω 33 ξ 0 3 + ω 03 ( ξ 0 0 + ξ 3 3 ) + ω 00 ξ 3 0 + ξ ∂ ω 03 = 0 Ω 02 ≡ ω 22 ξ 0 2 + ω 00 ξ 2 0 + ω 03 ξ 2 3 = 0 Ω 01 ≡ ω 11 ξ 0 1 + ω 00 ξ 1 0 + ω 03 ξ 1 3 = 0 Ω 00 ≡ 2 ( ω 00 ξ 0 0 + ω 03 ξ 0 3 ) + ξ ∂ ω 00 = 0
Now, we know that if R q ⊂ J q ( T ) is a system of infinitesimal Lie equations, then we have the algebroid bracket and its link with the prolongation/projection (PP) procedure ( [1] [2] [5] [10] ):
[ R q , R q ] ⊂ R q ⇒ [ R q + r ( s ) , R q + r ( s ) ] ⊂ R q + r ( s ) , ∀ q , r , s ≥ 0
As R 1 ( 1 ) = π 1 2 ( R 2 ) = R 1 , it follows that R 1 ( 2 ) = π 1 3 ( R 3 ) is such that [ R 1 ( 2 ) , R 1 ( 2 ) ] ⊂ R 1 ( 2 ) with d i m ( R 1 ( 2 ) ) = 20 − 16 = 4 because we have obtained a total of 6 new different first order equations. Using the first general diagram of the Introduction, we discover that the operator defining R 1 has 10 + 4 = 14 CC of order 2, a result obtained totally independently of any specific GR technical object like the Teukolski scalars or the Killing-Yano tensors introduced in ( [6] [7] [8] ).
Like in the case of the S metric, two prolongations allow to obtain 6 additional equations (instead of 5) that we set on the left side in the following list obtained m o d ( j 2 ( Ω ) ) :
We have on sections (care) the 16 (linear) equations m o d ( j 2 ( Ω ) ) of R 1 ( 2 ) as follows ( [23] ):
R 1 ( 2 ) ⊂ R 1 ⊂ J 1 ( T ) { ξ 1 = 0 , ξ 2 = 0 ⇒ ω 00 ξ 1 0 + ω 03 ξ 1 3 + ω 11 ξ 0 1 = 0 , ξ 1 1 = 0 , ξ 2 2 = 0 ξ 2 1 = 0 ⇒ ξ 1 2 = 0 ξ 3 1 + l i n ( ξ 0 1 , ξ 0 2 ) = 0 ⇒ ω 03 ξ 1 0 + ω 33 ξ 1 3 + ω 11 ξ 3 1 = 0 ξ 3 2 + l i n ( ξ 0 1 , ξ 0 2 ) = 0 ⇒ ω 00 ξ 2 0 + ω 03 ξ 2 3 + ω 22 ξ 0 2 = 0 , ω 03 ξ 2 0 + ω 33 ξ 2 3 + ω 22 ξ 3 2 = 0 ξ 3 0 = 0 ⇒ ξ 0 3 = 0 , ξ 0 0 = 0 , ξ 3 3 = 0
The coefficients of the linear equations l i n involved depend on the Riemann tensor as in ( [23] ). Accordingly, we may choose only the 2 parametric jets ( ξ 0 1 , ξ 0 2 ) among ( ξ 0 1 , ξ 3 1 , ξ 0 2 , ξ 3 2 ) to which we must add ( ξ 0 , ξ 3 ) in any case as they are not appearing in the Killing equations.
The system is not involutive because its symbol is finite type but non-zero.
Using one more prolongation, all the sections (care again) vanish but ξ 0 and ξ 3 , a result leading to d i m ( R 1 ( 3 ) ) = 2 in a coherent way with the only nonzero Killing vectors { ∂ t , ∂ ϕ } . We have indeed:
ξ 0 1 = 0 , ξ 0 2 = 0 ⇒ ξ 3 1 = 0 , ξ 3 2 = 0 ⇒ ξ 1 0 = 0 , ξ 1 0 = 0 , ξ 2 0 = 0 , ξ 2 3 = 0
Taking therefore into account that the metric only depends on ( x 1 = r , x 2 = cos ( θ ) ) we obtain after three prolongations the first order system:
R 1 ( 3 ) ⊂ R 1 ( 2 ) ⊂ R 1 ( 1 ) = R 1 ⊂ J 1 ( T ) { ξ 3 3 = 0 ξ 3 2 = 0 ξ 3 1 = 0 ξ 3 0 = 0 ξ 2 3 = 0 ξ 2 2 = 0 ξ 2 1 = 0 ξ 2 0 = 0 ξ 1 3 = 0 ξ 1 2 = 0 ξ 1 1 = 0 ξ 1 0 = 0 ξ 0 3 = 0 ξ 0 2 = 0 ξ 0 1 = 0 ξ 0 0 = 0 ξ 2 = 0 ξ 1 = 0 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 • 0 1 2 • 0 1 2 • 0 1 2 • 0 1 • • 0 1 • • 0 1 • • 0 1 • • 0 • • • 0 • • • 0 • • • 0 • • • • • • • • • • •
Surprisingly and contrary to the situation found for the S metric, we have now an involutive first order system with only solutions ( ξ 0 = c s t , ξ 1 = 0 , ξ 2 = 0 , ξ 3 = c s t ) and notice that R 1 ( 3 ) does not depend any longer on the parameters ( m , a ) ∈ K . The difficulty is to know what second members must be used along the procedure met for all the motivating examples. In particular, we have again identities to zero like d 0 ξ 1 − ξ 0 1 = 0 , d 0 ξ 2 − ξ 0 2 = 0 and thus at least 6 third order CC coming from the 6 following components of the Spencer operator, namely:
d 1 ξ 1 − ξ 1 1 = 0 , d 2 ξ 1 − ξ 2 1 = 0 , d 3 ξ 1 − ξ 3 1 = 0 , d 1 ξ 2 − ξ 1 2 = 0 , d 2 ξ 2 − ξ 2 2 = 0 , d 3 ξ 2 − ξ 3 2 = 0
a result that cannot be even imagined from ( [6] [7] [8] ). Of course, proceeding like in the motivating examples, we must substitute in the right members the values obtained from j 2 ( Ω ) and set for example ξ 1 1 = − 1 2 ω 11 ξ ∂ ω 11 while replacing ξ 1 and ξ 2 by the corresponding linear combinations of the Riemann tensor already obtained for the right members of the two zero order equations.
We have the fundamental diagram I no longer depending on ( m , a ) with fiber dimensions:
0 0 0 0 0 ↓ ↓ ↓ ↓ ↓ 0 → Θ → j 1 2 → D 1 8 → D 2 12 → D 3 8 → D 4 2 → 0 ↓ ↓ ↓ ↓ ↓ 0 → 4 → j 1 20 → D 1 40 → D 2 40 → D 3 20 → D 4 4 → 0 ∥ ↓ ↓ ↓ ↓ ↓ 0 → Θ → 4 → D 18 → D 1 32 → D 2 28 → D 3 12 → D 4 2 → 0 ↓ ↓ ↓ ↓ ↓ 0 0 0 0 0
providing the Euler-Poincaré characteristic 4 − 18 + 32 − 28 + 12 − 2 = 0 . However, the only intrinsic concepts associated with a differential sequence are the “extension modules” that only depend on the Kerr differential module but not on the differential sequence and it follows that ( [23] ):
THE ONLY IMPORTANT CONCEPT IS THE GROUP INVOLVED, NOT THE SEQUENCE.
In an equivalent way as g 2 = 0 ⇒ g r + 2 = 0 , ∀ r ≥ 0 , we obtain successively:
d i m ( R 1 ) = d i m ( J 1 ( T ) ) − d i m ( S 2 T * ) = 20 − 10 = 10,
d i m ( R 2 ) = d i m ( J 2 ( T ) ) − d i m ( J 1 ( S 2 T * ) ) = 60 − 50 = 10,
d i m ( R 3 ) = d i m ( R 2 ) − 6 = 4 ⇒ d i m ( R r + 4 ) = d i m ( R 4 ) = d i m ( R 3 ) − 2 = 4 − 2 = 2
and shall use these results from now on.
According to a cut of the preliminary diagram with now m = n = 4 , q = 1 , K = ℚ ( m , a ) , we obtain the following commutative and exact diagrams:
0 0 0 0 ↓ ↓ ↓ ↓ 0 → ρ 1 ( g ′ 2 ) → S 3 T * ⊗ F 0 → T * ⊗ Q 2 → h ′ 1 → 0 ↓ ↓ ↓ ↓ 0 → ρ 1 ( R ′ 2 ) → J 3 ( F 0 ) → J 1 ( Q 2 ) → Q ′ 1 → 0 ↓ ↓ ↓ ↓ 0 → R ′ 2 → J 2 ( F 0 ) → Q 2 → 0 ↓ ↓ ↓ 0 0 0
0 0 0 0 ↓ ↓ ↓ ↓ 0 → 144 + x → 200 → 56 → x → 0 ↓ ↓ ↓ ∥ 0 → 280 + x → 350 → 70 → x → 0 ↓ ↓ ↓ ↓ 0 → 136 → 150 → 14 → 0 ↓ ↓ ↓ 0 0 0
Denoting as before by y the number of additional CC of strict order 3 and by x the number d i m ( h ′ 1 ) = d i m ( Q ′ 1 ) , we discover from the above diagram that the sum of the number of second order CC (that is 14) and the number of differentially independent third order CC obtained by one prolongation of these second order CC is equal to 70 − x . As now d i m ( Q 3 ) = 72 , we obtain therefore 72 − y = 70 − x and thus y = x + 2 . However, as x ≥ 4 because of the 4 divergence conditions implied on the Einstein tensor by the 20 Bianchi identities, we must have y ≥ 6 . As we have already found effectively only 6 CC of order 3, we must have indeed x = 4 effectively and, in any case, we cannot have y = 4 as claimed in ( [6] [7] [8] ).
From the short exact sequence:
0 → R 4 → J 4 ( T ) → J 3 ( F 0 ) → Q 3 → 0
0 → 2 → 280 → 350 → 72 → 0
we obtain the commutative and exact diagrams:
0 ↓ 0 → R 4 → J 4 ( T ) → R ′ 3 → 0 ↓ ↓ ↓ 0 → R 3 → J 3 ( T ) → R ′ 2 → 0 ↓ ↓ 0 0 0 ↓ 0 → 2 → 280 → 278 → 0 ↓ ↓ ↓ 0 → 4 → 140 → 136 → 0 ↓ ↓ 0 0
As a byproduct, we have the commutative and exact diagrams:
0 0 ↓ ↓ 0 → g ′ 3 → ρ 1 ( g ′ 2 ) ↓ ↓ 0 → R ′ 3 → ρ 1 ( R ′ 2 ) ↓ ↓ 0 → R ′ 2 = R ′ 2 → 0 ↓ ↓ 0 0 0 0 0 ↓ ↓ ↓ 0 → 142 → 144 + x → y → 0 ↓ ↓ ∥ 0 → 278 → 280 + x → y → 0 ↓ ↓ ↓ 0 → 136 = 136 → 0 ↓ ↓ 0 0
leading thus to the strict inclusions g ′ 3 ⊂ ρ 1 ( g ′ 2 ) ⇔ R ′ 3 ⊂ ρ 1 ( R ′ 2 ) and to the formula y = x + 2 .
We obtain therefore the following most useful diagram with symbolic notations:
0 ↓ 0 0 y ↓ ↓ ↓ 0 → R ′ 3 → J 3 ( F 0 ) → Q 3 → 0 ↓ ∥ ↓ 0 → ρ 1 ( R ′ 2 ) → J 3 ( F 0 ) → J 1 ( Q 2 ) → Q ′ 1 → 0 ↓ ↓ ↓ ρ 1 ( R ′ 2 ) / R ′ 3 0 x ↓ ↓ 0 0
finally showing that x = d i m ( Q ′ 1 ) , y = d i m ( ρ 1 ( R ′ 2 ) / R ′ 3 ) = d i m ( ρ 1 ( R ′ 2 ) ) − d i m ( R ′ 3 ) , a result leading to the long exact connecting sequence of vector bundles:
0 → R ′ 3 → ρ 1 ( R ′ 2 ) → Q 3 → J 1 ( Q 2 ) → Q ′ 1 → 0.
in agreement with the main theorem of Section 2. We have the following dimensions:
0 ↓ 0 0 6 ↓ ↓ ↓ 0 → 278 → 350 → 72 → 0 ↓ ↓ ↓ 0 → 284 → 350 → 70 → 4 → 0 ↓ ↓ ↓ 6 0 4 ↓ ↓ 0 0
Prolonging once while taking into account that R 5 ≃ R 4 with common dimension 2, namely the dimension of the Kerr algebra generated by { ∂ t , ∂ ϕ } , we obtain the following commutative and exact diagram in which Q 2 and Q 3 are replaced by Q 3 and Q 4 :
0 0 0 ↓ ↓ ↓ 0 → g ′ 4 → S 4 T * ⊗ F 0 → h 4 → 0 ↓ ↓ ↓ 0 → R ′ 4 → J 4 ( F 0 ) → Q 4 → 0 ↓ ↓ ↓ 0 → R ′ 3 → J 3 ( F 0 ) → Q 3 → 0 ↓ ↓ ↓ 0 0 0.
showing that g ′ 4 ≃ S 5 T * ⊗ T with d i m ( R ′ 4 ) = 504 − 2 = 502 and d i m ( Q 4 ) = 700 − 502 = 198 . It follows that R ′ 4 = ρ 1 ( R ′ 3 ) is an involutive fourth order system allowing to construct a formally exact Janet sequence following the Killing operator as in ( [22] ), namely (exercise!!):
0 → Θ → 4 → 1 10 → 4 198 → 1 568 → 1 652 → 1 348 → 1 72 → 0
Of course, such a sequence is quite far from being minimum. However, as the Killing operator for the Kerr metric is not formally integrable as we saw, the corresponding free resolution of the Kerr differential module, namely:
0 → D 72 → 1 D 348 → 1 D 652 → 1 D 568 → 1 D 198 → 4 D 10 → 1 D 4 → p M → 0
is not strictly exact though we have indeed:
r k D ( M ) = 4 − 10 + 198 − 568 + 652 − 348 + 72 = 0
As the maximum size of the matrices involved is d i m ( J 4 ( 198 ) ) × d i m ( J 3 ( 568 ) ) , that is 13860 × 19880 , we hope to have convinced the reader that there is no hope for using computer algebra.
As R ′ 3 ⊂ ρ 1 ( R ′ 2 ) with a strict inclusion, the only possibility to escape from the above difficulty is to use only R ′ 3 and third order CC. However, as we have the strict inclusion S 4 T * ⊗ T ⊂ g ′ 3 with a strict inclusion because 140 < 142 . As for the S metric, we have the crucial theorem:
THEOREM 4.C.1: The operator D 1 : F 0 → j 3 J 3 ( F 0 ) → F 1 = Q 3 generates the CC of D : T → j 1 J 1 ( T ) → F 0 .
Proof: First of all, as R ′ 3 is strictly contained into ρ 1 ( R ′ 2 ) , we have at least one third order generating CC but we already know that we have the six ( d i ξ 1 − ξ i 1 = 0 , d i ξ 2 − ξ i 2 = 0 ) for i = 1 , 2 , 3 .
Collecting the previous results and applying the Spencer operator, we obtain the following commutative diagram in which the two central columns are known to be locally exact ( [1] [5] ):
0 0 0 ↓ ↓ ↓ 0 → Θ → T → D F 0 ↓ j 4 ↓ j 4 ↓ j 3 ↘ D 1 0 → R 4 → J 4 ( T ) → J 3 ( F 0 ) → Q 3 → 0 ↓ d ↓ d ↓ d ↓ 0 → T * ⊗ R 3 → T * ⊗ J 3 ( T ) → T * ⊗ J 2 ( F 0 ) → T * ⊗ Q 2 → 0 ↓ d ↓ d ↓ d ↓ 0 → ∧ 2 T * ⊗ R 2 → ∧ 2 T * ⊗ J 2 ( T ) → ∧ 2 T * ⊗ J 1 ( F 0 ) → 0
Chasing around the right upper commutative triangle, it follows from Theorem 2.A.3 that a section Ω ∈ F 0 with D 1 Ω = 0 is such that there exists ξ ∈ T with D ξ = Ω if and only if the left vertical Spencer sequence is locally exact at T * ⊗ R 3 .
However, one has isomorphisms R 4 ≃ R 5 ≃ R 6 because they have the same dimension equal to 2 and the map R 4 → R 3 is a monomorphism because d i m ( R 3 ) = 4 with parametric jets ( ξ 0 , ξ 3 , ξ 0 1 , ξ 0 2 ) and g 4 = 0 . Accordingly, as the Spencer sequence for the Killing algebroid is locally exact as it is isomorphic to the tensor product of the Poincaré sequence by a Lie algebra of dimension 2, it is locally exact. We have therefore the commutative diagram with exact columns:
0 0 0 0 ↓ ↓ ↓ ↓ 0 → Θ → j 5 R 5 → d T * ⊗ R 4 → d ∧ 2 T * ⊗ R 3 ∥ ↓ ↓ ↓ 0 → Θ → j 4 R 4 → d T * ⊗ R 3 → d ∧ 2 T * ⊗ R 2 ↓ ↓ ↓ ↓ 0 0 → T * ⊗ ( R 3 / R 3 ( 1 ) ) → d ∧ 2 T ⊗ ( R 2 / R 2 ( 1 ) ) ↓ ↓ 0 0
in which the upper row is locally exact. Chasing in this diagram, we discover that the central row is locally exact at T * ⊗ R 3 if the lower Spencer operator d is injective.
Indeed, we have the following representative parametric jets:
d i m ( R 2 ) = 10 ⇒ { ξ 0 , ξ 1 , ξ 2 , ξ 3 , ξ 0 1 , ξ 0 2 , ξ 3 0 , ξ 2 1 , ξ 3 1 , ξ 3 2 }
d i m ( R 3 ) = 4 ⇒ { ξ 0 , ξ 3 , ξ 0 1 , ξ 0 2 } , d i m ( R 4 ) = 2 ⇒ { ξ 0 , ξ 3 }
d i m ( R 3 / R 3 ( 1 ) ) = 2 ⇒ { ξ 0 1 , ξ 0 2 } , d i m ( R 2 / R 2 ( 1 ) ) = 6 ⇒ { ξ 1 , ξ 2 , ξ 3 0 , ξ 2 1 , ξ 3 1 , ξ 3 2 }
Accordingly, we may choose local coordinates ( ξ 0, i 1 , ξ 0, i 2 ) for a representative and a representative of the image by d is for example ( ξ , 0 i 1 = ξ 0 , i 1 − ξ i , 0 1 , ξ , 0 i 2 = ξ 0 , i 2 − ξ i , 0 2 ) . In the kernel of d, we have ( ξ 0 , i 1 = ξ i , 0 1 , ξ 0 , i 2 = ξ i , 0 2 ) , ∀ i = 1 , 2 , 3 but the situation is more tricky than for the S metric.
For i = 1 , we have ξ 1 1 = 0 , ξ 2 1 = 0 , ξ 1 2 = 0 , ξ 2 2 = 0 ⇒ ξ 0 , i 1 = 0 and similarly ξ 0 , i 2 = 0 , ∀ i = 1 , 2 .
Then, as { ξ 0 , ξ 3 } are among the parametric jets of R 3 , we have ξ , 13 0 = ξ 1 , 3 0 − ξ 3 , 1 0 = 0 ⇒ ξ 1 , 3 0 = ξ 3 , 1 0 = 0 and similarly ξ 1 , 3 3 = ξ 3 , 1 3 = 0 . Using the Lie Equations of R 1 ( 2 ) we obtain successively:
ω 00 ξ 1 , 3 0 + ω 03 ξ 1 , 3 3 + ω 11 ξ 0 , 3 1 = 0 ⇒ ξ 0 , 3 1 = 0 , ω 03 ξ 1 , 3 0 + ω 33 ξ 1 , 3 3 + ω 11 ξ 3 , 3 1 = 0 ⇒ ξ 3 , 3 1 = 0
Exchanging 1 and 2, we obtain similarly with the two other framed Lie equations the two relations ξ 0 , 3 2 = 0 , ξ 3 , 3 2 = 0 . Hence, when i = 3 , it follows that ξ 0 , 3 1 = ξ 3 , 0 1 = l i n ( ξ 0 , 0 1 , ξ 0 , 0 2 ) = 0 and ξ 0 , 3 2 = ξ 3 , 0 2 = l i n ( ξ 0 , 0 1 , ξ 0 , 0 2 ) = 0 , a result leading to ξ 0 , 0 1 = 0 , ξ 0 , 0 2 = 0 on one side and ξ 0 , 3 1 = 0 , ξ 0 , 3 2 = 0 on the other side because ξ 0 1 and ξ 0 2 are linearly independent jet coordinates, that is finally ξ 0 , i 1 = 0 , ξ 0 , i 2 = 0 , ∀ i = 0 , 1 , 2 , 3 .
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Like in the example of the S metric, the sequence constructed in the previous theorem for the K metric have the same “jumps” in the order of the successive operators. Indeed, using the same proof but now with d i m ( g ′ 3 ) = d i m ( S 4 T * ⊗ T ) + 2 = 142 , d i m ( A ) = 2 , d i m ( B ) = 6 , we have:
COROLLARY 4.C.2: The symbol of D 1 is not 2-acyclic and the CC operator D 2 is of order 2.
Following closely the procedure used for all the motivating examples, we have thus transformed the search for the generating CC of the Killing operator into a purely mathematical problem of formal integrability and diagram chasing, quite far away from any physical background ( [28] ).