I would like to find this paper in English

Question

I found the paper Sur les assemblages de lignes by Camille Jordan after reading a book by Frank Harary on graph theory (the paper can be found here). I am convinced that this paper contains a result I am very interested in; it is about centroids of trees (according to the book). But the paper is in French, and I don't understand it.

Does anyone know if there is a translation of the paper somewhere on the Internet (something I failed to find myself)? Maybe it is explained in another graph theory book (it is certainly not explained in the book I have): does anyone know in which?

EDIT: the DOI of the paper is: 10.1515/crll.1869.70.185 (it can be found using the link above).

Answer 1

Here is a translation of the text "Sur les assemblages de lignes" of Camille Jordan (1869) that can be found here, one of the first works prefigurating the theory of graphs.

I have chosen in particular to translate "assemblages" into "bindings" but I could have used the term "graph"...

In order to help, I have stuck as possible to the text. In some rare cases (noticeable by the use of parentheses), I have preferred to take some distance with it.

---

$$\text{ABOUT LINE BINDINGS}$$

$$\text{(by M. Camille Jordan in Paris)}$$

1. Let $x,y,z,u,$... be points in any quantity ; $xy,xz,yu...$ be arcs, straight or curved but not forked, each of those connecting two of these points. We will call such a system an arc binding having $x,y,z,u,...$ as its vertices. Such a binding can be in a symbolic way represented by the quadratic form $xy+xz+yu+...$ whose terms represent the different arcs of the binding. If, besides, $a$ of these arcs have the same endpoints $x y$, the representative form of the binding will contain $a$ times the term $xy$.

Let now $xy,xz,...$ be the terms of the quadratic form containing $x$ ; $yu,...$ those who have a common letter with some of these; $uv$ those who have a common letter with the previous ones. It is clear that the binding will be continuous if the set of these terms $xy,xz,...,yu,...,uv,...$ contains all the letters: Otherwise, no. The binding will be said $n+1$ times continuous if it is possible to cut $n$ of its arcs forming it without destroying its continuity.

Two bindings will be said identical if one can establish between the arcs and vertices a correspondence such that, to each arc $xy$ of one of the bindings, corresponds in the other an arc $x'y'$ joining vertices $x' y'$ corresponding respectively to $x$ and $y$. In this case if $axy+bxz...$ is the representative form of $A$, $ax'y'+bx'z'+...$ will be the representative form of $A'$. A binding can be considered identical to itself according to different points of view. In this case, $A'$ will be undistinguishable from $A$, and $x',y',z',...$ will be identified, up to the order, with $x,y,z,...$ Moreover the two forms representative of the same binding will necessarily be identical. Therefore the substitution replacing $x,y,z,...$ by $x',y',z',...$ doesn't change the form $axy+bxz+...$ Reciprocally, let $S$ be a substitution which doesn't change the form and let us assume that it replaces respectively $x,y,z,...$ by $x',y',z',...$ Let us place in correspondence vertices $x,y,z,...$ with vertices $x',y',z',...$ ; to the arcs joining $x$ and $y$ the corresponding arcs joining the corresponding vertices $x',y'$, the latter arcs being taken in any other ; the binding will remain similar to itself.

The order of symmetry of the binding, that is to say the number of different ways by which it is similar to itself, is therefore equal to $1.2....a 1.2.... b ...N$, $N$ being the number of substitutions of the type of $S$.

2. Problem What are the continuous bindings, non similar to themselves, that one can constitute with $m$ arcs? What are the degree of continuity and symmetry of each of them?

The attached array gives the solution of this problem for $m=2,3,4$. Each binding is represented by its representative form. (One can easily build the corresponding geometrical figure). In the second and third columns are indicated its degree of continuity (Roman numerals) and its order of symmetry (with Arabic numerals)

ARRAY

One could easily extend this array to the $m=5$, etc. cases but its complexity would increase very rapidly. One obtains more easily results having a certain generality by leaving $m$ undetermined, and examining successively bindings with simple, double, triple,... continuity.

3. Bindings with simple continuity. Let $x$ be any vertex of the binding. One can distinguish in the binding various branches, all of them constituted by one of the arcs $xy_1,xy_2,...$ reaching vertex $x$ and by the set of arcs wich are connected to x and by the set of arcs which are connected to $x$ through this (specific arc). Let $p_1,p_2,...$ be the number of arcs resp. contained in the various branches ; let $q$ be the biggest of them. The binding will contain a vertex for which $q \le m/2$, or two vertices, situated at the two endpoints of the same arc for which $q=(m+1)/2$. For all other vertices, $q$ will be $> (m+1)/2$. Let's indeed assume that $x$ is chosen in such a way that $q$ reaches its minimum, and for the sake of definiteness $q_1=q$. Let $xy_1, x'y_1, x''y_1$, be the arcs reaching vertex $y_1$ ; $r,r',r''$ the resp. number of arcs that are connected to $y_1$, through these arcs: we will evidently have $r=m-q+1, r'+r''+...=q-1$. Therefore $q \le (m+1)/2$ ; if this was not the case, we would have simultaneously $r,r',r'',... <q$, a contradiction with the hypothesis.

This said, let (us consider) first (the case) $q<(m+1)/2$. Let $u$ be any vertex other than $x$; $xy_{\mu},...,zu$ the successive arcs, the number of them being $k$, through which it is connected to $x$. The number of arcs that are connected to u through arc $zu$ is evidently greater or equal to $k+m-p_{\mu}$, which is bigger than $(m+1)/2$, $k$ being at least equal to 1, and $p_{\mu}$ being at most equal to $a$.

If $q=(m+1)/2$, the same proof can be applied, unless one has at the same time $k=1$ and $p_{\mu}=q=(m+1)/2$, out of which $\mu=1$ : in such a case vertex $u$ is reduced to $y_1$. One can call center of the binding the remarkable vertex or arc the existence of which we just established. It's clear that this center is in correspondence with itself in all cases for which the binding is similar to itself. Let us now consider a general binding: let us admit, for the sake of definiteness, that its center is a vertex, and that the branches issued from this point can be categorized in this way:

1° $k$ branches made of $n$ arcs, arcs among which $k$ are similar between themselves and different from the previous ones; etc. 2° $k'$ branches having $n'$ arcs, among which $k'_1$ are similar between themselves, and then $k'_2$ are similar between themselves, being different from the previous ones, etc. ; that among numbers $k_1',k_2',....$ there are $a_1'$ of them equal between themselves, and then $a_2'$ of them equal between themselves and different from the previous ones. Let $\delta$ be the number of quantities $a_1,a_2,...$, $\delta'$ the number of quantities $a_1',a_2',...$. Let finally $B_n$ be the number of (different) ways one can build a branch with $n$ lines. It is clear that the number of different bindings corresponding to the given values for $n, k, k_1, k_2, .... n', k', k_1', k_2', ....$ will be

$$ \begin{array}{llll} m=2&\biggr\{\begin{array}{ll|ll} ab+bc&I;2&2ab&II;4 \end{array} \\ & \\ m=3&\Biggr\{ \begin{array}{ll|ll}ab+bc+cd&I;2&2ab+bc&II;2\\ ab+bc+bd&I;6&3ab&III;12\\ ab+bc+ca&II;6&& \end{array} \\ & \\ m=4&\Biggr\{\begin{array}{{ll|ll}} ab+bc+cd+de&I;2&2ab+bc+bd&II;4\\ ab+bc+cd+ce&I;2&2ab+bc+ad&II;4\\ ab+bc+bd+be&I;24&2ab+bc+ca&III;4\\ ab+bc+cd+bd&II;2&2ab+2bc&III;8\\ ab+bc+cd+da&II;8&3ab+bc&III;6\\ 2ab+bc+cd&II;2&4ab&IV;48 \end{array} \end{array} $$

and we will get the total number of bindings that can be formed with $m$ arcs by summing up the expressions connected to the systems of values of these quantities satisfying relationships

(formulas 2 and 3)

the last one expressing that $x$ is the center of the binding.

The order of symmetry of each of the bindings obtained in this way will evidently be equal to

(formula 4)

$S_1, S_2,... S'_1, S'_2,...$ being the resp. orders of symmetry of the different branches it contains.

It remains to determine $B_n, S_1$ and analogous quantities. With this objective, let us remark that a branch made of $n$ arcs is composed 1° of a first arc $xy$, 2° of other $(n-1)$ arcs, constituting together a binding with $y$ one of its vertices. $B_n$ is evidently equal to the number of ways one can build this binding and $S_1$ its order of symmetry. One will get therefore these two numbers by formulas similar to the ones given upwards, with $m$ replaced by $n-1$.

Besides, the summation must be extended to all the values of the variables satisfying relationships analogous to (2), because point $y$ can be any vertex of the new binding, we have here no relationship analogous to (3). The case of bindings with a central arc could be treated as easily.

Let $A$ be any binding with simple continuity. Among its arcs, there are necessarily some of them connected to the rest of the binding by only one of their endpoints, the other one remaining free. Let us suppress all of them. In a similar way let us suppress in the remaining binding $A'$ all the arcs having a free end ; let us repeat this operation till the penultimate set (the next step would make disappear all the remaining arcs). Two cases can occur :

1° A single arc remains.

2° More than one arc remains, all of them with a free endpoint. Their other extremities are connected to a very same vertex.

Here is a new way to discriminate in a binding a remarkable arc or a remarkable vertex, with specific properties, whose position in the binding can be considered in some ways as central. We will name it center of second kind.

5. We will distinguish in $A$ three kinds of vertices 1° the ordinary ones, where two arcs are meeting, 2° the terminal ends with a single arc reaching it, 3° the branching ones, where more than two are meeting. This said, the binding can evidently be divided into sections, limited to terminal or branching vertices, or sequential arcs connecting themselves through ordinary vertices.

These various sections, each of them considered as a whole, form a binding, where one can determine two centers using the previous methods, each one being able to correspond to itself in all the cases where $A$ is similar to itself. In the second case, according to the case where the number of arcs is even or odd, its center will be a remarkable vertex or arc, being homologous to itself for all the cases where $A$ is similar to itself.

6. Bindings with double continuity. Let us operate as in n°4 as long as there remains in the binding some arcs with a free end. The remaining lines will constitute a closed contour, always homologous to itself. This observation done, one can set, as in n°3, formulas giving the solution to the problem.

7. Bindings with triple continuity. Let us operate as in n°4. The remaining arcs will constitute a system $A'$, with triple continuity. One will be able to determine the arcs constituting a closed contour $C$. Let us cut one of these arcs: the remaining binding will be with double continuity, on which one can repeat operations of n°4, until it remains only a closed contour $D$. Two cases will have to be considered here.

1°. If $C$ and $D$ have no common arc, $A'$ will be made of 2 contours and a connecting line L, polygonal in general, but able to be, in certain particular cases, reduced to a single arc, or even to a point. The middle of $L$ will be a center of the binding.

2°. If $C$ and $D$ have have a common part, this part and the other two non common parts constitute a system of three lines having their extremities in two same points $p$, $p'$ that we can call "poles of the binding": and it's clear that the system formed by these two poles will stay constantly homologous with itself.

8. One will continue in the same way with bindings having a quadruple continuity, etc. But here is a result dealing with general bindings.

Let $A$ be a binding, $S$ one of its vertices. Let us split the arcs with $S$ as their endpoint into two categories; then, let us assume that a disjunction is operated between the arcs of the first and second category, without breaking the existing link between the arcs in a same category. One can call "scission" (splitting ?) such a an operation. Moreover, a binding will be called "composed" or "primitive" according resp. to the fact that one can or cannot destroy its continuity by a "scission".

This being set, it is clear that any composed binding will be fractionable by a sequence of convenient "scissions" into partial bindings that will be primitive; and these fragments, being assumed connected again, one can see easily by reasonings similar to those of n° 3,4,5, that there exists 4 different ways to determine in the binding a central vertex or a central fragment, always homologous to itself.

The issue of the symmetry of the bindings boils down in this way to the case of primitive bindings.

Paris, 1869