# KILLING OPERATOR FOR THE KERR METRIC

J.-F. Pommaret
CERMICS, Ecole des Ponts ParisTech, Paris, France
<EMAIL_ADDRESS>
ORCID: 0000-0003-0907-2601

ABSTRACT

When ${\cal{D}}:E\rightarrow F$ is a linear differential operator of order $q$
between the sections of vector bundles over a manifold $X$ of dimension $n$,
it is defined by a bundle map $\Phi:J_{q}(E)\rightarrow F=F_{0}$ that may
depend, explicitly or implicitly, on constant parameters $a,b,c,...$. A
”direct problem ” is to find the generating compatibility conditions (CC) in
the form of an operator ${\cal{D}}_{1}:F_{0}\rightarrow F_{1}$. When
${\cal{D}}$ is involutive, that is when the corresponding system
$R_{q}=ker(\Phi)$ is involutive, this procedure provides successive first
order involutive operators ${\cal{D}}_{1},...,{\cal{D}}_{n}$ . Though
${\cal{D}}_{1}\circ{\cal{D}}=0$ implies $ad({\cal{D}})\circ
ad({\cal{D}}_{1})=0$ by taking the respective adjoint operators, then
$ad({\cal{D}})$ may not generate the CC of $ad({\cal{D}}_{1})$ and measuring
such ”gaps” led to introduce extension modules in differential homological
algebra. They may also depend on the parameters and such a situation is well
known in ordinary or partial control theory. When $R_{q}$ is not involutive, a
standard prolongation/projection (PP) procedure allows in general to find
integers $r,s$ such that the image $R^{(s)}_{q+r}$ of the projection at order
$q+r$ of the prolongation ${\rho}_{r+s}(R_{q})=J_{r+s}(R_{q})\cap
J_{q+r+s}(E)\subset J_{r+s}(J_{q}(E))$ is involutive but it may highly depend
on the parameters. However, sometimes the resulting system no longer depends
on the parameters and the extension modules do not depend on the parameters
because it is known that they do not depend on the differential sequence used
for their definition. The purpose of this paper is to study the above problems
for the Kerr $(m,a)$, Schwarzschild $(m,0)$ and Minkowski $(0,0)$ parameters
while computing the dimensions of the inclusions $R^{(3)}_{1}\subset
R^{(2)}_{1}\subset R^{(1)}_{1}=R_{1}\subset J_{1}(T(X))$ for the respective
Killing operators. Other striking motivating examples are also presented.

KEY WORDS
Differential operator; Adjoint operator; Differential sequence; Einstein
equations; Kerr metric; Differential module; Extension module; Contact
structure.

1) INTRODUCTION.

With standard notations of differential geometry, let $(E,F,...)$ be vector
bundles over a manifold $X$ with dimension $n$ with sections $(\xi,\eta,...)$,
tangent bundle $T$ and cotangent bundle $T^{*}$. We shall denote by $J_{q}(E)$
the $q$-jet bundle of $E$ with sections ${\xi}_{q}$ transforming like the
$q$-derivatives $j_{q}(\xi)$. If $\Phi:J_{q}(E)\rightarrow F$ is a bundle
morphism, we shall consider the system $R_{q}=ker(\Phi)\subset J_{q}(E)$ of
order $q$ on $E$. The $r$-prolongation ${\rho}_{r}(R_{q})=J_{r}(R_{q})\cap
J_{q+r}(E)\subset J_{r}(J_{q}(E))$ obtained by differentiating formally $r$
times the given ordinary (OD) or partial (PD) defining equations of $R_{q}$
will be the kernel of the composite morphism
${\rho}_{r}(\Phi):J_{q++r}(E)\rightarrow J_{r}(J_{q}(E))\rightarrow J_{r}(F)$.
The symbol $g_{q+r}=R_{q+r}\cap S_{q+r}T^{*}\otimes E\subset J_{q+r}(E)$ of
$R_{q+r}$ is the $r$-prolongation of the symbol $g_{q}$ of $R_{q}$ and the
kernel of the composite morphism ${\sigma}_{r}(\Phi):S_{q+r}T^{*}\otimes
E\rightarrow S_{r}T^{*}\otimes F$ obtained by restriction. The Spencer
operator $d:R_{q+1}\rightarrow T^{*} \otimes R_{q}:{\xi}_{q+1}\rightarrow
j_{1}({\xi}_{q})-{\xi}_{q+1}$ is obtained by using the fact that
$R_{q+1}=J_{1}(R_{q})\cap J_{q+1}(E)$ and that $J_{1}(R_{q})$ is an affine
vector bundle over $R_{q}$ modelled on $T^{*}\otimes R_{q}$. We shall always
suppose that $\Phi$ is an epimorphism and introduce the vector bundle
$F_{0}=J_{q}(E)/R_{q}$. The system $R_{q}$ is aid to be formally integrable
(FI) if $r+1$ prolongations do not bring new equations of order $q+r$ other
than the ones obtained after only $r$ prolongations, for any $r\geq 0$, that
is all the equations of order $q+r$ can be obtained by differentiating $r$
times only the given equations of order $q$ for any $r\geq 0$. The system is
said to be ” involutive ” if it is FI and the symbol $g_{q}$ is involutive, a
purely algebraic property [7, 9]. In that case, the successive CC operators
can only be at most ${\cal{D}}_{1},...,{\cal{D}}_{n}$ which are first order
and involutive operators.

The next problem is to define the CC operator ${\cal{D}}_{1}:F_{0}\rightarrow
F_{1}:\eta\rightarrow\zeta$ in such a way that the CC of ${\cal{D}}\xi=\eta$
is of the form ${\cal{D}}_{1}\eta=0$. As shown in many books [7-11, 14, 15]
and papers [16, 18, 22], such a problem may be quite difficult because the
order of the generating CC may be quite high. Proceeding in this way, we may
construct the CC ${\cal{D}}_{2}:F_{1}\rightarrow F_{2}$ of ${\cal{D}}_{1}$ and
so on. A second problem shown on the motivating examples is that ” jumps ” in
the successive orders may appear, even on elementary examples. Now, if the map
$\Phi$ depends on constant (or variable) parameters $(a,b,c,...)$, then the
study of the two previous problems becomes much harder because the ranks of
the matrices ${\rho}_{r}(\Phi)$ and/or ${\sigma}_{r}(\Phi)$ may also highly
depend on the parameters as we shall see. The case of the Killing operator for
the Minkowski metric is well known but the study of the Killing operator for
the Schwarzschild and Kerr metrics is rather recent and striking as it proves
that both the numbers of generating second order CC and the numbers of
generating third order CC may change [1-4].

The purpose of the present paper is to revisit these works by using new
homological techniques [10, 15, 20]. It is a matter of fact that they do not
agree with the previous ones for the third order CC. In order to escape from
such an unpleasant situation, we have written this paper in such a way that we
are only using elementary combinatorics and diagram chasing. However, an
equally important second purpose is to question the proper target of the
quoted publications. Indeed, important concepts such as differential extension
modules have been introduced in differential homological algebra and are
known, thanks to a quite difficult theorem [10, 25], to be the only intrinsic
results that could be obtained independently of the differential sequence that
could be used, provided that $\Theta=\ker({\cal{D}})$ is the same, that is
even if one is using another system on $E$ with the same solutions.
Equivalently, this amounts to say, in a few words but a more advanced
language, if we are keeping the same differential module $M$ but changing its
presentation.

However, as we shall see, there are even simple academic systems depending on
parameters but such that a convenient system (say involutive) may no longer
depend on the parameters. Also, in a totally independent way still not
acknowledged, E. Vessiot has shown that certain operators may depend on
geometric objects satisfying non-linear structure equations that are depending
on certain Vessiot structure constants $c$. The simplest example is the
condition of constant Riemannian curvature [7, 24] which is necessary in order
that the Killing system becomes FI but we shall prove that the case of contact
structures is similar. In such situations, we shall prove that the extension
modules only depend on these constants.

We now recall the main results and definitions that are absolutely needed for
the applications.
With canonical epimorphism ${\Phi}_{0}=\Phi:J_{q}(E)\Rightarrow
J_{q}(E)/R_{q}=F=F_{0}$, the various prolongations are described by the
following commutative and exact ”introductory diagram ” used in the sequel:

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&g_{q+r+1}&\rightarrow&S_{q+r+1}T^{*}\otimes
E&\stackrel{{\scriptstyle{\sigma}_{r+1}(\Phi)}}{{\longrightarrow}}&S_{r+1}T^{*}\otimes
F_{0}&\rightarrow&h_{r+1}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{q+r+1}&\rightarrow&J_{q+r+1}(E)&\stackrel{{\scriptstyle{\rho}_{r+1}(\Phi)}}{{\longrightarrow}}&J_{r+1}(F_{0})&\rightarrow&Q_{r+1}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{q+r}&\rightarrow&J_{q+r}(E)&\stackrel{{\scriptstyle{\rho}_{r}(\Phi)}}{{\longrightarrow}}&J_{r}(F_{0})&\rightarrow&Q_{r}&\rightarrow
0\\\ &&&\downarrow&&\downarrow&&\downarrow&\\\ &&&0&&0&&0&\end{array}$

Chasing along the diagonal of this diagram while applying the standard ”snake”
lemma, we obtain the useful ”long exact connecting sequence ” also often used
in the sequel:

$0\rightarrow g_{q+r+1}\rightarrow R_{q+r+1}\rightarrow R_{q+r}\rightarrow
h_{r+1}\rightarrow Q_{r+1}\rightarrow Q_{r}\rightarrow 0$

which is thus connecting in a tricky way FI (lower left) with CC (upper
right).
A key step in the procedure for constructing differential sequences will be to
use the following (difficult) theorems and corollary (For Spencer cohomology
and acyclicity or involutivity (See [7, 11] for details and compare to [5,
26]) (See [7, 9, 15, 22] for more details):

THEOREM 1.1: There is a finite Prolongation/Projection (PP) algorithm
providing two integers $r,s\geq 0$ by successive increase of each of them such
that the new system $R^{(s)}_{q+r}={\pi}^{q+r+s}_{q+r}(R_{q+r+s})$ has the
same solutions as $R_{q}$ but is FI with a $2$-acyclic or involutive symbol
and first order CC. The maximum order of ${\cal{D}}_{1}$ is thus equal to
$r+s+1$ as we used $r+s$ prolongations but it may be lower because certain CC
may generate the higher order ones as will be seen in the motivating examples.

DEFINITION 1.2: A differential sequence is said to be formally exact if it is
exact on the jet level composition of the prolongations involved. A formally
exact sequence is said to be strictly exact if all the operators/systems
involved are FI (See [7, 9, 13] for more details). A strictly exact sequence
is called canonical if all the operators/systems are involutive.

When $d:J_{q+1}(E)\rightarrow T^{*}\otimes J_{q}(E):{\xi}_{q+1}\rightarrow
j_{1}({\xi}_{q})-{\xi}_{q+1}$ is the Spencer operator, we have:

PROPOSITION 1.3: If $R_{q}\subset J_{q}(E)$ and $R_{q+1}\subset J_{q+1}(E)$
are two systems of respective orders $q$ and $q+1$, then
$R_{q+1}\subset{\rho}_{1}(R_{q})$ if and onlty if
${\pi}^{q+1}_{q}(R_{q+1})\subset R_{q}$ and $dR_{q+1}\subset T^{*}\otimes
R_{q}$.

DEFINITION 1.4: Let us ”cut” the preceding introductory diagram by means of a
central vertical line and define $R^{\prime}_{r}=im({\rho}_{r}(\Phi))\subseteq
J_{r}(F_{0})$ with $R^{\prime}_{0}=F_{0}$. Chasing in this diagram, we notice
that ${\pi}^{r+1}_{r}:J_{r+1}(F_{0})\rightarrow J_{r}(F_{0})$ induces an
epimorphism ${\pi}^{r+1}_{r}:R^{\prime}_{r+1}\rightarrow
R^{\prime}_{r},\forall r\geq 0$. However, a chase in this diagram proves that
the kernel of this epimorphism is not $im({\sigma}_{r+1}(\Phi))$ unless
$R_{q}$ is FI (care). For this reason, we shall define it to be exactly
$g^{\prime}_{r+1}$.

THEOREM 1.5: $R^{\prime}_{r+1}\subseteq{\rho}_{1}(R^{\prime}_{r})$ and
$dim({\rho}_{1}(R^{\prime}_{r}))-dim(R^{\prime}_{r+1})$ is the number of new
generating CC of order $r+1$ .

COROLLARY 1.6: The system $R^{\prime}_{r}\subset J_{r}(F_{0})$ becomes FI with
a $2$-acyclic or involutive symbol and
$R^{\prime}_{r+1}={\rho}_{1}(R^{\prime}_{r})\subset J_{r+1}(F_{0})$ when $r$
is large enough.

2) MOTIVATING EXAMPLES

EXAMPLE 2.1: Let $n=2,m=1$ and introduce the trivial vector bundle $E$ with
local coordinates $(x^{1},x^{2},\xi)$ for a section over the base manifold $X$
with local coordinates $(x^{1},x^{2})$. Let us consider the linear second
order system $R_{2}\subset J_{2}(E)$ defined by the two linearly independent
equations $d_{22}\xi=0,\,\,d_{12}\xi+ad_{1}\xi=0$ where $a$ is an arbitrary
constant parameter. Using crossed derivatives, we get the second order system
$R^{(1)}_{2}\subset R_{2}$ defined by the PD equations
$d_{22}\xi=0,d_{12}\xi+ad_{1}\xi=0,a^{2}d_{1}\xi=0$ which is easily seen not
to be involutive. Hence we have two possibilities:
$\bullet\,\,\,a=0$: We obtain the following second order homogeneous
involutive system:

$R^{(1)}_{2}=R_{2}\subset J_{2}(E)\hskip
56.9055pt\left\\{\begin{array}[]{rcl}d_{22}\xi&=&{\eta}^{2}\\\
d_{12}\xi&=&{\eta}^{1}\end{array}\right.\framebox{ $\begin{array}[]{ll}1&2\\\
1&\bullet\end{array}$ }$

with the only first order homogeneous involutive CC
$d_{1}{\eta}^{2}-d_{2}{\eta}^{1}=0$ leading to the Janet sequence:

$0\rightarrow\Theta\rightarrow
E\underset{2}{\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}}F_{0}\underset{1}{\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}}F_{1}\rightarrow
0$

We let the reader check as an easy exercise that $ad({\cal{D}})$ which is of
order $2$ does not generate the CC of $ad({\cal{D}}_{1})$ which is of order
$1$.

$\bullet\,\,\,a\neq 0$: We obtain the second order system $R^{(1)}_{2}$
defined by $d_{22}\xi=0,d_{12}\xi=0,d_{1}\xi=0$ with a strict inclusion
$R^{(1)}_{2}\subset R_{2}$ because $3<4$. We may define
$\eta=d_{1}{\eta}^{2}-d_{2}{\eta}^{1}+a{\eta}^{1}$ and obtain the involutive
and finite type system in $\delta$-regular coordinates:

$R^{(2)}_{2}\subset J_{2}(E)\hskip
56.9055pt\left\\{\begin{array}[]{rcl}d_{22}\xi&=&{\eta}^{2}\\\
d_{12}\xi&=&{\eta}^{1}-\frac{1}{a}\eta\\\
d_{11}\xi&=&\frac{1}{a^{2}}d_{1}\eta\\\
d_{1}\xi&=&\frac{1}{a^{2}}\eta\end{array}\right.\framebox{
$\begin{array}[]{ll}1&2\\\ 1&\bullet\\\ 1&\bullet\\\
\bullet&\bullet\end{array}$ }$

Counting the dimensions, we have the following strict inclusions by comparing
the dimensions:

$R^{(2)}_{2}\subset R^{1)}_{2}\subset R_{2}\subset J_{2}(E),\hskip
85.35826pt2<3<4<6$

LEMMA 2.1.1: The symbol $g_{2+r}$ is involutive with $dim(g_{2+r})=1,\forall
r\geq 0$. Moreover, we have $dim(R_{2+r})=4,\forall r\geq 0$.

Proof: Using jet notation, the $4$ parametric jets of $R_{2}$ are
$(\xi,{\xi}_{1},{\xi}_{2},{\xi}_{11})$. The $4$ parametric jets of $R_{3}$ are
now $(\xi,{\xi}_{2},{\xi}_{11},{\xi}_{111})$ and so on. Accordingly, the
dimension of $g_{2+r}$ is $1$ because the only parametric jet is
${\xi}_{1....1}$. We have the short exact sequence $0\rightarrow
g_{r+2}\rightarrow R_{r+2}\rightarrow R^{(1)}_{r+1}\rightarrow 0$ and the
symbol of $R_{2}$ is involutive. It follows from a delicate but crucial
theorem (See [7], Theorem 2.4.5 p 70 and proposition 2.5.1 p 76 with $q=2$)
that ${\rho}_{r}(R^{(1)}_{2})=R^{(1)}_{2+r}\,\,,\forall r\geq 0$ with
$dim(R^{(1)}_{r+1})=3$, a result leading to $dim(R_{r+2})=3+1=4$ by counting
the dimensions. As $R^{(1)}_{2}$ does not depend any longer on the parameter,
the general solution is easily seen to be of the form $\xi=cx^{2}+d$ and is
thus only depending on two arbitrary constants, contrary to what could be
imagined from this lemma but in a coherent way with the fact that
$dim(R^{(2)}_{2+r})=2,\forall r\geq 0$.
$\Box$

After differentiating twice, we could be waiting for CC of order $3$. However,
we obtain the $4$ CC:

$d_{2}{\eta}^{1}-\frac{1}{a}d_{2}\eta-
d_{1}{\eta}^{2}=0,\frac{1}{a^{2}}d_{12}\eta-
d_{1}{\eta}^{1}+\frac{1}{a}d_{1}\eta=0,\frac{1}{a^{2}}d_{2}\eta-{\eta}^{1}+\frac{1}{a}\eta=0,\frac{1}{a^{2}}(d_{1}\eta-
d_{1}\eta)=0$

The last CC that we shall call ”identity to zero ” must not be taking into
account. The second CC is just the derivative with respect to $x^{1}$ of the
third CC which amounts to

$(d_{12}{\eta}^{2}-d_{22}{\eta}^{1}+ad_{2}{\eta}^{1})-a^{2}{\eta}^{1}+a(d_{1}{\eta}^{2}-d_{2}{\eta}^{1}+a{\eta}^{1})=0\Leftrightarrow
d_{12}{\eta}^{2}-d_{22}{\eta}^{1}+ad_{1}{\eta}^{2}=0$

which is a second order CC amounting to the first. Hence we get the only
generating CC operator ${\cal{D}}_{1}:({\eta}^{1},{\eta}^{2})\rightarrow
d_{12}{\eta}^{2}-d_{22}{\eta}^{1}+ad_{1}{\eta}^{2}=\zeta$ which is thus
formally surjective.

For helping the reader, we recall that basic elementary combinatorics
arguments are giving $dim(S_{q}T^{*})=q+1$ while $dim(J_{q}(E))=(q+1)(q+2)/2$
because $n=2$ and $m=dim(E)=1$.

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&g_{2}&\rightarrow&S_{2}T^{*}\otimes
E&\longrightarrow&F_{0}&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\parallel&&&\\\
0\rightarrow&R_{2}&\rightarrow&J_{2}(E)&\longrightarrow&F_{0}&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&R_{1}&\rightarrow&J_{1}(E)&\longrightarrow&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\ &0&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&1&\longrightarrow&3&\longrightarrow&2&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\parallel&&&\\\
0\rightarrow&4&\rightarrow&6&\longrightarrow&2&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&3&\rightarrow&3&\longrightarrow&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\ &0&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&g_{3}&\rightarrow&S_{3}T^{*}\otimes E&\rightarrow&T^{*}\otimes
F_{0}&\rightarrow&h_{1}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{3}&\rightarrow&J_{3}(E)&\rightarrow&J_{1}(F_{0})&\rightarrow&Q_{1}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{2}&\rightarrow&J_{2}(E)&\rightarrow&F_{0}&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&1&\longrightarrow&4&\rightarrow&4&\rightarrow&1&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&10&\rightarrow&6&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&4&\rightarrow&6&\rightarrow&2&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&g_{4}&\rightarrow&S_{4}T^{*}\otimes
E&\rightarrow&S_{2}T^{*}\otimes F_{0}&\rightarrow&h_{2}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{4}&\rightarrow&J_{4}(E)&\rightarrow&J_{2}(F_{0})&\rightarrow&Q_{2}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{3}&\rightarrow&J_{3}(E)&\rightarrow&J_{1}(F_{0})&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&1&\rightarrow&5&\rightarrow&6&\rightarrow&2&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&15&\rightarrow&12&\rightarrow&1&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&10&\rightarrow&6&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

Using these diagrams, we obtain successively till we stop:

$R^{\prime}_{0}=F_{0}\Rightarrow
R^{\prime}_{1}=J_{1}(F_{0})\Rightarrow{\rho}_{1}(R^{\prime}_{1})=J_{2}(F_{0})\Rightarrow
R^{\prime}_{2}\subset{\rho}_{1}(R^{\prime}_{1})\Rightarrow
R^{\prime}_{3}={\rho}_{1}(R^{\prime}_{2})$

Hence, the number of generating CC of order $1$ is zero and the number of
generating CC of strict order $2$ is
$dim({\rho}_{1}(R^{\prime}_{1}))-dim(R^{\prime}_{2})=12-(15-4)=12-11=1$ in a
coherent way.
Setting $F_{1}=Q_{2}$ with $dim(Q_{2})=1$, we obtain the commutative diagram:

$\begin{array}[]{rcccccccl}&0&&0&&0&&0&\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&g_{5}&\rightarrow&S_{5}T^{*}\otimes E&\rightarrow&\framebox{
$S_{3}T^{*}\otimes F_{0}$}&\rightarrow&T^{*}\otimes F_{1}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{5}&\rightarrow&J_{5}(E)&\rightarrow&J_{3}(F_{0})&\rightarrow&J_{1}(F_{1})&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{4}&\rightarrow&J_{4}(E)&\rightarrow&J_{2}(F_{0})&\rightarrow&F_{1}&\rightarrow
0\\\ &&&\downarrow&&\downarrow&&\downarrow&\\\ &&&0&&0&&0&\end{array}$

with dimensions:

$\begin{array}[]{rcccccccl}&0&&0&&0&&0&\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&1&\rightarrow&6&\rightarrow&\framebox{
$8$}&\rightarrow&2&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&21&\rightarrow&20&\rightarrow&3&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&15&\rightarrow&12&\rightarrow&1&\rightarrow 0\\\
&&&\downarrow&&\downarrow&&\downarrow&\\\ &&&0&&0&&0&\end{array}$

The upper symbol sequence is not exact at $S_{3}T^{*}\otimes F_{0}$ even
though the two other sequences are exact on the jet level. As a byproduct we
have the exact sequences $\forall r\geq 0$:

$0\rightarrow R_{r+4}\rightarrow J_{r+4}(E)\rightarrow
J_{r+2}(F_{0})\rightarrow J_{r}(F_{1})\rightarrow 0$

Such a result can be checked directly through the identity:

$4-(r+5)(r+6)/2+2(r+3)(r+4)/2-(r+1)(r+2)/2=0$

We obtain therefore the formally exact sequence we were looking for, namely:

$0\rightarrow\Theta\rightarrow
E\underset{2}{\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}}F_{0}\underset{2}{\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}}F_{1}\rightarrow
0$

The surprising fact is that, in this case $ad({\cal{D}})$ generates the CC of
$ad({\cal{D}}_{1})$. Indeed, multiplying by the Lagrange multiplier test
function $\lambda$ and integrating by parts, we obtain the second order
operator $\lambda\rightarrow(-d_{22}\lambda={\mu}^{1},d_{12}\lambda-
ad_{1}\lambda={\mu}^{2})$ and thus
$-a^{2}d_{1}\lambda=d_{1}{\mu}^{1}+d_{2}{\mu}^{2}+a{\mu}^{2}$. Substituting,
we finally get the only second order CC
$d_{12}{\mu}^{1}+d_{22}{\mu}^{2}-ad_{1}{\mu}^{1}=0$.

In the differential module framework over the commutative ring
$D=K[d_{1},d_{2}]$ of differential operators with coefficients in the
trivially differential field $K=\mathbb{Q}(a)$, we have the free resolution:

$0\rightarrow
D\underset{2}{\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}}D^{2}\underset{2}{\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}}D\rightarrow
M\rightarrow 0$

of the differential module $M$ with Euler-Poincaré characteristic
$rk_{D}(M)=1-2+1=0$. We recall that $R=R_{\infty}=hom_{K}(M,K)$ is a
differential module for the Spencer operator $d:R\rightarrow T^{*}\otimes
R:R_{q+1}\rightarrow T^{*}\otimes R_{q}$ (See [15, 19] for more details).

The two following examples will show how the differential extension modules
may depend on the Vessiot structure constants.

EXAMPLE 2.2: With $m=n=2,q=1,K=\mathbb{Q}<\omega>$ and $\omega=(\alpha,\beta)$
with $\alpha\in T^{*},\beta\in{\wedge}^{2}T^{*}$, let us consider the Lie
operator
${\cal{D}}:T\rightarrow\Omega:\xi\rightarrow{\cal{L}}(\xi)\omega=(A={\cal{L}}(\xi)\alpha,B={\cal{L}}(\xi)\beta)$.
The corresponding first order system:

$A_{i}\equiv{\alpha}_{r}{\partial}_{i}{\xi}^{r}+{\xi}^{r}{\partial}_{r}{\alpha}_{i}=0,\,\,\,B\equiv\beta{\partial}_{r}{\xi}^{r}+{\xi}^{r}{\partial}_{r}\beta=0$

is involutive whenever $\beta\neq 0$ and  $d\alpha=c\beta$ where now $d$ is
the exterior derivative and $c=cst$.
We notice that $\bar{\omega}$ and $\omega$ provide the same system of Lie
equations if and only if $\bar{\alpha}=a\alpha,\bar{\beta}=\textpm b\beta$
with $a,b\neq 0$ and thus $\bar{c}=\frac{a}{b}c$, a result showing that the
only critical value of $c$ is $c=0$.
We have the formally exact differential sequence:

$0\rightarrow\Theta\rightarrow
T\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}T^{*}{\times}_{X}{\wedge}^{2}T^{*}\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}{\wedge}^{2}T^{*}\rightarrow
0$

or the resolution:

$0\rightarrow
D\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}D^{3}\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}D^{2}\stackrel{{\scriptstyle
p}}{{\longrightarrow}}M\rightarrow 0$

Multiplying $(A_{1},A_{2},B)$ respectively by
$({\mu}^{1},{\mu}^{2},{\mu}^{3})$, we obtain $ad({\cal{D}})$ in the form:

$-{\alpha}_{1}({\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2})-\beta({\partial}_{1}{\mu}^{3}-c{\mu}^{2})={\nu}^{1},\,\,-{\alpha}_{2}({\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2})-\beta({\partial}_{2}{\mu}^{3}+c{\mu}^{1})={\nu}^{2}$

Then, multiplying ${\partial}_{1}A_{2}-{\partial}_{2}A_{1}-cB$ by $\lambda$,
we obtain $ad({\cal{D}}_{1})$ as:

${\partial}_{2}\lambda={\mu}^{1},\,\,-{\partial}_{1}\lambda={\mu}^{2},\,\,-c\lambda={\mu}^{3}$

We have therefore to consider the two cases:
$\bullet$ $c=0$: We have the new CC
${\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2}=0$ and ${\mu}^{3}=0$. It
follows that the torsion module ${ext}^{1}(M)\neq 0$ is generated by the
residue of ${\mu}^{3}={\nu}^{\prime}$ because $\alpha\neq 0$ and we may thus
suppose that ${\alpha}_{1}\neq 0$. As for ${ext}^{2}(M)$, this torsion module
is just defined by the system
${\partial}_{2}\lambda=0,{\partial}_{1}\lambda=0$ for $\lambda$ and thus
${ext}^{2}(M)\neq 0$.
$\bullet$ $c\neq 0$: We must have the new CC:

${\partial}_{1}{\mu}^{3}-c{\mu}^{2}=0,{\partial}_{2}{\mu}^{3}+c{\mu}^{1}=0\Rightarrow{\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2}=0$

It follows that ${ext}^{1}(M)$ is now generated by the residue of
${\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2}={\nu}^{\prime}$. Finally,
$ker(ad({\cal{D}}_{1}))$ is defined by $\lambda=0$ and thus ${ext}^{2}(M)=0$.
Hence, both ${ext}^{1}(M)$ and ${ext}^{2}(M)$ highly depend on the Vessiot
structure constant $c$.

EXAMPLE 2.3: (Contact transformations)
With $m=n=3,q=1,K=\mathbb{Q}(x^{1},x^{2},x^{3})$ or simply $\mathbb{Q}(x)$, we
may introduce the $1$-form $\alpha=dx^{1}-x^{3}dx^{2}\in T^{*}$ and consider
the system of finite Lie equations defined by
$j_{1}(f)^{-1}(\alpha)=\rho(x)\alpha$. Eliminating the factor $\rho$ and
linearizing at the $q$-jet of the identity, we obtain a first order system
$R_{1}\subset J_{1}(T)$ made by ${\eta}^{1}$ and ${\eta}^{2}$ below which is
not even formally integrable:

$\left\\{\begin{array}[]{lcr}{\partial}_{3}{\xi}^{1}-x^{3}{\partial}_{3}{\xi}^{2}&=&{\eta}^{2}\\\
{\partial}_{2}{\xi}^{1}-x^{3}{\partial}_{2}{\xi}^{2}+x^{3}{\partial}_{1}{\xi}^{1}-(x^{3})^{2}{\partial}_{1}{\xi}^{2}-{\xi}^{3}&=&{\eta}^{1}\end{array}\right.\framebox{$\begin{array}[]{lll}1&2&3\\\
1&2&\bullet\end{array}$ }$

with one equation of class $3$ and one equation of class $2$. Strikingly, the
symbol $g_{1}$ is involutive with characters
${\alpha}^{1}_{1}=3,{\alpha}^{2}_{1}=2,{\alpha}^{3}_{1}=2$, a result leading
to $dim(g_{1})=3+2+2=7,dim(g_{2})=3+(2\times 2)+(3\times 2)=13$ and more
generally $dim(g_{r+1})=3+2(r+1)+(r+1)(r+2)=r^{2}+5r+7,\forall r\geq 0$.
However, the system $R_{1}$ is not formally integrable and thus not
involutive. The PP procedure brings
${\eta}^{3}={\partial}_{2}{\eta}^{2}-{\partial}_{3}{\eta}^{1}+x^{3}{\partial}_{1}{\eta}^{2}$
below and we get the involutive system $R^{(1)}_{1}\subset R_{1}$ of
infinitesimal Lie equations having the same solutions, which is already in
$\delta$-regular coordinates:

$\left\\{\begin{array}[]{lcr}{\partial}_{3}{\xi}^{3}+{\partial}_{2}{\xi}^{2}-{\partial}_{1}{\xi}^{1}+2x^{3}{\partial}_{1}{\xi}^{2}&=&{\eta}^{3}\\\
{\partial}_{3}{\xi}^{1}-x^{3}{\partial}_{3}{\xi}^{2}&=&{\eta}^{2}\\\
{\partial}_{2}{\xi}^{1}-x^{3}{\partial}_{2}{\xi}^{2}+x^{3}{\partial}_{1}{\xi}^{1}-(x^{3})^{2}{\partial}_{1}{\xi}^{2}-{\xi}^{3}&=&{\eta}^{1}\end{array}\right.\framebox{$\begin{array}[]{lll}1&2&3\\\
1&2&3\\\ 1&2&\bullet\end{array}$ }$

with two equations of class $3$, one equation of class $2$ and thus one CC of
order $1$ described by the trivially involutive first order operator
${\cal{D}}_{1}:\eta\rightarrow{\partial}_{3}{\eta}^{1}-{\partial}_{2}{\eta}^{2}-x^{3}{\partial}_{1}{\eta}^{2}+{\eta}^{3}=\zeta$.

The link existing between these two approaches is not evident as it highly
depends on the fact that ${\rho}_{r}(R^{(1)}_{1})=R^{(1)}_{r+1}$, according to
the (quite difficult) theorem 2.4.5 of (See [7], p 70) that must be compared
to [5, 26]. Setting $\Phi:J_{1}(T)\rightarrow J_{1}(T)/R^{(1)}_{1}=F_{0}$ with
$dim(F_{0})=3$ and ${\Phi}^{\prime}:J_{1}(T)\rightarrow
J_{1}(T)/R_{1}=F^{\prime}_{0}$ with $dim(F^{\prime}_{0})=2$, the inclusion
$R^{(1)}_{1}\subset R_{1}$ induces an epimorphism $F_{0}\rightarrow
F^{\prime}_{0}\rightarrow 0$ and we have the commutative diagram when $r\geq
2$ and $F_{1}=coker({\rho}_{1}(\Phi))$:

$\begin{array}[]{rcccccccl}&0&&0&&0&&&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&g_{r+1}&\rightarrow&S_{r+1}T^{*}\otimes
T&\stackrel{{\scriptstyle{\sigma}_{r}({\Phi}^{\prime})}}{{\longrightarrow}}&S_{r}T^{*}\otimes
F^{\prime}_{0}&&&\\\ &\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&R_{r+1}&\rightarrow&J_{r+1}(T)&\stackrel{{\scriptstyle{\rho}_{r}({\Phi}^{\prime})}}{{\longrightarrow}}&J_{r}(F^{\prime}_{0})&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&R^{(1)}_{r}&\rightarrow&J_{r}(T)&\stackrel{{\scriptstyle{\rho}_{r-1}(\Phi)}}{{\longrightarrow}}&J_{r-1}(F_{0})&\rightarrow&J_{r-2}(F_{1})&\rightarrow
0\\\ &\downarrow&&\downarrow&&&&&\\\ &0&&0&&&.&&\end{array}$

The operator ${\cal{D}}^{\prime}={\Phi}^{\prime}\circ j_{1}$ is formally
surjective, a snake chase proves that the columns are exact and the bottom
sequence is exact as it is just providing the formally exact Janet sequence:

$0\rightarrow\Theta\rightarrow
T\underset{1}{\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}}F_{0}\underset{1}{\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}}F_{1}\rightarrow
0$

Hence, it remains to check that the central row is also exact. For this, we
notice that:

$dim(R^{(1)}_{r})=(r+1)(r+2)(r+3)/2-r(r+1)(r+2)/2+r(r-1)(r+1)/6=(r+1)(r^{2}+8r+18)$
$dim(R_{r+1})=(r+2)(r+3)(r+4)/2-(r+1)(r+2)(r+3)/3=(r+2)(r+3)(r+10)/6$

and check that $dim(R_{r+1})-dim(R^{(1)}_{r})=dim(g_{r+1})$, a result leading
to the sequence:

$0\rightarrow\Theta\rightarrow
T\underset{1}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}}}{{\longrightarrow}}}F^{\prime}_{0}\rightarrow
0$ $\hskip
28.45274pt({\xi}^{1},{\xi}^{2},{\xi}^{3})\stackrel{{\scriptstyle{\cal{D}}^{\prime}}}{{\longrightarrow}}({\eta}^{1},{\eta}^{2})\rightarrow
0$

Now, it is well known that this contact operator ${\cal{D}}$ allowing to
define the contact module $M$ admits an injective parametrization by one
arbitrary potential function $\phi$ as follows:

${\cal{D}}_{-1}:\phi\rightarrow(-x^{3}{\partial}_{3}\phi+\phi={\xi}^{1},\,-{\partial}_{3}\phi={\xi}^{2},\,{\partial}_{2}\phi+x^{3}{\partial}_{1}\phi={\xi}^{3})\,\,\,\Rightarrow\,\,\,{\xi}^{1}-x^{3}{\xi}^{2}=\phi\,\,\,(lift)$

and thus $M\simeq D$. We have obtained the following locally exact
differential sequence ([8]):

$0\rightarrow\phi\stackrel{{\scriptstyle{\cal{D}}_{-1}}}{{\longrightarrow}}({\xi}^{1},{\xi}^{2},{\xi}^{3})\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}({\eta}^{1},{\eta}^{2},{\eta}^{3})\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}\zeta\rightarrow
0$

which is not a Janet sequence because the first operator is not involutive.
We obtain therefore a resolution of $M$ [13]:

$0\rightarrow
D\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}D^{3}\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}D^{3}\rightarrow
M\rightarrow 0$

but also a resolution which is not strictly exact [13] because $R_{1}$ is not
even formally integrable:

$0\rightarrow
D^{2}\stackrel{{\scriptstyle{\cal{D}}^{\prime}}}{{\longrightarrow}}D^{3}\rightarrow
M \rightarrow 0\hskip 28.45274pt\Rightarrow\hskip 28.45274ptM\oplus
D^{2}\simeq D^{3}$

As $M$ is free and thus projective, the respective adjoint sequences are also
exact. Applying $hom_{D}(\bullet,D)$, we obtain therefore a split exact
sequence of free right (care) differential modules. Passing from right to left
differential modules by ususing the side changing procedure $N_{D}\rightarrow
N={}_{D}N=hom_{K}({\wedge}^{n}T^{*},N_{D})$, it follows that the adjoint
sequence is exact too, though not strictly exact. As such a result does not
depend on the differential sequence used, according to a (difficult) general
theorem in homological algebra, we shall prove it directly through a
(delicate) technical explicit computation for the first of the two previous
ones and let the reader prove it similarly for the other.

For this, let us multiply $\zeta$ by a test function $\lambda$ and integrate
by parts in order to obtain:

$ad({\cal{D}}_{1}):\lambda\rightarrow\left\\{\begin{array}[]{lcr}-{\partial}_{3}\lambda&=&{\mu}^{1}\\\
{\partial}_{2}\lambda+x^{3}{\partial}_{1}\lambda&=&{\mu}^{2}\\\
\lambda&=&{\mu}^{3}\end{array}\right.$

This operator is injective but not involutive as the corresponding system
$R_{1}\subset J_{1}(\lambda)$ is far from being even formally integrable. In a
coherent way with the standard second Spencer sequence [ ], the corresponding
involutive operator is $j_{1}$ as follows, with 6 first order CC for itself
but first and second order for $\mu$ after substitution, that is a second
order system which is not FI.

$j_{1}:\lambda\rightarrow\left\\{\begin{array}[]{lcr}{\partial}_{3}\lambda&=&-{\mu}^{1}\\\
{\partial}_{2}\lambda&=&{\mu}^{2}-x^{3}{\partial}_{1}{\mu}^{3}\\\
{\partial}_{1}\lambda&=&{\partial}_{1}{\mu}^{3}\\\
\lambda&=&{\mu}^{3}\end{array}\right.\framebox{$\begin{array}[]{lll}1&2&3\\\
1&2&\bullet\\\ 1&\bullet&\bullet.\\\ \bullet&\bullet&\bullet\end{array}$ }$

$\bullet$ 1 identity to zero
${\partial}_{1}{\mu}^{3}-{\partial}_{1}{\mu}^{3}=0$.
$\bullet$ 2 first order CC, namely
$-{\nu}^{3}\equiv{\partial}_{3}{\mu}^{3}+{\mu}^{1}=0$,
$-({\nu}^{2}+x^{3}{\nu}^{1})\equiv{\partial}_{2}{\mu}^{3}+x^{3}{\partial}_{1}{\mu}^{3}-{\mu}^{2}=0$
which are differentially independent because, when ${\mu}^{3}$ is given
arbitrarily, the first is providing ${\mu}^{1}$ while the second is providing
${\mu}^{2}$.
$\bullet$ 2 second order CC that are just prolongations of the two previous
first order ones.
$\bullet$ 1 second order CC
${\partial}_{3}{\mu}^{2}-x^{3}{\partial}_{13}{\mu}^{3}-{\partial}_{1}{\mu}^{3}+{\partial}_{2}{\mu}^{1}=0$.
Using the previous second order CC
${\partial}_{13}{\mu}^{3}+{\partial}_{1}{\mu}^{1}=0$, we finally get the new
first order CC
$-{\nu}^{1}\equiv{\partial}_{3}{\mu}^{2}+x^{3}{\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{1}-{\partial}_{1}{\mu}^{3}=0$
which is a differential consequence of the two previous first order CC because
we have (by chance !) the involutive system:

$\left\\{\begin{array}[]{lcl}{\partial}_{3}{\mu}^{3}+{\mu}^{1}&=&-{\nu}^{3}\\\
{\partial}_{3}{\mu}^{2}+x^{3}{\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{1}-{\partial}_{1}{\mu}^{3}&=&-{\nu}^{1}\\\
{\partial}_{2}{\mu}^{3}+x^{3}{\partial}_{1}{\mu}^{3}-{\mu}^{2}&=&-({\nu}^{2}+x^{3}{\nu}^{1})\end{array}\right.\framebox{$\begin{array}[]{lll}1&2&3\\\
1&2&3\\\ 1&2&\bullet\end{array}$ }$

providing the only first order CC:

$ad({\cal{D}}_{-1}):({\nu}^{1},{\nu}^{2},{\nu}^{3})\rightarrow
x^{3}{\partial}_{3}{\nu}^{1}+2{\nu}^{1}+{\partial}_{3}{\nu}^{2}-{\partial}_{2}{\nu}^{3}-x^{3}{\partial}_{1}{
\nu}^{3}=\theta$

As a byproduct, we can find the formally exact differential sequence:

$0\leftarrow\theta\stackrel{{\scriptstyle
ad({\cal{D}}_{-1})}}{{\longleftarrow}}({\nu}^{1},{\nu}^{2},{\nu}^{3})\stackrel{{\scriptstyle
ad({\cal{D}})}}{{\longleftarrow}}({\mu}^{1},{\mu}^{2},{\mu}^{3})\stackrel{{\scriptstyle
ad({\cal{D}}_{1})}}{{\longleftarrow}}\lambda\leftarrow 0$

which is exactly the adjoint of the first differential sequence we have
provided.

Coming back to the Vessiot structure equations, we notice that $\alpha$ is not
invariant by the contact Lie pseudogroup and cannot be considered as an
associated geometric object. We have shown in ([8], p 684-691) that the
corresponding geometric object is a $1$-form density $\omega$ leading to the
system of infinitesimal Lie equations in Medolaghi form:

${\Omega}_{i}\equiv({\cal{L}}(\xi)\omega)_{i}\equiv{\omega}_{r}{\partial}_{i}{\xi}^{r}-\frac{1}{2}{\omega}_{i}{\partial}_{r}{\xi}^{r}+{\xi}^{r}{\partial}_{r}{\omega}_{i}=0$

that may be also written as:

${\partial}_{i}({\omega}_{r}{\xi}^{r})-\frac{1}{2}{\omega}_{i}{\partial}_{r}{\xi}^{r}+{\xi}^{r}({\partial}_{r}{\omega}_{i}-{\partial}_{i}{\omega}_{r})={\Omega}_{i}$

and to the only Vessiot structure equations, still not known today:

${\omega}_{1}({\partial}_{2}{\omega}_{3}-{\partial}_{3}{\omega}_{2})+{\omega}_{2}({\partial}_{3}{\omega}_{1}-{\partial}_{1}{\omega}_{3})+{\omega}_{3}({\partial}_{1}{\omega}_{2}-{\partial}_{2}{\omega}_{1})=c$

with the only structure constant $c$. In the present contact situation, we may
choose $\omega=(1,-x^{3},0)$ and get $c=1$ but we may also choose
$\omega=(1,0,0)$ and get $c=0$, these two choices both bringing an involutive
system. Let us prove that the situation becomes completely different with the
new system:

$-2{\Omega}_{1}\equiv{\partial}_{3}{\xi}^{3}+{\partial}_{2}{\xi}^{2}-{\partial}_{1}{\xi}^{1}=0,\,\,{\Omega}_{2}\equiv{\partial}_{2}{\xi}^{1}=0,\,\,{\Omega}_{3}\equiv{\partial}_{3}{\xi}^{1}=0$

having the only CC $d_{2}{\Omega}_{3}-d_{3}{\Omega}_{2}=0$.
Multilying the three previous equations by the three test functions $\mu$, the
only CC by the test function $\lambda$ and integrating by parts, we get the
adjoint operators:

$0={\mu}^{1},\,\,{\partial}_{3}\lambda={\mu}^{2},\,\,-{\partial}_{2}\lambda={\mu}^{3}$
${\partial}_{1}{\mu}^{1}-{\partial}_{2}{\mu}^{2}-{\partial}_{3}{\mu}^{3}={\nu}^{1},\,\,-{\partial}_{2}{\mu}^{1}={\nu}^{2},\,\,-{\partial}_{3}{\mu}^{1}={\nu}^{3}$

It follows that $0\neq D{\xi}^{1}=t(M)\subset M$ with a strict inclusion and
${ext}^{1}(M)\neq 0$. Similarly, $ker(ad({\cal{D}}_{1}))$ is defined by
${\partial}_{2}\lambda=0,{\partial}_{3}\lambda=0$ and thus ${ext}^{2}(M)\neq
0$.
Our problem will be now to construct and compare the differential sequences:

$\phi\stackrel{{\scriptstyle{\cal{D}}_{-1}}}{{\longrightarrow}}\xi\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}\Omega\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}C$
$\theta\stackrel{{\scriptstyle
ad({\cal{D}}_{-1})}}{{\longleftarrow}}\nu\stackrel{{\scriptstyle
ad({\cal{D}})}}{{\longleftarrow}}\mu\stackrel{{\scriptstyle
ad({\cal{D}}_{1})}}{{\longleftarrow}}\lambda$

For this, linearizing the only Vessiot structure equation, we get the CC
operator ${\cal{D}}_{1}$ and the corresponding system ${\cal{D}}_{1}\Omega=C$
in the form:

${\omega}_{1}({\partial}_{2}{\Omega}_{3}-{\partial}_{3}{\Omega}_{2})+{\omega}_{2}({\partial}_{3}{\Omega}_{1}-{\partial}_{1}{\Omega}_{3})+{\omega}_{3}({\partial}_{1}{\Omega}_{2}-{\partial}_{2}{\Omega}_{1})$
$+({\partial}_{2}{\omega}_{3}-{\partial}_{3}{\omega}_{2}){\Omega}_{1}+({\partial}_{3}{\omega}_{1}-{\partial}_{1}{\omega}_{3}){\Omega}_{2}+({\partial}_{1}{\omega}_{2}-{\partial}_{2}{\omega}_{1}){\Omega}_{3}=C$

Multiplying on the left by the test function $\lambda$ and integrating by
parts, we get the operator $ad({\cal{D}}_{1})$ in the form:

$\left\\{\begin{array}[]{lccl}{\Omega}_{1}\rightarrow&{\omega}_{3}{\partial}_{2}\lambda-{\omega}_{2}{\partial}_{3}\lambda+2({\partial}_{2}{\omega}_{3}-{\partial}_{3}{\omega}_{2})\lambda&={\mu}^{1}\\\
{\Omega}_{2}\rightarrow&{\omega}_{1}{\partial}_{3}\lambda-{\omega}_{3}{\partial}_{1}\lambda+2({\partial}_{3}{\omega}_{1}-{\partial}_{1}{\omega}_{3})\lambda&={\mu}^{2}\\\
{\Omega}_{3}\rightarrow&{\omega}_{2}{\partial}_{1}\lambda-{\omega}_{1}{\partial}_{2}\lambda+2({\partial}_{1}{\omega}_{2}-{\partial}_{2}{\omega}_{1})\lambda&={\mu}^{3}\\\
\end{array}\right.$

We obtain therefore the crucial formula $2c\,\lambda={\omega}_{i}{\mu}^{i}$
showing how the previous sequences are essentially depending on the Vessiot
structure constant $c$. Indeed, if $c\neq 0$, then $\mu=0\Rightarrow\lambda=0$
and the operator $ad({\cal{D}}_{1})$ is injective. This is the case when
$\omega=(1,-x^{3},0)\Rightarrow c=1\Rightarrow\lambda=0$. On the contrary, if
$c=0$, then the operator $ad({\cal{D}}_{1})$ may not be injective as can be
seen by choosing $\omega=(1,0,0)$. Indeed, in this case we get a kernel
defined by ${\partial}_{3}\lambda=0,{\partial}_{2}\lambda=0$.
Finally, in order to exhibit the generating CC of $ad({\cal{D}}_{1})$ when
$c\neq 0$, we just need to substitute $\lambda=(1/2c){\omega}_{i}{\mu}^{i}$ in
the previous equations $ad({\cal{D}}_{1})\lambda=\mu$. On the other side,
multiplying the equations ${\cal{D}}\xi=\Omega$ by test functions ${\mu}^{i}$
and integrating by parts, we get $ad({\cal{D}})\mu=\nu$ in the form:

$\left\\{\begin{array}[]{lccl}{\xi}^{1}\rightarrow&&-{\partial}_{i}({\omega}_{1}{\mu}^{i})+\frac{1}{2}{\partial}_{1}({\omega}_{i}{\mu}^{i})+({\partial}_{1}{\omega}_{i}){\mu}^{i}&={\nu}^{1}\\\
{\xi}^{1}\rightarrow&&-{\partial}_{i}({\omega}_{2}{\mu}^{i})+\frac{1}{2}{\partial}_{2}({\omega}_{i}{\mu}^{i})+({\partial}_{2}{\omega}_{i}){\mu}^{i}&={\nu}^{2}\\\
{\xi}^{1}\rightarrow&&-{\partial}_{i}({\omega}_{3}{\mu}^{i})+\frac{1}{2}{\partial}_{3}({\omega}_{i}{\mu}^{i})+({\partial}_{3}{\omega}_{i}){\mu}^{i}&={\nu}^{3}\end{array}\right.$

We let the reader check, as a tricky exercise, that we have indeed
$ad({\cal{D}})\circ ad({\cal{D}}_{1})\equiv 0$ and it remains to prove that
$ad({\cal{D}})$ generates the CC of $ad({\cal{D}}_{1})$. It is in such a
situation that we can measure the usefulness of homological algebra and we
only prove this result directly when $\omega=(1,-x^{3},0)\Rightarrow c=1$. In
this case, the kernel of $ad({\cal{D}})$ is easily seen to be defined by:

$\left\\{\begin{array}[]{ll}\frac{1}{2}{\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{2}+{\partial}_{3}{\mu}^{3}+\frac{1}{2}x^{3}{\partial}_{1}{\mu}^{2}&=0\\\
-(x^{3})^{2}{\partial}_{1}{\mu}^{2}+x^{3}{\partial}_{1}{\mu}^{1}+{\partial}_{2}{\mu}^{1}-x^{3}{\partial}_{2}{\mu}^{2}+2{\mu}^{3}&=0\\\
{\partial}_{3}{\mu}^{1}-x^{3}{\partial}_{3}{\mu}^{2}-3{\mu}^{2}&=0\end{array}\right.$

while, setting now $2\lambda={\mu}^{1}-x^{3}{\mu}^{2}$ and substituting, the
CC of $ad({\cal{D}}_{1})$ seem to be only defined by the two PD equations:

$\left\\{\begin{array}[]{ll}{\partial}_{3}{\mu}^{1}-x^{3}{\partial}_{3}{\mu}^{2}-3{\mu}^{2}&=0\\\
{\partial}_{2}{\mu}^{1}-x^{3}{\partial}_{2}{\mu}^{2}-(x^{3})^{2}{\partial}_{1}{\mu}^{2}+x^{3}{\partial}_{1}{\mu}^{1}+2{\mu}^{3}&=0\end{array}\right.$

The strange fact is that such a system is not formally integrable and one has
to differentiate the second PD equation with respect to $x^{3}$ and substract
the first PD equation differentiated with respect to $x^{2}$ in order to get
the additional PD equation:

${\partial}_{3}{\mu}^{3}+{\partial}_{2}{\mu}^{2}+\frac{1}{2}x^{3}{\partial}_{1}{\mu}^{2}+\frac{1}{2}{\partial}_{1}{\mu}^{1}=0$

and find an isomorphic involutive system, a result showing that the
differential sequence and its formal adjoint are both formally exact though
not strictly exact. We conclude this example with the following striking
result [arXiv:1803.09610]:

THEOREM 2.3.1: The contact differential sequence and its formal adjoint are
both split long exact sequences of free and thus projective modules, if and
only if $c\neq 0$.

Proof: As we have just proved that $ad({\cal{D}})$ was generating the CC of
$ad({\cal{D}}_{1})$, we may look for the CC of $ad({\cal{D}})$ in order to
recover the parametrization of ${\cal{D}}$ given at the beginning of this
example.
First of all, as proved in [GB2], two geometric objects $\omega$ and
$\bar{\omega}$ provide the same system $R_{1}\subset J_{1}(T)$ if and only if
$\bar{\omega}=a\omega$ for a constant parameter $a=cst$, a result providing
$\bar{c}=a^{2}c$. Accordingly, the only possible critical value of the Vessiot
structure constant is $c=0$.
Now, as already noticed, we have ${\omega}_{i}{\mu}^{i}=2c\lambda$ and
$ad({\cal{D}}_{1})$ is injective if and only if $c\neq 0$. It follows that the
differential module defined by ${\cal{D}}_{1}$ is projective and the sequences
split if we are able to construct ${\cal{D}}_{-1}$ and to prove that it is an
injective operator. For this, changing slightly our previous notations, we
notice that the symbol map of $ad({\cal{D}})$ is:

$-{\omega}_{i}d_{r}{\mu}^{r}+\frac{1}{2}{\omega}_{r}d_{i}{\mu}^{r}+...={\nu}_{i}$

Using standard notations of classical geometry, we may rewrite it as:

$-A\vec{\omega}+\vec{\nabla}.B+...=\vec{\nu}\Rightarrow\vec{\omega}\wedge(\vec{\nabla}.A)+...=\vec{\nabla}\wedge\vec{\nu}\Rightarrow\vec{\omega}.(\vec{\nabla}\wedge\vec{\nu})+...=0$

Hence, after one prolongation on the symbol level, we get the only CC:

${\omega}_{1}(d_{3}{\nu}_{2}-d_{2}{\nu}_{3})+{\omega}_{2}(d_{1}{\nu}_{3}-d_{3}{\nu}_{1})+{\omega}_{3}(d_{2}{\nu}_{1}-d_{1}{\nu}_{2})+...=0$

After substitution and a tedious computation, one finally obtains:

$\begin{array}[]{rcl}{\omega}_{1}(d_{3}{\nu}_{2}-d_{2}{\nu}_{3})+{\omega}_{2}(d_{1}{\nu}_{3}-d_{3}{\nu}_{1})+{\omega}_{3}(d_{2}{\nu}_{1}-d_{1}{\nu}_{2})&&\\\
+2({\partial}_{2}{\omega}_{3}-{\partial}_{3}{\omega}_{2}){\nu}_{1}+2({\partial}_{3}{\omega}_{1}-{\partial}_{1}{\omega}_{3}){\nu}_{2}+2({\partial}_{1}{\omega}_{2}-{\partial}_{2}{\omega}_{1}){\nu}_{3}&=&\theta\end{array}$

We may thus obtain ${\cal{D}}_{-1}$ in the form:

$\left\\{\begin{array}[]{lccl}{\omega}_{2}{\partial}_{3}\phi-{\omega}_{3}{\partial}_{2}\phi+({\partial}_{2}{\omega}_{3}-{\partial}_{3}{\omega}_{2})\phi&={\xi}^{1}\\\
{\omega}_{3}{\partial}_{1}\phi-{\omega}_{1}{\partial}_{3}\phi+({\partial}_{3}{\omega}_{1}-{\partial}_{1}{\omega}_{3})\phi&={\xi}^{2}\\\
{\omega}_{1}{\partial}_{2}\phi-{\omega}_{2}{\partial}_{1}\phi+({\partial}_{1}{\omega}_{2}-{\partial}_{2}{\omega}_{1})\phi&={\xi}^{3}\\\
\end{array}\right.$

and this operator is injective whenever $c\neq 0$ because we have the lift
${\omega}_{r}{\xi}^{r}=c\phi$.
The situation is similar in arbitrary dimension $n=2p+1$ with $1$-form
$\alpha=dx^{n}-{\sum}^{p}_{\alpha=1}x^{\bar{\alpha}}dx^{\alpha}$ as we have
again one Vessiot structure constant $c$ and the injective parametrization :

$\phi-x^{\bar{\beta}}\frac{\partial\phi}{\partial x^{\beta}}={\xi}^{n},\hskip
17.07164pt-\frac{\partial\phi}{\partial
x^{\bar{\alpha}}}={\xi}^{\alpha},\hskip 17.07164pt\frac{\partial\phi}{\partial
x^{\alpha}}+x^{\bar{\alpha}}\frac{\partial\phi}{\partial
x^{n}}={\xi}^{\bar{\alpha}}\hskip 11.38109pt\Rightarrow\hskip
11.38109pt\phi=\alpha(\xi)$

We have the locally exact split differential sequence:

$0\rightarrow{\wedge}^{0}T^{*}\stackrel{{\scriptstyle{\cal{D}}_{-1}}}{{\longrightarrow}}T\stackrel{{\scriptstyle\cal{D}}}{{\longrightarrow}}F_{0}\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}F_{1}\stackrel{{\scriptstyle{\cal{D}}_{2}}}{{\longrightarrow}}.....\stackrel{{\scriptstyle{\cal{D}}_{n-2}}}{{\longrightarrow}}F_{n-2}\rightarrow
0$

with $dim(F_{r})=n!/(r+2)!(n-r-2)!$ and we refer again to [8, 21] for more
details because, when $n\geq 5$, we have to use a $2$-contravariant
skewsymmetric tensor density.
It follows that contact geometry must be entirely revisited in the light of
these new results.
$\Box$

EXAMPLE 2.4: (Macaulay example).
With $n=3,m=1,q=2,K=\mathbb{Q},D=K[d_{1},d_{2},d_{3}]$, let us consider the
linear homogeneous second order operator
${\cal{D}}:\xi\rightarrow(d_{33}\xi={\eta}^{1},d_{23}\xi-
d_{11}\xi={\eta}^{2},d_{22}\xi={\eta}^{3})$ with corresponding system
$R_{2}\subset J_{2}(E)$ which is trivially formally integrable (FI). It is
easy to check that $dim(g_{2})=3,dim(g_{3})=1,g_{4}=0$. Hence, $R_{2}$ and
thus $R_{3}$ are FI but not involutive. Then $R_{4}$ is trivially involutive
with $dim(R_{3+r})=2^{n}=8,\forall r\geq 0$ and a full basis of parametric
jets is
$(\xi,{\xi}_{1},{\xi}2,{\xi}_{3},{\xi}_{11},{\xi}_{12},{\xi}_{13},{\xi}_{111})$.
We have proved in [14] that there are three second order CC with only one
second order CC. The corresponding formally exact differential sequence is:

$0\rightarrow\Theta\rightarrow\xi\underset{2}{\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}}\eta\underset{2}{\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}}\zeta\underset{2}{\stackrel{{\scriptstyle{\cal{D}}_{2}}}{{\longrightarrow}}}\tau\rightarrow
0$

and the minimum free differential resolution of the corresponding differential
module $M$ is:

$0\rightarrow
D\stackrel{{\scriptstyle{\cal{D}}_{2}}}{{\longrightarrow}}D^{3}\stackrel{{\scriptstyle{\cal{D}}_{1}}}{{\longrightarrow}}D^{3}\stackrel{{\scriptstyle{\cal{D}}}}{{\longrightarrow}}D\stackrel{{\scriptstyle
p}}{{\longrightarrow}}M\rightarrow 0$

where $p$ is the canonical projection. The situation is much more tricky if we
consider now the new third order differential operator
${\cal{D}}^{\prime}=\Phi\circ j_{3}:E\stackrel{{\scriptstyle
j_{3}}}{{\longrightarrow}}J_{3}(E)\stackrel{{\scriptstyle\Phi}}{{\longrightarrow}}J_{3}(E)/R_{3}$
with $R_{3}={\rho}_{1}(R_{2})$. Indeed, it is easy to check that the sequence:

$0\rightarrow{\wedge}^{2}T^{*}\otimes
g_{3}\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}{\wedge}^{3}T^{*}\otimes
g_{2}\rightarrow 0$

is exact because both bundles have the same dimension equal to $3$. It follows
that $g_{3}$ is $3$-acyclic and it follows that the $21$ CC are described by
an operator of order $1$. However, the corresponding system has a symbol which
is not $2$-acyclic. We have proved in [14] that the next CC are of order $2$,
and so on. By chance, it happens that the next operator
${\cal{D}}^{\prime}_{2}$ is involutive and we may start a Janet sequence with
first order involutive operators
${\cal{D}}^{\prime}_{3},{\cal{D}}^{\prime}_{4},{\cal{D}}^{\prime}_{6}$ in the
differential sequence:

$0\rightarrow\Theta\rightarrow
1\underset{3}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}}}{{\longrightarrow}}}12\underset{1}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}_{1}}}{{\longrightarrow}}}21\underset{2}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}_{2}}}{{\longrightarrow}}}46\underset{1}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}_{3}}}{{\longrightarrow}}}72\underset{1}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}_{4}}}{{\longrightarrow}}}48\underset{1}{\stackrel{{\scriptstyle{\cal{D}}^{\prime}_{5}}}{{\longrightarrow}}}12\rightarrow
0$

in which we have specified the dimensions of the bundles and the orders of the
successive operators. We finally notice again that
$rk_{D}(M)=1-12+21-46+72-48+12=0$.
We invite the reader to consider the similar example proposed by F. S.
Macaulay in $1916$ ([6], p 79) when $n=2$. Indeed, if we consider the second
order system $d_{22}\xi=0,d_{12}\xi-d_{11}\xi=0$, we have
$dim(g_{2})=1,g_{3}=0$ and $dim(R_{2})=dim(R_{3})=2^{n}=4$ with the basis of
parametric jets $(\xi,{\xi}_{1},{\xi}_{2},{\xi}_{11})$ . In this case, $R_{2}$
is FI but not involutive while $R_{3}$ is trivially involutive with Janet
sequence $0\rightarrow\Theta\rightarrow
1\underset{3}{\stackrel{{\scriptstyle}}{{\longrightarrow}}}6\underset{1}{\stackrel{{\scriptstyle}}{{\longrightarrow}}}7\underset{1}{\stackrel{{\scriptstyle}}{{\longrightarrow}}}2\rightarrow
0$ and $1-6+7-2=0$.

3) KERR METRIC

Let us recall a few facts from Riemannian geometry. A metric
$\omega=({\omega}_{ij})\in S_{2}T^{*}$ with $det(\omega)\neq 0$ is providing
the Christoffel symbols $\gamma=({\gamma}^{k}_{ij})$ as geometric objects
according to the forgotten work of E. Vessiot in 1903 [7, 24, 27]. The Riemann
tensor ${\rho}^{k}_{l,ij}\in{\wedge}^{2}T^{*}\otimes
g_{1}\subset{\wedge}^{2}T^{*}\otimes T^{*}\otimes T$ is a section of the
Spencer $\delta$-chomology vector bundle $H^{2}(g_{1})$ because it is also
killed by the Spencer surjective map $\delta:{\wedge}^{2}T^{*}\otimes
g_{1}\rightarrow{\wedge}^{3}T^{*}\otimes T^{*}\otimes T$ and has thus
$n^{2}(n-1)^{2}/4-n^{2}(n-1)(n-2)/6=n^{2}(n^{2}-1)/12$ components because
$dim(g_{1})=n(n-1)/2$ and $g_{2}=0$. Needless to say that these definitions
are far from the ones that can be found in any place (books or papers) dealing
with GR ([12, 15]).
Now, the linearization $\Omega=({\Omega}_{ij})\in S_{2}T^{*}$ of $\omega$
induces a linearization $\Gamma={\Gamma}^{k}_{ij}\in S_{2}T^{*}\otimes T$ and
a linearization $R=(R^{k}_{l,ij})\in{\wedge}^{2}T^{*}\otimes T$. With more
details, we have:

${\rho}^{k}_{l,ij}={\partial}_{i}{\gamma}^{k}_{lj}-{\partial}_{j}{\gamma}^{k}_{li}+{\gamma}^{r}_{lj}{\gamma}^{k}_{ri}-{\gamma}^{r}_{li}{\gamma}^{k}_{rj}$

and thus, because $\Gamma\in S_{2}T^{*}\otimes T$ is a tensor::

$\begin{array}[]{rcl}R^{k}_{l,ij}&=&d_{i}{\Gamma}^{k}_{lj}-d_{j}{\Gamma}^{k}_{li}+{\gamma}^{r}_{lj}{\Gamma}^{k}_{ri}-{\gamma}^{r}_{li}{\Gamma}^{k}_{rj}+{\gamma}^{k}_{ri}{\Gamma}^{r}_{lj}-{\gamma}^{k}_{rj}{\Gamma}^{r}_{li}\\\
&=&(d_{i}{\Gamma}^{k}_{lj}-{\gamma}^{r}_{li}{\Gamma}^{k}_{rj}+{\gamma}^{k}_{ri}{\Gamma}^{r}_{lj})-(d_{j}{\Gamma}^{k}_{li}-{\gamma}^{r}_{lj}{\Gamma}^{k}_{ri}+{\gamma}^{k}_{rj}{\Gamma}^{r}_{li})\\\
&=&{\nabla}_{i}{\Gamma}^{k}_{lj}-{\nabla}_{j}{\Gamma}^{k}_{li}\end{array}$

by introducing the covariant derivative $\nabla$. We recall that
${\nabla}_{r}{\omega}_{ij}=0,\forall r,i,j$ or, equivalently, that
$(id,-\gamma):\xi\in
T\rightarrow({\xi}^{k},{\xi}^{k}_{i}=-{\gamma}^{k}_{ir}{\xi}^{r})\in R_{1}$ is
a $R_{1}$-connection with
${\omega}_{sj}{\gamma}^{s}_{ir}+{\omega}_{is}{\gamma}^{s}_{jr}={\partial}_{r}{\omega}_{ij}$,
a result allowing to move down the index $k$ in the previous formulas (See
[17, 18, 22] for more details).
We may thus take into account the Bianchi identities implied by the cyclic
sums on $(ijr)$

${\beta}_{kl,ijr}\equiv{\nabla}_{r}{\rho}_{kl,ij}+{\nabla}_{i}{\rho}_{kl,jr}+{\nabla}_{j}{\rho}_{kl,ri}=0\hskip
28.45274pt\Leftrightarrow\hskip
28.45274pt\beta\equiv\underset{cycl}{\Sigma}(\partial\rho-\gamma\rho)=0$

and their respective linearizations $B_{kl,ijr}=0$ as described below. Both
$\beta$ and $B$ are sections of the Spencer cohomogy vector bundle
$H^{3}(g_{1})$ defined by the short exact sequence:

$0\rightarrow H^{3}(g_{1})\rightarrow{\wedge}^{3}T^{*}\otimes
g_{1}\stackrel{{\scriptstyle\delta}}{{\longrightarrow}}{\wedge}^{4}T^{*}\otimes
T\rightarrow 0$
$\begin{array}[]{lcl}dim(H^{3}(g_{1}))&=&(n(n-1)(n-2)/6)(n(n-1)/2)-(n(n-1)(n-2)(n-3)/24)n\\\
&=&n^{2}(n^{2}-1)(n-2)/24\end{array}$

that is $20$ when $n=4$ because $dim(g_{1})=n(n-1)/2$ and the coboundary
bundle is $B^{3}(g_{1})=0$.
Such results cannot be even imagined by somebody not aware of the
$\delta$-acyclicity ([7, 9, 11, 12]).
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:

$\begin{array}[]{rcl}B_{kl,rij}\equiv{\nabla}_{r}R_{kl,ij}+{\nabla}_{i}R_{kl,jr}+{\nabla}_{j}R_{kl,ri}=0\hskip
2.84526ptmod(\Gamma)&\Leftrightarrow&\underset{cycl}{\Sigma}(dR-\gamma
R-\rho\Gamma)=0\\\ &\Leftrightarrow&B\equiv\underset{cycl}{\Sigma}(\nabla
R)=\underset{cycl}{\Sigma}(\rho\Gamma)\end{array}$

In order to recapitulate these new concepts obtained after one, two or three
prolongations, we have successively
$\omega\longrightarrow\gamma\longrightarrow\rho\longrightarrow\beta$ and the
respective linearizations $\Omega\longrightarrow\Gamma\longrightarrow
R\longrightarrow B$.

The purpose of this section will be to consider the Killing operator
${\cal{D}}:T\rightarrow S_{2}T^{*}:\xi\rightarrow{\cal{L}}(\xi)\omega$ and the
corresponding Killing system $R_{1}\subset J_{1}(T)$ when $\omega$ is the Kerr
metric. In particular, we shall exhibit the successive inclusions
$R^{(3)}_{1}\subset R^{(2)}_{1}\subset R^{(1)}_{1}=R_{1}\subset J_{1}(T)$ with
dimensions $2<4<10=10<20$ by means of elementary combinatorics and diagram
chasing exactly like in the previous motivating examples.

We now write the Kerr metric in Boyer-Lindquist coordinates:

$ds^{2}=\frac{{\rho}^{2}-mr}{{\rho}^{2}}dt^{2}-\frac{{\rho}^{2}}{\Delta}dr^{2}-{\rho}^{2}d{\theta}^{2}\\\
-\frac{2amrsin^{2}(\theta)}{{\rho}^{2}}dtd\phi-(r^{2}+a^{2}+\frac{ma^{2}rsin^{2}(\theta)}{{\rho}^{2}})sin^{2}(\theta)d{\phi}^{2}$

where $\Delta=r^{2}-mr+a^{2},\,\,{\rho}^{2}=r^{2}+a^{2}cos^{2}(\theta)$ as
usual and we recover the Schwarschild metric when $a=0$. We notice that $t$ or
$\phi$ do not appear in the coefficients of the metric. We shall simplify the
coordinate system by using the so-called ” rational polynomial ” coefficients
as follows:

$(x^{0}=t,\,x^{1}=r,\,x^{2}=c=cos(\theta),\,x^{3}=\phi)\Rightarrow
dx^{2}=-sin(\theta)d\theta\Rightarrow(dx^{2})^{2}=(1-c^{2})d\theta^{2}$

We obtain over the differential field
$K=\mathbb{Q}(a,m)(t,r,c,\phi)=\mathbb{Q}(a,m)(x)$:

$ds^{2}=\frac{{\rho}^{2}-mr}{{\rho}^{2}}dt^{2}-\frac{{\rho}^{2}}{\Delta}dr^{2}-\frac{{\rho}^{2}}{1-c^{2}}dc^{2}-\frac{2amr(1-c^{2})}{{\rho}^{2}}dtd\phi-(1-c^{2})(r^{2}+a^{2}+\frac{ma^{2}r(1-c^{2})}{{\rho}^{2}}){d\phi}^{2}$

with now $\Delta=r^{2}-mr+a^{2}$ and ${\rho}^{2}=r^{2}+a^{2}c^{2}$. For a
later use, it is also possible to set
${\omega}_{33}=-(1-c^{2})((r^{2}+a^{2})^{2}-a^{2}((1-c^{2})(a^{2}-mr+r^{2}))/(r^{2}+a^{2}c^{2})$
and $det(\omega)=-(r^{2}+a^{2}c^{2})^{2}$.

Framing the leading derivatives and setting
$\xi\partial={\xi}^{i}{\partial}_{i}$, we obtain:

$R_{1}\subset
J_{1}(T)\,\,\,\left\\{\begin{array}[]{lcl}{\Omega}_{33}&\equiv&2({\omega}_{33}\framebox{${\xi}^{3}_{3}$}+{\omega}_{03}{\xi}^{0}_{3})+\xi\partial{\omega}_{33}=0\\\
{\Omega}_{23}&\equiv&{\omega}_{33}\framebox{${\xi}^{3}_{2}$}+{\omega}_{03}{\xi}^{0}_{2}+{\omega}_{22}{\xi}^{2}_{3}=0\\\
{\Omega}_{22}&\equiv&2{\omega}_{22}\framebox{${\xi}^{2}_{2}$}+\xi\partial{\omega}_{22}=0\\\
{\Omega}_{13}&\equiv&{\omega}_{33}\framebox{${\xi}^{3}_{1}$}+{\omega}_{03}{\xi}^{0}_{1}+{\omega}_{11}{\xi}^{1}_{3}=0\\\
{\Omega}_{12}&\equiv&{\omega}_{22}\framebox{${\xi}^{2}_{1}$}+{\omega}_{11}{\xi}^{1}_{2}=0\\\
{\Omega}_{11}&\equiv&2{\omega}_{11}\framebox{${\xi}^{1}_{1}$}+\xi\partial{\omega}_{11}=0\\\
{\Omega}_{03}&\equiv&{\omega}_{33}\framebox{${\xi}^{3}_{0}$}+{\omega}_{03}({\xi}^{0}_{0}+{\xi}^{3}_{3})+{\omega}_{00}{\xi}^{0}_{3}+\xi\partial{\omega}_{03}=0\\\
{\Omega}_{02}&\equiv&{\omega}_{22}\framebox{${\xi}^{2}_{0}$}+{\omega}_{00}{\xi}^{0}_{2}+{\omega}_{03}{\xi}^{3}_{2}=0\\\
{\Omega}_{01}&\equiv&{\omega}_{11}\framebox{${\xi}^{1}_{0}$}+{\omega}_{00}{\xi}^{0}_{1}+{\omega}_{03}{\xi}^{3}_{1}=0\\\
{\Omega}_{00}&\equiv&2({\omega}_{00}\framebox{${\xi}^{0}_{0}$}+{\omega}_{03}{\xi}^{3}_{0})+\xi\partial{\omega}_{00}=0\end{array}\right.$

Looking at the symbol $g_{1}\subset T^{*}\otimes T$, elementary linear
combinatorics allow to prove [18, 22]:

${\omega}_{33}\framebox{${\xi}^{3}_{3}$}+{\omega}_{03}{\xi}^{0}_{3}=0\,\,mod(\xi),\,\,\,{\omega}_{33}\framebox{${\xi}^{3}_{0}$}+{\omega}_{00}{\xi}^{0}_{3}=0\,\,mod(\xi),\,\,\,{\omega}_{33}\framebox{${\xi}^{0}_{0}$}-{\omega}_{03}{\xi}^{0}_{3}=0\,\,mod(\xi)$

Then, multiplying ${\Omega}_{22}$ by ${\omega}_{11}$, ${\Omega}_{11}$ by
${\omega}_{22}$ and adding, we finally obtain:

$2({\omega}_{11}{\omega}_{22})({\xi}^{1}_{1}+{\xi}^{2}_{2})+\xi\partial({\omega}_{11}{\omega}_{22})=0$

Using the rational coefficients in the differential field
$K=\mathbb{Q}(m,a)(r,c)$, the nonzero components of the corresponding Riemann
tensor can be found in textbooks. Among them, we have:

$\left\\{\begin{array}[]{rcl}{\rho}_{03,03}&=&\frac{mr(1-c^{2})(r^{2}-mr+a^{2})(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{12,12}&=&-\frac{mr(r^{2}-3a^{2}c^{2})}{2(1-c^{2})(r^{2}+a^{2}c^{2})(r^{2}-mr+a^{2})}\\\
\\\
{\rho}_{13,13}&=&\frac{-(1-c^{2})mr(r^{4}-2a^{2}c^{2}r^{2}+4a^{2}r^{2}-2a^{4}c^{2}+3a^{4}-2a^{2}mr(1-c^{2}))(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}(r^{2}-mr+a^{2})}\\\
\\\
{\rho}_{23,23}&=&\frac{mr(2r^{4}-a^{2}c^{2}r^{2}+5a^{2}r^{2}-a^{4}c^{2}+3a^{4}-a^{2}mr(1-c^{2}))(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{01,23}&=&\frac{amc(2r^{2}-a^{2}c^{2}+3a^{2})(3r^{2}-a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{02,31}&=&-\frac{amc(r^{2}-2a^{2}c^{2}+3a^{2})(3r^{2}-a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{03,12}&=&-\frac{amc(3r^{2}-a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{2}}\end{array}\right.$

As they will be used in the sequel, we also provide a few components vanishing
for the S-metric:

$\left\\{\begin{array}[]{rcl}{\rho}_{02,10}&=&\frac{3a^{2}mc(3r^{2}-a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{02,32}&=&\frac{amr(3r^{2}-mr+3a^{2})(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{13,23}&=&-\frac{3a^{2}mc(1-c^{2})(r^{2}+a^{2})(3r^{2}-a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}\\\
\\\
{\rho}_{01,13}&=&\frac{amr(1-c^{2})(3r^{2}+3a^{2}-2mr)(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}(r^{2}-mr+a^{2})}\end{array}\right.$

We finally exhibit the $6$ vanishing components of the Riemann tensor:

${\rho}_{01,03}=0,{\rho}_{01,12}=0,{\rho}_{02,03}=0,{\rho}_{02,12}=0,{\rho}_{03,13}=0,{\rho}_{03,23}=0,{\rho}_{12,13}=0,{\rho}_{12,23}=0$

Introducing the formal Lie derivative $R=L({\xi}_{1})\rho$ and using the fact
that $\rho\in{\wedge}^{2}T^{*}\otimes T^{*}\otimes T^{*}$ is a tensor, the
system $R^{(2)}_{1}\subset R^{(1)}_{1}=R_{1}\subset J_{1}(T)$ contains the new
equations:

$R_{kl,ij}\equiv{\rho}_{rl,ij}{\xi}^{r}_{k}+{\rho}_{kr,ij}{\xi}^{r}_{l}+{\rho}_{kl,rj}{\xi}^{r}_{i}+{\rho}_{kl,ir}{\xi}^{r}_{j}+{\xi}^{r}{\partial}_{r}{\rho}_{kl,ij}=0$

but we have to take into account the ten identities to zero obtained because
${\rho}_{ij}=0$, namely:

$R_{ij}\equiv{\rho}_{rj}{\xi}^{r}_{i}+{\rho}_{ir}{\xi}^{r}_{J}+{\xi}^{r}{\partial}_{r}{\rho}_{ij}=0$

We check the identity
${\rho}_{01,23}+{\rho}_{02,31}+{\rho}_{03,12}=0\Rightarrow
R_{01,23}+R_{02,31}+R_{03,12}=0$ and we have:

$\left\\{\begin{array}[]{lcl}R_{03,03}&\equiv&2{\rho}_{03,03}({\xi}^{0}_{0}+{\xi}^{3}_{3})+\xi\partial{\rho}_{03,03}=0\\\
R_{12,12}&\equiv&2{\rho}_{12,12}({\xi}^{1}_{1}+{\xi}^{2}_{2})+\xi\partial{\rho}_{12,12}=0\\\
R_{01,23}&\equiv&{\rho}_{01,23}({\xi}^{0}_{0}+{\xi}^{1}_{1}+{\xi}^{2}_{2}+{\xi}^{3}_{3})+\xi\partial{\rho}_{01,23}=0\end{array}\right.$

We obtain therefore
$\xi\partial({\rho}_{12,12}/({\omega}_{11}{\omega}_{22}))=0$ but we have also
$\xi\partial({\rho}_{03,03}{\rho}_{12,12}/det(\omega))=0$.
The following invariants are obtained successively in a coherent way:

${\omega}_{11}{\omega}_{22}=\frac{(r^{2}+a^{2}c^{2})^{2}}{(1-c^{2})(r^{2}-mr+a^{2})}\,\,\,\Rightarrow\,\,\,\mid{\rho}_{12,12}\mid/({\omega}_{11}{\omega}_{22})=\frac{mr(r^{2}-3a^{2}c^{2})}{2(r^{2}+a^{2}c^{2})^{3}}$

Also, as $a\in K$, then ${\rho}_{01,23}$ and ${\rho}_{02,13}$ can be both
divided by $a$ and we get the new invariant:

${\rho}_{01,23}/{\rho}_{03,12}=\frac{2r^{2}-a^{2}c^{2}+3a^{2}}{r^{2}+a^{2}c^{2}}$

These results are leading to ${\xi}^{1}=0$, ${\xi}^{2}=0$, thus to
${\xi}^{1}_{1}=0$, ${\xi}^{2}_{2}=0$ and ${\xi}^{0}_{0}+{\xi}^{3}_{3}=0$ after
substitution. In the case of the S-metric, we have only ${\xi}^{1}=0$ as an
equation of zero order.
The following relations allow to keep only the four parametric jets
$({\xi}^{1}_{0},{\xi}^{2}_{0},{\xi}^{1}_{3},{\xi}^{2}_{3})$ on the right side:

$\left\\{\begin{array}[]{lcl}{\omega}^{11}{\xi}^{0}_{1}+{\omega}^{00}{\xi}^{1}_{0}+{\omega}^{03}{\xi}^{1}_{3}=0,&&{\omega}^{11}{\xi}^{3}_{1}+{\omega}^{03}{\xi}^{1}_{0}+{\omega}^{33}{\xi}^{1}_{3}=0\\\
{\omega}^{22}{\xi}^{0}_{2}+{\omega}^{00}{\xi}^{2}_{0}+{\omega}^{03}{\xi}^{2}_{3}=0,&&{\omega}^{22}{\xi}^{3}_{2}+{\omega}^{03}{\xi}^{2}_{0}+{\omega}^{33}{\xi}^{2}_{3}=0\end{array}\right.$

if we use the fact that
${\omega}^{03}=-{\omega}_{03}/({\omega}_{00}{\omega}_{33}-({\omega}_{03})^{2})$
in the inverse metric.

Taking now into account that $R_{ij}={\omega}^{rs}R_{ri,sj}\,\,\,mod(\Omega)$,
we have proved in [18] that the $6$ second order equations $R_{ij,ij}=0$ are
only depending on the $3$ equations $R_{01,01}=0,R_{02,02}=0,R_{01,13}=0$ by
using $R_{ii}=0$ and $R_{03}=0$. We obtain in particular the two equations:

$\left\\{\begin{array}[]{lcl}\frac{1}{2}R_{01,01}&\equiv&{\rho}_{01,01}({\xi}^{0}_{0}+{\xi}^{1}_{1})+{\rho}_{01,31}{\xi}^{3}_{0}+{\rho}_{01,02}{\xi}^{2}_{1}=0\\\
\\\
\frac{1}{2}R_{02,02}&\equiv&{\rho}_{02,02}({\xi}^{0}_{0}+{\xi}^{2}_{2})+{\rho}_{02,32}{\xi}^{3}_{0}+{\rho}_{01,02}{\xi}^{1}_{2}=0\end{array}\right.$

Using the fact that we have now:

${\omega}_{22}{\xi}^{2}_{1}+{\omega}_{11}{\xi}^{1}_{2}=0\,\,\,\Leftrightarrow\,\,\,{\omega}^{11}{\xi}^{2}_{1}+{\omega}^{22}{\xi}^{1}_{2}=0$

we may multiply the first equation by ${\omega}^{11}$, the second by
${\omega}^{22}$ and sum in order to obtain:

$({\omega}^{11}{\rho}_{01,01}+{\omega}^{22}{\rho}_{02,02}){\xi}^{0}_{0}+({\omega}^{11}{\rho}_{01,31}+{\omega}^{22}{\rho}_{02,32}){\xi}^{3}_{0}=0$

Using the previous identities for $R_{00}=0$ and $R_{03}=0$, we obtain
therefore:

${\omega}^{33}{\rho}_{03,03}{\xi}^{0}_{0}-{\omega}^{03}{\rho}_{03,03}{\xi}^{3}_{0}=0\,\,\Rightarrow\,\,{\omega}^{33}{\xi}^{0}_{0}-{\omega}^{03}{\xi}^{3}_{0}=0\,\,\Leftrightarrow\,\,{\omega}_{00}{\xi}^{0}_{0}+{\omega}_{03}{\xi}^{3}_{0}=0$

As this relation was satisfied $mod(\xi)$, we must introduce the new equation:

$R_{01,13}\equiv{\rho}_{01,13}({\xi}^{0}_{0}+2{\xi}^{1}_{1}+{\xi}^{3}_{3})-{\rho}_{13,13}{\xi}^{3}_{0}-{\rho}_{01,01}{\xi}^{0}_{3}+({\rho}_{01,23}+{\rho}_{02,13}){\xi}^{2}_{1}=0$

As we have ${\xi}^{0}_{0}+{\xi}^{3}_{3}=0$ and ${\xi}^{1}_{1}=0$, we obtain
therefore a linear equation of the form:

${\rho}_{13,13}{\xi}^{3}_{0}+{\rho}_{01,01}{\xi}^{0}_{3}-({\rho}_{01,23}+{\rho}_{02,13}){\xi}^{2}_{1}=0$

Similarly, we have also:

$R_{01,02}\equiv{\rho}_{01,02}(2{\xi}^{0}_{0}+{\xi}^{1}_{1}+{\xi}^{2}_{2})-({\rho}_{01,23}+{\rho}_{02,13}){\xi}^{3}_{0}+{\rho}_{01,02}{\xi}^{1}_{2}+{\rho}_{01,02}{\xi}^{2}_{1}=0$

and we obtain therefore a linear equation of the form:

$2{\rho}_{01,02}{\xi}^{0}_{0}-({\rho}_{01,23}+{\rho}_{02,13}){\xi}^{3}_{0}+{\rho}_{01,01}{\xi}^{1}_{2}+{\rho}_{01,02}{\xi}^{2}_{1}=0$

In the case of the S-metric, that is when $a=0$, we obtain respectively
${\xi}^{0}_{3}=0$ and ${\xi}^{1}_{2}=0$ as in [18] because
${\xi}^{0}_{0}\simeq{\xi}^{0}_{3}$. The previous linear system has thus a rank
equal to $2$ and we obtain therefore because
${\xi}^{3}_{0}\simeq{\xi}^{0}_{3},{\xi}^{2}_{1}\simeq{\xi}^{1}_{2}$:

$\framebox{${\xi}^{0}_{3}=0$},\framebox{${\xi}^{1}_{2}=0$}\,\,\Leftrightarrow\,\,\framebox{${\xi}^{3}_{0}=0$},\framebox{${\xi}^{2}_{1}=0$},\framebox{${\xi}^{0}_{0}=0$},\framebox{
${\xi}^{3}_{3}=0$}$

It remains to study the following $4$ linear equations, namely [18]:

$R_{01,03}=0,\,\,R_{03,23}=0,\,\,R_{03,13}=0,\,\,R_{02,03}=0$

THEOREM 3.1: The rank of the previous system with respect to the $4$ jet
coodinates $({\xi}^{1}_{0},{\xi}^{2}_{0},{\xi}^{1}_{3},{\xi}^{2}_{3})$ is
equal to $2$, for both the S and K-metrics. We have the two striking
identities:

$R_{03,13}+a(1-c^{2})R_{01,03}=0,\,\,\,\,\,\,R_{02,03}+\frac{a}{(r^{2}+a^{2})}R_{03,23}=0$

We may define the so-called algebroid brackety for sections
${\xi}_{q},{\eta}_{q}\in J_{q}(T)$ by setting ([7, 8, 9]:

$[{\xi}_{q},{\eta}_{q}]=\\{{\xi}_{q+1},{\eta}_{q+1}\\}+i(\xi)d{\eta}_{q+1}-i(\eta)d{\xi}_{q+1}\
J_{q}(T)$

In this formula, ” $i()$” is the interior multiplication of a $1$-form by a
vector and we let the reader check that the right member does not depend on
the respective lifts when the algebraic bracket is defined by linearity from
the standard bracket of vector fields by the formula:

$\\{j_{q+1}(\xi),j_{q+1}(\eta)\\}=j_{q}([\xi,\eta])\in J_{q}(T)$

Now, we know that if $R_{q}\subset J_{q}(T)$ is a system of infinitesimal Lie
equations, then we have the algebroid bracket and its link with the
prolongation/projection (PP) procedure ([7, 8, 14, 15]):

$[R_{q},R_{q}]\subset R_{q}\Rightarrow[R^{(s)}_{q+r},R^{(s)}_{q+r}]\subset
R^{(s)}_{q+r},\forall q,r,s\geq 0$

As $R^{(1)}_{1}={\pi}^{2}_{1}(R_{2})=R_{1}$, it follows that
$R^{(2)}_{1}={\pi}^{3}_{1}(R_{3})$ is such that
$[R^{(2)}_{1},R^{(2)}_{1}]\subset R^{(2)}_{1}$ with $dim(R^{(2)}_{1})=20-16=4$
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 the specific GR technical objects (Teukolski scalars,
Killing-Yano tensors) introduced in ([1-4]).

We have on sections (care) the $16$ (linear) equations $mod(j_{2}(\Omega))$ of
$R^{(2)}_{1}$ as follows ([22]):

$R^{(2)}_{1}\subset R_{1}\subset
J_{1}(T)\,\left\\{\begin{array}[]{lcl}{\xi}^{1}=0,{\xi}^{2}=0&\Rightarrow&\framebox{
${\omega}_{00}{\xi}^{0}_{1}+{\omega}_{03}{\xi}^{3}_{1}$}+{\omega}_{11}{\xi}^{1}_{0}=0,\,\,{\xi}^{1}_{1}=0,\,\,{\xi}^{2}_{2}=0\\\
{\xi}^{1}_{2}=0&\Rightarrow&{\xi}^{2}_{1}=0\\\
{\xi}^{1}_{3}+lin({\xi}^{1}_{0},{\xi}^{2}_{0})=0&\Rightarrow&\framebox{
${\omega}_{03}{\xi}^{0}_{1}+{\omega}_{33}{\xi}^{3}_{1}$}+{\omega}_{11}{\xi}^{1}_{3}=0\\\
{\xi}^{2}_{3}+lin({\xi}^{1}_{0},{\xi}^{2}_{0})=0&\Rightarrow&\framebox{
${\omega}_{00}{\xi}^{0}_{2}+{\omega}_{03}{\xi}^{3}_{2}$}+{\omega}_{22}{\xi}^{2}_{0}=0,\\\
&&\framebox{
${\omega}_{03}{\xi}^{0}_{2}+{\omega}_{33}{\xi}^{3}_{2}$}+{\omega}_{22}{\xi}^{2}_{3}=0\\\
{\xi}^{0}_{3}=0&\Rightarrow&{\xi}^{3}_{0}=0,\,\,{\xi}^{0}_{0}=0,\,\,{\xi}^{3}_{3}=0\end{array}\right.$

The coefficients of the linear equations $lin$ involved depend on the Riemann
tensor as in ([22]). Accordingly, we may choose only the $2$ parametric jets
$({\xi}^{1}_{0},{\xi}^{2}_{0})$ among
$({\xi}^{1}_{0},{\xi}^{1}_{3},{\xi}^{2}_{0},{\xi}^{2}_{3})$ to which we must
add $({\xi}^{0},{\xi}^{3})$ 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
${\xi}^{0}$ and ${\xi}^{3}$, a result leading to $dim(R^{(3)}_{1})=2$ in a
coherent way with the only nonzero Killing vectors
$\\{{\partial}_{t},{\partial}_{\phi}\\}$. We have indeed:

$\framebox{${\xi}^{1}_{0}=0$}\,,\,\,\,\framebox{
${\xi}^{2}_{0}=0$}\,\,\Rightarrow\,\,{\xi}^{1}_{3}=0,\,\,{\xi}^{2}_{3}=0\,\,\Rightarrow\,\,{\xi}^{0}_{1}=0,\,\,{\xi}^{3}_{1}=0,\,\,{\xi}^{0}_{2}=0,\,\,{\xi}^{3}_{2}=0$

Taking therefore into account that the metric only depends on
$(x^{1}=r,x^{2}=c=cos(\theta))$ we obtain after three prolongations the first
order system $R^{(3)}_{1}\subset R^{(2)}_{1}\subset R^{(1)}_{1}=R_{1}\subset
J_{1}(T)$ which is trivially involutive because all the first order jets
vanish but $({\xi}^{0},{\xi}^{3})$.

Surprisingly and contrary to the situation found for the S metric, we have now
an involutive first order system with only solutions
$({\xi}^{0}=cst,{\xi}^{1}=0,{\xi}^{2}=0,{\xi}^{3}=cst)$ and notice that
$R^{(3)}_{1}$ does not depend any longer on the parameters $(m,a)\in K$. 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 $d_{0}{\xi}^{1}-{\xi}^{1}_{0}=0,d_{0}{\xi}^{2}-{\xi}^{2}_{0}=0$ and
thus at least $6$ third order CC coming from the $6$ following components of
the Spencer operator, namely:

$d_{1}{\xi}^{1}-{\xi}^{1}_{1}=0,\,\,d_{2}{\xi}^{1}-{\xi}^{1}_{2}=0,\,\,d_{3}{\xi}^{1}-{\xi}^{1}_{3}=0,\,\,d_{1}{\xi}^{2}-{\xi}^{2}_{1}=0,\,\,d_{2}{\xi}^{2}-{\xi}^{2}_{2}=0,\,\,d_{3}{\xi}^{2}-{\xi}^{2}_{3}=0$

a result that cannot be even imagined from ([1-4]). Of course, proceeding like
in the motivating examples, we must substitute in the right members the values
obtained from $j_{2}(\Omega)$ and set for example
${\xi}^{1}_{1}=-\frac{1}{2{\omega}_{11}}\xi\partial{\omega}_{11}$ while
replacing ${\xi}^{1}$ and ${\xi}^{2}$ by the corresponding linear combinations
of the Riemann tensor already obtained for the right members of the two zero
order equations.

Along with the examples, we recall below all the diagrams that must be
introduced successively:

$\begin{array}[]{rccccccl}&&&0&&0&&\\\ &&&\downarrow&&\downarrow&&\\\
&0&\rightarrow&S_{2}T^{*}\otimes
T&\stackrel{{\scriptstyle{\sigma}_{1}(\Phi)}}{{\longrightarrow}}&T^{*}\otimes
S_{2}T^{*}&\rightarrow 0\\\ &\downarrow&&\downarrow&&\downarrow&&\\\
0\rightarrow&R_{2}&\rightarrow&J_{2}(T)&\stackrel{{\scriptstyle{\rho}_{1}(\Phi)}}{{\longrightarrow}}&J_{1}(S_{2}T^{*})&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\\\
0\rightarrow&R_{1}&\rightarrow&J_{1}(T)&\stackrel{{\scriptstyle\Phi}}{{\longrightarrow}}&S_{2}T^{*}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\\\ &0&&0&&0&&\end{array}$

$\begin{array}[]{rcccccccl}&&&0&&0&&\\\ &&&\downarrow&&\downarrow&&\\\
&0&\rightarrow&40&\rightarrow&40&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\\\
0\rightarrow&10&\rightarrow&60&\rightarrow&50&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\\\
0\rightarrow&10&\rightarrow&20&\rightarrow&10&\rightarrow 0\\\
&&&\downarrow&&\downarrow&&\\\ &&&0&&0&&\end{array}$

$\begin{array}[]{rcccccccl}&&&0&&0&&&\\\ &&&\downarrow&&\downarrow&&&\\\
&0&\rightarrow&S_{3}T^{*}\otimes
T&\stackrel{{\scriptstyle{\sigma}_{2}(\Phi)}}{{\longrightarrow}}&S_{2}T^{*}\otimes
S_{2}T^{*}&\rightarrow&h_{2}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{3}&\rightarrow&J_{3}(T)&\stackrel{{\scriptstyle{\rho}_{2}(\Phi)}}{{\longrightarrow}}&J_{2}(S_{2}T^{*})&\rightarrow&Q_{2}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{2}&\rightarrow&J_{2}(T)&\stackrel{{\scriptstyle{\rho}_{1}(\Phi)}}{{\longrightarrow}}&J_{1}(S_{2}T^{*})&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&&&0&&0&&&\\\ &&&\downarrow&&\downarrow&&&\\\
&0&\rightarrow&80&\rightarrow&100&\rightarrow&20&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&140&\rightarrow&150&\rightarrow&14&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&10&\rightarrow&60&\rightarrow&50&\rightarrow&0&\\\
&&&\downarrow&&\downarrow&&&\\\ &&&0&&0&&&\end{array}$

$\begin{array}[]{rcccccccl}&&&0&&0&&&\\\ &&&\downarrow&&\downarrow&&&\\\
&0&\rightarrow&S_{4}T^{*}\otimes
T&\stackrel{{\scriptstyle{\sigma}_{3}(\Phi)}}{{\longrightarrow}}&S_{3}T^{*}\otimes
S_{2}T^{*}&\rightarrow&h_{3}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{4}&\rightarrow&J_{4}(T)&\stackrel{{\scriptstyle{\rho}_{3}(\Phi)}}{{\longrightarrow}}&J_{3}(S_{2}T^{*})&\rightarrow&Q_{3}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R_{3}&\rightarrow&J_{3}(T)&\stackrel{{\scriptstyle{\rho}_{2}(\Phi)}}{{\longrightarrow}}&J_{2}(S_{2}T^{*})&\rightarrow&Q_{2}&\rightarrow
0\\\ &&&\downarrow&&\downarrow&&\downarrow&\\\ &&&0&&0&&0&\end{array}$

$\begin{array}[]{rcccccccl}&&&0&&0&&&\\\ &&&\downarrow&&\downarrow&&&\\\
&0&\rightarrow&140&\rightarrow&200&\rightarrow&60&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&2&\rightarrow&280&\rightarrow&350&\rightarrow&72&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&4&\rightarrow&140.&\rightarrow&150&\rightarrow&14&\rightarrow
0\\\ &&&\downarrow&&\downarrow&&\downarrow&\\\ &&&0&&0&&0&\end{array}$

Applying the Spencer $\delta$-cohomology to the top sequence, we get the
commutative diagram in which the two central columns are exact:

$\begin{array}[]{rcccccccl}&&&0&&0&&0&\\\
&&&\downarrow&&\downarrow&&\downarrow&\\\ &0&\rightarrow&S_{4}T^{*}\otimes
T&\rightarrow&S_{3}T^{*}\otimes F_{0}&\rightarrow&h_{3}&\rightarrow 0\\\
&&&\downarrow&&\downarrow&\searrow&\downarrow&\\\ &0&\rightarrow&T^{*}\otimes
S_{3}T^{*}\otimes T&\rightarrow&T^{*}\otimes S_{2}T^{*}\otimes
F_{0}&\rightarrow&\framebox{$T^{*}\otimes h_{2}$ }&\rightarrow 0\\\
&&&\downarrow&&\downarrow&&&\\\ &0&\rightarrow&{\wedge}^{2}T^{*}\otimes
S_{2}T^{*}\otimes T&\rightarrow&{\wedge}^{2}T^{*}\otimes T^{*}\otimes
F_{0}&\rightarrow&0\\\ &&&\downarrow&&\downarrow&&&\\\
0\rightarrow&\framebox{${\wedge}^{3}T^{*}\otimes g_{1}$
}&\rightarrow&{\wedge}^{3}T^{*}\otimes T^{*}\otimes
T&\rightarrow&{\wedge}^{3}T^{*}\otimes F_{0}&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\
0\rightarrow&{\wedge}^{4}T^{*}\otimes T&=&{\wedge}^{4}T^{*}\otimes
T&\rightarrow&0&&\\\ &\downarrow&&\downarrow&&&&&\\\ &0&&0&&&&&\end{array}$

Using a (delicate) snake chase, we obtain the short exact sequence:

$0\longrightarrow h_{3}\longrightarrow T^{*}\otimes
h_{2}\longrightarrow{\wedge}^{3}T^{*}\otimes
g_{1}\longrightarrow{\wedge}^{4}T^{*}\otimes T\longrightarrow 0,\hskip
14.22636pt0\rightarrow 60\rightarrow 80\rightarrow 24\rightarrow 4\rightarrow
0$

and the long exact connecting diagram:

$\begin{array}[]{rcccccccccl}&&0&&0&&&&&&\\\ &&\downarrow&&\downarrow&&&&&&\\\
0&\longrightarrow&S_{4}T^{*}\otimes T&\longrightarrow&S_{3}T^{*}\otimes
F_{0}&\longrightarrow&T^{*}\otimes
h_{2}&\longrightarrow&H^{3}(g_{1})&\longrightarrow&0\\\
&&\downarrow&&\parallel&&\downarrow&&\downarrow&&\\\
0&\longrightarrow&{\rho}_{1}(g^{\prime}_{2})&\longrightarrow&S_{3}T^{*}\otimes
F_{0}&\longrightarrow&T^{*}\otimes
Q_{2}&\longrightarrow&Q^{\prime}_{1}&\longrightarrow&0\\\
&&&&\downarrow&&\downarrow&&\downarrow&&\\\ &&&&0&&0&&0&&\end{array}$

allowing to compare the geometry of the Kerr metric (lower sequence) to the
classical geometry of the Minkowski metric (upper sequence). In the later case
we recall that the Bianchi indentities provide a section $B\in H^{3}(g_{1})$.
Moreover, we have $dim(Q^{\prime}_{1})\leq dim(H^{3}(g_{1}))=20$ if we define
$Q^{\prime}_{1}$ by the following commutative and exact diagram obtained by
cutting the the fundamental diagram and going one step further on in the
sequences in order to replace the $R$ and $Q$ by $R^{\prime}$ and
$Q^{\prime}$:

$\begin{array}[]{rccccccccl}&0&&0&&0&&0&\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&{\rho}_{1}(g^{\prime}_{2})&\rightarrow&S_{3}T^{*}\otimes
F_{0}&\stackrel{{\scriptstyle{\sigma}_{1}(\Psi)}}{{\longrightarrow}}&T^{*}\otimes
Q_{2}&\rightarrow&h^{\prime}_{1}&\rightarrow 0\\\
&\downarrow&&\downarrow&&\downarrow&\searrow&\downarrow&\\\
0\rightarrow&{\rho}_{1}(R^{\prime}_{2})&\rightarrow&J_{3}(F_{0})&\stackrel{{\scriptstyle{\rho}_{1}(\Psi)}}{{\longrightarrow}}&J_{1}(Q_{2})&\rightarrow&Q^{\prime}_{1}&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&R^{\prime}_{2}&\rightarrow&J_{2}(F_{0})&\stackrel{{\scriptstyle\Psi}}{{\longrightarrow}}&Q_{2}&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\ &0&&0&&0&&&\end{array}$

with dimensions:

$\begin{array}[]{rccccccccl}&0&&0&&0&&0&\\\
&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&144+x&\rightarrow&200&\longrightarrow&56&\rightarrow&x&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&\searrow&\parallel&\\\
0\rightarrow&280+x&\rightarrow&350&\longrightarrow&70&\rightarrow&x&\rightarrow
0\\\ &\downarrow&&\downarrow&&\downarrow&&\downarrow&\\\
0\rightarrow&136&\rightarrow&150&\longrightarrow&14&\rightarrow&0&\\\
&\downarrow&&\downarrow&&\downarrow&&&\\\ &0&&0&&0&&&\end{array}$

Accordingly, the number $14$ of second order CC plus the number of
differentially independent third order CC obtained by one prolongation of
these second order CC is equal to $70-x$. However, if we denote by $y$ the
number of new generating CC of third order that are not differential
consequences of the CC of second order already found, we must have $72-y=70-x$
and thus $y=x+2$.
As we have $4$ divergence identities for the Einstein operator, we obtain
therefore $y=4+2=6$ contrary to [1-4]. We have provided these new third order
CC as $6$ components of the Spencer operator.

Similarly, in the case of the S-metric, we have $R^{(3)}_{1}\subset
R^{(2)}_{1}\subset R^{(1)}_{1}=R_{1}\subset J_{1}(T)$ with respective
dimensions $4<5<10=10<20$ and we obtain:

$Q_{1}=0,dim(Q_{2})=15,dim(Q_{3})=74\Rightarrow 74-y=75-x\Rightarrow y=x-1=3$
$d_{1}{\xi}^{1}-{\xi}^{1}_{1}=0,\,\,\,d_{2}{\xi}^{1}-{\xi}^{1}_{2}=0,\,\,\,d_{3}{\xi}^{1}-{\xi}^{1}_{3}=0$

In the case of the M-metric (See [7, 14, 15] for more details), we have
$R^{(3)}_{1}=R^{(2)}_{1}=R^{(1)}_{1}=R_{1}\subset J_{1}(T)$ with respective
dimensions $10=10=10<20$ and we obtain:

$Q_{1}=0,\,\,\,dim(Q_{2})=20,\,\,\,dim(Q_{3})=80,dim(Q^{\prime}_{1})=20\rightarrow
x=20,y=0\rightarrow 80-y=100-x$

in a coherent way with the number of components of the Riemann tensor ($20$)
and the number of Bianchi identities ($20$) when $n=4$. The corresponding
Killing system is FI if and only if the following Vessiot structure equations
${\rho}^{k}_{l,ij}=c({\delta}^{k}_{i}{\omega}_{lj}-{\delta}^{k}_{j}{\omega}_{li})$
with the only constant $c$ are satisfied, that is the well known constant
Riemannian curvature condition.

4) CONCLUSION

E. Vessiot discovered the so-called Vessiot structure equations as early as in
1903 and, only a few years later, E. Cartan discovered the so-called Maurer-
Cartan structure equations. Both are depending on a certain number of
constants like the single geometric structure constant of the constant
Riemannian curvature for the first and the many algebraic structure constants
of Lie algebra for the second. However, Cartan and followers never
acknowledged the existence of another approach which is therefore still
totally ignored today, in particular by physicists ([22-24]). Now, it is well
known that the structure constants of a Lie algebra play a fundamental part in
the Chevalley-Eilenberg cohomology of Lie algebras and their deformation
theory ([14]). It was thus a challenge to associate the Vessiot structure
constants with other homological properties related to systems of Lie
equations, namely the extension modules determined by Lie operators. As a
striking consequence, such a possibility opens a new way to understand and
revisit the various contradictory works done during the last fifty years or so
by different groups of researchers, using respectively Cartan, Gröbner or
Janet bases while looking for a modern interpretation of the work done by C.
Lanczos from 1938 to 1962. However, the reader must not forget that the Weyl
tensor was not known by Lanczos, even as late as in 1967, and that it was not
possible to discover any solution of the parametrization problem by potentials
through double duality before $1990/1995$, that is too late for the many
people already engaged in this type of research ([20]). As for the study of
the Killing operator, the intrinsic understanding of such a technical problem
that we achieved has strictly nothing to do with GR. Indeed, thanks to a
difficult theorem of homological algebra, the intrinsic properties of the
resolution of a differential module $M$ only depend on $M$ itself but not on
the resolution. Roughly, the only important thing is the group, not the
metric. In actual practice, the resolutions by themselves no longer depend on
the respective Kerr, Schwarzschild or Minkowski parameters. The same comment
can be done on the link existing between electromagnetism and the conformal
group ([15, 21, 23]). We finally hope that this paper will open a new domain
for applying computer algebra and we are offering a collection of useful but
tricky test examples.

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