Let $\mathcal{D}:\xi \to \eta$ be a linear differential operator of order $q$. A direct problem is to look for a differential operator ${\mathcal{D}}_1:\eta \to \zeta$ such that ${\mathcal{D}}_1\eta =0$ is generating the compatibility conditions (CC). For example, starting with $q=2$ and the operator $d_{22}\xi ={\eta }^2$, $d_{12}\xi ={\eta }^1$ we obtain at once the single first order CC $d_1{\eta }^2-d_2{\eta }^1=0$. Now, multiplying on the left by a test function $\lambda$ and integrating by parts, the adjoint operator becomes $-d_1\lambda ={\mu }^2,d_2\lambda ={\mu }^1$ with the single first order CC $d_1{\mu }^1+d_2{\mu }^2=0$. However, multiplying now $\left({\eta }^1,{\eta }^2\right)$ by two test functions $\left({\mu }^1,{\mu }^2\right)$, adding and integrating by parts, we obtain the second order adjoint operator $d_{22}{\mu }^2+d_{12}{\mu }^1$ which is NOT generating the CC of $ad\left({\mathcal{D}}_1\right)$ even if it is well known that $ad\left(\mathcal{D}\right)\circ ad\left({\mathcal{D}}_1\right)=ad\left({\mathcal{D}}_1\circ \mathcal{D}\right)=0$. This result is just showing that, when ${\mathcal{D}}_1$ generates the CC of $\mathcal{D}$, then $ad\left(\mathcal{D}\right)$ does not generate in general ALL the CC of $ad\left({\mathcal{D}}_1\right)$. If $M$ is the differential module that can be defined by $\mathcal{D}$, the above “ defect “ is measured by a module called $ex{t}^1\left(M\right)$ that only depends on $M$ and not of the way to define it as we shall see. We shall denote by $N$ the differential module which is defined by $ad\left(\mathcal{D}\right)$ and introduce the torsion submodule $t\left(N\right)$ made by all the elements satisfying at least one OD or PD equation FOR ITSELF. In the present example, $N$ is defined by $d_{12}{\eta }^1+d_{22}{\eta }^2=d_2\left(d_1{\eta }^1+d_2{\eta }^2\right)=0$ and $t\left(N\right)$ is generated by $\nu =d_1{\eta }^1+d_2{\eta }^2$. The following formula:

$\boxed{t\left(N\right)\simeq ex{t}^1\left(M\right)\iff t\left(M\right)\simeq ex{t}^1\left(N\right)}$

is a cornerstone of homological algebra and the equivalence exists because $ad\left(ad\left(\mathcal{D}\right)\right)=\mathcal{D}$. All the second section will be presenting in a rather self-contained way all the definitions and homological results needed for understanding these new concepts and refer the reader to the Zentralblatt review Zbl 1079.93001 for understanding how difficult it is to collect in the literature all the references needed. However, sometimes it may work, that is $ad\left(\mathcal{D}\right)$ may generate the CC of $ad\left({\mathcal{D}}_1\right)$ or, equivalently, $ex{t}_1\left(M\right)=0$ as we shall see in the following motivating examples that we set up together in this Introduction. For the moment, we ask the reader to stop for a few minutes in order to imagine any link that could exist between this formula and Kalman test in control theory on one side or between this formula and Einstein or Maxwell equations on the other side.

Being a former student of A. Lichnerowicz, specialist of systems of PDE in group theory and control theory, it has been a challenge for me to apply the new methods of Differential Homological Algebra introduced around 1990 by U. Oberst [1] and E. Zerz [2] for systems with constant coefficients or by M. Kashiwara [3] for systems with variable coefficients in order to study gravitational waves. The three last papers published in 2024 [4] - [6] with a short summary in [7] could be roughly summarized by the single formula:

$\boxed{t\left(N\right)\simeq ex{t}^1\left(M\right)=0}$

where $M$ is the differential module defined by the Killing operator and $N$ is the differential module defined by the $Cauchy=ad\left(Killing\right)$ adjoint operator with torsion submodule $t\left(N\right)$, as extension modules are torsion modules that do not depend on the resolution of $M$ that MUST be used, namely the differential sequence in which the order of an operator is under its arrow, which is exact in the sense that any operator generates the CC of the previous one:

$\boxed{0\to \text{Θ}\to n\underset{1}{\stackrel{Killing}{\to }}\frac{n\left(n+1\right)}{2}\underset{2}{\stackrel{Riemann}{\to }}\frac{n^2\left(n^2-1\right)}{12}\underset{1}{\stackrel{Bianchi}{\to }}\frac{n^2\left(n^2-1\right)\left(n-2\right)}{24}}$

Such a differential sequence sequence can be found, from a purely computational way, in any textbook of general relativity. However, its adjoint differential sequence made by the respective adjoint operators going therefore “backwards” (that is from right to left) and totally unknown, is also surprisingly exact, the stress $\left({\sigma }^{ij}\right)\in {\wedge }^n{T}^{\text{*}}\otimes S_{2}T$ being a tensor density and not only a tensor [4] :

$\boxed{0\leftarrow n\underset{1}{\stackrel{Cauchy}{\leftarrow }}\frac{n\left(n+1\right)}{2}\underset{2}{\stackrel{Beltrami}{\leftarrow }}\frac{n^2\left(n^2-1\right)}{12}\underset{1}{\stackrel{Lanczos}{\leftarrow }}\frac{n^2\left(n^2-1\right)\left(n-2\right)}{24}}$

This result points out the confusion (the word is weak !) done by Einstein (1915) for space-time while following the work of Beltrami (1892) for space only, both using the Einstein operator but ignoring that it was was self-adjoint in the framework of differential double duality. The Cauchy operator can be thus parametrized (backwards !) by $n\left(n+1\right)/2$ stress functions having strictly nothing to do with the metric, exactly like in the case of the single Airy stress function for plane elasticity, because the Airy parametrization is only the adjoint of the Riemann operator when $n=2$. Though unpleasant it is, this result questions the origin and existence of gravitational waves, not because of a problem of detection but because of a problem of equations as we shall prove that Einstein did confuse the $Cauchy=ad\left(Killing\right)$ operator with the so-called “divergence” operator induced from the Bianchi operator. In a rough way, it is like claiming that the single Airy (for plane) or the six Beltrami (for space) stress functions in elasticity should have something to do with earthquakes.

One striking byproduct of our claim is provided by the next example [8] but we advise the reader to read all the following examples that are presented totally independently.

Example 1.1: (Kalman system): The previous formula is the Kalman test in classical OD control theory. As any operator is the adjoint of its own adjoint because $ad\left(ad\left(\mathcal{D}\right)\right)=\mathcal{D}$ as we already said and one can thus exchange $M$ and $N$ in the formula, that is $t\left(N\right)\simeq ex{t}^1\left(M\right)\iff t\left(M\right)\simeq ex{t}^1\left(N\right)$. Hence, if $M$ is the differential module defined by the formally surjective Kalman operator $\left(y^k,u^r\right)\to \left(-d{y}^k+A_l^k{y}^l+B_r^k{u}^r\right)$ with time derivative $d$, inputs $u$ and outputs $y$ while $N$ is the differential module defined by its adjoint operator with torsion submodule $t\left(N\right)$, then we prove that the Kalman controllability test amounts to say that the given control system is controllable if and only if $N=0$. Indeed, introducing Lagrange multipliers $\lambda =\left({\lambda }_k\right)$, the kernel of the adjoint operator is defined by the OD equations ( $y^k\to d{\lambda }_k+{\lambda }_l{A}_k^l=0$, $u^r\to {\lambda }_k{B}_r^k=0$) with all their differential consequences, namely:

$\boxed{d\lambda +\lambda A=0,\lambda B=0\Rightarrow d\left(\lambda B\right)=\left(d\lambda \right)B=-\left(\lambda A\right)B=0\Rightarrow \lambda AB=0,\cdots }$

and so on, as a way to recover the well known controllability matrix $\left(B,AB,A^{2}B,\cdots \right)$. It follows that $t\left(N\right)=N$ is already a torsion module and that the Kalman system is controllable if and only if $N=t\left(N\right)=0$ as claimed. Moreover, it is well known that a control system is controllable if and only if it is parametrizable, that is $M$ can be embedded into a free differential module. In fact, when $n=1$, $D=K\left[d\right]$ is a principal ideal domain, that is any ideal can be generated by a single element, and it is well known that any torsion-free module over $D$ is indeed free. Accordingly, the kernel of the projection of $Dy+Du\simeq D^2$ onto $M$ is free too and there is no loss of generality by supposing that the control system is made by differentially independent equations. The controllability of an OD control system is thus the purely structural property $t\left(M\right)=0$ independently of the presentation, a fact amounting to the impossibility to find any torsion element, that is any linear combination of the the control variables that could be a solution of an autonomous OD equation for itself.

Example 1.2: With $m=3,n=1$ and a parameter $a=a\left(x\right)\in K$, let us consider the formally surjective first order operator with $d=d_x$:

$\boxed{{\mathcal{D}}_1:\left({\eta }^1,{\eta }^2,{\eta }^3\right)\to \left(d{\eta }^1-a{\eta }^2-d{\eta }^3={\zeta }^1,{\eta }^1-d{\eta }^2+d{\eta }^3={\zeta }^2\right)}$

Multiplying on the left by two test functions $\left({\lambda }^1,{\lambda }^2\right)$ and integrating by parts , we obtain:

$\boxed{ad\left({\mathcal{D}}_1\right):\left({\lambda }^1,{\lambda }^2\right)\to \left(-d{\lambda }^1+{\lambda }^2={\mu }^1,-a{\lambda }^1+d{\lambda }^2={\mu }^2,d{\lambda }^1-d{\lambda }^2={\mu }^3\right)}$

In order to look for the CC of this operator, we obtain:

${\lambda }^2-a{\lambda }^1={\mu }^1+{\mu }^2+{\mu }^3\Rightarrow -\left(\partial a+a^2-a\right){\lambda }^1=d\left({\mu }^1+{\mu }^2+{\mu }^3\right)+\left(a-1\right){\mu }^2+a{\mu }^3$

but we have also $\left(d-a\right){\lambda }^1={\mu }^2+{\mu }^3$ and may easily eliminate ${\lambda }^1$.

Introducing the notation $j_q\left(\mu \right)$ for all the derivatives of $\mu$ up to order $q$, we obtain therefore:

$\boxed{\left(\partial a+a^2-a\right)\lambda \in j_1\left(\mu \right)}$

When the structural controllability condition is satisfied, that is when $a$ is not a solution of the Riccati equation in the bracket, we may obtain a second order CC operator of the form:

$\boxed{ad\left(\mathcal{D}\right):\left({\mu }^1,{\mu }^2,{\mu }^3\right)\to d^2{\mu }^1+d^2{\mu }^2+d^2{\mu }^3+\cdots =\nu }$

Multiplying on the left by a test function $\xi$ and integrating by parts, we obtain the second order injective parametrization, provided that $\partial a+a^2-a\ne 0$:

$\boxed{\mathcal{D}:\xi \to \left(d^2\xi +\cdots ={\eta }^1,d^2\xi +\cdots ={\eta }^2,d^2\xi -\cdots ={\eta }^3\right)}$

We have the long exact (splitting) sequence and its adjoint (splitting) sequence which is also exact:

$\boxed{\begin{array}{ccccccccc}0& \to & \xi & \underset{2}{\stackrel{\mathcal{D}}{\to }}& \eta & \underset{1}{\stackrel{{\mathcal{D}}_1}{\to }}& \zeta & \to & 0\\ 0& \leftarrow & \nu & \underset{2}{\stackrel{ad\left(\mathcal{D}\right)}{\leftarrow }}& \mu & \underset{1}{\stackrel{ad\left({\mathcal{D}}_1\right)}{\leftarrow }}& \lambda & \leftarrow & 0\end{array}}$

At no moment one has to decide about the choice of inputs and outputs and we advise the reader to effect any choice for applying the Kalman test when $a$ is a constant parameter. Of course $a=cst\Rightarrow a\left(a-1\right)\ne 0$ in a coherent way with a classical approach but we have thus been able to extend the controllability test even for variable coefficients, a result still not known because it essentially depends on a systematic use of the adjoint operators.

Example 1.3: (Double pendulum): Many examples can be found in classical ordinary differential control theory because it is known that a linear control system is controllable if and only if it is parametrizable [6] . In our opinion, the best and simplest one is the so-called double pendulum in which a rigid bar is able to move horizontally with reference position $x$ and we attach two pendulums with respective length $l_1$ and $l_2$ making the (small) angles ${\theta }_1$ and ${\theta }_2$ with the vertical, the corresponding control system does not depend on the mass of each pendulum and the two equations easily follow by projection from the Newton laws:

$\boxed{{\mathcal{D}}_1\eta =0\iff d^{2}x+l_1{d}^2{\theta }^1+g{\theta }^1=0,d^{2}x+l_2{d}^2{\theta }^2+g{\theta }^2=0}$

where $g$ is the gravity. A first result, still not acknowledged by the control community, is to prove that this control system is controllable if and only if $l_1\ne l_2$ without using a tedious computation through the standard Kalman test but, equivalently, to prove that the corresponding second order operator $ad\left({\mathcal{D}}_1\right)$ is injective. Though this is not evident, such a result comes from the fact $D$ is a principal ideal ring when $n=1$ and thus, if the differential module $M_1$ is torsion-free, then $M_1$ is also free and has a basis allowing to split the short exact resolution $0\to D^2\stackrel{{\mathcal{D}}_1}{\to }{D}^3\to M_1\to 0$ with $M_1\simeq D$ in this case. When learning control theory, it has also been a surprise to be unable to find examples in which the controllability was explicitly shown not to depend on the choice of inputs and outputs among the system variables, like in such an example as we shall see.

Hence, multiplying on the left the first OD equation by ${\lambda }^1$, the second by ${\lambda }^2$, then adding and integrating by parts, we get:

$\boxed{ad\left({\mathcal{D}}_1\right)\lambda =\mu \iff \{\begin{array}{lllll}x\hfill & \to \hfill & d^2{\lambda }^1+d^2{\lambda }^2\hfill & =\hfill & {\mu }^1\hfill \\ {\theta }^1\hfill & \to \hfill & l_1{d}^2{\lambda }^1+g{\lambda }^1\hfill & =\hfill & {\mu }^2\hfill \\ {\theta }^2\hfill & \to \hfill & l_2{d}^2{\lambda }^2+g{\lambda }^2\hfill & =\hfill & {\mu }^3\hfill \end{array}}$

The main problem is that the operator $ad\left({\mathcal{D}}_1\right)$ is not formally integrable because we have:

$l_2{\lambda }^1+l_1{\lambda }^2=\frac{1}{g}\left(l_2{\mu }^2+l_1{\mu }^3-l_1{l}_2{\mu }^1\right)$

and is thus injective if and only if $l_1\ne l_2$ because, differentiating twice this equation, we also get:

$\left(l_2/l_1\right){\lambda }^1+\left(l_1/l_2\right){\lambda }^2\in j_2\left(\mu \right)$

Hence, as the determinant of this $2\times 2$ matrix is exactly $l_1-l_2$, if $l_1\ne l_2$, we finally obtain $\lambda \in j_2\left(\mu \right)$ and, after tricky substitutions, a single fourth order CC for $\mu$, namely:

$\boxed{-l_1{l}_2{d}^4{\mu }^1-g\left(l_1+l_2\right)d^2{\mu }^1-g^2{\mu }^1+l_2{d}^4{\mu }^2+g{d}^2{\mu }^2+l_1{d}^4{\mu }^3+g{d}^2{\mu }^3=\nu }$

showing that $ad\left(\mathcal{D}\right)$ and thus $\mathcal{D}$ is indeed a fourth order operator a result not evident at first sight. It follows that we have thus been able to work out the parametrizing operator $\mathcal{D}$ of order 4, namely:

$\boxed{\mathcal{D}\varphi =\eta \iff \{\begin{array}{ccc}-l_1{l}_2{d}^4\varphi -g\left(l_1+l_2\right)d^2\varphi -g^2\varphi & =& x\\ l_2{d}^4\varphi +g{d}^2\varphi & =& {\theta }_1\\ l_1{d}^4\varphi +g{d}^2\varphi & =& {\theta }_2\end{array}}$

This parametrization is injective iff $l_1\ne l_2$ because we have successively with $g\ne 0$:

$l_2{d}^2\varphi +g\varphi =0\Rightarrow l_1{d}^2\varphi +g\varphi =0\Rightarrow g\left(l_1-l_2\right)\varphi =0\Rightarrow \varphi =0$

We have the short exact splitting sequence and its adjoint splitting sequence which is also exact:

$\boxed{\begin{array}{ccccccccc}0& \to & 1& \underset{4}{\stackrel{\mathcal{D}}{\to }}& 3& \underset{2}{\stackrel{{\mathcal{D}}_1}{\to }}& 2& \to & 0\\ 0& \leftarrow & 1& \underset{4}{\stackrel{ad\left(\mathcal{D}\right)}{\leftarrow }}& 3& \underset{2}{\stackrel{ad\left({\mathcal{D}}_1\right)}{\leftarrow }}& 2& \leftarrow & 0\end{array}}$

We now study the way to split these sequences. As any operator is the adjoint of its own adjoint, we define the lift of the lower sequence as follows:

$\boxed{g^2\left(l_1-l_2\right){\lambda }^1=g\left(l_1-l_2\right){\mu }^2-g{\left(l_1\right)}^2{\mu }^1+l_1{l}_2{d}^2{\mu }^2+{\left(l_1\right)}^2{d}^2{\mu }^3-{\left(l_1\right)}^2{l}_2{d}^2{\mu }^1}$

$\boxed{g^2\left(l_1-l_2\right){\lambda }^2=g\left(l_1-l_2\right){\mu }^3+g{\left(l_2\right)}^2{\mu }^1-{\left(l_2\right)}^2{d}^2{\mu }^2-l_1{l}_2{d}^2{\mu }^3+l_1{\left(l_2\right)}^2{d}^2{\mu }^1}$

obtain the lift of the upper sequence, up to a factor $g^2\left(l_1-l_2\right)$, namely:

$\boxed{-g{\left(l_1\right)}^2{\zeta }^1+g{\left(l_2\right)}^2{\zeta }^2-{\left(l_1\right)}^2{l}_2{d}^2{\zeta }^1+l_1{\left(l_2\right)}^2{\zeta }^2=x}$

$\boxed{g\left(l_1-l_2\right){\zeta }^1+l_1{l}_2{d}^2{\zeta }^1-{\left(l_2\right)}^2{d}^2{\zeta }^2={\theta }^1}$

$\boxed{g\left(l_1-l_2\right){\zeta }^2+{\left(l_1\right)}^2{d}^2{\zeta }^1-l_1{l}_2{d}^2{\zeta }^2={\theta }^2}$

We finally consider the case $l_1=l_2=l$. Subtracting the two OD equations, we discover that $z={\theta }^1-{\theta }^2$ is an observable quantity that satisfies the autonomous system $l{d}^{2}z+gz=0$ existing for a single pendulum. It follows that $z$ is a torsion element and the system cannot be controllable. When $z=0\Rightarrow {\theta }^1={\theta }^2=\theta$ we let the reader prove that the remaining OD equation $d^{2}x+l{d}^2\theta +g\theta =0$ can be parametrized by $l{d}^2\xi +g\xi =x$, $-d^2\xi =\theta$.

At this stage of the reading, we invite the reader to realize this experiment with a few dollars, check how the controllability depends on the lengths and wonder how this example may have anything to do with the Cosserat, Einstein or Maxwell equations !.

Example 1.4 (RLC electrical circuit) As we shall prove below, we do believe that the standard control theory of electrical circuits does not allow at all to study the structure of the various underlying differential modules defined by the corresponding systems (torsion submodules, extension modules, resolutions, ...), in particular if some of the RLC components depend on time.

Let us consider a RLC electrical circuit made up by a battery with voltage $u$ delivering a current $y$ to a parallel subsystem with a branch containing a capacity $C$ with voltage $x^1$ between its two plates and a resistance $R_1$ while the other branch, crossed by a current $x^2$, is containing a coil $L$ and a resistance $R_2$. The corresponding OD equations are easily seen to be:

$\boxed{R_{1}Cd{x}^1+x^1=u,Ld{x}^2+R_2{x}^2=u,x^1-R_1{x}^2=u-R_{1}y}$

Such a system can be set up at once in the standard matrix form $\stackrel{\dot{}}{x}=Ax+Bu$, $y=Cx+Eu$ with input $u$, state $\left(x^1,x^2\right)$ and output $y$, but we shall avoid the corresponding Kalman criterion that could not be used if $R_1,R_2,L$ or $C$ should depend on time. The two first order OD equations for $\left(u,x^1,x^2\right)$ are defining a differential module $N$ over the differential field $K=\mathbb{Q}\left(R_1,R_2,L,C\right)$ while the elimination of $\left(x^1,x^2\right)$ is providing the input submodule $Du=L\subset N$ and the output submodule $Dy=M\subset N$ with $\left(L,M\right)\subseteq N$. Nothing can be said as long as the prolongation/projection (PP) procedure has not been achieved like in the previous example but it has never been used in control theory, in particular for electrical circuit. The idea is to forget about the state by eliminating it but also to forget about the distinction between the input $u$ and the output $y$ because we already know that controllability is a structural property of the control system for $\left(u,y\right)$ that we have now to work out.

Differentiating the third zero order OD equation, we get

$d{x}^1-R_{1}d{x}^2=du-R_{1}dy$. Using the two first order OD equations, we obtain:

$-x^1+\frac{{\left(R_1\right)}^2{R}_{2}C}{L}{x}^2=-{\left(R_1\right)}^{2}Cdy-R_{1}Cdu+\left(\frac{{\left(R_1\right)}^{2}C}{L}-1\right)u$ and thus

$\left(L-R_1{R}_{2}C\right)x^2=R_{1}CLdy+Ly-CLdu-R_{1}Cu$ with a similar result for $x^1$. Hence, in the present situation, we have to distinguish carefully between two cases:

$•$ If $\boxed{R_1{R}_{2}C\ne L}$, we have $\left(x^1,x^2\right)\in j_1\left(u,y\right)$ and obtain a single second order CC for $\left(u,y\right)$:

$\boxed{R_{1}CL{d}^{2}y+\left(L+R_1{R}_{2}C\right)dy+R_{2}y-CL{d}^{2}u-\left(R_1+R_2\right)Cdu-u=0}$

The system is observable, that is we have indeed the strict equality $\left(L,M\right)=N$. Surprisingly, multiplying the first OD equation by ${\lambda }^1$ the second by ${\lambda }^2$, adding and integrating by parts, the kernel of the adjoint system is:

$\boxed{\begin{array}{l}{x}^1\to R_{1}Cd{\lambda }^1-{\lambda }_1=0,x^2\to Ld{\lambda }^2-R_2{\lambda }^2=0,u\to {\lambda }^1+{\lambda }^2=0\\ \Rightarrow Ld{\lambda }^1-R_2{\lambda }^1=0\end{array}}$

and thus ${\lambda }^1={\lambda }^2=0$ because the determinant of the system for the couple $\left(d{\lambda }^1,{\lambda }^1\right)$ is just $L-R_1{R}_{2}C\ne 0$, a result showing that $N$ is a torsion-free module, that is $t\left(N\right)=0$. The reader may compare this approach to the Kalman procedure. For example, if $R_1=C=L=1$, $R_2=2$, we get the system $d^{2}y+3dy+2y-d^{2}u-3du-u=0$ which is easily seen to be controllable. It is quite difficult to find such examples.

$•$ If $\boxed{R_1{R}_{2}C=L}$, we have only a single first order CC equation for $\left(u,y\right)$, namely:

$\boxed{Ldy+R_{2}y-R_{2}Cdu-u=0}$

Multiplying by a test function $\lambda$ and integrating by parts, we have to solve the two equations $-Ld\lambda +R_2\lambda =0$ and $R_{2}Cd\lambda -\lambda =0$. Hence the system is controllable if and only if the only possible solution is $\lambda =0$, that is when $L\ne {\left(R_2\right)}^{2}C$ or $\boxed{R_1\ne R_2}$ when $C\ne 0$ and we have the strict inclusion

$\left(L,M\right)\subset N$. Indeed, setting $z=y-\frac{1}{R_1}u$, we get $R_{1}Cdz+z=0$ and $z$ is a

torsion element. For example, if $R_1=R_2=L=C=1$, we get the system $dy+y-du-u=0$ which is not controllable because $z=y-u$ is a torsion element with $dz+z=0$.

Remark 1.5: In the general situation, one can use the differential submodules $t\left(N\right)$, $L$ and $M$ of $N$ both with the new differential modules $L^{\prime }=L+t\left(N\right)$ and $M\text{'}=M+t\left(N\right)$ in order to study all the problems concerning poles and zeros of control systems. As we are only interested by controllability, we have just to study the differential submodules of the torsion-free differential module $N/t\left(N\right)$. If we suppose that $L\cap M=0$, we have the following commutative diagram of inclusions, in which the upper commutative square is the so-called minimum controllable realization used as a logo for our ERCIM courses (1990 to 1996):

$\boxed{\begin{array}{ccccc}& & N& & \\ & \nearrow & & \nwarrow & \\ L^{\prime }& & & & M^{\prime }\\ \uparrow & \nwarrow & & \nearrow & \uparrow \\ \uparrow & & t\left(N\right)& & \uparrow \\ L& & \uparrow & & M\\ & \nwarrow & \uparrow & \nearrow & \\ & & 0& & \end{array}}$

Studying the differential correspondence between $\left(x^1,x^2\right)$ and $\left(u,y\right)$, we have to eliminate $\left(x^1,x^2\right)$ in order to find the resolvent system for $\left(u,y\right)$. These results could be extended to time dependent electrical components and open a large domain for future research on electrical circuits. We notice that at no moment we have used the fact that $u$ is called input and $y$ is called output !

This paper is also a kind of Summary Note sketching in a rather self-contained but condensed way the results presented through a series of lectures at the Albert Einstein Institute (AEI, Berlin/Potsdam), October, 23-27, 2017 [9] . The initial motivation for studying the methods used in this paper has been a 1000$ challenge proposed in 1970 by J. Wheeler in the physics department of Princeton University while the author of this paper was a visiting student of D.C. Spencer in the close-by mathematics department, namely:

Is it possible to express the generic solutions of Einstein equations in vacuum by means of the derivatives of a certain number of arbitrary functions, like the potentials for Maxwell equations ?.

After recalling the negative answer we already provided in 1995 [10] [11] , the main purpose of this paper is to use again these new techniques of differential double duality in order to revisit the mathematical foundations of the concepts and equations involved in general relativity and gauge theory that are leading to gravitational waves. At the same time, we point out the fact that the above parametrization problem is equivalent to the controllability property of a control system, such a result showing for the first time that it is a structural property, that is a property that does not depend on the choice of inputs and outputs or even on the presentation of the system, that is on a change of all the independent variables used to describe the system, contrary to the commonly accepted point of view of the control community. Many explicit examples are illustrating the paper, ranging from ordinary differential (OD) or partial differential (PD) control theory to mathematical physics, explaining in particular why the mathematical foundations of both gravitation and electromagnetism only depend on the structure of the conformal group of space-time. Accordingly, the foundations of control theory, engineering and mathematical physics must be revisited within this new framework, though striking it may sometimes look like. Of course, it is rather easy to study systems involving OD equations as we saw and we shall need new tools for studying systems of PD equations, though these new methods can also be used for OD equations. An additional difficulty will be met when dealing with operators having variable coefficients.

Example 1.6: While using Kalman test in control theory, it is often useful to transform a second order system $d^{2}y=0$ into a first order system $d{z}^1-z^2=0$, $d{z}^2=0$ by setting $y=z^1$, $dy=z^2$, transforming one OD equation for one unknown to two OD equations for two unknown. However, the mathematical community is not aware that, more generally, this has been exactly the procedure followed by Spencer from transforming ANY system of PD equations of order $q$ with $n$ independent variables and $m$ unknowns to a new system of PD equations of order one. One of the best elementary examples to be met in the literature has been provided by F. S. Macaulay in 1916 [12] [13] . With $n=2,m=1,q=2$ and using the jet notation $d_{23}y=y_{23}$, let us consider the second order homogeneous system $Py\equiv y_{22}=0$, $Qy\equiv y_{12}-y_{11}=0$. Differentiating once, we notice that all the derivatives of order 3 vanish. We obtain therefore four arbitrary parametric jets:

$\left\{z^1=y,z^2=y_1,z^3=y_2,z^4=y_{11}\right\}$

satisfying the non-homogeneous first order “equivalent system”, called “First Spencer operator”:

$\begin{array}{l}{d}_2{z}^1-z^3=0,d_2{z}^2-z^4=0,d_2{z}^3=0,d_2{z}^4=0,\\ d_1{z}^1-z^2=0,d_1{z}^2-z^4=0,d_1{z}^3-z^4=0,d_1{z}^4=0\end{array}$

In the present situation, we can integrate the system explicitly. We have indeed at once a basis of four solutions, namely $\left\{{\theta }_{\tau }\left(x\right)|1\le \tau \le 4\right\}=\left\{1,x^1,x^2,x^1{x}^2+\frac{1}{2}{\left(x^1\right)}^2\right\}$ and the space of solutions is a vector space $\mathcal{V}$ of dimension 4 over the field $\mathbb{Q}$ of constants of $K$.

Changing slightly the notations along with the other examples, we may consider the second order system $R_2\subset J_2\left(E\right)$ written $d_{22}\xi ={\eta }^1$, $d_{12}\xi -d_{11}\xi ={\eta }^2$ or $\mathcal{D}\xi =\eta$. Differentiating it once, we obtain the trivially involutive third order system $R_3\subset J_3\left(E\right)$ with corresponding Janet tabular:

$\{\begin{array}{lll}{d}_{222}\xi \hfill & =\hfill & d_2{\eta }^1\hfill \\ d_{122}\xi \hfill & =\hfill & d_1{\eta }^1\hfill \\ d_{112}\xi \hfill & =\hfill & d_1{\eta }^1-d_2{\eta }^2\hfill \\ d_{111}\xi \hfill & =\hfill & d_1{\eta }^1-d_2{\eta }^2-d_1{\eta }^2\hfill \\ d_{22}\xi \hfill & =\hfill & {\eta }^1\hfill \\ d_{12}\xi -d_{11}\xi \hfill & =\hfill & {\eta }^2\hfill \end{array}\boxed{\begin{array}{ll}1\hfill & 2\hfill \\ 1\hfill & •\hfill \\ 1\hfill & •\hfill \\ 1\hfill & •\hfill \\ •\hfill & •\hfill \\ •\hfill & •\hfill \end{array}}$

with the CC $d_{22}{\eta }^2-d_{12}{\eta }^1+d_{11}{\eta }^1=\left(PQ-QP\right)\xi =0$ or ${\mathcal{D}}_1\eta =0$ in the exact differential sequence:

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

which is nevertheless far from being a Janet sequence because $\mathcal{D}$ is not involutive. In the present situation, we let the reader check that $ad\left(\mathcal{D}\right)$ indeed generates the CC of $ad\left({\mathcal{D}}_1\right)$.

Example 1.7: With two independent variables $\left(x^1,x^2\right)$ and one unknown $y$, let us consider the following second order system with constant coefficients:

$\{\begin{array}{lll}Py\equiv \hfill & d_{22}y\hfill & =u\hfill \\ Qy\equiv \hfill & d_{12}y-y\hfill & =v\hfill \end{array}$

where now $P$ and $Q$ are PD operators with coefficients in the subfield $k=\mathbb{Q}$ of constants of the differential field $K=k\left(x^1,x^2\right)$. We obtain at once from a first use of crossed derivatives:

$d_{2}y=d_{1}u-d_{2}v$

and from a second use:

$y=d_{11}u-d_{12}v-v$

and could hope to obtain the $4^{th}$-order generating compatibility conditions (CC) by substitution, that is to say:

$\{\begin{array}{lll}A\equiv \hfill & d_{1122}u-d_{1222}v-d_{22}v-u\hfill & =0\hfill \\ B\equiv \hfill & d_{1112}u-d_{11}u-d_{1122}v\hfill & =0\hfill \end{array}$

with the only generating CC : $w\equiv d_{11}A-d_{12}B-B=0$.

However, in this particular case, there is an unexpected unique second order generating CC:

$C\equiv d_{12}u-u-d_{22}v=0$

as we now have indeed $PQ-QP=0$ both with $A\equiv d_{12}C+C$ and $B\equiv d_{11}C$, a result leading to $C\equiv d_{22}B-d_{12}A+A$. Accordingly, the systems $A=0,B=0$ on one side and $C=0$ on the other side are completely different though they have the same solutions in $u,v$ which can be parametrized injectively by $y$.

Finally, setting $u=0,v=0$, we notice that the preceding homogeneous system can be written in the form $\mathcal{D}y=0$ and admits the only solution $y=0$. More precisely, if a linear system $R_q\subset J_q\left(E\right)$ of order $q$ on $E$ is given we may find two integers $\left(r,s\right)$ such that, prolonging $r+s$ times to obtain $R_{q+r+s}$ and keeping only the equations of order $q+r$, we obtain a system $R_{q+r}^{\left(s\right)}$ providing all the informations on the solutions up to any order (prolongation/projection (PP) procedure) [14] [15] . In the present case, we get successively:

$0=R_2^{\left(4\right)}\subset R_2^{\left(3\right)}\subset R_2^{\left(2\right)}\subset R_2^{\left(1\right)}\subset R_2\subset J_2\left(E\right)$

with strict inclusions and respective dimensions: $0<1<2<3<4<6$.

Example 1.8: Denoting by $y_i^k=d_i{y}^k$ for $i=1,2$ and $k=1,2,3$ the formal derivatives of the three differential indeterminates $y^1,y^2,y^3$, we consider the system of three PD equations for 3 unknowns and 2 independent variables $\left(x^1,x^2\right)$ which is defining a differential module $M$ over the non-commutative ring $D=\mathbb{Q}\left(a\right)\left(x^1,x^2\right)\left[d_1,d_2\right]$ of differential operators with coefficients in $\mathbb{Q}\left(a\right)\left(x^1,x^2\right)$ when $a$ is a constant parameter:

$\{\begin{array}{lll}{\Phi }^1\equiv \hfill & y_2^3-y_2^2\hfill & =0\hfill \\ {\Phi }^2\equiv \hfill & y_2^2+y_1^1-a{x}^2{y}^1\hfill & =0\hfill \\ {\Phi }^3\equiv \hfill & y_1^3-y_1^2\hfill & =0\hfill \end{array}$

No one among $\left(y^1,y^2\right)$ can be given arbitrarily and that there is a unique generating CC, namely:

$\Psi \equiv d_2{\Phi }^3-d_1{\Phi }^1=0$

Also, setting $z=y^3-y^2$, we get both $z_1=0,z_2=0$ and $z$ is an autonomous element. Then one can easily prove that any other autonomous element can be expressible by means of a differential operator acting on $z$ which is therefore a generator of the torsion module $t\left(M\right)\subset M$. Accordingly, in the present situation, any autonomous element is a constant multiple of $z$.

Finally, setting $z=0$ and thus $y^3=y^2$, we obtain for $\left(y^1,y^2\right)$, after substitution:

${\Phi }^{\prime }\equiv y_2^2+y_1^1-a{x}^2{y}^1=0$

which is defining an operator ${\mathcal{D}}^{\prime }:\left(y^1,y^2\right)\to {\Phi }^{\prime }$ and a torsion-free module $M^{\prime }\simeq M/t\left(M\right)$ in the short exact sequences:

$0\to D\to D^2\to M^{\prime }\to 0,0\to t\left(M\right)\to M\to M^{\prime }\to 0$

Multiplying the previous operator by a test function $\lambda$ and integrating by parts, the kernel of the adjoint operator $ad\left({\mathcal{D}}^{\prime }\right):\lambda \to \left({\mu }^1,{\mu }^2\right)$ is defined by:

$y^2\to -d_2\lambda ={\mu }^2,y^1\to -d_1\lambda -a{x}^2\lambda ={\mu }^1\Rightarrow d_1{\mu }^2-d_2{\mu }^1-a{x}^2{\mu }^2=a\lambda$

We have thus two quite different situations:

$•$ $a=0$: The adjoint operator is not injective and we are in the situation of the $div$ operator when $n=2$ that can be parametrized by the $curl$ operator in such a way that $M^{\prime }$ is neither free nor projective but $M^{\prime }\subset D$ with a strict inclusion.

$•$ $a\ne 0$, say $a=1$: The adjoint operator is injective and, using the fact that any operator can be written as the adjoint of an operator, we have obtained a lifting operator such that . We shall prove later on that $M^{\prime }$ is not free but projective, thus torsion-free, because this lift provides an isomorphism $D^2\simeq M^{\prime }\oplus D$ and an isomorphism $M\simeq t\left(M\right)\oplus M^{\prime }$ which may not exist in general.

Example 1.9: (Elasticity) A first striking result that does not seem to have been even noticed by mechanicians up till now, let us consider the situation of classical elasticity theory where $\mathcal{D}$ is the Killing operator for the euclidean metric, namely $\Omega \equiv \mathcal{D}\xi =ℒ\left(\xi \right)\omega \in {\mathcal{S}}_2{\mathcal{T}}^{*}$ and ${\mathcal{D}}_1$ the corresponding CC, namely the linearized Riemann curvature with $n^2\left(n^2-1\right)/12$ components that can be found in any textbooks of elasticity theory or general relativity. In that case, as it is well known that the Poincaré sequence for the exterior derivative is self-adjoint up to sign (for $n=3$ the adjoints of $grad,curl,div$ are respectively $-div,curl,-grad$) then the first extension module does not depend on the differential sequence used and therefore vanishes. Accordingly, $ad\left(\mathcal{D}\right)$ generates the CC of $ad\left({\mathcal{D}}_1\right)$. Hence, in order to parametrize the Cauchy stress equations, that is $ad\left(\mathcal{D}\right)$, namely:

$\boxed{{\sigma }^{12}={\sigma }^{21},d_1{\sigma }^{11}+d_2{\sigma }^{21}=0,d_1{\sigma }^{12}+d_2{\sigma }^{22}=0}$

one just needs to compute $ad\left({\mathcal{D}}_1\right)$. For $n=2$, we get:

${\Omega }_{11}=2d_1{\xi }_1,{\Omega }_{12}=d_1{\xi }_2+d_2{\xi }_1,{\Omega }_{22}=2d_2{\xi }_2\Rightarrow d_{11}{\Omega }_{22}+d_{22}{\Omega }_{11}-2d_{12}{\Omega }_{12}=0$

$\varphi \left(d_{11}{\Omega }_{22}+d_{22}{\Omega }_{11}-2d_{12}{\Omega }_{12}\right)=\left(d_{22}\varphi \right){\Omega }_{11}-2\left(d_{12}\varphi \right){\Omega }_{12}+\left(d_{11}\varphi \right){\Omega }_{22}+d_i{(\cdots )}^i$

and recover the parametrization by means of the Airy function in a rather unexpected way:

$\boxed{{\sigma }^{11}=d_{22}\varphi ,{\sigma }^{12}={\sigma }^{21}=-d_{12}\varphi ,{\sigma }^{22}=d_{11}\varphi }$

Exhibiting a parametrization for $n\ge 3$ thus becomes an exercise of computer algebra, the number of (pseudo)-potentials being the number $n^2\left(n^2-1\right)/12$ of components of the Riemann tensor.

We now treat the case of Cosserat equations with zero second members, namely [16] [17] :

$\boxed{d_1{\sigma }^{1,1}+d_2{\sigma }^{2,1}=0,d_1{\sigma }^{1,2}+d_2{\sigma }^{2,2}=0,d_i{\mu }^{i,12}+{\sigma }^{1,2}-{\sigma }^{2,1}=0}$

For this, instead of using the Janet sequence as before, we now use the Spencer sequence which is isomorphic to the gauge sequence, namely the tensor product of the Poincaré sequence by a lie algebra. However, according to the general theorems of homological algebra, the existence of a parametrization does not depend on the differential sequence used and therefore follows again, like in the previous example, from the fact that the Poincaré sequence is self-adjoint up to the sign. In the present situation, we have with $dim\left(\mathcal{G}\right)=n\left(n+1\right)/2$. We have shown that the Cosserat equations were just $ad\left(D_1\right)$, their first order parametrization is thus described by $ad\left(D_2\right)$ and needs $dim\left(C_2\right)=n^2\left(n^2-1\right)/4$ (pseudo)-potentials. We provide the details when $n=2$ but we know at once that we must use 3 (pseudo)-potentials only. The case $n=3$ could be treated similarly and is left as an exercise.

In fact, for constructing the adjoint of $D_1$, we have just to integrate by parts the duality summation ${\sigma }_j^i\left({\partial }_i{\xi }^j-{\xi }_i^j\right)+{\mu }_k^{ij}\left({\partial }_i{\xi }_j^k-{\xi }_{ij}^k\right)$ while taking into account that ${\xi }_{ij}^k=0$. Lowering the indices by means of the constant Euclidean metric, the Spencer operator $D_1$ is described by the equations:

${\partial }_1{\xi }_1=A_{11},{\partial }_1{\xi }_2-{\xi }_{1,2}=A_{12},{\partial }_2{\xi }_1-{\xi }_{2,1}=A_{21},{\partial }_2{\xi }_2=A_{22},{\partial }_1{\xi }_{1,2}=B_1,{\partial }_2{\xi }_{1,2}=B_2$

because $R_1$ is defined by the equations ${\xi }_{1,1}=0,{\xi }_{1,2}+{\xi }_{2,1}=0,{\xi }_{2,2}=0$.

Accordingly the 3 CC describing the Spencer operator $D_2$ are:

${\partial }_2{A}_{11}-{\partial }_1{A}_{21}+B_1=0,{\partial }_2{A}_{12}-{\partial }_1{A}_{22}+B_2=0,{\partial }_2{B}_1-{\partial }_1{B}_2=0$

Multiplying these equations respectively by ${\varphi }^1,{\varphi }^2,{\varphi }^3$, summing and integrating by part, we get $ad\left(D_2\right)$ and the first order parametrization in the form [16] [17] :

$\boxed{{\sigma }^{1,1}=-{\partial }_2{\varphi }^1,{\sigma }^{1,2}=-{\partial }_2{\varphi }^2,{\sigma }^{2,1}={\partial }_1{\varphi }^1,{\sigma }^{2,2}={\partial }_1{\varphi }^2,{\mu }^{1,12}=-{\partial }_2{\varphi }^3+{\varphi }^1,{\mu }^{2,12}={\partial }_1{\varphi }^3+{\varphi }^2}$

as announced previously. As we are dealing with PD equations with constant coefficients, it is important to notice that such a parametrization could also have been obtained by localization later on. When the stress is symmetric, that is ${\sigma }^{1,2}={\sigma }^{2,1}$, the Airy parametrization can be recovered if we cancel the couple-stress with ${\varphi }_1={\partial }_2{\varphi }_3,{\varphi }_2=-{\partial }_1{\varphi }_3$ and set ${\varphi }_3=-\varphi$.

Changing the presentation will be studied later on as we shall need a lot of additional tools.

Example 1.10: (Electromagnetism) A similar comment can be done for electromagnetism through the exterior derivative as the first set of Maxwell equations can be parametrized by the EM potential 1-form while the second set of Maxwell equations, adjoint of this parametrization, can be parametrized by the EM pseudo-potential. With more details, the beginning of the classical Poincaré sequence for the exterior derivative is:

$\boxed{{\wedge }^0{T}^{\text{*}}\stackrel{d}{\to }{\wedge }^1{T}^{\text{*}}\stackrel{d}{\to }{\wedge }^2{T}^{\text{*}}\stackrel{d}{\to }{\wedge }^3{T}^{\text{*}}}$

Using standard notations, we denote by $A\in {\wedge }^1{T}^{\text{*}}=T^{\text{*}}$ the EM potential, by $F=\left(F_{ij}\right)\in {\wedge }^2{T}^{\text{*}}$ the EM field and the first set of Maxwel equations, namely $dF=0$, is parametrized by $dA=F$. Denoting by $ℱ=\left({ℱ}^{ij}\right)\in {\wedge }^4{T}^{\text{*}}\otimes {\wedge }^{2}T\simeq \wedge$ the EM induction, a tensorial density, the second set of Maxwell equations is usually written as ${\partial }_i{ℱ}^{ij}={\mathcal{J}}^j$ and thus $ad\left(d\right)ℱ=\mathcal{J}\in {\wedge }^4{T}^{\text{*}}\otimes T$, with the only CC ${\partial }_j{\mathcal{J}}^j=0$ describing the so-called conservation of current. The problem that we faced while teaching EM during twenty years, is that only tensors are used in most textbooks and the above formulas, if they are used by physicists, are not correct at all from a mathematical point of view. When $E$ is any vector bundle over a manifold $X$ of dimension $n$, the idea is to introduce the adjoint vector bundle $ad\left(E\right)={\wedge }^n{T}^{\text{*}}\otimes E^{\text{*}}$ with $E^{\text{*}}$ defined by patching the inverse transition matrices, exactly like $T^{\text{*}}$ is obtained from $T$. Such a formal approach, totally lacking in the literature, allows to describe both the second set of Maxwell equations and the conservation of current in the following dual sequence existing when $n=4$:

$\boxed{0\leftarrow ad\left({\wedge }^0{T}^{\text{*}}\right)\stackrel{ad\left(d\right)}{\leftarrow }ad\left({\wedge }^1{T}^{\text{*}}\right)\stackrel{ad\left(d\right)}{\leftarrow }ad\left({\wedge }^2{T}^{\text{*}}\right)\stackrel{ad\left(d\right)}{\leftarrow }ad\left({\wedge }^3{T}^{\text{*}}\right)}$

in which $ad\left(d\right)$ is going “backwards”, that is from right to left. For the reader knowing more mathematics, such a procedure may be simplified by using Hodge duality with the volume form $dx=d{x}^1\wedge \cdots \wedge d{x}^n$ as a natural way to obtain the dual sequence when $n=4$ in the form:

$\boxed{0\leftarrow {\wedge }^4{T}^{\text{*}}\stackrel{d}{\leftarrow }{\wedge }^3{T}^{\text{*}}\stackrel{d}{\leftarrow }{\wedge }^2{T}^{\text{*}}\stackrel{d}{\leftarrow }{\wedge }^1{T}^{\text{*}}}$

Such a confusing procedure has in fact to do with the so-called side changing functor in differential homological algebra but is far out of the purpose of this paper. Of course, in the actual practice of computer algebra and electromagnetism, the two dual sequences can be written, up to sign, as:

$\boxed{\begin{array}{lllllllll}\hfill & \hfill & 1\hfill & \stackrel{d}{\to }\hfill & 4\hfill & \stackrel{d}{\to }\hfill & 6\hfill & \stackrel{MI}{\to }\hfill & 4\hfill \\ 0\hfill & \leftarrow \hfill & 1\hfill & \stackrel{d}{\leftarrow }\hfill & 4\hfill & \stackrel{MII}{\leftarrow }\hfill & 6\hfill & \stackrel{d}{\leftarrow }\hfill & 4\hfill \end{array}}\begin{array}{c}MaxwellI\\ \stackrel{}{}MaxwellII\end{array}$

Let us finally simply say that it is a way to transform a left differential module into a right differential module and vice-versa, one of the most difficult concepts that must be used when studying differential extension modules and the reason for which an adjoint operator must always be written “backward” as we saw (See [18] [19] for more details and examples).

We end this Introduction with one of the best academic examples we know in order to understand that working out differential sequences is not an easy task, even on elementary examples.

Example 1.11: (Macaulay) Let us revisit Example 1.6 using more advanced methods. With $m=1,n=2,q=2,D=\mathbb{Q}\left[d_1,d_2\right]$, let us consider the linear second order system $R_2\subset J_2\left(E\right)$ with $dim\left(E\right)=1$ defined by the two PD equations $P\xi \equiv d_{22}\xi =0$, $Q\xi \equiv d_{12}\xi -d_{11}\xi =0$. We let the reader check easily that $g_3=0$ with $d_{ijk}\xi =0$ and thus $dim\left(R_2\right)=4$ with parametric jets $\left(z^1=\xi ,z^2={\xi }_1,z^3={\xi }_2,z^4={\xi }_{11}\right)$, a result leading to $dim\left(R_{2+r}\right)=4$ if we differentiate $r$ times. We recall the dimensions:

$\begin{array}{cccccccc}q& \to & 0& 1& 2& 3& 4& 5\\ S_q{T}^{\text{*}}& \to & 1& 2& 3& 4& 5& 6\\ J_q\left(E\right)& \to & 1& 3& 6& 10& 15& 21\end{array}$

both with the commutative and exact diagram allowing to construct inductively the Spencer bundles $C_r\subset C_r\left(E\right)$ and the Janet bundles $F_r$ for $r=0,1,\cdots ,n$ with $F_0=J_q\left(E\right)/R_q$ and $C_0=R_q\subset J_q\left(E\right)=C_1\left(E\right)$ while replacing the system $R_q\subset J_q\left(E\right)$ of order $q$ on $E$ by the system $R_{q+1}\subset J_1\left(R_q\right)$ of order 1 on $R_q$ when $q$ is large enough, that is $q=3$ in the present example because $g_3=0$. The following diagram allows to start the induction:

$\boxed{\begin{array}{ccccccccccc}& & & & & & 0& & 0& & \\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & R_{q+1}& \to & J_1\left(R_q\right)& \to & C_1& \to & 0\\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & J_{q+1}\left(E\right)& \to & J_1\left(J_q\left(E\right)\right)& \to & C_1\left(E\right)& \to & 0\\ & & & & \parallel & & \downarrow & & \downarrow & & \\ 0& \to & R_{q+1}& \to & J_{q+1}\left(E\right)& \to & J_1\left(F_0\right)& \to & F_1& \to & 0\\ & & & & & & \downarrow & & \downarrow & & \\ & & & & & & 0& & 0& & \end{array}}$

In the present situations with $q=3$ we have the dimensions:

$\boxed{\begin{array}{ccccccccccc}& & & & & & 0& & 0& & \\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & 4& \to & 12& \to & 8& \to & 0\\ & & & & & & \downarrow & & \downarrow & & \\ & & 0& \to & 15& \to & 30& \to & 15& \to & 0\\ & & & & \parallel & & \downarrow & & \downarrow & & \\ 0& \to & 4& \to & 15& \to & 18& \to & 7& \to & 0\\ & & & & & & \downarrow & & \downarrow & & \\ & & & & & & 0& & 0& & \end{array}}$