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From what I can tell, a linear substitution is an operation on a set of variables $x_1,\ldots,x_n$ which sends them to a new set of variables $y_1,\ldots, y_n$ via a linear transformation

$$\vec{y} = \mathbf{A}\vec{x}$$

Provided the matrix $\mathbf{A}$ is invertible, the transformation $\vec{x}\rightarrow \vec{y}$ is invertible. Because these linear substitutions are composable, invertible, and have unit, it is possible to form group structures out of them.

In a paper, Painlevé considers an arbitrary group $\alpha$ of linear substitutions for two variables. From what I understand, Painlevé asks us to consider a pair of functions $\varphi$ and $\psi$ which constitute an invariant with respect to the group operation; that is to say,

$$\langle x, y\rangle =\langle \varphi(t,u), \psi(t,u)\rangle$$

and whenever $\langle T, U\rangle$ are a solution to this equation, the complete set of solutions is exactly obtained by applying all the transformations in the group $\alpha$ to $T$ and $U$: $\{ f(T,U) : f\in \alpha\}$.

I am mystified, however, when Painlevé subsequently treats $t$ and $u$ as functions of $x$ and $y$, forming all second-order partial derivatives of

$$ \begin{align*} t(aT + bU + c) = a^\prime T + b^\prime U + c^\prime\\ u(aT + bU + c) = a^{\prime\prime} T + b^{\prime\prime}U + c^{\prime\prime}\\ \end{align*} $$

My questions are:

  1. Where did these equations come from?
  2. How is it that we treat $t$ and $u$ as functions of $x$ and $y$?
  3. Why are there constant terms $c$? (Surely a "linear substitution" cannot include constant terms or else the transformation wouldn't be invertible?)
  4. What does it mean to call $\langle \phi, \psi\rangle$ invariants? They look like a description of an arbitrary member $f\in \alpha$.

Perhaps something was lost in translation? The relevant excerpt, in French (which I don't speak), is included below:

Painlevé's description of finite groups

I surmise that it says something like:

Consider a finite group $\alpha$ of linear substitutions involving two variables and two fundamental invariant functions which correspond to them: $$x=\phi(t,u),\qquad y=\psi(t,u);$$ If, for a system $(x,y)$, the values $(T,U)$ satisfy the equations (1), all the other solutions of these equations are obtained by applying to the values $(T,U)$ all of the substitutions in the group $\alpha$. We differentiate the equations

$$ \begin{align*} t(aT + bU + c) = a^\prime T + b^\prime U + c^\prime\\ u(aT + bU + c) = a^{\prime\prime} T + b^{\prime\prime}U + c^{\prime\prime}\\ \end{align*} $$

with respect to $x$ and $y$ up to second order, inclusive.

In this way, we form twelve equations [i.e. $\partial_x, \partial_y, \partial_{xx}, \partial_{xy}, \partial_{yx}, \partial_{yy}$ for the two equations?], homogeneous and linear in $a$, $b$, $c$, $a^\prime$, ...., and, if we eliminate those constants, the remaining four equations involve partial derivatives (first and second) of $t$ and of $u$.


I discovered that in Transformations of the fundamental equations of thermodynamics, Buckley describes a similar kind of (reversible differentiable but not necessarily linear?) substitution which might shed light on this one. To paraphrase:

Suppose the state of a physical system can be described by variables $x_1,\ldots x_n$. Then there exists a characteristic function $E$, which is the solution of a particular exact differential equation $$\partial E - \sum_i x_i\,dx_i = 0,$$ such that the behavior of the system in any state can be conveniently described in terms of $E$ and its partial derivatives with respect to the variables $x_1,\ldots, x_n$. Specifically, in thermodynamics, we can use the state variables volume $V$ and entropy $S$; all general thermodynamic formulas are derivable from $E$ and the basic equations $$\begin{align*}H &\equiv E - V\, \partial_V E\\ F &\equiv E - S\partial_S E\\G &\equiv E - V\,\partial_V E - S\partial_S E = H-F+E\end{align*}$$

"Legitimate operations" (?) on these equations generate a collection $C$ of thermodynamic formulas which are valid with respect to the state variables $V, S$.

If we change variables $\mathbf{\mathsf t}:(V,S)\mapsto (V^\prime, S^\prime)$, the new variables

\begin{align*}V^\prime = \varphi(V)\\S^\prime = \psi(S)\\\end{align*}

are functionally independent and hence soluble for the originals. Under this change of variables, the functions $EHFG$ and formulas $C$ will, generally speaking, change form, but they must remain identically true.

Suppose, however, that the functions $EHFG$ retain the same form under a particular change of variables. Then the formulas $C$ must also retain the same form. If the collection of all such transformations forms a group $\mathsf{G}$, then the fundamental formulas will remain unchanged under $\mathsf{G}$, and hence the group characterizes a fundamental arbitrariness in the thermodynamic description of all systems.

This is analogous to the fact that, in relativity, the description of the laws of electromagnetic radiation remain unchanged under a Lorentz(-Einstein) transform.

But I guess I still don't understand formally what it means to say that an equation has "the same form" under a transformation $T$, e.g. expressed as a composition of functions. Perhaps it means that unlike in the general case where of course a change of variables doesn't affect the equation

$$(f\circ T^{-1})[T(x)] = (g\circ T^{-1})[T(x)]\;\iff\; f(x) = g(x),$$

a change of variables has the same functional form if not only that, but also (?):

$$f(T(x)) = g(T(x)) \iff f(x) = g(x)$$

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  • $\begingroup$ I think a parital answer to "where did those two equations come from?" is if you have an equation $\vec{x} = f(\vec{u})$ and you say that all other solutions are of the form $\vec{x} = f(g(\vec{U}))$, and $f$ and $g$ are both linear substitutions, you might get an equation of that form. $\endgroup$ – user326210 Jun 18 '17 at 20:18
  • $\begingroup$ No, it is not that simple! It deals with Lie groups. $\endgroup$ – Jean Marie Jun 18 '17 at 20:31
  • $\begingroup$ @JeanMarie But can't I just think of a Lie group as a group in the category of Manifolds and smooth maps? It seems that the only thing Lie groups provide here is differentiability of the transformations in the group (?). $\endgroup$ – user326210 Jun 19 '17 at 1:58
  • $\begingroup$ Well, certainly Painlevé didn't think of them that way, since he died a couple of decades before categories were born! I suspect you have to read with a 19th century mind (which I have an extremely hard time with, in addition to knowing nothing about any of this). For example, I think "linear" in today's linear algebra sense is not meant (and possibly hadn't come about yet), but instead was used to mean a polynomial of degree 1 (as is relatively common even today). $\endgroup$ – pjs36 Jun 19 '17 at 2:45
  • $\begingroup$ @pjs36, Thanks! That's helpful. Also, from what I understand, a general group theory hadn't been invented yet either; substitution transforms are a precursor. $\endgroup$ – user326210 Jun 20 '17 at 2:05

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