By analogy with Maxwell equations for electromagnetism, deciding about the existence of a potential for Einstein equations in vacuum has been proposed in 1970 as a $1000 challenge by J. Wheeler while the author of this paper was a visiting student of D. C. Spencer in Princeton University. No progress has been done during the next 25 years, till I gave a negative answer, contrary to what the GR community was believing and Wheeler sent me back a letter with a one-dollar bill attached, refusing to admit this result. Indeed, while teaching elasticity, I proposed an exercise explaining why a dam made with concrete is always vertical on the water-side with a slope of about 42 degrees on the other free side in order to obtain a minimum cost and the auto-stability under cracking of the surface under water (See the Introduction of [1] and Zbl 1079.93001). The main tool was the approximate computation of the Airy function inside the dam. I discovered that the Airy parametrization was just the adjoint of the (linearized) Riemann operator used as generating compatibility condition (CC) for the deformation tensor by any engineer. Being involved in GR with A. Lichnerowicz, I got the idea of using the adjoint of an operator in a systematic way.

Then I found the recently published master thesis of the Japanese student M. Kashiwara [2] . It has been a shock to discover this mixing up of differential geometry and homological algebra, ending with the use of the Differential Extension Modules. In particular, if $\mathcal{D}\xi =\eta$ has the generating CC ${\mathcal{D}}_1\eta =0$, then $ad\left(\mathcal{D}\right)$ may not generate all the CC of $ad\left({\mathcal{D}}_1\right)$ and $ex{t}^1\left(M\right)$ “ measures” this gap only depending on the differential module $M$ defined by $\mathcal{D}$ [1] [3] . The simplest example with $ex{t}^1\left(M\right)\ne 0$ is provided by the system $d_{22}\xi ={\eta }^2$, $d_{12}\xi ={\eta }^1$ with a first order CC, but we have indeed $ex{t}^1\left(M\right)=0$ if we consider the system $d_{22}\xi ={\eta }^2$, $d_{12}\xi -d_{11}\xi ={\eta }^1$ with a second order CC. Hence, exactly like homological algebra brought a revolution in mathematics, it will bring a revolution in physics. I also noticed that GR could be considered as “a” way to parametrize the $Cauchy=ad\left(Killing\right)$ operator, leading to Gravitational Waves (GW). It follows that the same confusion has been done by E. Beltrami (1892) for space and A. Einstein (1915) for space-time because they both used the Einstein operator, not knowing that it was self-adjoint, confusing the Cauchy operator with the “div” operator induced from the Bianchi operator on one side and the stress functions with the deformation of the metric on another side [4] - [6] .

Accordingly and until now, the GR community has never wanted to take these new tools into account and [7] provides a good example of such a poor situation both with the reason for which no other reference can be given. By chance, the control community has been interested during a while by these new techniques for studying OD or PD control systems with constant coefficients, thanks to U. Oberst [8] . Hence, the impossibility to parametrize Einstein equations in vacuum can only be found in books of control theory (See Springer LNCIS 256, 2000 [9] and 311, 2005 [10] ).

Studying the Lanczos problems in 2001 [11] , I noticed that the $Beltrami=ad\left(Riemann\right)$ operator can be parametrized by the $Lanczos=ad\left(Bianchi\right)$ operator in the adjoint sequence. As a byproduct, the purpose of this paper is to explain the two previous confusions without the need of any computation. In fact, according to H. Poincaré, the geometrical and physical long exact dual differential sequences of operators acting on tensors, giving order of operators and number of components, are:

$\boxed{\begin{array}{ccccccc}n& \underset{1}{\stackrel{Killing}{\to }}& \frac{n\left(n+1\right)}{2}& \underset{2}{\stackrel{Riemann}{\to }}& \frac{n^2\left(n^2-1\right)}{12}& \underset{1}{\stackrel{Bianchi}{\to }}& \frac{n^2\left(n^2-1\right)\left(n-2\right)}{24}\\ n& \underset{1}{\stackrel{Cauchy}{\leftarrow }}& \frac{n\left(n+1\right)}{2}& \underset{2}{\stackrel{Beltrami}{\leftarrow }}& \frac{n^2\left(n^2-1\right)}{12}& \underset{1}{\stackrel{Lanczos}{\leftarrow }}& \frac{n^2\left(n^2-1\right)\left(n-2\right)}{24}\end{array}}$

The first sequence is known but can only be found in textbooks from a purely computational point of view because it only depends on the use of the Spencer $\delta$-cohomology. We do believe that the adjoint sequence is not even known today. We point out that both sequences are formally exact in the sense that each operator is generating the CC of the preceding one.

A jigsaw puzzle, contrary to a headache puzzle, is a tiling puzzle that requires the assembly of often irregularly shaped interlocking and mosaicked pieces. Typically each piece brings a portion of a picture, which is completed by solving the puzzle and thus not known as long as you have not finished to set up all the pieces. Of course, the more pieces you have, the more difficult is the puzzle.

We recall that the linear Spencer sequence for a Lie group of transformations $G\times X\to X$, which essentially depends on the action because infinitesimal generators are needed, is locally isomorphic to the linear gauge sequence which does not depend on the action any longer as it is the tensor product of the Poincaré sequence by the Lie algebra $\mathcal{G}=T_e\left(G\right)$. Indeed, we present a few basic definitions leading to the Spencer sequence and the gauge sequence used in the present paper. If the linear system $R_q\subset J_q\left(T\right)$ has a symbol $g_q=R_q\cap S_q{T}^{\text{*}}\otimes T$ with $r$-prolongations $g_{q+r}\subset S_{q+r}{T}^{\text{*}}\otimes T$, it is well known that when $g_q$ is finite type, that is $g_{q+r}=0$ fo $r$ large enough, then $R_q$ is involutive if and only if $g_q=0$ that will be the situation considered in this paper. Writing the action in the form $y=f\left(x,a\right)$ and differentiating $q$ times, high enough as needed for eliminating the parameters $a$, we obtain a non-linear system ${ℛ}_q\subset {\Pi }_q\left(X\times X\right)$ of order $q$ for the $q$-jets $\left(x,y_q\right)$ with the assumption that $det\left(y_i^k\right)\ne 0$. For example, if $y=\left(ax+b\right)/\left(cx+1\right)$ one needs to differentiate 3 times in order to get the Schwarzian OD equation $\left(y_{xxx}/y_x\right)-\frac{3}{2}{\left(y_{xx}/y_x\right)}^2=0$. We have indeed successively:

$\begin{array}{l}{y}_x=\left(a-bc\right)/{\left(cx+1\right)}^2,\\ y_{xx}=-2\left(a-bc\right)c/{\left(cx+1\right)}^3,\\ y_{xxx}=6c^2\left(a-bc\right)/{\left(cx+1\right)}^4\end{array}$

Linearizing this system at the $q$-jet of the identity $y=x$ by setting $\left(y^k=x^k+t{\xi }^k,\cdots ,y_{\mu }^k=t{\xi }_{\mu }^k\right)$, dividing by $t$ and setting $t\to 0$, we obtain the system $R_q\subset J_q\left(T\right)$ of infinitesimal Lie equations, for example ${\xi }_{xxx}=0$ with $q=3$ in the present example.

Using the fundamental theorems of Lie, we may exhibit a basis $\left({\theta }_{\tau }^k\left(x\right)\right)$ of infinitesimal generators of the action with $1\le \tau \le p=dim\left(G\right)$ and an isomorphism $X\times \mathcal{G}\to R_q:{\lambda }^{\tau }\left(x\right)\to {\xi }_{\mu }^k\left(x\right)={\lambda }^{\tau }\left(x\right){\partial }_{\mu }{\theta }_{\tau }^k\left(x\right)$ between sections for all multi-indices $\mu =\left({\mu }_1,\cdots ,{\mu }_n\right)$ of length $0\le \left|\mu \right|={\mu }_1+\cdots +{\mu }_n\le q$. For a later use, we set as usual $\mu +1_i=\left({\mu }_1,\cdots ,{\mu }_i+1,\cdots ,{\mu }_n\right)$ and may introduce the operator $j_q:\xi \left(x\right)\to j_q\left(\xi \right)=\left({\partial }_{\mu }\xi \left(x\right),0\le \left|\mu \right|\le q\right)$ in place of sections ${\xi }_q=\left({\xi }_{\mu }\left(x\right)\right)$ [12] - [15] .

DEFINITION 2.1: The Spencer operator $d:R_{q+1}\to T^{\text{*}}\otimes R_q:{\xi }_{q+1}\to {\partial }_i{\xi }_{\mu }^k\left(x\right)-{\xi }_{\mu +1_i}^k\left(x\right)$ can be extended to $d:{\wedge }^r{T}^{\text{*}}\otimes R_{q+1}\to {\wedge }^{r+1}{T}^{\text{*}}\otimes R_q$ by introducing a multi-index $I=\left(i_1<\cdots <i_r\right)$ for $r$-forms with basis $d{x}^I=d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_r}$ while setting $d:{\left({\xi }_{q+1}\right)}_{\mu ,I}^{k}d{x}^I\to \left({\partial }_i{\xi }_{\mu ,I}^k-{\xi }_{\mu +1_i,I}^k\right)d{x}^i\wedge d{x}^I$ and we check easily that $d^2=d\circ d=0$, a result leading to the Spencer operators $\left(D_1,\cdots ,D_n\right)$. The restriction to the symbols is minus the Spencer map $\delta :{\wedge }^r{T}^{\text{*}}\otimes g_{q+1}\to {\wedge }^{r+1}{T}^{\text{*}}\otimes g_q$ such that ${\delta }^2=\delta \circ \delta =0$ and we have $\delta :{\left({\xi }_{q+1}\right)}_{\mu ,I}^{k}d{x}^I\to {\xi }_{\mu +1_i,I}^{k}d{x}^i\wedge d{x}^I$ with ${\xi }_{\mu +1_i+1_j,I}^{k}d{x}^i\wedge d{x}^j=0$.

In the present situation of Lie group actions, using the chain rule for derivations, we obtain:

$\boxed{D_1{\xi }_q=d{\xi }_{q+1}=\left({\partial }_i{\lambda }^{\tau }\left(x\right)\right)\left({\partial }_{\mu }{\theta }_{\tau }^k\left(x\right)\right)d{x}^i}$

As $dim\left(R_q\right)=dim\left(\mathcal{G}\right)=dim\left(G\right)=p$, we obtain the following essential commutative and exact diagram relating the upper gauge sequence, which is the tensor product of the Poincaré sequence for the exterior derivative by the Lie algebra $\mathcal{G}$, with the lower Spencer sequence. Of course, though the dimensions are equal, the respective operators are completely different as we shall see.

$\boxed{\begin{array}{ccccccccccccc}& & & & {\wedge }^0{T}^{*}\otimes \mathcal{G}& \stackrel{d}{\to }& {\wedge }^1{T}^{*}\otimes \mathcal{G}& \stackrel{d}{\to }& \cdots & \stackrel{d}{\to }& {\wedge }^n{T}^{*}\otimes \mathcal{G}& \to & 0\\ & & & & \downarrow & & \downarrow & & & & \downarrow & & \\ 0& \to & \Theta & \stackrel{j_q}{\to }& {\wedge }^0{T}^{*}\otimes R_q& \stackrel{D_1}{\to }& {\wedge }^1{T}^{*}\otimes R_q& \stackrel{D_2}{\to }& \cdots & \stackrel{D_n}{\to }& {\wedge }^n{T}^{*}\otimes R_q& \to & 0\end{array}}$

The upper differential sequence is just the tensor product of the Poincaré sequence (in France!) for the exterior derivative by the Lie algebra $\mathcal{G}$ [16] . It is not evident at first sight to discover any relation between this diagram and its dual obtained by using adjoint operators (thus only containing first order operators) with the previous diagram (containing second order operators).

Accordingly, using standard notations, the main idea will be to introduce and compare the three following Lie groups of transformations but other subgroups of the conformal group may be considered, like the optical subgroup which is a maximal subgroup with 10 parameters, contrary to the Poincaré subgroup which is not maximal though it has also 10 parameters [17] :

${\Omega }_{ij}\equiv {\left(L\left({\xi }_1\right)\omega \right)}_{ij}\equiv {\omega }_{rj}\left(x\right){\xi }_i^r+{\omega }_{ir}\left(x\right){\xi }_j^r+{\xi }^r{\partial }_r{\omega }_{ij}\left(x\right)=0$

${\Gamma }_{ij}^k\equiv {\left(L\left({\xi }_2\right)\gamma \right)}_{ij}^k\equiv {\xi }_{ij}^k+{\gamma }_{rj}^k\left(x\right){\xi }_i^r+{\gamma }_{ir}^k\left(x\right){\xi }_j^r-{\gamma }_{ij}^r\left(x\right){\xi }_r^k+{\xi }^r{\partial }_r{\gamma }_{ij}^k\left(x\right)=0$

${\Omega }_{ij}\equiv {\left(L\left({\xi }_1\right)\omega \right)}_{ij}\equiv {\omega }_{rj}\left(x\right){\xi }_i^r+{\omega }_{ir}\left(x\right){\xi }_j^r+{\xi }^r{\partial }_r{\omega }_{ij}\left(x\right)=2A\left(x\right){\omega }_{ij}\left(x\right)$

${\Gamma }_{ij}^k\equiv {\left(L\left({\xi }_2\right)\gamma \right)}_{ij}^k\equiv {\xi }_{ij}^k+{\gamma }_{rj}^k\left(x\right){\xi }_i^r+{\gamma }_{ir}^k\left(x\right){\xi }_j^r-{\gamma }_{ij}^r\left(x\right){\xi }_r^k+{\xi }^r{\partial }_r{\gamma }_{ij}^k\left(x\right)=0$

${\Omega }_{ij}\equiv {\left(L\left({\xi }_1\right)\omega \right)}_{ij}\equiv {\omega }_{rj}\left(x\right){\xi }_i^r+{\omega }_{ir}\left(x\right){\xi }_j^r+{\xi }^r{\partial }_r{\omega }_{ij}\left(x\right)=2A\left(x\right){\omega }_{ij}\left(x\right)$

$\begin{array}{c}{\Gamma }_{ij}^k\equiv {\left(L\left({\xi }_2\right)\gamma \right)}_{ij}^k\\ \equiv {\xi }_{ij}^k+{\gamma }_{rj}^k\left(x\right){\xi }_i^r+{\gamma }_{ir}^k\left(x\right){\xi }_j^r-{\gamma }_{ij}^r\left(x\right){\xi }_r^k+{\xi }^r{\partial }_r{\gamma }_{ij}^k\left(x\right)\\ ={\delta }_i^k{A}_j\left(x\right)+{\delta }_j^k{A}_i\left(x\right)-{\omega }_{ij}\left(x\right){\omega }^{kr}\left(x\right)A_r\left(x\right)\end{array}$

where one has to eliminate the arbitrary function $A\left(x\right)$ and 1-form $A_i\left(x\right)d{x}^i$ for finding sections, replacing the ordinary Lie derivative $ℒ\left(\xi \right)$ by the formal Lie derivative $L\left({\xi }_q\right)$, that is replacing $j_q\left(\xi \right)$ by ${\xi }_q$ when needed. According to the structure of the above Medolaghi equations, it is important to notice that $\text{Ω}=L\left({\xi }_1\right)\omega \in S_2{T}^{\text{*}}$ and that $\text{Γ}=L\left({\xi }_2\right)\gamma \in S_2{T}^{\text{*}}\otimes T$. As another way to consider the Christoffel symbols, $\left(\delta ,-\gamma \right)=\left({\delta }_i^k,-{\gamma }_{ij}^k\right)$ is a $R_1$-connection and thus also a ${\hat{R}}_1$-connection because $R_1\subset {\tilde{R}}_1={\hat{R}}_1$, in such a way that $-{\omega }_{kj}{\gamma }_{jr}^k-{\omega }_{ik}{\gamma }_{jr}^k+{\partial }_r{\omega }_{ij}=0$.

It is well known that ${\gamma }_{ri}^r=\frac{1}{2}{\omega }^{rs}{\partial }_i{\omega }_{rs}\Rightarrow {\partial }_i{\gamma }_{rj}^r={\partial }_j{\gamma }_{ri}^r$ by introducing the inverse matrix of the metric. Accordingly, we have successively in the conformal case:

${\partial }_i{\xi }_r^r+{\gamma }_{rs}^r{\partial }_i{\xi }^s+{\xi }^s{\partial }_i{\gamma }_{rs}^r=n{\partial }_{i}A,{\xi }_{ri}^r+{\gamma }_{rs}^r{\xi }_i^s+{\xi }^s{\partial }_s{\gamma }_{ri}^r=n{A}_i$

Subtracting, we obtain $\left({\partial }_i{\xi }_r^r-{\xi }_{ri}^r\right)+{\gamma }_{rs}^r\left({\partial }_i{\xi }^s-{\xi }_i^s\right)=n\left({\partial }_{i}A-A_i\right)$ as a way to produce a restricted Spencer operator $d:\left(A,A_i\right)\to {\partial }_{i}A-A_i$.

We make a few comments on the relationship existing between these systems.

First of all, when $\omega =\left({\omega }_{ij}\left(x\right)={\omega }_{ji}\left(x\right)\right)$ is a non-degenerate metric with $det\left(\omega \right)\ne 0$, the corresponding Christoffel symbols are $\gamma =\left({\gamma }_{ij}^k\left(x\right)=\frac{1}{2}{\omega }^{kr}\left(x\right)\left({\partial }_i{\omega }_{rj}\left(x\right)+{\partial }_j{\omega }_{ri}\left(x\right)-{\partial }_r{\omega }_{ij}\left(x\right)\right)={\gamma }_{ji}^k\left(x\right)\right)$. We have the relations $R_1\subset {\tilde{R}}_1={\hat{R}}_1$ and obtain therefore $R_2={\rho }_1\left(R_1\right)$, ${\tilde{R}}_2\subset {\rho }_1\left({\tilde{R}}_1\right)$, ${\hat{R}}_2={\rho }_1\left({\hat{R}}_1\right)$, a result leading to the strict inclusions $R_2\subset {\tilde{R}}_2\subset {\hat{R}}_2$ with respective fiber dimensions $10<11<15$ when $n=4$ and $\omega$ is the Minkowski metric with signature $\left(1,1,1,-1\right)$.

Secondly, if we want to deal with geometric objects in both cases, we have to introduce the symmetric tensor density ${\hat{\omega }}_{ij}={\omega }_{ij}/{\left|det\left(\omega \right)\right|}^{1/n}$ and the second order object ${\hat{\gamma }}_{ij}^k={\gamma }_{ij}^k-\frac{1}{n}\left({\delta }_i^k{\gamma }_{rj}^r+{\delta }_j^k{\gamma }_{ri}^r-{\omega }_{ij}{\omega }^{ks}{\gamma }_{rs}^r\right)$ such that $\left|det\left(\hat{\omega }\right)\right|=1$, ${\hat{\gamma }}_{ri}^r=0$, in such a way that ${\widehat{ℛ}}_2=\left\{f_2\in {\Pi }_2|f_1^{-1}(\hat{\omega }=\hat{\omega },f_2^{-1}\left(\hat{\gamma }\right)=\hat{\gamma }\right\}$. It follows that ${\hat{g}}_1$ is defined by the equations ${\omega }_{rj}{\xi }_j^r+{\omega }_{rj}{\xi }_i^r-\frac{2}{n}{\omega }_{ij}{\xi }_r^r=0$ while ${\hat{g}}_2$ is defined by the equations ${\xi }_{ij}^k-\frac{1}{n}\left({\delta }_i^k{\xi }_{rj}^r+{\delta }_j^k{\xi }_{ri}^r-{\omega }_{ij}{\omega }^{ks}{\xi }_{rs}^r\right)=0$ which only depend on $\omega$ and no longer on $\hat{\omega }$. Only the first of the three following technical lemmas is known:

LEMMA 2.2: ${\hat{g}}_1$ is finite type with ${\hat{g}}_3=0$ when $n\ge 3$.

Proof: The symbol ${\hat{g}}_3$ is defined by the equations ${\xi }_{ijt}^k-\frac{1}{n}\left({\delta }_i^k{\xi }_{jrt}^r+{\delta }_j^k{\xi }_{irt}^r-{\omega }_{ij}{\omega }^{ks}{\xi }_{rst}^r\right)=0$. Summing on $k$ and $t$, we get ${\xi }_{rij}^r-\frac{1}{n}\left(2{\xi }_{rij}^r-{\omega }_{ij}{\omega }^{st}{\xi }_{rst}^r\right)=0$. Multiplying by ${\omega }^{ij}$ and summing on $i$ and $j$, we get ${\omega }^{ij}{\xi }_{rij}^r-\frac{2}{n}{\omega }^{ij}{\xi }_{rij}^r+{\omega }^{ij}{\xi }_{rij}^r=0$, that is to say $\left(n-1\right){\omega }^{ij}{\xi }_{rij}^r=0$ and thus ${\omega }^{ij}{\xi }_{rij}^r=0$ whenever $n\ge 2$. Substituting, we obtain $\left(n-2\right){\xi }_{rij}^r=0$ and thus ${\xi }_{rij}^r=0$ when $n\ge 3$, a result finally leading to ${\xi }_{ijt}^k=0$ and thus ${\hat{g}}_3=0,\forall n\ge 3$. In this case, it is important to notice that the third order jets only vanish when $\gamma =0$ locally or, equivalently, when $\omega$ is locally constant, for example when $n=4$ and $\omega$ is the Minkowski metric of space-time.

$\square$

LEMMA 2.3: ${\hat{g}}_2$ is 2-acyclic when $n\ge 4$.

Proof: As ${\hat{g}}_{3+r}=0,\forall r\ge 0$, we have only to prove the injectivity of the map $\delta$ in the sequence:

$\boxed{0\to {\wedge }^2{T}^{\text{*}}\otimes {\hat{g}}_2\stackrel{\delta }{\to }{\wedge }^3{T}^{\text{*}}\otimes {\hat{g}}_1\subset {\wedge }^3{T}^{\text{*}}\otimes T^{\text{*}}\otimes T}$

and thus to solve the linear system:

${\xi }_{i\alpha ,\beta \gamma }^k+{\xi }_{i\beta ,\gamma \alpha }^k+{\xi }_{i\gamma ,\alpha \beta }^k=0$

Substituting, we get the alternate sum over the cycle, where $\delta$ is again the Kronecker symbol:

$\mathcal{C}\left(\alpha \beta \gamma \right)\left({\delta }_i^k{\xi }_{r\alpha ,\beta \gamma }^r+{\delta }_{\alpha }^k{\xi }_{ri,\beta \gamma }^r-{\omega }_{i\alpha }{\omega }^{ks}{\xi }_{rs,\beta \gamma }^r\right)=0$

Summing on $k$ and $i$, we get:

$\mathcal{C}\left(\alpha \beta \gamma \right){\xi }_{r\alpha ,\beta \gamma }^r=0\Rightarrow \mathcal{C}\left(\alpha \beta \gamma \right)\left({\delta }_{\alpha }^k{\xi }_{ri,\beta \gamma }^r-{\omega }_{i\alpha }{\omega }^{ks}{\xi }_{rs,\beta \gamma }^r\right)=0$

that is to say:

$\left({\delta }_{\alpha }^k{\xi }_{ri,\beta \gamma }^r+{\delta }_{\beta }^k{\xi }_{ri,\gamma \alpha }^r+{\delta }_{\gamma }^k{\xi }_{ri,\alpha \beta }^r\right)-{\omega }^{ks}\left({\omega }_{i\alpha }{\xi }_{rs,\beta \gamma }^r+{\omega }_{i\beta }{\xi }_{rs,\gamma \alpha }^r+{\omega }_{i\gamma }{\xi }_{rs,\alpha \beta }^r\right)=0$

Summing now on $k$ and $\alpha$, we get:

$\left(n-3\right){\xi }_{ri,\beta \gamma }^r-{\omega }^{st}\left({\omega }_{i\beta }{\xi }_{rs,\gamma t}^r+{\omega }_{i\gamma }{\xi }_{rs,t\beta }^r\right)=0$

Multiplying by ${\omega }^{ij}$ and summing on $i$, we get:

$\left(n-3\right){\omega }^{ij}{\xi }_{ri,\beta \gamma }^r-{\omega }^{st}\left({\delta }_{\beta }^j{\xi }_{rs,\gamma t}^r+{\delta }_{\gamma }^j{\xi }_{rs,t\beta }^r\right)=0$

Summing on $j$ and $\beta$, we finally obtain:

$2\left(n-2\right){\omega }^{ij}{\xi }_{ri,j\gamma }^r=0\Rightarrow {\xi }_{ri,\beta \gamma }^r=0\Rightarrow {\xi }_{ij,\beta \gamma }^k=0,\forall n\ge 4$

Accordingly, the linear system has the only zero solution and ${\hat{g}}_2$ is thus $2$-acyclic $\forall n\ge 4$, a quite deep reason for which space-time has formal properties that are not satisfied by space alone.

$\square$

LEMMA 2.4: ${\hat{g}}_2$ is 3-acyclic when $n\ge 5$.

Proof: As ${\hat{g}}_{3+r}=0,\forall r\ge 0$, we have only to prove the injectivity of the map $\delta$ in the sequence:

$\boxed{0\to {\wedge }^3{T}^{\text{*}}\otimes {\hat{g}}_2\stackrel{\delta }{\to }{\wedge }^4{T}^{\text{*}}\otimes {\hat{g}}_1\subset {\wedge }^4{T}^{\text{*}}\otimes T^{\text{*}}\otimes T}$

and thus to solve the linear system:

${\xi }_{i\alpha ,\beta \gamma \delta }^k-{\xi }_{i\beta ,\gamma \delta \alpha }^k+{\xi }_{i\gamma ,\delta \alpha \beta }^k-{\xi }_{i\delta ,\alpha \beta \gamma }^k=0$

Substituting, we get the alternate sum over the cycle in which $\delta$ must not be confound with the Kronecker symbol $\delta$:

$\mathcal{C}\left(\alpha \beta \gamma \delta \right)\left({\delta }_i^k{\xi }_{r\alpha ,\beta \gamma \delta }^r+{\delta }_{\alpha }^k{\xi }_{ri,\beta \gamma \delta }^r-{\omega }_{i\alpha }{\omega }^{ks}{\xi }_{rs,\beta \gamma \delta }^r\right)=0$

Contracting in $k$ and $i$ the previous formula, we get:

$\mathcal{C}\left(\alpha \beta \gamma \delta \right){\xi }_{r\alpha ,\beta \gamma \delta }^r=0\Rightarrow \mathcal{C}\left(\alpha \beta \gamma \delta \right)\left({\delta }_{\alpha }^k{\xi }_{ri,\beta \gamma \delta }^r-{\omega }_{i\alpha }{\omega }^{ks}{\xi }_{rs,\beta \gamma \delta }^r\right)=0$

Contracting now in $k$ and $\alpha$, we get:

$\left(n{\xi }_{ri,\beta \gamma \delta }^r-{\xi }_{ri,\gamma \delta \beta }^r+{\xi }_{ri,\delta \gamma \beta }^r-{\xi }_{ri,\delta \beta \gamma }^r\right)-{\xi }_{ri,\beta \gamma \delta }^r+{\omega }^{st}\left({\omega }_{i\beta }{\xi }_{rs,\gamma \delta t}^r-{\omega }_{i\gamma }{\xi }_{rs,\delta t\beta }^r+{\omega }_{i\delta }{\xi }_{rs,t\beta \gamma }^r\right)=0$

and thus:

$\left(n-4\right){\xi }_{ri,\beta \gamma \delta }^r+{\omega }^{st}\left({\omega }_{i\beta }{\xi }_{rs,\gamma \delta t}^r+{\omega }_{i\gamma }{\xi }_{rs,\delta \beta t}^r+{\omega }_{i\delta }{\xi }_{rs,\beta \gamma t}^r\right)=0$

that we may transform into:

$\left(n-4\right){\omega }^{ij}{\xi }_{ri,\beta \gamma \delta }^r+{\omega }^{st}\left({\delta }_{\beta }^j{\xi }_{rs,\gamma \delta t}^r+{\delta }_{\gamma }^j{\xi }_{rs,\delta \beta t}^r+{\delta }_{\delta }^j{\xi }_{rs,\beta \gamma t}^r\right)=0$

Contracting in $j$ and $\beta$, we finally obtain:

$2\left(n-3\right){\omega }^{ij}{\xi }_{ri,j\gamma \delta }^r=0\Rightarrow \left(n-4\right){\xi }_{ri,\beta \gamma \delta }^r=0$

and ${\hat{g}}_2$ is thus 3-acyclic for $n\ge 5$.

$\square$

It now remains to explain what must be done for $n=1,2$ in order to preserve the number of parameters of the conformal group which is indeed $\left(n+1\right)\left(n+2\right)/2$ whenever $n\ge 3$ as we saw because, in this case, we have n translations + n(n − 1)/2 rotations + 1 dilatation + n elations, the latter introduced by E. Cartan in 1922 [17] .

EXAMPLE 3.1: (Weyl and Conformal groups for $n=1$)

Let $K$ be a differential field with differentials $\partial$ and $y$ be differential indeterminates with formal derivatives $d$ such that $d|K=\partial$. In order to adapt differential algebra with differential geometry, let us set $dy=y_x$ and introduce the differential field $K\langle y\rangle =K\left(y,y_x,y_{xx},y_{xxx},\cdots \right)$ with $d_x{y}_x=y_{xx}$ and so on. We may consider the chain of inclusions of differential fields [18] :

$\boxed{K\subset K\langle \frac{y_{xxx}}{y_x}-\frac{3}{2}{\left(\frac{y_{xx}}{y_x}\right)}^2\rangle \subset K\langle \frac{y_{xx}}{y_x}\rangle \subset K\langle \frac{y_x}{y}\rangle \subset K\langle y\rangle }$

but we may also replace $y_x/y$ by $y_x$ if we want to replace dilatation by translation. Setting indeed for example $\Phi \equiv y_{xx}/y_x$, we obtain $\left(y_{xxx}/y_x\right)-\frac{3}{2}{\left(y_{xx}/y_x\right)}^2=d_x\Phi -\frac{1}{2}{\Phi }^2$ while, setting $\Psi \equiv y_x/y$, we have $y_{xx}/y_x=\left(1/\Psi \right)d_x\Psi +{\Psi }^2$ and so on.

However, a main point is that, if the affine group $y=ax+b$, analogue of the Weyl group but considered as a Lie pseudogroup, is defined by the second order infinitesimal Lie equation ${\xi }_{xx}=0$ with jet notation, the projective group $y=\left(ax+b\right)/\left(cx+1\right)$, analogue to the conformal group but considered as a Lie pseudogroup, is defined by the third order infinitesimal Lie equation ${\xi }_{xxx}=0$ while the elation becomes $\frac{1}{y}=\frac{1}{x}+c$ and such a situation will be even worst when $n=2$.

The general section of ${\hat{R}}_3\subset J_3\left(T\right)$ with solution $\hat{\Theta }$ will be thus defined by:

$\boxed{\left(\begin{array}{c}\xi \left(x\right)\\ {\xi }_x\left(x\right)\\ {\xi }_{xx}\left(x\right)\end{array}\right)=\left(\begin{array}{ccc}1& x& \frac{1}{2}{x}^2\\ 0& 1& x\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{\lambda }^1\left(x\right)\\ {\lambda }^2\left(x\right)\\ {\lambda }^3\left(x\right)\end{array}\right)}$

and we must add ${\xi }_{xxx}=0$. The components of the Spencer operator $D_1$ thus become when = 1:

$\boxed{\left(\begin{array}{c}{\partial }_x\xi \left(x\right)-{\xi }_x\left(x\right)\\ {\partial }_x{\xi }_x\left(x\right)-{\xi }_{xx}\left(x\right)\\ {\partial }_x{\xi }_{xx}\left(x\right)\end{array}\right)=\left(\begin{array}{ccc}1& x& \frac{1}{2}{x}^2\\ 0& 1& x\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{\partial }_x{\lambda }^1\left(x\right)\\ {\partial }_x{\lambda }^2\left(x\right)\\ {\partial }_x{\lambda }^3\left(x\right)\end{array}\right)}$

which is a rather tricky result. We obtain the fundamental diagram I:

$\boxed{\begin{array}{ccccccccccc}& & & & & & 0& & 0& & \\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & \hat{\Theta }& \stackrel{j_3}{\to }& 3& \stackrel{D_1}{\to }& 3& \to & 0\\ & & & & & & \downarrow & & \parallel & & \\ & & 0& \to & 1& \stackrel{j_3}{\to }& 4& \stackrel{D_1}{\to }& 3& \to & 0\\ & & & & \parallel & & \downarrow & & \downarrow & & \\ 0& \to & \hat{\Theta }& \to & 1& \stackrel{\hat{\mathcal{D}}}{\to }& 1& \to & 0& & \\ & & & & \downarrow & & \downarrow & & & & \\ & & & & 0& & 0& & & & \end{array}}$

Using now ${\xi }_{xx}=0$ and ${\xi }_{xxx}=0$, a similar diagram can be obtained for the second order system with solutions $\tilde{\Theta }$, namely:

$\boxed{\begin{array}{ccccccccccc}& & & & & & 0& & 0& & \\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & \tilde{\Theta }& \stackrel{j_3}{\to }& 2& \stackrel{D_1}{\to }& 2& \to & 0\\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & 1& \stackrel{j_3}{\to }& 4& \stackrel{D_1}{\to }& 3& \to & 0\\ & & & & \parallel & & \downarrow & & \downarrow & & \\ 0& \to & \tilde{\Theta }& \to & 1& \stackrel{\tilde{\mathcal{D}}}{\to }& 2& \stackrel{{\tilde{\mathcal{D}}}_1}{\to }& 1& \to & 0\\ & & & & \downarrow & & \downarrow & & \downarrow & & \\ & & & & 0& & 0& & 0& & \end{array}}$

As $\tilde{\Theta }\subset \hat{\Theta }$, the upper Spencer sequence for $\tilde{\Theta }$ can be injected into the upper Spencer sequence for $\hat{\Theta }$ and the co-kernel of this injection is the sequence $1\stackrel{d}{\to }1\to 0$ that only depends on the unique elation. Similarly, the lower Janet sequence for $\tilde{\Theta }$ can be projected onto the lower Janet sequence for $\hat{\Theta }$ and the kernel of this projection is an isomorphic sequence that thus only depends on the unique elation. As neither the Spencer nor the Janet sequences have ever been used in physics and the Fundamental Diagram I is not known (Compare to [7] ), we refer the reader to [19] for discovering why black holes cannot exist because the only important object associated with a metric is thus its group of invariance which can be of quite low dimension 2 for the Kerr metric while the Janet sequence can be awfully complicated with a mixture of second and third order CC!.

Using finally geometric objects, we explain the reason for which we have associated the so-called projective group with the conformal group. Indeed, writing down the Lie form of the OD equation defining the affine group, we have $\Phi \equiv \frac{y_{xx}}{y_x}=\gamma \left(x\right)$. Changing the source by setting $\bar{x}=\phi \left(x\right)$, with $y_x=y_{\bar{x}}\frac{\partial \phi }{\partial x}$, $y_{xx}=y_{\bar{x}\bar{x}}{\left(\frac{\partial \phi }{\partial x}\right)}^2+y_{\bar{x}}\frac{{\partial }^2\phi }{\partial x^2}$ and so on, we notice that $\gamma$ is transformed like a Christoffel symbol and obtain the corresponding Medolaghi OD equation $\text{Γ}\equiv {\xi }_{xx}+\gamma \left(x\right){\xi }_x+\xi {\partial }_x\gamma \left(x\right)=0$. Setting $\nu ={\partial }_x\gamma -\frac{1}{2}{\gamma }^2$ and linearizing, we obtain the Medolaghi OD equation for the Lie form $\left(y_{xxx}/y_x\right)-\frac{3}{2}{\left(y_{xx}/y_x\right)}^2=\nu \left(x\right)$, namely:

$\boxed{d_x\Gamma -\gamma \Gamma \equiv {\xi }_{xxx}+2\nu \left(x\right){\xi }_x+\xi {\partial }_x\nu \left(x\right)=0}$

that only depends on $\nu$ and no longer on $\gamma$. Up to our knowledge, third order geometric object have never been considered.

The link with the Spencer operator may be easily obtained by considering the first order system ${\hat{R}}_{q+1}\subset J_1\left({\hat{R}}_q\right)$ in general, setting now $\left\{z^1=y,z^2=y_x,z^3=y_{xx}\right\}$ and considering the first order system $z_x^1-z^2=0$, $z_x^2-z^3=0$, $z_x^3=0$ for constructing the Spencer sequence with $q=3$.

EXAMPE 3.2: (Weyl group for $n=2$) With $m=n=2,q=3$, consider the third order involutive system ${\tilde{R}}_2\subset J_2\left(T\right)$ defined by 8 PD equations with corresponding Janet tabular:

$\{\begin{array}{lll}{d}_{22}{\xi }^k\hfill & =\hfill & 0\hfill \\ d_{12}{\xi }^k\hfill & =\hfill & 0\hfill \\ d_{11}{\xi }^k\hfill & =\hfill & 0\hfill \\ d_2{\xi }^2-d_1{\xi }^1\hfill & =\hfill & 0\hfill \\ d_2{\xi }^1+d_1{\xi }^2\hfill & =\hfill & 0\hfill \end{array}\boxed{\begin{array}{ll}1\hfill & 2\hfill \\ 1\hfill & •\hfill \\ 1\hfill & •\hfill \\ •\hfill & •\hfill \\ •\hfill & •\hfill \end{array}}$

We have $2+2+4=8$ first order CC related by 2 first order CC in the exact Janet sequence:

$\boxed{0\to \Theta \to 2\underset{2}{\stackrel{\mathcal{D}}{\to }}8\underset{1}{\stackrel{{\mathcal{D}}_1}{\to }}8\underset{1}{\stackrel{{\mathcal{D}}_2}{\to }}2\to 0}$

Multiplying the successive CC by the test functions $\lambda ,\mu ,\nu$ and integrating by parts, we obtain the dual adjoint differential sequence:

$\boxed{0\leftarrow 2\underset{2}{\stackrel{ad\left(\mathcal{D}\right)}{\leftarrow }}8\underset{1}{\stackrel{ad\left({\mathcal{D}}_1\right)}{\leftarrow }}\boxed{8}\underset{1}{\stackrel{ad\left({\mathcal{D}}_2\right)}{\leftarrow }}2\leftarrow 0}$

which is not exact at $\boxed{8}$ though $ad\left({\mathcal{D}}_2\right)$ is injective, a result not evident that can be checked by computer algebra but which it coming from deep results of homological algebra as we shall see.

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

$\begin{array}{cccccccccc}& & 0& & 0& & 0& & 0& \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & \\ 0& \to & g_3& \to & S_3{T}^{*}\otimes E& \to & T^{*}\otimes F_0& \to & F_1& \to 0\\ & & \downarrow & & \downarrow & & \downarrow & \searrow & \parallel & \\ 0& \to & R_3& \to & J_3\left(E\right)& \to & J_1\left(F_0\right)& \to & F_1& \to 0\\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & \\ 0& \to & R_2& \to & J_2\left(E\right)& \to & F_0& \to & 0& \\ & & \downarrow & & \downarrow & & \downarrow & & & \\ & & 0& & 0& & 0& & & \end{array}$

$\begin{array}{cccccccccccc}& & & & 0& & 0& & 0& & & \\ & & & & \downarrow & & \downarrow & & \downarrow & & & \\ & & 0& \to & 8& \to & 16& \to & 8& \to & 0& \\ & & \downarrow & & \downarrow & & \downarrow & & \parallel & & & \\ 0& \to & 4& \to & 20& \to & 24& \to & 8& \to & 0& \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & & \\ 0& \to & 4& \to & 12& \to & 8& \to & 0& & & \\ & & \downarrow & & \downarrow & & \downarrow & & & & & \\ & & 0& & 0& & 0& & & & & \end{array}$

$\begin{array}{ccccccccccccc}& & 0& & 0& & 0& & 0& & 0& & \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0& \to & g_4& \to & S_4{T}^{\text{*}}\otimes E& \to & S_2{T}^{\text{*}}\otimes F_0& \to & T^{\text{*}}\otimes F_1& \to & F_2& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & \searrow & \parallel & & \\ 0& \to & R_4& \to & J_4\left(E\right)& \to & J_2\left(F_0\right)& \to & J_1\left(F_1\right)& \to & F_2& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0& \to & R_3& \to & J_3\left(E\right)& \to & J_1\left(F_0\right)& \to & F_1& \to & 0& & \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & & & \\ & & 0& & 0& & 0& & 0& & & & \end{array}$

$\begin{array}{ccccccccccccc}& & & & 0& & 0& & 0& & 0& & \\ & & & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0& \to & 10& \to & 24& \to & 16& \to & 2& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \parallel & & \\ 0& \to & 4& \to & 30& \to & 48& \to & 24& \to & 2& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \\ 0& \to & 4& \to & 20& \to & 24& \to & 8& \to & 0& & \\ & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & & & \\ & & 0& & 0& & 0& & 0& & & & \end{array}$

Using these diagrams, we obtain successively, till we stop, all the successive generating CC.

As a byproduct we have the exact sequences : $\forall r\ge 0$:

$0\to R_{r+4}\to J_{r+4}\left(T\right)\to J_{r+2}\left(F_0\right)\to J_{r+1}\left(F_1\right)\to J_r\left(F_2\right)\to 0$

Such a result can be checked directly through the identity:

$4-\left(r+5\right)\left(r+6\right)+4\left(r+3\right)\left(r+4\right)-4\left(r+2\right)\left(r+3\right)+\left(r+1\right)\left(r+2\right)=0$

We obtain therefore the formally exact sequence we were looking for.

In the differential module framework over the commutative ring $D=K\left[d_1,d_2\right]=K\left[d\right]$ of differential operators with coefficients in the trivially differential field $K=\mathbb{Q}$, we have the free resolution:

$\boxed{0\to D^2\underset{1}{\stackrel{{\mathcal{D}}_2}{\to }}{D}^8\underset{1}{\stackrel{{\mathcal{D}}_1}{\to }}{D}^8\underset{2}{\stackrel{\mathcal{D}}{\to }}{D}^2\to M\to 0}$