Differential Galois Theory and Hopf Algebras for Lie Pseudogroups

Abstract

According to a clever but rarely quoted or acknowledged work of E. Vessiot that won the prize of the Académie des Sciences in 1904, “Differential Galois Theory” (DGT) has mainly to do with the study of “Principal Homogeneous Spaces” (PHSs) for finite groups (classical Galois theory), algebraic groups (Picard–Vessiot theory) and algebraic pseudogroups (Drach–Vessiot theory). The corresponding automorphic differential extensions are such that $di{m}_K\left(L\right)=\mid L/K\mid <\infty$, the transcendence degree $trd(L/K)<\infty$ and $trd(L/K)=\infty$ with $difftrd(L/K)<\infty$, respectively. The purpose of this paper is to mix differential algebra, differential geometry and algebraic geometry to revisit DGT, pointing out the deep confusion between prime differential ideals (defined by J.-F. Ritt in 1930) and maximal ideals that has been spoiling the works of Vessiot, Drach, Kolchin and all followers. In particular, we utilize Hopf algebras to investigate the structure of the algebraic Lie pseudogroups involved, specifically those defined by systems of algebraic OD or PD equations. Many explicit examples are presented for the first time to illustrate these results, particularly through the study of the Hamilton–Jacobi equation in analytical mechanics. This paper also pays tribute to Prof. A. Bialynicki-Birula (BB) on the occasion of his recent death in April 2021 at the age of 90 years old. His main idea has been to notice that an algebraic group G acting on itself is the simplest example of a PHS. If G is connected and defined over a field K, we may introduce the algebraic extension $L=K\left(G\right)$; then, there is a Galois correspondence between the intermediate fields $K\subset K^{\prime }\subset L$ and the subgroups $e\subset G^{\prime }\subset G$, provided that $K^{\prime }$ is stable under a Lie algebra $\Delta$ of invariant derivations of $L/K$. Our purpose is to extend this result from algebraic groups to algebraic pseudogroups without using group parameters in any way. To the best of the author’s knowledge, algebraic Lie pseudogroups have never been introduced by people dealing with DGT in the spirit of Kolchin; that is, they have only been considered with systems of ordinary differential (OD) equations, but never with systems of partial differential (PD) equations.

1. Introduction

Before describing the content of this paper, we start by providing a few personal historical details explaining the motivation for writing it. We separate this historical background into three parts:

(A) The story started in 1970, when the author of this paper was a visiting student of D.C. Spencer at Princeton University. As the mathematical library was opened day and night while being well furnished in French publications, he had the chance to discover the original publication of M. Janet in 1920 [1] on systems of ordinary differential (OD) or partial differential (PD) equations and of E. Vessiot in 1903 [2] on the so-called (at that time) finite and infinite groups of transformations, now respectively called Lie groups and Lie pseudogroups of transformations. This last reference had also been used one year later by Vessiot to study “Differential Galois Theory” (DGT) in a short but clever paper published in 1904 [3] introducing, for the first time, “Principal Homogeneous Spaces” (PHSs) for algebraic pseudogroups defined by automorphic systems of OD or PD equations and the corresponding “automorphic differential extensions”. The consideration of these results furnished the main content of the first Gordon and Breach (GB) book of 1978 [4]. However, he also discovered a rarely quoted long paper published as a doctoral thesis by J. Drach in 1898 [5], the interesting fact being that the jury was made by G. Darboux, E. Picard and H. Poincaré, the three most famous French mathematicians existing at that time, a good reason for reading it in detail.

Roughly, one may say that Vessiot divided his study of DGT into three parts of increasing difficulty, each one being illustrated by original examples. Using standard notations of differential algebra, if K is a differential field of characteristic zero, that is, $\mathbb{Q}\subset K$ with n derivations ${\partial }_i$ for $i=1,\dots ,n$ and L with derivations $d_i$ for $i=1,\dots ,n$, it is in such a way that $K\subset L$ and $d_i{\mid }_K={\partial }_i$.

(a) 

Classical Galois Theory: ${dim}_{K}L=\mid L/K\mid <\infty$.

In this case, L being a finite-dimensional vector space over K means that any element of L is algebraic over K, and we refer the reader to the work of E. Galois, who died in a duel in 1821, and to the extensive literature on this subject. The purpose, roughly speaking, is to study systems of algebraic equations that are invariant under finite groups, such as groups of permutations.

(b) 

Picard-Vessiot Theory: $trd(L/K)<\infty ,difftrd(L/K)=0$.

In this case, the maximum number of elements of L that are not algebraic over K or the transcendence degree of L over K is finite, but any element of L must satisfy at least one OD or PD algebraic equation over K. The problem, roughly, is to study systems of algebraic OD or PD equations that are invariant under an algebraic group of matrices. The application to the study of shells, chains and analytical mechanics is quite recent [6].

(c) 

Drach-Vessiot Theory: $trd(L/K)=\infty ,difftrd(L/K)<\infty$.

In this case, the maximum number of elements of L that are not algebraic over K may be infinite. Still, one can find a well-defined intrinsic maximum number of elements of L that can be given arbitrarily, such a set being called a differential transcendence basis. The problem is, thus, to study systems of algebraic OD or PD equations that are invariant under an algebraic pseudogroup, namely, a group of transformation solutions of a system of algebraic OD or PD equations that can be of high order. The best example is provided by the study of various reductions of the group of invariance of the Hamilton–Jacobi equation, depending on the Hamiltonian function of the considered mechanical system. It can indeed be described by sub-pseudogroups of the pseudogroup of contact transformations, the most important one being the one made up of unimodular transformations that preserve the volume form [6,7].

The clever idea of Vessiot has been to prove that, in all these cases, the corresponding Galois theory is just a theory of PHS, respectively, for finite groups, Lie groups, or Lie pseudogroups. Of course, when a Lie group G is acting on a manifold X with an action law $X\times G\to X:(x,a)\to y=ax=f(x,a)$, one can introduce the corresponding graph $X\times G\to X\times X:(x,a)\to (x,y)$ for the action, and X is a PHS for the action of G when such a graph is an isomorphism. The main idea of Vessiot has been to extend such a definition to systems of OD or PD equations and Lie pseudogroups $\Gamma \subset aut\left(X\right)$ made by invertible transformations in such a way that, whenever $y=f\left(x\right)$ and $\overline{y}=\overline{f}\left(x\right)$ are two solutions of this system, then one can find one and only one invertible transformation $\overline{y}=g\left(y\right)$ such that $\overline{f}=g\circ f$, at least, locally, and such a system is called an automorphic system by Vessiot. Of course, the difficulty, not solved by Drach or Vessiot, was to establish an effective test for checking such a property. In view of the many specific examples presented, the first and second criteria for automorphic systems in Section 3 will be absolutely unavoidable to decide whether a given differential extension will be automorphic or not (see [6,8] for details).

(B) Having at that time no idea of how to use the “Spencer machine” in mathematical physics (see the Conclusion!), the author tried to translate the work of Drach and Vessiot into a modern differential geometric language, and it is at this precise moment that the true difficulties started. Indeed, though the work of Drach was introducing, for the first time, crucial concepts such as differential fields and other founding stones of differential algebra (only sketched by J.F. Ritt more than thirty years later [9]), the author did not succeed, after months of hard work, to understand the thesis of Drach or even to justify the so-called “ reducibility” concept that is the heart of the Drach–Vessiot theory. Even worse, it became clear that the thesis was based on a fundamental confusion between a maximal ideal and a prime ideal, the latter concept only being introduced by Ritt in 1930 [9]. To illustrate this difficulty, let us consider the following elementary example, which will be revisited later.

Example 1.

With $n=1$ independent variable x and $m=1$ dependent variable y, let us consider the second-order linear system $y_{xx}=0$ while using standard jet notations [4,10,11,12]. With $K=\mathbb{Q}$ a trivial differential field, we may introduce, at once, the linear differential ideal $\mathfrak{p}\subset K\left\{y\right\}=K[y,y_x,y_{xx},\dots ]$, which is, of course, trivially a prime differential ideal generated by the differential polynomial $P=y_{xx}$, providing, successively, $y_{xx}=0,y_{xxx}=0,\dots$. As a byproduct, taking the full ring of quotients, we may define the differential extension $L=Q(K\left\{y\right\}/\mathfrak{p})=K(y,y_x)$ with $d_{x}y=y_x,d_x{y}_x=0$. It is seen at once to be a differential automorphic extension for the affine group of the real line, namely $\overline{y}=a^{1}y+a^2,{\overline{y}}_x=a^1{y}_x$ with two infinitesimal generators $\Theta =\{{\theta }_1={\displaystyle \frac{\partial }{\partial y}},{\theta }_2=y{\displaystyle \frac{\partial }{\partial y}}+y_x{\displaystyle \frac{\partial }{\partial y_x}}$ and an ordinary bracket $[{\theta }_1,{\theta }_2]={\theta }_1$. Indeed, we have $a^1={\displaystyle \frac{{\overline{y}}_x}{y_x}},a^2=\overline{y}-{\displaystyle \frac{{\overline{y}}_x}{y_x}}y$ in such a way that the graph of the action $((y,y_x),(a^1,a^2))\to ((y,y_x),(\overline{y},{\overline{y}}_x))$ is an isomorphism. Now, by setting $\Delta =\{{\delta }_1=y_x{\displaystyle \frac{\partial }{\partial y}},{\delta }_2=y_x{\displaystyle \frac{\partial }{\partial y_x}}\}$, we obtain $[\Theta ,\Delta ]=0$, that is, each $\theta \in \Theta$ commutes with each $\delta \in \Delta$. Accordingly, we may find intermediate fields $K\subset K^{\prime }\subset L$ that are not differential fields because they are not stable under ${\delta }_1$ and are not fields of invariants of subgroups, whereas others are. For example, $K^{\prime }=\mathbb{Q}\left(y{y}_x\right)$ is an intermediate field but not a differential field because $d_x\left(y{y}_x\right)=y_x^2\notin K^{\prime }$. The subgroup of invariance is $\overline{y}=\pm y$, which lets invariant $K^{″}=\mathbb{Q}(y{y}_x,y_x^2)=\mathbb{Q}(y{y}_x,y^2)=\mathbb{Q}({\displaystyle \frac{y_x}{y}},y^2)$ with the strict inclusion $K^{\prime }\subset K^{″}$ as pure fields. In fact, ${\delta }_1\left(y{y}_x\right)=y_x^2$, while ${\delta }_2\left(y{y}_x\right)=y{y}_x$ and $y_x^2/\left(y{y}_x\right)=y_x/y$, that is, $\overline{y}=a^{1}y$, but we have also $y_x^2/{\left({\displaystyle \frac{y_x}{y}}\right)}^2=y^2$, that is, $a^1=\pm 1$. In contrast, $K^{\prime }=\mathbb{Q}\left({\displaystyle \frac{y_x}{y}}\right)$ is indeed a differential field because $d_x\left({\displaystyle \frac{y_x}{y}}\right)=-{\left({\displaystyle \frac{y_x}{y}}\right)}^2\in K^{\prime }$, which is also stable according to $\Delta$ because ${\delta }_1\left({\displaystyle \frac{y_x}{y}}\right)=-{\left({\displaystyle \frac{y_x}{y}}\right)}^2,{\delta }_2\left({\displaystyle \frac{y_x}{y}}\right)={\displaystyle \frac{y_x}{y}}$, and the biggest group of invariance is $\overline{y}=ay$, which allows invariance exactly $K^{\prime }$ in a coherent way with [13], Theorem 3. We let the reader treat the more general situation with the chain of strict inclusions of differential fields with $\mathbb{Q}<y>=\mathbb{Q}(y,y_x,y_{xx},\dots )$:

$$\mathbb{Q}\subset \mathbb{Q}<{\displaystyle \frac{y_{xxx}}{y_x}}-{\displaystyle \frac{3}{2}}{\left({\displaystyle \frac{y_{xx}}{y_x}}\right)}^2>\subset \mathbb{Q}<{\displaystyle \frac{y_{xx}}{y_x}}>\subset \mathbb{Q}<{\displaystyle \frac{y_x}{y}}>\mathbb{Q}<y>$$

Example 2.

When K is a differential field containing $\mathbb{Q}$ and $n=1$ and $m=2$, setting $K<y^1,y^2>=K(y^1,y^2,y_x^1,y_x^2,\dots )$ and $K<\Phi >=K(\Phi ,d_x\Phi ,d_{xx}\Phi ,\dots )$, for any $\Phi \in K<y>$, we may consider the chain of differential extensions $\mathbb{Q}\subset \mathbb{Q}<y^2{y}_x^1>\subset \mathbb{Q}<y^1,y^2>$. Using the chain rule for derivatives on the jet level, we easily obtain that the biggest Lie pseudogroup preserving $\Phi =y^2{y}_x^1$ is $\Gamma =\{{\overline{y}}^1=g\left(y^1\right),{\overline{y}}^2=y^2/{\displaystyle \frac{\partial g}{\partial y^1}}\}$, which is defined by the Pfaffian system ${\overline{y}}^2{\overline{y}}_x^1=y^2{y}_x^1\Rightarrow {\overline{y}}^1\wedge d{\overline{y}}^2=d{y}^1\wedge d{y}^2$; this is not formally integrable, namely,

$${\overline{y}}^2{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^1}}=y^2,{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^2}}=0,\Rightarrow {\displaystyle \frac{\partial ({\overline{y}}^1,{\overline{y}}^2)}{\partial (y^1,y^2)}}=1\Rightarrow {\displaystyle \frac{1}{{\overline{y}}^2}}{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^2}}={\displaystyle \frac{1}{y^2}}$$

and we obtain, therefore, a well-defined algebraic pseudogroup. Considering the chain of differential fields $K=\mathbb{Q}<y^2{y}_x^1>\subset K^{\prime }=\mathbb{Q}<y^2{y}_x^1,y_x^2>\subset \mathbb{Q}<y^1,y^2>=L$, we notice that the biggest pseudogroup preserving $K^{\prime }$ is the sub-pseudogroup ${\Gamma }^{\prime }=\{{\overline{y}}^1=y^1+a,{\overline{y}}^2=y^2\}\subset \Gamma$, which is preserving the differential field $K^{″}=\mathbb{Q}<y^2{y}_x^1,y^2>=\mathbb{Q}<y_x^1,y^2>$ with the strict inclusion $K^{\prime }\subset K^{″}$. Once again, it does not seem possible to establish a Galois correspondence between intermediate differential fields and sub-pseudogroups of Γ.

(C) It is at this precise moment (1979) that the author of this paper discovered that M. Janet was still alive, living a few blocks away in Paris 16° and in rather good condition, as he died quite a bit later on in 1983 at the age of 96. He was a close friend of Vessiot and became quite pleased by the fact that the 1978 GB book had been dedicated to him for his 88th Birthday. As a byproduct, he gave him a bunch of private documents, now stored as a deposit in the main Library of Ecole Normale Supérieure (ENS Paris), where they can be consulted [7,14]. We quote below a few lines from a letter from P. Painlevé to Vessiot about the thesis of J. Drach (Paris, 17 October 1898):

“Dear “Friend, I just read Drach’s thesis and I agree entirely with you about the inaccuracy of the two fundamental theorem and their proofs. The mistake is so big that I can hardly conceive that it has been overlooked by the author and the jury. …I said to Picard that I had received a letter from you about this question; after a few minutes of explanation, he was astonished to have missed this. I do believe, indeed, there cannot be the slightest shadow of a doubt for anybody making his mind on this thing …" (see [14], Appendix for the original letters given by Janet).

Then a “Mathematical Affair” started in a rather unpleasant way because Drach, supported by E. Borel, never accepted having made a mistake. Janet also confirmed that the only paper written by E. Cartan on PD equations, taken from his letters to A. Einstein on Absolute Parallelism in 1930 [15], never indicated that it was directly coming from the classical similar paper of Janet published in 1920 [1], without quoting his source. Similarly, Cartan never said that the Vessiot structure equations could compete with or even supersede the Cartan structure equations, although both had been created about the same time, around 1905. Last but not least, Janet confirmed that everybody in 1930 knew that the future of DGT should pass through algebraic pseudogroups and not through the so-called “differential groups” of J. F. Ritt [9,16,17,18] that were considered as leading to a dead end because only trivial examples had been exhibited.

As a result, after losing almost half a year for nothing, the author lectured for a month at Columbia University in New York, courtesy of an invitation from E. Kolchin, who, despite discovering these facts, still refused to provide a preface for the 1983 DGT book [8]. It is easy for any reader to check the list of the differential algebra community, provided by J. Kovacic, including himself as a former student of Kolchin at that time, that not a single of these more than one hundred persons even once quoted this DGT book for about forty years, despite the fact that Hopf algebras had been used for the first time in this book [19,20]. Finally, to be fully fair, we may say that, after being invited to lecture at King’s College, London, about the Backlund problem and jet theory by F. A. E. Pirani, D. C. Robinson and W.F. Shadwick—just after the publication of the 1978 GB book—it happened that the only visitor’s room left was the one occupied by M. E. Sweedler [21,22] (on leave for a week), where bunches of papers on Hopf algebras were on display on the table, a huge chance indeed.

We finally explain, using a simple example, the content of the paper written by A. Bialynicki-Birula in 1961 [23], which, for the first time, utilizes tensor products of rings and fields to study DGT. The main idea is that a Lie group acting on itself is the simplest example of a PHS.

Example 3.

Let us consider, again, the group of affine transformations on the real line. If $y=a^{1}x+a^2,z=b^{1}y+b^2$, we obtain, by composition, $z=\left(b^1{a}^1\right)x+(b^1{a}^2+b^2)$, and the group G with composition $({\overline{a}}^1,{\overline{a}}^2)=(b^1,b^2)(a^1,a^2)=(b^1{a}^1,b^1{a}^2+b^2)$ and inversion ${(a^1,a^2)}^{-1}=({\displaystyle \frac{1}{a^1}},-{\displaystyle \frac{a^2}{a^1}})$. The two reciprocal commuting left and right distributions on G are, respectively, generated by

$$\begin{array}{ccc}\Theta \hfill & =& \{{\theta }_1=a^1{\displaystyle \frac{\partial }{\partial a^1}}+a^2{\displaystyle \frac{\partial }{\partial a^2}},{\theta }_2={\displaystyle \frac{\partial }{\partial a^2}}\}\hfill \\ \hfill & & \\ \Delta \hfill & =& \{{\delta }_1=a^1{\displaystyle \frac{\partial }{\partial a^1}},{\delta }_2=a^1{\displaystyle \frac{\partial }{\partial a^2}}\}\hfill \end{array}$$

in such a way that $[\Theta ,\Delta ]=0$; that is, $[{\theta }_i,{\delta }_j]=0,\forall i,j=1,2$. Of course, with the group G being defined on $k=\mathbb{Q}$, we have $k\left[G\right]=k[a^1,a^2]$, and we may introduce the fields $K=\mathbb{Q}$, $L=\mathbb{Q}(a^1,a^2)$ with $K\subset L$. Introducing the intermediate field $K^{\prime }=\mathbb{Q}(a^2/a^1)$ with $K\subset K^{\prime }\subset L$, we obtain, at once,

$${\displaystyle \frac{b^1{a}^2+b^2}{b^1{a}^1}}={\displaystyle \frac{a^2}{a^1}}+{\displaystyle \frac{b^2}{b^1{a}^1}}={\displaystyle \frac{a^2}{a^1}}⟺b^2=0$$

The subgroup preserving $K^{\prime }$ has parameters $(b^1,0)$ in such a way that $(b^1,0)(a^1,a^2)=(b^1{a}^1,b^1{a}^2)$, and the only invariant subfield is, again, $K^{\prime }$. In contrast, choosing $K^{\prime }=\mathbb{Q}\left(a^1{a}^2\right)$, we obtain

$$\left(b^1{a}^1\right)(b^1{a}^2+b^2)={\left(b^1\right)}^2\left(a^1{a}^2\right)+\left(b^1{b}^2{a}^1\right)=a^1{a}^2⟷{\left(b^1\right)}^2=1,b^2=0$$

and the invariant subfield is now $K^{″}=\mathbb{Q}({\left(a^1\right)}^2,a^1{a}^2,{\left(a^2\right)}^2)$, with the strict inclusion $K\subset K^{\prime }\subset K^{″}$.

BB did notice that only the intermediate fields stabilized by $\Delta$ do provide a Galois correspondence $K\subset K^{\prime }\subset L\leftrightarrow e\subset G^{\prime }\subset G$, as we have indeed ${\delta }_1\left({\displaystyle \frac{a^2}{a^1}}\right)=0,{\delta }_2\left({\displaystyle \frac{a^2}{a^1}}\right)=1$; however,

$${\delta }_1\left(a^1{a}^2\right)=a^1{a}^2,{\delta }_2\left(a^1{a}^2\right)={\left(a^1\right)}^2\Rightarrow {\left(a^2\right)}^2={\left(a^1{a}^2\right)}^2/{\left(a^1\right)}^2$$

To recapitulate, in his first paper, BB discovered that only intermediate fields stable under $\Delta$ generally provide a Galois correspondence [13]. Such a comment is at the origin of his second paper, providing a new approach to the Picard–Vessiot theory by using “fields with operators”, differential fields being only a specific example. Roughly speaking, his first idea has been to introduce the tensor product $L{\otimes }_{K}L$ with $L=\mathbb{Q}(a^1,a^2)$ on the left and $L=\mathbb{Q}({\overline{a}}^1,{\overline{a}}^2)$ on the right. His second idea has been to enlarge Δ to the new derivations on the tensor product, namely,

$$\Delta =\{{\delta }_1=a^1{\displaystyle \frac{\partial }{\partial a^1}}+{\overline{a}}^1{\displaystyle \frac{\partial }{\partial {\overline{a}}^1}},{\delta }_2=a^1{\displaystyle \frac{\partial }{\partial a^2}}+{\overline{a}}^1{\displaystyle \frac{\partial }{\partial {\overline{a}}^2}}\}$$

He then noticed that the only quantities killed by the new ${\delta }_1$ and ${\delta }_2$ are

$$\{b^1={\displaystyle \frac{{\overline{a}}^1}{a^1}}={\displaystyle \frac{1}{a^1}}\otimes a^1,b^2={\overline{a}}^2-\left({\displaystyle \frac{{\overline{a}}^1}{a^1}}\right)a^2=1\otimes a^2-{\displaystyle \frac{a^2}{a^1}}\otimes a^1\}$$

It follows that $\mathbb{Q}[b^1,b^2]=\mathbb{Q}\left[G\right]=cst\left(L{\otimes }_{K}L\right)$, where the “constants” are now the quantities killed by the extension of the derivations $\Delta$ to $L{\otimes }_{K}L$. We do consider that the upper “bar” notation supersedes the tensorial notation in the differential geometric framework. The “novelty” (we may even say “revolution") is that the group parameters are no longer “differential constants", contrary to the standard approach of Kolchin. Such a FACT can be sketched as follows in Example 1, as we indeed obtain $(a^1={\displaystyle \frac{{\overline{y}}_x}{y_x}},a^2=\overline{y}-y{\displaystyle \frac{{\overline{y}}_x}{y_x}})\Rightarrow (d_x{a}^1=0,d_x{a}^2=0)$ when $y_{xx}=0,{\overline{y}}_{xx}=0$.

The content of the paper is now clear from this long Introduction. In Section 2, we revisit the classical Galois theory. In Section 3, we recall the definition of rings and corings in a purely algebraic framework. In Section 4, we present the basic concepts of jet theory and Lie pseudogroups to facilitate an understanding of the concepts of automorphic systems and automorphic differential extensions. Finally, in Section 5, we provide motivating examples to illustrate the main results obtained by using Hopf algebras for Differential Galois Theory (DGT), preceding the conclusion in Section 6.

We hope that the comparison of our two books [6,8] with the paper published by J. Kovacic (1941–2009) in 2005 [19] requires no comment on the anteriority of using Hopf algebras in DGT. We also point out that it is difficult to understand the conceptual confusion that has existed for almost fifty years between algebraic pseudogroups and differential algebraic groups, given that such a comment was already made in 1930. We believe that the reason mainly stems from the fact that Ritt and Kolchin were involved in analysis differential geometry, always focusing on solutions of systems of OD or PD equations. In any case, the future will judge!

2. Classical Galois Theory Revisited

Let us start this section by explaining the clever idea of E. Vessiot in the first chapter of his 1904 paper [3], along with the deep confusion that is spoiling it. The following example will prove that the standard link existing between the classical Galois theory and group theory is not at all as evident as one might imagine from the extensive literature on the subject [24,25,26,27,28,29,30,31]. We also point out the fact that tensor products of rings and fields have rarely been used [32,33,34,35].

Example 4.

First of all, with ground field $K=\mathbb{Q}$ and one indeterminate y, to understand the distinction that may exist between general and special algebraic equations, let us consider the general cubic polynomial equation $P\equiv y^3-{\omega }^1{y}^2+{\omega }^{2}y-{\omega }^3=0$ with upper indices on ω to agree with the next sections. In algebra, giving special values in K to ω, the main problem has always been about knowing the three roots $({\eta }_1,{\eta }_2,{\eta }_3)$, that is to say, to exhibit a splitting field $K({\eta }_1,{\eta }_2,{\eta }_3)$ of P. Comparing, now, P to the product $(y-{\eta }_1)(y-{\eta }_2)(y-{\eta }_3)$, we obtain the three symmetric functions of the roots:

$${\eta }_1+{\eta }_2+{\eta }_3={\omega }^1,{\eta }_1{\eta }_2+{\eta }_1{\eta }_3+{\eta }_2{\eta }_3={\omega }^2,{\eta }_1{\eta }_2{\eta }_3={\omega }^3$$

The roots are different if and only if $\delta =({\eta }_1-{\eta }_2)({\eta }_1-{\eta }_3)({\eta }_2-{\eta }_3)\ne 0$. Introducing the derivative $P^{\prime }=dP/dy$, it is easy to construct the resultant of P and $P^{\prime }$, which only depends on ω, and is equal to ${\delta }^2$ up to a factor in $\mathbb{Q}$. Modifying slightly standard notations of textbooks, we obtain

$${\delta }^2=-27{\left({\omega }^3\right)}^2+18{\omega }^1{\omega }^2{\omega }^3-4{\left({\omega }^2\right)}^3-4{\left({\omega }^1\right)}^3{\omega }^3+{\left({\omega }^1{\omega }^2\right)}^2$$

We invite the reader to consider the special polynomial $P\equiv y^3-3y+1=0$ before reading ahead to discover that it is not so natural to associate the permutation subgroup with the Galois group:

$$A_3=\{e=\left(\begin{array}{c}123\\ 123\end{array}\right),\left(\begin{array}{c}123\\ 231\end{array}\right),\left(\begin{array}{c}123\\ 312\end{array}\right)\}⊲S_3=\{e=\left(\begin{array}{c}123\\ 123\end{array}\right),\left(\begin{array}{c}123\\ 231\end{array}\right),\left(\begin{array}{c}123\\ 312\end{array}\right),\left(\begin{array}{c}123\\ 132\end{array}\right),\left(\begin{array}{c}123\\ 321\end{array}\right),\left(\begin{array}{c}123\\ 213\end{array}\right)\}$$

with $\mid A_3\mid =3<\mid S_3\mid =6$ when studying the Galois extension $L=Q\left(K\right[y]/(P\left)\right)/K$ generated by the polynomial P, which is irreducible over K, or, equivalently, P has no root in K.

Indeed, let us introduce the generic zero $\eta$ via the specialization $K\left[y\right]\to K\left[y\right]/\left(P\right):y\to \eta$ to get ${\eta }^3-3\eta +1=0$. Any automorphism $\sigma \in aut(L/K)$ is of the form $\sigma \left(\eta \right):\alpha 1+\beta \eta +\gamma {\eta }^2$ with $\alpha ,\beta ,\gamma \in K$. A straightforward but quite tedious substitution left to the reader proves that, if $\eta$ is a root of P, then $\sigma \left(\eta \right)={\eta }^2-2$ is another root, also with ${\sigma }^2\left(\eta \right)={\eta }^4-4{\eta }^2+2=-{\eta }^2+\eta +2$, and that ${\sigma }^3\left(\eta \right)=\eta \Rightarrow {\sigma }^3=e$ is the identity. It follows that the splitting field of $L/K$ is just $L/K$, which is, therefore, a Galois extension, as we already said. The main problem is that the definitions of $\eta$ and $\sigma$ only depend on $Ł/K$ and do not provide an algebraic group “per se”. The following “trick” allows one to get rid of such a problem [30]. For this, we notice that the equation can be written as $y^3=3y-1\iff y(y^2-2)=y-1$. We may, therefore, consider the purely rational transformation $y\to \sigma \left(y\right)=\overline{y}=1-{\displaystyle \frac{1}{y}}$ defined over $\mathbb{Q}$, thus obtaining ${\sigma }^2\left(y\right)=1-(1/(1-{\displaystyle \frac{1}{y}}))={\displaystyle \frac{1}{1-y}}$ and ${\sigma }^3\left(y\right)=y$ in a way totally independent of the equation. Such a new way will permit the presentation of the classical Galois theory along the way, initiated by Vessiot, through the so-called general equations.

We have a finite algebraic group $\Gamma$ of transformations of the real line Y, which is defined over the ground field $k=\mathbb{Q}$. Consider the following rational function:

$$y\to \Phi \left(y\right)\equiv y+\sigma \left(y\right)+{\sigma }^2\left(y\right)=y+(1-{\displaystyle \frac{1}{y}})+\left({\displaystyle \frac{1}{1-y}}\right)={\displaystyle \frac{y^3-3y+1}{y^2-y}}\in k\left(y\right)=L$$

Let us prove that such a function, which is, of course, invariant by $\sigma$, is the generating invariant of the rational action of $\Gamma$ on $k\left(y\right)$ as follows. Writing $\Phi \left(\overline{y}\right)=\Phi \left(y\right)$, we successively get

$$(y^2-y)({\overline{y}}^3-3\overline{y}+1)-({\overline{y}}^2-\overline{y})(y^3-3y+1)=0\Rightarrow (\overline{y}-y)(\overline{y}-(1-{\displaystyle \frac{1}{y}}))(\overline{y}-{\displaystyle \frac{1}{1-y}})=0$$

with $k=\mathbb{Q},K=k(\Phi ),L=k\left(y\right)$, we obtain the chain of inclusions $k\subset K\subset L$ with $\mid L/K\mid =3$ and k algebraically closed in L, and thus, in K. We discover that Y is a Principal Homogeneous Space (PHS) for $\Gamma$ with graph $Y\times Y\simeq Y\times \Gamma$ with a sight abuse of language but with a well-defined isomorphism $L{\otimes }_{K}L\simeq L{\otimes }_{k}k[\Gamma ]$; this can be extended to the rings of quotients $Q\left(L{\otimes }_{K}L\right)\simeq Q\left(L{\otimes }_{k}k[\Gamma ]\right)$, although each side is already a direct sum of 3 fields, as we saw. Introducing what we shall call the Lie form $\Phi \left(y\right)=\omega$ of the action, we finally notice that the left term is defined by using the so-called general equations $y^3-\omega y^2+(\omega -3)y+1=0$ and ${\overline{y}}^3-\omega {\overline{y}}^2+(\omega -3)\overline{y}+1=0$. Indeed, subtracting the first from the second, we get

$$(\overline{y}-y)(({\overline{y}}^2+y\overline{y}+y^2)-\omega (\overline{y}+y)+\omega -3)=(\overline{y}-y)({\overline{y}}^2-(\omega -y)\overline{y}-{\displaystyle \frac{1}{y}})=0$$

because, when dividing the first general equation by y, we get $y^2-\omega y+\omega -3=-{\displaystyle \frac{1}{y}}$ and obtain the same factorization (as before) as $(1-{\displaystyle \frac{1}{y}})+{\displaystyle \frac{1}{1-y}}=\omega -y,(1-{\displaystyle \frac{1}{y}})\left({\displaystyle \frac{1}{1-y}}\right)=-{\displaystyle \frac{1}{y}}$.

We finally explain, for this example, how Vessiot got into his mind the fact that classical Galois theory is only a theory of PHS for groups of permutations. For this, let us introduce three indeterminates $y=(y^1,y^2,y^3)$ or $y^k$ for $k=1,2,3$ and consider the three equations $P\left(y^k\right)=0$ as a linear system for $({\omega }^1,{\omega }^2,{\omega }^3)$ with a Van der Mond determinant that is homogeneous of degree 3 in $\mathbb{Q}\left[y\right]$, thus equal up to a sign to $\Delta \left(y\right)=(y^1-y^2)(y^1-y^3)(y^2-y^3)$. We obtain three general equations in Lie form:

$${\Phi }^1\left(y\right)\equiv y^1+y^2+y^3={\omega }^1,{\Phi }^2\left(y\right)\equiv y^1{y}^2+y^1{y}^3+y^2{y}^3={\omega }^2,{\Phi }^3\left(y\right)\equiv y^1{y}^2{y}^3={\omega }^3$$

We have proved in [8], p. 151, that the ideal $\mathfrak{a}$ generated by the equations $\Phi -\omega =0$ is perfect if and only if $\delta \ne 0$, that is, when the three roots of the equation $P\left(y\right)=0$ are different. The main definition of Vessiot has been the following:

Definition 1.

A system of equations is called an automorphic system if any solution may be obtained from a given one by one and only one transformation of a (finite or eventually infinite) group of transformations acting on the variables, with Lie pseudogroups being called infinite groups at that time.

In the present situation, we have, indeed, a PHS for the group $S_3$ of permutations in three variables. However, if $\mathfrak{a}\subset \mathbb{Q}\left[y\right]$ is prime for the general situation and for certain special situations, such as for $P\equiv y^3+y+1=0$ with ${\delta }^2=-31$, it may not be prime for others; in particular, the present version $P\equiv y^3-3y+1=0$ with ${\delta }^2=81=9^2$ because $\mathfrak{a}$ surely contains the product $(\Delta (y)-9)(\Delta (y)+9)$, although each of the factors does not belong to $\mathfrak{a}$. Adding the equation $\Delta \left(y\right)=9$, we get a prime ideal, reducing the group of invariance from $S_3$ to $A_3$. However, Vessiot was writing in 1904, and the concept of a prime ideal was only introduced in 1930 by J.F. Ritt when he created differential algebra [8,9]. Roughly speaking, Vessiot had been confusing “prime ideal” with “maximal ideal” (thus prime) in his (personal) definition of irreducibility. The worst fact is that the whole Picard–Vessiot theory has also been based on this confusion, one that was not known or even acknowledged by E.R. Kolchin, despite what we told him in front of his students while lecturing at Columbia University in New York (8 lectures of 2 h in April 1981). The reason is also that Kolchin was engaged in a kind of “dead end" with his so-called differential algebraic groups, along with a confusing definition provided by J.-F. Ritt in the last few papers he wrote around 1950, just before he died [16,17,18]. We invite the reader to similarly treat the general quadratic equation $y^2-\omega y+1=0$ with group $\Gamma =\{\overline{y}=y,\overline{y}={\displaystyle \frac{1}{y}}\}$ while setting $\Phi \left(y\right)\equiv y+{\displaystyle \frac{1}{y}}=\omega$.

In characteristic 0, let us recall a few technical results on tensor products of rings and fields that are not so well known, following closely [8,34]. In this section, we shall only deal with finitely generated field extensions, contrary to the next sections.

Definition 2.

Let A be a ring with unit 1 and elements $a,b,c,\dots$ If $\mathfrak{a}\subset A$ is an ideal, we may introduce its radical as $rad\left(\mathfrak{a}\right)=\{a\in A\mid \exists n\in \mathbb{N},a^n\in \mathfrak{a}\}$, which is also sometimes simply denoted by $\sqrt{\mathfrak{a}}$. An ideal $\mathfrak{a}$ is said to be perfect if $rad\left(\mathfrak{a}\right)=\mathfrak{a}$, and the residue ring $k\left[y\right]/\mathfrak{a}=A$ is said to be reduced in this case. An ideal $\mathfrak{p}\in A$ is said to be prime if $ab\in \mathfrak{p}\Rightarrow a\in \mathfrak{p}orb\in \mathfrak{p}$, and the residue ring $A/\mathfrak{p}$ is, thus, an integral domain in this case because $\overline{ab}=0\Rightarrow \overline{a}=0$ or $\overline{b}=0$ by denoting a residue with a bar.

Proposition 1.

Any perfect ideal $\mathfrak{a}$ in a polynomial ring is the intersection $\mathfrak{a}={\mathfrak{p}}_1\cap \dots \cap {\mathfrak{p}}_r$ of a finite number r of prime polynomial ideals.

Proposition 2.

A maximal ideal $\mathfrak{m}\subset max\left(A\right)$ is prime, and an ideal $\mathfrak{m}$ is maximal if and only if the residue ring $A/\mathfrak{m}$ is a field.

Proof. 

If $ab\in \mathfrak{m},a\notin \mathfrak{m}$, then $\mathfrak{m}+aA=A\Rightarrow \exists c\in A,ac-1\in \mathfrak{m}\Rightarrow b\in \mathfrak{m}$. Then, let us first prove that $\mathfrak{m}\in max\left(A\right)\Rightarrow A/\mathfrak{m}$ is a field. If $a\in A,a\notin \mathfrak{m}$, then ∃ an ideal $\mathfrak{a}=\{ab+c\in A\mid b\in A,c\in \mathfrak{m}\}\subset A$. Indeed, we have successively $u(ab+c)=a\left(ub\right)+uc\in \mathfrak{a},(a{b}_1+c_1)+(a{b}_2+c_2)=a(b_1+b_2)+(c_1+c_2)\in \mathfrak{a}$. Moreover, choosing $b=0,c\in \mathfrak{m}$, we get $\mathfrak{m}\subset \mathfrak{a}$ and a contradiction unless $\mathfrak{a}=A$. We may thus find b and c such that $ab+c=1$. Passing to the residue $A\to A/\mathfrak{m}:a\to \overline{a}$, we get $\overline{a}\ne 0$ and $\overline{a}\overline{b}=1$; that is, $A/\mathfrak{m}$ is a field.

Conversely, let us imagine that $A/\mathfrak{p}$ is a field; that is, $\mathfrak{p}$ is at least a prime ideal, but $\mathfrak{p}\notin max\left(A\right)$. Then, we may find ideals $\mathfrak{p}⫋\mathfrak{q}\subset A$ and choose $a\in \mathfrak{q},a\notin \mathfrak{p}$ in such a way that $\mathfrak{p}⫋aA+\mathfrak{p}\subseteq \mathfrak{q}\subset A$. Now, as $A/\mathfrak{p}$ is a field, $\exists b\in A,\overline{a}\overline{b}=1\Rightarrow (a+\mathfrak{p})(b+\mathfrak{p})=1+\mathfrak{p}\Rightarrow \exists c\in \mathfrak{p},ab+c=1$, and thus, $aA+\mathfrak{p}=A\Rightarrow \mathfrak{q}=A\Rightarrow \mathfrak{p}=\mathfrak{m}\in max\left(A\right)$. □

Definition 3.

When R is a ring, the subrings A and B containing a subfield k are said to be linearly disjoint over k in R if, whenever $a_1,\dots ,a_r\in A$ are linearly independent over k and $b_1,\dots ,b_s\in B$ are linearly independent over k, then the $rs$ products $a_i{b}_j$ are linearly independent over k in R. We shall denote by $[A,B]$ the smallest subring of R containing both A and B.

Similarly, when N is a field, the subfields K and L, both containing a subfield k, are said to be linearly disjoint over k in N if, whenever $a_1,\dots ,a_r\in K$ are linearly independent over k and $b_1,\dots ,b_s\in L$ are linearly independent over k, then the $rs$ products $a_i{b}_j$ are linearly independent over k in N. We shall denote by $(K,L)$ the smallest subfield of N containing both K and L.

The two following propositions will be quite useful, along with the following diagram [13,23]:

$$\begin{array}{ccccc}M& ⟶& (K,M)& ⟶& (L,M)\\ \uparrow & & \uparrow & & \uparrow \\ k& ⟶& K& ⟶& L\end{array}$$

Proposition 3.

If $k\subset K\subset L$ and $k\subset M$ are subfields of a bigger field N, then L and M are linearly disjoint over k in $(L,M)$ if and only if L and $(K,M)$ are linearly disjoint over K in $(L,M)$ and K and M are linearly disjoint over k in $(K,M)$.

Proof. 

As vector spaces, let $\left\{{\lambda }_r\right\}$ be a basis of K over k, $\left\{{\mu }_s\right\}$ be a basis of L over K and $\left\{{\nu }_t\right\}$ be a basis of M over k. Then, $\left\{{\lambda }_r{\mu }_s\right\}$ is a basis of L over k. If L and M were not linearly disjoint over k in $(L,M)$, we may find linear relations of the form ${\Sigma }_{r,s}\left({\Sigma }_t{c}_{rst}{\nu }_t\right){\lambda }_r{\mu }_s=0$ that we could write as ${\Sigma }_s\left({\Sigma }_{r,t}{c}_{rst}{\lambda }_r{\mu }_t\right){\mu }_s=0$, contradicting the linear disjointness of L and $(K,M)$ over K. The converse is similar. □

Proposition 4.

In the situation of the last proposition, we have $(K,M)\cap L=K$.

Proof. 

Again, let $\left\{{\nu }_t\right\}$ be a basis of $M/k$. Then, any element $\mu \in (K,M)$ may be written as $\mu ={\Sigma }_t{a}_t{\nu }_t/{\Sigma }_t{b}_t{\nu }_t$, with $a_t,b_t\in K$. If $\mu \in L$, then ${\Sigma }_t(a_t-\mu b_t){\nu }_t=0$ with $(a_t,\mu b_t)\in L$. However, according to the preceding proposition, $(K,M)$ and L are linearly disjoint over K in $(L,M)$, and we must, therefore, have $\mu b_t=a_t$. Now, one of the $b_t$ must at least be different from zero because, otherwise, ${\Sigma }_t{b}_t{\nu }_t$ could not be used as a denominator. Hence, for some t, we have $\mu =a_t/b_t\in K$. Finally, we have, of course, $K\subset (K,M),K\subset L\Rightarrow K\subseteq (K,M)\cap L$, and this ends the proof. □

Definition 4.

An extension $L/K$ is called regular if K is algebraically closed in L.

Theorem 1.

If K and L are two fields containing a field k, and $L/k$ is regular, then $K{\otimes }_{k}L$ is an integral domain.

Proof. 

Let us decompose the extension $K/k$ by introducing a transcendence basis $\{s_i\mid i\in I\}$ of $K/k$ in such a way that K becomes algebraic over $k\left(s\right)=k(s_1,\dots ,s_m)$. We have the commutative diagram of inclusions:

$$\begin{array}{ccccc}L& \to & L\left(s\right)& \to & Q\left(K{\otimes }_{k}L\right)\\ \uparrow & & \uparrow & & \uparrow \\ k& \to & k\left(s\right)& \to & K\end{array}$$

By induction on m, one can prove ([8], Lemma 4.47) that $L\left(y\right)/k\left(y\right)$ is regular for any indeterminate y, and thus, $L\left(s\right)/k\left(s\right)$ is regular. Moreover, when $P\in K\left[y\right]$ is irreducible over K, then it is also irreducible over L when $L/K$ is regular ([8], Proposition 4.48). Hence, we only have to prove that $K{\otimes }_{k\left(s\right)}L\left(s\right)$ is a field. However, this fact follows because K may be generated by a single primitive element $\eta$ over $k\left(s\right)$, a generic root of an irreducible polynomial $P\in k\left(s\right)\left[y\right]$ that remains irreducible over $L\left(s\right)$. If we require that K and L be linearly disjoint over k in $(K,L)$, we need the homomorphism $K{\otimes }_{k}L\to (K,L)$ to be a monomorphism. In this case, there exists an isomorphism $K{\otimes }_{k}L\simeq [K,L]$, a reason for which we need $K{\otimes }_{k}L$ to be an integral domain to be able to set $[K,L]=K{\otimes }_{k}L\Rightarrow (K,L)=Q\left(K{\otimes }_{k}L\right)$ exactly; that is, independently of any bigger field N, as before. □

Example 5.

Let us consider the following automorphic system ([8], Example 8.57, p. 177):

$$\mathcal{A}\Phi \equiv {\displaystyle \frac{y^4-1}{y^2}}=\omega$$

or, equivalently, the general quartic equation $y^4-\omega y^2-1=0$, which is known to have the Galois group $D_8$ with $\mid D_8\mid =8$. For this, considering the new equation ${\overline{y}}^4-\omega {\overline{y}}^2-1=0$ and substracting the previous equation, we get $(\overline{y}-y)(\overline{y}+y)({\overline{y}}^2+y^2-\omega )=0$; however, the last term is easily seen to be $({\overline{y}}^2+{\displaystyle \frac{1}{y^2}})=(\overline{y}-{\displaystyle \frac{i}{y}})(\overline{y}+{\displaystyle \frac{i}{y}})$, and we obtain the finite group $\Gamma (L/K)=\{y\to y,-y,{\displaystyle \frac{i}{y}},-{\displaystyle \frac{i}{y}}\}$ with $\mid L/K\mid =4$, along with the following diagram in which $split(L/K)=L\left(i\right)$ is a smallest Galois extension of K containing L with $\mid split(L/K)/K\mid =\mid D_8\mid =8$ in the following picture:

$$\begin{array}{ccc}L=\mathbb{Q}\left(y\right)\hfill & \stackrel{2}{⟶}& L\left(i\right)=split(L/K)\hfill \\ 4\uparrow \hfill & 8\nearrow & \uparrow 4\hfill \\ K=\mathbb{Q}(\Phi )\hfill & \stackrel{2}{⟶}& K\left(i\right)\hfill \\ \uparrow \hfill & & \uparrow \hfill \\ k=\mathbb{Q}\hfill & \stackrel{2}{⟶}& k\left(i\right)\hfill \end{array}$$

Now, $L\left(i\right)$ is a Galois extension of $K\left(i\right)$ with $\mid L\left(i\right)/K\left(i\right)\mid =4$, and we may decompose it into two quadratic extensions. First, we have the subgroup $\{y\to y,{\displaystyle \frac{i}{y}}\}$ with generating invariant $\Psi =y+{\displaystyle \frac{i}{y}}$, satisfying ${\Psi }^2-\Phi -2i=0$. If we consider the intermediate field $K(i,\Psi )$ between $K\left(i\right)$ and $L\left(i\right)$, we successively get $K(i,\Psi )\cap L=K\to (K,k(i\left)\right)=K\left(i\right)\subset K(i,\Psi )$ with a strict inclusion. However, if we consider the other subgroup $\{y\to y,-y\}$, the only generating invariant is $\Psi =y^2$, and we now have $K(i,\Psi )\cap L=K(\Psi )=K^{\prime }$. We invite the reader to revisit this example by using a linear group Γ of $2\times 2$-matrices preserving the two invariants ${\Phi }^1\equiv y^2-z^2,{\Phi }^2\equiv yz$, proving first that

$${(ay+bz)}^2-{(cy+dz)}^2=y^2-z^2,(ay+bz)(cy+dz)=yz$$

provides an ideal $\mathfrak{a}=(a^2-c^2=1,b^2-d^2=-1,ab-cd=0,ac=0,bd=0,ad+bc=1)$, showing that the algebraic group can be decomposed into the three isolated prime components:

$$(a=1,b=0,c=0,d=1)\cup (a=-1,b=0,c=0,d=-1)\cup (a=0,b^2=-1,b+c=0,d=0)$$

In this case, we have, thus, $k[\Gamma ]\simeq \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}\left(i\right)$. This example proves that one cannot hope to refer to a classical Galois extension to study intermediate fields.

We invite the reader to similarly treat the case ${\Phi }^1\equiv y^2+z^2,{\Phi }^2\equiv yz$ and compare these. Equivalently, this amounts to considering the irreducible general equation $y^4-\omega y^2+1=0$ and the Galois extension $L/K$ with $K=\mathbb{Q}(y^2+{\displaystyle \frac{1}{y^2}})=\mathbb{Q}(\Phi )\subset L=\mathbb{Q}\left(y\right)$ and $\mid L/K\mid =4$. Then, $L{\otimes }_{K}L$ is defined by the factorization $(\overline{y}-y)(\overline{y}+y)({\overline{y}}^2+y^2-\omega )$, with the last term equal to $({\overline{y}}^2-{\displaystyle \frac{1}{y^2}})=(\overline{y}-{\displaystyle \frac{1}{y}})(\overline{y}+{\displaystyle \frac{1}{y}})$. Hence, $Ł/K$ is a Galois extension with a Galois group in the Klein group $V_4$ such that $\mid V_4\mid =4$, and thus, $k\left(V_4\right)\simeq \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}$. The subgroup ${\Gamma }^{\prime }=\{\overline{y}=y,\overline{y}={\displaystyle \frac{1}{y}}\}$ is defined by the additional invariant $\Psi =y+{\displaystyle \frac{1}{y}}$ with ${\Psi }^2-\Phi -2=0$. In contrast, the other subgroup ${\Gamma }^{\prime }=\{\overline{y}=y,\overline{y}=-y\}$ is defined by the additional invariant $\Psi =y^2$ with ${\Psi }^2-\Phi \Psi +1=0$. In both cases, we have ${\Gamma }^{\prime }⊲\Gamma$, and $K^{\prime }/K$ is a Galois extension.

Example 6.

Coming back to the Vessiot point of view, we may choose $k=\mathbb{Q}\subset K=k({\Phi }^1,{\Phi }^2,{\Phi }^3)=k(\Phi )\subset K^{\prime }=k(\Phi ,\Delta )\subset L=k(y^1,y^2,y^3)=k\left(y\right)$ with

$${\Phi }^1\equiv y^1+y^2+y^3,{\Phi }^2\equiv y^1{y}^2+y^1{y}^3+y^2{y}^3,{\Phi }^3\equiv y^1{y}^2{y}^3,\Delta \equiv (y^1-y^2)(y^1-y^3)(y^2-y^3)$$

to revisit the classical Galois theory for $A_3⊲S_3$ while understanding that the identification of $\Gamma (L/K)$ with $aut(L/K)$ is just a pure coincidence; this is because Galois extensions are just examples of automorphic extensions for groups of permutations represented by square matrices with entries in $(0,1)$. Of course, dealing with normal subgroups in the algebraic framework also remains, as we shall see later on.

Definition 5.

Two ideals $\mathfrak{a},\mathfrak{b}\subset A$ are said to be comaximal if $\mathfrak{a}+\mathfrak{b}=A$.

Lemma 1.

Let us consider ideals ${\mathfrak{a}}_1,\dots ,{\mathfrak{a}}_r\subset A$ that are comaximal at two by two; that is, ${\mathfrak{a}}_i+{\mathfrak{a}}_j=A$, $\forall i\ne j$. Then, we have ${\mathfrak{a}}_1\dots {\mathfrak{a}}_r={\mathfrak{a}}_1\cap \dots \cap {\mathfrak{a}}_r$.

Proof. 

For simplicity, we only consider the case $r=2$ with $\mathfrak{a}$ and $\mathfrak{b}$, two comaximal ideals in A, only proving that $\mathfrak{a}\mathfrak{b}=\mathfrak{a}\cap \mathfrak{b}$. The inclusion $\mathfrak{a}\mathfrak{b}\subseteq \mathfrak{a}\cap \mathfrak{b}$ being evident, we may find elements $a\in \mathfrak{a},b\in \mathfrak{b}$ with $a+b=1$. Hence, for any $x\in \mathfrak{a}\cap \mathfrak{b}$, we have $x=ax+bx$, with both $ax$ and $bx$ in $\mathfrak{a}\mathfrak{b}$; that is, $x\in \mathfrak{a}\mathfrak{b}$. The general situation can be proved by induction [8]. □

Let $A_1,\dots ,A_r$ be rings, and consider all the sequences $\{a_1,\dots ,a_r\mid a_i\in A_i\}$. We may provide a structure of a ring to this set by setting

$$(a_1,\dots ,a_r)+(b_1,\dots ,b_r)=(a_1+b_1,\dots ,a_r+b_r),(a_1,\dots ,a_r)(b_1,\dots ,b_r)=(a_1{b}_1,\dots ,a_r{b}_r)$$

This ring is called the direct sum of $A_1,\dots ,A_r$ and is denoted by $A_1\oplus \dots \oplus A_r$.

Proposition 5.

(Chinese remainder theorem) With the same assumption as in the last lemma, we have an isomorphism:

$$A/({\mathfrak{a}}_1\cap \dots \cap {\mathfrak{a}}_r)\simeq (A/{\mathfrak{a}}_1)\oplus \dots \oplus (A/{\mathfrak{a}}_r)$$

An Artinian ring is a ring satisfying the descending chain condition on ideals; that is, no infinite descending sequence of ideals may exist. An Artinian ring has only a finite number of prime ideals, and each of them is maximal. As above, we shall denote them by ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$.

Theorem 2.

Any Artinian ring with zero nilradical, that is, with ${\mathfrak{m}}_1\dots {\mathfrak{m}}_r=0$, is isomorphic to a direct sum of fields.

Proof. 

Any reduced residue ring of a polynomial ring in one indeterminate over a field k is an Artinian ring. In fact, let $A=k\left[y\right]/\mathfrak{a}$, where the principal ideal $\mathfrak{a}$ is generated by the polynomial $P\in k\left[y\right]$, which can be written as $P=P_1\dots P_r$, with each $P_i\ne P_j$ irreducible in $k\left[y\right]$ over k and relatively prime to $P_j,\forall i\ne j$. If ${\mathfrak{p}}_i$ is the prime, and thus, maximal principal ideal generated in $k\left[y\right]$ by $P_i$, then we have $\mathfrak{a}={\mathfrak{p}}_1\cap \dots \cap {\mathfrak{p}}_r$, and thus,

$$A=k\left[y\right]/\mathfrak{a}\simeq (k\left[y\right]/{\mathfrak{p}}_1)\oplus \dots \oplus (k\left[y\right]/{\mathfrak{p}}_r)$$

Taking the residue ${\mathfrak{m}}_i$ of ${\mathfrak{p}}_i$ with respect to $\mathfrak{a}$, we obtain ${\mathfrak{m}}_1\dots {\mathfrak{m}}_r=0$, and thus (make a picture),

$${\mathfrak{m}}_1\cap \dots \cap {\mathfrak{m}}_r=0\Rightarrow A\simeq (A/{\mathfrak{m}}_1)\oplus \dots \oplus (A/{\mathfrak{m}}_r)$$

The commutative and exact diagram

$$\begin{array}{ccccccccc}\hfill & & 0& & 0& & 0& & \\ \hfill & & \downarrow & & \downarrow & & \downarrow & & \\ \hfill 0& \to & \mathfrak{a}& \to & {\mathfrak{p}}_i& \to & {\mathfrak{m}}_i& \to & 0\hfill \\ \hfill & & \Vert & & \downarrow & & \downarrow & & \\ \hfill 0& \to & \mathfrak{a}& \to & k\left[y\right]& \to & A& \to & 0\hfill \\ \hfill & & \downarrow & & \downarrow & & \downarrow & & \\ \hfill & & 0& \to & k\left[y\right]/{\mathfrak{p}}_i& \to & A/{\mathfrak{m}}_i& \to & 0\hfill \\ \hfill & & & & \downarrow & & \downarrow & & \\ \hfill & & & & 0& & 0& \end{array}$$

finally proves the isomorphism $A/{\mathfrak{m}}_i\simeq k\left[y\right]/{\mathfrak{p}}_i$, $\forall i=1,\dots ,r$. □

Corollary 1.

$A\simeq A_1\oplus \dots \oplus A_r\Rightarrow Q\left(A\right)\simeq Q\left(A_1\right)\oplus \dots \oplus Q\left(A_r\right)$.

Example 7.

With $k=\mathbb{Q},r=2$, let us consider the principal ideals $\mathfrak{a},{\mathfrak{p}}_1,{\mathfrak{p}}_2$ in $k\left[y\right]$ generated, respectively, by $P=y^3-1,P_1=y-1,P_2=y^2+y+1$. Denoting by η the residue of y in $A=k\left[y\right]/\mathfrak{a}$, we have $\mathfrak{a}={\mathfrak{p}}_1\cap {\mathfrak{p}}_2$, and thus, $A\simeq (k\left[y\right]/{\mathfrak{p}}_1)\oplus (k\left[y\right]/{\mathfrak{p}}_2)$, thanks to the Bezout identity:

$$-{\displaystyle \frac{1}{3}}(y+2)(y-1)+{\displaystyle \frac{1}{3}}((y^2+y+1)=1$$

Introducing the complex imaginary quantity i with $i^2+1=0$ and the complex cube root of unity $j=(-1+i\sqrt{3})/2$, we may set $A\simeq (\mathbb{Q}.1)\oplus (\mathbb{Q}.1+\mathbb{Q}.j)\simeq \mathbb{Q}\oplus \mathbb{Q}\left(j\right)$ with $j^2+j+1=0\Rightarrow j^3=1$. Accordingly, any element of A may be written $a=(\lambda ,\mu +\nu j)$ with $\lambda ,\mu ,\nu \in \mathbb{Q}$.

Recapitulating the results obtained so far, we have two procedures for computing the tensor product $K{\otimes }_{k}L$ of two fields that contain a field k.

(1)

Assuming, for simplicity, that the extension $K/k$ is finitely generated, we may exhibit a finite transcendence basis $\left(s\right)$ of $K/k$ in such a way that K is algebraic over $k\left(s\right)$, while $k\left(s\right)$ is regular over k. This is the method that has been used in Theorem 1.

(2)

We may also introduce the algebraic closure $K_0$ of k in K in such a way that $K_0$ is algebraic over k and K is regular over $K_0$. In this case, we have $K{\otimes }_{k}L=K{\otimes }_{K_0}\left(K_0{\otimes }_{k}L\right)$. According to the previous Theorem, $K_0{\otimes }_{k}L$ is isomorphic to a direct sum of fields, say $M_0\oplus \dots \oplus M_r$, with each $M_i$ containing both $K_0$ and L. Thus, $K{\otimes }_{k}L\simeq K{\otimes }_{K_0}(M_1\oplus \dots \oplus M_r)\simeq \left(K{\otimes }_{K_0}{M}_1\right)\oplus \dots \oplus \left(K{\otimes }_{K_0}{M}_r\right)$. Because $K/K_0$ is a regular extension, then each $K{\otimes }_{K_0}{M}_i$ is an integral domain, and each $Q\left(K{\otimes }_{K_0}{M}_i\right)$ is a field.

Coming back to the classical Galois theory, let us consider an algebraic extension $L/K$ defined by $L=K\left[y\right]/\left(P\right)$, where the polynomial P is irreducible over K, and the principal ideal $\left(P\right)$ is, thus, prime in $K\left[y\right]$. We recall that $L/K$ is a Galois extension if L is a splitting field of P; that is, $\left(P\right)$ decomposes over L into the maximal principal ideals $(y-{\eta }_i)$ in $L\left[y\right]$ in which the ${\eta }_i\in L$ are the roots of P. As $L\left[y\right]/(y-{\eta }_i)\simeq L,\forall i=1,\dots ,\mid L/K\mid$, we obtain the following (compare with existing textbooks!):

Proposition 6.

$L/K$ is a Galois extension if and only if $L{\otimes }_{K}L\simeq L\oplus \dots \oplus L$ with $\mid L/K\mid$ terms or, equivalently, $L{\otimes }_{K}L\simeq L{\otimes }_{\mathbb{Q}}(\mathbb{Q}\oplus \dots \oplus \mathbb{Q})$ with $\mid L/K\mid$ terms, as $\mathbb{Q}\subseteq K\subset L$.

Definition 6.

If $L/K$ is a finite algebraic extension, we shall denote by $split(L/K)$ a (care) smallest Galois extension of K containing L, which is defined up to an isomorphism over L.

Example 8.

If $P=y^3-2\Rightarrow K=\mathbb{Q}\subset L=\mathbb{Q}\left(\sqrt[3]{2}\right)\Rightarrow split(L/K)=L\left(j\right)$ with an imaginary quantity $j=(-1+i\sqrt{3})/2$ and $M=K\left(j\right)=K\left[z\right]/(z^2+z+1)$, we see the following diagram:

$$\begin{array}{ccc}L=\mathbb{Q}\left(\sqrt[3]{2}\right)& \stackrel{2}{⟶}& split(L/K)=(L,M)\\ 3\uparrow & \nearrow 6& \uparrow 3\\ K=\mathbb{Q}& \underset{2}{⟶}& \mathbb{Q}\left(j\right)=M\end{array}$$

We notice that L and M are linearly disjoint over K in $split(L/K)=(L,M)$.

In contrast, by setting $\eta =\sqrt[3]{2}$, $L=K\left(\eta \right)$ and $L^{\prime }=K\left(j\eta \right)$, then L and $L^{\prime }$ are not linearly disjoint over K in $split(L/K)=(L,L^{\prime })$ because $(1,\eta ,{\eta }^2)$ is a basis of L over K while $(1,j\eta ,{\left(j\eta \right)}^2)$ is a basis of $L^{\prime }$ over K; however, we have the linear relation ${\left(j\eta \right)}^2\times 1+\left(j\eta \right)\times \eta +1\times {\eta }^2=0$ even though $L\cap L^{\prime }=K$. As η is a root of $y^3-2=0$, while ${\eta }^{\prime }=j\eta$ is a root of ${\left(y^{\prime }\right)}^3-2=0$ and ${\eta }^{\prime }-\eta \ne 0$, we get ${\left({\eta }^{\prime }\right)}^2+\eta {\eta }^{\prime }+{\eta }^2=0$ because $j^2+j+1=0$.

More generally, if $\eta$ is a generic zero of a prime ideal $\mathfrak{p}$ such that $L=K\left(\eta \right)=Q\left(K\right[y]/\mathfrak{p})$, then the perfect ideal $M\mathfrak{p}\subset M\left[y\right]$ is a finite intersection ${\mathfrak{p}}_1\cap \dots \cap {\mathfrak{p}}_r$ of prime ideals, and we may define $Q\left(L{\otimes }_{K}M\right)=Q(M\left[y\right]/M\mathfrak{p})$ as a direct sum of fields. The situation of $L{\otimes }_{K}L$ is slightly different, as we saw it is a reduced Artinian ring having only a finite number of prime ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$ that are also maximal with zero intersection, and we have the direct sum of fields:

$$Q\left(L{\otimes }_{K}L\right)\simeq Q(L{\otimes }_{K}L/{\mathfrak{m}}_1)\oplus \dots \oplus Q(L{\otimes }_{K}L/{\mathfrak{m}}_r)$$

As we do not need to take the full rings of quotients, using the commutative diagram, we have

$$\begin{array}{ccccccccc}& & & & 0& & 0& & \\ & & & & \downarrow & & \downarrow & & \\ & & 0& \to & L& =& L& \to & 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& {\mathfrak{m}}_i& ⟶& L{\otimes }_{K}L& ⟶& L{\otimes }_{K}L/{\mathfrak{m}}_i& ⟶& 0\end{array}$$

We shall call ${\eta }_i$ the image of $\eta$ under the composite monomorphism:

$$L\to 1\otimes L\to L{\otimes }_{K}L\to L{\otimes }_{K}L/{\mathfrak{m}}_i$$

setting $L_i=K\left({\eta }_i\right)$ and identifying $\eta$ with its image under the composite monomorphism:

$$L\to L\otimes 1\to L{\otimes }_{K}L\to L{\otimes }_{K}L/{\mathfrak{m}}_i$$

In actual practice, if P is the minimal unitary polynomial in $K\left[y\right]$ of a primitive element of $L/K$, we may denote by $\overline{P}$ the image of P under the isomorphism $K\left[y\right]\to K\left[\overline{y}\right]:y\to \overline{y}$ to obtain $Q\left(L{\otimes }_{K}L\right)\simeq Q(K[y,\overline{y}]/(P,\overline{P}))$. As $\overline{P}-P$ is divisible by $(\overline{y}-y)$, we may label the ${\mathfrak{m}}_i$ in such a way that ${\eta }_1=\eta \Rightarrow L_1=K\left({\eta }_1\right)=K\left(\eta \right)=L$.

We obtain $Q(L{\otimes }_{K}L/{\mathfrak{m}}_i)=K(\eta ,{\eta }_i)=L\left({\eta }_i\right)=L_i\left(\eta \right)=(L,L_i)=N_i$ and the following useful formula:

$${\Sigma }_i\mid (L,L_i)/L\mid =\mid (L,L_1)/L\mid +\dots +\mid (L,L_r)/L\mid =\mid L/K\mid$$

We finally define a finite family ${\sigma }_1,\dots ,{\sigma }_r\in iso(L/K)$ with ${\sigma }_1=i{d}_L$ by using the formula ${\sigma }_i\left(\eta \right]={\eta }_i$ in such a way that $\sigma \in iso(L/K)\Rightarrow K(\sigma \left(\eta \right)\simeq L_i$ for some $i=1,\dots ,r$. This definition only depends on the finite extension $L/K$.

Definition 7.

We shall say that an isomorphism σ is conjugate to ${\sigma }_i$ if $\sigma \left(\eta \right)$ and ${\eta }_i$ are roots of the same minimum polynomial over L. When dealing with finite extensions, any $\sigma \in iso(L/K)$ is the conjugate of one and only one ${\sigma }_i$, as already defined. For this reason, we shall say that each isomorphism ${\sigma }_i$ is an isolated isomorphism.

Lemma 2.

One has $inv({\sigma }_1,\dots ,{\sigma }_r)=K$.

Proof. 

Let $\eta$ be a primitive element of $L/K$, with a minimum unitary polynomial P of degree m. Accordingly, every element of L can be written as a polynomial in $\eta$ of degree $\le (m-1)$. Thus, let $\zeta =a_1{\eta }^{m-1}+\dots +a_m$ with $a_1,\dots ,a_m\in K$ be such an element satisfying ${\sigma }_i\left(\zeta \right)=\zeta ,\forall i=1,\dots ,r$. As we are dealing with principal ideals, we have the prime decomposition $L\left(P\right)=\left(P_1\right)\cap \dots \cap \left(P_r\right)$. Now, the polynomial $R\left(y\right)=a_1{y}^{m-1}+\dots +a_{m-1}y-(a_1{\eta }^{m-1}+\dots +a_{m-1}\eta )\in L\left[y\right]$ is such that $R\left({\eta }_1\right)=\dots =R\left({\eta }_r\right)=0$ by assumption. According to the Hilbert theorem, we must have $R\in L\left(P\right)$, and this is impossible unless $a_1=\dots =a_{m-1}=0$, a result leading to $\zeta =a_m\in K$. □

Definition 8.

We say that $L/K$ is an automorphic extension if a model variety Σ of $L/K$ is a Principal Homogeneous Space (PHS) for a finite algebraic group $\Gamma =\Gamma (L/K)$ defined over $k\subset K$ in such a way that each component of $k[\Gamma ]$ is linearly disjoint from L over k. We have the fundamental isomorphism $Q\left(L{\otimes }_{K}L\right)\simeq Q\left(L{\otimes }_{k}k[\Gamma ]\right)$, but the full rings of the quotient may not be needed. For simplicity, we shall suppose that $L/k$ is a regular extension to be more coherent with the point of view adopted by Vessiot, and we shall say that $L/K$ is regular over k.

Example 9.

If we consider the finite extension $L/K$ with $K=\mathbb{Q}\subset L=\mathbb{Q}\left(\eta \right)$, with $\eta =\sqrt[8]{2}$, then the (real) generic zero of the underlying irreducible equation is $y^8-2=0$; thus, we have $\sqrt{2}={\eta }^4\subset L$, and we successively obtain the following (see any textbook for cyclotomic fields):

$$\begin{array}{ccc}\hfill {\overline{y}}^8-y^8& =& (\overline{y}-y)(\overline{y}+y)({\overline{y}}^2+y^2)({\overline{y}}^4+y^4)\hfill \\ & =& (\overline{y}-y)(\overline{y}+y)(\overline{y}-iy)(\overline{y}+iy)(\overline{y}-{\displaystyle \frac{1+i}{\sqrt{2}}}y)(\overline{y}+{\displaystyle \frac{1+i}{\sqrt{2}}}y)(\overline{y}-{\displaystyle \frac{1-i}{\sqrt{2}}}y)(\overline{y}+{\displaystyle \frac{1-i}{\sqrt{2}}}y)\hfill \end{array}$$

Setting $\alpha =(1+i)/\sqrt{2}\Rightarrow {\alpha }^2=i\Rightarrow \sqrt{2}=\alpha +{\displaystyle \frac{1}{\alpha }}$, we notice that $\mathbb{Q}\left(\alpha \right)=Q(\mathbb{Q}\left[a\right]/(a^4+1))$ and L are not linearly disjoint in $split(L/K)=L\left(\alpha \right)$ over $\mathbb{Q}$ because we have $1\times ({\alpha }^2+1)-\sqrt{2}\times \alpha =0$ for bases $(1,\eta ,{\eta }^2,\dots ,{\eta }^4=\sqrt{2},\dots ,{\eta }^7)$, and we have $L/K$ and $(1,\alpha ,{\alpha }^2,{\alpha }^3)$ for $\mathbb{Q}\left(\alpha \right)/\mathbb{Q}$.

In contrast, with $k=\mathbb{Q}\subset K=k\left(y^8\right)\subset L=k\left(y\right)$, $\mathbb{Q}\left(\alpha \right)$ and L are linearly disjoint in $L\left(\alpha \right)$ over k. In the present situation, the group of invariance is the cyclic group $\Gamma$ with generator $\overline{y}=ay$ and $k[\Gamma ]=k\left[a\right]/(a^8-1)=k\oplus k\oplus k\left[a\right]/(a^2+1)\oplus k\left[a\right]/(a^4+1)=M_1\oplus M_2\oplus M_3\oplus M_4$ with $M_1=M_2=k,M_3=k\left(i\right),M_4=k\left(\alpha \right)$.

We finally consider the irreducible equation $y^{12}-3=0$ over $\mathbb{Q}$ by using the fact that

$${\overline{y}}^{12}-y^{12}=(\overline{y}-y)(\overline{y}+y)({\overline{y}}^2+y\overline{y}+y^2)({\overline{y}}^2-y\overline{y}+y^2)({\overline{y}}^2+y^2)({\overline{y}}^4-y^2{\overline{y}}^2+y^4)$$

It follows that $\mathbb{Q}\left(\eta \right)$ with $\eta =\sqrt[12]{3}$ is linearly disjoint over both $M_3=\mathbb{Q}\left(j\right)$ and $M_5=\mathbb{Q}\left(i\right)$ or even $M_6=\mathbb{Q}\left(ij\right)$ but not over $\mathbb{Q}(i,j)$, as this later field contains ${\eta }^6=\sqrt{3}=(1+2j)/i$.

Developing the tensor products that appear in the fundamental isomorphism, we obtain the following using linear disjointness:

Proposition 7.

One has $Q\left(L{\otimes }_{K}L\right)/{\mathfrak{m}}_i=(L,L_i)=(L,M_i)=Q\left(L{\otimes }_k{M}_i\right)$, $\forall i=1,\dots ,r$:

$$\begin{array}{ccc}\hfill Q\left(L{\otimes }_{K}L\right)& \simeq & (L,L_1)\oplus \dots \oplus (L,L_r)\hfill \\ & \simeq & Q\left(L{\otimes }_k{M}_1\right)\oplus \dots \oplus Q\left(L{\otimes }_k{M}_r\right)\hfill \\ & \simeq & Q\left(L{\otimes }_k(M_1\oplus \dots \oplus M_r)\right)\hfill \\ & \simeq & Q\left(L{\otimes }_{k}k[\Gamma ]\right)\hfill \end{array}$$

and the commutative diagram is

$$\begin{array}{ccccc}L& ⟶& N_i& ⟶& split(L/K)\\ \uparrow & & \uparrow & & \uparrow \\ K& ⟶& (K,M_i)& ⟶& (K,M)\\ \uparrow & & \uparrow & & \uparrow \\ k& ⟶& M_i& ⟶& M\end{array}$$

in which $M=(M_1,\dots ,M_r)$ and $split(L/K)=(L,M)=(L_1,L_2,\dots ,L_r)$ because $L_1=L$.

From now on, to simplify the proofs while showing the usefulness of these methods, as we shall only use fields, we shall suppress the rings of quotients $Q\left(\right)$.

Proposition 8.

$k[\Gamma ]\subset L{\otimes }_{K}L$ is the ring of polynomial functions on an algebraic finite group $\Gamma =\Gamma (L/K)$ defined over k and has, thus, an induced structure of Hopf algebra because $k\subset k[\Gamma ]$.

Proof. 

Let us introduce the two monomorphisms:

$$\left\{\begin{array}{ccc}\tilde{\alpha }\hfill & :\hfill & L⟶L{\otimes }_{K}L:a\to a\otimes 1\hfill \\ \tilde{\beta }\hfill & :\hfill & L⟶L{\otimes }_{K}L:b\to 1\otimes b\hfill \end{array}\right.$$

called, respectively, source inclusion and target inclusion, with the exact double arrow sequence:

$$0⟶K⟶L\underset{\tilde{\beta }}{\stackrel{\tilde{\alpha }}{\rightrightarrows }}L{\otimes }_{K}L$$

Now, we have the following isomorphisms:

$$L{\otimes }_{K}L{\otimes }_{K}L\simeq L{\otimes }_{K}L{\otimes }_{k}k[\Gamma ]\simeq L{\otimes }_{k}k[\Gamma ]{\otimes }_{k}k[\Gamma ]$$

and the top row of the commutative diagram

$$\begin{array}{ccccc}0& ⟶& L{\otimes }_{k}k[\Gamma ]& ⟶& L{\otimes }_{k}k[\Gamma ]{\otimes }_{k}k[\Gamma ]\\ & & \downarrow & & \downarrow \\ 0& ⟶& L{\otimes }_{K}L& ⟶& L{\otimes }_{K}L{\otimes }_{K}L\\ & & & & \\ & & a\otimes b& ⟶& a\otimes 1\otimes b\end{array}$$

This induces the diagonal comorphism $\widetilde{ϵ}:k[\Gamma ]\to k[\Gamma ]{\otimes }_{k}k[\Gamma ]$ by linear disjointness over k. In actual practice, we have, for example, $\overline{y}=ay,\overline{\overline{y}}=b\overline{y}\Rightarrow \overline{\overline{y}}=bay$ for the multiplicative group; that is, $a={\displaystyle \frac{\overline{y}}{y}}={\displaystyle \frac{1}{y}}\otimes y\otimes 1,b={\displaystyle \frac{\overline{\overline{y}}}{\overline{y}}}=1\otimes {\displaystyle \frac{1}{y}}\otimes y\Rightarrow ba={\displaystyle \frac{\overline{\overline{y}}}{\overline{y}}}={\displaystyle \frac{1}{y}}\otimes 1\otimes y$ (compare with [13], Th 1, p. 97).

Let us study this monomorphism in more detail. When substituting, we get

$$k[\Gamma ]={\oplus }_i{M}_i\to k[\Gamma ]{\otimes }_{k}k[\Gamma ]=\left({\oplus }_i{M}_i\right){\otimes }_k\left({\oplus }_j{M}_j\right)={\oplus }_{i,j}\left(M_i{\otimes }_k{M}_j\right)$$

Then, the top row of the following commutative diagram:

$$\begin{array}{ccccc}L{\otimes }_{k}k[\Gamma ]& \to & L& \to & 0\hfill \\ \downarrow & & \Vert & & \\ L{\otimes }_{K}L& \to & L& \to & 0\hfill \\ & & & & \\ a\otimes b& \to & ab& \end{array}$$

induces the augmentation comorphism $\widetilde{id}:k[\Gamma ]\to k\to k[\Gamma ]$ by linear disjointness over k. In actual practice, we have $\overline{y}=ay\Rightarrow a={\displaystyle \frac{\overline{y}}{y}}={\displaystyle \frac{1}{y}}\otimes y\Rightarrow {\displaystyle \frac{1}{y}}\otimes y\to 1$ for the identity $\overline{y}=1y=y$.

Finally, the top row of the following commutative diagram:

$$\begin{array}{ccccc}L{\otimes }_{k}k[\Gamma ]& \to & L{\otimes }_{k}k[\Gamma ]& \to & 0\hfill \\ \downarrow & & \downarrow & & \\ L{\otimes }_{K}L& \to & L{\otimes }_{K}L& \to & 0\hfill \\ & & & & \\ a\otimes b& \to & b\otimes a& \end{array}$$

induces the antipode comorphism $\tilde{\iota }:k[\Gamma ]\to k[\Gamma ]$ by linear disjointness over k, which sends each $M_i$ to itself, as can be easily seen in each preceding example. In actual practice, we have $\overline{y}=ay\Rightarrow a={\displaystyle \frac{\overline{y}}{y}}={\displaystyle \frac{1}{y}}\otimes y\Rightarrow a^{-1}={\displaystyle \frac{1}{a}}={\displaystyle \frac{y}{\overline{y}}}=y\otimes {\displaystyle \frac{1}{y}}$. □

Theorem 3.

If $L/K$ is an automorphic extension regular over k for a group $\Gamma$, and if $K\subset K^{\prime }\subset L$ is an intermediate field, then $L/K^{\prime }$ is an automorphic extension regular over k for a subgroup ${\Gamma }^{\prime }\subset \Gamma$.

Proof. 

As $iso(L/K^{\prime })\subset iso(L/K)$, we may use convenient labelling such that the isomorphisms ${\sigma }_1,\dots ,{\sigma }_s\in iso(L/K^{\prime })$ constructed as before are among the isomorphisms ${\sigma }_1,\dots ,{\sigma }_r\in iso(L/K)$ with $s<r$ in such a way that ${\sigma }_1=i{d}_L$. In actual practice, if $\mathfrak{r}$ is the ideal of $L{\otimes }_{K}L$ generated by all the elements of the form $a\otimes 1-1\otimes a$ with $a\in K^{\prime }$, then $L{\otimes }_{K}L/\mathfrak{r}\simeq L{\otimes }_{K^{\prime }}L$. Now, if ${\mathfrak{m}}_i^{\prime }$ is a prime ideal of $L{\otimes }_{K^{\prime }}L$, the inverse image of ${\mathfrak{m}}_i^{\prime }$ under the canonical epimorphism $L{\otimes }_{K}L\to L{\otimes }_{K^{\prime }}L$ is a prime ideal of $L{\otimes }_{K}L$ and must, therefore, be equal to some ${\mathfrak{m}}_i$. Using a chase in the following commutative and exact diagram:

$$\begin{array}{ccccccccc}& & 0& & 0& & & & \\ & & \downarrow & & \downarrow & & & & \\ 0& ⟶& \mathfrak{r}& =& \mathfrak{r}& ⟶& 0& & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& {\mathfrak{m}}_i& ⟶& L{\otimes }_{K}L& ⟶& \left(L{\otimes }_{K}L\right)/{\mathfrak{m}}_i& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& {\mathfrak{m}}_i^{\prime }& ⟶& L{\otimes }_{K^{\prime }}L& ⟶& \left(L{\otimes }_{K^{\prime }}L\right)/{\mathfrak{m}}_i^{\prime }& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0& & 0& & 0& \end{array}$$

we obtain an isomorphism $L{\otimes }_{K^{\prime }}L/{\mathfrak{m}}_i^{\prime }\simeq L{\otimes }_{K}L/{\mathfrak{m}}_i=N_i$.

Now, according to Proposition 7, we have $N_i=Q\left(L{\otimes }_k{M}_i\right)=(L,M_i)\Rightarrow M_i=N_i\cap M$. In addition, we have $s<r\to split(L/K^{\prime })\subseteq split(L/K)$ in the following commutative diagram:

$$\begin{array}{ccccccc}L& ⟶& N_i& ⟶& split(L/K^{\prime })& ⟶& split(L/K)\\ \uparrow & & \uparrow & & \uparrow & & \uparrow \\ K^{\prime }& ⟶& (K^{\prime },M_i)& ⟶& (K^{\prime },M^{\prime })& ⟶& (K^{\prime },M)\\ \uparrow & & \uparrow & & \uparrow & & \uparrow \\ K& ⟶& (K,M_i)& ⟶& \uparrow & ⟶& (K,M)\\ \uparrow & & \uparrow & & \uparrow & & \uparrow \\ k& ⟶& M_i& ⟶& M^{\prime }& ⟶& M\end{array}$$

with $M^{\prime }=(M_1,\dots ,M_s)\subset (M_1,\dots ,M_r)=M$, and still, L is linearly disjoint over $M^{\prime }$ in $split(L/K^{\prime })$.

It remains to be proven that ${\Gamma }^{\prime }$ can be constructed like $\Gamma$ through its Hodge algebra, as in [13] (Lemma 1, p. 93 and Lemma 5, p. 101). For this, by introducing the ideal $\mathfrak{s}=\mathfrak{r}\cap k[\Gamma ]\subset L{\otimes }_{k}L$ in the following commutative and exact diagram:

$$\begin{array}{ccccc}& & 0& & 0\\ & & \downarrow & & \downarrow \\ \hfill 0& ⟶& \mathfrak{s}& ⟶& \mathfrak{r}\\ & & \downarrow & & \downarrow \\ \hfill 0& ⟶& k[\Gamma ]& ⟶& L{\otimes }_{K}L\\ & & \downarrow & & \downarrow \\ \hfill 0& ⟶& k\left[{\Gamma }^{\prime }\right]& ⟶& L{\otimes }_{K^{\prime }}L\\ & & \downarrow & & \downarrow \\ & & 0& & 0\end{array}$$

we obtain the specialization epimorphism $k[\Gamma ]\to k\left[{\Gamma }^{\prime }\right]\to 0$. If ${\mathfrak{n}}_i={\mathfrak{m}}_i\cap k[\Gamma ]$ and ${\mathfrak{n}}_i^{\prime }={\mathfrak{m}}_i^{\prime }\cap k\left[{\Gamma }^{\prime }\right]$, then ${\mathfrak{n}}_i$ is the inverse image of ${\mathfrak{n}}_i^{\prime }$ according to this epimorphism. A chase in the following commutative and exact diagram:

$$\begin{array}{ccccccccc}& & 0& & 0& & & & \\ & & \downarrow & & \downarrow & & & & \\ 0& ⟶& \mathfrak{s}& =& \mathfrak{s}& ⟶& 0& & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& {\mathfrak{n}}_i& ⟶& k[\Gamma ]& ⟶& k[\Gamma ]/{\mathfrak{n}}_i& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0& ⟶& {\mathfrak{n}}_i^{\prime }& ⟶& k\left[{\Gamma }^{\prime }\right]& ⟶& k\left[{\Gamma }^{\prime }\right]/{\mathfrak{n}}_i^{\prime }& ⟶& 0\\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0& & 0& & 0& \end{array}$$

finally proves the isomorphism $k\left[{\Gamma }^{\prime }\right]/{\mathfrak{n}}_i^{\prime }\simeq k[\Gamma ]/{\mathfrak{n}}_i=M_i$, and thus, we have $k\left[{\Gamma }^{\prime }\right]\simeq M_1\oplus \dots \oplus M_s$. Therefore, we obtain

$$\begin{array}{ccc}L{\otimes }_{k}k\left[{\Gamma }^{\prime }\right]& \simeq & L{\otimes }_k(M_1\oplus \dots \oplus M_s)\hfill \\ & \simeq & \left(L{\otimes }_k{M}_1\right)\oplus \dots \oplus \left(L{\otimes }_k{M}_s\right)\hfill \\ & \simeq & N_1\oplus \dots \oplus N_s\hfill \\ & \simeq & L{\otimes }_{K^{\prime }}L\hfill \end{array}$$

and the following commutative and exact diagram:

$$\begin{array}{ccccccc}0& ⟶& L{\otimes }_{k}k[\Gamma ]& ⟶& L{\otimes }_{K}L& ⟶& 0\\ & & \downarrow & & \downarrow & & \\ 0& ⟶& L{\otimes }_{k}k\left[{\Gamma }^{\prime }\right]& ⟶& L{\otimes }_{K^{\prime }}L& ⟶& 0\\ & & \downarrow & & \downarrow & & \\ & & 0& & 0& \end{array}$$

proving that $L/K^{\prime }$ is an automorphic extension for ${\Gamma }^{\prime }=\Gamma (L/K^{\prime })\subset \Gamma (L/K)=\Gamma$. □

Let us now consider an automorphic extension $L/K$ regular over k for an algebraic finite group $\Gamma =\Gamma (L/K)$ defined over k. We have proved that, if $K\subset K^{\prime }\subset L$ for an intermediate field $K^{\prime }$, then $L/K^{\prime }$ is an automorphic extension for an algebraic subgroup ${\Gamma }^{\prime }=\Gamma (L/K^{\prime })\subset \Gamma$ defined over k. Thus, the remaining problem is to study the normality condition ${\Gamma }^{\prime }⊲\Gamma$ by finding a criterion involving only the three field extensions $K,K^{\prime },L$ of k; however, such a result is not intuitive at all.

Definition 9.

The composite translation comorphism $\tilde{\tau }:L\stackrel{\tilde{\beta }}{⟶}L{\otimes }_{K}L⟶L{\otimes }_{k}k[\Gamma ]$ is the comorphism of a rational action of $\Gamma$ on a model variety Σ of $L/K$.

Lemma 3.

One has ${\Gamma }^{\prime }⊲\Gamma$ if and only if the action of $\Gamma$ on $L/K$ over k induces an action of $\Gamma$ on $K^{\prime }/K$ over k according to the following commutative and exact diagram for the translation comorphism $\tilde{\tau }$:

$$\begin{array}{ccccc}0& & 0& & 0\\ \downarrow & & \downarrow & & \downarrow \\ K^{\prime }& \stackrel{\tilde{\tau }}{⟶}& K^{\prime }{\otimes }_{k}k[\Gamma ]& & K(\Omega )=K^{\prime }\\ \downarrow & & \downarrow & & \downarrow \\ L& \stackrel{\tilde{\tau }}{⟶}& L{\otimes }_{k}k[\Gamma ]& & K(\Sigma )=L\end{array}$$

Proof. 

If ${\Gamma }^{\prime }⊲\Gamma$ and $\eta ,{\eta }^{\prime }$ are two generic points of $\Sigma$ such that ${\eta }^{\prime }=h\eta$ for a certain $h\in {\Gamma }^{\prime }$ and $g\in \Gamma$, then $\exists h^{\prime }\in {\Gamma }^{\prime }$ such that $g{\eta }^{\prime }=gh\eta =h^{\prime }g\eta$. Conversely, if the action of $\Gamma$ passes to the quotient on $\Omega =\Sigma /{\Gamma }^{\prime }$, then $g{\eta }^{\prime }=gh\eta =h^{\prime }g\eta$ for a certain $h^{\prime }\in {\Gamma }^{\prime }$, and thus, $gh=h^{\prime }g$ for any $g\in \Gamma$. This is because the action of $\Gamma$ is free; that is to say, ${\Gamma }^{\prime }⊲\Gamma$. We have the following picture:

$$\begin{array}{ccccc}& & \eta & \stackrel{g}{⟶}& g\eta \\ \Sigma & & h\downarrow & & \downarrow h^{\prime }\\ & & {\eta }^{\prime }& \stackrel{g}{⟶}& g{\eta }^{\prime }\\ & & ⋮& & ⋮\\ \Omega & & •& \stackrel{g}{⟶}& .•\end{array}.$$

It is finally sufficient to notice that $\Omega$ is a model variety for $K^{\prime }/K$.

In a more practical way, the chain of inclusions $K\subset K^{\prime }\subset L$ provides a chain of inclusions $K^{\prime }{\otimes }_K{K}^{\prime }\subset L{\otimes }_K{K}^{\prime }\subset L{\otimes }_{K}L$. Accordingly, any prime and, thus, maximal ideal ${\mathfrak{m}}_i^{\prime }\subset K^{\prime }{\otimes }_K{K}^{\prime }$ can be extended to a perfect ideal of $L{\otimes }_K{K}^{\prime }$ because $K^{\prime }\subset L$, which is an intersection of prime ideals in $L{\otimes }_K{K}^{\prime }$, and each such prime ideal can be similarly extended to a perfect ideal in $L{\otimes }_{K}L$. Hence, each ${\mathfrak{m}}_i^{″}$ can be extended to a certain prime and, thus, maximal ideal ${\mathfrak{m}}_i\subset L{\otimes }_{K}L$. It follows that each isolated isomorphism ${\sigma }_i^{\prime }\in iso(K^{\prime }/K)$ can be extended to an isolated isomorphism ${\sigma }_i\in iso(L/K)$. □

Definition 10.

If $L/K$ is an automorphic extension over a field k, an intermediate automorphic extension $K^{\prime }/K$ is said to be admissible if one has the following commutative and exact diagram of the reciprocal image for $G=\Gamma (K^{\prime }/K)$:

$$\begin{array}{ccccc}& & 0& & 0\\ & & \downarrow & & \downarrow \\ 0& ⟶& k\left[G\right]& ⟶& K^{\prime }{\otimes }_K{K}^{\prime }\\ & & \downarrow & & \downarrow \\ 0& ⟶& k[\Gamma ]& \to & L{\otimes }_{K}L\end{array}.$$

If $L/K$ is an automorphic extension over a field k for an algebraic group $\Gamma =\Gamma (L/K)$, we have proved that there exists a bijective dual correspondence between the algebraic subgroups of Γ defined over k and the intermediate fields between K and L. The following example will prove that $K^{\prime }/K$ may be an automorphic extension, even if ${\Gamma }^{\prime }$ is not normal in Γ when $K^{\prime }/K$ is not admissible, contrary to the classical Galois theory.

Example 10.

With $k=K=\mathbb{Q},K^{\prime }=\mathbb{Q}\left(\sqrt[3]{2}\right)=\mathbb{Q}\left(\eta \right),L=\mathbb{Q}\left(\sqrt[3]{2}\right),j)$ and with $j^2+j+1=0\Rightarrow j^3=1$, we already know that $\mid L/K\mid =\mid L/K^{\prime }\mid \times \mid K^{\prime }/K\mid =2\times 3=6$ and that $L/K$ is a Galois extension for $S_3$; thus, it is an automorphic extension, as in Example 4, with ${\omega }^1={\omega }^2=0,{\omega }^3=2$. We may define $\sigma \left(\eta \right)=j\eta ,{\sigma }^2\left(\eta \right)=j^2\eta ,\sigma \left(j\right)=j$ and $\tau \left(\eta \right)=\eta ,\tau \left(j\right)=j^2$; that is, $\Gamma =\{e,\sigma ,\tau \}$ has 3 generators but $\mid \Gamma \mid =6$ indeed, with $\Gamma =\{e,\sigma ,{\sigma }^2,\tau ,\sigma \tau ,\tau \sigma \}$ and with $\sigma \tau \ne \tau \sigma$, as $\tau \sigma \left(\eta \right)=j^2\eta ={\sigma }^2\tau \left(\eta \right)$ while $\sigma \tau \left(\eta \right)=j\eta$. Finally, we notice that ${\sigma }^3=e,{\tau }^2=e$. However, we also know that $K^{\prime }/K$ is not a Galois extension because ${\Gamma }^{\prime }=\{e,\tau \}$ is not normal in $\Gamma$, as we have ${\sigma }^{-1}\tau \sigma ={\sigma }^2\tau \sigma ={\sigma }^4\tau =\sigma \tau \notin {\Gamma }^{\prime }$. Finally, we have $k[\Gamma ]\simeq \mathbb{Q}\oplus \dots \oplus \mathbb{Q}$ with six terms, while $k\left[G\right]\simeq \mathbb{Q}\oplus \mathbb{Q}\left(j\right)$; that is, we have $K^{\prime }{\otimes }_K{K}^{\prime }\subset L{\otimes }_{K}L$ indeed, but $k\left[G\right]\notin k[\Gamma ]$. Indeed, we have $y^3-2=0,{\overline{y}}^3-2=0\Rightarrow (\overline{y}-y)({\overline{y}}^2+y\overline{y}+y^2)=0\Rightarrow \overline{y}=y,\overline{y}=jy$. In contrast, we choose $K^{\prime }=K\left(j\right)$, which is a Galois extension of K with $\mid K^{\prime }/K\mid =2$, both with $L/K^{\prime }$, which is a Galois extension for $A_3⊲S_3$ with $\mid L/K^{\prime }\mid =3$ and ${\Gamma }^{\prime }=\{e,\sigma ,{\sigma }^2\}$. Now, we have $K^{\prime }{\times }_K{K}^{\prime }$ defined by $z^2+z+1=0,{\overline{z}}^2+\overline{z}+1=0\Rightarrow (\overline{z}-z)(\overline{z}+z+1)$, a result leading to $k\left[G\right]\simeq \mathbb{Q}\oplus \mathbb{Q}\subset k[\Gamma ]$ because $\overline{z}=z=j$ or $\overline{z}=-(z+1)=-(j+1)=j^2$ or, equivalently, $\tau \left(j\right)=j,{\tau }^2\left(j\right)=j^2=-(j+1)$ for $G\simeq S_3/A_3=\{e,\tau \}$. Such an example explains why only admissible extensions $K^{\prime }/K$ must be considered when normality is involved. In this case, we have the following:

Theorem 4.

An intermediate field $K^{\prime }$ is an automorphic extension of K for a group G defined over k if and only if ${\Gamma }^{\prime }=\Gamma (L/K^{\prime })$ is a normal subgroup of Γ.

Proof. 

Composing with the morphism $\tilde{\beta }$, we obtain the following commutative composite diagram:

$$\begin{array}{ccccc}{K}^{\prime }& \stackrel{\tilde{\beta }}{⟶}& K^{\prime }{\otimes }_K{K}^{\prime }& ⟶& K^{\prime }{\otimes }_{k}k\left[G\right].\\ \downarrow & & \downarrow & & \downarrow \\ L& \stackrel{\tilde{\beta }}{⟶}& L{\otimes }_{K}L& ⟶& L{\otimes }_{k}k[\Gamma ]\end{array}$$

in which the vertical arrows are monomorphisms, and the inclusion $k\left[G\right]\subset k[\Gamma ]$ is induced by the inclusion $K^{\prime }{\otimes }_K{K}^{\prime }\subset L{\otimes }_{K}L$, while the specialization $k[\Gamma ]\to k\left[{\Gamma }^{\prime }\right]\to 0$ is induced by the specialization $L{\otimes }_{K}L\to L{\otimes }_{K^{\prime }}L\to 0$. According to the preceding lemma, we, therefore, obtain ${\Gamma }^{\prime }⊲\Gamma$ and $\Gamma /{\Gamma }^{\prime }\simeq G$.

Conversely, as $k[\Gamma ]\subset L{\otimes }_{K}L$, introducing the Hopf algebra $k\left[G\right]=\left(K^{\prime }{\otimes }_K{K}^{\prime }\right)\cap k[\Gamma ]\subset L{\otimes }_{K}L$ as in the last definition, we have the commutative composite diagram

$$\begin{array}{ccccc}{K}^{\prime }{\otimes }_{k}k\left[G\right]& ⟶& K^{\prime }{\otimes }_K{K}^{\prime }& ⟶& K^{\prime }{\otimes }_{k}k[\Gamma ]\\ \downarrow & & \downarrow & lemma& \downarrow \\ L{\otimes }_{k}k[\Gamma ]& ⟶& L{\otimes }_{K}L& ⟶& L{\otimes }_{k}k[\Gamma ]\end{array}$$

in which the three vertical arrows are monomorphisms.

By composition, the morphism $K^{\prime }{\otimes }_{k}k\left[G\right]\to K^{\prime }{\otimes }_K{K}^{\prime }$ is, thus, a monomorphism because $k\left[G\right]\subset k[\Gamma ]$. However, as $K^{\prime }\subset L$ and $k\left[G\right]\subset k[\Gamma ]$, the left diagram is just an inverse image, and the latter morphism is also an epimorphism because $L/K$ is an automorphic extension. Needless to say that, when $L/K,L/K^{\prime },K^{\prime }/K$ are Galois extensions, this well-known result of normality becomes evident because $L{\otimes }_{K}L$ is a direct sum of $\mid L/K\mid$ copies of L, $L{\otimes }_{K^{\prime }}L$ is a direct sum of $\mid L/K^{\prime }\mid$ copies of L while $K^{\prime }{\otimes }_K{K}^{\prime }$ is a direct sum of $\mid K^{\prime }/K\mid$ copies of $K^{\prime }$ with $\mid L/K\mid =\mid L/K^{\prime }\mid \times \mid K^{\prime }/K\mid$, a result leading to $\mid \Gamma \mid =\mid {\Gamma }^{\prime }\mid \times \mid G\mid$. □

Remark 1.

Though we have an epimorphism $k[\Gamma ]\to k\left[{\Gamma }^{\prime }\right]\to 0$ and a monomorphism $0\to k\left[G\right]\to k[\Gamma ]$, the later does not define the kernel of the previous one, as can be seen in the last example by using the commutative diagram

$$\begin{array}{ccc}{K}^{\prime }{\otimes }_K{K}^{\prime }& ⟶& K^{\prime }\\ \downarrow & & \downarrow \\ L{\otimes }_{K}L& ⟶& L{\otimes }_{K^{\prime }}L\end{array}$$

in which the left vertical arrow is a monomorphism, the upper arrow is the epimorphism $a\otimes b\to ab$ while the right vertical arrow is the monomorphism $a\to a\otimes 1=1\otimes a$. In any case, we have the recapitulating diagram

$$\begin{array}{ccccccccc}L& \to & L{\otimes }_{K}L& \simeq & N_1& \oplus & \dots & \oplus & N_r\\ \uparrow & & \uparrow & & \uparrow & & & & \uparrow \\ k& \to & k[\Gamma ]& \simeq & M_1& \oplus & \dots & \oplus & M_r\end{array}$$

in which we recall that $N_1=L$. We have, in particular, $r=6$ in the last example because $L/K$ is a Galois extension for $S_3$. We check at once that $\Gamma \left(K^{\prime }\right)\subset K^{\prime }$ when $K^{\prime }=K\left(j\right)$, as we have $(e\left(j\right)=j,\sigma \left(j\right)=j,\dots \tau \sigma \left(j\right)=j^2=-(j+1))$, and thus, ${\Gamma }^{\prime }⊲\Gamma$. In contrast, we have $\Gamma \left(K^{\prime }\right)\notin K^{\prime }$ when $K^{\prime }=K\left(\eta \right)$ because $\sigma \left(\eta \right)=j\eta$, and thus, ${\Gamma }^{\prime }$ is not normal in Γ.

3. Algebraic Tools

In all that follows, we shall consider unitary rings $A,B,\dots$ with elements $1,a,b,c,\dots$ and full rings of quotients $K=Q\left(A\right),L=Q\left(B\right),\dots$, which are fields when $A,B,\dots$ are integral domains; that is, they do not have divisors of 0, namely $a,b\in A,ab=0\Rightarrow a=0$ or $b=0$. When X is an algebraic set defined over a field k of characteristic zero, that is, $\mathbb{Q}\subset k$, we may introduce, as usual, the ring $A=k\left[X\right]$ of polynomial functions on X. In particular, if G is an algebraic group defined over k, we may introduce the ring $R=k\left[G\right]$ as an algebra over k. We recall the basic operations on G:

$$\left\{\begin{array}{cccccccc}ϵ& :& G\times G& \to & G& :& (a,b)\to ab& \left(\mathit{composition}\right)\hfill \\ \iota & :& G& \to & G& :& a\to a^{-1}& \left(\mathit{inverse}\right)\hfill \\ e& :& G& \to & G& :& a\to e& \left(\mathit{identity}\right)\hfill \end{array}\right.$$

with the following axioms and corresponding commutative diagrams:

$$a,b,c\in G\to \left(ab\right)c=a\left(bc\right)=abc\in G.$$

$$\begin{array}{ccccc}& G\times G\times G& \stackrel{(ϵ,{id}_G)}{⟶}& G\times G& \\ \hfill ({id}_G,ϵ)& \downarrow & & \downarrow & ϵ\hfill \\ & G\times G& \stackrel{ϵ}{⟶}& G\end{array}$$

$$a\in G\to a^{-1}\in G,a{a}^{-1}=a^{-1}a=e$$

$$\begin{array}{ccccc}& G& \stackrel{(\iota ,{id}_G)}{⟶}& G\times G& \\ \hfill ({id}_G,\iota )& \downarrow & \searrow e& \downarrow & ϵ\hfill \\ & G\times G& \stackrel{ϵ}{⟶}& G\end{array}$$

$$ea=ae=a$$

$$\begin{array}{ccccc}& G& \stackrel{(e,{id}_G)}{⟶}& G\times G& \\ \hfill ({id}_G,e)& \downarrow {id}_G& \searrow {id}_G& \downarrow & ϵ\hfill \\ & G\times G& \stackrel{ϵ}{⟶}& G\end{array}$$

where we have set $G\stackrel{e}{\to }G:a\to e$ for the map, sending any element $a\in G$ to $e\in G$.

When R is an algebra over k, we may consider the comorphisms of the morphisms appearing in the preceding diagrams, and such a ring R will be called a Hopf algebra over k. We may define

$$\left\{\begin{array}{ccccccc}\widetilde{ϵ}& :& R& \to & R{\otimes }_{k}R& :& \left(\mathit{diagonal}\right)\hfill \\ \tilde{\iota }& :& R& \to & R& :& \left(\mathit{antipode}\right)\hfill \\ \tilde{e}& :& R& \to & k\subset R& :& \left(\mathit{augmentation}\right)\hfill \end{array}\right.$$

with the following commutative diagrams:

$$\begin{array}{ccccc}& R& \stackrel{\widetilde{ϵ}}{⟶}& R{\otimes }_{k}R& \\ \hfill \widetilde{ϵ}& \downarrow & & \downarrow & {id}_R\otimes \widetilde{ϵ}\hfill \\ & R{\otimes }_{k}R& \stackrel{\widetilde{ϵ}\otimes {id}_R}{⟶}& R{\otimes }_{k}R{\otimes }_{k}R\end{array}$$

$$\begin{array}{ccccc}& R& \stackrel{\widetilde{ϵ}}{⟶}& R{\otimes }_{k}R& \\ \hfill \widetilde{ϵ}& \downarrow & \searrow \tilde{e}& \downarrow & ({id}_R,\tilde{\iota })\hfill \\ & R{\otimes }_{k}R& \underset{(\tilde{\iota },{id}_R)}{⟶}& R\end{array}$$

$$\begin{array}{ccccc}& R& \stackrel{\widetilde{ϵ}}{⟶}& R{\otimes }_{k}R& \\ \hfill \widetilde{ϵ}& \downarrow & \searrow {id}_R& \downarrow & ({id}_R,\tilde{e})\hfill \\ & R{\otimes }_{k}R& \underset{(\tilde{e},{id}_R)}{⟶}& R\end{array}$$

where we have set $(\tilde{\iota },{id}_R):R{\otimes }_{k}R\stackrel{\tilde{\iota }\otimes {id}_R}{⟶}R{\otimes }_{k}R\to R$, and the last morphism is $a\otimes b\to ab$.

These definitions are just those of the corresponding comorphisms when $R=k\left[G\right]$. However, the main difficulty, in general, is that one cannot find any generic solution to the defining finite Lie equations that could depend on certain arbitrary functions. The two examples of algebraic pseudogroups provided by the Pfaffian system ${\overline{y}}^{2}d{\overline{y}}^1=y^{2}d{y}^1$ or by the Schwarzian OD equation $({\displaystyle \frac{{\partial }^3\overline{y}}{{\partial y}^3}}/{\displaystyle \frac{\partial \overline{y}}{\partial y}})-{\displaystyle \frac{3}{2}}{({\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}/{\displaystyle \frac{\partial \overline{y}}{\partial y}})}^2=0$ are well known. The following key theorem provides the tricky differential geometric counterpart of the results in [13,23], which will be essential in the next sections.

Theorem 5.

If the manifold X is a PHS for G, that is, the graph of the action $X\times G\to X\times Y$ is an isomorphism when Y is a copy of X, the group parameters are constants on $X\times Y$ for the reciprocal distribution of the infinitesimal action of G on X. Of course, the simplest example is that of a Lie group G acting on itself, as we already observed.

Proof. 

In the purely algebraic framework, when counting the dimensions, we have $n=dim\left(X\right)=dim\left(G\right)=p$, and we may exhibit p functions $a=\phi (x,y)$ such that we have the n identities $y\equiv f(x,\phi (x,y\left)\right)$. If $\delta ={\xi }^i\left(x\right){\displaystyle \frac{\partial }{\partial x^i}}\in \Delta$ is a transformation commuting with all the infinitesimal generating transformations $\Theta =({\theta }_1,\dots ,{\theta }_p)$, we can extend each $\delta$ to $X\times Y$ by setting $\delta ={\xi }^i\left(x\right){\displaystyle \frac{\partial }{\partial x^i}}+{\xi }^k\left(y\right){\displaystyle \frac{\partial }{\partial y^k}}$. Applying such an extended $\delta$ to the previous identities, we obtain

$${\xi }^k\left(y\right)={\xi }^i\left(x\right){\displaystyle \frac{\partial f^k}{\partial x^i}}(x,a)+\left(\delta a^{\tau }\right){\displaystyle \frac{\partial f^k}{\partial a^{\tau }}}(x,a)$$

whenever $a=\phi (x,y)$. As $[\delta ,{\theta }_{\tau }]=0,\forall \tau =1,\dots ,p$, we have $\xi \left(y\right)=\xi \left(x\right){\displaystyle \frac{\partial f}{\partial x}}$. Moreover, as X is a PHS for G, we have $rk\left({\displaystyle \frac{\partial f}{\partial a}}\right)=n=p$, and thus, $\delta a^{\tau }=0$. Finally, when $\Phi$ is an invariant of G, that is, $\theta \Phi =0$, as $[\Delta ,\Theta ]=0$, we get $\theta (\delta \Phi )=\delta (\theta \Phi )=0$, and thus, $\delta \Phi$must be one invariant of $\Theta$.

In the purely algebraic framework, let X be an irreducible variety defined over the field K, and let G be an algebraic group defined over the field $k\subset K$. The group parameters must, therefore, be “constants” for $\Delta$; that is, $k\left[G\right]\subset cst\left(L{\otimes }_{K}L\right)\subset cst\left(Q\left(L{\otimes }_{K}L\right)\right)$. Accordingly, by setting $A=K\left[X\right]\Rightarrow L=Q\left(A\right)=K\left(X\right)$ and introducing the subfield $K=L^G\subset L$ invariant by G, we have, thus, an isomorphism $Q\left(L{\otimes }_{k}k\left[G\right]\right)\simeq Q\left(L{\otimes }_{K}L\right)$. In more detail, $L{\otimes }_{K}L$ is a direct sum of integral domains acted on separately by $\Delta$, while $Q\left(L{\otimes }_{K}L\right)$ is a direct sum of the fields. In the case of a Lie group acting on itself, one only has to use the fact that the left invariant distribution commutes with the right invariant distribution. □

Recapitulating these results while introducing the two injective comorphisms $L\to L{\otimes }_{K}L$, respectively, defined by $\tilde{\alpha }:a\to a\otimes 1$ and $\tilde{\beta }:a\to 1\otimes a$ by analogy with the source projection $\alpha :X\times Y\to X:(x,y)\to x$ and the target projection $\beta :X\times Y\to Y:(x,y)\to y$, we discover that $L{\otimes }_{K}L$ is the simplest possible cogroupoid. We indeed obtain the following (see Example 3):

• Diagonal: $k\left[G\right]\to k\left[G\right]{\otimes }_{k}k\left[G\right]$ is induced by

  $$L{\otimes }_{K}L\to \left(L{\otimes }_{K}L\right){\otimes }_L\left(L{\otimes }_{K}L\right)\simeq L{\otimes }_{K}L{\otimes }_{K}L:a\otimes b\to a\otimes 1\otimes b$$
• Antipode: $k\left[G\right]\to k\left[G\right]$ is induced by

  $$L{\otimes }_{K}L\to L{\otimes }_{K}L:a\otimes b\to b\otimes a$$
• Augmentation: $k\left[G\right]\to k\subset k\left[G\right]$ is induced by

  $$L{\otimes }_{K}L\to L:a\otimes b\to ab$$

if we identify $a\in L$ with $a\otimes 1$ and $b\in L$ with $1\otimes b$ in $L{\otimes }_{K}L$.

Example 11.

With $n=1,m=2,k=\mathbb{Q}$, if K is a differential field containing k, we may consider the differential automorphic extension $L/K$ defined by $L=K(y^1,y^2,y_x^1,y_x^2)$ in such a way that $d_x{y}^1=y_x^1,d_x{y}^2=y_x^2,d_x{y}_x^1=0,d_x{y}_x^2=0$. This amounts to considering the automorphic system $y_{xx}^1=0,y_{xx}^2=0$ under the standard action of $GL\left(2\right)$, which cannot be treated using the classical approach of the Picard–Vessiot theory introduced by Kolchin. In the present situation, we indeed have ${\overline{y}}^1=a{y}^1+b{y}^2,{\overline{y}}^2=c{y}^1+d{y}^2$, and we easily obtain the four functions φ with the following (see Example 18):

$$a=(y_x^2{\overline{y}}^1-y^2{\overline{y}}_x^1)/(y^1{y}_x^2-y^2{y}_x^1)$$

The four infinitesimal generators of the action are

$$\Theta =\{y^l{\displaystyle \frac{\partial }{\partial y^k}}+y_x^l{\displaystyle \frac{\partial }{\partial y_x^k}}\mid k,l=1,2\}$$

and the four infinitesimal generators of the reciprocal distribution are

$$\Delta =\{{\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y^k}},{\delta }_2=y_x^k{\displaystyle \frac{\partial }{\partial y^k}},{\delta }_3=y^k{\displaystyle \frac{\partial }{\partial y_x^k}},{\delta }_4=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}$$

such that we may extend to ${\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y^k}}+{\overline{y}}^k{\displaystyle \frac{\partial }{\partial {\overline{y}}^k}}$ and so on. We let the reader check that ${\delta }_{i}a=0,\forall i=1,\dots ,4$, a result that is not evident at all. Of course, $L/K$ is not a PV extension in the sense of Kolchin because $y_x^1,y_x^2\in L$ are “differential constants” that are not in K. Accordingly, if we consider the intermediate differential field $K^{\prime }=K\left(y_x^2\right)$ with the strict inclusions $K\subset K^{\prime }\subset L$, the subgroup of invariance preserving $y_x^2$ is defined by ${\overline{y}}_x^2=c{y}_x^1+d{y}_x^2$; that is, we must have $c=0,d=1$, and this group also preserves the intermediate differential field $K^{″}=K(y^2,y_x^2)$ with the strict inclusion $K^{\prime }\subset K^{″}$. It follows that we cannot have a Galois-type correspondence. Of course, such a result is coherent with the fact that $L/K^{\prime }$ is NOT a PV differential extension in the sense of Kolchin because L contains new differential constants, as we indeed have $y_x^1\in L,d_x{y}_x^1=0$. In contrast, if we adopt the point of view of BB, we have ${\delta }_1{y}_x^2=0,{\delta }_2{y}_x^2=0,{\delta }_3{y}_x^2=y^2,{\delta }_4{y}_x^2=y_x^2$, and $K^{\prime }$ is not stable under Δ. It is a pity that the work of BB has never been acknowledged by Kolchin, who had never even been able to use the tensor products of rings and fields. Any reader can check this directly by looking through all his books and papers. Additionally, in our personal opinion, and justified by private letters, we do believe that Kolchin never did read any paper of Vessiot, in particular, the ones we quote.

4. Differential Tools

Our purpose is now to use the fact that the Lie groups of transformations are just examples of Lie pseudogroups of transformations in such a way that $trd(L/K)=\infty$. However, when differentiating the group law $y=f(x,a)$ to obtain the prolongations $y_x={\partial }_{x}f(x,a),y_{xx}={\partial }_{xx}f(x,a)$, and so on, eliminating the parameters may be quite a hard task, as in the case of the projective group of transformations of the real line of the form $y=(ay+b)/(cy+d)$, which is generated by one translation $\overline{y}=y+b$, one dilatation $\overline{y}=ay$ and one elation ${\displaystyle \frac{1}{\overline{y}}}={\displaystyle \frac{1}{y}}+c$ with a basis of infinitesimal generators $\Theta =\{{\displaystyle \frac{\partial }{\partial x}},x{\displaystyle \frac{\partial }{\partial x}},{\displaystyle \frac{1}{2}}{x}^2{\displaystyle \frac{\partial }{\partial x}}$ }. Moreover, no classical method known for Lie groups can work for Lie pseudogroups because one cannot find generic solutions in most cases when $trd(L/K)=\infty$.

We start with a few basic technical results and formulas that are not well known, as they involve the Spencer operator. For simplicity, we shall deal with trivial fibered manifolds $\mathcal{E}=X\times Y$ such that $dim\left(X\right)=n,dim\left(Y\right)=m$ and local coordinates $(x^i,y^k)$ with $i=1,\dots ,n,k=1,\dots ,m$. The q-jet bundle $J_q\left(\mathcal{E}\right)$ of $\mathcal{E}$ will be a fibered manifold with local coordinates $(x^i,y_{\mu }^k)$ for a multi-index $\mu =({\mu }_1,\dots ,{\mu }_n)$ of length $0\le \mid \mu \mid ={\mu }_1+\dots +{\mu }_n\le q$; we shall set $\mu +1_i=({\mu }_1,\dots ,{\mu }_{i-1},{\mu }_i+1,{\mu }_{i+1},\dots ,{\mu }_n)$ or simply $(x,y_q)$ with projection $\pi$ to X. The tangent bundle $T\left(\mathcal{E}\right)$ may be described by means of local coordinates $(x,y;u,v)$, while the vertical bundle $V\left(\mathcal{E}\right)$ will be obtained by setting $u=0$ in the short exact sequence of vector bundles pulled back over $\mathcal{E}$ with successive local coordinates $(x,y;0,v),\left(\right(x,y;u,v),(x,y;u)$:

$$0\to V\left(\mathcal{E}\right)\to T\left(\mathcal{E}\right)\stackrel{T\left(\pi \right)}{⟶}T{\times }_X\mathcal{E}\to 0$$

Introducing the formal derivatives $d_i={\partial }_i+y_{\mu +1_i}^k{\displaystyle \frac{\partial }{\partial y_{\mu }^k}}$, we have the following (see [4,8,10,12,14] for details):

Lemma 4.

Prolongation of vertical vector fields:

$$\eta ={\eta }^k\left(y\right){\displaystyle \frac{\partial }{\partial y^k}}\to {\rho }_q\left(\eta \right)=d_{\mu }{\eta }^k{\displaystyle \frac{\partial }{\partial y_{\mu }^k}}$$

$$d_i{\eta }^k\left(y\right)={\displaystyle \frac{\partial {\eta }^k}{\partial y^r}}{y}_i^r,d_{ij}{\eta }^k\left(y\right)={\displaystyle \frac{\partial {\eta }^k}{\partial y^r}}{y}_{ij}^r+{\displaystyle \frac{{\partial }^2{\eta }^k}{\partial y^r\partial y^s}}{y}_i^r{y}_j^s,\dots .$$

By introducing a section ${\eta }_q$ of $J_q\left(T\left(Y\right)\right)$ over the target and replacing derivatives by sections, we get

$$♯\left({\eta }_q\right)={\eta }^k{\displaystyle \frac{\partial }{\partial y^k}}+{\eta }_r^k{y}_i^r{\displaystyle \frac{\partial }{\partial y_i^k}}+({\eta }_r^k{y}_{ij}^r+{\eta }_{rs}^k{y}_i^r{y}_j^s){\displaystyle \frac{\partial }{\partial y_{ij}^k}}+\dots$$

Lemma 5.

Prolongation of horizontal vector fields:

$$\xi ={\xi }^i\left(x\right){\partial }_i\to {\rho }_q\left(\xi \right)={\xi }^i{\partial }_i+{\zeta }_{\mu }^k{\displaystyle \frac{\partial }{\partial y_{\mu }^k}}$$

$${\zeta }_{\mu +1_i}^k=d_i{\zeta }_{\mu }^k-y_{\mu +1_r}^k{\partial }_i{\xi }^r\left(x\right)$$

$${\zeta }^k=0,{\zeta }_i^k=-y_r^k{\partial }_i{\xi }^r,{\zeta }_{ij}^k=-(y_r^k{\partial }_{ij}{\xi }^r+y_{rj}^k{\partial }_i{\xi }^r+y_{ri}^k{\partial }_j{\xi }^r),\dots .$$

By introducing a section ${\xi }_q$ of $J_q\left(T\right)$ over the source and replacing derivations by sections, we get

$$♭\left({\xi }_q\right)={\xi }^i\left(x\right){\partial }_i-y_r^k{\xi }_i^r\left(x\right){\displaystyle \frac{\partial }{\partial y_i^k}}-(y_r^k{\xi }_{ij}^r\left(x\right)+y_{rj}^k{\xi }_i^r\left(x\right)+y_{ri}^k{\xi }_j^r\left(x\right)){\displaystyle \frac{\partial }{\partial y_{ij}^k}}+\dots$$

Theorem 6.

There exists a bracket for sections of $J_q\left(T\right)$, generalizing the standard bracket of vector fields of T.

Proof. 

We recall that ${\left([\xi ,\eta ]\right)}^i={\xi }^r{\partial }_r{\eta }^i-{\eta }^r{\partial }_r{\xi }^i,\forall \xi ,\eta \in T$. Taking the q-derivation by applying the operator $j_q$, we obtain a bilinear combination of $j_{q+1}\left(\xi \right)$ and $j_{q+1}\left(\eta \right)$. We may thus define the so-called algebraic bracket $\{{\xi }_{q+1},{\eta }_{q+1}\}$ with a value in $J_q\left(T\right)$ and obtain on $J_q\left(T\right)$ the algebroid bracket where d is the Spencer operator ${\left(d{\xi }_{q+1}\right)}_{\mu ,i}^k\left(x\right)={\partial }_i{\xi }_{\mu }^k\left(x\right)-{\xi }_{\mu +1_i}^k\left(x\right)$ or simply $d{\xi }_{q+1}=j_1\left({\xi }_q\right)-{\xi }_{q+1}$:

$$[{\xi }_q,{\eta }_q]=\{{\xi }_{q+1},{\eta }_{q+1}\}+i\left(\xi \right)d{\eta }_{q+1}-i\left(\eta \right)d{\xi }_{q+1}$$

which does not depend any longer on the jets of strict order $q+1$ whenever ${\xi }_{q+1},{\eta }_{q+1}\in J_{q+1}\left(T\right)$ are over ${\xi }_q,{\eta }_q\in J_q\left(T\right)$. When ${\xi }_{q+1}=j_{q+1}\left(\xi \right)$ and ${\eta }_{q+1}=j_{q+1}\left(\eta \right)$, we have $d{\xi }_{q+1}=0,d{\eta }_{q+1}=0$, and thus, $[j_q\left(\xi \right),j_q\left(\eta \right)]=j_q\left([\xi ,\eta ]\right)$. Finally, it is not at all evident to verify the Jacobi identity:

$$[{\xi }_q,[{\eta }_q,{\zeta }_q]]+[{\eta }_q,[{\zeta }_q,{\xi }_q]]+[{\zeta }_q,[{\xi }_q,{\eta }_q]]=0,\forall {\xi }_q,{\eta }_q,{\zeta }_q\in J_q\left(T\right)$$

□

Definition 11.

A sub-bundle $R_q\subset J_q\left(T\right)$ is a Lie algebroid of order q if $[R_q,R_q]\subset R_q$, that is, $[{\xi }_q,{\eta }_q]\subset R_q,\forall {\xi }_q,{\eta }_q\in R_q$. We say that $R_q$ is transitive if the morphism $R_q\to T$ induced by the canonical epimorphism ${\pi }_0^q:J_q\left(T\right)\to T$ is also an epimorphism. We shall introduce the isotropy Lie algebra bundle $R_q^0$ by the short exact sequence $0\to R_q^0\to R_q\to T\to 0$. We have $[R_q^0,R_q^0]\subset R_q^0$ fiber by fiber, and an $R_q$-connection is a map ${\chi }_q:T\to R_q$ such that ${\pi }_0^q\circ {\chi }_q=i{d}_T$.

Corollary 2.

One has $[♯\left({\eta }_q\right),♯\left({\eta }_q^{\prime }\right]=♯\left([{\eta }_q,{\eta }_q^{\prime }]\right)$ over the target and $[♭\left({\xi }_q\right),♭\left({\xi }_q^{\prime }\right)]=♭\left([{\xi }_q,{\xi }_q^{\prime }]\right)$ over the source.

Corollary 3.

One has $[♭\left({\xi }_q\right),♯\left({\eta }_q\right)]=0,\forall {\xi }_q\in J_q\left(T\right)$ over the source and $\forall {\eta }_q\in J_q\left(T\left(Y\right)\right)$ over the target as a generalization of the fact that any source transformation commutes with any target transformation, namely, that $[\xi ,\eta ]=0,\forall \xi \in T=T\left(X\right),\forall \eta \in T\left(Y\right)$.

In actual practice, apart from [4,10,36,37,38], we do not know any reference for these results, which crucially depend on the use of the Spencer operator. We invite the reader to check the fact that a direct proof of these formulas is rather easy to obtain when $q=1$, but this becomes quite tricky when $q=2$. The same comment applies to the following two formulas, which are among the most challenging yet also the most useful ones.

Theorem 7.

For any function $\Phi (x,y_q)$, we have the formula

$$♯\left({\eta }_{q+1}\right)d_i\Phi =d_i(♯\left({\eta }_q\right)\Phi )-y_i^k\left(d_Y{\eta }_{q+1}\left({\displaystyle \frac{\partial }{\partial y^k}}\right)\right)\Phi$$

when ${\eta }_{q+1}$ projects onto ${\eta }_q$ over the target and, with a similar formula over the source:

$$♭\left({\xi }_{q+1}\right)d_i\Phi =d_i(♭\left({\xi }_q\right)\Phi )-{\xi }_i^r{d}_r\Phi -♭\left(d{\xi }_{q+1}\left({\partial }_i\right)\right)\Phi$$

Corollary 4.

If $\Phi =\Phi \left(y_q\right)$ is a differential invariant of strict order q and ${\eta }_{q+1}\in R_{q+1}\left(Y\right)$ over the target, then $d_Y{\eta }_{q+1}\in T^{*}\left(Y\right)\otimes R_q\left(Y\right)$ over the target, and $d_i\Phi$ is a differential invariant of strict order $q+1$.

Example 12.

To help the reader deal with sections instead of solutions, we consider the case $n=m=1$ with ${\xi }_2=(\xi ,{\xi }_x,{\xi }_{xx})$, and we have

$$♭\left({\xi }_2\right)=\xi {\partial }_x+0{\displaystyle \frac{\partial }{\partial y}}-y_x{\xi }_x{\displaystyle \frac{\partial }{\partial y_x}}-(y_x{\xi }_{xx}+2y_{xx}{\xi }_x){\displaystyle \frac{\partial }{\partial y_{xx}}}$$

We let the reader prove, as a tricky exercise, that

$$♭\left({\xi }_2\right)d_x\Phi =d_x(♭\left({\xi }_1\right)\varphi )-{\xi }_x{d}_x\varphi -♭\left(d{\xi }_2\left({\partial }_x\right)\right)\Phi$$

When ${\xi }_2=(0,-1,0)\Rightarrow d{\xi }_2=(1,0)$ and $\Phi =\Phi (y,y_x)$, then one has

$$(y_x{\displaystyle \frac{\partial }{\partial y_x}}+2y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}})(y_x{\displaystyle \frac{\partial \Phi }{\partial y}}+y_{xx}{\displaystyle \frac{\partial \Phi }{\partial y_x}})=d_x\left(y_x{\displaystyle \frac{\partial \Phi }{\partial y_x}}\right)+d_x\Phi$$

Remark 2.

If η is an infinitesimal generator of the action of a Lie group on Y, choosing ${\eta }_{q+1}=j_{q+1}\left(\eta \right)$ which projects onto ${\eta }_q=j_q\left(\eta \right)$ over the target, we obtain ${\rho }_{q+1}\left(\eta \right)d_i\Phi =d_i({\rho }_q\left(\eta \right)\Phi )$. It follows that $d_i$ commutes with the prolongations of target transformations, as we already checked in the examples.

Theorem 8.

We have the following formula for the bracket of sections of $J_{q+1}\left(T\right)$:

$$i\left(\zeta \right)d[{\xi }_{q+1},{\eta }_{q+1}]=[i\left(\zeta \right)d{\xi }_{q+1},{\eta }_{q+1}]+[{\xi }_q,i\left(\zeta \right)d{\eta }_{q+1}]$$

for any $\zeta \in T$ (see [4,12] for more details).

Using an induction starting with $r=1$, we obtain the following:

Corollary 5.

If $[R_q,R_q]\subset R_q$, then $[R_{q+r},R_{q+r}]\subset R_{q+r},\forall r\ge 0$, and the prolongations of a Lie algebroid are Lie algebroids even if $R_q\subset J_q\left(T\right)$ is not formally integrable. Moreover, if $R_{q+r}^{\left(s\right)}\subset R_{q+r}$ is the projection of $R_{q+r+s}$ in $R_{q+r}$, we have $[R_{q+r}^{\left(s\right)},R_{q+r}^{\left(s\right)}]\subset R_{q+r}^{\left(s\right)},\forall r,s\ge 0$. The case of the killing system for the Schwarzschild and Kerr metrics provides a good example, in general, of relativity, showing that the only important object for a metric is its group of invariance, which can be of a very low dimension (2 for the Kerr metric) [39,40].

Example 13.

When $m=n=2$, the algebraic Lie pseudogroup defined by the Pfaffian system $y^{2}d{y}^1=x^{2}d{x}^1$ is equivalently defined by the nonlinear system $y^2{\displaystyle \frac{\partial y^1}{\partial x^1}}=x^2,{\displaystyle \frac{\partial y^1}{\partial x^2}}=0$. The corresponding Lie algebroid $R_1\subset J_1\left(T\right)$ is defined by the linearized system $x^2{\xi }_1^1+{\xi }^2=0,{\xi }_2^1=0$, which is not involutive and not even formally integrable because, when using crossed derivatives, we obtain, at once, ${\xi }_1^1+{\xi }_2^2=0$. Hence, we can use two sections ${\xi }_1,{\eta }_1\in R_1$ with ${\xi }_1^1+{\xi }_2^2\ne 0,{\eta }_1^1+{\eta }_2^2\ne 0$ and check that $[{\xi }_1,{\eta }_1]\in R_1$ in such a way that $({\xi }^1=0,{\xi }^2=-x^2,{\xi }_1^1=1,{\xi }_2^1=0,{\xi }_1^2=0,{\xi }_2^2=0)$ and $({\eta }^1=1,{\eta }^2=-0,{\eta }_1^1=0,{\eta }_2^1=0,{\eta }_1^2=1,{\eta }_2^2=1)$ with no relation at all with solutions, and we obtain the involutive Lie algebroid $R_1^{\left(1\right)}\subset R_1$ with $[R_1^{\left(1\right)},R_1^{\left(1\right)}]\subset R_1^{\left(1\right)}$.

We now consider an algebraic pseudogroup defined over the target by a nonlinear system ${\mathcal{R}}_q\left(Y\right)\subset {\Pi }_q(Y,Y)\subset J_q(Y\times Y)$, with local coordinates simply denoted by $(y,\overline{y},{\displaystyle \frac{\partial \overline{y}}{\partial y}},{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y\partial y}},\dots )$ such that $det\left({\displaystyle \frac{\partial \overline{y}}{\partial y}}\right)\ne 0$. The jet composition $\left((\overline{y},\overline{\overline{y}},{\displaystyle \frac{\partial \overline{\overline{y}}}{\partial \overline{y}}},\dots )(y,\overline{y},{\displaystyle \frac{\partial \overline{y}}{\partial y}},\dots )\right)\to (y,\overline{\overline{y}},{\displaystyle \frac{\partial \overline{\overline{y}}}{\partial \overline{y}}}{\displaystyle \frac{\partial \overline{y}}{\partial y}},\dots )$ is obtained by using the chain rule for derivatives while the inversion is $(y,\overline{y},{\displaystyle \frac{\partial \overline{y}}{\partial y}},\dots )\to (\overline{y},y,{\left({\displaystyle \frac{\partial \overline{y}}{\partial y}}\right)}^{-1},\dots )$.

The composition $J_q(X\times Y){\times }_Y{\mathcal{R}}_q\left(Y\right)\to J_q(X\times Y)$ may be similarly defined by using

$$(x,y,y_x,y_{xx},\dots )(y,\overline{y},{\displaystyle \frac{\partial \overline{y}}{\partial y}},{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y\partial y}},\dots )\to (x,\overline{y},{\displaystyle \frac{\partial \overline{y}}{\partial y}}{y}_x,{\displaystyle \frac{\partial \overline{y}}{\partial y}}{y}_{xx}+{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y\partial y}}{y}_x{y}_x,\dots )$$

By using a free generic action, we understand that the morphism $♯:R_q\left(Y\right)\to J_q\left(V(X\times Y)\right)\simeq V\left(J_q(X\times Y)\right)$ is a monomorphism, and thus, we can set the following:

Definition 12.

A nonlinear system ${\mathcal{A}}_q\subset J_q(X\times Y)$ over the source is said to be a PHS for the Lie groupoid ${\mathcal{R}}_q\left(Y\right)\subset {\Pi }_q(Y,Y)$ over the target if the corresponding graph

$${\mathcal{A}}_q{\times }_Y{\mathcal{R}}_q\left(Y\right)\to {\mathcal{A}}_q{\times }_X{\mathcal{A}}_q$$

is an isomorphism while caring about the respective fibered products. Setting $A_q=V\left({\mathcal{A}}_q\right)$, the action is said to be free (transitive, simply transitive) when the morphism $♯:{\mathcal{A}}_q{\times }_Y{R}_q\left(Y\right)\to A_q$ of vector bundles over ${\mathcal{A}}_q$ is a monomorphism (an epimorphism, an isomorphism). The system is said to be an automorphic system if the r-prolongation ${\mathcal{A}}_{q+r}$ is a PHS for the r-prolongation ${\mathcal{R}}_{q+r}\left(Y\right)$, $\forall r\ge 0$.

We have already proved and illustrated in [6,8] the two following criteria:

Theorem 9.

(First criterion for automorphic systems) If an involutive system ${\mathcal{A}}_q\subset J_q(X\times Y)$ is a PHS for a Lie groupoid ${\mathcal{R}}_q\left(Y\right)\subset {\Pi }_q(Y,Y)$ and if ${\mathcal{A}}_{q+1}={\rho }_1\left({\mathcal{A}}_q\right)\subset J_{q+1}(X\times Y)$ is a PHS for the Lie groupoid ${\mathcal{R}}_{q+1}\left(Y\right)={\rho }_1\left({\mathcal{R}}_q\left(Y\right)\right)\subset {\Pi }_{q+1}(Y,Y)$, then ${\mathcal{R}}_q$ is an involutive system over the target with the same non-zero characters, and ${\mathcal{A}}_q$ is an automorphic system.

Theorem 10.

(Second criterion for automorphic systems) If ${\mathcal{R}}_q\subset {\Pi }_q(Y,Y)$ is an involutive system of finite Lie equations such that the action of ${\mathcal{R}}_q\left(Y\right)$ on $J_q(X\times Y)$ is generically free, then the action of ${\mathcal{R}}_{q+r}\left(Y\right)$ on $J_{q+r}(X\times Y)$ is generically free $\forall r\ge 0$, and all the differential invariants are generated by a fundamental set of order $q+1$ (care).

Theorem 11.

When $n=1,m=2,q=2$ and $k=\mathbb{Q}$, we may consider the Lie pseudogroup $\Gamma$ to be defined as a Lie group by the action $\overline{y}=Ay+B$ with $det\left(A\right)=1$. The only generating differential invariant at order 2 is $\Phi =y_x^1{y}_{xx}^2-y_x^2{y}_{xx}^1$; however, at order 3, we must use $d_x\Phi =y_x^1{y}_{xxx}^2-y^{2}x{y}_{xxx}^1$, of course, but we also have to add $\Psi =y_{xx}^1{y}_{xxx}^2-y_{xx}^2{y}_{xxx}^1$. Thus, we have the strict inclusion ${\mathcal{A}}_3\subset {\rho }_1\left({\mathcal{A}}_2\right)$, and the symbol of ${\mathcal{A}}_3$ vanishes if and only if $y_x^1{y}_{xx}^2-y_x^2{y}_{xx}^1\ne 0$. We shall meet a similar condition with the non-zero Wronskian determinant $y^1{y}_x^2-y^2{y}_x^1\ne 0$ at order $q=1$ in the Picard–Vessiot theory if we consider the action $\overline{y}=Ay$ with $det\left(A\right)=1$. Indeed, we must consider the only first-order differential invariant $\Phi =y^1{y}_x^2-y^2{y}_x^1$ and use the second-order differential invariant $d_x\Phi =y^1{y}_{xx}^2-y^2{y}_{xx}^1$, to which we must add $\Psi =y_x^1{y}_{xx}^2-y_x^2{y}_{xx}^1$. We notice that the symbol of order 2 is vanishing if and only if $y^1{y}_x^2-y^2{y}_x^1\ne 0$. We let the reader check that all these results are coherent with the two previous criteria.

We are now able to recognize whether a nonlinear system of algebraic OD or PD equations with n independent variables and m unknowns is an automorphic system for its biggest pseudogroup of invariance. Thus, the transition to differential algebra remains to be shown. For this, keeping in mind the difference existing between special and general relativity in physics, we shall explain using an example of the difference existing between a special and a general automorphic system.

Example 14.

With $n=1,m=2,q=1,k=\mathbb{Q}$, let us consider the algebraic pseudogroup of target transformations preserving the 1-form $y^{2}d{y}^1$ and, thus, also the 2-form $d{y}^1\wedge d{y}^2$. On one side, we may start with a given differential field K containing k and consider the special system $y^2{y}_x^1=\omega \in K$. Looking for the biggest Lie pseudogroup of invariance of this OD equation, we must have

$${\overline{y}}^2{\overline{y}}_x^1={\overline{y}}^2{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^1}}{y}_x^1+{\overline{y}}^2{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^2}}{y}_x^2=y^2{y}_x^1\Rightarrow {\overline{y}}^2{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^1}}=y^2,{\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^2}}=0$$

Such a system is not involutive, as it is not even formally integrable, and we must add ${\displaystyle \frac{\partial ({\overline{y}}^1,{\overline{y}}^2)}{\partial (y^1,y^2)}}=1$. We may, thus, start with the linear involutive system $R_1\left(Y\right)$ defined by the first-order system of infinitesimal Lie equations $y^2{\eta }_1^1+{\eta }^2=0,{\eta }_2^1=0,{\eta }_1^1+{\eta }_2^2=0$. Introducing the prime differential ideal $\mathfrak{p}\subset K\left\{y\right\}=K[y,y_x,y_{xx},\dots ]$, we may define the special differential automorphic extension $L/K$ with $L=Q(K\left\{y\right\}/\mathfrak{p})\simeq K(y^1,y^2,y_x^2,y_{xx}^2,\dots )$. However, we may also start with the differential extension $L=Q\left(k\right\{y\left\}\right)=k<y>$ and consider the general differential automorphic extension $L/K$ with $K=k<\Phi >$ by introducing the generating differential invariant $\Phi =y^2{y}_x^1$. In actual practice, caring about the fibered products in the defining formula of Definition 12, we have $q=1$ by counting (the number of jet coordinates) − (number of equations) for each member:

$${dim}_X\left({\mathcal{A}}_1{\times }_Y{\mathcal{R}}_1\left(Y\right)\right)=(4-1)+(6-3)=3+3=6,{dim}_X\left({\mathcal{A}}_1{\times }_X{\mathcal{A}}_1\right)=3+3=6$$

with parametric jets $(y^1,y^2,y_x^2)$ for ${\mathcal{A}}_1$, $({\overline{y}}^1,{\overline{y}}^2,{\overline{y}}_1^2)$ for ${\mathcal{R}}_1\left(Y\right)$ and a similar equality for $q=2$.

With a slight abuse of language, we may set $\Theta \left(q\right)=♯\left(R_q\left(Y\right)\right)$, and this distribution can be studied without introducing the space of solutions of the Lie algebroid $R_q\left(Y\right)$ that may not be known explicitly, for example, if one replaces $y^{2}d{y}^1$ with $y^{2}d{y}^1-y^{1}d{y}^2$. Using the Frobenius theorem for the involutive distribution $\Theta \left(q\right)$ on $V(J_q(X\times Y)$, we may find a reciprocal distribution $\Delta \left(q\right)$ of maximal rank such that $[\Theta (q),\Delta (q\left)\right]=0$; that is, $[\theta ,\delta ]=0,\forall \theta \in \Theta ,\delta \in \Delta$.

Lemma 6.

$\delta y=0,\forall \delta \in \Delta$.

Proof. 

As the sections of $R_q\left(Y\right)$ are only defined up to a linear combination, with coefficients only depending on y, we successively have

$${\theta }^{\prime }=\Sigma \lambda \left(y\right)\theta \Rightarrow [\delta ,{\theta }^{\prime }]=\Sigma \left(\delta \lambda \right)\theta +\Sigma \lambda [\delta ,\theta ]=\Sigma \left(\delta \lambda \right)\theta =0\Rightarrow \delta \lambda =0$$

Quantities killed by $\Delta$ will be called “constants” according to [13,23], and thus, we have $cst\left(L\right)=k\left(y\right)$. The next example will illustrate this new concept, which was already introduced forty years ago in [28] but was never acknowledged [19,20]. These results provide the need to entirely revisit the Picard–Vessiot theory as it is known today. □

Example 15.

Coming back to the previous example, with $q=1$ and the parametric sections $\{{\eta }^1,{\eta }^2,{\eta }_1^2\}$ of $R_1\left(Y\right)$, we get

$$\Theta \left(1\right)\left\{\begin{array}{ccccc}{\eta }_1^2& \to & {\theta }_3& =& y_x^1{\displaystyle \frac{\partial }{\partial y_x^2}}\hfill \\ {\eta }^2& \to & {\theta }_2& =& y^2{\displaystyle \frac{\partial }{\partial y^2}}-y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}\hfill \\ {\eta }^1& \to & {\theta }_1& =& {\displaystyle \frac{\partial }{\partial y^1}}\hfill \end{array}\right.$$

We notice that the relation $y^1{\theta }^1-{\theta }^2={\rho }_1(y^1{\displaystyle \frac{\partial }{\partial y^1}}-y^2{\displaystyle \frac{\partial }{\partial y^2}})$ proves that the pseudogroup of invariance contains finite transformations of the form $({\overline{y}}^1=a{y}^1,{\overline{y}}^2={\displaystyle \frac{1}{a}}{y}^2)$ whenever $a=cst$.

A reciprocal distribution could a priori be

$$\left\{\begin{array}{ccc}{\delta }_1& =& y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}\\ {\delta }_2& =& y^2{\displaystyle \frac{\partial }{\partial y_x^2}}\\ {\delta }_3& =& y^2{\displaystyle \frac{\partial }{\partial y^2}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}\\ {\delta }_4& =& {\displaystyle \frac{\partial }{\partial y^1}}\end{array}\right.$$

However, taking into account the previous lemma, we must push out ${\delta }_3$ and ${\delta }_4$ to get

$$\Delta \left(1\right)\left\{\begin{array}{ccc}{\delta }_1& =& \hfill y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}\\ {\delta }_2& =& \hfill y^2{\displaystyle \frac{\partial }{\partial y_x^2}}\end{array}\right.$$

which is of rank 2 if and only if $y^2{y}_x^1\ne 0$. We notice that ${\delta }^1$ is the factor of $-{\xi }_x$ in $♭\left({\xi }_1\right)$. Similarly, yet after a tricky computation, we should obtain the following at order 2:

$$\Theta \left(2\right)\left\{\begin{array}{ccccc}{\eta }_{11}^2& \to & {\theta }_4& =& {\left(y_x^1\right)}^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}\\ {\eta }_1^2& \to & {\theta }_3& =& y^2{y}_x^1{\displaystyle \frac{\partial }{\partial y_x^2}}-{\left(y_x^1\right)}^2{\displaystyle \frac{\partial }{\partial y_{xx}^1}}+(y^2{y}_{xx}^1+2y_x^1{y}_x^2){\displaystyle \frac{\partial }{\partial y_{xx}^2}}\\ {\eta }^2& \to & {\theta }_2& =& y^2{\displaystyle \frac{\partial }{\partial y^2}}-y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}-y_{xx}^1{\displaystyle \frac{\partial }{\partial y_{xx}^1}}+y_{xx}^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}\\ {\eta }^1& \to & {\theta }_1& =& {\displaystyle \frac{\partial }{\partial y^1}}\end{array}\right.$$

and we now have $y^1{\theta }_1-{\theta }_2={\rho }_2(y^1{\displaystyle \frac{\partial }{\partial y^1}}-y^2{\displaystyle \frac{\partial }{\partial y^2}})$. A reciprocal distribution could be

$$\Delta \left(2\right)\left\{\begin{array}{ccc}{\delta }_1& =& y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_x^2}}+2(y_{xx}^1{\displaystyle \frac{\partial }{\partial y_{xx}^1}}+y_{xx}^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}})\\ {\delta }_2& =& y^2{\displaystyle \frac{\partial }{\partial y_x^2}}+2y_x^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}\\ {\delta }_3& =& y^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}\\ {\delta }_4& =& y_x^1{\displaystyle \frac{\partial }{\partial y_{xx}^1}}+y_x^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}\end{array}\right.$$

It is of rank 4 in $(y_x^1,y_x^2,y_{xx}^1,y_{xx}^2)$, and ${\delta }_4$ is now the factor of $-{\xi }_{xx}$ in $♭\left({\xi }_2\right)$.

Finally, one can easily check that $\Delta \left(2\right)$ stabilizes both $\Phi =y^2{y}_x^1$ and $d_x\Phi =y^2{y}_{xx}^1+y_x^1{y}_x^2$.

When passing from groups to pseudogroups while using the results of [13,23], we have the following:

Proposition 9.

Only stable intermediate differential fields can provide a Galois differential correspondence.

Proof. 

Let us consider the chain of inclusions $k\subset K\subset K^{\prime }\subset L$, where the differential fields involved are, respectively, $K=k<\Phi >$, $K^{\prime }=k<\Phi ,\Psi >$, $L=k<y>$ for simplicity. The pseudogroup ${\Gamma }^{\prime }$ preserving $\Phi$ and  $\Psi$ is, thus, a subpseudogroup of the pseudogroup $\Gamma$ preserving only $\Phi$. Hence, any new infinitesimal target transformation is of the form ${\theta }^{\prime }=\Sigma \lambda \left(y\right)\theta$ with ${\theta }^{\prime }\Psi =0$. Therefore, from the previous lemma, we obtain $[{\theta }^{\prime },\delta ]=\Sigma \lambda \left(y\right)[\theta ,\delta ]=0$, and thus, ${\theta }^{\prime }(\delta \Psi )=\delta ({\theta }^{\prime }\Psi )=0$; that is to say, $\delta \Psi$ is necessarily a differential invariant of ${\Gamma }^{\prime }$ and, thus, a differential function of $\Phi$ and $\Psi$. □

Example 16.

If one chooses $k=\mathbb{Q}$ and $K^{\prime }=k<\Phi ,\Psi >$, as in the previous example, that is, with $\Phi =y^2{y}_x^1$ and $\Psi =y_x^2$, we obtain ${\delta }_1\Psi =\Psi$ but ${\delta }_2\Psi =y^2$, meaning $K^{\prime }$ is not stable. Indeed, the pseudogroup preserving Φ and Ψ also preserves $y^2$, that is, it becomes ${\overline{y}}^1=y^1+b,{\overline{y}}^2=y^2$. In contrast, if we choose $\Psi ={\displaystyle \frac{y_x^2}{y^2}}$, then ${\delta }_1\Psi ==\Psi ,,{\delta }_2\Psi ==1$ and ${\Gamma }^{\prime }$ become ${\overline{y}}^1=a{y}^1+b,{\overline{y}}^2={\displaystyle \frac{1}{a}}{y}^2$.

Last but not least, we study the condition for which not only do we have ${\Gamma }^{\prime }\subset \Gamma$, but also ${\Gamma }^{\prime }⊲\Gamma$. First of all, to the best of the author’s knowledge, formal conditions do not exist. Indeed, if $g\in \Gamma ,h\in {\Gamma }^{\prime }$, then $g^{-1}\circ h\circ g=h^{\prime }\in {\Gamma }^{\prime }$cannot be checked in general.

Now, the vector bundle $J_q\left(T\right)$ is also a bundle of geometric objects associated with ${\Pi }_{q+1}(X,X)$ in such a way that ${\eta }_q=f_{q+1}\left({\xi }_q\right)\in J_q\left(T\right),\forall {\xi }_q\in J_q\left(T\right),\forall f_{q+1}\in {\Pi }_{q+1}$ along the formulas deduced from the transformations of a vector field and its various derivatives, namely [10],

$${\eta }^k\left(f\left(x\right)\right)=f_r^k\left(x\right){\xi }^r\left(x\right),{\eta }_u^k{f}_i^u=f_r^k{\xi }_i^r+f_{ri}^k{\xi }^r,{\eta }_{uv}^k{f}_i^u{f}_j^v+{\eta }_u^k{f}_{ij}^u=f_r^k{\xi }_{ij}^r+f_{ri}^k{\xi }_j^r+f_{rj}^k{\xi }_i^r+f_{rij}^k{\xi }^r$$

and so on. We may define the finite normalizer ${\tilde{\mathcal{R}}}_{q+1}=\{{\tilde{f}}_{q+1}\mid {\tilde{f}}_{q+1}\left({\xi }_q\right)\in R_q\left(Y\right),\forall {\xi }_q\in R_q=R_q\left(X\right)\}$ when Y is a copy of X in this framework.

Passing to the infinitesimal point of view, we may define the formal Lie derivative:

$$L\left({\xi }_{q+1}\right){\eta }_q=[{\xi }_q,{\eta }_q]+i\left(\eta \right)d{\xi }_{q+1}=\{{\xi }_{q+1},{\eta }_{q+1}\}+i\left(\xi \right)d{\eta }_{q+1}$$

and define the infinitesimal normalizer by the condition ${\tilde{R}}_{q+1}=\{{\tilde{\xi }}_{q+1}\mid L\left({\tilde{\xi }}_{q+1}\right){\xi }_q\in R_q,\forall {\xi }_q\in R_q\}$ We obtain, in particular, the following quite difficult result ([4], pp. 390–393):

Theorem 12.

If $R_q\subset J_q\left(T\right)$ is a formally integrable and transitive system of infinitesimal Lie equations with a symbol $g_q=R_q\cap S_q{T}^{*}\otimes T$ that is 2-acyclic (involutive), then ${\tilde{R}}_{q+1}$ is formally integrable (involutive) such that ${\tilde{g}}_{q+1}=g_{q+1}$.

Example 17.

Again, with $n=1,m=2,k=\mathbb{Q}$, let us consider the inclusions $k\subset K=k<\Psi >\subset K^{\prime }=k<\Phi >\subset k<y^1,y^2>$ with $\Phi =y^2{y}_x^1$ and $\Psi =(y^{2}y{y}_{xx}^1+y_x^1{y}_x^2)/y^2{y}_x^1=d_x\Phi /\Phi$. By shear luck, we can exhibit ${\Gamma }^{\prime }=\{{\overline{y}}^1=g\left(y^1\right),{\overline{y}}^2=y^2/\left({\displaystyle \frac{\partial g}{\partial y^1}}\right)\}⊲\Gamma =\{{\overline{y}}^1=g\left(y^1\right),{\overline{y}}^2=a{y}^2/\left({\displaystyle \frac{\partial g}{\partial y^1}}\right)\}$. According to the previous results, a necessary condition for having $\Gamma ⊲{\Gamma }^{\prime }$ is that $difftrd(K^{\prime }/K)=0$ because $difftrd(L/K)=difftrd(L/K^{\prime })=1$ and $K^{\prime }/K$ represent a PV-extension for the multiplicative group of the real line.

In view of the preceding results, even the general Picard–Vessiot theory must be entirely revisited in a coherent way with BB as follows.

Remark 3.

The classical textbook definition $N(\Theta )=\{\xi \in T\mid [\xi ,\theta ]\in \theta ,\forall \eta \in \Theta \}$ of the normalizer of Θ in T is useless in actual practice and must be replaced by the formal definition of ${\tilde{R}}_{q+1}$, as given previously through the formal Lie derivative. As such a definition crucially depends on the Spencer operator, it is still not acknowledged. Moreover, as $d{\eta }_{q+1}\in T^{*}\otimes R_q$, the normalizer can, thus, be obtained in a purely algebraic way by the condition $\{{\tilde{\xi }}_{q+1},{\eta }_{q+1}\}\in R_q$. In particular, when $n=4$, the Poincaré group is of codimension 1 in its normalizer, which is the Weyl group.

Example 18.

Following Vessiot exactly, as we did in the Introduction, with $n=1,m=2,k=\mathbb{Q}$, let us consider the generic second-order OD equation $y_{xx}-{\omega }^1{y}_x+{\omega }^{2}y=0$ and copy it twice with $k=1,2$ to obtain the linear second order automophic system $y_{xx}^k-{\omega }^1{y}_x^k+{\omega }^2{y}^k=0$ for the standard action of $GL\left(2\right)$ on $(y^1,y^2)$ when $k=1,2$. Such a system admits the well-known generating Lie form as a quotient of determinants:

$${\Phi }^1\equiv {\displaystyle \frac{\mid \begin{array}{cc}{y}^1\hfill & y_{xx}^1\hfill \\ y^2\hfill & y_{xx}^2\hfill \end{array}\mid }{\mid \begin{array}{cc}{y}^1\hfill & y_x^1\hfill \\ y^2\hfill & y_x^2\hfill \end{array}\mid }}={\omega }^1,{\Phi }^2\equiv {\displaystyle \frac{\mid \begin{array}{cc}{y}_x^1\hfill & y_{xx}^1\hfill \\ y_x^2\hfill & y_{xx}^2\hfill \end{array}\mid }{\mid \begin{array}{cc}{y}^1\hfill & y_x^1\hfill \\ y^2\hfill & y_x^2\hfill \end{array}\mid }}={\omega }^2$$

In matrix form, the prolongation of the action up to order $q=2$ is

$$\left(\begin{array}{ccc}{\overline{y}}^1\hfill & {\overline{y}}_x^1\hfill & {\overline{y}}_{xx}^1\hfill \\ {\overline{y}}^2\hfill & {\overline{y}}_x^2\hfill & {\overline{y}}_{xx}^2\hfill \end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{ccc}{y}^1\hfill & y_x^1\hfill & y_{xx}^1\hfill \\ y^2\hfill & y_x^2\hfill & y_{xx}^2\hfill \end{array}\right)$$

We have the automorphic extension $k\subset K\subset L$ with $K=<{\Phi }^1,{\Phi }^2>$ and $L=k<y^1,y^2>$.

Therefore, we obtain $A=\overline{M}{M}^{-1}$ in the matrix form:

$$\overline{M}=AM\Rightarrow \left(\begin{array}{cc}a\hfill & b\hfill \\ c\hfill & d\hfill \end{array}\right)=\left(\begin{array}{cc}{\overline{y}}^1\hfill & {\overline{y}}_x^1\hfill \\ {\overline{y}}^2\hfill & {\overline{y}}_x^2\hfill \end{array}\right){\left(\begin{array}{cc}{y}^1\hfill & y_x^1\hfill \\ y^2\hfill & y_x^2\hfill \end{array}\right)}^{-1}$$

on the condition of having the non-zero Wronskian determinant condition $\Psi \equiv y^1{y}_x^2-y^2{y}_x^1\ne 0$. The four infinitesimal generators of the target action are

$$\Theta =\{{\theta }_1=y^1{\displaystyle \frac{\partial }{\partial y^1}},{\theta }_2=y^2{\displaystyle \frac{\partial }{\partial y^1}},{\theta }_3=y^1{\displaystyle \frac{\partial }{\partial y^2}},{\theta }_4=y^2{\displaystyle \frac{\partial }{\partial y^2}}\}=\left\{y^l{\displaystyle \frac{\partial }{\partial y^k}}\right\}$$

which can be prolonged at any order, such as ${\theta }_1=y^1{\displaystyle \frac{\partial }{\partial y^1}}+y_x^1{\displaystyle \frac{\partial }{\partial y_x^1}}+y_{xx}^1{\displaystyle \frac{\partial }{\partial y_{xx}^1}}+\dots$ and so on.

One easily obtains the reciprocal distribution at order 1:

$$\Delta =\{{\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y^k}},{\delta }_2=y_x^k{\displaystyle \frac{\partial }{\partial y^k}},{\delta }_3=y^k{\displaystyle \frac{\partial }{\partial y_x^k}},{\delta }_4=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}\}$$

in such a way that the rank with respect to $(y^1,y^2)$ and $(y_x^1,y_x^2)$ is indeed maximum equal to 2 provided that the Wronskian determinant does not vanish. Accordingly, one can extend each δ to $L{\otimes }_{K}L$ by setting, for example, ${\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y^k}}+{\overline{y}}^k{\displaystyle \frac{\partial }{\partial {\overline{y}}^k}},\dots ,{\delta }_4=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}+{\overline{y}}_x^k{\displaystyle \frac{\partial }{\partial {\overline{y}}_x^k}}$, and we have

$$\Psi =det\left(M\right)\ne 0,\delta \overline{M}=\left(\delta A\right)M+A\left(\delta M\right)=A\left(\delta M\right)\Rightarrow \delta A=0$$

In particular, $a={\displaystyle \frac{y_x^2}{\Psi }}{\overline{y}}^1-{\displaystyle \frac{y^2}{\Psi }}{\overline{y}}_x^1\in cst\left(L{\otimes }_{K}L\right)$ because ${\delta }_1\Psi =\Psi ,{\delta }_2\Psi =0,{\delta }_3\Psi =0,{\delta }_4\Psi =\Psi$ for $SL\left(2\right)⊲GL\left(2\right)$ when $ad-bc=1$, which is a result that is not evident at first sight. We finally notice that $A=\left({\displaystyle \frac{\partial \overline{y}}{\partial y}}\right)$ and ${\displaystyle \frac{{\partial }^2\overline{y}}{\partial y\partial y}}=0$ for the differential automorphic extension $L/K$, and we must add $det\left(A\right)=1$ for the differential automorphic extension $L/K^{\prime }$ now with $K^{\prime }=K<\Psi >$. In the next section, we shall explain why we must take out ${\delta }_1$ and ${\delta }_2$ in the differential algebraic framework if we consider a Lie group as a Lie pseudogroup, that is, if we no longer introduce the parameters $(a,b,c,d)$. The reader can spend a few minutes now to imagine how to manage such a target.

Example 19.

An example provided by Vessiot, when $n=1,m=3,q=3$, is given by showing the link existing between DGT and formal integrability (see [8], pp. 378–380, pp. 340–342 for details). Let us consider the trivially involutive finite type system of ODE $y_{xxx}^k-p\left(x\right)y_{xx}^k+q\left(x\right)y_x^k-r\left(x\right)y^k=0$ with a zero symbol. Exactly like in the previous example, and when the second-order Wronskian W does not vanish, we obtain a third-order involutive automorphic system for $GL\left(3\right)$ for any choice of $(p,q,r)$ through the three similar third-order differential invariants $({\Phi }^1,{\Phi }^2,{\Phi }^3)$. Now, suppose we add the constraint $\Psi \equiv {\left(y^3\right)}^2-y^1{y}^2=0$. In that case, it is not evident how to discover the differential condition that must be satisfied by $(p,q,r)$ to obtain a third-order automorphic system for the subgroup $GL\left(2\right)\subset GL\left(3\right)$ that can be introduced by setting $(u,v)\to (\overline{u}=au+bv,\overline{v}=cu+dv)$ for the parametrization $(y^1={\left(u\right)}^2,y^2={\left(v\right)}^2,y^3=uv)$. As a first comment, we notice that we obtain ${\left({\overline{y}}^3\right)}^2-{\overline{y}}^1{\overline{y}}^2={(ad-bc)}^2({\left(y^3\right)}^2-y^1{y}^2)$ and $\overline{W}={(ad-bc)}^{3}W$. Then, it is clear that the system ${\mathcal{A}}_3$ considered is not formally integrable. We let the reader check that the system ${\mathcal{A}}_3^{\left(5\right)}={\mathcal{A}}_3^{\left(4\right)}$ is an involutive automorphic system for the action of $GL\left(2\right)$, which is defined by $3+5=8$ OD equations and has, therefore, the fiber dimension $(3+3+3+3)-8=4$, provided that (!)

$$9{\partial }_{xx}p-18p{\partial }_{x}p+27{\partial }_{x}q+4{\left(p\right)}^3-18pq+54r=0$$

We advise the reader to treat the restriction to $SL\left(2\right)$ obtained similarly by setting $ad-bc=1$ and considering now W and Ψ as new added differential invariants.

Remark 4.

According to the famous theorems of S. Lie, any infinitesimal transformation of a Lie group action is a linear combination of a finite basis of p infinitesimal generators ${\theta }_{\tau }={\theta }_{\tau }^k\left(x\right){\partial }_k$ in the form ${\xi }^k={\lambda }^{\tau }{\theta }_{\tau }^k$ with p constant coefficients λ. More generally, setting ${\xi }^k\left(x\right)={\lambda }^{\tau }\left(x\right){\theta }_{\tau }^k\left(x\right)$ and introducing the section ${\xi }_q\in J_q\left(T\right)$ defined by ${\xi }_{\mu }^k\left(x\right)={\lambda }^{\tau }\left(x\right){\partial }_{\mu }{\theta }_{\tau }^k\left(x\right)$, for an arbitrary section $\lambda =\left({\lambda }^{\tau }\left(x\right)\right)\in {\wedge }^0{T}^{*}\otimes \mathcal{G}$ with the Lie algebra $\mathcal{G}=T_e\left(G\right)$, for a q that is large enough, we obtain

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

as a way to prove that the Spencer sequence, in this case, is isomorphic to the tensor product of the Poincaré sequence for the exterior derivative d by the Lie algebra $\mathcal{G}$ in the diagram

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

in which Θ is the set of solutions of $R_q$, and the vertical maps just described are isomorphisms.

Example 20.

Examples 2 and 14 provide the best examples of the differences existing between a Lie group and a Lie pseudogroup of transformations, while showing why the concept of “constant” must be entirely revisited. Let us consider the involutive system of infinitesimal Lie equations $R_1\subset J_1\left(T\right)$ on T when $m=n=2,q=1$, defined by the three linear first-order PD equations $x^2{\xi }_1^1+{\xi }^2=0,{\xi }_2^1=0,{\xi }_1^1+{\xi }_2^2=0$ with solutions $\Theta =\{{\xi }^1=\lambda \left(x^1\right),{\xi }^2=-x^2{\partial }_1\lambda \left(x^1\right)$, where λ is an arbitrary function of $x^1$. It follows that $dim\left(R_q\right)=q+2,\forall q\ge 0$, and we may introduce the equivalent Spencer system $R_2\subset J_1\left(R_1\right)$ on $R_1$ by using the three new variables $\{z^1={\xi }^1,z^2={\xi }^2,z^3={\xi }_1^2\}$ defined by the first Spencer operator $D_1$ with five first-order equations killed by the second first-order Spencer operator $D_2$ with two first-order CCs:

$$z_2^1=0,z_2^2-{\displaystyle \frac{1}{x^2}}{z}^2=0,z_2^3-{\displaystyle \frac{1}{x^2}}{z}^3=0,z_1^1+{\displaystyle \frac{1}{x^2}}{z}^2=0,z_1^2-z^3=0$$

We obtain the Spencer sequence: $0\to \Theta \stackrel{j_1}{⟶}{C}_0\stackrel{D_1}{⟶}{C}_1\stackrel{D_2}{⟶}{C}_2\to 0$, in which we now have $dim\left(C_0\right)=dim\left(R_1\right)=3,dim\left(C_1\right)=5,dim\left(C_2\right)=2$ and a Euler–Poincaré characteristic $3-5+2=0$. There is no longer any way to relate such an example to the Poincaré sequence for the exterior derivative d because, now, we have one arbitrary function of one variable instead of a few constants.

5. Differential Algebra

The purpose of this section is to revisit the theory of algebraic pseudogroups by using Hopf rings in a way similar to the one pioneered by Bialynicki-Birula in [13,23] (see also [8] for more details and examples). For simplicity, we shall restrict our study to the general situation, as the special situation can be treated by restricting the various distributions. For example, the Picard–Vessiot version can be achieved for the single OD equation $y_{xx}=0$ by transforming it into the OD automorphic system $y_{xx}^1=0,y_{xx}^2=0$. In this case, we may choose $k=K=\mathbb{Q}$ and $L=\mathbb{Q}(y^1,y^2,y_x^1,y_x^2)$, considered as a differential field, by setting $d_x{y}^k=y_x^k,d_x{y}_x^k=0$ for $k=1,2$, as in the Introduction.

The first comment is to notice that $cst\left(L\right)=k\left(y\right)$ when $L=k<y>$, and to exhibit a link between algebraic pseudogroups and the constants of $L{\otimes }_{K}L$ for $\Delta$, such as in the last examples. As no reference can be quoted, we provide a motivating example.

Example 21.

With $n=1,m=2,q=1,k=\mathbb{Q}$ and $\Phi \equiv y^2{y}_x^1$, let us consider the differential automorphic extension $L/K$ with $K=k<\Phi >\subset L=k<y^1,y^2>$. The corresponding algebraic pseudogroup is defined by ${\overline{y}}^2{\overline{y}}_x^1=y^2{y}_x^1\Rightarrow d{\overline{y}}^1\wedge d{\overline{y}}^2=d{y}^1\wedge d{y}^2$ or, equivalently, by the first-order involutive system of finite Lie equations:

$${\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^1}}={\displaystyle \frac{y^2}{{\overline{y}}^2}}={\displaystyle \frac{{\overline{y}}_x^1}{y_x^1}},{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^2}}={\displaystyle \frac{{\overline{y}}^2}{y^2}}={\displaystyle \frac{y_x^1}{{\overline{y}}_x^1}},{\overline{y}}_x^2={\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}{y}_x^1+{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^2}}{y}_x^2\Rightarrow {\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}={\displaystyle \frac{{\overline{y}}_x^2}{y_x^1}}-{\displaystyle \frac{y_x^2}{{\overline{y}}_x^1}}$$

Starting with ${\delta }_1=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}$ and ${\delta }_2=y^2{\displaystyle \frac{\partial }{\partial y_x^2}}$, and extending these derivations from L to $L{\otimes }_{K}L$, we discover that ${\delta }_1$ and ${\delta }_2$ kill the four elements of the matrix ${\displaystyle \frac{\partial \overline{y}}{\partial y}}$, although this fact is not evident for

$${\delta }_2\left({\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}\right)={\displaystyle \frac{{\overline{y}}^2}{y_x^1}}-{\displaystyle \frac{y^2}{{\overline{y}}_x^1}}={\displaystyle \frac{{\overline{y}}^2{\overline{y}}_x^1-y^2{y}_x^1}{y_x^1{\overline{y}}_x^1}}=0$$

We obtain, therefore, $k\left(y\right){\otimes }_{k}k\left(\overline{y}\right)\in cst\left(L{\otimes }_{K}L\right)$ and ${\displaystyle \frac{\partial \overline{y}}{\partial y}}\in cst\left(L{\otimes }_{K}L\right)$.

Prolonging to order 2 is even more tricky with, for example,

$${\overline{y}}_{xx}^1={\displaystyle \frac{\partial {\overline{y}}^1}{\partial y^1}}{y}_{xx}^1+{\displaystyle \frac{{\partial }^2{\overline{y}}^1}{\partial y^1\partial y^1}}{\left(y_x^1\right)}^2\Rightarrow {\displaystyle \frac{{\partial }^2{\overline{y}}^1}{\partial y^1\partial y^1}}={\displaystyle \frac{1}{{\left(y_x^1\right)}^2}}{\overline{y}}_{xx}^1-{\displaystyle \frac{y^2{y}_{xx}^1}{{\left(y_x^1\right)}^2}}{\displaystyle \frac{1}{{\overline{y}}^2}}$$

and we let the reader check that this term is killed by both ${\delta }_1,\dots ,{\delta }_4$. Similarly, we get

$${\overline{y}}_{xx}^2={\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}{y}_{xx}^1+{\displaystyle \frac{{\overline{y}}^2}{y^2}}{y}_{xx}^2+{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1\partial y^1}}{\left(y_x^1\right)}^2+{\displaystyle \frac{1}{y^2}}{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}{y}_x^1{y}_x^2+d_x\left({\displaystyle \frac{{\overline{y}}^2}{y^2}}\right)y_x^2$$

Applying ${\delta }_3=y^2{\displaystyle \frac{\partial }{\partial y_x^2}}+2y_x^2{\displaystyle \frac{\partial }{\partial y_{xx}^2}}$ while taking into account the previous result at order 1, we get

$$2{\overline{y}}_x^2={\delta }_3\left({\displaystyle \frac{{\partial }^2{\overline{y}}^2}{\partial y^1\partial y^1}}\right){\left(y_x^1\right)}^2+{\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}{y}_x^1+{\displaystyle \frac{{\overline{y}}^2}{y^2}}{y}_x^2+{\overline{y}}_x^2$$

Using the relation ${\displaystyle \frac{\partial {\overline{y}}^2}{\partial y^1}}{y}_x^1+{\displaystyle \frac{{\overline{y}}^2}{y^2}}{y}_x^2={\overline{y}}_x^2$, we finally obtain ${\delta }_3\left({\displaystyle \frac{{\partial }^2{\overline{y}}^2}{\partial y^1\partial y^1}}\right)=0$, and so on.

Though the extension of reciprocal distributions can be carried out by induction on the order (see [8] for details), we do not know any reference on their explicit computation, which may be quite difficult, as shown by the next example.

Example 22.

With $m=n=1,k=\mathbb{Q}$ but $q=3$, we have

$${\overline{y}}_x={\displaystyle \frac{\partial \overline{y}}{\partial y}}{y}_x,{\overline{y}}_{xx}={\displaystyle \frac{\partial \overline{y}}{\partial y}}{y}_{xx}+{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}{\left(y_x\right)}^2,{\overline{y}}_{xxx}={\displaystyle \frac{\partial \overline{y}}{\partial y}}{y}_{xxx}+3{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}{y}_x{y}_{xx}+{\displaystyle \frac{{\partial }^3\overline{y}}{\partial y^3}}{\left(y_x\right)}^3$$

$$\begin{array}{ccc}\eta \hfill & \to & {\displaystyle \frac{\partial }{\partial y}}\hfill \\ {\eta }_y\hfill & \to & y_x{\displaystyle \frac{\partial }{\partial y_x}}+y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}}+y_{xxx}{\displaystyle \frac{\partial }{\partial y_{xxx}}}\hfill \\ {\eta }_{yy}\hfill & \to & {\left(y_x\right)}^2{\displaystyle \frac{\partial }{\partial y_{xx}}}+3y_x{y}_{xx}{\displaystyle \frac{\partial }{\partial y_{xxx}}}\hfill \\ {\eta }_{yyy}\hfill & \to & {\left(y_x\right)}^3{\displaystyle \frac{\partial }{\partial y_{xxx}}}\hfill \end{array}$$

The reciprocal distribution is

$$\begin{array}{ccc}-{\xi }_x\hfill & \to & \hfill y_x{\displaystyle \frac{\partial }{\partial y_x}}+2y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}}+3y_{xxx}{\displaystyle \frac{\partial }{\partial y_{xxx}}}\\ -{\xi }_{xx}\hfill & \to & \hfill y_x{\displaystyle \frac{\partial }{\partial y_{xx}}}+3y_{xx}{\displaystyle \frac{\partial }{\partial y_{xxx}}}\\ -{\xi }_{xxx}\hfill & \to & \hfill y_x{\displaystyle \frac{\partial }{\partial y_{xxx}}}\end{array}$$

Collecting these results, we obtain the following crucial theorem, bringing the need to revisit DGT, as in the following [8,13]:

Theorem 13.

The groupoid components $({\displaystyle \frac{\partial \overline{y}}{\partial y}},{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}},{\displaystyle \frac{\partial 3\overline{y}}{\partial y^3}},\dots )$ up to any order q are constants for the reciprocal distribution up to order q, which can be expressed as rational functions of all the jet components $(y,y_x,y_{xx},y_{xxx},\dots )$ and $(\overline{y},{\overline{y}}_x,{\overline{y}}_{xx},{\overline{y}}_{xxx},\dots )$ up to order q. Therefore, we obtain the Hopf ring $k[\Gamma ]\subset cst\left(L{\otimes }_{K}L\right)$ with the ring of fractions $k(\Gamma )=cst\left(Q\left(L{\otimes }_{K}L\right)\right)$ as a direct sum of differential fields for the target derivative $d_y$, although $Q\left(L{\otimes }_{K}L\right)$ is a direct sum of differential fields for the source derivative $d_x$ with an isomorphism $Q\left(L{\otimes }_{k\left(y\right)}k[\Gamma ]\right)\simeq Q\left(L{\otimes }_{K}L\right)$.

Proof. 

$L/k\left(y\right)$ is a regular extension because $k\left(y\right)$ is algebraically closed in L. □

The proof of the next computational lemmas is left to the reader (see [8], p. 404 for details):

Lemma 7.

If the vector field $W(q+1)={\Sigma }_{1\le \mid \nu \mid \le q+1}{a}_{\nu }^k\left(y_{q+1}\right){\displaystyle \frac{\partial }{\partial y_{\nu }^k}}$ is commuting with vertical vector fields $V(q+1)\in \Theta$, we have the following relations:

$$W(q+1)d_i\Phi =d_i(W\left(q\right)\Phi )-{\Sigma }_{0\le \mid \mu \mid \le q}(d_i{a}_{\mu }^k-a_{\mu +1_i}^k){\displaystyle \frac{\partial }{\partial y_{\mu }^k}}$$

$$[V(q+1),W(q+1)]d_i\Phi =V(q+1)(W(q+1)d_i\Phi )-W(q+1)(V(q+1)d_i\Phi )=V(q+1)(W(q+1)d_i\Phi )$$

$$\Rightarrow W((q+1)d_i\Phi =L_i(\Phi ,d\Phi )$$

Lemma 8.

One has the useful formula for prolongations of source transformations:

$$♭\left({\xi }_{q+1}\right)d_i\Phi =d_i(♭\left({\xi }_q\right)\Phi )-{\xi }_i^r{d}_r\Phi -♭\left(d{\xi }_{q+1}\left({\partial }_i\right)\right)\Phi$$

The following examples explain the origin of the well-known Wronskian determinant, which exists in the classical Picard–Vessiot theory, albeit in a completely different setting (see [8], p. 401).

Example 23.

When $m=n=1$ and $k=\mathbb{Q}$, let us consider the differential automorphic extension $L/K$ with $K=k<\Phi >\subset L=k<y>$ and $\Phi ={\displaystyle \frac{y_x}{y}}$. The underlying Lie group action is $\overline{y}=ay\Rightarrow {\overline{y}}_x=a{y}_x$ when $a=cst$, and it is clear that the word “constant” is not well-defined. In contrast, the underlying Lie pseudogroup is defined over k by the nonlinear first-order system ${\displaystyle \frac{1}{\overline{y}}}{\displaystyle \frac{\partial \overline{y}}{\partial y}}={\displaystyle \frac{1}{y}}$ in Lie form or, equivalently, by the differential algebraic OD equation $y{\displaystyle \frac{\partial \overline{y}}{\partial y}}-\overline{y}=0$ over the standard target differential field $k\left(y\right)$. Differentiating with respect to y, we obtain the linear second-order equation ${\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}=0$ that does not need to be integrated. As for the PHS law, we have, at once, the two relations ${\displaystyle \frac{\partial \overline{y}}{\partial y}}={\displaystyle \frac{{\overline{y}}_x}{y_x}}={\displaystyle \frac{\overline{y}}{y}},{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}=0$ for second-order jets.

Now, we have the target transformations:

$$\theta =y{\displaystyle \frac{\partial }{\partial y}}\Rightarrow ♯\left(j_2\left(\theta \right)\right)={\rho }_2\left(\theta \right)=y{\displaystyle \frac{\partial }{\partial y}}+y_x{\displaystyle \frac{\partial }{\partial y_x}}+y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}}$$

because $j_2\left(\theta \right)=(y,1,0)$ over the target.

For the commuting distribution used in the last lemmas, we may use

$$\delta \left(1\right)=y_x{\displaystyle \frac{\partial }{\partial y_x}},\delta \left(2\right)=y_x{\displaystyle \frac{\partial }{\partial y_x}}+2y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}}$$

because $♭\left({\xi }_2\right)=\xi {\displaystyle \frac{\partial }{\partial x}}-y_x{\xi }_x{\displaystyle \frac{\partial }{\partial y_x}}-(y_x{\xi }_{xx}+2y_{xx}{\xi }_x){\displaystyle \frac{\partial }{\partial y_{xx}}}$ (care with the factor 2), and thus,

$$-{\xi }_x\to y_x{\displaystyle \frac{\partial }{\partial y_x}}+2y_{xx}{\displaystyle \frac{\partial }{\partial y_{xx}}},-{\xi }_{xx}\to y_x{\displaystyle \frac{\partial }{\partial y_{xx}}}$$

and we indeed have $det\left(\begin{array}{cc}{y}_x& 2y_{xx}\\ 0& y_x\end{array}\right)={\left(y_x\right)}^2\ne 0$.

With $\Phi ={\displaystyle \frac{y_x}{y}}\Rightarrow d_x\Phi ={\displaystyle \frac{y_{xx}}{y}}-{\left({\displaystyle \frac{y_x}{y}}\right)}^2={\displaystyle \frac{y_{xx}}{y}}-{\Phi }^2$, we therefore obtain

$$\delta \left(1\right)\Phi =\Phi \Rightarrow \delta \left(2\right)d_x\Phi =d_x\Phi +(y_x{\displaystyle \frac{\partial \Phi }{\partial y}}+y_{xx}{\displaystyle \frac{\partial \Phi }{\partial y_x}}=d_x\Phi )+(-{\Phi }^2+{\displaystyle \frac{y_{xx}}{y}})=2d_x\Phi$$

We finally notice that $d_x\left({\displaystyle \frac{\partial \overline{y}}{\partial y}}\right)=d_x\left({\displaystyle \frac{\overline{y}}{y}}\right)={\displaystyle \frac{{\overline{y}}_x}{y}}-{\displaystyle \frac{\overline{y}{y}_x}{y^2}}={\displaystyle \frac{y_x}{y}}({\displaystyle \frac{\partial \overline{y}}{\partial y}}-{\displaystyle \frac{\overline{y}}{y}})=0$.

It follows from the chain rule for derivatives that we also have $d_x\left({\displaystyle \frac{\partial \overline{y}}{\partial y}}\right)=y_x{\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}=0$ and, thus, ${\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}=0$ whenever $y_x\ne 0$. This result is coherent with the fact that the OD equation $y_x=0$ is invariant under any diffeomorphism $\overline{y}=g\left(y\right)$ contrary to a linear OD equation of the form $y_x-\omega y=0$ with $\omega \ne 0$ in a differential field.

Example 24.

When $n=1,m=2$, and $k=\mathbb{Q}$, let us consider the Picard–Vessiot differential automorphic extension $L/K$ with $K=k<{\Phi }^1,{\Phi }^2>\subset L=k<y^1,y^2>$ along Example 18. The underlying Lie group action is ${\overline{y}}^1=a{y}^1+b{y}^2,{\overline{y}}^2=c{y}^1+d{y}^2$, or simply $\overline{y}=Ay$ when $(a,b,c,d)$ are constants, and it is clear that the word “constant” is not well defined, as in the previous example. Moreover, the underlying Lie pseudogroup is surely defined over k by the nonlinear first-order system ${\overline{y}}^u={\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}{y}^k$ or simply $\overline{y}={\displaystyle \frac{\partial \overline{y}}{\partial y}}y$ over the standard target differential field $k\left(y\right)$. However, by differentiating these equations with respect to $y^1$ and $y^2$, as in the previous example, we cannot obtain the linear second-order equations ${\displaystyle \frac{{\partial }^2{y}^u}{\partial y^k\partial y^l}}=0$ or simply ${\displaystyle \frac{{\partial }^2\overline{y}}{\partial y^2}}=0$ as before, and these second-order PD equations must be added independently. The underlying Lie pseudogroup must, therefore, be defined by second-order PD equations, and we are no longer allowed to exhibit solutions in the classical form ${\displaystyle \frac{\partial \overline{y}}{\partial y}}=A$ with $A=cst$.

Now, we have the 4 target transformations ${\theta }_k^l=y^l{\displaystyle \frac{\partial }{\partial y^k}}$ for $k,l=1,2$ and their 4 prolongations up to any order.

As for the commuting distribution, we have the two ${\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y_x^k}},{\delta }_2=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}$ at the order of one, which can be completed by the two ${\delta }_3=y^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}},{\delta }_4=y_x^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}}$ at the order of two. By prolonging source transformations, as in the previous example, we obtain (take care, again, with the factor 2):

$$-{\xi }_x\to y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}+2y_{xx}^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}}={\delta }_2-2{\Phi }^2{\delta }_3+2{\Phi }^1{\delta }_4,-{\xi }_{xx}\to {\delta }_4=y_x^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}}$$

It follows from the chain rule for derivatives that we also have ${\overline{y}}_x^u=d_x\left({\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}{y}^k\right)={\displaystyle \frac{{\partial }^2{\overline{y}}^u}{\partial y^k\partial y^l}}{y}^k{y}_x^l+{\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}{y}_x^k$ and, thus, both ${\overline{y}}^u={\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}{y}^k,{\overline{y}}_x^u={\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}{y}_x^k$.

By extending ${\delta }_1$ to the derivation ${\delta }_1=y^k{\displaystyle \frac{\partial }{\partial y_x^k}}+{\overline{y}}^k{\displaystyle \frac{\partial }{\partial {\overline{y}}_x^k}}$ of $L{\otimes }_{K}L$ while applying it, we successively get

$$y^k{\delta }_1\left({\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}\right)=0,y_x^k{\delta }_1\left({\displaystyle \frac{\partial {\overline{y}}^u}{\partial y^k}}\right)=0$$

The determinant of this $2\times 2$ linear system for each u is the Wronskian determinant:

$$det\left(\begin{array}{cc}{y}^1& y^2\\ y_x^1& y_x^2.\end{array}\right)=y^1{y}_x^2-y^2{y}_x^1\ne 0$$

We may similarly extend ${\delta }_2$ to the derivation ${\delta }_2=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}}+{\overline{y}}_x^k{\displaystyle \frac{\partial }{\partial {\overline{y}}_x^k}}$ of $L{\otimes }_{K}L$ to check that ${\delta }_2$ also kills ${\displaystyle \frac{\partial \overline{y}}{\partial y}}$ whenever $y^1{y}_x^2-y^2{y}_x^1\ne 0$.

The extension of these results to an arbitrary m is elementary and left to the reader.

Example 25.

With $k=\mathbb{Q},m=2,n=1$, let us consider the general automorphic extension $k\subset K\subset K_0\subset L$ for the Lie groupoid of isometries of the Euclidean metric for ${\mathbb{R}}^2$. In this case, taking the determinant of the isometry, we obtain ${\Delta }^2=1$ with $\Delta ={\displaystyle \frac{\partial ({\overline{y}}^1,{\overline{y}}^2)}{\partial (y^1,y^2)}}$, that is, $\Delta =\pm 1$. The prolongations at order two of the infinitesimal target transformations are

$${\theta }_1={\displaystyle \frac{\partial }{\partial y^1}},{\theta }_2={\displaystyle \frac{\partial }{\partial y^2}},{\theta }_3=y^1{\displaystyle \frac{\partial }{\partial y^2}}-y^2{\displaystyle \frac{\partial }{\partial y^1}}+y_x^1{\displaystyle \frac{\partial }{\partial y_x^2}}-y_x^2{\displaystyle \frac{\partial }{\partial y_x^1}}+y_{xx}^1{\displaystyle \frac{\partial }{\partial y_{xx}^2}}-y_{xx}^2{\displaystyle \frac{\partial }{\partial y_{xx}^1}}$$

Accordingly, the only generating differential invariant of order one is $\Omega ={\left(y_x^1\right)}^2+{\left(y_x^2\right)}^2$, while the generating differential invariants at strict order two are $\Gamma ={\displaystyle \frac{1}{2}}{d}_x\Omega =y_x^1{y}_{xx}^1+y_x^2{y}_{xx}^2$ and ${\rm Y}={\left(y_{xx}^1\right)}^2+{\left(y_{xx}^2\right)}^2$. Therefore, we obtain the general differential automorphic extension $K\subset L$ with $K=k<\Omega ,\Gamma ,{\rm Y}>$ and $L=k<y^1,y^2>$. By now setting $\Sigma =y_x^1{y}_{xx}^2-y_x^2{y}_{xx}^1$, we may introduce the other intermediate differential field $K_0=k<\Omega ,\Gamma ,\Sigma >$ with $di{m}_K\left(K_0\right)=\mid K_0/K\mid =2$.

Indeed, using jet coordinates, we have

$$\left(\begin{array}{cc}{\overline{y}}_x^1& {\overline{y}}_{xx}^1\\ {\overline{y}}_x^2& {\overline{y}}_{xx}^2\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}{y}_x^1& y_{xx}^1\\ y_x^2& y_{xx}^2\end{array}\right)$$

Hence, when taking the determinants, we finally obtain $\overline{\Sigma }=(ad-bc)\Sigma$ with Jacobian $\Delta =(ad-bc)$. Meanwhile, we have the algebraic relation ${\Sigma }^2+{\Gamma }^2-\Omega {\rm Y}=0$, and $K_0$ is the algebraic closure of K in L. In the present situation, we have an action of the Lie group G on $(y^1,y^2)$ with ${\overline{y}}^1=a{y}^1+b{y}^2+\alpha ,{\overline{y}}^2=c{y}^1+d{y}^2+\beta$, where $(a,b,c,d,\alpha ,\beta )=(1rotation+2translations)$ are ordinary constants. The connected component $G^0$ of the identity determines the differential automorphic extension $L/K_0$. In the case of G, the symbol $g_2$ of order two is defined by

$$y_x^1{v}_{xx}^1+y_x^2{v}_{xx}^2=0,y_{xx}^1{v}_{xx}^1+y_{xx}^2{v}_{xx}^2=0$$

and $g_2=0$ if and only if $\Sigma \ne 0$. In the case of $G^0$, we have the two linear equations:

$$y_x^1{v}_{xx}^1+y_x^2{v}_{xx}^2=0,y_x^2{v}_{xx}^1-y_x^1{v}_{xx}^2=0$$

and $g_2=0$ if and only if ${\left(y_x^1\right)}^2+{\left(y_x^2\right)}^2\ne 0$ in a coherent way with the linearization of the previous algebraic equation. The reciprocal distribution at order two is generated by

$${\delta }_1=y_x^k{\displaystyle \frac{\partial }{\partial y_x^k}},{\delta }_2=y_{xx}^k{\displaystyle \frac{\partial }{\partial y_x^k}},{\delta }_3=y_x^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}},{\delta }_4=y_{xx}^k{\displaystyle \frac{\partial }{\partial y_{xx}^k}}$$

and is easily seen to stabilize both K and even $K_0$ because we have

$${\delta }_1\Sigma =\Sigma ,{\delta }_2\Sigma =0,{\delta }_3\Sigma =0,{\delta }_4\Sigma =\Sigma$$

and is full rank, equal to 4, whenever $\Sigma \ne 0$. However, $K_0$ is algebraically closed in L, and thus, $k\left[G^0\right]\subset cst\left(L{\otimes }_{K_0}L\right)$ is an integral domain because $L{\otimes }_{K_0}L$ is an integral domain. In contrast, $L{\otimes }_{K}L$ is the direct sum of two integral domains because $K_0{\otimes }_K{K}_0\simeq K_0\oplus K_0\simeq K_0{\otimes }_{\mathbb{Q}}(\mathbb{Q}\oplus \mathbb{Q})$ is a direct sum of two fields, as $K_0/K$ is even a classical Galois extension with a Galois group defined by ${\Delta }^2=1$, and $L/K_0$ is a regular extension, that is, $K_0$ is algebraically closed in L. In more detail, we have $\overline{\Sigma }=\Delta \Sigma \Rightarrow {\overline{\Sigma }}^2={\Sigma }^2$, which can also be obtained by substraction from the previous algebraic identity, and thus, ${\overline{\Sigma }}^2-{\Sigma }^2=(\overline{\Sigma }-\Sigma )(\overline{\Sigma }+\Sigma )=0$, as a way to split the tensor product when ${\Sigma }^2\in K$ ([8], Remark 4.57, p. 115; [12], pp. 268–271; [6], p. 122). Extending each reciprocal distribution δ of $L/K$ to $L{\otimes }_{K}L$ as usual, we let the reader prove directly that $\Delta =\overline{\Sigma }/\Sigma \Rightarrow \delta \Delta =0$. We also let the reader repeat the previous computations in the special case by introducing the nonlinear equations $\Omega =\omega \in K,\Gamma =\gamma \in K,{\rm Y}=\upsilon \in K,\Sigma =\sigma \in K_0$ with CC $\gamma ={\displaystyle \frac{1}{2}}{\partial }_x\omega ,{\sigma }^2=\omega \upsilon -{\gamma }^2\in K$ whenever K is a given differential field of characteristic zero and $L=Q(Ky/\mathfrak{p})$, where $\mathfrak{p}$ is a prime differential ideal defining a differential automorphic system along the ideas of Vessiot.

Remark 5.

It is quite difficult to discover the confusion carried by Drach, Kolchin and their followers between “maximum ideals” and “prime ideals” when dealing with the so-called concept of reductibility in the Picard–Vessiot or Drach–Vessiot theories. For the interested reader, we only indicate how to recover it by reading the book of Kolchin ([17]) backwards as follows:

VI.5: Proposition 13, p. 412 ⇒ IV; 5: Corollary 2 to Proposition 2, p. 152 ⇒ III.10: Propositions 6 and 7, p. 142 ⇒ II.1: Theorem 1, p. 86.

To define a Picard–Vessiot differential extension, the basic idea is to exhibit a prime differential ideal that is maximally constrained with respect to the fact that it must not contain the Wronskian determinant W. In actual practice, for an ODE of order m like $d^{m}y+a_{n-1}{d}^{m-1}y+\dots +a_{0}y=0$ with coefficients in K, we have $dW+a_{m-1}W=0$, and the localization of $K\{y^1,\dots ,y^m\}$ at W is a differential ring, in which we may look for a maximal radical ideal that is, therefore, prime as a way to construct the PV extension $L/K$. This may indeed be a challenging task, given the technical content of these results and their proofs.

Moreover, the assumption made by Kolchin that the field C of differential constants, which must be the same for K and L, must be algebraically closed, is in contradiction with the spirit of the Galois theory. Another way to grasp the importance of maximal ideals is to look at reference [19] while comparing it with [13]. It is also interesting to compare this paper with [41] and other Hopf/Galois tentatives, such as those in [42,43]. When $m=3$, copying (three times) the Riccati OD equation $y_x={\omega }^1{\left(y\right)}^2+{\omega }^{2}y+{\omega }^3$ for $(y^1,y^2,y^3)$ with $\omega \in K$, the Wronskian determinant is replaced by the Vandermonde determinant $(y^1-y^2)(y^1-y^3)(y^2-y^3)$, and $\Delta$ is known (see [8] pp. 443–444).

Example 26.

(Hamilton–Jacobi Equation Revisited) Let $z=f(t,x)$ be a solution to the nonlinear PD equation $z_t+H(t,x,z,p)=0$, written with jet notations, while setting $z_x=p$ when t is time, x is space, z is the action and p is the momentum. A complete integral $z=f(t,x;a,b)$ is a family of solutions depending on two constant parameters $(a,b)$ in such a way that the Jacobian condition is $\partial (z,p)/\partial (a,b)\ne 0$ whenever $p={\partial }_{x}f(t,x;a,b)$. It has been shown by Vessiot (in 1915) that the search for a complete solution to the HJ equation is equivalent to the search for a single solution to the automorphic system for the algebraic pseudogroup Γ of contact transformations, preserving $dZ-PdX$ up to a function factor, obtained by eliminating the factor $\rho (t,x,z,p)$ in the Pfaffian system:

$$dz-pdx+H(t,x,z,p)dt={\displaystyle \frac{\partial f}{\partial a}}da+{\displaystyle \frac{\partial f}{\partial b}}db=\rho (dZ-PdX)$$

Setting $a=X(t,x,z,p),b=Z(t,x,z,p),\rho (t,x,z,p)={\displaystyle \frac{\partial f}{\partial b}},P(t,x,z,p)={\displaystyle \frac{\partial f}{\partial a}}/{\displaystyle \frac{\partial f}{\partial b}}$, we may recover the complete integral by using the implicit function theorem. The final involutive automorphic system ${\mathcal{A}}_1^{\left(1\right)}$ for ${\mathcal{R}}_1^{\left(1\right)}$ is defined by six algebraic PDEs only, and no restriction may be imposed on $H(t,x,z,p)$.

Similarly, the search for a complete integral of the form $z=f(t,x;a)+b$ is equivalent to the search for a single solution to the automorphic system for the algebraic pseudogroup ${\Gamma }^{\prime }\subset \Gamma$ of unimodular contact transformations, preserving $dZ-PdX$, obtained by using the Pfaffian system:

$$dz-pdx+H(t,x,p)dt=dZ-PdX\Rightarrow dx\wedge dp+dH\wedge dt=dX\wedge dP$$

For $dZ\wedge dX\wedge dP$, we obtain

$$(dz-pdx+Hdt)\wedge (dx\wedge dp+dH\wedge dt)=dz\wedge dx\wedge dp+Hdt\wedge dx\wedge dp+dz\wedge dH\wedge dt+pdH\wedge dx\wedge dt$$

By again closing this exterior system, we obtain $2dH\wedge dt\wedge dx\wedge dp=0\Rightarrow {\displaystyle \frac{\partial H}{\partial z}}=0$.

Alternatively, we have $p={\displaystyle \frac{\partial f}{\partial x}}(t,x;a)\Rightarrow a=X(t,x,p)\Rightarrow {\displaystyle \frac{\partial X}{\partial z}}=0$, $b=z-f(t,x;a)=Z(t,x,z,p)\Rightarrow {\displaystyle \frac{\partial Z}{\partial z}}\ne 0$, ${\displaystyle \frac{\partial f}{\partial a}}=P(t,x,p)\Rightarrow {\displaystyle \frac{\partial P}{\partial z}}=0,\rho =1$ and, thus, ${\displaystyle \frac{\partial P}{\partial t}}{\displaystyle \frac{\partial X}{\partial z}}-{\displaystyle \frac{\partial P}{\partial z}}{\displaystyle \frac{\partial X}{\partial t}}={\displaystyle \frac{\partial H}{\partial z}}=0$. Accordingly, the only compatibility condition needed on H is

$${\displaystyle \frac{\partial H}{\partial z}}=0$$

Similarly, the search for a complete integral of the form $z=u(t;a)+v(x;a)+b$ is equivalent to the search for a single solution to the automorphic system for the algebraic pseudogroup ${\Gamma }^{″}\subset {\Gamma }^{\prime }$ obtained by using the Pfaffian system:

$$dz-pdx+H(t,x,p)dt=dZ-PdX\Rightarrow {\displaystyle \frac{\partial X}{\partial x}}/{\displaystyle \frac{\partial X}{\partial p}}={\displaystyle \frac{\partial H}{\partial x}}/{\displaystyle \frac{\partial H}{\partial p}}$$

Indeed, we have $p=g(x;a)\Rightarrow a=X(x,p)\Rightarrow \partial X/\partial t=0,\partial X/\partial z=0$, but we also have $\partial (X,P)/\partial (t,x)=-\partial H/\partial x$,  $\partial (X,P)/\partial (t,p)=-\partial H/\partial p$, and thus,

$${\displaystyle \frac{\partial X}{\partial t}}=0\Rightarrow {\displaystyle \frac{\partial H}{\partial p}}.{\displaystyle \frac{\partial X}{\partial x}}-{\displaystyle \frac{\partial H}{\partial x}}.{\displaystyle \frac{\partial X}{\partial p}}=0$$

The corresponding Lie sub-pseudogroup is

$${\Gamma }^{″}=\{\overline{X}=g\left(X\right),\overline{Z}=Z+h\left(X\right),\overline{P}=(P+\partial h/\partial X)/(\partial g/\partial X)\}\subset {\Gamma }^{\prime }$$

The above necessary conditions for separating the variables in the integration of the HJ equation were found by T. Levi-Civita in 1904 [44] and are

$${\displaystyle \frac{\partial H}{\partial z}}=0,{\displaystyle \frac{\partial }{\partial t}}({\displaystyle \frac{\partial H}{\partial x}}/{\displaystyle \frac{\partial H}{\partial p}})=0$$

Finally, the search for a complete integral of the form $z=u\left(t\right)+v(x;a)+at+b$ is equivalent to the search for a single solution to the automorphic system for the algebraic Lie pseudogroup ${\Gamma }^{‴}=\{\overline{X}=X+c,\overline{Z}=Z+h\left(X\right),\overline{P}=P+\partial h/\partial X\}$, provided that ${\displaystyle \frac{\partial H}{\partial z}}=0,{\displaystyle \frac{{\partial }^{2}H}{\partial t\partial x}}=0,{\displaystyle \frac{{\partial }^{2}H}{\partial t\partial p}}=0$, and so on. In particular, if we ask for a complete integral of the form $z=v(x;a)+at+b$, we have to look for one solution to the automorphic system for the intransitive Lie pseudogroup ${\Gamma }^{⁗}=\{\overline{X}=X,\overline{Z}=Z+h\left(X\right),\overline{P}=P+\partial h/\partial X\}$ because we must add the equation $X=-H$, provided that $H=H(x,p)$. Indeed, we have $p=\partial z/\partial x=g(x;a)\Rightarrow a=X(x,p)\Rightarrow \partial X/\partial t=0$, and the automorphic system is defined by 12 algebraic PDEs, including 1 zero-order equation.

As an explicit example, by setting $w=z_t$, we may consider the general HJ equation $wp-txz=0$, and we let the reader prove that $z=({\displaystyle \frac{1}{2}}{t}^2+a)({\displaystyle \frac{1}{2}}{x}^2+b)$ is a complete integral.

To recapitulate, if K is a differential field for $\partial =\{{\partial }_t,{\partial }_x,{\partial }_z,{\partial }_p\}$ containing p and H, we may introduce the finitely generated differential extension $K\subset L=K<X,Z,P>$ and apply DGT to the five successive intermediate differential fields corresponding to the algebraic Lie pseudogroups:

$$\{\overline{X}=X,\overline{Z}=Z,\overline{P}=P\}\subset {\Gamma }^{⁗}\subset {\Gamma }^{‴}\subset {\Gamma }^{″}\subset {\Gamma }^{\prime }\subset \Gamma$$

We invite the reader to find a basis of algebraic differential invariants defined over $\mathbb{Q}$ to understand the technical/mathematical difficulty to test the effective application of the two criteria for automorphic systems already provided to discover the importance of the work of Vessiot, which is still not acknowledged up to now!

Remark 6.

We prove that DGT cannot be separated from the formal theory of systems of OD or PD equations, a reason for which Kolchin has not achieved it along the work lines of Spencer, as we explained in the Introduction. We consider the second situation of the Pfaffian automorphic system $dz-pdx-H(t,x,p)dt=dZ-PdX$ for the algebraic Lie pseudogroup ${\Gamma }^{\prime }$ defined by $d\overline{Z}-\overline{P}d\overline{X}=dZ-PdX\Rightarrow d\overline{X}\wedge d\overline{P}=dX\wedge dP$. First of all, we notice that we have ${\displaystyle \frac{\partial (\overline{X},\overline{P})}{\partial (P,Z)}}=0,{\displaystyle \frac{\partial (\overline{X},\overline{P})}{\partial (Z,X)}}=0\Rightarrow {\displaystyle \frac{\partial \overline{X}}{\partial Z}}=0,{\displaystyle \frac{\partial \overline{P}}{\partial Z}}=0$ whenever ${\displaystyle \frac{\partial (\overline{X},\overline{P})}{\partial (X,P)}}=1$, and thus, ${\displaystyle \frac{\partial \overline{Z}}{\partial Z}}=1$. Now, using the jet notations, the infinitesimal transformations of the pseudogroup are defined by the system $R_1\subset J_1\left(T\left(Y\right)\right)$ over the target Y with local coordinates $(X,P,Z)=(y^1,y^2,y^3)$ and with the specific ordering $y^1\prec y^2\prec y^3$, namely $\{{\eta }_1^3-y^2{\eta }_1^1-{\eta }^2=0,{\eta }_2^3-y^2{\eta }_2^1=0,{\eta }_3^3-y^2{\eta }_3^1=0\}$; this is not even formally an integrable, as three crossed derivatives allow for three additional first-order PD equations, providing the new first-order system $R_1^{\left(1\right)}\subset J_1\left(T\left(Y\right)\right)$:

$$\left\{\begin{array}{c}{\eta }_3^3=0\hfill \\ {\eta }_3^2=0\hfill \\ {\eta }_3^1=0\hfill \\ {\eta }_2^2+{\eta }_1^1=0\hfill \\ {\eta }_2^3-y^2{\eta }_2^1=0\hfill \\ {\eta }_1^3-y^2{\eta }_1^1-{\eta }^2=0\hfill \end{array}\right.\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\\ 1& 2& 3\\ 1& 2& •\\ 1& 2& •\\ 1& •& •\end{array}$$

With three equations of class 3, two equations of class 2 and one equation of class 1, this system is seen to be involutive through its Janet tabular; thus, we have $dim\left(R_1^{\left(1\right)}\right)=(3+3\times 3)-6=12-6=6$. The corresponding Janet sequence is a resolution of ${\Theta }^{\prime }\left(Y\right)$ with a Euler–Poincaré characteristic $3-6+4-1=0$ as follows:

$$0\to {\Theta }^{\prime }\left(Y\right)\to 3\underset{1}{⟶}6\underset{1}{⟶}4\underset{1}{⟶}1\to 0$$

Now, the set of equations of the invariant automorphic system surely contains

$${\displaystyle \frac{\partial Z}{\partial t}}-P{\displaystyle \frac{\partial X}{\partial t}}=H(t,x,p),{\displaystyle \frac{\partial Z}{\partial p}}-P{\displaystyle \frac{\partial X}{\partial p}}=0,{\displaystyle \frac{\partial Z}{\partial x}}-P{\displaystyle \frac{\partial X}{\partial x}}=-p,{\displaystyle \frac{\partial X}{\partial z}}=0,{\displaystyle \frac{\partial Z}{\partial z}}=1,{\displaystyle \frac{\partial P}{\partial z}}=0$$

$${\displaystyle \frac{\partial (X,P)}{\partial (x,p)}}=1,{\displaystyle \frac{\partial (X,P)}{\partial (x,t)}}={\displaystyle \frac{\partial H}{\partial x}},{\displaystyle \frac{\partial (X,P)}{\partial (p,t)}}={\displaystyle \frac{\partial H}{\partial p}}$$

As before, we obtain nine equations, including ${\displaystyle \frac{\partial Z}{\partial z}}=1,{\displaystyle \frac{\partial X}{\partial z}}=0,{\displaystyle \frac{\partial X}{\partial z}}=0$. Using the specific ordering $t\prec p\prec x\prec z$ for the corresponding Janet tabular, we obtain, at once, these three equations of class z, but also, three equations of class x, two equations of class p and only one equation of class t, namely, the only one containing $H(t,x,p)$; the resulting 10 CCs are satisfied because ${\displaystyle \frac{\partial H}{\partial z}}=0$ (see [6], p. 178). Accordingly, setting $(t,x,p,z)=(x^1,x^2,x^3,x^4)$ for the source X, for the corresponding automorphic system, we obtain $dim\left({\mathcal{A}}_1^{\left(1\right)}\right)=(3+3\times 4)-9=6$, providing an automorphic system for ${\Gamma }^{\prime }\subset aut\left(Y\right)$ because the same equality is still valid at any order $q\ge 1$. Such a situation is exactly similar to that of Example 14 with ${\mathcal{A}}_1$ and ${\mathcal{R}}_1^{\left(1\right)}$.

We may finally wonder how to extend such a result to arbitrary source/target dimensions $n=2p,m=2p+1$. As explained in ([8], pp. 684–691), the geometric object allowing us to study a contact structure is no longer a 1-form, but becomes a 2-contravariant skew-symmetric tensor density satisfying first-order, highly nonlinear Vessiot structure equations, and there is no work in this direction.

6. Conclusions

In 1983, the author of this paper was correcting the proofs of his GB book “Differential Galois Theory”, in particular, the pages of the second part, dealing with the linear and nonlinear Spencer operators. He suddenly remembered the content of page 137 in a book called “Théorie des Corps Déformables”, written by the brothers E. and F. Cosserat in 1909 [45,46,47], in which the two brothers were describing the so-called “Cosserat Couple-Stress Equations” for continuum mechanics as a better substitute for the “Cauchy Stress Equations”. In a few seconds, the comparison provided the third GB book “Lie Pseudogroups and Mechanics”, published in 1988 [14].

This chance has also given him the motivation of applying Lie pseudogroups to mathematical physics [42,43], and this, contrary to what could be imagined, is surely not out of the scope of this paper, as it brings heavy doubts as to the usefulness of introducing the “Differential Groups” of Ritt, Kolchin and all followers [48], despite the early advertising of the author, as we already said in the Introduction. This is also the reason for which we did not speak at all about the many followers of Kolchin (Casale, Ramis, Singer, Magid, Malgrange, Van der Put, etc.), who used a definition of the Galois group based on a confusion between prime differential ideals defined by Ritt in 1930 and maximal ideals used by Picard, Vessiot and Drach at the beginning of the last century. Such a definition is not conceptually correct, and it is a pity both that Drach never agreed to having made a mistake and also that Painlevé went along with these ideas, even though he was perfectly aware of this confusion; this can be seen from the letters that Janet gave to the author of this paper, which can be consulted in the ENS library in Paris. Additionally, when I met BB in Warsaw, during a period so bleak that only empty iron boxes filled the windows of shops, he held an important official position at the University. He told me that he had sent his papers to Kolchin, who never answered, and that he had other things to do rather than fight for recognition from someone unable to use tensor products of fields. Needless to say, Kolchin never understood what a Lie pseudogroup was, even at the level of Vessiot’s 1903 paper (according to private letters!), and I do not speak of the work of his colleague Spencer in Princeton. As a rather astonishing fact, it is a pity that Spencer and coworkers, in particular, H. Goldshmidt or A. Kumpera, never produced explicit examples apart from a few academic ones. The reader may discover in a few minutes that the examples presented in the introduction of [38] have no relation at all with the remaining core of the book. They also never understood that the only constant c of the constant curvature condition in Riemannian geometry, which has strictly nothing to do with the structure constants of any Lie algebra, is only to be found in the Vessiot structure equations that we have studied in detail throughout all our books and particularly in [7].

Thus, we have proved that in agreement with Vessiot, the interest of using DGT is to transform a problem with no evident group theoretical framework, for example, the study of a polynomial or the integration of a given Hamilton–Jacobi (HJ) equation, into a problem of looking for one solution to an automorphic system for an algebraic Lie group or Lie pseudogroup. As a result, the Galois correspondence existing between intermediate differential fields and algebraic Lie sub-pseudogroups provides, for the first time, a link between the various properties of the HJ equation and the form of the Hamiltonian, which only appears in the second members of the various corresponding automorphic systems successively considered. It appears that this is the first time DGT has been introduced in analytical mechanics, and Theorem 3.1 is unavoidable. We finally point out that the problem of defining precisely the meaning of “constant” has to do with the comparison of the two possible differential sequences that can be constructed when a linear differential operator is given, namely the “Janet sequence” and the “Spencer sequence”. In Remark 4, we have only sketched the solution to this important question, which should become a challenging domain of research regarding modern methods of studying DGT through differential homological algebra (more details can be found in the recent reference [49], which just appeared).

We hope these new methods will provide open fields of applied research in the future, based on the use of Hopf algebras and being compatible with modern computer algebra packages, even if they largely contradict the present ones accepted up to now by the mathematical community, despite many intensive European courses and summer schools (see AACA 2009 for more details).