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% "On the 3-dimensional spaces which admit a continous group of motions"
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% http://www.homepage.villanova.edu/robert.jantzen

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{\huge{On the three-dimensional spaces which admit a continuous group of
motions}}\footnote{Original title: Sugli spazi a tre dimensioni che ammettono
un gruppo continuo di movimenti, {\it Memorie  di  Matematica  e  di  Fisica
della Societa Italiana delle Scienze, Serie Terza}, {\bf Tomo XI}, pp.~267--352
(1898). Printed with the kind permission of the Accademia Nazionale delle
Scienze, detta dei XL, in Rome, the current copyright owner. Translated by
Robert Jantzen, Department of Mathematical Sciences, Villanova University,
Villanova, Pa 19085, USA.}$^,$%
\footnote{This paper was also reprinted in: {\it
Opere} [The Collected Works of Luigi Bianchi], Rome, Edizione Cremonese, 1952,
vol.{\bf ~9}, pp.~17-109.}

\bigskip

\bigskip

{\Large {Essay by member Luigi Bianchi}}

\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\typeout{**** Begin section Preface}

\section*{Preface.}

We define the metric of a
space of $n$ dimensions in the manner of Riemann
by giving the expression for the square of its line
element:
$$
ds^2 = \sum_{i,k}^{1\ldots n} \, a_{ik} \, dx_i \, dx_k \ ,
\eqno(1)
$$
namely the law by which we measure infinitesimal arclengths in the
space $S_n$, from which the law of measure for finite
arclengths follows.

We consider $n$ independent real variables
$x_1,x_2,\ldots,x_n$ and assume that the coefficients $a_{ik}$
of the quadratic differential form (1) as well as their first and
second derivatives are real, finite and continuous functions of
the $x$ for the entire range of values which we consider.
We also assume that the discriminant
of expression (1) is always nonzero and that the
coefficients $a_{ik}$ fulfill the well known inequalities which
make this differential form {\it positive-definite}.

It is well known how the law for measuring angles and the entire
geometry of the space $S_n$ is determined by equation (1).  If two
spaces $S_n$, $S_n^\prime$ can be put into a one-to-one
correspondence in such a way that the line elements are the
same, the two spaces will be called {\it isometric\/}
and the two geometries will be identical.
When the line elements of the
two spaces only differ by a constant factor or can be reduced to this
relationship by a transformation of coordinates, the
two spaces will be called {\it similar},
and we will consider them as belonging to the same type.
Their geometries are essentially identical; the only thing which changes
from one to the other is the unit of linear measure.

An isometry of a space $S_n$ into itself will be called a {\it motion\/} of
this space.  We will consider the spaces which admit {\it continuous\/} motions
into themselves, namely, such that in the corresponding equations of the
transformation appear some arbitrary parameters. The set of all these motions
for a given $S_n$ clearly forms a {\it group}.  Simple geometrical
considerations show that the number of parameters of this group is necessarily
finite, which is in fact easily demonstrated analytically as we will see. If
$r$ is the number of these parameters in the complete group of motions, in
every case this group will consist of a {\it finite-dimensional continuous\/}
Lie\footnote{ S. Lie-F. Engel, {\it Theorie der Transformationsgruppen\/},
Vol.~I (1888), Chap.~18, p.~310 and Vol.~III (1893), p.~575. [In the
bibliographical footnotes, authors' first initials and, wherever missing,
authors' names, have been added by the Editor. Also, journal titles were
corrected and details of the citations were added wherever necessary.
(Editor)]} group $G_r$ generated by $r$ infinitesimal transformations $X_1 f,
X_2 f, \ldots, X_r f$.

The problem of determining which spaces possess a continuous
group of motions reduces therefore essentially to the
classification of all possible forms of $ds^2$ which possess a
Lie group $G_r\equiv(X_1 f,\ldots,X_n f)$
which transforms $ds^2$ into itself.

While the fundamental equations for the solution of this problem
are already known from the work of Lie himself and of Killing,
the problem has not been treated in complete form as far as I know.
Indeed for arbitrary $n$, attention has been limited to the case
in which figures in the space
$S_n$ can be transported with the {\it maximum\/} number of degrees of
freedom: then the space is of constant curvature and the group
possesses $r=n(n+1)/2$ parameters. Only for $n=2$, namely for
ordinary surfaces, do we know the complete solution of the
problem, and it is known that
the number of parameters can only fall into the two
cases $r=1$, $r=3$. The surfaces of the first family are the one
and only ones which are isometric to a surface of revolution;
those of the second are exactly the surfaces of constant curvature.

In the present work I propose to study completely the case $n=3$,
in other words to classify all types of 3-dimensional spaces
in which it is possible to transport figures along a certain
degree of freedom.  Apart from the extreme case of spaces of
constant curvature which have a group $G_6$ of motions, there
exist, as we will show, many intermediary types for which the
number of parameters of the group can assume one of the four values
$r=1,2,3,4$, while there do not exist spaces with groups of
motions (or with partial subgroups) of 5 parameters.

To point out the main difference between the case of the surfaces
$n=2$ and that of $n=3$, we remind ourselves that a surface which
admits a transitive group of motions is necessarily of constant
curvature, namely, if a point can be transported anywhere, it can
also be rotated around every point.  On the other hand there
exist spaces of 3 dimensions in which we can transport any point
of the space everywhere with a transformation, but the space is
not of constant curvature; these spaces admit a transitive group
of transformations with 3 or 4 parameters.  In the spaces which
admit only a group $G_3$ the entire space is fixed if we fix a single
point.  In the ones which admit a group $G_4$, it is still
possible to have a continuous rotation $G_1$ around any
arbitrary point $P$; however, together with $P$ all the points of a
certain geodesic through $P$ remain fixed, so that these groups
$G_4$ belong, according to the nomenclature of Lie, to the class of
{\it systatic\/} groups.  The space is then lined with a double
infinity of such geodetic axes which completely fill the space,
and besides the transformations (translations) which permit a
point of a figure to be transported everywhere, there are still
arbitrary rotations possible around any of these axes.
Moreover, spaces which
admit a group $G_3$ and those admitting a group $G_4$ are further
distinguished into different irreducible types as we will see.

In the treatise of this problem I present here, I have constantly
made use of the theorems and notations contained in the great
work of Lie and more particularly his results on the composition of
groups.  They allow us to completely solve the question which
approached directly would present great difficulties. Naturally
the same method could be applied to a space of a larger number of
dimensions, but as soon as $n>3$, the investigation seems to
get complicated very quickly.

\typeout{**** Begin section 1}

\section{%1.
The Killing Equations.\protect\footnotemark}
\footnotetext{W. Killing, {\it\"Uber die Grundlagen der Geometrie\/}, Journ.
f\"ur die r. und ang.  Math. (Crelle), 109 (1892), 121--186.}


Given a quadratic differential form in $n$ variables:
$$
ds^2 = \sum_{i,k}^{1..n} a_{ik} \, dx_i \, dx_k \ ,
\eqno(1)$$
we look for the conditions which this form must satisfy in order to
admit the group $G_1$ generated by the infinitesimal
transformation $X f = \sum_r^{1..n} \xi_r \, \partial
f/\partial x_r$.

It will therefore be {\it necessary and sufficient\/} that the operation
$X f$ acting on the form (1) give an identically null result.
Now we have:
$$
X (ds^2) = \sum_{i,k} X(a_{ik}) \, dx_i \, dx_k
          +\sum_{r,k} a_{rk} \, dX(x_r) \, dx_k
          +\sum_{i,r} a_{ir} \, dX(x_r) \, dx_i \ ,
$$
namely
\begin{eqnarray}
X (ds^2) &=&
           \sum_{i,k,r} \xi_r \frac{\partial a_{ik}}{\partial x_r}
                \, dx_i \, dx_k
          +\sum_{r,k} a_{rk} \, d\xi_r \, dx_k
          +\sum_{i,r} a_{ir} \, d\xi_r \, dx_i
\ \nonumber\\ &=&
    \sum_{i,k} \left\{ \sum_r \left(
      \xi_r \frac{\partial a_{ik}}{\partial x_r}
      +a_{rk} \frac{\partial \xi_r}{\partial x_i}
      +a_{ir} \frac{\partial \xi_r}{\partial x_k}
    \right)\right\} dx_i \, dx_k\ .
\nonumber
\end{eqnarray}

The $n$ functions  $\xi_1,\xi_2,\ldots,\xi_n$  therefore will
have to fulfill the $n(n+1)/2$ linear homogeneous first order
partial differential equations:
\begin{eqnarray}
&&
\sum_r \left\{ \xi_r \frac{\partial a_{ik}}{\partial x_r}
+ a_{ir} \frac{\partial \xi_r}{\partial x_k}
+ a_{kr} \frac{\partial \xi_r}{\partial x_i}
\right\} =0\ ,
\seteqno{A}\\ && \qquad
i, k = 1, 2, 3,\ldots, n
\ .
\nonumber
\end{eqnarray}
Because the determinant of the $a_{ik}$ is different from zero,
these linear homogeneous equations in $\xi$ and their first
derivatives are linearly independent; moreover it is immediately
seen that they are also linearly independent with respect to the
$n^2$  first derivatives of  $\xi$ so that they can be solved for
$n(n+1)/2$ of these derivatives, chosen conveniently. It is
important to observe with Killing ({\it ibid.\/}, p.168) that by again
differentiating the fundamental equations (A) all the
second derivatives of the $\xi$ can be obtained expressed
linearly and homogeneously as functions of the first derivatives
and the $\xi$ themselves.  In fact, we differentiate (A) with
respect to $x_l$ , obtaining:
\begin{eqnarray}
&&
\sum_r \left\{
\frac{\partial^2 a_{ik}}{\partial x_r \partial x_l} \xi_r
+ \frac{\partial a_{ik}}{\partial x_r} \frac{\partial \xi_r}{\partial x_l}
+ \frac{\partial a_{ir}}{\partial x_l} \frac{\partial \xi_r}{\partial x_k}
+ \frac{\partial a_{kr}}{\partial x_l} \frac{\partial \xi_l}{\partial x_i}
\right.
\nonumber\\ &&\qquad \left.
+ a_{ir} \frac{\partial^2 \xi_r}{\partial x_k \partial x_l}
+ a_{kr} \frac{\partial^2 \xi_r}{\partial x_i \partial x_l}
\right\} =0
\ .
\nonumber
\end{eqnarray}

We then write the equations obtained from this last one by first
interchanging $k$ with $l$ , then $i$ with $k$ , namely:
\begin{eqnarray}
&&
\sum_r \left\{
\frac{\partial^2 a_{il} }{ \partial x_r \partial x_k} \xi_r
+ \frac{\partial a_{il}}{\partial x_r} \frac{\partial \xi_r}{\partial x_k}
+ \frac{\partial a_{ir}}{\partial x_k} \frac{\partial \xi_k}{\partial x_l}
+ \frac{\partial a_{rl}}{\partial x_k} \frac{\partial \xi_r}{\partial x_i}
\right.
\nonumber\\ &&\qquad \left.
+ a_{ir} \frac{\partial^2 \xi_r}{\partial x_k \partial x_l}
+ a_{lr} \frac{\partial^2 \xi_r}{\partial x_i \partial x_k}
\right\} =0
\ ,\nonumber\\
&&
\sum_r \left\{
\frac{\partial^2 a_{kl} }{ \partial x_r \partial x_i} \xi_r
+ \frac{\partial a_{kl}}{\partial x_r} \frac{\partial \xi_r}{\partial x_i}
+ \frac{\partial a_{kr}}{\partial x_i} \frac{\partial \xi_r}{\partial x_l}
+ \frac{\partial a_{rl}}{\partial x_i} \frac{\partial \xi_r}{\partial x_k}
\right.
\nonumber\\ &&\qquad \left.
+ a_{kr} \frac{\partial^2 \xi_r}{\partial x_i \partial x_l}
+ a_{lr} \frac{\partial^2 \xi_r}{\partial x_i \partial x_k}
\right\} =0
\ .\nonumber
\end{eqnarray}


Subtracting the first from the sum of these last two and dividing
the result by 2, we obtain:
%%
\begin{eqnarray}
&&\kern-.4cm
\sum_r\left\{
 a_{rl} \frac{\partial^2 \xi_r}{\partial x_i \partial x_k}
+ \frac{\partial }{ \partial x_r} \chrisbot{l}{ik} \xi_r
\right.
\nonumber\\ &&\kern-.4cm \qquad
\left.
+\chrisbot{r}{ik} \frac{\partial \xi_r}{\partial x_l}
+\chrisbot{l}{ir} \frac{\partial \xi_r}{\partial x_k}
+\chrisbot{l}{kr} \frac{\partial \xi_r}{\partial x_i}
\right\}
=0\ ,
\nonumber
\\ && \kern-.4cm
\qquad
l,i,k = 1, 2,3,\ldots, n
\ ,
\seteqno{2}
\end{eqnarray}
where the Christoffel symbol $\chrisbot{l}{ik}$ has the well known meaning
$$
\chrisbot{l}{ik} = \frac12 \left(
 \frac{\partial a_{il}}{\partial x_k}
+\frac{\partial a_{kl}}{\partial x_i}
-\frac{\partial a_{ik}}{\partial x_l}
\right)
\ .
$$

If in (2) we fix $i,k$ and let $l$ take all the values from 1 to $n$, the
equations thus obtained, since the determinant of $a_{ik}$ is nonzero, can be
solved for the second derivatives of $\xi$. To write down the solution we
indicate by $A_{ik}$ the adjoint of $a_{ik}$ divided by the latter's
determinant.\footnote{Namely, the inverse [Translator].} Multiplying (2) by
$A_{lv}$ and summing from $l=1$ to $l=n$ we thus obtain the required equations:
\begin{eqnarray}
&&
\frac{\partial^2 \xi_v}{\partial x_i \partial x_k}
+\sum_{r,l} A_{lv} \xi_r \frac{\partial }{\partial x_r} \chrisbot{l}{ik}
+\sum_{r,l} A_{lv} \chrisbot{r}{ik} \frac{\partial \xi_r}{\partial x_l}
\nonumber\\&&\qquad
+\sum_r \christop{v}{ir} \frac{\partial \xi_r}{\partial x_k}
+\sum_r \chrisbot{v}{kr} \frac{\partial \xi_r}{\partial x_i}
=0
\seteqno{B}
\\&&\qquad
(i, k= 1, 2, 3,\ldots, n)
\ ,
\nonumber
\end{eqnarray}
where the Christoffel symbol of the second kind
$\christop{v}{ir}$   has the meaning
$$
\christop{v}{ir} = \sum_k A_{kv} \chrisbot{k}{ir}
\ .
$$

Equations (B) show us that the general integral of (A) contains
the maximum number of arbitrary constants. In fact assuming
the $n(n+1)/2$ functions
$$
\xi_r\ , \frac{\partial \xi}{\partial x_i}
\ ,
\qquad
(i,r=1,2,\ldots,n)
$$
as unknowns, using (B)
we can express all their first derivatives as (linear
and homogeneous) functions of the same unknowns and we therefore
have a system of linear homogeneous total differential equations,
the unknowns then being related by the $n(n+1)/2$ equations (A).
The {\it maximum\/} number of arbitrary constants that can appear in the
general integral of (A) will therefore be given
by:\footnote{The integral system is in fact specified if we give at
one point of space the initial values of the $n(n+1)$ unknown
functions which are, however, constrained by $n(n+1)/2$
independent relations.}
$$ r =n(n + 1)  - n(n + 1)/2 = n(n + 1)/2
\ .
$$

If this maximal number is reached we will have the case of
{\it complete\/} integrability and the space $S_n$, as is well known, will
then be of constant curvature.  In each case, the number $r$ of
independent infinitesimal transformations that the
differential form (1) admits will be a finite number $r \leq n(n+
1)/2$, and these $r$ transformations $X_1 f, X_2 f, \ldots, X_r f$
will generate the continuous group $G_r$ of motions of the space
$S_n$.

\typeout{**** Begin section 2}

\section{%2.
  Spaces which admit a group $G_1$.}

{}From equations (A) we immediately deduce a consequence
which is important to note; we can state: {\it two infinitesimal
transformations of the space $S_n$ cannot have common
trajectories without coinciding}.  And indeed we show immediately
that if $\xi_1,\xi_2,\ldots,\xi_n$ satisfy equations (A) and
$\lambda\xi_1, \lambda \xi_2,\ldots, \lambda\xi_n$ is a new
set of solutions, the factor $\lambda$ must necessarily be
constant. In fact replacing  $\xi_r$ by $\lambda\xi_r$ in
(A) gives
$$
\sum_r \left( a_{ir}\xi_r \frac{\partial \lambda}{\partial x_k}
+ a_{kr}\xi_r \frac{\partial \lambda}{\partial x_i}\right) = 0
\ ,
\eqno(3)
$$
from which, setting $i=k$:
$$
\sum_r a_{rs}\xi_r \frac{\partial \lambda}{\partial x_i} = 0
\ .
$$


Assuming that $\partial \lambda/\partial x_s\neq0$, it
follows that    $\sum_r a_{rs}\xi_r =0$, and from (3),
setting $k=s$, we deduce that $\sum_r a_{rs}\xi_r=0$; but the
determinant of the $a$    is nonzero and this will imply that all the
$\xi$ are zero.

We now assume that the space $S_n$ admits a group of motions
$G_1$ generated by the infinitesimal transformation  $X f =
\sum_i \xi_i \partial f/\partial x_i$.
We can simplify the computations by assuming the trajectories
of the group as the coordinate lines
$(x_1)$, so that we have
$\xi_2=\xi_3=\cdots\xi_n=0$, and by changing the parameters
conveniently we can make $\xi_1=1$, namely $Xf=  \partial f/\partial
x_1$.\footnote{It is sufficient to assume as new
variables   $y_1, y_2,\ldots,y_n$ an integral of the equation
$X(y_1)=1$ and $n-1$  independent integrals of the equation
$X(y)= 0$.}

Then (A) gives us simply
$$
\frac{\partial a_{ik}}{\partial x_1} =  0
\qquad    (i, k = 1 , 2,\ldots , n)
\  ,
$$
which shows that the coefficients $a_{ik}$ are independent
of $x_1$.
Viceversa it is clear that if in (1) the coefficients $a_{ik}$
do not depend on $x_1$, the transformation $x_1^\prime = x_1 + constant$
gives a continuous group $G_1$ of transformations in the space.
And as long as the $a_{ik}$ remain arbitrary functions of the other
variables $x_2,x_3,\ldots,x_n$,  this group $G$ will be the
{\it complete\/} group of motions.

In the case $n=2$ we then recover the well known result that the
surface is isometric to a surface of rotation.

\typeout{**** Begin section 3}

\section{%3.
Surfaces with a group $G_2$.}

We now study the types of $ds^2$ which admit a group $G_2$ of motions, assuming
that the number of variables is $n= 2$ or $n= 3$. The result for $n=2$ is well
known but it is worthwhile to rederive it again here.

So let us first assume $n=2$ and indicate by $X_1f$, $X_2f$ the two
infinitesimal transformation generators of the group $G_2$ under consideration.
Replacing $X_1f$, $X_2f$ by  new convenient linear combinations of them, we can
always reduce ourselves to the case in which we have for the composition
equations\footnote{S. Lie-F. Engel, Vol.~III, p.~713}
$$(a)\qquad [X_1,X_2]f=0\
,\ {\rm or}\qquad (b)\qquad [X_1,X_2]f= X_1 f\ . $$

The trajectories of the two infinitesimal transformation generators being in
each case distinct (\S2), we can assume them respectively as coordinate lines
and we then have     $X_1 f = \xi \, \partial f / \partial x_1$, $X_2 f = \eta
\, \partial f / \partial x_2$. In case $(a)$ it follows that $\partial \xi/
\partial x_2 = 0$, $\partial \eta/ \partial x_1 = 0$, so that by making a
change of parameters, we can assume $ \xi=\eta=1$.

Since the equations (A) have to be satisfied
either with $\xi_1=1 $, $\xi_2=0 $
or with $\xi_1=0 $, $\xi_2=1 $,
it follows that the coefficients of the differential form
\begin{displaymath}
ds^2=a_{11} \, dx_1{}^2 + 2 a_{12} \, dx_1 dx_2 + a_{22} \, dx_2{}^2
\end{displaymath}
are constants and with a (linear) change of variables we can therefore have
$ ds^2 = dx_1{}^2 + dx_2{}^2$,
hence the surface has zero curvature.  The complete group of motions is the
$G_3$ generated by the three infinitesimal transformations
\begin{displaymath}
X_1 f = \frac{\partial f}{\partial x_1} \ ,\
X_2 f = \frac{\partial f}{\partial x_2} \ ,\
X_3 f =  x_2 \frac{\partial f}{\partial x_1}
- x_1 \frac{\partial f}{\partial x_2} \ .
\end{displaymath}

In case $(b)$ we must have $\partial \eta / \partial x_1=0 $,
$-\eta \, \partial \xi / \partial x_2=\xi $,
and by changing the parameters $x_1$, $x_2$, we can set
$\eta=1 $, $\xi=e^{-x_2} $,
so  that
$$
 X_1 f = e^{-x_2} \, \partial f/\partial x_1\ ,\
 X_2 f =  \partial f/\partial x_2\ .
$$

Equations (A), assuming successively $\xi_1=0 $, $\xi_2=1 $
and $\xi_1=e^{-x_2} $, $\xi_2=0 $, give us
\begin{eqnarray}
&&
\frac{\partial a_{11}}{\partial x_2}
 = \frac{\partial a_{12}}{\partial x_2}
 = \frac{\partial a_{22}}{\partial x_2} = 0
\ ,\nonumber\\
&&
\frac{\partial a_{11}}{\partial x_1} = 0\ ,\
\frac{\partial a_{12}}{\partial x_1} = a_{11}\ ,\
\frac{\partial a_{22}}{\partial x_1} = 2 a_{12}\ ,
\nonumber
\end{eqnarray}
from which by integration we have   $a_{11}=\alpha $,
$a_{12}=\alpha  x_1 + \beta$, $a_{22}=\alpha  x_1{}^2 + 2 \beta x_1 + \gamma$,
with  $\alpha,\beta,\gamma$  constants.  Without loss of generality we can assume
$\alpha=1 $
(by absorbing it into $x_1$), and writing $x_1$ in place of  $x_1+\beta$, we
will have
\begin{displaymath}
ds^2 = dx_1^2 +2 x_1 \, dx_1 dx_2 + (x_1{}^2 + R^2) \, dx_2{}^2 \ .
\end{displaymath}

If we set  $ x_1=-v e^{u/R}$, $x_2=u/R $,
we obtain the typical (parabolic) form
$$
  ds^2=du^2+e^{2u/R} dv^2
$$
of the line element of the pseudo-spherical surface.
The complete group of motions is the $G_3$ generated by the
infinitesimal transformations:
\begin{eqnarray}
&&
X_1 f = e^{-x_2} \frac{\partial f}{\partial x_1}\ ,\
X_2 f = \frac{\partial f}{\partial x_2}\ ,\nonumber\\
&&
X_3 f = \frac12 e^{x_2}(x_1{}^2+R^2) \frac{\partial f}{\partial x_1}
        -x_1 e^{x_2} \frac{\partial f}{\partial x_2}\ .
\nonumber
\end{eqnarray}
\typeout{IF THESE EQUATIONS CAN FIT ON ONE LINE, DO IT!}

The subgroup $G_2$ under consideration  consists of all those groups $G_1$
which have as trajectories the geodetic circles (with ideal center) inclined at
a constant angle to the parallel oricycles\footnote{In Italian: ``oricicli"
[Translator].}
$x_2 = constant $.\footnote{%(7)
If one
represents these surfaces as pseudo-spheres these trajectories are
loxodromes of the surfaces.}

In the analysis of the present \S\ only the surfaces of constant zero or
negative curvature have appeared, not those of constant positive curvature. The
reason for this is the fact that the latter surfaces admit a group $G_3$ of
motions, but never a {\it real\/} 2-parameter subgroup.

\typeout{**** Begin section 4}

\section{%4.
Spaces of 3 dimensions with a group $G_2$.}

We now turn our attention to 3-dimensional spaces which admit a 2-parameter
group of motions.  The trajectories of the two infinitesimal transformation
generators of this $G_2$ being distinct (\S2), each point of the space will be
moved over a surface by the transformations of $G_2$.  We have therefore a
family of surfaces $\Sigma$ which represent for our group what Lie calls the
{\it minimum invariant varieties\/}. For a given transformation of the $G_2$,
any one of the $\Sigma$  is transformed into itself and consequently any
surface geodesically parallel to a $\Sigma$ as well; we deduce from this that
the $\infty^1$ surfaces   $\Sigma$ are geodesically
parallel;\footnote{%(8)
We can deduce the same conclusion from the fundamental equations (A).  Let us
assume in fact $\Sigma$ for the $x_1$ coordinate surfaces and for the $x_1$
coordinate lines [translator note: second $x_1$ corrected from Bianchi typo
$x_3$ here and in the text as well] their orthogonal trajectories; we will have
\begin{displaymath}
ds^2=a_{11} \, dx_1{}^2 + a_{22} \, dx_2{}^2
       + 2 a_{23} \, dx_2 dx_3 + a_{33} \, dx_3{}^2 \ .
\end{displaymath}
If  $X_1 f $,  $X_2 f $ are their infinitesimal transformation generators, we
have to have $X_1(x_1)=0 $, $X_2(x_1)=0 $ and consequently $X_1 f = \xi_2 \,
\partial f/\partial x_2 + \xi_3 \, \partial f/\partial x_3$,\ $X_2 f =
\eta_2 \, \partial f/\partial x_2 + \eta_3 \, \partial f/\partial x_3$. Now
applying (A) successively to $X_1f$, $X_2 f $ setting $i=k=1$ we deduce
\begin{displaymath}
 \frac{\partial a_{11}}{\partial x_2} \xi_2
+  \frac{\partial a_{11}}{\partial x_3} \xi_3 = 0\ ,\
 \frac{\partial a_{11}}{\partial x_2} \eta_2
+  \frac{\partial a_{11}}{\partial x_3} \eta_3 = 0\ ,
\end{displaymath}
from which, since $\left|\matrix{\xi_2&\xi_3\cr \eta_2& \eta_3}\right|\neq0 $
from \S2, it follows that $\partial a_{11}/\partial x_2= \partial
a_{11}/\partial x_3=0$. Changing the parameter  $x_1$, one can therefore make
$a_{11}=1$ which gives to the line element the geodetic form of the text.}
moreover, any each of them, admitting a group $G_2$ of transformations, will be
of constant zero or negative curvature (\S3).  If we take the surfaces $\Sigma$
as coordinate surfaces $x_1 = constant$ and their orthogonal trajectories for
coordinate lines\footnote{In the original text, ``coordinate lines $x_3$",
which is incorrect [Editor].} $x_1$, we put the line element into the geodetic
form: $$ ds^2 = dx_1{}^2 + a_{22} \, dx_2{}^2
  + 2 a_{23} \, dx_2 dx_3 + a_{33} \, dx_3{}^2\ .
\eqno(4)
$$

In each of the infinitesimal transformations $X_1 f $, $X_2 f $,
since $\xi_1=0 $, the equations $(A)$, setting $i=1$,
$k=2,3$, give
\begin{displaymath}
a_{22} \frac{\partial\xi_2}{\partial x_1}
+a_{23} \frac{\partial\xi_3}{\partial x_1} = 0\ ,
\
a_{23} \frac{\partial\xi_2}{\partial x_1}
+a_{33} \frac{\partial\xi_3}{\partial x_1} = 0\ ,
\end{displaymath}
from which
since $a_{22} a_{33} - a_{23}{}^2 \neq0$,
we conclude that $\partial \xi_2/\partial x_1 = \partial \xi_3/\partial x_1=0 $,
namely that the coefficients of  $X_1 f $, $X_2 f $ are independent of $x_1$.

Assuming this to be true, we take the
respective (distinct) trajectories of $X_1 f $, $X_2 f $
as coordinate lines over one of the surfaces  $x_1 = constant$ and we will have
$ X_1 f = \xi \, \partial f/\partial x_2$,
$ X_2 f = \eta \, \partial f/\partial x_3$.

We now distinguish again the two cases
$$
(a) \qquad [X_1,X_2]f =0 \qquad {\rm and}\qquad
(b) \qquad [X_1,X_2]f = X_2 f\ .
$$
In the first case, as in the preceding \S, we can make
$$
  X_1 f = \partial f/\partial x_2\ ,\
  X_2 f = \partial f/\partial x_3
$$
and the line element of the space will take the form
$$
ds^2 = dx_1{}^2 +  \alpha \, dx_2{}^2 + 2 \beta \, dx_2 dx_3
       + \gamma \, dx_3{}^2 \ ,
\eqno(a^\ast)
$$
with $\alpha, \beta,\gamma$ being functions only of $x_1$.


In case (b) we take
$$
  X_1 f = \partial f/\partial x_3\ ,\
  X_2 f = e^{x_3} \, \partial f/\partial x_2
$$
and we will then have
$$
ds^2 = dx_1{}^2 +  \alpha \, dx_2{}^2
       + 2(\beta-\alpha x_2) \, dx_2 dx_3
 + (\alpha x_2{}^2 - 2\beta x_2 +\gamma) \, dx_3{}^2 \  ,
\eqno(b^\ast)
$$
where $\alpha, \beta,\gamma$  are still functions only of $x_1$.

Vice versa, whatever are the functions $\alpha, \beta,\gamma$ of
$x_1 $ in $(a^\ast)$ or $(b^\ast)$,
we will have a space which admits the 2-parameter group of
motions $(\partial f /\partial x_2, \partial f/\partial x_3)$
in the first case and another
$(\partial f /\partial x_3, e^{x_3}\partial f/\partial x_2)$
in the second case.
If $\alpha, \beta,\gamma$ remain arbitrary functions of $x_1 $,
this $G_2$ is the complete group of motions, as will
be shown by the analysis in the following \S\S.

\typeout{**** Begin section 5}

\section{%5.
Spaces with an intransitive group $G_r$ of motions  ($r \geq 3$).}

We now pass to the treatment of 3-dimensional spaces which admit a group of
motions with more than two parameters, beginning with the case in which this
group $G_r$ is intransitive.

{}From the considerations of the preceding \S\ the minimum invariant
varieties
with respect to the group will be geodesically parallel surfaces,
and because each of these has to admit a group $G_r$ with $r\geq3$
parameters,\footnote{%(8)
That on any surface $\Sigma$ the group $G_r$ retains $r$ parameters follows
immediately from what we have seen in \S4 because if we take the line element
in the geodetic form (4), in every single infinitesimal transformation of $G_r$
one has $\xi_1= 0$ and  $\xi_2$, $\xi_3$ are independent of $x_1$. Of course
this is also clear geometrically since if all the points of a surface $\Sigma$
remain fixed, the entire space is immobilized.} one necessarily must have
$r=3$. To the line element of the space we then give the geodetic form $$
ds^2=dx_1^2 + a_{22} \, dx_2^2 + 2 a_{23} \, dx_2 dx_3 + a_{33} \, dx_3^2
\eqno(4) $$
and the geodesically parallel surfaces $x_1 = constant$ will be of
constant curvature.


Arbitrarily selecting one of these, say $x_1 = 0$, we distinguish three cases
characterized by the curvature $K$ being zero, positive or negative.
By substituting for this space a similar space, we can assume
successively
$$
 K_0=0\ ,\
 K_0= 1\ ,\
 K_0= -1
$$
and correspondingly we can change the coordinate lines of $x_2$, $x_3$
on the surface $x_1 = 0$
so that the line element $ds_0^2$ of $x_1=0$ assumes the
respective typical forms:
\begin{eqnarray}
K_0=0:\qquad
ds_0^2 &=& dx_2^2 + dx_3^2\ ,\seteqno{$\alpha$}\\
K_0=1:\qquad
ds_0^2 &=& dx_2^2 + \sin^2 x_3\,dx_3^2\ ,\seteqno{$\beta$}\\
K_0=-1:\qquad
ds_0^2 &=& dx_2^2 + e^{2x_3} dx_3^2\ .\seteqno{$\gamma$}
\end{eqnarray}


The group $G_3$ of motions of  $x_1 = 0$ into itself will be generated respectively
by the three infinitesimal transformations:
%%
\begin{eqnarray}
\kern-.5cm
X_1 f = \frac{\partial f}{\partial x_2}
\ ,\
&&
X_2 f = \frac{\partial f}{\partial x_3}
\ ,\
X_3 f = x_3 \frac{\partial f}{\partial x_2}
- x_2 \frac{\partial f}{\partial x_3}
\ ;\seteqno{$\alpha^\ast$}\\
\kern-.5cm
X_1 f = \frac{\partial f}{\partial x_3}
\ ,\
&&
X_2 f = \sin x_3 \frac{\partial f}{\partial x_2}
  + \cot x_2 \cos x_3 \frac{\partial f}{\partial x_3}
\ ,\nonumber\\
&&
X_3 f = \cos x_3 \frac{\partial f}{\partial x_2}
- \cot x_2 \sin x_3 \frac{\partial f}{\partial x_3}
\ ;\seteqno{$\beta^\ast$}\\
\kern-.5cm
X_1 f = \frac{\partial f}{\partial x_3}
\ ,\
&&
X_2 f =  \frac{\partial f}{\partial x_2}
  - x_3 \frac{\partial f}{\partial x_3}
\ ,\nonumber\\
&&
X_3 f =  x_3 \frac{\partial f}{\partial x_2}
+ \frac12(e^{-2x_2}- x_3^2) \frac{\partial f}{\partial x_3}
\ .\seteqno{$\gamma^\ast$}
\end{eqnarray}

In all three cases these are also the infinitesimal transformations of the
group of motions of the whole space.  Now if for the line element (4)
we write the three equations which result from the
fundamental equations (A) setting
$\xi_1=0$   and successively     $(i,k)   = (2,2), (2,3), (3,3)$,
we find
%%
\begin{eqnarray}
&&\kern-.5cm
 \frac{\partial a_{22}}{\partial x_2} \xi_2
+\frac{\partial a_{22}}{\partial x_3} \xi_3
+ 2 a_{22} \frac{\partial \xi_2}{\partial x_2}
+ 2 a_{23} \frac{\partial \xi_3}{\partial x_2}
 =0
\ ,\nonumber\\
&&\kern-.5cm
 \frac{\partial a_{23}}{\partial x_2} \xi_2
+\frac{\partial a_{23}}{\partial x_3} \xi_3
+  a_{22} \frac{\partial \xi_2}{\partial x_3}
+  a_{23} \left(
\frac{\partial \xi_2}{\partial x_2}+\frac{\partial \xi_3}{\partial x_3}\right)
+  a_{33} \frac{\partial \xi_3}{\partial x_2}
 =0
\ ,\nonumber\\
&&\kern-.5cm
 \frac{\partial a_{33}}{\partial x_2} \xi_2
+\frac{\partial a_{33}}{\partial x_3} \xi_3
+ 2 a_{23} \frac{\partial \xi_2}{\partial x_3}
+ 2 a_{33} \frac{\partial \xi_3}{\partial x_3}
 =0
\ .\seteqno{C}
\end{eqnarray}


These must be satisfied when for $\xi_2$, $\xi_3$ in the three
respective cases we substitute the three pairs of values which belong
respectively to the 3 generator substitutions $(\alpha^\ast)$, $(\beta^\ast)$
or $(\gamma^\ast)$.

\typeout{**** Begin section 6}

\section{%6.
Discussion of the system (C).}

We begin with case $(\alpha^\ast)$ and putting into (C) first\footnote{The
original paper had $(\xi_1, \xi_2)$ instead of $(\xi_2, \xi_3)$ here, which was
an obvious typo [Editor].} $\xi_2 = 1$, $\xi_3=0$ and then $\xi_2 = 0$,
$\xi_3=1$, we deduce from this that $\partial a_{ik}/\partial x_2 = \partial
a_{ik}/\partial x_3 = 0\ (i,k=2,3)$, from which it follows that the
coefficients $a_{ik}$ here are functions only of $x_1$. If we now introduce
into (C) the values $\xi_2 = x_3$, $\xi_3=-x_2$ which belong to the third
infinitesimal transformation, we have $a_{23}= 0$, $a_{22}= a_{33}$ and
therefore for the line element of the space $$ ds^2= dx_1^2 + \varphi^2(x_1) \,
(dx_2^2+dx_3^2)\ , \eqno(5) $$ where $\varphi(x_1)$ indicates an (arbitrary)
function  of $x_1$.

In case $(\beta)$, first setting in (C) $\xi_2 = 0$, $\xi_3=1$, the values
which correspond to $X_1 f$, we see that $a_{22}$, $a_{23}$, $a_{33}$ do not
depend on $x_3$. Then substituting the values $\xi_2 = \sin x_3$, $\xi_3= \cot
x_2 \cos x_3$ corresponding to $X_3 f$, the first of (C) gives us $\sin x_3
\,\partial a_{22}/\partial x_2 = 2\cos x_3/ \sin^2 x_2 \, a_{23}$, and since
neither $a_{22}$ nor $a_{23}$ depend on $x_3$,  it follows that $a_{23}=0$,
$\partial a_{22}/\partial x_2 =0$ and consequently $a_{22}=\varphi^2(x_1)$.

The second of (C) then gives immediately
$a_{33}=\sin^2 x_2\, \varphi^2(x_1)$,
so that the line element of the space has the form
$$
ds^2= dx_1^2 + \varphi^2(x_1) \, (dx_2^2+ \sin^2 x_2 \, dx_3^2)\ .
\eqno(6)
$$

Finally in case $(\gamma)$, equations (C) with the values $\xi_2 = 0$,
$\xi_3=1$ belonging to $X_1 f$ show that $a_{22}$, $a_{23}$, $a_{33}$ are again
independent of  $x_3$. Substituting next the values $\xi_2 = 1$, $\xi_3=-x_3$
corresponding to $X_2 f$ we find\footnote{In the original paper, the third
equation was $\partial a_{33} / \partial x_3 = 2a_{33}$, which was incorrect
[Editor].}: $\partial a_{22}/\partial x_2 = 0$, $\partial a_{23}/\partial x_2 =
a_{23}$, $\partial a_{33}/\partial x_2 = 2 a_{33}$, and finally with the values
$\xi_2 = x_3$, $\xi_3=\frac12 (e^{-2 x_2}-x_3^2)$ belonging to $X_3f$:
$a_{23}=0$, $a_{22}=a_{33} \, e^{-2 x_2}$, from which we arrive at the line
element $$ ds^2= dx_1^2 + \varphi^2(x_1) \, (dx_2^2+ e^{2 x_2}  dx_3^2)\ .
\eqno(7)$$

Vice versa for any function $\varphi(x_1)$ the spaces of the line elements (5),
(6), (7) admit  the respective intransitive group $G_3$ of motions
$(\alpha^\ast)$, $(\beta^\ast)$ or $(\gamma^\ast)$.

We must now discover for which special forms of the function
$\varphi(x_1)$  it will happen that the
complete group of motions of the space will be larger.

\typeout{**** Begin section 7}

\section{%7.
The complete group of motions of the space:\\
$ ds^2 = dx_1^2 + \varphi^2(x_1)\, (dx_2^2+dx_3^2)$.}

In order to determine the most general infinitesimal motion
$X f = \eta_1 \, \partial f/\partial x_1
+ \eta_2 \, \partial f/\partial x_2
+\eta_3 \partial f/\partial x_3$
of the present space, the fundamental equations (A), setting successively
$(i,k)$ = (1,1), (2,2), (3,3), (1,2), (1,3), (2,3)
give the following 6 equations:\footnote{The prime
indicates the derivative with respect to $x_1$.}
\begin{eqnarray}
&&
 \frac{\partial \eta_1}{\partial x_1} = 0\ ,\seteqno{8}\\
&&
 \frac{\partial \eta_2}{\partial x_2}
  + \frac{\varphi^\prime}{\varphi} \eta_1 = 0\ ,\seteqno{9}\\
&&
 \frac{\partial \eta_3}{\partial x_3}
  + \frac{\varphi^\prime}{\varphi} \eta_1 = 0\ ,\seteqno{10}\\
&&
 \frac{\partial \eta_1}{\partial x_2}
  + \varphi^2(x_1)  \frac{\partial \eta_2}{\partial x_1} = 0
  \ ,\seteqno{11}\\
&&
 \frac{\partial \eta_1}{\partial x_3}
  + \varphi^2(x_1)  \frac{\partial \eta_3}{\partial x_1} = 0
  \ ,\seteqno{12}\\
&&
 \frac{\partial \eta_2}{\partial x_3}
 + \frac{\partial \eta_3}{\partial x_2} = 0\ . \seteqno{13}
\end{eqnarray}

By taking $\eta_1=0$ naturally one has only the three transfomations
$(\alpha^\ast)$ and the question to be examined is therefore this: if the above
equations can be satisfied with $\eta_1 \neq 0$.

Differentiating (9) with respect to $x_1$, (11) with respect $x_2$ and
comparing, with the observation that by (8), $\eta_1$ does not
depend on $x_1$, we find that
$$
\frac{\partial^2 \eta_1}{\partial x_2^2}
= (\varphi^{\prime\prime} \varphi -\varphi^{\prime\, 2}) \, \eta_1 \ ,
\eqno(14)$$
and similarly from (10), (12)
$$
\frac{\partial^2 \eta_1}{\partial x_3^2}
= (\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2}) \, \eta_1 \ .
\eqno(15)$$

Since $\eta_1$ is different from zero and does not depend on $x_1$,
while $\varphi$ is a function only of $x_1$,
the resulting equations (14), (15) show that one will have:
\begin{eqnarray}
&&
\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2} = c
   \ ,\seteqno{16}\\
&&
\frac{\partial^2 \eta_1}{\partial x_2^2}
  = \frac{\partial^2 \eta_1}{\partial x_3^2} = c \eta_1
\ ,\seteqno{17}
\end{eqnarray}
where $c$ is a constant.  Integrating (11), (12) we find
\begin{eqnarray}
&&
\eta_2 =
-\frac{\partial \eta_1}{\partial x_2}
    \int \frac{d x_1}{\varphi^2(x_1)} + \psi(x_2,x_3) \ ,\nonumber\\
&&
\eta_3 =
-\frac{\partial \eta_1}{\partial x_3}
    \int \frac{d x_1}{\varphi^2(x_1)} + \chi(x_2,x_3)
\ ,\seteqno{18}
\end{eqnarray}
where $\psi$, $\chi$ are two functions only of $x_2$, $x_3$.
By substituting these into (13)
it follows that
$$
2 \frac{\partial^2 \eta_1}{\partial x_2 \partial x_3}
    \int \frac{d x_1}{\varphi^2(x_1)}
   = \frac{\partial \psi}{\partial x_3}
     +\frac{\partial \psi}{\partial x_2} \ ,
$$
from which, since $\eta_1$, $\psi$, $\chi$ are independent of $x_1$
while the integral necessarily contains it, we have
$$
\frac{\partial^2 \eta_1}{\partial x_2 \partial x_3} =0 \ .
\eqno(17^\ast)$$
Comparing with (17), we have immediately
$c \, \partial\eta_1/\partial x_2 =0$,
$c \, \partial\eta_1/\partial x_3 =0$.

If $c\neq0$ we will therefore have $\eta_1= constant$,
$\eta_2=\psi(x_2,x_3)$, $\eta_3=\chi(x_2,x_3)$,
from which (9) or (10) shows that one has $\varphi^\prime / \varphi =constant$.
But this last result follows even
if $c = 0$, since then by (17) and (17$^\ast$), $\eta_1$
is a linear function of   $x_2$, $x_3$   and since by (18)
$\partial \eta_2/\partial x_2=\partial \psi/\partial x_2$,
(9) gives us:
$\varphi^\prime / \varphi =-1/\eta_1\, \partial \psi/\partial x_2 \ ,$
from which we can conclude again that $\varphi^\prime / \varphi =constant$.
Therefore if the present space admits a larger group of motions
(with $r>3$ parameters) we necessarily have
$\varphi^\prime = k \varphi$ ($k$ constant).

If $k = 0$ one can make $\varphi(x_1)= 1$ and have ordinary Euclidean space.
If $k\neq0$ one can assume that  $\varphi(x_1)=e^{k x_1}$,
and have the space of constant negative curvature $K=-k^2$.

In both cases the complete group of motions has 6 parameters.  The result being
well known, we do not concern ourselves with giving the actual 6 infinitesimal
transformation generators, which are obtained by integrating the above
equations.

\typeout{**** Begin section 8}

\section{%8.
The complete group of motions of the space:\\
$ ds^2 = dx_1^2 + \varphi^2(x_1)\, (dx_2^2+\sin^2{x_2}\,dx_3^2)$.}

We proceed as in the previous \S, writing first the equations which follow from
(A) in order to find the most general infinitesimal motion of the space under
consideration. We thus find
\begin{eqnarray}
&&
 \frac{\partial \eta_1}{\partial x_1} = 0\ ,\seteqno{19}\\
&&
 \frac{\partial \eta_2}{\partial x_1}
  = - \frac{1}{\varphi^2} \frac{\partial\eta_1}{\partial x_2}
  \ ,\seteqno{20a}\\
&&
 \frac{\partial \eta_2}{\partial x_2}
  = - \frac{\varphi^\prime}{\varphi} \eta_1\ ,\seteqno{20b}\\
&&
 \frac{\partial \eta_3}{\partial x_1}
  = -\frac{1}{\varphi^2\sin^2 x_2}  \frac{\partial \eta_1}{\partial x_3}
  \ ,\seteqno{21a}\\
&&
 \frac{\partial \eta_3}{\partial x_3}
  = - \frac{\varphi^\prime}{\varphi} \eta_1 -\cot x_2\, \eta_2
  \ ,\seteqno{21b}\\
&&
 \frac{\partial \eta_2}{\partial x_3}
 + \sin^2 x_2 \, \frac{\partial \eta_3}{\partial x_2} = 0
  \ . \seteqno{22}
\end{eqnarray}

Eliminating by differentiation $\eta_2$ from (20) and $\eta_3$ from (21), we
find
\begin{eqnarray}
&&
\frac{\partial^2 \eta_1}{\partial x_2^2}
= (\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2}) \, \eta_1
  \ ,\nonumber\\
&&
\frac{\partial^2 \eta_1}{\partial x_3^2}
= (\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2}) \sin^2 x_2\, \eta_1
  - \sin x_2 \cos x_2 \frac{\partial\eta_1}{\partial x_2}
 \ ,\nonumber
\end{eqnarray}
from which, since $\eta_1\neq0$  doesn't depend on $x_1$, we conclude that
$$
\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2} = c
\quad {\rm(constant)}
   \ ,
$$
%%
$$
\kern-.5cm
\frac{\partial^2 \eta_1}{\partial x_2^2} = c \eta_1\ ,\
 \frac{\partial^2 \eta_1}{\partial x_3^2}
= c \sin^2 x_2\, \eta_1 - \sin x_2  \cos x_2 \frac{\partial \eta_1}{\partial x_2}
\ .\eqno(23)
$$

Integrating the first of (20) and the first of (21) with respect to $x_1$
we have:
\begin{eqnarray}
&&
\eta_2 =
-\frac{\partial \eta_1}{\partial x_2}
    \int \frac{d x_1}{\varphi^2(x_1)} + \psi(x_2,x_3)
   \ ,\nonumber\\
&&
\eta_3 =
-\frac{1}{\sin^2 x_2} \frac{\partial \eta_1}{\partial x_3}
    \int \frac{d x_1}{\varphi^2(x_1)} + \chi(x_2,x_3)
\ ,\seteqno{24}
\end{eqnarray}
and substituting into (22) we obtain
$$
2 \left(\frac{\partial^2 \eta_1}{\partial x_2 \partial x_3}
       - \cot x_2 \frac{\partial \eta_1}{\partial x_3}\right)
    \int \frac{d x_1}{\varphi^2(x_1)}
   = \frac{\partial \psi}{\partial x_3}
     + \sin^2 x_2 \frac{\partial \chi}{\partial x_2} \ .
$$

Since $x_1$ appears here only in the integrals, we
necessarily have
$$
\frac{\partial^2 \eta_1}{\partial x_2 \partial x_3}
=\cot x_2  \frac{\partial \eta_1}{\partial x_3}  \ ,
$$
and if we differentiate this with respect to $x_2$ and the first of (23) with
respect to $x_3$, we conclude that
$(c+1) \, \partial \eta_1/\partial x_3=0$,
and consequently $c=-1$ or $\partial \eta_1/\partial x_3=0$.

We consider in this \S\ the first case $c = -1$; then from
$$
\varphi^{\prime\prime} \varphi-\varphi^{\prime\,2} = -1\ ,
\eqno(25)$$
it follows by differentiation that
$\varphi^{\prime\prime\prime}- \varphi^\prime \varphi^{\prime\prime} = 0$,
so that $\varphi^{\prime\prime}= k \varphi$, ($k$ constant)
and (25) becomes  $\varphi^{\prime\,2} = 1+ k \varphi^2$.

If $k=0$, neglecting the additive constant in $x_1$ we have $\varphi(x_1)=x_1$.
If $k$     is negative, we put  $k=-1/R^2$ and we will have
$\varphi(x_1)=R \sin(x_1/R)$;
finally if $k$    is positive, let $k=1/R^2$ and it will be
$\varphi(x_1)=R \sinh(x_1/R)$.

We have as a consequence the following three forms of the line element
of the space:
\begin{eqnarray}
&&
 ds^2=dx_1^2+x_1^2 \, (dx_2^2+\sin^2{x_2}\,dx_3^2)
\ ,\nonumber\\
&&
 ds^2=dx_1^2+R^2\sin^2(x_1/R) \, (dx_2^2+\sin^2{x_2}\,dx_3^2)
\ ,\nonumber\\
&&
 ds^2=dx_1^2+R^2\sinh^2(x_1/R) \, (dx_2^2+\sin^2{x_2}\,dx_3^2)
\ .\nonumber
\end{eqnarray}

The first form belongs to ordinary Euclidean space (in polar coordinates),
the second and third respectively to spaces of constant positive or negative
curvature     $K=\pm 1/R^2$.
In all three cases the complete group of motions has 6 parameters.

\typeout{**** Begin section 9}

\section{%9.
The group $G_3$ of motions of the space:\\
$ds^2=dx_1^2+ dx_2^2+\sin^2{x_2}\,dx_3^2$.}

In order to complete the discussion of the previous \S\ there remains to be
treated the case in which we have $\partial \eta_1/\partial x_3 = 0$. Equations
(23) then become\footnote{In the second equation, the original paper has a
second derivative, which is incorrect [Editor].}

$$ \frac{\partial^2 \eta_1}{\partial x_2^2} = c \eta_1 \ ,\ \frac{\partial
\eta_1}{\partial x_2} = c  \tan x_2 \, \eta_1 \ , $$
from
which by differentiating the second with respect to $x_2$ and comparing with
the first we conclude (since by assumption $\eta_1\neq0$): $c(c+1)=0$.

Since the case $c=-1$ has already been discussed in the previous \S,
there remains for us here only to assume $c=0$   so that $\eta_1=a$ (constant).
Then (24) become $\eta_2=\psi(x_2,x_3) $, $\eta_3=\chi(x_2,x_3) $
and (20), (21), (22) give us
\begin{eqnarray}
&&
\frac{\partial \psi}{\partial x_2}
    + a \frac{\varphi^\prime}{\varphi} =0
\ ,\seteqno{26}\\
&&
\frac{\partial \chi}{\partial x_3}
    + a \frac{\varphi^\prime}{\varphi} + \cot x_2 \, \psi =0
\ ,\seteqno{27}\\
&&
\frac{\partial \psi}{\partial x_3}
+ \sin^2 x_2\, \frac{\partial \chi}{\partial x_2} = 0
\ .\seteqno{28}
\end{eqnarray}

In (26), (27) $x_1$ should appear only in   $\varphi^\prime/\varphi $
and therefore $\varphi^\prime/\varphi=k$ (constant), so that
$\psi=-ak x_2 + \theta(x_3) $, with  $\theta$ a function only of $x_3$.
After this (27), (28) become:
$$
\frac{\partial\chi}{\partial x_2}
  = - \frac{\theta'(x_3)}{\sin^2 x_2}\ ,\
\frac{\partial\chi}{\partial x_3}
  = -ak + ak x_2 \cot x_2 -\cot x_2\, \theta(x_3)\ ,
\eqno(29)
$$

Forming the integrability condition for these last two equations, we conclude
that $\theta''(x_3)+\theta(x_3) = ak(x_2-\cos x_2 \sin x_2)$, so that  $a k =
0$   and since $ \eta_1=a\neq0$, we must have $k=0$. So  one therefore has
$\varphi = constant$ and without loss of generality (by substituting a similar
space), we can make  $\varphi(x_1) = 1 $, which gives us the line element $$
ds^2=dx_1^2+ dx_2^2+\sin^2{x_2} \, dx_3^2 $$ indicated in the title of the
section.

As a consequence we must have
$\theta''(x_3)+\theta(x_3)  = 0$, from which
$\theta(x_3)=b\cos x_3 + c\sin x_3$
with $b,c$ (arbitrary) constants.
Then integrating (29), we have
$\chi =  -\cot x_2 (b \sin x_3 - c\cos x_3) + d$,
where $d$ is a new arbitrary constant.

The most general way of satisfying the fundamental equations in the present
case is therefore given by the formula
$$
\eta_1=a\ ,\
\eta_2=b\cos x_3 + c\sin x_3\ ,\
\eta_3=\cot x_2 (-b \sin x_3 + c\cos x_3) + d\ ,
$$
with $a,b,c,d$ arbitrary constants.

Thus the complete group of motions of the present space is
the 4-parameter group generated by the infinitesimal transformations
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_3}\ ,\
X_2 f = \sin x_3 \frac{\partial f}{\partial x_2}
          + \cot x_2 \cos x_3 \frac{\partial f}{\partial x_3}\ ,\nonumber\\
&&
X_3 f = \cos x_3 \frac{\partial f}{\partial x_2}
          - \cot x_2 \sin x_3 \frac{\partial f}{\partial x_3}\ ,\
X_4 f = \frac{\partial f}{\partial x_1}\ ,\nonumber
\end{eqnarray}
whose composition is given therefore in the equations
%%
\begin{eqnarray}
&&
  [X_1,X_2]f  = X_3 f\ ,\
  [X_1,X_3]f  = -X_2 f\ ,\
  [X_2,X_3]f  = X_1 f\ ,\nonumber\\\relax
&&
  [X_1,X_4]f  =
  [X_2,X_4]f  =
  [X_3,X_4]f  = 0\ .\nonumber
\end{eqnarray}

The form of the line element
$ds^2=dx_1^2+ dx_2^2+\sin^2{x_2}\,dx_3^2$
already renders {\it a priori\/} evidence that, other than the $\infty^3$
motions which correspond to the sliding of each surface $x_1= constant$
into itself, there exists here a group $G_1$ with finite equations
$x_1'=x_1+constant $, $x_2'=x_2 $, $x_3'=x_3 $.

But our calculations show that this $G_4$ is also the {\it complete\/}
group of motions.
Such a group $G_4$ is clearly transitive; furthermore it is
{\it systatic\/} since
the motions that leave a point of the space fixed also leave fixed all the points
of that geodesic ($x_1$) which passes through it, so that these geodesics are the
{\it systatic varieties\/}
of the group.     The whole space can be
freely rotated around each one of these, but no other rotation is possible.

\typeout{**** Begin section 10}

\section{%10.
The group of motions of the space:\\
$ds^2=dx_1^2+ \varphi^2(x_1) \, (dx_2^2+ e^{2x_2} dx_3^2)$.}

The fundamental equations (A) are translated by the present space into
the following:\footnote{In the original paper, eq. (31b) had $\partial \eta_2 /
\partial x_3$ on the l.h.s., and eq. (32a) had $\partial \eta_1 / \partial x_2$
on the r.h.s., both of which were incorrect. Correction after the {\it Opere}
[Editor].}
\begin{eqnarray}
&&
 \frac{\partial \eta_1}{\partial x_1} = 0\ ,\seteqno{30}\\
&&
 \frac{\partial \eta_2}{\partial x_1}
  = - \frac{1}{\varphi^2} \frac{\partial\eta_1}{\partial x_2}
  \ ,\seteqno{31a}\\
&&
 \frac{\partial \eta_2}{\partial x_2}
  = - \frac{\varphi^\prime}{\varphi} \eta_1\ ,\seteqno{31b}\\
&&
 \frac{\partial \eta_3}{\partial x_1}
  = -\frac{e^{-2 x_2}}{\varphi^2}  \frac{\partial \eta_1}{\partial x_3}
  \ ,\seteqno{32a}\\
&&
 \frac{\partial \eta_3}{\partial x_3}
  = - \frac{\varphi^\prime}{\varphi} \eta_1 - \eta_2
  \ ,\seteqno{32b}\\
&&
 \frac{\partial \eta_2}{\partial x_3}
 + e^{2 x_2} \, \frac{\partial \eta_3}{\partial x_2} = 0
  \ . \seteqno{33}
\end{eqnarray}
Eliminating by differentiation $\eta_2$ from (31) and $\eta_3$ from (32)
we find
\begin{eqnarray}
&&
\frac{\partial^2 \eta_1}{\partial x_2^2}
= (\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2}) \, \eta_1
  \ ,\nonumber\\
&&
\frac{\partial^2 \eta_1}{\partial x_3^2}
= e^{2 x_2}(\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2}) \, \eta_1
  - e^{2 x_2} \frac{\partial\eta_1}{\partial x_2}
 \ ,\nonumber
\end{eqnarray}
from which (assuming $\eta_1\neq0$) one derives as usual
$(\varphi^{\prime\prime} \varphi -\varphi^{\prime\,2})
=c $ (constant),
so that
\begin{eqnarray}
\frac{\partial^2 \eta_1}{\partial x_2^2}
  &=& c \eta_1\ ,\seteqno{34}\\
 \frac{\partial^2 \eta_1}{\partial x_3^2}
  &=& e^{2 x_2} \left(c \eta_1 - \frac{\partial \eta_1}{\partial x_2}\right)
\ .\seteqno{35}
\end{eqnarray}

Integrating (31a) and  (32a) with respect to $x_1$ we obtain:
\begin{eqnarray}
&&
\eta_2 =
-\frac{\partial \eta_1}{\partial x_2}
    \int \frac{d x_1}{\varphi^2(x_1)} + \psi(x_2,x_3)
   \ ,\seteqno{36}\\
&&
\eta_3 =
- e^{-2 x_2} \frac{\partial \eta_1}{\partial x_3}
    \int \frac{d x_1}{\varphi^2(x_1)} + \chi(x_2,x_3)
\ ,\seteqno{37}
\end{eqnarray}
and substituting into (33) we have
$$
2 \left(\frac{\partial^2 \eta_1}{\partial x_2 \partial x_3}
       -  \frac{\partial\eta_1}{\partial x_3}\right)
    \int \frac{d x_1}{\varphi^2(x_1)}
   = \frac{\partial \psi}{\partial x_3}
     + e^{2 x_2} \frac{\partial \chi}{\partial x_2} \ .
$$
Applying the usual observation, we deduce from this
$$
\frac{\partial^2 \eta_1}{\partial x_2 \partial x_3}
= \frac{\partial \eta_1}{\partial x_3}\ .
$$

Differentiating this with respect to $x_2$ and comparing with (34)
differentiated with respect to $x_3$, it follows that
$(c-1) \, \partial \eta_1/\partial x_3=0$,
from which it follows that $c=1$ or
$\partial \eta_1/\partial x_3=0$.

We treat the first case in this section. The equation
$
\varphi^{\prime\prime} \varphi-\varphi^{\prime\,2} = 1
$
differentiated gives $\varphi^{\prime\prime}= k \varphi$,
($k$ constant), so that $\varphi^{\prime\,2} = k \varphi^2-1$.
The constant $k$ will necessarily be positive and, putting
$k=1/R^2 $
and neglecting the additive constant in $x_1$, we will have
$\varphi(x_1)=R \cosh(x_1/R) $.
In such a case the space has the line element
$$
ds^2=dx_1^2 + R^2\cosh^2(x_1/R) \, (dx_2^2+e^{2 x_2} \, dx_3^2)
$$
and is of constant negative curvature $K= -1/R^2$.
Its complete group of motions is a  $G_6$.

\typeout{**** Begin section 11}

\section{%11.
 The group $G_4$ of motions of the space:\\
$ds^2=dx_1^2+ dx_2^2 + e^{2x_2} \, dx_3^2$.}

We continue the discussion of the previous section assuming now
$\partial\eta_1/\partial x_3=0 $.

Equations (34), (35) give\footnote{Bianchi used the ordinary derivative $d$
instead of $\partial$ in both equations here, correction by the Editor.}
$\partial^2\eta_1/\partial x_2{}^2 = c \eta_1$, $\partial\eta_1/\partial x_2 =
c \eta_1 $ from which $c^2=c$ and consequently $c=0$, the case $c=1$ having
already been discussed in \S 10. So we then have $\eta_1=a $ (constant), and
(36), (37) become $\eta_2=\psi(x_2,x_3) $, $\eta_3=\chi(x_2,x_3) $, while the
equations at the beginning of \S 10 give $$ \frac{\partial\psi}{\partial x_2} +
a \frac{\varphi'}{\varphi}=0\ ,\ \frac{\partial\chi}{\partial x_3} + a
\frac{\varphi'}{\varphi} +\psi=0\ ,\ \frac{\partial\psi}{\partial x_3}
        + e^{2 x_2} \frac{\partial\chi}{\partial x_2}=0\ .
$$

We conclude from this that $\varphi'=k\varphi $ ($k$ constant),
from which it follows that
\begin{eqnarray}
&&
 \psi=-ak x_2+\theta(x_3)\ ,\nonumber\\
&&
 \frac{\partial\chi}{\partial x_2}=-e^{2 x_2} \theta'(x_3)\ ,\
 \frac{\partial\chi}{\partial x_3}=-ak +ak x_2 - \theta(x_3)\ .
\nonumber
\end{eqnarray}

Writing the integrability condition  for these last two equations, we
find $e^{-2 x_2} \theta''(x_3)+ak=0 $,
from which
$k=0 $, $\theta''(x_3)=0 $
establishing the most general values of $\eta_1,\eta_2,\eta_3 $ to be:
$$
  \eta_1=a\ ,\
  \eta_2=b x_3 +c\ ,\
  \eta_3=\frac{b}2 (e^{-2 x_2} - x_3^2) -c x_3 +d\ ,
$$
with $a,b,c,d $ arbitrary constants.
By replacing the space with a similar space,
one can make $\varphi(x_1)=1 $ as in \S 9
and one therefore has the line element
$$
ds^2=dx_1^2+ dx_2^2 + e^{2x_2} \, dx_3^2\ .
$$

Therefore here also as in \S 9, the complete group of motions is a $G_4$.
Its infinitesimal transformation generators are:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_3}\ ,\
X_2 f = - \frac{\partial f}{\partial x_2}
          + x_3 \frac{\partial f}{\partial x_3}\ ,\nonumber\\
&& X_3 f =  x_3 \frac{\partial f}{\partial x_2}
          +\frac12(e^{-2 x_2}-x_3^2)\frac{\partial f}{\partial x_3}\ ,\
X_4 f = \frac{\partial f}{\partial x_1}\ ,\nonumber
\end{eqnarray}
and have the composition
\begin{eqnarray}
&&
  [X_1,X_2]f  = X_1 f\ ,\
  [X_2,X_3]f  = X_3 f\ ,\
  [X_3,X_1]f  = X_2 f\ ,\nonumber\\\relax
&&
  [X_1,X_4]f  =
  [X_2,X_4]f  =
  [X_3,X_4]f  = 0\ .\nonumber
\end{eqnarray}

The properties of the group are entirely similar to those already
described for the group in \S 10.  However, the two corresponding
spaces belong to essentially different types, a fact established by
the observation that the surfaces orthogonal to the systatic
geodesics ($x_1$) are surfaces
of constant positive curvature for the space of \S 10,
while for the present space they are of constant negative curvature.

We summarize these last results obtained here in the theorem:

{\it
If a space of three dimensions admits an intransitive group
$G_3$ of motions, its line element is reducible to one of the
3 standard forms:
\begin{eqnarray}
&&
 ds^2=dx_1^2+\varphi^2(x_1) \, (dx_2^2+dx_3^2)
\ ,\nonumber\\
&&
 ds^2=dx_1^2+\varphi^2(x_1) \, (dx_2^2+\sin^2{x_2} \, dx_3^2)
\ ,\nonumber\\
&&
 ds^2=dx_1^2+\varphi^2(x_1) \, (dx_2^2+e^{2 x_2} \, dx_3^2)
\nonumber
\end{eqnarray}
and in general the complete group of motions is exactly a
3-parameter group.  The only exceptions are the two special spaces
\begin{eqnarray}
&&
 ds^2=dx_1^2+ dx_2^2+\sin^2{x_2} \, dx_3^2
\ ,\nonumber\\
&&
 ds^2=dx_1^2+ dx_2^2+e^{2 x_2} \, dx_3^2\ ,
\nonumber
\end{eqnarray}
each with a 4-parameter group of motions, and the spaces of constant
curvature with 6-parameter groups}.

\typeout{**** Begin section 12}

\section{%12.
Spaces with a transitive group $G_3$ of motions.}

         Having exhausted the study of
spaces which admit an intransitive $G_3$ of motions in the
previous sections, let us
now turn to the treatment of the spaces with a transitive
group of motions.

In this section, we begin to establish in general that given any group $G_3$
whatsoever, transitive over 3 variables $x_1,x_2,x_3$, there always exist some
spaces of 3 dimensions which admit it as a group of motions. In fact we
establish more generally the analogous result for any number $n$  of dimensions
with the theorem:

{\it
Given any transitive group of $n$ parameters
over $n$ variables:
$$
 G_n\equiv (X_1 f, X_2 f, \ldots, X_n f)\ ,
$$
it is always possible to find spaces of $n$ dimensions which admit it
as a group of motions.\footnote{%(10)
If the group is not simply transitive the theorem does not hold in
general as is already shown by the theorem at the beginning of \S 2.}
}

To avoid confusion, however,  we state immediately that the
spaces $S_n $ so determined may very well admit a larger group as
the {\it complete\/} group of motions, as the case
$n=2$ has already shown (see \S3).

    We assume in general
$$
 X_\alpha f = \sum_i^{1\ldots n} \xi_i^{(\alpha)}
           \frac{\partial f}{\partial x_i}\ ,\
(\alpha=1,2,\ldots,n)\ ,
$$
and one will have:
$$
[X_\alpha,X_\beta]f
= \sum_\gamma c_{\alpha\beta\gamma} \, X_\gamma f\ ,
\eqno(38)
$$
where $c_{\alpha\beta\gamma}$ are the
{\it constants of composition\/}.
Furthermore, since the group is
assumed to be transitive, the determinant
$$
  |\xi^{\alpha)}|
  = \left| \begin{array}{cccc}
      \xi_1^{(1)} & \xi_2^{(1)} & \cdots & \xi_n^{(1)}\\
      \xi_1^{(2)} & \xi_2^{(2)} & \cdots & \xi_n^{(2)}\\
       \cdot & \cdot & \cdot & \cdot\\
      \xi_1^{(n)} & \xi_2^{(n)} & \cdots & \xi_n^{(n)}
        \end{array}\right|
$$
will be different from zero.

Here the coefficients $\xi_i^{(\alpha)} $ are given as
functions of $x$   and we have
to determine the coefficients $a_{ik} $
of the differential form
$ds^2=\sum_{i,k} a_{ik} \, dx_i dx_k $
so that it admits the group $G_n$,
in other words so that
the fundamental equations (A) are satisfied by all the
$n$ transformations $X_\alpha f$.
To determine the $a_{ik}$ we therefore have the
$n^2(n+1)/2$
partial differential equations
\begin{eqnarray}
&&
X_\alpha (a_{ik}) + \sum_r \left(
      a_{ir} \frac{\partial\xi_r^{(\alpha)}}{\partial x_k}
     + a_{kr} \frac{\partial\xi_r^{(\alpha)}}{\partial x_i}
    \right) = 0\ ,\seteqno{D}\\
&&
 (\alpha,i,k,=1,2,3,\ldots,n)\ .
\nonumber
\end{eqnarray}

If in (D) we fix $i,k$ and let $\alpha$ take the $n$ values $1,2,\ldots,n$, we
can solve the resulting equations for the $n$ first derivatives of $a_{ik}$
since by hypothesis $|\xi_i^{(\alpha)}|\neq0 $. We therefore have a system of
linear and homogeneous total differential equations for our unknowns $a_{ik}$.
We show that this system is completely integrable, for which it suffices to
prove that by writing two of the equations (D) for the same unknown $a_{ik}$:
\begin{eqnarray}
&&
X_\alpha (a_{ik})
   + \sum_r a_{ir} \frac{\partial\xi_r^{(\alpha)}}{\partial x_k}
   + \sum_r a_{kr} \frac{\partial\xi_r^{(\alpha)}}{\partial x_i}
     = 0\ ,\nonumber\\
&&
X_\beta (a_{ik})
   + \sum_s a_{is} \frac{\partial\xi_s^{(\beta)}}{\partial x_k}
   + \sum_s a_{ks} \frac{\partial\xi_s^{(\beta)}}{\partial x_i}
    = 0\ ,\nonumber
\end{eqnarray}
and if the operation $X_\alpha$ is performed on the second of these, the
operation  $X_\beta$ on the first of these, and one subtracts the results
making use of the same equation (D), the result is an identity.  Using (38) on
this relation one obtains in this way first\footnote{In the last line of eq.
(39), $\sum_i$ was corrected to $\sum_r$ [Editor].}
\begin{eqnarray}
&& \sum_\gamma c_{\alpha\beta\gamma} X_\gamma (a_{ik})
   + \sum_s X_\alpha (a_{is}) \frac{\partial\xi_s^{(\beta)}}{\partial x_k}
   + \sum_s X_\alpha (a_{ks}) \frac{\partial\xi_s^{(\beta)}}{\partial x_i}
\nonumber\\
&&\quad
  -\sum_r X_\beta (a_{ir}) \frac{\partial\xi_r^{(\alpha)}}{\partial x_k}
  -\sum_r X_\beta (a_{kr}) \frac{\partial\xi_r^{(\alpha)}}{\partial x_i}
\nonumber\\
&&\quad
  + \sum_r a_{ir}
    \left[ X_\alpha \left(\frac{\partial\xi_r^{(\beta)}}{\partial x_k}\right)
         - X_\beta \left(\frac{\partial\xi_r^{(\alpha)}}{\partial x_k}\right)
    \right]\nonumber\\
&&\quad
  + \sum_r a_{kr}
    \left[ X_\alpha \left(\frac{\partial\xi_r^{(\beta)}}{\partial x_i}\right)
         - X_\beta \left(\frac{\partial\xi_r^{(\alpha)}}{\partial x_i}\right)
    \right]=0\ .
\seteqno{39}
\end{eqnarray}

Now from (38) itself one has
$$
   X_\alpha(\xi_r^{(\beta)})
 -  X_\beta(\xi_r^{(\alpha)})
  = \sum_r c_{\alpha\beta\gamma} \, \xi_r^{(\gamma)} \ ,
$$
which by differentiating with respect to $x_k$
becomes
%%
\begin{eqnarray}
&&
   X_\alpha\left( \frac{\partial\xi_r^{(\beta)}}{\partial x_k}\right)
 -  X_\beta\left( \frac{\partial\xi_r^{(\alpha)}}{\partial x_k} \right)
\nonumber\\
&&\qquad
= \sum_\gamma c_{\alpha\beta\gamma}
        \frac{\partial\xi_r^{(\gamma)}}{\partial x_k}
 +\sum_s\left( \frac{\partial\xi_r^{(\alpha)}}{\partial x_s}
                 \frac{\partial\xi_s^{(\beta)}}{\partial x_k}
              -  \frac{\partial\xi_s^{(\alpha)}}{\partial x_k}
                 \frac{\partial\xi_r^{(\beta)}}{\partial x_s}
         \right)
\ ,
\nonumber
\end{eqnarray}
and similarly
%%
\begin{eqnarray}
&&
   X_\alpha\left( \frac{\partial\xi_r^{(\beta)}}{\partial x_i}\right)
 -  X_\beta\left( \frac{\partial\xi_r^{(\alpha)}}{\partial x_i} \right)
\nonumber\\
&&\qquad
= \sum_\gamma c_{\alpha\beta\gamma}
        \frac{\partial\xi_r^{(\gamma)}}{\partial x_i}
 +\sum_s\left( \frac{\partial\xi_r^{(\alpha)}}{\partial x_s}
                 \frac{\partial\xi_s^{(\beta)}}{\partial x_i}
              -  \frac{\partial\xi_r^{(\beta)}}{\partial x_s}
                 \frac{\partial\xi_s^{(\alpha)}}{\partial x_i}
         \right)
\ .
\nonumber
\end{eqnarray}

If in the first 5 terms of (39) we introduce
the values of $X(a)$ given
by (D) and in the last 2 terms the values
calculated above, we see that
it is converted into an identity.
We conclude from this that the system
of total differential equations for the $a_{ik}$
is completely integrable and
we can therefore give  the initial values of the
$a_{ik}$ arbitrarily at a point
of the space $S_n$.
So if we choose them in such a way
that the conditions (of inequality) making the differential form
positive definite are {\it initially\/}
satisfied, they will remain so in a certain
neighborhood of that point and we will therefore have
defined a space of $n$ dimensions which admits the
group $G_n$ as a group of motions.

\typeout{**** Begin section 13}

\section{%13.
Preliminary classification of the various types of $G_3$.}

With the general considerations of the previous
sections we are
assured that to any $G_3$ transitive over 3 variables
always correspond spaces
of 3 dimensions which admit it as a group of motions.
It is not true,  however,
and is not even true in all cases, that the
{\it complete\/} group of motions of the space obtained is
indeed the given $G_3$. It will be seen
instead that there are certain compositions
of the $G_3$ which
necessarily imply the existence of a larger
group of motions.\footnote{%(11)
This happens for the groups $G_3$ of types I, II, III, V in the classification
of the present section.} Furthermore we wish to establish for any possible type
of $G_3$ a corresponding canonical form for the line element, by performing the
integration which we have only described in the previous section. As the basis
of our calculations we take the classification given by Lie of the possible
compositions of groups of 3 par\-a\-me\-ters.\footnote{%(12)
S. Lie-F. Engel, Vol.~III, p.~713 and
S. Lie-G. Scheffers, {\it Vorlesungen \"uber continuierliche
Gruppen\/} (1893), p.~565.}
But here an essential warning is necessary for us.
In the classification of Lie there is no way
for us to distinguish between real and complex,
whereas in this study we wish to report only on
real groups and their real subgroups: we will
therefore have to subdivide into more types some types
which are a single type from the general point of view of Lie.

Without repeating the discussion given by Lie ({\it ibid.\/}),
it will suffice to point out that,
considering first the {\it integrable\/} groups,
to the 6 types classified by Lie according to the following
compositions
\begin{eqnarray}
{\rm(Type\ I)}   &&
[X_1,X_2]f  =
[X_1,X_3]f  =
[X_2,X_3]f  = 0\ ,
\nonumber\\
{\rm(Type\ II)}  &&
[X_1,X_2]f  =
[X_1,X_3]f  = 0\ ,\
[X_2,X_3]f  = X_1 f\ ,
\nonumber\\
{\rm(Type\ III)} &&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_1 f\ ,\
\nonumber\\ && \qquad
[X_2,X_3]f  = 0\ ,
\nonumber\\
{\rm(Type\ IV)}  &&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_1 f\ ,\
\nonumber\\ && \qquad
[X_2,X_3]f  = X_1 f + X_2 f\ ,
\nonumber\\
{\rm(Type\ V)}   &&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_1 f\ ,\
\nonumber\\ && \qquad
[X_2,X_3]f  = X_2 f\ ,
\nonumber\\
{\rm(Type\ VI)}  &&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_1 f\ ,\
\nonumber\\ && \qquad
[X_2,X_3]f  = h X_2 f\ ,\ (h\neq0,1)\ ,
\nonumber
\end{eqnarray}
we must add a seventh type with the composition
\begin{eqnarray}
{\rm(Type\ VII)}
&&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_2 f\ ,\
\nonumber\\ && \qquad
[X_2,X_3]f  = -X_1 f + h X_2 f\ ,
\nonumber
\end{eqnarray}
where the constant $h$ satisfies the inequality
$0\leq h<2$.\footnote{%(13)
The sign of $h$ is not essential, as one sees by simultaneously changing the
signs of $X_2 f $, $X_3 f $.}

{}From our real point of view this composition in effect differs from all of
the previous ones in that, while in the first 6 types one has at least a real
{\it invariant\/} subgroup $G_1$, in type VII,
however, no such {\it real\/} subgroup exists.\footnote{%(14)
If  $ Y f =
\alpha_1 X_1 f + \alpha_2 X_2 f +\alpha_3 X_3 f $
were the infinitesimal transformation
generator of such a subgroup, the three
infinitesimal transformations
$[Y,X_1]f  $, $[Y,X_2]f  $, $[Y,X_3]f  $,
would have to differ from $Y f $
only by a constant factor.
It follows immediately from this that $\alpha_3=0 $, and
then from
$$
  [Y,X_3]f  = \alpha_1 X_2 f + \alpha_2(-X_1 f + h X_2 f)
  = \rho(\alpha_1 X_1 f + \alpha_2 X_2 f)\ ,
$$
we obtain
$\rho \alpha_1 + \alpha_2 = 0$,
$\rho \alpha_2 - \alpha_1 - h \alpha_2 = 0$,
so that
$ \rho^2-h\rho+1=0$,
an equation with complex roots since $ h^2<4$.}

Furthermore, it is necessary to observe that
in the new composition VII
the constant $h$ is truly {\it essential\/}, namely that
if there is a second group
$(Y_1 f, Y_2 f, Y_3 f)$ of composition
\begin{eqnarray}
&& \kern-1cm
[Y_1,Y_2]f  = 0\ ,\
[Y_1,Y_3]f  = Y_2 f\ ,
[Y_2,Y_3]f  = -Y_1 f + k Y_2 f\ ,
\nonumber\\
&&\qquad
 (0\leq k<2)\ ,
\end{eqnarray}
if $k\neq h$, then the two groups cannot be put
into an isomorphic correspondence.
Indeed if this occurred and we indicate
by $\bar X_1 f$, $\bar X_2 f$, $\bar X_3 f$,
the infinitesimal
transformations of the first group
which correspond respectively to
$Y_1 f$, $Y_2 f$, $Y_3 f$
in the second, then
$\bar X_1 f$, $\bar X_2 f$
must be constructed only with
$X_1 f$, $X_2 f$
since
both pairs of transformations
belong to the derived group.
We assume therefore:
\begin{eqnarray}
&&
\bar X_1 f = a X_1 f + \beta X_2 f \ ,\
\bar X_2 f = \gamma X_1f + \delta X_2 f \ ,
\nonumber\\
&&\qquad
\bar X_3 f = a X_1 f + b X_2 f + c X_3 f\ ,
\nonumber
\end{eqnarray}
and from the assumed relations of composition
we find the following relations among the constants
$ \alpha,\beta,\gamma,\delta,c$:
\begin{eqnarray}
&& \gamma+\beta c=0\ ,\
 \delta-\alpha c-h\beta c =0\ ,
\nonumber\\
&& \alpha-k\gamma-c\delta =0\ ,\
\beta-k\delta + c\gamma +hc\delta =0\ ,
\nonumber
\end{eqnarray}
so that
$$
\alpha=c(\delta-k\beta)\ ,\
\gamma=-\beta c\ ,\
\cases{\beta(1-c^2)+(hc-k)\delta=0\ ,&\cr
       \beta c(h-kc)+ (c^2-1)\delta=0\ .&\cr}
$$

{}From these last two equations, since both $\beta$
and  $\delta$ cannot be simultaneously zero,
it follows that $c$ satisfies the   4th
degree equation
$$
c^4-hkc^3+(h^2+k^2-2) c^2 -hk c+1=0\ ;
$$
but then the determinant $\alpha\delta-\beta\gamma $
(since $c^2\neq1 $ because $k\neq\pm h $)
would have to be zero, but that is absurd.

There remains finally to consider the case
in which the group $G_3$ is not integrable.
For these groups Lie assigned the single type
\begin{eqnarray}
{\rm(Type\ VIII)}
&&
[X_1,X_2]f  = X_1 f\ ,\
[X_1,X_3]f  = 2 X_2 f\ ,\
\nonumber\\ &&\qquad
[X_2,X_3]f  = X_3 f\ ,
\nonumber
\end{eqnarray}
but we must add another:
\begin{eqnarray}
{\rm(Type\ IX)}
&&
[X_1,X_2]f  = X_3 f\ ,\
[X_2,X_3]f  = X_1 f\ ,\
\nonumber\\ &&\qquad
[X_3,X_1]f  = X_2 f\ ,
\nonumber
\end{eqnarray}
which differs from the previous one only in
that there does not exist a {\it real\/}
2-parameter subgroup in this last case.%
\footnote{%(15)
In the geometrical representation given by Lie
on p.~718 of Vol.~III,
according to reciprocity in the plane with respect to a conic,
one case
is distinguished from the other by type VIII  having a real conic
and type IX a complex conic.
}

\typeout{**** Begin section 14}

\section{%14.
The groups of type I.}

In the first seven types the group $G_3$
contains the Abelian
2-parameter subgroup of
motions   $G_2\equiv (X_1 f,X_2 f)$.
The considerations of \S4 show that with
respect to this $G_2$ the minimum invariant
varieties are geodesically parallel surfaces
of zero curvature. By
assuming these as the coordinate surfaces $x_1=constant$,
we can furthermore make
$X_1 f = \partial f/\partial x_2 $,
$X_2 f = \partial f/\partial x_3 $
and the line element of the space will
take the form
$$
 ds^2=dx_1^2 + \alpha \, dx_2^2 + 2\beta \, dx_2 dx_3 + \gamma \, dx_3^2\ ,
\eqno(40) $$ with $\alpha$, $\beta$, $\gamma$ functions only of $x_1$. To
determine the most general infinitesimal motion of this space the fundamental
equations (A) give us the system
\begin{eqnarray}
&&
\frac{\partial\eta_1}{\partial x_1} = 0
\ ,\nonumber\\
&&
\frac{\partial\eta_1}{\partial x_2}
  + \alpha \frac{\partial\eta_2}{\partial x_1}
  + \beta  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\nonumber\\
&&
\frac{\partial\eta_1}{\partial x_3}
  + \beta \frac{\partial\eta_2}{\partial x_1}
  + \gamma  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\nonumber\\
&& \seteqno{E}\\
&&
\frac12 \alpha' \eta_1
  + \alpha \frac{\partial\eta_2}{\partial x_2}
  + \beta  \frac{\partial\eta_3}{\partial x_2}= 0
\ ,\nonumber\\
&&
\frac12 \gamma' \eta_1
  + \beta \frac{\partial\eta_2}{\partial x_3}
  + \gamma  \frac{\partial\eta_3}{\partial x_3}= 0
\ ,\nonumber\\
&&
 \beta' \eta_1
  + \alpha \frac{\partial\eta_2}{\partial x_3}
  + \beta \left( \frac{\partial\eta_2}{\partial x_2}
       + \frac{\partial\eta_3}{\partial x_3} \right)
  + \gamma  \frac{\partial\eta_3}{\partial x_2}= 0
\ .\nonumber
\end{eqnarray}

Now if we assume that there exists a third
infinitesimal transformation
$
  X_3 f = \xi_1 \partial f/\partial x_1
        + \xi_2 \partial f/\partial x_2
        + \xi_3 \partial f/\partial x_3
$,
which with  $X_1 f, X_2 f $
generates a group $G_3$, we have in general
\begin{eqnarray}
&&
[X_1,X_3]f  = a X_1 f + b X_2 f + c X_3 f\ ,
\nonumber\\
&&
[X_2,X_3]f  = a' X_1 f + b' X_2 f + c' X_3 f\ ,
\nonumber
\end{eqnarray}
with  $a,b,c,a',b',c'$ constants, and therefore
the following 6 equations hold
\begin{eqnarray}
&&
\frac{\partial\xi_1}{\partial x_2} = c \xi_1\ ,\
\frac{\partial\xi_2}{\partial x_2} = c \xi_2 + a\ ,\
\frac{\partial\xi_3}{\partial x_2} = c \xi_3 + b\ ,
\seteqno{41a}\\
&&
\frac{\partial\xi_1}{\partial x_3} = c' \xi_1\ ,\
\frac{\partial\xi_2}{\partial x_3} = c' \xi_2 + a'\ ,\
\frac{\partial\xi_3}{\partial x_3} = c' \xi_3 + b'\ ,
\seteqno{41b}
\end{eqnarray}
and since we furthermore assume that the group
$ (X_1 f,X_2 f,X_3 f)$
is transitive, we will have $\xi_1\neq0$.

Now the system (E) has to be satisfied when the $\eta$ are replaced by the
$\xi$ and so we will therefore have\footnote{In the first line of the equation,
the original paper has $\partial \xi_1 / \partial x_2$; correction based on the
{\it Opere} [Editor].}
\begin{eqnarray}
&&
\frac{\partial\xi_1}{\partial x_1} = 0
\ ,\nonumber\\
&&
c \xi_1
  + \alpha \frac{\partial\xi_2}{\partial x_1}
  + \beta  \frac{\partial\xi_3}{\partial x_1}= 0
\ ,\nonumber\\
&&
c' \xi_1
  + \beta \frac{\partial\xi_2}{\partial x_1}
  + \gamma  \frac{\partial\xi_3}{\partial x_1}= 0
\ ,\nonumber\\
&& \seteqno{F}\\
&&
\frac12 \alpha' \xi_1
  + \alpha (c \xi_2+a)
  + \beta  (c \xi_3+b)= 0
\ ,\nonumber\\
&&
\frac12 \gamma' \xi_1
  + \beta (c' \xi_2+ d')
  + \gamma (c' \xi_3+b')= 0
\ ,\nonumber\\
&&
 \beta' \xi_1
  + \alpha (c' \xi_2+a')
  + \beta ( c \xi_2 + c' \xi_3+a+b')
  + \gamma (c \xi_3+b)= 0
\ ,\nonumber
\end{eqnarray}

These are the equations which will serve to solve for us the problem posed for
the groups of the first seven types.

Meanwhile for type I, since the constants $a,b,c,a',b',c'$ are all zero, the
last three equations of (F), remembering that $\xi_1\neq0 $, show that
$\alpha,\beta,\gamma $ are constants and so the space is of zero curvature.
Since then there do not exist spaces with an Abelian intransitive $G_3$ of
motions, as results from the discussion of the previous sections and also if
one wishes, from the same system (F) and from (41), we can state the result:
{\it If a space of 3 dimensions admits a 3-parameter Abelian group of motions,
it is of zero curvature and the group is the translation group\/}.

\typeout{**** Begin section 15}

\section{%15.
Digressions relative to spaces of $n$ dimensions.}

It will not be useless to observe that the preceding
theorem holds for spaces of
any number of dimensions, namely:

{\it A space of $n$ dimensions which admits an
$n$-parameter Abelian group of
translations is necessarily of zero curvature
and the group is the translation group.}

To show this it is sufficient to appeal
to the result established by
Lie\footnote{%16
See S. Lie-F. Engel, Vol.~I, p.~339.}
namely the theorem that if $r$ infinitesimal
transformations
$X_1 f,X_2 f,\ldots,X_r f $ over $n$ variables
$x_1,x_2,\ldots,x_n$
commute,
i.e., one has $[X_i,X_j]f  =0$, $(i,k=1,2,\ldots,r)$
and among the $r$ $X f $ does not exist any
linear identity of the form
$$
  \sum_{i=1}^r \alpha_i(x_1,x_2,\ldots,x_n) X_i f = 0 \ ,
$$
where the $\alpha $ are functions of the $x $,
with a convenient transformation of
variables they can be reduced to the
form:
$$
X_1 f=\frac{\partial f}{\partial x_1}\ ,\
X_2 f=\frac{\partial f}{\partial x_2}\ ,\
\ldots\ ,\
X_r f=\frac{\partial f}{\partial x_r}\ ,
$$

Therefore with     $ G_n\equiv (X_1 f,X_2 f,\ldots,X_n f)  $
the hypothetical group,
it will be enough to show that there does not
exist among the  $X_1 f,X_2 f,\ldots,X_n f  $
an identity of the above form, namely that $G_n$
is transitive, since then having reduced
the group of motions to the canonical form
$(\partial f/\partial x_1, \partial f/\partial x_2,\ldots,
\partial f/\partial x_n)$
by the fundamental equations (A) the
coefficients $a_{ik}$ of the line element will be
independent of all the $x$, namely absolute constants,
and so we will have a space of zero
curvature.  Now we assume that among the
first $s$ of the $X_i f $: $X_1 f,X_2 f,\ldots,X_s f  $
does not exist any linear
identity of the
form mentioned above (and we will have by the
theorem of \S2: $s\geq2 $ ), while one has
$X_{s+1} f= \xi_1 X_1 f+  \xi_2 X_2 f+ \cdots +\xi_s X_s f$,
the $\xi $ being functions of the $x $
which are {\it not all\/} constants.  By the cited theorem
of Lie we can assume
$$
X_1 f=\frac{\partial f}{\partial x_1}\ ,\
X_2 f=\frac{\partial f}{\partial x_2}\ ,\
\ldots\ ,\
X_s f=\frac{\partial f}{\partial x_s}\ ,
$$
and we will have
$$
X_{s+1} f= \xi_1 \frac{\partial f}{\partial x_1}
         +  \xi_2 \frac{\partial f}{\partial x_2}
         + \cdots
        +   \xi_s \frac{\partial f}{\partial x_s}\ .
$$

First the conditions
$$
 [X_{s+1},X_1]f  =0\ ,\
 [X_{s+1},X_2]f  =0\ ,\
\ldots\ ,
 [X_{s+1},X_s]f  =0
$$
show that the $\xi $ do not depend on the first
$s$ variables  $x_1,x_2,\ldots, x_s$.
Secondly, the fundamental equations (A),
where one fixes $k $  and sets
$i=1,2,\ldots,s $,
give
$$
\sum_{r=1}^s a_{ir} \frac{\partial\xi_r}{\partial x_k} =0\ ,
(i=1,2,3,\ldots,s)\ .
$$

Now the determinant
$\left|\begin{array}{cccc}
 a_{11}&a_{12}&\ldots&a_{1s}\\
 \cdot & \cdot & \cdot & \cdot\\
 a_{s1}&a_{s2}&\ldots&a_{ss}
 \end{array}\right|
 $
is different from zero, and
also positive since the differential form
$ \sum_{i,k} a_{ik} \, dx_i dx_k$ is positive-definite, so that we
have the result that
$\xi_1,\xi_2,\ldots,\xi_s $ are absolute
constants, which is absurd.

\typeout{**** Begin section 16}

\section{%16.
The groups of type II:\\
  $[X_1,X_2]f  = [X_1,X_3]f  = 0,\ [X_2,X_3]f  = X_1 f $.}

Applying the general method described in \S14,
we must now set $a=b=c=0$, $a'=1$, $b'=c'=0$.

{}From (41) and the first of (F) one then sees that $\xi_1 $ must be a
constant, so we set $\xi_1=-1/h $, and the last three equations of (F) give us
$\alpha'=0$, $\beta'=h\alpha$, $\gamma'=2 h\beta$, from which by integrating $$
\alpha = k^2\ ,\ \beta = h k^2 x_1 + l\ ,\ \gamma = h^2k^2 x_1^2 + 2hl x_1 +m\
, $$
with    $k,l,m$ new constants.\footnote{%(17)
We have indicated the value of $\alpha$ by $k^2$ since it  must be positive.}
The line element of the space therefore has the form\footnote{The $h^2k^2x_1^2$
in the coefficient of $dx_3^2$ is a correction based on the {\it Opere}, the
original had $h^2k^2x_1$ here [Editor].}
$$ ds^2= dx_1^2 +
k^2 \, dx_2^2 + 2(h k^2 x_1 +l) \, dx_2 dx_3 +(h^2 k^2 x_1^2 + 2hl x_1 +m) \, dx_3^2\ .
$$

Replacing $x_2$, $x_3$ respectively by
$ x_2/k$, $x_3/k$, we can write
$$
ds^2= dx_1^2 + dx_2^2 + 2(h x_1 +l/k^2) \, dx_2 dx_3
+[(h x_1 + l/k^2)^2 +n^2]\, dx_3^2\ ,
\eqno(42)$$
having set $n^2=m/k^2-l^2/k^4 $, a constant necessarily positive since
$\alpha\gamma-\beta^2>0 $.

If we put $ h x_1+l/k^2=n y_1$, $ x_2=n/h\, y_2$, $ x_3=1/h\, y_3$, (42)
becomes $$ ds^2= n^2/h^2\, [ dy_1^2 + dy_2^2 + 2 y_1 \, dy_2 dy_3 +(y_1^2+1) \,
dx_3^2]\ . $$

By substituting a similar space, we can therefore
assume as the standard form for the line element:
$$
ds^2 = dx_1^2 + dx_2^2 + 2x_1 \, dx_2 dx_3 +(x_1^2+1) \, dx_3^2\ .
\eqno(43)$$

This space certainly admits a transitive group $G_3$
of motions of type II, but as we now show, its complete
group of motions is a $G_4$ of which the original
$G_3$
is not the derived subgroup.

To determine the most general infinitesimal motion
$ X f =
 \eta_1 \partial f/\partial x_1
+\eta_2 \partial f/\partial x_2
+\eta_3 \partial f/\partial x_3$
of the space (43) it suffices to apply
the equations (E) of \S14, which here become:
\begin{eqnarray}
&&
\frac{\partial\eta_1}{\partial x_1} = 0
\ ,\seteqno{44}\\
&&
\frac{\partial\eta_1}{\partial x_2}
  +  \frac{\partial\eta_2}{\partial x_1}
  + x_1  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\seteqno{45a}\\
&&
\frac{\partial\eta_1}{\partial x_3}
  + x_1 \frac{\partial\eta_2}{\partial x_1}
  + (x_1^2+1)  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\seteqno{45b}\\
&&
\frac{\partial\eta_2}{\partial x_3}
  + x_1  \frac{\partial\eta_3}{\partial x_2}= 0
\ ,\seteqno{46}\\
&&
x_1 \eta_1
  + x_1 \frac{\partial\eta_2}{\partial x_3}
  + (x_1^2+1)  \frac{\partial\eta_3}{\partial x_3}= 0
\ ,\seteqno{47}\\
&&
 \eta_1
  + \frac{\partial\eta_2}{\partial x_3}
  + x_1 \left( \frac{\partial\eta_2}{\partial x_2}
       + \frac{\partial\eta_3}{\partial x_3} \right)
  + (x_1^2+1)  \frac{\partial\eta_3}{\partial x_2}= 0
\ .\seteqno{48}
\end{eqnarray}

    Solving (45) for $\partial\eta_2/\partial x_1$
and $\partial\eta_3/\partial x_1$ and integrating
with respect to $x_1 $ with the observation that
by (44) $\eta_1$ does not depend on $x_1$, we have
\begin{eqnarray}
&&
\eta_2 = \frac{x_1^2}{2} \frac{\partial\eta_1}{\partial x_3}
    - \left( \frac{x_1^2}{3}+x_1 \right)
             \frac{\partial\eta_1}{\partial x_2}
    + \psi(x_2,x_3)\ , \nonumber\\
&&
\eta_3 = \frac{x_1^2}{2} \frac{\partial\eta_1}{\partial x_2}
    - x_1\frac{\partial\eta_1}{\partial x_3}
    + \chi(x_2,x_3)\ . \nonumber
\end{eqnarray}

By substituting these values of $\eta_2,\eta_3 $ into (46) we
obtain a 3rd degree polynomial in $x_1 $ which must
be identically zero; from this we then deduce:
$$
  \frac{\partial^2\eta_1}{\partial x_2{}^2}
 =\frac{\partial^2\eta_1}{\partial x_2\partial x_3} =0\ ,\
\frac{\partial \psi}{\partial x_2}
 = \frac{\partial \chi}{\partial x_2}=0\ .
$$

Proceeding similarly with (47) we finally find $$
  \frac{\partial^2\eta_1}{\partial x_3{}^2} =0\ ,\
\frac{\partial \psi}{\partial x_3}= -\eta_1\ ,\
  \frac{\partial \chi}{\partial x_3}=0\ ,
$$
so that
$$
  \frac{\partial\eta_1}{\partial x_2}
 =-\frac{\partial^2\psi}{\partial x_2\partial x_3} =0\ .
$$

Therefore $\eta_1 $ will be a linear function
depending only on $x_3 $, so we set $\eta_1=a x_3+b$,
and we have $\psi = -\frac12 a x_3^2 -b x_3 +c$, $\chi=d$,
with  $a,b,c,d  $  arbitrary constants.
With the  corresponding values of
$\eta_1,\eta_2,\eta_3 $:
$$
\eta_1 = a x_3 +b\ ,\
\eta_2 = \frac12 a x_1^2 -\frac12 a x_3^2 -b x_3+c\ ,\
\eta_3 = -a x_1 +d\ ,
$$
(48) is also satisfied no matter what values $a,b,c,d $ take.
So the complete group of motions of the space
(43) is the $G_4$ generated by the four infinitesimal
transformations
\begin{eqnarray}
&& X_1 f = \frac{\partial f}{\partial x_2}\ ,\
X_2 f = \frac{\partial f}{\partial x_3}\ ,\
X_3 f = - \frac{\partial f}{\partial x_1}
          + x_3 \frac{\partial f}{\partial x_2}\ ,\nonumber\\
&& X_4 f =  x_3 \frac{\partial f}{\partial x_1}
          +\frac12(x_1^2-x_3^2)\frac{\partial f}{\partial x_2}
          - x_1 \frac{\partial f}{\partial x_3}\ ,\nonumber
\end{eqnarray}
whose composition is expressed by the equations
\begin{eqnarray}
&&
  [X_1,X_2]f  = 0\ ,\
  [X_1,X_3]f  = 0\ ,\
  [X_1,X_4]f  = 0\ ,\nonumber\\
&&
  [X_2,X_3]f  = X_1 f\ ,\
  [X_2,X_4]f  = -X_3 f\ ,\
  [X_3,X_4]f  = X_2 f\ .\nonumber
\end{eqnarray}

As one can see, its derived group is the transitive group
$G_3\equiv (X_1 f,X_2 f,X_3 f)$
of type II.  The three transformations
$X_1 f$, $X_2 f$, $X_3 f$
are not related by any linear identity while one has
$$
  X_4 f = \frac12(x_2^2+x_3^2) X_1 f - x_1 X_2 f - x_3 X_3 f\ ,
$$
and since the coefficients of this relation are
functions only of  $x_1$, $x_3 $,
we conclude from this\footnote{%(18)
S. Lie-F.  Engel, Vol.~I, p.~502.}
that the group $ $ is systatic and the systatic varieties
are the coordinate lines ($x_2$). It is clear geometrically
that these systatic lines are geodesics of the
space,\footnote{%(19)
In fact take two arbitrary points   $P,Q$ on a coordinate line ($x_2 $).
Those transformations of the space which leave $P$
fixed also leave $Q$ fixed and consequently
all the points of the geodesic which joins $P$ to $Q$,
which therefore must coincide with the coordinate line ($x_2$).}
and this statement also follows immediately
from the form (43) of the line element of the space.
The properties of the
group are similar to those described in \S9, \S11
for the groups of the spaces:
\begin{eqnarray}
&&
 ds^2=dx_1^2+dx_2^2+\sin^2 x_2\,dx_3^2
\ ,\nonumber\\
&&
 ds^2=dx_1^2+dx_2^2+e^{2 x_2}\,dx_3^2\ .
\nonumber
\end{eqnarray}

However, the different nature of these spaces
follows immediately upon examining the compositions
of their groups of motions.  While for these latter
spaces the derived group is an intransitive and simple
$G_3$, for the space (43) the derived group
is a transitive and integrable $G_3$. We also observe
an essential difference geometrically since for those
spaces discussed previously the systatic geodesics admit a
family of orthogonal surfaces, which does not occur
for the space (43).\footnote{ %20
To determine the possible surfaces
orthogonal to the geodesic ($x_2 $) one
would have the total differential equation
$d x_2 + x_1 dx_3=0 $
which is not integrable.}

Finally we observe that it is easy to write the
equations of the present group $G_4$ in finite terms.
Those of the derived subgroup are given by the
equations:
$$
x_1'=x_1+a_1\ ,\
x_2'=x_2-a_1 x_3 + a_2-a_1 a_3\ ,\
x_3'=x_3+a_3\ ,
$$
with  parameters $a_1,a_2,a_3 $.  It now suffices to
associate with these $\infty^3$ motions the group $G_1$ generated
by the infinitesimal transformation
whose finite equations are
\begin{eqnarray}
&& x_1'=x_1 \cos t + x_3 \sin t\ ,\
x_3'=-x_1 \sin t +x_3 \cos t\ ,\nonumber\\
&& x_2'=\frac14(x_1^2-x_3^2)\sin(2t)-\frac12 x_1 x_3 \cos(2t)
     +x_2-\frac12 x_1 x_3\nonumber
\end{eqnarray}
and which represents a rotation around the geodesic
$x_1=0$, $x_3=0 $
by an angle easily seen to be $t$.

\typeout{**** Begin section 17}

\section{%17.
The groups of type III:\\
 $[X_1,X_2]f  = 0\ ,\ [X_1,X_3]f  = X_1 f,\ [X_2,X_3]f  = 0 $.}

For the above composition we must set $a=1$, $b=c=0$, $a'=b'=c'=0$ in the
equations of \S14, from which it again follows that $\xi_1 $ is constant, so we
set $\xi_1=-1/h $, and the last 3 equations of (F) give us $\alpha'=2h\alpha$,
$\gamma'=0$, $\beta'=h\beta $.

Integrating and choosing conveniently the
variables $x_2,x_3 $
we can make
$\alpha=e^{2hx_1}$, $\beta=n e^{hx_1}$, $\gamma=1 $,
with $n$ a new constant, and by replacing the space
by a similar one, we can set $h=1$ and have as the
standard form of the line element of the present space:
$$
 ds^2=dx_1^2+ e^{2x_1}\, dx_2^2+2ne^{x_1}\, dx_2 dx_3 +dx_3^2\ .
\eqno(49)$$

One will observe that if $n=0$ one again obtains
the space of \S11.
Since  $\alpha\gamma-\beta^2$ has to be positive,
we will have $n^2<1$,
and since the
sign of $n$ is not essential (as one sees by changing
$x_2$ into $-x_2$, for example),
we can assume $0<n<1$.

We will see that also in the case   $n>0$   as for $n=0$, the space (49) has a
4-parameter group of motions.

The equations (E) \S14 here become
\begin{eqnarray}
&&
\frac{\partial\eta_1}{\partial x_1} = 0
\ ,\seteqno{50}\\
&&
\frac{\partial\eta_1}{\partial x_2}
  +  e^{2x_1}\frac{\partial\eta_2}{\partial x_1}
  + n e^{x_1}  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\seteqno{51a}\\
&&
\frac{\partial\eta_1}{\partial x_3}
  + n e^{x_1} \frac{\partial\eta_2}{\partial x_1}
  +  \frac{\partial\eta_3}{\partial x_1}= 0
\ ,\seteqno{51b}\\
&&
e^{x_1}\eta_1 +e^{x_1}\frac{\partial\eta_2}{\partial x_2}
  + n  \frac{\partial\eta_3}{\partial x_2}= 0
\ ,\seteqno{52}\\
&&
 n e^{x_1}\frac{\partial\eta_2}{\partial x_3}
  + \frac{\partial\eta_3}{\partial x_3}= 0
\ ,\seteqno{53}\\
&&
 n e^{x_1}\eta_1
  + e^{2x_1}\frac{\partial\eta_2}{\partial x_3}
  + n e^{x_1}\left( \frac{\partial\eta_2}{\partial x_2}
       + \frac{\partial\eta_3}{\partial x_3} \right)
  + \frac{\partial\eta_3}{\partial x_2}= 0
\ .\seteqno{54}
\end{eqnarray}

Solving (51) and integrating with respect to $x_1 $
we obtain:
\begin{eqnarray}
&&
\eta_2 = \frac{-n e^{-x_1}}{1-n^2}\frac{\partial\eta_1}{\partial x_3}
    +\frac{e^{-2x_1}}{2(1-n^2)}\frac{\partial\eta_1}{\partial x_2}
    + \psi(x_2,x_3)\ , \nonumber\\
&&
\eta_3 = \frac{-n e^{-x_1}}{1-n^2}\frac{\partial\eta_1}{\partial x_2}
    -\frac{x_1}{1-n^2}\frac{\partial\eta_1}{\partial x_3}
    + \chi(x_2,x_3)\ . \nonumber
\end{eqnarray}

Substituting into (52), (53), (54)   we conclude that
$$
 \frac{\partial\eta_1}{\partial x_3}=0\ ,\
 \frac{\partial^2\eta_1}{\partial x_2{}^2} =0\ ,\
 \frac{\partial \psi}{\partial x_3}
  = \frac{\partial \chi}{\partial x_2}
  = \frac{\partial \chi}{\partial x_3}=0\ ,\
 \frac{\partial\psi}{\partial x_2}=-\eta_1\ ,
$$
from which
$$
\eta_1 = a x_2 +b\ ,\
\eta_2 = \frac{a e^{-2x_1}}{2(1-n^2)} -\frac12 a x_2^2 -b x_2+c\ ,\
\eta_3 = \frac{-an e^{-x_1}}{1-n^2} +d\ ,
$$
with    $a,b,c,d$ arbitrary constants.
The group of motions of the space
(49) is therefore the $G_4$ generated by
the 4 infinitesimal transformations:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2}\ ,\
X_2 f = \frac{\partial f}{\partial x_3}\ ,\
X_3 f = \frac{\partial f}{\partial x_1}
          - x_2 \frac{\partial f}{\partial x_2}\ ,\nonumber\\
&&
X_4 f =  x_2 \frac{\partial f}{\partial x_1}
          +\frac12 \left( \frac{e^{-2x_1}}{1-n^2} -x_2^2\right)
                 \frac{\partial f}{\partial x_2}
          -\frac{n e^{-x_1}}{1-n^2}\frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}
with the composition:
\begin{eqnarray}
&&
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = - X_1 f\ ,\
[X_1,X_4]f  = X_3 f\ ,
\nonumber\\
&&
[X_2,X_3]f  = 0\ ,\
[X_2,X_4]f  = 0\ ,\
[X_3,X_4]f  = - X_4 f \ .\nonumber
\end{eqnarray}

The relation
$$
X_4 f =
    \frac12 \left( \frac{e^{-2x_1}}{1-n^2} +x_2^2\right) X_1 f
          -\frac{n e^{-x_1}}{1-n^2} X_2 f
  + x_2 X_3 f
$$
shows that the group is systatic
and that the systatic varieties are the geodesics
($x_3$). These geodesics do not admit orthogonal
surfaces except in the case $n=0$ already considered
in \S11.
We observe that the derived group is
here the group
$G_3\equiv(X_1 f,X_3 f,X_4 f) $, which is simply transitive
and belongs to type VIII.
In the space (49) we therefore also have an
example of spaces corresponding to this type.
To this purpose and for a better comparison with
the results that we will establish in \S28, we note the following
transformation of the line element (49).
Set:
$$
x_1=y_1\ ,\
x_2=e^{-y_1} (y_2-n y_3)\ ,\
x_3=y_3
$$
and one will obtain
$$
ds^2=[1+(y_2-ny_3)^2]\, dy_1^2 +dy_2^2 + (1-n^2) \, dy_3^2
        - 2(y_2-ny_3) \, dy_1 dy_2 \ .
\eqno(49*)$$

\typeout{**** Begin section 18}

\section{%18.
Similarities of the groups of motions of two spaces of the type (49).}

The line element (49) of the space of the previous section contains a constant
$n$ and we propose to demonstrate that this constant is truly essential, namely
that to two distinct values of $n $ ($0<n<1$) correspond two spaces which are
neither
isometric nor similar.\footnote{%(21)
See the preface.}

Assuming therefore a second line element of the form
$$
  ds^2=dy_1{}^2 + e^{2 y_1} \, dy_2{}^2
            + 2m e^{y_1} \, dy_2 dy_3 + dy_3{}^2\ ,
\eqno(55)
$$
where $m\neq n$, we must prove that it cannot be
transformed into the line element (49) nor into
one which differs from it by a constant factor.
In our investigation we will make use of the well known
criteria for the transformability of two differential
quadratic forms established by Christoffel and
Lipshitz, but most of all we utilize here the
circumstance that the two forms to be compared
admit two respective 4-parameter groups $G_4$,
$\Gamma_4$ of transformations into themselves, making
available for us the general theorems of Lie. Therefore we
make the following observation that we will equally apply
to the analogous research of the following
sections.  The supposed equations of transformation
$$
  x_1= \varphi_1(y_1,y_2,y_3)\ ,\
  x_2= \varphi_2(y_1,y_2,y_3)\ ,\
  x_3= \varphi_3(y_1,y_2,y_3)
$$
must obviously transform the group of motions $G_4$
of the one space into the $\Gamma_4$ of the other.  First it
is necessary to see if the two groups $G_4$, $\Gamma_4$ are
similar.
When this {\it necessary\/} condition is satisfied
the assumed transformability of the two
line elements still does not follow from it, but there will
remain only to see if the equation found by
transforming $G_4$ into $\Gamma_4$ can be specialized
so that it also puts the two spaces into the
relation of similarity.

To see if the two groups $G_4$, $\Gamma_4$ are similar, according to the
general criteria of Lie\footnote{%22
S. Lie-F. Engel, Vol.~I, p.~327.}
we must
first of all get the groups into an
isomorphic correspondence\footnote{In Italian: ``isomorfismo oleodrico"
[Translator].}
in the most general way.
Therefore with (\S17)
\begin{eqnarray}
&&
Y_1 f = \frac{\partial f}{\partial y_2}\ ,\
Y_2 f = \frac{\partial f}{\partial y_3}\ ,\
Y_3 f = \frac{\partial f}{\partial y_1}
          - y_2 \frac{\partial f}{\partial y_2}\ ,\nonumber\\
&&
Y_4 f =  y_2 \frac{\partial f}{\partial y_1}
          +\frac12 \left( \frac{e^{-2y_1}}{1-m^2} -y_2^2\right)
                 \frac{\partial f}{\partial y_2}
          -\frac{m e^{-y_1}}{1-m^2}\frac{\partial f}{\partial y_3}
\ ,\nonumber
\end{eqnarray}
as the generating transformations of $\Gamma_4$,  with the same composition as
the generators $X_1 f, X_2 f, X_3 f, X_4 f$ of $G_4$, it will be useful to
choose in $G_4$ (in the most general way) four other generators $\bar X_1 f,
\bar X_2 f, \bar X_3 f, \bar X_4 f$ so that they still have the same
composition, namely one has:
\begin{eqnarray}
&&
[\bar X_1,\bar X_2]f  = 0 \ ,\
[\bar X_1,\bar X_3]f  = -   \bar X_1 f\ ,\
[\bar X_1,\bar X_4]f  = \bar X_3 f\ ,
\nonumber\\
&&
[\bar X_2,\bar X_3]f  = 0\ ,\
[\bar X_2,\bar X_4]f  = 0\ ,\
[\bar X_3,\bar X_4]f  = - \bar X_4 f \ .\nonumber
\end{eqnarray}

If one observes first that the derived group of $G_4$ coincides either with
$(\bar X_1 f, \bar X_3 f,$\typeout{bad in-line math eqn linebreak} $\bar X_4 f)
$ or with $(X_1 f, X_3 f, X_4 f) $, it follows from this that $ X_1 f, X_3 f,
X_4 f$ must be composed of only $\bar X_1 f, \bar X_3 f,\bar X_4 f $. Moreover
since $\bar X_2 f $, like $X_2 f$, is the only transformation in $G_4$ which
commutes with every other in the group, $X_2 f$ must differ from $\bar X_2 f$
by a constant factor $\lambda$; we have therefore
\begin{eqnarray}
&&
X_1 f
= \alpha_1 \bar X_1 f
+ \alpha_2 \bar X_3 f
+ \alpha_3 \bar X_4 f\ ,\nonumber\\
&&
X_3 f
= \beta_1 \bar X_1 f
+ \beta_2 \bar X_3 f
+ \beta_3 \bar X_4 f\ ,\nonumber\\
&&
X_4 f
= \gamma_1 \bar X_1 f
+ \gamma_2 \bar X_3 f
+ \gamma_3 \bar X_4 f\ ,\nonumber\\
&&
X_2 f
= \lambda \bar X_2 f\ ,\seteqno{56}
\end{eqnarray}
with $\alpha,\beta,\gamma,\lambda $   being constants. The
composition relations translate  into the following equations
for $\alpha,\beta,\gamma$;
the $\alpha,\gamma$ must be constrained by the relations
$$
\alpha_2^2+2\alpha_1\alpha_3 =0\ ,\
\gamma_2^2+2\gamma_1\gamma_3 =0\ ,\
\alpha_1\gamma_3 +\alpha_3\gamma_1 +\alpha_2\gamma_2 =1
\eqno(57)$$
and the $\beta $
must be expressed in terms of these by the formulas
$$
\beta_1 = \alpha_2 \gamma_1 - \alpha_1 \gamma_2\ ,\
\beta_2 = \alpha_1 \gamma_3 - \alpha_3 \gamma_1\ ,\
\beta_3 = \alpha_3 \gamma_2 - \alpha_2 \gamma_3\ .
\eqno(58)$$

In order to check what follows, it is worth noting
that the following relations are a consequence of the ones above
\begin{eqnarray}
&&
\alpha_3\beta_1 +\alpha_1\beta_3 +\alpha_2\beta_2 =0\ ,\
\gamma_3\beta_1 +\gamma_1\beta_3 +\gamma_2\beta_2 =0\ ,\nonumber\\
&&
\beta_2^2 + 2\beta_1\beta_3=1\ .
\seteqno{59}
\end{eqnarray}

As a consequence the determinant
$\left|\begin{array}{ccc}
        \alpha_1 & \alpha_2 & \alpha_3 \\
        \beta_1 & \beta_2 & \beta_3 \\
        \gamma_1 & \gamma_2 & \gamma_3 \end{array}\right|
$
is equal to $+1$, and solving for the $\bar X f $ one has\footnote{The second
and third equation in the original paper had $\bar X_2 f$ and $\bar X_3 f$ on
the l.h.s., respectively; correction based on the {\it Opere} [Editor].}
\begin{eqnarray}
&&
\bar X_1 f
= \gamma_3 X_1 f
+ \beta_3  X_3 f
+ \alpha_3 X_4 f\ ,\nonumber\\
&&
\bar X_3 f
= \gamma_2 X_1 f
+ \beta_2  X_3 f
+ \alpha_2 X_4 f\ ,\nonumber\\
&&
\bar X_4 f
= \gamma_1 X_1 f
+ \beta_1  X_3 f
+ \alpha_1 X_4 f\ ,\
\nonumber
\end{eqnarray}
by which all the minors of second order of this
determinant are equal to one element, for example,
$\alpha_3=\alpha_2\beta_3-\alpha_3\beta_2 $, etc.

So having put the two groups
$G_4\equiv  (\bar X_1 f, \bar X_2 f, \bar X_3 f, \bar X_4 f)$,
$\Gamma_4\equiv (Y_1 f,$ $Y_2 f,$ $Y_3 f,$ $Y_4 f) $
into an isomorphic correspondence, it is
necessary to identify the relationship among
$Y_1 f, Y_2 f, Y_3 f, Y_4 f $,
namely:
$$
Y_4 f = \frac12 \left( \frac{e^{-2y_1}}{1-m^2} +y_2^2\right) Y_1 f
          -\frac{m e^{-y_1}}{1-m^2} Y_2 f
          +y_2 Y_3 f
\eqno(60)$$
with the one which correspondingly relates $\bar X_4 f $
to $\bar X_1 f, \bar X_2 f, \bar X_3 f $.
Now substituting into the relation
$$
X_4 f =
    \frac12 \left( \frac{e^{-2x_1}}{1-n^2} +x_2^2\right) X_1 f
          -\frac{n e^{-x_1}}{1-n^2} X_2 f
  + x_2 X_3 f
$$ the values (56), we find:\footnote{The last term on the r.h.s.\ was preceded
by a minus sign in the original, now corrected to a positive sign after the
{\it Opere\/} [Editor].}
\begin{eqnarray}
&&
\left\{
   \frac{\alpha_3}{2} \left( \frac{e^{-2x_1}}{1-n^2} +x_2^2\right)
   +\beta_3 x_2 -\gamma_3\right\} \bar X_4 f \nonumber\\
&&
   = \left\{ \gamma_1-\beta_1 x_2
         -\frac{\alpha_1}{2}\left( \frac{e^{-2x_1}}{1-n^2} +x_2^2\right)\right
             \} \bar X_1 f \seteqno{60*}\\
&&\qquad
+\left\{\gamma_2-\beta_2 x_2-\frac{\alpha_2}{2}
       \left( \frac{e^{-2x_1}}{1-n^2} +x_2^2\right) \right\} \bar X_3 f
          +\frac{n \lambda e^{-x_1}}{1-n^2} \bar X_2 f\ .
\nonumber
\end{eqnarray}

We introduce the abbreviations $$ \frac{e^{-x_1}}{\sqrt{1-n^2}}=\xi\ ,\
\frac{e^{-y_1}}{\sqrt{1-m^2}}=\eta\ ,\ \lambda \frac{n \sqrt{1-m^2}}{m
\sqrt{1-n^2}} = \mu \eqno(61)$$ and identifying the coefficients of (60), (60*)
we find the three equations\footnote{The r.h.s.\ of (62b) lacked the minus sign
in the original, now corrected after the {\it Opere\/} [Editor].}
\begin{eqnarray}
&&
y_2 = \frac{\gamma_2-\beta_2 x_2-\alpha_2(\xi^2+x_2^2)/2}
       {\alpha_3 (\xi^2+ x_2^2)/2 + \beta_3 x_2 -\gamma_3}\ ,
 \seteqno{62a}\\
&&
\eta = \frac{-\mu\xi}{\alpha_3 (\xi^2+ x_2^2)/2
  + \beta_3 x_2 -\gamma_3 }\ ,
 \seteqno{62b}\\
&&
\frac12 (\eta^2+y_2^2)
     \left\{\alpha_3(\xi^2+x_2^2)/2+\beta_3 x_2-\gamma_3\right\}
\nonumber\\
&&\qquad
   + \frac{\alpha_1}{2}(\xi^2+x_2^2) +\beta_1 x_2 -\gamma_1=0\ .
\seteqno{62*}
\end{eqnarray}

If these three equations are compatible the two
groups are similar and equations (62) then give
in the corresponding equations of transformation
$y_1,y_2 $
expressed in terms of $x_1,x_2 $ (Lie, {\it ibid.\/}).
Now by substituting the values of
$y_2,\eta $
given by (62) into (62*) and completing the square,
we obtain
\begin{eqnarray}
&&\kern-0.8cm
\mu^2\xi^2 +\mu^2\left[\frac{\alpha_2}{2} \xi^2+\frac{\alpha_2}{2} x_2^2
  + \beta_2 x_2 -\gamma_2 \right]^2 \nonumber\\
&&\kern-0.8cm
  + 2\left(\frac{\alpha_3}{2} \xi^2+\frac{\alpha_3}{2} x_2^2
  + \beta_3 x_2 -\gamma_3 \right)
    \left(\frac{\alpha_1}{2} \xi^2+\frac{\alpha_1}{2} x_2^2
  + \beta_1 x_2 -\gamma_1 \right) =0 \nonumber
\end{eqnarray}
which must therefore be an identity in $\xi,x_2 $.
Taking into account the relations (57), (58), (59)
among the constants $\alpha,\beta,\gamma $
one immediately
finds it necessary and sufficient for this to be true that one have
$\mu^2=1 $, namely
$\lambda^2=\frac{m^2(1-n^2)}{n^2(1-m^2)} $.

One concludes from this that the two groups $G_4$, $\Gamma_4 $ are indeed
similar and for the most general equations which transform the one group into
the other, one necessarily has\footnote{The original paper had $\alpha_2$
instead of $\alpha_3$ in (63a), which was a typo [Editor].}
\begin{eqnarray}
&&
y_2 = \frac{\gamma_2-\beta_2 x_2-\alpha_2(\xi^2+x_2^2)/2}
       {\alpha_3 (\xi^2+ x_2^2)/2 + \beta_3 x_2 -\gamma_3}\ ,
\seteqno{63a}\\
&&
\eta = \frac{\pm \xi}{\alpha_3 (\xi^2+ x_2^2)/2
  + \beta_3 x_2 -\gamma_3 }\ ,
\seteqno{63b}
\end{eqnarray}
from which, as one sees,
$y_1,y_2 $
are independent of $x_3$.

\typeout{**** Begin section 19}

\section{%19.
The constant $n$ is essential in\\
 $ds^2=dx_1^2 + e^{2x_1} \, dx_2^2 + 2n e^{x_1} \, dx_2 dx_3 + dx_3^2$.}

To demonstrate this claim we observe
finally that since the equations of transformation
must change $X_1 f,X_2 f,X_3 f, X_4 f$
respectively into
\begin{eqnarray}
&& \alpha_1 Y_1 f + \alpha_2 Y_3 f + \alpha_3 Y_4 f\ ,\
   \lambda Y_2 f\ ,\nonumber\\
&& \beta_1 Y_1 f  + \beta_2 Y_3 f + \beta_3 Y_4 f\ ,\
   \gamma_1 Y_1 f + \gamma_2 Y_3 f + \gamma_3 Y_4 f\ ,\nonumber
\end{eqnarray}
from these follow the values of all
the first partial derivatives
of the $y$ with respect to the $x$.\footnote{%(23)
In general $X_i f $ is changed into
$\partial f/\partial y_1 \, X_i(y_1)
+\partial f/\partial y_2 \, X_i(y_2)
+\partial f/\partial y_3 \, X_i(y_3)$,
hence the formulas indicated in the text.}
Of these equations it is enough for
us to write the following ones:
$$
 \frac{\partial y_1}{\partial x_2} = \alpha_2+\alpha_3 y_2\ ,
$$
$$
   \frac{\partial y_3}{\partial x_1}
     = - \frac{m e^{-y_1}}{1-m^2} (\alpha_3 x_2+\beta_3)\ ,\
   \frac{\partial y_3}{\partial x_2}
     = - \frac{m \alpha_3 e^{-y_1}}{1-m^2}\ ,\
   \frac{\partial y_3}{\partial x_3}
     = \lambda\ .
\eqno(64)$$

By substituting the expression (63) for $y_2$ into the value of
$\partial y_1/\partial x_2 $ one has
$$
\frac{\partial y_1}{\partial x_2}
 = \frac{\alpha_3 x_2+\beta_3}
       {\alpha_3 (\xi^2+ x_2^2)/2 + \beta_3 x_2 -\gamma_3}\ .
\eqno(65)$$

Given this,
from the assumed transformability
of the two line elements we will have:
\begin{eqnarray}
 & & dy_1^2+e^{2 y_1} \, dy_2^2 + 2me^{y_1} \, dy_2 dy_3+ dy_3^2 \nonumber\\
    &=&\lambda^2\{ dx_1^2+e^{2 x_1} \, dx_2^2
         + 2ne^{x_1} dx_2 dx_3+ dx_3^2 \}\ .
\seteqno{66}
\end{eqnarray}

We now apply the equations of Christoffel
$$
\frac{\partial^2 y_\nu}{\partial x_r \partial x_s}
  + \sum_{i,k} \christop{\nu}{ik}_y
      \frac{\partial y_i}{\partial x_r}  \frac{\partial y_k}{\partial x_s}
  = \sum_\mu \christop{\mu}{rs}_x \frac{\partial y_\nu}{\partial x_\mu}\ ,\
  (\nu,r,s=1,2,3)\ ,
$$
the index $x$ or $y$
attached to the Christoffel symbol indicating whether
it is constructed in terms of the form of the $x$
or that of the $y$.\footnote{%24
It is useful to note that the Christoffel symbols
of the second kind $\christop{l}{ik}$
are not changed in value by
multiplying the line element by a constant factor.}
Setting $\nu=2$, $r=2$, $s=3$   and replacing the Christoffel
symbols by their actual values one obtains
$ \frac{m^2\lambda}{1-m^2} \partial y_1/\partial x_2
=n e^{x_1} \partial y_3/\partial x_1$,
or using (64), (65)
$$
 \frac{m^2\lambda e^{-x_1}}{1-m^2}
 \frac{\alpha_3 x_2+\beta_3}
       {\alpha_3 (\xi^2+ x_2^2)/2 + \beta_3 x_2 -\gamma_3}
  + \frac{nm e^{-y_1}}{1-m^2} (\alpha_3 x_2+\beta_3) =0\ .
$$

Therefore, if  $\alpha_3=0$, $\beta_3=0 $ does not hold, one must have
$$
  m\lambda\sqrt{1-n^2} \xi + n\sqrt{1-m^2} \eta \,
  \{ \alpha_3 (\xi^2+ x_2^2)/2 + \beta_3 x_2 -\gamma_3 \} =0\ ,
$$
namely by (62b)
$$
  m\lambda\sqrt{1-n^2} - n\sqrt{1-m^2}\mu =0\ ,
$$
or equivalently
$$
 m^2(1-n^2) - n^2(1-m^2)=0\ ,
$$
or equivalently
$$
 m^2(1-n^2) + n^2(1-m^2)=0\ ,
$$
which is absurd since $n^2<1$, $m^2<1$.
Therefore we will have
$\alpha_3=\beta_3=0$ implying $\alpha_2=0$, $\beta_2^2=1$,
from which (63) tells  us that $y_2$ is only a function of $x_2$,
and $y_1$ differs from $x_1$ only by an additive constant.
After this (66) immediately gives $\lambda^2=1$
(comparing  the terms in  $d x_1^2$), namely
$$
 m^2(1-n^2)=n^2(1-m^2)\ ,
$$
and consequently $n^2=m^2$ as indeed we wished to show.\footnote{The signs
preceding the second terms on the l.h.s.\ in the previous two displayed
equations were plus signs in the original (propagating from sign errors noted
above), requiring a further short argument to obtain the desired result now
deleted after the {\it Opere\/}.
The deleted material after the second displayed equation was:

\hangindent 10pt
``which is absurd since $n^2<1$, $m^2<1$.
Therefore we will have
$\alpha_3=\beta_3=0$ implying $\alpha_2=0$, $\beta_2^2=1$,
from which (63) tells  us that $y_2$ is only a function of $x_2$,
and $y_1$ differs from $x_1$ only by an additive constant.
After this (66) immediately gives $\lambda^2=1$
(comparing  the terms in  $d x_1^2$), namely
$$
 m^2(1-n^2)=n^2(1-m^2)\ ,"
$$

\noindent
[Editor].}

\typeout{**** Begin section 20}

\section{%20.
The groups of type IV:\\
 $[X_1,X_2]f  = 0\ ,\ [X_1,X_3]f  = X_1 f,\ [X_2,X_3]f  = X_1 f +X_2 f$.}

To apply the equations of \S14 to the present composition we must set
$a=1$, $b=0$, $c=0$, $a'=1$, $b'=1$, $c'=0$.
{}From this it follows that $\xi_1 $ is constant, so we set $\xi_1=-2/h $
and the last three of (F) \S14 become
$$
 \alpha'=h\alpha\ ,\
 \beta'=h(\alpha+2\beta)/2\ ,\
 \gamma'=h(\beta+\gamma)\ ,
$$ from which by integrating and conveniently disposing of a factor independent
of $x_2,x_3$, we can assume\footnote{$hx_1$ in the original paper was corrected
to $hx_1/2$ after the {\it Opere} [Editor].} $$ \alpha=e^{hx_1}\ ,\
\beta=e^{hx_1}(h x_1/2 +l)\ ,\ \gamma=e^{hx_1}[(h x_1/2 +l)^2+m^2]\ , $$ with
$l,m $ constants.    Changing $x_1 $ into $x_1+constant$ and replacing the
space with a similar space, we have for the standard form of the present case
$$ ds^2 = dx_1{}^2 + e^{x_1}\{dx_2^2 + 2x_1 \, dx_2 dx_3 + (x_1^2+n^2) \,
dx_3^2 \}\ . \eqno(67) $$

Applying equations (E) \S14 to determine the most general infinitesimal motion
of the  space, we find the following equations:
\begin{eqnarray}
&&\kern-1.2cm
  \frac{\partial\eta_1}{\partial x_1}=0\ ,
\seteqno{68}\\
&&\kern-1.2cm
 e^{-x_1} \frac{\partial\eta_1}{\partial x_2}
      + \frac{\partial\eta_2}{\partial x_1}
      + x_1 \frac{\partial\eta_3}{\partial x_1}=0\ ,
\seteqno{69a}\\
&&\kern-1.2cm
 e^{-x_1} \frac{\partial\eta_1}{\partial x_3}
      + x_1 \frac{\partial\eta_2}{\partial x_1}
      + (x_1{}^2+n^2) \frac{\partial\eta_3}{\partial x_1}=0\ ,
\seteqno{69b}\\
&&\kern-1.2cm
  \frac12 \eta_1 + \frac{\partial\eta_2}{\partial x_2}
     + x_1 \frac{\partial\eta_3}{\partial x_2}=0\ ,
\seteqno{70}\\
&&\kern-1.2cm
  \frac12 (x_1^2+n^2+2x_1) \eta_1
   + x_1\frac{\partial\eta_2}{\partial x_3}
     + (x_1^2 + n^2) \frac{\partial\eta_3}{\partial x_3}=0\ ,
\seteqno{71}\\
&&\kern-1.2cm
 (x_1+1) \eta_1
   + \frac{\partial\eta_2}{\partial x_3}
     + x_1 \left(\frac{\partial\eta_2}{\partial x_2}
              +\frac{\partial\eta_3}{\partial x_3}\right)
     + (x_1^2 + n^2) \frac{\partial\eta_3}{\partial x_2}=0\ .
\seteqno{72}
\end{eqnarray}

Solving (69) we obtain
\begin{eqnarray}
&&
 \frac{\partial\eta_2}{\partial x_1}
  = \frac{e^{-x_1}}{n^2} \left[ x_1 \frac{\partial\eta_1}{\partial x_3}
       - (x_1^2+n^2) \frac{\partial\eta_1}{\partial x_2} \right]\ ,
\nonumber\\
&&
 \frac{\partial\eta_3}{\partial x_1}
  = \frac{e^{-x_1}}{n^2} \left[ x_1 \frac{\partial\eta_1}{\partial x_2}
       - \frac{\partial\eta_1}{\partial x_3} \right]\ ,
\nonumber
\end{eqnarray}
and integrating with respect to $x_1$, of  which $\eta_1$ is independent by
(68),
we have
\begin{eqnarray}
&&\kern-.7cm
\eta_2 = \frac{e^{-x_1}}{n^2}
       \left[(x_1^2+n^2+2x_1+2) \frac{\partial\eta_1}{\partial x_2}
               -(x_1+1) \frac{\partial\eta_1}{\partial x_3}
         \right]+ \psi(x_2,x_3)\ , \nonumber\\
&&\kern-.7cm
\eta_3 = \frac{e^{-x_1}}{n^2}
       \left[ \frac{\partial\eta_1}{\partial x_3}
            -(x_1+1) \frac{\partial\eta_1}{\partial x_2}
       \right]+ \chi(x_2,x_3)\ . \nonumber
\end{eqnarray}

Substituting into (70), we immediately deduce from this $$
 \frac{\partial^2\eta_1}{\partial x_2{}^2}
   = \frac{\partial^2\eta_1}{\partial x_2\partial x_3} =0\ ,\
 \frac{\partial \chi}{\partial x_2} = 0\ ,\
 \frac{\partial\psi}{\partial x_2}=-\frac12 \eta_1\ ,
$$
so that the substitution in (71) gives
$$
 \frac{\partial^2\eta_1}{\partial x_3{}^2} =0\ ,\
 \frac{\partial \chi}{\partial x_3} = -\frac12 \eta_1\ ,\
 \frac{\partial\psi}{\partial x_3}=-\eta_1\ ,
$$
hence
$$
 \frac{\partial\eta_1}{\partial x_2}
   = -2\frac{\partial^2\chi}{\partial x_2\partial x_3} =0\ ,\
 \frac{\partial\eta_1}{\partial x_3}
   = -2\frac{\partial^2\psi}{\partial x_2\partial x_3}
  = 2 \frac{\partial\eta_1}{\partial x_2} =0\ .
$$
In this way we find for the most general values of $\eta_1,\eta_2,\eta_3 $
the formulas
$$
\eta_1 = a\ ,\
\eta_2 = -\frac{a}{2}x_2 -a x_3+b\ ,\
\eta_3 = -\frac{a}{2}x_3 +c\ ,
$$
with $a,b,c$ arbitrary constants.  Therefore the complete group of motions
is only a $G_3$ here; it is generated by the 3 infinitesimal transformations
\begin{eqnarray}
&&
X_1 f= \frac{\partial f}{\partial x_2}\ ,\
X_2 f= \frac{\partial f}{\partial x_3}\ ,\nonumber\\
&&
X_3 f= -\frac{\partial f}{\partial x_1}
     + \left(\frac{x_2}{2}+x_3\right)\frac{\partial f}{\partial x_2}
     + \frac{x_3}{2} \frac{\partial f}{\partial x_3}
\seteqno{73}
\end{eqnarray}
and has the composition\footnote{%(25)
To have the canonical composition it would suffice to double $X_1 f, X_3 f$.}
$$
[X_1, X_2]f  = 0\ ,\
[X_1, X_3]f  = \frac12 X_1 f\ ,\
[X_2, X_3]f  = X_1 f + \frac12 X_2 f \ .
$$

\typeout{**** Begin section 21}

\section{%21.
The constant $n$ is essential in\\
 $ds^2=dx_1^2 + e^{x_1} [dx_2^2 + 2x_1 \, dx_2 dx_3 + (x_1^2+n^2) \, dx_3^2]$.}

Analogously to what we have done for the spaces of \S17, we also want to see
here if the constant $n$ of the present line element (67) is essential.  We
respond affirmatively to the question by showing that a second line
element\footnote{Equation number "(74)" is missing in the original, added by
the Editor.} $$ ds^2=dy_1^2 + e^{y_1} \, [dy_2^2 + 2y_1 \, dy_2 dy_3 +
(y_1^2+m^2) \, dy_3^2] \ , \eqno(74)$$ where  $m^2\neq n^2$, cannot be
identified with the original nor be proportional to it.  Proceeding exactly as
in \S18 we first compare the two respective groups of motions $G_3$,
$\Gamma_3$, the first generated by the transformations (73), the second instead
by\footnote{The second term on the r.h.s. of the 3rd equation had $\partial f/
\partial y_3$ in the original, which was incorrect [Editor].} $$ Y_1 f =
\frac{\partial f}{\partial y_2}\ ,\ Y_2 f= \frac{\partial f}{\partial y_3}\ ,\
Y_3 f= -\frac{\partial f}{\partial y_1}
     + \left(\frac{y_2}{2}+y_3\right)\frac{\partial f}{\partial y_2}
     + \frac{y_3}{2} \frac{\partial f}{\partial y_3}
$$
with the same composition
$$
[Y_1,Y_2]f  = 0\ ,\
[Y_1, Y_3]f  = \frac12 Y_1 f\ ,\
[Y_2, Y_3]f  = Y_1 f + \frac12 Y_2 f
\ .
$$

We must find the most general transformation which changes the one group into
the other and see if it can give rise to the hypothesized transformation of the two
line elements.  We therefore take three other transformation generators of
$\Gamma_3$, let them be $\bar Y_1 f,\bar Y_2 f,\bar Y_3 f $,
which have the same composition as above; we therefore have
\begin{eqnarray}
&&
\bar Y_1 f = \alpha Y_1 f + \beta Y_2 f\ ,
\bar Y_2 f = \gamma Y_1 f + \delta Y_2 f\ ,\nonumber\\
&&
\bar Y_3 f = a Y_1 f + b Y_2 f + c Y_3 f\ ,
\nonumber
\end{eqnarray}
and among the constants $\alpha,\beta,\gamma,\delta,c$ the relations
$$
\alpha c+2\beta c = \alpha\ ,\
\beta c =\beta\ ,\
\gamma c + 2\delta c = 2 \alpha +\gamma\ ,\
\delta c = 2\beta + \delta
\ ,
$$
from which it follows that $\beta=0 $, $c=1$, $\delta=\alpha$
and consequently:
\begin{eqnarray}
&&
\bar Y_1 f = \alpha Y_1 f \ ,
\bar Y_2 f = \gamma Y_1 f + \alpha Y_2 f\ ,\nonumber\\
&&
\bar Y_3 f = a Y_1 f + b Y_2 f +  Y_3 f\ .
\nonumber
\end{eqnarray}

There certainly exist transformations which change
$X_1 f$, $X_2 f$, $X_3 f $
respectively into $\bar Y_1 f,\bar Y_2 f,\bar Y_3 f $
because the two simply transitive groups
are isomorphic.\footnote{%26
S. Lie-F. Engel, Vol.~I, p.~340.}
For one such transformation the partial derivatives of the $y$
with respect to the $x$ must assume the following values:
%%
\begin{eqnarray}
&&\kern-.6cm
\frac{\partial y_1}{\partial x_1} =1\ ,\
\frac{\partial y_1}{\partial x_2} =0\ ,\
\frac{\partial y_1}{\partial x_3} =0\ ,
\nonumber\\
&&\kern-.6cm
\frac{\partial y_2}{\partial x_1} =
\alpha\left(\frac{x_2}{2} +x_3\right) +\gamma\frac{x_3}{2}
  -a -\frac{y_2}{2}-y_3
\ ,\
\frac{\partial y_2}{\partial x_2} =\alpha\ ,\
\frac{\partial y_2}{\partial x_3} =\gamma\ ,
\nonumber\\
&&\kern-.6cm
\frac{\partial y_3}{\partial x_1} =
    \alpha\frac{x_3}{2} -b -\frac{y_3}{2}
   \ ,\
\frac{\partial y_3}{\partial x_2} =0\ ,\
\frac{\partial y_3}{\partial x_3} =\alpha
\ .
\nonumber
\end{eqnarray}

Integrating we have the actual equations in finite terms
\begin{eqnarray}
&& y_1 = x_1 +h\ ,\nonumber\\ && y_2 = \alpha x_2 +\gamma x_3 +k e^{-x_1/2} -
cx_1e^{-x_1/2} -2a +4b\ ,\nonumber\\ && y_3 = \alpha x_3 + c e^{-x_1/2} -2b\
,\nonumber
\end{eqnarray}
with $h,c,k$ new constants.  The line element (74) therefore becomes:
%%
\begin{eqnarray}
&&
dx_1^2 + e^h \left\{ \left(\frac{cx_1}{2} -c -\frac{k}{2}\right) dx_1
  +\alpha e^{\frac{x_1}{2}} dx_2 + \gamma e^{\frac{x_1}{2}} dx_3
   \right\}^2
\nonumber\\
&&
  +2 e^h (x_1+h) \left\{\left(\frac{cx_1}{2} -c -\frac{k}{2}\right) dx_1
     + \alpha e^{\frac{x_1}{2}} dx_2 + \gamma e^{\frac{x_1}{2}} dx_3\right\}
\nonumber\\
&&\qquad
     \times \left\{-\frac{c}{2} dx_1 + \alpha e^{\frac{x_1}{2}} dx_3\right\}
\nonumber\\
&&
  + e^h\left\{(x_1+h)^2 +m^2\right\} \cdot
    \left\{-\frac{c}{2} dx_1 + \alpha e^{\frac{x_1}{2}} dx_3\right\}^2 \ .
\nonumber
\end{eqnarray}

Comparing with the line element (67) we must set the coefficients of $dx_1dx_2$
and $dx_1dx_3$ to zero; we immediately find $c = 0$, $k = 0$; and then
comparing the terms in $dx_2^2$, $dx_2dx_3$ and $dx_3^2$ one deduces
\footnote{This partial sentence is the translator's interpretation of Bianchi's
intended meaning. Bianchi's original phrase in which he meant to refer to
equation (67) ``Comparing this with the line element (64), we must set the
coefficients of $dx_1dx_2$ and $dx_1dx_3$ to zero, which leads to..." has a
proof correction at the end of his article (implemented by the Editors of {\it
Opere}) stating ``The penultimate line on p. 312 should read: equating the
coefficients of $dx_1dx_2$ and $dx_1dx_3$ we find immediately", but this omits
the necessary ``to zero" and removes any equation number [Editor].}
\begin{eqnarray}
&&
\alpha^2 e^h = 1\ ,\
\{\gamma\alpha +\alpha^2(x_1+h)\} e^h = x_1\ ,
\nonumber\\
&&
\{ \gamma^2 + 2\alpha\gamma(x_1+h) + \alpha^2(x_1+h)^2
    + \alpha^2 m^2 \} e^h
  = x_1^2+n^2
\ .
\nonumber
\end{eqnarray}

{}From this it follows that
$\alpha^2 e^h=1$, $\gamma + \alpha h = 0$,
so that $ n^2=m^2$, Q.E.D.

\typeout{**** Begin section 22}

\section{%22.
The groups of type V:\\
$[X_1,X_2]f = 0\ ,\ [X_1,X_3]f =X_1 f,\ [X_2,X_3]f = X_2 f$.}

The constants $a,b,c,a',b',c'$ of \S14 here take the values $a=1$, $b=0$,
$c=0$, $a'=0$, $b'=1$, $c'=0$, from which it follows that $\xi_1 $ is again
constant, so we set $\xi_1=-2/h$, and the last three of (F) \S14  give $$
\alpha'=h\alpha\ ,\ \gamma'=h\gamma\ ,\ \beta'=h\beta\ , $$ from which by
integrating we have $$ \alpha=le^{hx_1}\ ,\ \beta=me^{hx_1}\ ,\
\gamma=ne^{hx_1}\ , $$ with $l,m,n$ constants.  Changing (linearly) the
parameters $ x_2,x_3$ we obtain $$
  ds^2= dx_1^2 + e^{2hx_1} \, (dx_2^2+dx_3^2)
\ ,
$$
the line element which belongs to the space of constant negative curvature.

In this case the existence of the transitive group $G_3$ of motions of the
designated type implies a complete group of motions (non-Euclidean) of 6
parameters.

\typeout{**** Begin section 23}

\section{%23.
The groups of type VI:\\
$[X_1,X_2]f = 0\ ,\ [X_1,X_3]f =X_1 f,\ [X_2,X_3]f = h X_2 f$\ ,\
$h\neq0,1$.}

For the groups of this type we must set $a=1$, $b=c=0$, $a'=0$, $b'=h$, $c'=0$
in the equations. {}From this it follows that $\xi_1 $ is constant, so we set
$\xi_1=-2/k $ and the usual equations (F) \S14 give us $$ \alpha'=k\alpha\ ,\
2 \beta'=k(h+1)\beta\ ,\ \gamma'=hk\gamma\ , $$ from which by integrating and
absorbing two of the constants of integration into $x_2,x_3 $, we find $$
\alpha=e^{kx_1}\ ,\ \beta=n e^{k(h+1)x_1/2}\ ,\ \gamma=e^{hk x_1}\ , $$ where
$n $ is a constant which can clearly be assumed positive, so that (because $
\alpha \gamma-\beta^2>0$) we will have $0<n<1 $.

By passing to a similar space we can make  $k=2$,
so
$$
ds^2= dx_1^2 + e^{2x_1} \, dx_2^2 + 2n e^{(h+1)x_1} \, dx_2 dx_3
     + e^{2hx_1} \, dx_3^2
\ .
\eqno(76)
$$

One will observe that for $h=0$  this reduces to type III and the line element
(76) is then changed into the one (49) of the spaces of
\S17. \footnote{%27
We have not been able to treat the particular case $h=0$ together with the
general case, because only for $h=0$ does one have a 4-parameter group
of motions.}

The most general infinitesimal motion of this space is determined, according to
(E) \S14, by the following equations:\footnote{The factor $n$ in the last term
of (79) is absent in the original, correction after the {\it Opere} [Editor].}
\begin{eqnarray}
&&
  \frac{\partial\eta_1}{\partial x_1}=0\ ,
\seteqno{77}\\
&&
 \frac{\partial\eta_1}{\partial x_2}
      + e^{2x_1}\frac{\partial\eta_2}{\partial x_1}
      + n e^{(h+1)x_1} \frac{\partial\eta_3}{\partial x_1}=0\ ,
\seteqno{78a}\\
&&
 \frac{\partial\eta_1}{\partial x_3}
     + n e^{(h+1)x_1} \frac{\partial\eta_2}{\partial x_1}
      + e^{2hx_1}\frac{\partial\eta_3}{\partial x_1} =0\ ,
\seteqno{78b}\\
&&
  e^{x_1} \left(\eta_1 + \frac{\partial\eta_2}{\partial x_2}\right)
     + ne^{hx_1}  \frac{\partial\eta_3}{\partial x_2}=0\ ,
\seteqno{79}\\
&&
h \eta_1 + n e^{(1-h)x_1} \frac{\partial\eta_2}{\partial x_3}
  + \frac{\partial\eta_3}{\partial x_3}=0\ ,
\seteqno{80}\\
&&
n(h+1)e^{(h+1)x_1} \eta_1
  + e^{2x_1} \frac{\partial\eta_2}{\partial x_3}
  + n e^{(h+1)x_1} \left( \frac{\partial\eta_2}{\partial x_2}
    + \frac{\partial\eta_3}{\partial x_3} \right)
\nonumber\\ && \qquad
   + e^{2hx_1} \frac{\partial\eta_3}{\partial x_2}=0\ .
\seteqno{81}
\end{eqnarray}

Solving (78) for
$\partial\eta_2/\partial x_1 $, $\partial\eta_3/\partial x_1  $, we obtain
\begin{eqnarray}
&&
 \frac{\partial\eta_2}{\partial x_1}
  = \frac{ne^{-(h+1)x_1}}{1-n^2} \frac{\partial\eta_1}{\partial x_3}
       -   \frac{e^{-2x_1}}{1-n^2}\frac{\partial\eta_1}{\partial x_2}
\ ,\nonumber\\
&&
 \frac{\partial\eta_3}{\partial x_1}
  = \frac{ne^{-(h+1)x_1}}{1-n^2} \frac{\partial\eta_1}{\partial x_2}
       -   \frac{e^{-2hx_1}}{1-n^2}\frac{\partial\eta_1}{\partial x_3}
\ .\seteqno{82}
\end{eqnarray}

The integration of these last two with respect to $x_1 $ leads us to separate the
two cases
$$
a)\ h=-1\ ,\qquad
b)\ h\neq-1 \ .
$$
In case $a)$ by integrating we obtain
\begin{eqnarray}
&&
\eta_2
= \frac{n x_1}{1-n^2}  \frac{\partial\eta_1}{\partial x_3}
  +\frac{e^{-2x_1}}{2(1-n^2)}\frac{\partial\eta_1}{\partial x_2}
  + \psi(x_2,x_3)\ , \nonumber\\
&& \eta_3 = \frac{n x_1}{1-n^2}  \frac{\partial\eta_1}{\partial x_2}
  -\frac{e^{2x_1}}{2(1-n^2)} \frac{\partial\eta_1}{\partial x_3}
  + \chi(x_2,x_3)\ , \nonumber
\end{eqnarray}
and substituting into the successive equations (79), (80), (81),
we find
\begin{eqnarray}
&&
 \frac{\partial^2\eta_1}{\partial x_2{}^2}
  = \frac{\partial^2\eta_1}{\partial x_2\partial x_3}
  =  \frac{\partial^2\eta_1}{\partial x_3{}^2} =0\ ,\nonumber\\
&&
 \frac{\partial\psi}{\partial x_3}=0\ ,\
 \frac{\partial\psi}{\partial x_2}=-\eta_1\ ,\
 \frac{\partial \chi}{\partial x_2} = 0\ ,\
 \frac{\partial\chi}{\partial x_3}=\eta_1\ .
\nonumber
\end{eqnarray}
from which it follows that
$$
\eta_1 = a\ ,\
\eta_2 = -a x_2 + b\ ,\
\eta_3 = a x_3 +c\ ,
$$
with $a,b,c$ constants.
In this case therefore the group of motions is only a $G_3$ with the
infinitesimal transformation generators:
$$
X_1 f = \frac{\partial f}{\partial x_2}\ ,\
X_2 f = \frac{\partial f}{\partial x_3}\ ,\
X_3 f = -\frac{\partial f}{\partial x_1}
        + x_2 \frac{\partial f}{\partial x_2}
        - x_3 \frac{\partial f}{\partial x_3}\ .
$$

    Case $b)$.
Now let $h\neq-1 $.
Equations (82) integrated give
%%
\begin{eqnarray}
&&
\eta_2
= -\frac{n e^{(h+1)x_1}}{(h+1)(1-n^2)} \frac{\partial\eta_1}{\partial x_3}
   +\frac{e^{-2x_1}}{2(1-n^2)}\frac{\partial\eta_1}{\partial x_2}
    + \psi(x_2,x_3)\ , \nonumber\\
&& \eta_3 = - \frac{n e^{(h+1)x_1}}{(h+1)(1-n^2)}
\frac{\partial\eta_1}{\partial x_2}
   +\frac{e^{-2hx_1}}{2h(1-n^2)} \frac{\partial\eta_1}{\partial x_3}
    + \chi(x_2,x_3)\ , \nonumber
\end{eqnarray}
and substituting into (79) remembering that $h $   is different from
$0,1,-1 $,
we find
$$
 \frac{\partial^2\eta_1}{\partial x_2{}^2}=0\ ,\
 n\,\frac{\partial^2\eta_1}{\partial x_2\partial x_3} =0\ ,\
 \frac{\partial \chi}{\partial x_2} = 0\ ,\
 \frac{\partial\psi}{\partial x_2}=-\eta_1\ .
$$
Substituting into (80) therefore gives
$$
 \frac{\partial^2\eta_1}{\partial x_3{}^2} =0\ ,\
 \frac{\partial \psi}{\partial x_3} = 0\ ,\
 \frac{\partial\chi}{\partial x_3}=-h\eta_1\ ,
$$
and from this follow for $\eta_1,\eta_2,\eta_3 $ the values
$$
\eta_1 = a\ ,\
\eta_2 = -a x_2+b\ ,\
\eta_3 = -ha x_3 +c\ ,
$$
with $a,b,c$ arbitrary constants.  We therefore have as the complete group of
motions the $G_3$ generated by the three infinitesimal transformations
$$
X_1 f = \frac{\partial f}{\partial x_2}\ ,\
X_2 f = \frac{\partial f}{\partial x_3}\ ,\
X_3 f = -\frac{\partial f}{\partial x_1}
        + x_2 \frac{\partial f}{\partial x_2}
        + h x_3 \frac{\partial f}{\partial x_3}\ .
$$
which indeed has the composition
$$
[X_1,X_2]f =0\ ,
[X_1,X_3]f =X_1 f\ ,
[X_2,X_3]f =h X_2 f\ .
$$
We now see that the result obtained above for $h=-1$ is included in the
general case.

\typeout{**** Begin section 24}

\section{%24.
The constant $n$ is essential in\protect\footnotemark \\ $ds^2 = dx_1^2 +
e^{2x_1} dx_2^2 + 2n e^{(h+1)x_1} dx_2 dx_3 + e^{2hx_1} dx_3^2$. }
\footnotetext{ Bianchi's obvious typo was corrected here, there was a "+"
between $e^{2x_1}$ and $dx_2^2$ [Editor].}

That the constant $h$ is essential in this line element is clear since it is
already essential in the composition of its group of motions; but now we wish to
show that the constant $n$ (apart from sign) is also essential.  Therefore let
there be the two spaces
%%
\begin{eqnarray}
&&\kern-.6cm ds^2 = dx_1^2 + e^{2x_1} dx_2^2 + 2n e^{(h+1)x_1} dx_2 dx_3 +
e^{2hx_1} dx_3^2 \seteqno{$\alpha$}\\ &&\kern-.6cm ds^2 = dy_1^2 + e^{2y_1}
dy_2^2 + 2m e^{(h+1)y_1} dy_2 dy_3 + e^{2hy_1} dy_3^2 \ ; \seteqno{$\beta$}
\end{eqnarray}
we wish to prove that assuming the two spaces are similar implies $n^2=m^2$.

The group $\Gamma_3$ of motions of the second space is generated by the three
infinitesimal transformations
$$
Y_1 f = \frac{\partial f}{\partial y_2}\ ,\
Y_2 f = \frac{\partial f}{\partial y_3}\ ,\
Y_3 f = -\frac{\partial f}{\partial y_1}
        + y_2 \frac{\partial f}{\partial y_2}
        + h y_3 \frac{\partial f}{\partial y_3}
$$
and if, along with the hypothesized equations of correspondence between the two
spaces, we assume that\footnote{The original paper has $Y_1f$, $Y_2f$ and
$Y_3f$ here; correction based on the {\it Opere} [Editor].}
$X_1 f,X_2 f,X_3 f $ are changed respectively into
$\bar Y_1 f,\bar Y_2 f,\bar Y_3 f $,
then these latter ones must be combinations of
$Y_1 f,Y_2 f,Y_3 f $
and have the same composition
$$
[\bar Y_1, \bar Y_2]f  = 0\ ,\
[\bar Y_1, \bar Y_3]f  = \bar Y_1 f\ ,\
[\bar Y_2, \bar Y_3]f  = h \bar Y_2 f\ .
$$

It is clear in the first place that $\bar Y_1 f,\bar Y_2 f $
must be combinations of
$Y_1 f,Y_2 f$ only, so that we will have
\begin{eqnarray}
&&
\bar Y_1 f = \alpha Y_1 f + \beta Y_2 f\ ,\
\bar Y_2 f = \gamma Y_1 f + \delta Y_2 f\ ,\nonumber\\
&&
\bar Y_3 f = a Y_1 f + b Y_2 f + c Y_3 f\ .
\nonumber
\end{eqnarray}

Taking into account the composition equations we see that for
$h \neq -1$ one necessarily has $\beta=\gamma=0 $, $c=1$,
so that it follows that
$$
\bar Y_1 f = \alpha Y_1 f \ ,\
\bar Y_2 f = \delta Y_2 f\ ,\
\bar Y_3 f = a Y_1 f + b Y_2 f +  Y_3 f\ .
$$
while for $h=-1$ there is also possible another case
$$
\bar Y_1 f = \beta Y_2 f\ ,\
\bar Y_2 f = \gamma Y_1 f \ ,\
\bar Y_3 f = a Y_1 f + b Y_2 f - Y_3 f\ ,
$$
which does not differ from the previous one, however, apart from the exchange of
$y_2 $ and $y_3 $ and the change of $y_1 $ into $-y_1 $
(this clearly does not change the
line element).  We can therefore limit ourselves to the first
case, in which by integrating the equations of the transformations we find
$$
y_1 = x_1+k\ ,\
y_2 = \alpha x_2 + l e^{-x_1} - a\ ,\
y_3 = \delta x_3 + p e^{-hx_1} -b/h\ ,\
$$
where $k,l,p $ indicate new constants.  Substituting into the
line element ($\beta$) we obtain
\begin{eqnarray}
&&
dx_1^2 + e^{2x_1+2k} (\alpha \, dx_2- l e^{-x_1} \, dx_1)^2
\nonumber\\
&&
  + 2m e^{(h+1)(x_1+k)} (\alpha \, dx_2- l e^{-x_1} \, dx_1)
      (\delta \, dx_3 -hp e^{-hx_1} \, dx_1)
\nonumber\\
&&
  + 2e^{2h(x_1+k)} (\delta \, dx_3-hp e^{-hx_1} \, dx_1)^2
\ .
\nonumber
\end{eqnarray}
Expressing the fact that this differential form differs from
$(\alpha)$ only by a constant factor,
it suffices to compare the coefficients of $dx_2^2 $, $dx_2 dx_3 $, $dx_3^2 $
to find
$\alpha^2 e^{2k}
= \delta^2 e^{2hk}
= (m/n) e^{(h+1)k} \alpha\delta
\ , $
from which it indeed follows that $n^2=m^2$, Q.E.D.

\typeout{**** Begin section 25}

\section{%25.
The groups of type VII$_1$:\\
$[X_1,X_2]f  =0,\ [X_1,X_3]f  =X_2 f,\ [X_2,X_3]f  = -X_1 f$.}

Treating in general the case of the groups of type VII of composition
$$
[X_1,X_2]f  =0,\
[X_1,X_3]f  =X_2 f,\
[X_2,X_3]f  = -X_1 f + h X_2 f\ ,
$$
we must give to the constants $a,b,c,a',b',c' $ of \S14 the values
$a=0$, $b=1$, $c=0$, $a'=-1$, $b'=h$, $c'=0$,
from which one has $\xi_1= constant$, so we set $\xi_1=1/k $ and
equations (F) ({\it ibid.\/})
give us
$$
\alpha' + 2k\beta=0\ ,\
\gamma' - 2k\beta +2hk\gamma=0\ ,\
\beta' - k\alpha + hk\beta +k\gamma  =0\ .
\eqno(83)
$$

For the integration it is convenient to separate the case $h=0$ from the
general case.  We assume $h=0$ in this section and integrating (83) we will
have
\begin{eqnarray}
&&
\alpha = c_1 \sin(2kx_1) + c_2 \cos(2kx_1) + c_3\ ,\nonumber\\
&&
\beta = -c_1 \cos(2kx_1) + c_2 \sin(2kx_1) \ ,\nonumber\\
&&
\gamma= -c_1 \sin(2kx_1) - c_2 \cos(2kx_1) + c_3\ ,\nonumber
\end{eqnarray}
where $c_1,c_2,c_3$ are three constants.
We exclude the case in which the first two are both zero
since then the space would be of zero curvature.
Changing $x_1$ into $x_1+constant$, we can make $c_1=0$
and varying the parameters $x_2,x_3$ proportionally
we can make $c_2=1$; finally by replacing the space with
a similar space, we will have the following standard form
for the line element:
$$
  ds^2 = dx_1^2 +(n+\cos x_1) \, dx_2^2 + 2\sin x_1 \, \, dx_2 dx_3
          + (n-\cos x_1) \, dx_3^2
\ ,
\eqno(84)
$$
where the constant $n$ will be positive and $ > 1$ since
$\alpha$, $\gamma$, $\alpha\gamma-\beta^2 $ must be positive.
The equations (E) \S14 to determine $\eta_1,\eta_2,\eta_3 $
become:
\begin{eqnarray}
&&
  \frac{\partial\eta_1}{\partial x_1}=0\ ,
\seteqno{85}\\
&&
 \frac{\partial\eta_1}{\partial x_2}
      + (n+ \cos x_1)\frac{\partial\eta_2}{\partial x_1}
      + \sin x_1 \frac{\partial\eta_3}{\partial x_1}=0\ ,
\seteqno{86a}\\
&&
 \frac{\partial\eta_1}{\partial x_3}
     +      \sin x_1  \frac{\partial\eta_2}{\partial x_1}
      + (n - \cos x_1)\frac{\partial\eta_3}{\partial x_1} =0\ ,
\seteqno{86b}\\
&&
  - \frac12 \sin x_1 \cdot \eta_1
   + (n+ \cos x_1)\frac{\partial\eta_2}{\partial x_2}
      + \sin x_1  \frac{\partial\eta_3}{\partial x_2}=0\ ,
\seteqno{87}\\
&&
   \frac12 \sin x_1 \cdot \eta_1
     + \sin x_1  \frac{\partial\eta_2}{\partial x_3}
%\nonumber\\ && \qquad
     +(n - \cos x_1) \frac{\partial\eta_3}{\partial x_3}=0\ ,
\seteqno{88}\\
&&
   \cos x_1 \eta_1 + (n+ \cos x_1)\frac{\partial\eta_2}{\partial x_3}
     +\sin x_1 \left( \frac{\partial\eta_2}{\partial x_2}
                     + \frac{\partial\eta_3}{\partial x_3} \right)
\nonumber\\ && \qquad
     +(n - \cos x_1) \frac{\partial\eta_3}{\partial x_2}=0\ .
\seteqno{89}
\end{eqnarray}

Solving (86) for $\partial \eta_2/\partial x_1 $, $\partial \eta_3/\partial
x_1$ and integrating we have
%%
\begin{eqnarray}
&&\kern-.6cm
\eta_2 = \frac{1}{n^2-1}  \left\{
          (\sin x_1 - nx_1)\frac{\partial\eta_1}{\partial x_2}
               -\cos x_1 \frac{\partial\eta_1}{\partial x_3} \right\}
    + \psi(x_2,x_3)\ , \nonumber\\
&&\kern-.6cm
\eta_3 = \frac{1}{n^2-1}  \left\{
        -(\sin x_1 + nx_1)\frac{\partial\eta_1}{\partial x_3}
            -\cos x_1 \frac{\partial\eta_1}{\partial x_2} \right\}
    + \chi(x_2,x_3)\ , \nonumber
\end{eqnarray}
and substituting these values into the successive equations
we see that one must have
\begin{eqnarray}
&&
 \frac{\partial^2\eta_1}{\partial x_2{}^2}
  = \frac{\partial^2\eta_1}{\partial x_2\partial x_3}
  =  \frac{\partial^2\eta_1}{\partial x_3{}^2} =0\ ,\nonumber\\
&&
 \frac{\partial\psi}{\partial x_2}=0\ ,\
 \frac{\partial\psi}{\partial x_3}=-\frac12 \eta_1\ ,\
 \frac{\partial \chi}{\partial x_2} = \frac12 \eta_1\ ,\
 \frac{\partial\chi}{\partial x_3}=0\ ,
\nonumber
\end{eqnarray}
so that we obtain: $$ \eta_1 = a\ ,\ \eta_2 = -\frac{a}{2} x_3 +b\ ,\ \eta_3 =
\frac{a}{2} x_2 +c\ , $$ with $a,b,c$ constants. Here too the group of motions
is of three parameters and its generating transformations are $$
 X_1 f = \frac{\partial f}{\partial x_2}\ ,\
 X_2 f = \frac{\partial f}{\partial x_3}\ ,\
 X_3 f = 2\frac{\partial f}{\partial x_1}
          - x_3 \frac{\partial f}{\partial x_2}
          + x_2 \frac{\partial f}{\partial x_3}\ ,
\eqno(90)
$$
with the composition
$$
[X_1,X_2]f  = 0\ ,\
[X_1,X_3]f  = X_2 f\ ,\
[X_2,X_3]f  = - X_1 f\ .
$$

\typeout{**** Begin section 26}

\section{%26.
The groups of type VII$_2$:\\
$[X_1,X_2]f =0\ ,\ [X_1,X_3]f =X_2 f\ ,\ [X_2,X_3]f = -X_1 f +h X_2 f
\ ,\ h\neq0\ (0<h<2)$.}

The equations (83)\footnote{In the original paper, ``equations (84)", which is
incorrect [Editor].} give us
$$ \beta=-\frac{\alpha'}{2k}\ ,\ \gamma=\alpha+
\frac{\alpha''}{2k^2} + \frac{h\alpha'}{2k}\ , $$
and hence to determine
$\alpha $, the linear and homogeneous constant coefficient equation $$
  \alpha''' + 3hk\alpha'' + 2k^2(h^2+2)\alpha' +4hk^3\alpha = 0
\ ,
$$
whose characteristic equation
$$
  \rho^3 + 3hk\rho^2 + 2k^2(h^2+2)\rho +4hk^3= 0
\ ,
$$
setting $\rho = kr$, becomes
$$
  r^3 + 3hr^2 + 2(h^2+2)r +4h= 0
\ .
$$.

One root of this equation is $r_1=-h $ and the other two
$r_2$, $r_3$,
since $h^2<4 $, are complex conjugates:
$$
 r_2 = -h + i \sqrt{4-h^2}\ ,\
r_3 = -h - i \sqrt{4-h^2}\ .
$$

If for brevity we set $v=\sqrt{4-h^2} $,
we have for $\alpha $ the expression:
$$
\alpha = c_1 e^{-hkx_1} + c_2 e^{-hkx_1} \cos (kvx_1)
                        + c_3 e^{-hkx_1} \sin (kvx_1)\ ,
$$
where $c_1,c_2,c_3 $ are three arbitrary constants.

We exclude the case in which one takes $c_2=c_3=0 $
because then the space would be of constant negative curvature.
By adding a constant to $x_1$ we can make
(if $c_2\neq 0$) $c_3=0$,
and passing to a similar space we will obtain
\begin{eqnarray}
&& \alpha=e^{-hx_1}(n + \cos vx_1) \ ,\nonumber\\ && \beta= \frac12
e^{-hx_1}(h\cos vx_1  + v\sin vx_1 +nh) \ ,\nonumber\\ && \gamma=
e^{-hkx_1}\left(\frac{2-v^2}{2} \cos vx_1
                   + \frac{hv}{2} \sin vx_1 + n\right)
\ .\nonumber
\end{eqnarray}
We note that from this follows the result
$$
  \alpha\gamma-\beta^2=v(n^2-1) e^{-2h x_1}/4\ ,
$$
from which  $|n|>1 $ so that $n>0$  since $\alpha > 0$.

The first equations  of (E) \S14, solved for
$\partial \eta_2/\partial x_1 $, $\partial \eta_3/\partial x_1 $,
give
\begin{eqnarray}
&&
\frac{\partial \eta_2}{\partial x_1}
   = \frac{4e^{2hx_1}}{(n^2-1)v^2}
         \left\{ \beta\frac{\partial \eta_1}{\partial x_3}
                - \gamma \frac{\partial \eta_1}{\partial x_2} \right\}
\ ,\nonumber\\
&&
\frac{\partial \eta_3}{\partial x_1}
   = \frac{4e^{2hx_1}}{(n^2-1)v^2}
         \left\{ \beta\frac{\partial \eta_1}{\partial x_2}
                - \alpha \frac{\partial \eta_1}{\partial x_3} \right\}\ .
\nonumber
\end{eqnarray}

Substituting the values of $\alpha,\beta,\gamma $
and integrating with respect to $x_1$ we obtain
\begin{eqnarray}
&&
\eta_2
= \left\{ \frac{h^2-v^2}{2v^2(n^2-1)} \cos vx_1
    + \frac{h}{(n^2-1)v} \sin vx_1 + \frac{2n}{v^2(n^2-1)} \right\}
e^{hx_1} \frac{\partial\eta_1}{\partial x_3} \nonumber\\ && -\left\{
\frac{h(1-v^2)}{v^2(n^2-1)} \cos vx_1
    + \frac{h^2-1}{(n^2-1)v} \sin vx_1 + \frac{4n}{v^2h(n^2-1)}
\right\} e^{hx_1} \frac{\partial\eta_1}{\partial x_2} \nonumber\\&&\qquad
  +\psi(x_2,x_3)\ ,
\nonumber\\
&&
\eta_3 =
\left\{ \frac{h^2-v^2}{2v^2(n^2-1)} \cos vx_1
    + \frac{h}{(n^2-1)v} \sin vx_1 + \frac{2n}{v^2(n^2-1)}
\right\} e^{hx_1} \frac{\partial\eta_1}{\partial x_2} \nonumber\\ && -\left\{
\frac{h}{v^2(n^2-1)} \cos vx_1
    + \frac{1}{(n^2-1)v} \sin vx_1 + \frac{4n}{v^2h(n^2-1)}
 \right\} e^{hx_1} \frac{\partial\eta_1}{\partial x_3}
\nonumber\\&&\qquad +\chi(x_2,x_3)\ . \nonumber
\end{eqnarray}

If we now take the other three equations (E) \S14:
%%
\begin{eqnarray}
&&\kern-1cm
\gamma' \eta_1 + 2\beta \frac{\partial\eta_2}{\partial x_3}
   + 2\gamma \frac{\partial\eta_3}{\partial x_3}= 0\ ,\
\alpha'\eta_1 + 2\alpha \frac{\partial\eta_2}{\partial x_2}
   + 2\beta \frac{\partial\eta_3}{\partial x_2}=0\ ,\
\nonumber\\
&&\kern-1cm
\beta'\eta_1 + 2\alpha \frac{\partial\eta_2}{\partial x_3}
   + \beta \left(\frac{\partial\eta_2}{\partial x_2}
            + \frac{\partial\eta_3}{\partial x_3}\right)
   + \gamma \frac{\partial\eta_3}{\partial x_2}= 0
\nonumber
\end{eqnarray}
and substitute the values of    $\alpha,\beta,\gamma,\eta_2,\eta_3$
into them,
it suffices to equate the coefficients of the terms in
$e^{-hx_1} $, $e^{-hx_1}\cos vx_1 $, $e^{-hx_1}\sin vx_1 $,
to find
$$
\frac{\partial \psi}{\partial x_2}=0\ ,\
\frac{\partial \psi}{\partial x_3}=-\eta_1\ ,\
\frac{\partial \chi}{\partial x_2}=\eta_1\ ,\
\frac{\partial \chi}{\partial x_3}= h\eta_1\ ,
$$
from which it follows that
$$
\eta_1 = a\ ,\
\eta_2 = -a x_3 +b\ ,\
\eta_3 = a x_2 + ah x_3 +c\ ,
$$
with $a, b, c$ arbitrary constants.
Therefore in the present case the space has as a group
of motions the $G_3$ generated by the three infinitesimal transformations:
$$
  X_1 f = \frac{\partial f}{\partial x_2} \ ,\
  X_2 f = \frac{\partial f}{\partial x_3}\ ,\
  X_3 f = \frac{\partial f}{\partial x_1}
          - x_3 \frac{\partial f}{\partial x_2}
          +(x_2+hx_3) \frac{\partial f}{\partial x_3}\ ,
$$ with the composition $$ [X_1,X_2]f  = 0\ ,\ [X_1,X_3]f  = X_2 f\ ,\
[X_2,X_3]f  = -X_1 f + h X_2 f\ . $$ We see that setting $h=0$ one returns to
the results of the previous section, changing the notation in a very simple
way.

\typeout{**** Begin section 27}

\section{%27.
The constant $n$ is essential in the line elements of the two previous
sections.}

In the line element of the spaces of the previous section appear the two
constants $h,n $, the first of which is essential, already being so by the
composition of the group (\S13).  We now show that the constant $n$ is
essential, and with this result the same thing will also be proved for the
spaces of \S25 which correspond to $h=0$.

We must show that two spaces of respective line elements:
\begin{eqnarray}
&&
ds^2 = dx_1^2 + e^{-hx_1} (n+\cos vx_1) \, dx_2^2
\nonumber\\
&&\quad
         + e^{-hx_1} (h\cos vx_1 +v \sin v x_1 +nh) \, dx_2 dx_3
\nonumber\\
&&\quad
      + e^{-hx_1} \left( \frac{2-v^2}{2} \cos vx_1
           + \frac{hv}{2} \sin v x_1 +n \right) \, dx_3^2
\ , \nonumber\\ && ds^2 = dy_1^2 + e^{-hy_1} (m+\cos vy_1) \, dy_2^2
\nonumber\\ &&\quad
         + e^{-hy_1} (h\cos vy_1 +v \sin v y_1 +mh) \, dy_2 dy_3
\nonumber\\
&&\quad
      + e^{-hy_1} \left( \frac{2-v^2}{2} \cos vy_1
           + \frac{hv}{2} \sin v y_1 +m \right) \, dy_3^2
\ ,
\nonumber
\end{eqnarray}
cannot be similar unless $n^2=m^2$.

The group $G_3$ of motions of the first space is generated by the infinitesimal
transformations (90) and the $\Gamma_3$ of the second by the three
$$
Y_1 f = \frac{\partial f}{\partial y_2}\ ,
Y_2 f = \frac{\partial f}{\partial y_3}\ ,
Y_3 f = \frac{\partial f}{\partial y_1}
        -y_3 \frac{\partial f}{\partial y_2}
        +(y_2 +hy_3) \frac{\partial f}{\partial y_3} \ ,
$$
with the same composition.  Suppose that in the
hypothesized transformation
$X_1 f,X_2 f,X_3 f $ are changed into
$\bar Y_1 f,\bar Y_2 f,\bar Y_3 f $; we will have:
\begin{eqnarray}
&&
\bar Y_1 f = \alpha Y_1 f + \beta Y_2 f \ ,
\bar Y_2 f = \gamma Y_1 f + \delta Y_2 f \ ,\nonumber\\
&&
\bar Y_3 f = a Y_1 f + b Y_2 f + c Y_3 f \ .
\nonumber
\end{eqnarray}

{}From the composition equations
$$
 [\bar Y_1,\bar Y_2]f  = 0\ ,\
 [\bar Y_1,\bar Y_3]f  = \bar Y_2 f\ ,\
 [\bar Y_2,\bar Y_3]f  = -\bar Y_1 f + h\bar Y_2 f\ ,
$$
it immediately follows that
$c=1 $, $\gamma=-\beta $, $\delta=\alpha+h\beta $,
so
\begin{eqnarray}
&&
\bar Y_1 f = \alpha Y_1 f + \beta Y_2 f \ ,\
\bar Y_2 f = -\beta Y_1 f + (\alpha + h\beta) Y_2 f \ ,\nonumber\\
&&
\bar Y_3 f = a Y_1 f + b Y_2 f + Y_3 f \ .
\nonumber
\end{eqnarray}

When the $y$ are expressed in terms of the $x$,
they must consequently satisfy the following equations:
\begin{eqnarray}
 \frac{\partial y_1}{\partial x_1}=1\ ,\
&&
\frac{\partial y_1}{\partial x_2}=0\ ,\
\frac{\partial y_1}{\partial x_3}=0\ ,\nonumber\\
 &&
\frac{\partial y_2}{\partial x_2}=\alpha\ ,\
\frac{\partial y_2}{\partial x_3}=-\beta\ ,\nonumber\\
 &&
\frac{\partial y_3}{\partial x_2}=\beta\ ,\
\frac{\partial y_3}{\partial x_3}=\alpha+ h\beta\ .\nonumber
\end{eqnarray}

It suffices to compare the terms in $dx_2^2 $
in the two line elements (91), (92)
to obtain the following equation, in which $\lambda $
denotes a constant factor:
\begin{eqnarray}
&&
\alpha^2 (\cos vy_1 +m) +\alpha\beta(h\cos vy_1 + v\sin vx_1 +hm)
\nonumber\\
&&\qquad
+\beta^2 \left( \frac{2-v^2}{2}\cos vy_1
            +\frac{hv}{2} \sin vy_1 +m\right)
\nonumber\\
&&
= \lambda (\cos vx_1 + n)
\ .
\nonumber
\end{eqnarray}

This must be converted into an identity in $x_1 $ by setting
$y_1=x_1+k $ ($k$ constant).
Setting $ vk=\sigma$ (constant), and comparing corresponding terms in the above
equations, we derive the three relations
$$
\alpha^2 +h\alpha\beta + \beta^2= \lambda n/m
\eqno(93)
$$
\begin{eqnarray}
&&
\alpha^2 \cos\sigma + \alpha\beta (h\cos\sigma + v\sin\sigma)
\nonumber\\
&&\qquad
       + \beta^2 \left( \frac{2-v^2}{2} \cos\sigma
                 +\frac{hv}{2} \sin\sigma \right)= \lambda
\ ,\nonumber\\
&&
-\alpha^2 \sin\sigma + \alpha\beta (-h\sin\sigma + v\cos\sigma)
\nonumber\\
&&\qquad
       + \beta^2 \left( -\frac{2-v^2}{2} \sin\sigma
                 +\frac{hv}{2} \cos\sigma \right)= 0
\ .\nonumber
\end{eqnarray}

Multiplying respectively the last two equations, first by $\cos\sigma $,
$-\sin\sigma $ then by $\sin\sigma $, $\cos\sigma $, and each time summing, we
obtain $$ \alpha^2 + h\alpha\beta + \frac{2-v^2}{2} \beta^2
  = \lambda \cos\sigma\ ,\
v\alpha\beta + \frac{hv}{2} \beta^2
  = \lambda \sin\sigma\ ,
$$
which squared and summed, remembering that $v^2+h^2=4 $ give
$(\alpha^2+h\alpha\beta+\beta^2)^2 =\lambda^2 $,
from which by (93) $n^2=m^2$, Q.E.D.

\typeout{**** Begin section 28}

\section{%28.
The groups of type VIII:\\
 $[X_1,X_2]f  =X_1 f,\ [X_1,X_3]f  = 2 X_2 f,\ [X_2,X_3]f  = X_3 f$.}

Having exhausted the research on spaces which admit an integrable
transitive $G_3$ of motions, we now turn to the case of a
simple transitive $G_3$, beginning with type VIII.

We consider in $G_3$ the $G_2$ generated by $X_2 f, X_3f$ and proceed as in \S4
by assuming the geodesically parallel surfaces invariant with respect to the
subgroup $G_2$ as the coordinate surfaces $x_1 = constant$, and we furthermore
give to $X_2 f,X_3 f$ the canonical form ({\it ibid.\/}) $$ X_2 f =\partial
f/\partial x_3\ ,\ X_3 f =e^{x_3}\partial f/\partial x_2\ . $$

For the line element of the space we therefore have
$$
ds^2= dx_1^2 + \alpha \, dx_2^2 +2(\beta-\alpha x_2) \, dx_2 dx_3
       + (\alpha x_2^2 -2\beta x_2 +\gamma) \, dx_3^2
\ ,
\eqno(94)
$$
with $\alpha,\beta,\gamma $ functions of $x_1$.

Now let
$ X_1 f
= \xi_1 \, \partial f/\partial x_1
+ \xi_2 \, \partial f/\partial x_2
+ \xi_3 \, \partial f/\partial x_3$
be the third generating transformation of $G_3$, in which,
the group being transitive, we will have $\xi_1\neq0 $.
Because the  composition equations
$[X_1,X_2]f =X_1 f$,
$[X_1,X_3]f =2X_2 f$
hold, the $\xi $ must satisfy the following equations:
\begin{eqnarray}
&& \frac{\partial \xi_1}{\partial x_3} = -\xi_1 \ ,\ \frac{\partial
\xi_2}{\partial x_3} = -\xi_2 \ ,\ \frac{\partial \xi_3}{\partial x_3} = -\xi_3
\ , \nonumber\\ && \frac{\partial \xi_1}{\partial x_2} = 0 \ ,\ \frac{\partial
\xi_2}{\partial x_2} = \xi_3 \ ,\ \frac{\partial \xi_3}{\partial x_2} = -2
e^{-x_3} \ , \nonumber
\end{eqnarray}
from which integrating leads to $$ \xi_1= A e^{-x_3} \ ,\ \xi_2= (Bx_2-x_2^2+C)
e^{-x_3} \ ,\ \xi_3= (B-2x_2) e^{-x_3} \ , \eqno(95) $$ with $A, B, C$
functions only of $x_1 $.

Expressing the fact that, with the values (95) of the $\xi $ and assuming
\begin{eqnarray}
&&
a_{11}=1\ ,\
a_{12}= a_{13}=0\ ,\nonumber\\
&&
a_{22}=\alpha\ ,a_{23}=\beta-\alpha x_2\ ,\
a_{33}=\alpha x_2^2-2\beta x_2+\gamma\ ,\
\nonumber
\end{eqnarray}
the fundamental equations (A) \S1 are satisfied, we find
among the unknown functions $\alpha,\beta,\gamma, A, B, C$
of $x_1 $ the 6 following equations:
\begin{eqnarray}
&&
A'=0\ ,\nonumber\\
&&
\alpha C'+\beta B'=0\ ,\ \beta C'+\gamma B'=A\ ,
\nonumber\\
&&
 \frac12 A \alpha' + \alpha B - 2\beta = 0\ ,\
 \frac12 A \beta' - \alpha C - \gamma = 0\ ,\
 \frac12 A \gamma' - 2\beta C - \gamma B = 0\ .
\nonumber
\end{eqnarray}
The first tells us that $A$ is a constant, different from zero
by hypothesis;
then multiplying the last
three respectively by $\gamma$, $-2\beta$, $\alpha $ and summing
leads to
$ A(\alpha'\gamma+\alpha \gamma'-2\beta\beta')=0$
so that $\alpha\gamma-\beta^2=constant$.

We therefore set
$$
 A= 2k\ ,\
 \alpha\gamma-\beta^2=n^2
\eqno(96)
$$
and it follows that $\alpha,\beta,\gamma$ are expressed in terms
of $B',C'$ by the formulas
$$
  \alpha= \frac{n^2}{2k} B'\ ,\
  \beta= -\frac{n^2}{2k} C'\ ,\
  \gamma= \frac{2k}{B'} + \frac{n^2}{2k} \frac{C'{}^2}{B'}\ ,
\eqno(97)
$$
while
$B,C$ must satisfy the simultaneous second order
differential equations:
$$
  k B'' + B B' + 2 C'=0\ ,\
  n^2 B' C'' + \frac{n^2}{k} B'{}^2 C
             + \frac{n^2}{k} C'{}^2 +4k =0\ .
$$
The first of these is
immediately integrable, and indicating by $2a $ the constant
of integration,
we find
$$
C=a-\frac{k}{2} B'- \frac14 B^2\ .
\eqno(98)
$$
Finally by substituting this into the last one we have,
to determine $B(x_1) $, the third order differential equation
$$
 -\frac{kn^2}{4} B'B''' + \frac{kn^2}{8} B''{}^2
   -\frac{n^2}{2} B'{}^3 +\frac{an^2}{2k} B'{}^2 +2k =0\ .
$$
Having integrated this, (98) gives us the value of $C$
and (97) those of $\alpha,\beta,\gamma$
in the line element (94) of the space.

We treat in this section the particular case in which $B' $ is constant, namely
$B''=0$, a case which returns us to the spaces already considered in \S17. We
will have $$
 B'=l\ ,\
 B=lx_1 +m\ ,\
 C'=-l(l x_1+m)/2\ ,
$$
with $l,m $ constants,\footnote{%(28)
One observes that the constant $l$ cannot be zero because
then we would have $B'=C'=0 $, and consequently $A=0$.}
so by (97):
$$
\alpha = \frac{n^2l}{2k}\ ,\
\beta = \frac{n^2l}{2k} \frac{l x_1 +m}{2}\ ,\
\gamma = \frac{2k}{l} + \frac{n^2l}{2k}
       \left( \frac{l x_1 +m}{2} \right)^2
\ .
$$

Setting $(l x_1+m)/2=y_1 $, the line element (94) becomes
\begin{eqnarray}
ds^2 &=& \frac{4}{l^2} \, dy_1^2 + \frac{n^2l}{2k}
        \left\{ dx_2^2 + 2(y_1-x_2) \, dx_2 dx_3
%\right. \nonumber\\&&\qquad
     + \left[
    (y_1-x_2)^2+ \frac{4k^2}{n^2l^2} \right] \, dx_3^2
%\left.\phantom{(}\kern-4pt
\right\}
\ ,
\nonumber
\end{eqnarray}
and passing to a similar space by dividing by $n^2 l / 2k$:
$$
ds^2= a^2 \, dy_1^2 + dx_2^2 + 2(y_1-x_2) \, dx_2 dx_3 +
     \left\{(y_1-x_2)^2+ b^2\right\} \, dx_3^2
\ ,
$$
$a,b$ being constants.
We now set
$y_1=b/a \, z_1 $, $x_2=b y_2$
and dividing by $b^2 $
leads to
$$
ds^2=  dz_1^2 + dy_2^2 + 2\left(\frac{z_1}{a}-y_2\right) \, dy_2 dx_3
    + \left\{\left(\frac{z_1}{a}-y_2\right)^2+ 1\right\} \, dx_3^2
\ ,
$$
a formula which differs only in notation from
(49*) of \S17.  Therefore, in the case
$B''=0 $,
the group of motions of the space is a $G_4$ of
composition already examined.

\typeout{**** Begin section 29}

\section{%29.
Integration in the general case by elliptic functions.}

We now treat the general case in which $B' $ is variable,
therefore $B''$ as well because of the differential equation (99).

We immediately reduce this equation to a quadrature,
assuming as
the independent variable $B'=s $ and taking
$ B''{}^2=t$ for the unknown function.
In this way (99) becomes
$$
 - s \frac{dt}{ds} + t
  = \frac{4s^3}{k} - \frac{4as^2}{k^2} -\frac{16}{n^2}
\ ,
$$
from which by integrating
$$
  t= -\frac{2s^3}{k} + \frac{4as^2}{k^2} + cs -\frac{16}{n^2}
\ ,
$$
with $c$ a new arbitrary constant.
We have therefore
$$
  B'' = \frac{ds}{dx_1}
    = \sqrt{-\frac{2s^3}{k} + \frac{4as^2}{k^2} + cs -\frac{16}{n^2}}
\ ,
$$
namely
$$
x_1 = \int \frac{ds}
       {\sqrt{ -\frac{2s^2}{k} +\frac{4as^2}{k^2} +cs-\frac{16}{n^2}}}
\ .
$$

We integrate this by introducing the Weierstrass elliptical
function\footnote{Bianchi's notation ${\cal P} x_1$, $\zeta x_1$ was changed to
the now-common ${\cal P} (x_1)$, $\zeta (x_1)$ [Editor].} ${\cal P}(x_1) $ with
the invariants $$
  g_2 = \frac{4a^2}{3k^4} + \frac{c}{2k}\ ,\
  g_3 = \frac{4}{n^2k^2} - \frac{8a^3}{27k^6} - \frac{ac}{6k^3}\ ,
\eqno(100)
$$
and neglecting the additive constant in $x_1 $
as is permissible, we will have
$$
  B'= s = \frac{2a}{3k} - 2k {\cal P}(x_1)
\ .
\eqno(101)
$$
Integrating again we introduce the Weierstrass
function $\zeta(x_1)=\frac{\sigma'(x_1)}{\sigma(x_1)} $
and one has
$$
 B= \frac{2a}{3k} x_1 - 2k \zeta(x_1) + h
\ ,
\eqno(102)
$$
with $h $ a new constant, so that from (98) we have
%%
\begin{eqnarray}
&&\kern-1cm
C= \frac{2a}{3} + k^2 {\cal P}(x_1)
    -\frac14 \left\{\frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right\}^2
\ ,
\seteqno{103a}\\
&&\kern-1cm
C'= k^2 {\cal P}'(x_1)
    -\left(\frac{a}{3k}-k {\cal P}(x_1) \right)
\left\{\frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right\}
\ .
\nonumber\\
&&
\seteqno{103b}
\end{eqnarray}
Equations (97) then give us immediately for the values of
$\alpha,\beta$ which appear in the line element
%%
\begin{eqnarray}
&&\kern-.7cm
\alpha= \frac{an^2}{3k^2} - n^2 {\cal P}(x_1)
\ ,
\seteqno{104a}\\
&&\kern-.7cm
\beta= - \frac{kn^2}{2} {\cal P}'(x_1)
       + \frac{n^2}{2k} \left(\frac{a}{3k} - k {\cal P}(x_1)\right)
       \left(\frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right)
\ .
\nonumber\\
&&
\seteqno{104b}
\end{eqnarray}

The value of $\gamma $ appears above instead in fractional form
with the denominator
$ B'=2k(a/(3k^2)-{\cal P}(x_1))$,
but if we transform it, taking into account the relation
${\cal P}'{}^2(x_1) = 4{\cal P}^3(x_1)-g_2 {\cal P}(x_1) -g_3 $                                               and applying it to the values (100) of the invariants we find
\begin{eqnarray}
&&
\gamma=  -k^2n^2 {\cal P}^2(x_1) - \frac{an^2}{3} {\cal P}(x_1)
     +\frac{2a^2n^2}{9k^2} + \frac{ck^2n^2}{8}
\nonumber\\
&&
+ \frac{n^2}{4k}\left(\frac{a}{3k} - k {\cal P}(x_1) \right)
   \left( \frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right)^2
\nonumber\\
&&
   - \frac{kn^2}{2} {\cal P}'(x_1)
        \left(\frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right)
\ .
\seteqno{105}
\end{eqnarray}
It is worth noting that, in view of the relation ${\cal P}''(x_1)=6 {\cal
P}^2(x_1)-g_2/2 $, the derivative of $\beta $ has the following
value:\footnote{ The preceding formulas can be greatly simplified by observing
that without loss of generality one can set $h=0$, $n=k=1$, as follows from
simple considerations. Then $e=a/3$ is a root of the equation $4 e^3-g_2 e
-g_3=0$ and one has
\begin{eqnarray}
&&
\alpha = e- {\cal P}(x_1)\ ,\
\beta = -\frac12 {\cal P}(x_1)
         + (e- {\cal P}(x_1)) (ex_1+\zeta(x_1))\ ,
\nonumber\\
&&
\gamma = -{\cal P}^2(x_1) - e {\cal P}(x_1) + e^2 + \frac{g_2}{4}
         + (e- {\cal P}(x_1)) (e x_1+\zeta(x_1))^2
\ .
\nonumber
\end{eqnarray}
}
\begin{eqnarray}
&& \beta'= \frac{cn^2}{8} +\frac{4a^2n^2}{9k^3}
        -\frac{2an^2}{3k} {\cal P}(x_1) - 2 k n^2 {\cal P}^2(x_1)
\nonumber\\ && \qquad
        - \frac{n^2}{2} {\cal P}'(x_1)
        \left(\frac{2a}{3k} x_1 + 2k \zeta(x_1) +h \right)
\ .
\seteqno{106}
\end{eqnarray}


\typeout{**** Begin section 30}

\section{%30.
The most general group of motions of the space of the previous section.}


To find the most general infinitesimal motion
$ Xf =
  \xi_1 \, \partial f/\partial x_1
+ \xi_2 \, \partial f/\partial x_2
+ \xi_3 \, \partial f/\partial x_3
$
of our space we recall the fundamental equations
(A) \S1.
First setting
$i=k=1 $
we have $\partial \xi_1/\partial x_1=0 $,
which shows that $\xi_1 $ does not depend on $x_1 $.
The remaining equations give us
\begin{eqnarray}
&&\kern-1.5cm
\frac{\partial \xi_1}{\partial x_2}
  + \alpha \frac{\partial \xi_2}{\partial x_1}
  + (\beta -\alpha x_2) \frac{\partial \xi_3}{\partial x_1} = 0
\ ,
\seteqno{107a}\\
&&\kern-1.5cm
\frac{\partial \xi_1}{\partial x_3}
    + (\beta -\alpha x_2) \frac{\partial \xi_2}{\partial x_1}
    + (\alpha x_2^2 - 2\beta x_2 +\gamma)
            \frac{\partial \xi_3}{\partial x_1} = 0
\ ,
\seteqno{107b}\\
&&\kern-1.5cm
\frac12 \alpha'\xi_1 + \alpha \frac{\partial \xi_2}{\partial x_2}
+ (\beta -\alpha x_2) \frac{\partial \xi_3}{\partial x_2} = 0
\ ,
\seteqno{108}\\
&&\kern-1.5cm
\frac12 (\alpha' x_2^2 - 2\beta' x_2 +\gamma') \xi_1
  - (\beta -\alpha x_2) \xi_2
  + (\beta -\alpha x_2) \frac{\partial \xi_2}{\partial x_3}
\nonumber\\
&&\kern-1.5cm\qquad
 + (\alpha x_2^2 - 2\beta x_2 +\gamma)
         \frac{\partial \xi_3}{\partial x_3} = 0
\ ,
\seteqno{109}\\
&&\kern-1.5cm
(\beta'-\alpha' x_2) \xi_1 - \alpha \xi_2
       + \alpha \frac{\partial \xi_2}{\partial x_3}
       + (\beta - \alpha x_2)\left(\frac{\partial \xi_2}{\partial x_2}
            +\frac{\partial \xi_3}{\partial x_3} \right)
\nonumber\\
&&\kern-1.5cm\qquad
 + (\alpha x_2^2 - 2\beta x_2 +\gamma)
         \frac{\partial \xi_3}{\partial x_2} = 0
\ .
\seteqno{110}
\end{eqnarray}

Solving (107) for
$\partial \xi_2/\partial x_1$,
$\partial \xi_3/\partial x_1$
we have:
%%
\begin{eqnarray}
&&\kern-1.3cm
\frac{\partial \xi_2}{\partial x_1}
 = \frac{1}{n_2} \left\{
   (\beta-\alpha x_2) \frac{\partial \xi_1}{\partial x_3}
   - (\alpha x_2^2 - 2\beta x_2 +\gamma)
           \frac{\partial \xi_1}{\partial x_2} \right\}
\ ,
\seteqno{111a}\\
&&\kern-1.3cm
\frac{\partial \xi_3}{\partial x_1}
= \frac{1}{n_2} \left\{
   (\beta-\alpha x_2) \frac{\partial \xi_1}{\partial x_2}
   - \alpha \frac{\partial \xi_1}{\partial x_3} \right\}
\ .
\seteqno{111b}
\end{eqnarray}

We integrate the preceding equations with respect to $x_1 $,
and for brevity set
$\alpha_0 = \int \alpha \,dx_1 $,
$\beta_0 = \int \beta \,dx_1 $,
$\gamma_0 = \int \gamma \,dx_1 $,
fixing, however, the additive constants in
$\alpha_0$, $\beta_0$, $\gamma_0 $: we assume, according to (104)
%%
\begin{eqnarray}
&&\kern-1cm
\alpha_0 = \frac{an^2}{3k^2} x_1 + n^2 \zeta(x_1)\ ,
\seteqno{112a}\\
&&\kern-1cm
\beta_0 = -\frac{kn^2}{2} {\cal P}(x_1)
       + \frac{n^2}{8k}
        \left( \frac{2a}{3k} x_1 +2k \zeta(x_1) + h \right)^2
\ .
\seteqno{112b}
\end{eqnarray}

Regarding the value of $\gamma_0 $, we need only observe that by formula (105)
it contains terms that cannot in any way be eliminated with those arising from
$\alpha_0,\beta_0 $. Given this, integrating (111) we have
\begin{eqnarray}
&&\kern-1.5cm
\xi_2 = \frac{1}{n^2} \left\{ (\beta_0-\alpha_0 x_2)
     \frac{\partial \xi_1}{\partial x_3}
   - (\alpha_0 x_2^2 - 2\beta_0 x_2 +\gamma_0)
         \frac{\partial \xi_1}{\partial x_2} \right\} + \psi(x_2,x_3)
\ ,
\seteqno{113}\\
&&\kern-1.5cm
\xi_3 = \frac{1}{n^2} \left\{ (\beta_0-\alpha_0 x_2)
     \frac{\partial \xi_1}{\partial x_2}
   - \alpha_0 \frac{\partial \xi_1}{\partial x_3} \right\}
   + \chi(x_2,x_3)
\ .
\nonumber
\end{eqnarray}

Substituting into (108), (109), (110) all those terms
which contain
$\gamma_0 $
must be zero separately
by the observation made above; from this we
then obtain
$\partial^2 \xi_1/\partial x_2{}^2=0 $,
$\partial^2 \xi_1/\partial x_2 \partial x_3
  =\partial \xi_1/\partial x_2$,
so that (108) becomes:
%%
\begin{eqnarray}
&&\kern-1cm
\frac12 \alpha' \xi_1
  + \frac{3\alpha}{n^2} (\beta_0-\alpha_0 x_2)
           \frac{\partial \xi_1}{\partial x_2}
  - \frac{2\alpha_0}{n^2} (\beta-\alpha x_2)
           \frac{\partial \xi_1}{\partial x_2}
  - \frac{\alpha\alpha_0}{n^2}
           \frac{\partial \xi_1}{\partial x_3}
\nonumber\\
&&\kern-1cm\qquad
+ (\beta-\alpha x_2)
           \frac{\partial \chi}{\partial x_2}
+ \alpha \frac{\partial \psi}{\partial x_2} = 0
\ .
\nonumber
\end{eqnarray}
If we observe that in this the term in $\zeta^2(x_1)$, arising from $\beta_0$,
cannot be cancelled by any other, we see that we must have
$$
\frac{\partial \xi_1}{\partial x_2}=0\ , \eqno(114)
$$
after which the previous equation
becomes
%%
\begin{eqnarray}
&&\kern-1cm
-\frac{n^2}{2} {\cal P}'(x_1) \xi_1
 +\left\{ \frac{n^2}{2k} \left(\frac{a}{3k} -k{\cal P}(x_1)\right)
               \left( \frac{2a}{3k} x_1 +2k \zeta(x_1) + h \right)
          - \frac{kn^2}{2} {\cal P}'(x_1)\right\}
           \frac{\partial \chi}{\partial x_2}
\nonumber\\
&&\kern-1cm
-\frac{n^2}{k} \left(\frac{a}{3k} -k{\cal P}(x_1)\right) x_2
      \frac{\partial \chi}{\partial x_2}
  +\frac{n^2}{k} \left(\frac{a}{3k} -k{\cal P}(x_1)\right)
      \frac{\partial \psi}{\partial x_2}
\nonumber\\
&&\kern-1cm
-\frac{n^2}{k} \left(\frac{a}{3k} -k{\cal P}(x_1)\right)
      \left(\zeta(x_1)+\frac{a}{3k^2} x_1 \right)
               \frac{\partial \xi_1}{\partial x_3} = 0
\ .
\nonumber
\end{eqnarray}

Equating to zero the terms in ${\cal P}'(x_1)$, ${\cal P}(x_1) \zeta(x_1)$
leads to
$$
k \frac{\partial \chi}{\partial x_2} = -\xi_1\ ,\
\frac{\partial \xi_1}{\partial x_3} = -\xi_1
\eqno(115)
$$
and subsequently
$$
\frac{\partial \psi}{\partial x_2}
  = \frac{h-2x_2}{2k} \,\xi_1
\ .
\eqno(116)
$$
Taking into account the equations obtained so far,
(113) become
$$
\xi_2 = \frac{\alpha_0 x_2-\beta_0}{n^2} \,\xi_1 +\psi(x_2,x_3)
\ ,\
\xi_3 = \frac{\alpha_0}{n^2} \xi_1 +\chi(x_2,x_3)
$$
and so one has\footnote{In the original paper, the second denominator on the
r.h.s of the first equation is just $k$. Correction made after the {\it Opere}
[Editor].}
%%
\begin{eqnarray}
&&\kern-1cm
\frac{\partial \xi_2}{\partial x_2}
= \left( \frac{\alpha_0}{n^2} + \frac{h-2x_2}{2k} \right)  \xi_1
\ ,\
\frac{\partial \xi_2}{\partial x_3}
= -\frac{\alpha_0 x_2-\beta_0}{n^2} \,\xi_1
     + \frac{\partial \psi}{\partial x_3}
\ ,
\nonumber\\
&&\kern-1cm
\frac{\partial \xi_3}{\partial x_2}
= \frac{\partial \chi}{\partial x_2}
\ ,\
\frac{\partial \xi_3}{\partial x_3}
=-\frac{\alpha_0}{n^2} \xi_1
     + \frac{\partial \chi}{\partial x_3}
\ .
\nonumber
\end{eqnarray}


Substituting into (109), we then find
$$
\frac{\partial \chi}{\partial x_2}
 = \frac{x_2}{k} \xi_1
\ ,\
\frac{\partial \psi}{\partial x_3}
 = \psi + \left(\frac{x_2^2}{k}-\frac{h x_2}{k}
          -\frac{2a}{3k}\right) \xi_1
$$
and finally
we find for the most general values of
$\xi_1,\xi_2,\xi_3$:\footnote{The term $hx_2/2k$ in the second equation is
absent in the original paper; correction made after the {\it Opere} [Editor].}
\begin{eqnarray}
&&
\xi_1= c_1 e^{-x_3}\ ,\
\xi_2= c_1 \left\{ \frac{\alpha_0 x_2-\beta_0}{n^2}
           -\frac{x_2^2}{2k} +\frac{h x_2}{2k}
         +\frac{a}{3k}\right\} e^{-x_3}\ ,
\nonumber\\
&&
\xi_3= c_1 \frac{\alpha_0}{n^2} e^{-x_3}
             -\frac{c_1 x_2}{k} e^{-x_3} +c_3\ ,
\nonumber
\end{eqnarray}
with three arbitrary constants $c_1,c_2,c_3 $. Therefore in the general case
considered in the present section the complete group of motions is only a
$G_3$.

\typeout{**** Begin section 31}

\section{%31.
Another method for the groups of type VIII.}

In the work of the previous sections on the spaces
which admit a transitive group $G_3$ of motions of type VIII
we have seen the elliptical functions introduced.
This depends on having wished to establish the geodesic
form of the line element, making evident a family
of pseudospherical surfaces, geodesically parallel and
invariant with respect to a subgroup of two parameters.
But, if we aim only to establish any form whatsoever for
the line element, we can proceed much more directly
by applying the general method described in \S12 to a
simple form of the group $G_3$.  We now discuss this
second way of treating the problem. In any event,
we necessarily have to apply it in the last
case of the groups of type IX, because there (real)
2-parameter subgroups do not exist.

    We start from the theorem of Lie that two
simply transitive and equally composed groups are
always similar.
Therefore if we take any particular
form whatsoever of a group $G_3$ transitive over three
variables with the composition
of type VIII and determine the {\it most general\/}
3-dimensional
spaces which admit it as a group of motions, any other
space with a transitive group of motions of the same composition
will necessarily be identical with one of these.

Given this,  referring ourselves
to the calculations made at
the beginning of \S28, we choose for the type of $G_3$
transitive over three variables of composition
$$
 [X_1,X_2]f = X_1 f\ ,\
 [X_1,X_3]f = 2 X_2 f\ ,\
 [X_2,X_3]f = X_3 f
$$ the one which is generated by the following three infinitesimal
transformations:\footnote{The original paper has $\frac{\partial f}{\partial
x_3}$ instead of $\frac{\partial f}{\partial x_2}$ in the first equation;
correction after the {\it Opere} [Editor].}
\begin{eqnarray}
&&
  X_1 f =
  e^{-x_3}       \frac{\partial f}{\partial x_1}
 -x_2^2 e^{-x_3} \frac{\partial f}{\partial x_2} % typo in subscript
 -2x_2 e^{-x_3}  \frac{\partial f}{\partial x_3}
\ ,\nonumber\\
&&
  X_2 f = \frac{\partial f}{\partial x_3}\ ,\
  X_3 f = e^{x_3} \frac{\partial f}{\partial x_2}
\nonumber
\end{eqnarray}
and we determine, in the most general way, the
coefficients of the line element of the space
$$
 ds^2 = \sum_{i,k} a_{ik} \, dx_i dx_k
$$
so that it admits the group $G_3$.

For this we must make use of the fundamental equations (A),
or equivalently (D) \S12, applying them to the above three
transformations.  Beginning with $X_2 f$ we see that
{\it the 6 coefficients have to be independent of $x_3$}.
Then applying them to $X_3 f$,  we find:
\begin{eqnarray}
&& \frac{\partial a_{11}}{\partial x_2} = 0\ ,\ \frac{\partial a_{22}}{\partial
x_2} = 0\ ,\ \frac{\partial a_{33}}{\partial x_2} + 2 a_{23} = 0\ , \nonumber\\
&& \frac{\partial a_{12}}{\partial x_2} = 0\ ,\ \frac{\partial a_{13}}{\partial
x_2} + a_{12} = 0\ ,\ \frac{\partial a_{23}}{\partial x_2} + a_{22} = 0 \
,\nonumber
\end{eqnarray}
from which by integrating
%%
\begin{eqnarray}
&&\kern-1cm
 a_{11}=A\ ,\
 a_{12}=B\ ,\
 a_{22}=C\ ,
\nonumber\\
&&\kern-1cm
 a_{13}=D - B x_2\ ,\
 a_{23}=E - C x_2\ ,\
 a_{33}=C x_2^2 - 2E x_2 + F\ ,\nonumber
\end{eqnarray}
where $A, B, C, D, E, F$ are functions only of $x_1 $.
Finally if we apply them to $X_1 f$,
taking into account the preceding values of the
$\partial a_{ik}/\partial x_2 $, we obtain
\begin{eqnarray}
&&
\frac{\partial a_{11}}{\partial x_1} = 0\ ,\
\frac{\partial a_{22}}{\partial x_1} = 4 a_{22} x_2 + 4 a_{23}\ ,
\nonumber\\ && \qquad
\frac{\partial a_{33}}{\partial x_1}
  = 2 a_{13} -4 a_{23} x_2^2 - 4 a_{33} x_2 \ ,
\nonumber\\
&&
\frac{\partial a_{12}}{\partial x_1} = 2 a_{12} x_2 + 2 a_{13}\ ,\
\frac{\partial a_{13}}{\partial x_1}
         = a_{11} - 2 a_{12} x_2^2 - 2 a_{13} x_2\ ,
\nonumber\\ && \qquad
\frac{\partial a_{23}}{\partial x_1}
         = a_{12} + 2 a_{33} - 2 a_{22} x_2^2\ ,
\nonumber
\end{eqnarray}
from which we derive
\begin{eqnarray}
&&
 A=constant\ ,\
 C'=4E\ ,\
 E'=B+2F\ ,
\nonumber\\
&&
 F'=2D\ ,\
 B'=2D\ ,\
 D'=A\ ,
\nonumber
\end{eqnarray}
and so
%%
\begin{eqnarray}
&&\kern-1.5cm A = a^2\ ,\ B = a^2 x_1^2 + 2b x_1 +c\ ,\ \seteqno {117}\\
&&\kern-1.5cm C = a^2 x_1^4 + 4b x_1^3 + 2(c+2d) x_1^2 + 4e x_1 + f\ ,\ D = a^2
x_1+b\ , \nonumber\\ &&\kern-1.5cm E = a^2 x_1^3 + 3b x_1^2 +(c+2d) x_1 + e\ ,\
F = a^2 x_1^2 + 2b x_1 + d\ ,\nonumber
\end{eqnarray}
with $a, b, c, d, e, f$ six arbitrary constants.
In conformity with the general theorem of \S12, we
verify in this way that our system of total differential
equations is completely integrable, the initial values
of the 6 coefficients  $a_{ik} $ remaining arbitrary for
$x_1=x_2=0 $.

We observe that from equations (117) it follows that $C$ is an arbitrary fourth
degree polynomial in $x_1 $; say $Q(x_1)$, with the first coefficient positive
(or zero), and one then has\footnote{Bianchi's $Q^{IV}$ was replaced by the
more familiar $Q^{(4)}$ [Editor].}
\begin{eqnarray}
&&
  A=\frac{Q^{(4)}(x_1)}{24}\ ,\
  B=\frac{Q''(x_1)}{12}+h\ ,\
  C=Q(x_1)\ ,\nonumber\\
&&
  D=\frac{Q'''(x_1)}{24}\ ,\
  E=\frac{Q'(x_1)}{4}\ ,\
  F=\frac{Q''(x_1)}{12}-\frac{h}{2}\ ,
\nonumber
\end{eqnarray}
with $h $ an arbitrary constant.

The surfaces invariant with respect to the subgroup
$(X_2 f ,X_3 f)$ are
$x_1 = constant$; these are geodesically parallel,
as follows from the general theorem and as we
confirm here by calculating the differential
parameter of the first order for $x_1$,
which has the value
$$
  \Delta_1 x_1 = a_{22} a_{33} - a_{23}^2
     = CF-E^2\ .
$$

{}From the expressions (117) for $C, E, F$, the binomial
$CF - E^2 $
is a fourth degree polynomial $P(x_1)$  in $x_1 $.
The arclength $s$ of the geodesics
orthogonal to the surfaces $x_1= constant$
is given, as one knows, by
$$
  s= \int \frac{dx_1}{\sqrt{\Delta_1 x_1}} =
     \int \frac{dx_1}{\sqrt{P(x_1)}}\ ,
$$ from which we again see the elliptic functions introduced here, confirming
what we have said at the beginning of the present section.

\typeout{**** Begin section 32}

\section{%32.
The groups of type IX:\\
$[X_1,X_2]f =X_3 f\ , [X_2,X_3]f =X_1 f\ , [X_3,X_1]f =X_2 f$.}

According to the method described at the beginning of the previous section, we
must first choose here the form of a group $G_3$ transitive over three
variables of the desired composition.  We fix as the general type the group
generated by the following three infinitesimal transformations:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2}\ ,\nonumber\\
&&
X_2 f = \cos x_2 \frac{\partial f}{\partial x_1}
        -\cot x_1 \sin x_2 \frac{\partial f}{\partial x_2}
        +\frac{\sin x_2}{\sin x_1} \frac{\partial f}{\partial x_3}
\ ,\nonumber\\
&&
X_3 f = -\sin x_2 \frac{\partial f}{\partial x_1}
        -\cot x_1 \cos x_2 \frac{\partial f}{\partial x_2}
        +\frac{\cos x_2}{\sin x_1} \frac{\partial f}{\partial x_3}
\ ,\nonumber
\end{eqnarray}
which is clearly transitive and offers the composition IX.
If a space of line element
$$
 ds^2=\sum_{i,k} a_{ik} \, dx_i dx_k
$$
is to admit this group as a group of motions, first of all
the coefficients $a_{ik}$ must be independent of $x_2 $
because of the form of  $X_1 f $.
Secondly, expressing by means of (D) \S12   the fact
that the space admits the infinitesimal transformation
$X_2 f$  (or the other $X_3 f$)\footnote{%
It suffices that it admit the two $X_1 f$, $X_2 f$ in order for it to also
admit the third  since $[X_1,X_2]f =X_3 f$.} we find the following system of
partial differential equations\footnote{The original paper had $a_{23}$ on the
l.h.s. of the 4th equation and $\partial/\partial x_2$ on the l.h.s. of the 8th
equation, both of which were incorrect [Editor].} for the 6 coefficients
$a_{ik}$:
%%
\begin{eqnarray}
&&\kern-1cm
\frac{\partial a_{11}}{\partial x_1} = 0\ ,\
\frac{\partial a_{11}}{\partial x_3}
   = 2 a_{13} \cot x_1 - \frac{2 a_{12}}{\sin x_1} \ ,\nonumber\\
&&\kern-1cm
\frac{\partial a_{22}}{\partial x_1}
   = 2 a_{22} \cot x_1 - \frac{2 a_{23}}{\sin x_1} \ ,\
\frac{\partial a_{22}}{\partial x_3}
   = 2 a_{12} \sin x_1  \ ,\nonumber\\
&&\kern-1cm
\frac{\partial a_{33}}{\partial x_1} = 0\ ,\
\frac{\partial a_{33}}{\partial x_3} = 0 \ ,\nonumber\\
&&\kern-1cm
\frac{\partial a_{12}}{\partial x_1}
  = 2 a_{12} \cot x_1 - \frac{a_{13}}{\sin x_1}\ ,\
\frac{\partial a_{12}}{\partial x_3}
  = a_{11} \sin x_1 + a_{23} \cot x_1 - \frac{a_{22}}{\sin x_1}\ ,\nonumber\\
&&\kern-1cm
\frac{\partial a_{13}}{\partial x_1} = 0\ ,\
\frac{\partial a_{13}}{\partial x_3}
  = a_{33}\cot x_1 - \frac{a_{23}}{\sin x_1}\ ,\nonumber\\
&&\kern-1cm
\frac{\partial a_{23}}{\partial x_1}
  = a_{23} \cot x_1 - \frac{a_{33}}{\sin x_1}\ ,\
\frac{\partial a_{23}}{\partial x_3}
  = a_{13} \sin x_1  \ .\nonumber
\end{eqnarray}

We observe that $a_{33} $ is a constant and so
$$
\frac{\partial^2 a_{13}}{\partial x_3{}^2}
  = - \frac 1{\sin x_1} \frac{\partial a_{23}}{\partial x_3}
  = -a_{13}\ ,
$$
we then have
$$
  a_{33}= a^2\ ,\
  a_{13}= b \cos x_3 + c \sin x_3\ ,
$$
with $a, b, c$ constants.
Substituting into the formula which gives
$\partial a_{13}/\partial x_3$
we obtain
$$
   a_{23}= a^2 \cos x_1 +  \sin x_1 \,(b\sin x_3 -c\cos x_3)\ .
$$

Now since $a_{11} $ is a function only of $x_3 $,   we set $
a_{11}=2\varphi(x_3)$, so that from the formula which gives $\partial
a_{11}/\partial x_3$ it follows that $$
   a_{12}= \cos x_1 \,(b\cos x_3 +c\sin x_3)
         -  \sin x_1 \, \varphi'(x_3)\ .
$$

Then integrating the two equations for $a_{22}$ we have
%%
\begin{eqnarray}
&&\kern-1cm a_{22}= 2\sin x_1 \cos x_1 \,(b\sin x_3 -c\cos x_3)
         - 2\sin^2 x_1 \, \varphi(x_3) \nonumber\\
&&\kern-1cm\qquad\qquad
+ a^2 +d \sin^2 x_1 \ ,
\nonumber
\end{eqnarray}
with $d$ a new constant.
Finally by substituting into the formula
which gives $\partial a_{12}/\partial x_3 $,
we find for $\varphi(x_3)$ the differential equation
$$
  \varphi''(x_3) = -4 \varphi(x_3) + a^2 + d\ ,
$$
and so by integration\footnote{The original paper has $x_3/2$ instead of
$2x_3$; correction after the {\it Opere} [Editor].}
$$
  \varphi(x_3) = e \cos(2 x_3) + f\sin(2 x_3) +(a^2 + d)/4\ ,
$$
with $e, f$ new constants.
With the values thus determined for the
6 coefficients $a_{ik} $, the above stated equations are
actually satisfied,
whatever the 6 constants $a, b, c, d, e, f$ are.

We can then directly show, making use of the usual fundamental
equations (A), that the complete group of motions is
the given $G_3$, except when the
four constants $b, c, e, f$ are simultaneously zero.
We prefer to treat this problem in another way, taking advantage
of the theorem of Lie on the composition of groups, which makes
the work simpler.  We add that we can also apply the same
method to the groups of type VIII to derive again the results
of \S28, \S30.

\typeout{**** Begin section 33}

\section{%33.
Spaces which admit as a subgroup of motions a group $G_3$ of type IX.}

Suppose that we have a space which admits a transitive $G_3$ of type IX
as a subgroup of motions,
but that its group of motions is larger.
If we exclude the case of spaces of constant
curvature, this larger group cannot be other
than a 4-parameter group, a fact which we state here
postponing its demonstration to \S36.

Given the hypothesized $G_4$ containing the simple subgroup $G_3$ of
composition\footnote{The original paper had $X_1, X_3$ on the left in the first
commutator, an obvious typo [Editor].} $$
 [X_1,X_2]f =X_3 f\ ,\ [X_2,X_3]f =X_1 f\ ,\ [X_3,X_1]f =X_2 f\ ,
$$
by the indicated theorem of Lie,\footnote{%
See S. Lie-F.  Engel,  Vol.~III, p.~723 and S. Lie-C.
Scheffers, p.~574, Theorem 9. ---
It is worth noting that the theorems used here depend only on the
relationships among the constants of composition $c_{iks}$ and
do not lose their validity by limiting them to the consideration of
real groups and subgroups, as we do here.}
we can choose the fourth infinitesimal generating transformation
of $G_4$ so that one has
$$
    [X_1, X_4]f  = [X_2,X_4]f  = [X_3,X_4]f  = 0\ .
$$

We consider in $G_4$ the $G_2$ of Abelian motions generated
for example by $X_1 f,X_4 f $
and  as in \S14 we choose as coordinate surfaces $x_1 = constant $
the surfaces invariant with respect to the group $G_2$.
Proceeding as
in the cited section we can furthermore assume
$X_1 f=\partial f/\partial x_2$,
$X_4 f=\partial f/\partial x_3$,
and give the line element of the space the form
$$
 ds^2 = dx_1^2+\alpha \, dx_2^2 + 2\beta \, dx_2 dx_3 + \gamma \, dx_3^2\ ,
$$
with  $\alpha,\beta,\gamma $ functions only of $x_1 $.
Now let
$ X_2 f =
 \eta_1 \partial f/\partial x_1
+\eta_2 \partial f/\partial x_2
+\eta_3 \partial f/\partial x_3
$;
because of
$
[X_1,X_2]f  = X_3 f
$,
it follows that
$$
 X_3 f =
 \frac{\partial \eta_1}{\partial x_2} \frac{\partial f}{\partial x_1}
+\frac{\partial \eta_2}{\partial x_2} \frac{\partial f}{\partial x_2}
+\frac{\partial \eta_3}{\partial x_2} \frac{\partial f}{\partial x_3}\ .
$$

Since on the other hand one must also have
$[X_2, X_4]f  = 0$,
$[X_3, X_1]f  = X_2 f$,
$\eta_1,\eta_2,\eta_3$ must satisfy the conditions:
\begin{eqnarray}
&&
 \frac{\partial \eta_1}{\partial x_3}
=\frac{\partial \eta_2}{\partial x_3}
=\frac{\partial \eta_3}{\partial x_3}=0\ ,\nonumber\\
&&
 \frac{\partial^2 \eta_1}{\partial x_2{}^2} +\eta_1=0\ ,
 \frac{\partial^2 \eta_2}{\partial x_2{}^2} +\eta_2=0\ ,
 \frac{\partial^2 \eta_3}{\partial x_2{}^2} +\eta_3=0\ ,
\nonumber
\end{eqnarray}
from which we will have
\begin{eqnarray}
&& \eta_1 =A \sin x_2 + B \cos x_2\ ,\ \eta_2 =C \sin x_2 + D \cos x_2\
,\nonumber\\ && \eta_3 =E \sin x_2 + F \cos x_2\ , \nonumber
\end{eqnarray}
where $A, D, C, D, E, F$ are functions only of $x_1$.
{}From the first of equations (E) \S14, it follows
that $\eta_1$ does not depend on $x_1$  and so
$A, B$ are absolute constants.
Finally from the composition equation
$ [X_2,X_3]f =X_1 f$
we get the following three equations
\begin{eqnarray}
&&
  AC+BD = 0\ ,\
  BC'-AD'= C^2+D^2+1\ ,\nonumber\\
&&
  AF'-BE'+CE+DF=0\ .
\nonumber
\end{eqnarray}

There is no loss of generality in adding a constant to $x_2 $ in such
a way that $B=0$ \footnote{
This assumes $A\neq0 $; if this is not true, one would change
$x_2$  into $\pi/2+x_2 $.}
and since one cannot simultaneously have $A=0$,
as can be seen from the second of the equations ($\beta$),
we will also have $C=0$, so
$ -AD'=1+D^2$, $AF'+DF=0$.

Integrating the first equation and ignoring the additive
constant in $x_1 $, as is allowed, we will have
$D=-\tan(x_1/A)$, $F=k/\cos(x_1/A)$,
with $k $ an arbitrary constant.

If we now apply the other equations (E) \S14, the relation
$$
  \frac12 \gamma' \eta_1
     + \beta \frac{\partial \eta_2}{\partial x_3}
     + \gamma \frac{\partial \eta_3}{\partial x_3} = 0
$$
shows us that $\gamma $ is constant, so we set $\gamma=h^2 $
and the remaining equations give us
\begin{eqnarray}
&&
E=0
\ ,\nonumber\\
&&
\alpha D'+\beta F' = -A\ ,\
\beta D' + h^2 F' = 0\ ,
\nonumber\\
&&
A\alpha'/2=\alpha D +\beta F\ ,\
A \beta'= \beta D + h^2 F\ ,
\nonumber
\end{eqnarray}
from which we get
$$
  \alpha = A^2 \cos\left(\frac{x_1}{A}\right)
        + h^2 k^2 \sin^2\left(\frac{x_1}{A}\right)\ ,\
  \beta = h^2 k \sin\left(\frac{x_1}{A}\right)\ ,\
  \gamma = h^2\ .
$$

If we set
$x_1=A y_1 $,
$x_3=A y_3/h$,
$n=hk/A$,
then dividing the line element by $h^2$,
we find the standard form
$$
  ds^2= dy_1^2 + (\cos^2 y_1 + n^2\sin^2 y_1) \, dx_2^2
         + 2n \sin y_1 \, dx_2 dy_3 + dy_3^2
$$
which, by changing the notation, we can write as
$$
  ds^2 = dx_1^2 + (\sin^2 x_1 + n^2\cos^2 x_1) \, dx_2^2
         + 2n \cos x_1 \, dx_2 dx_3 + dx_3^2 \ .
\eqno(118)
$$

One sees that for $n=0$ we obtain the space already
considered in \S9. This case must be excluded here
though because it would lead to $\eta_3=0 $ and
the derived group $(X_1 f,X_2 f,X_3 f)$ is then intransitive.

We also exclude the case $n=1 $ because the
line element then becomes
$$
  ds^2 = dx_1^2 +  dx_2^2 + dx_3^2 + 2 \cos x_1 \, dx_2 dx_3
$$
and belongs to the space of constant positive curvature
$K=1/4 $.
In fact let
$x_1=2 y_1$,
$x_2=y_2+y_3$,
$x_3=y_2-y_3$
and one has
$$
  ds^2 = 4( dy_1^2 + \cos^2 y_1 \, dy_2^2 + \sin^2 y_1 \, dy_3^2)
$$
which indeed belongs to one such space.
The geodesically parallel surfaces
$x_1=constant $
are in this case Clifford surfaces of zero curvature.

\typeout{**** Begin section 34}

\section{%34.
The complete group of motions of the space:\\
$ds^2 = dx_1^2 + (\sin^2 x_1 + n^2 \cos^2 x_1) \, dx_2^2
        + 2n \cos x_1 \, dx_2 dx_3 + dx_3^2$.}

To determine the most general infinitesimal motion
$$X f =
 \eta_1 \, \partial f/\partial x_1
+\eta_2 \, \partial f/\partial x_2
+\eta_3 \, \partial f/\partial x_3
$$
of the space of the line element (118), the equations (E)
\S14 give us the following equations
%%
\begin{eqnarray}
&&\kern-1.2cm
  \frac{\partial\eta_1}{\partial x_1}=0\ ,
\seteqno{119}\\
&&\kern-1.2cm
 \frac{\partial\eta_1}{\partial x_2}
      + (\sin^2 x_1 + n^2 \cos^2 x_1)\frac{\partial\eta_2}{\partial x_1}
      + n\cos x_1 \frac{\partial\eta_3}{\partial x_1}=0\ ,
\seteqno{120a}\\
&&\kern-1.2cm
 \frac{\partial\eta_1}{\partial x_3}
     + n\cos x_1  \frac{\partial\eta_2}{\partial x_1}
      + \frac{\partial\eta_3}{\partial x_1} =0\ ,
\seteqno{120b}\\
&&\kern-1.2cm
   (1-n^2) \sin x_1 \cos x_1 \,\eta_1
  + (\sin^2 x_1 + n^2 \cos^2 x_1)\frac{\partial\eta_2}{\partial x_2}
\nonumber\\ &&\kern-1.2cm \qquad
     + n\cos x_1  \frac{\partial\eta_3}{\partial x_2}=0\ ,
\seteqno{121}\\
&&\kern-1.2cm
  n \cos x_1  \frac{\partial\eta_2}{\partial x_3}
     + \frac{\partial\eta_3}{\partial x_3}=0\ ,
\seteqno{122}\\
&&\kern-1.2cm
   -n\sin x_1 \,\eta_1
  + (\sin^2 x_1 + n^2 \cos^2 x_1)\frac{\partial\eta_2}{\partial x_3}
\nonumber\\ &&\kern-1.2cm \qquad
     +n\cos x_1 \left( \frac{\partial\eta_2}{\partial x_2}
                     + \frac{\partial\eta_3}{\partial x_3} \right)
     + \frac{\partial\eta_3}{\partial x_2}=0\ .
\seteqno{123}
\end{eqnarray}

Solving (120) for
$\partial \eta_2/\partial x_1  $,
$\partial \eta_3/\partial x_1  $,
and integrating with
respect to $x_1 $, on which $\eta_1 $ does not depend by (119),
leads to the result:
\begin{eqnarray}
&&\kern-1cm
\eta_2 = \cot x_1 \frac{\partial\eta_1}{\partial x_2}
               -\frac{n}{\sin x_1} \frac{\partial\eta_1}{\partial x_3}
    + \psi(x_2,x_3)\ , \seteqno{124a}\\
&&\kern-1cm
\eta_3 = \left\{ n^2\cot x_1 -(1-n^2)x_1 \right\}
                 \frac{\partial\eta_1}{\partial x_3}
            -\frac{n}{\sin x_1} \frac{\partial\eta_1}{\partial x_2}
            + \chi(x_2,x_3)\ . \seteqno{124b}
\end{eqnarray}

Substituting into (122), we have
$$
  -n\sin x_1 \frac{\partial^2\eta_1}{\partial x_2\partial x_3}
  +n\cos x_1 \frac{\partial\psi}{\partial x_3}
   + (n^2-1) x_1 \frac{\partial^2\eta_1}{\partial x_3{}^2}
  + \frac{\partial \chi}{\partial x_2} = 0\ ,
$$
and so since $n^2-1\neq0 $:
$$
 \frac{\partial^2\eta_1}{\partial x_2\partial x_3}
  = \frac{\partial^2\eta_1}{\partial x_3{}^2}=0  \ ,\
\frac{\partial\psi}{\partial x_3}
 =\frac{\partial\chi}{\partial x_3}=0\ .
$$

Substituting now into (121), we then obtain $$ \frac{\partial\psi}{\partial
x_2}
 =\frac{\partial\chi}{\partial x_2}=0\ ,\
\frac{\partial^2\eta_1}{\partial x_2{}^2}=-\eta_1\ ,
$$
and with these equations (123) is also satisfied.
It follows next that
$ \partial \eta_1/\partial x_3=0$, so
\begin{eqnarray}
&&
\eta_1 = a\cos x_2 + b\sin x_2\ ,\nonumber\\
&&
\eta_2 = \cot x_1( -a\sin x_2 + b\cos x_2) + c\ ,\nonumber\\
&&
\eta_3 = \frac{n}{\sin x_1}( a\sin x_2 - b\cos x_2) + d\ ,
\nonumber
\end{eqnarray}
with $a, b, c, d$ arbitrary constants.

Therefore the complete group of motions of the space
(118) is the $G_4$ generated by the four infinitesimal
transformations:
\begin{eqnarray}
&& X_1 f = \frac{\partial f}{\partial x_2}\ ,\ \nonumber\\ && X_2 f = \cos x_2
\frac{\partial f}{\partial x_1}
      - \cot x_1\sin x_2 \frac{\partial f}{\partial x_2}
      + \frac{n\sin x_2}{\sin x_1} \frac{\partial f}{\partial x_3}\ ,
\nonumber\\
&&
X_3 f = -\sin x_2 \frac{\partial f}{\partial x_1}
      - \cot x_1\cos x_2 \frac{\partial f}{\partial x_2}
      + \frac{n\cos x_2}{\sin x_1} \frac{\partial f}{\partial x_3}\ ,
\nonumber \\ && X_4 f = \frac{\partial f}{\partial x_3}\ ,\ \nonumber
\end{eqnarray}
which has the composition:
%%
\begin{eqnarray}
&&\kern-1.7cm
 [X_1,X_2]f = X_3 f\ ,\
 [X_2,X_3]f = X_1 f\ ,\
 [X_3,X_1]f = X_2 f\ ,
\nonumber\\
&&\kern-1.7cm
 [X_1,X_4]f
 =[X_2,X_4]f
 =[X_3,X_4]f = 0\ .
\nonumber
\end{eqnarray}

This group is systatic and the systatic varieties are the
geodesics ($x_3$), which however, except in the case
$n=0 $,
do not admit orthogonal trajectories.

\typeout{**** Begin section 35}

\section{%35.
The constant $n$ is essential in\\
$ds^2 = dx_1^2 + (\sin^2 x_1 + n^2 \cos^2 x_1) \, dx_2^2
        + 2n \cos x_1 \, dx_2 dx_3 + dx_3^2$.}

We wish to show  finally that in the line element (113)
the constant $n$, apart from sign,\footnote{%
Changing the sign of either $x_2 $ or $x_3 $ changes
the sign of $n$.}
is actually essential and namely that if a
second space
$$
  ds^2= dy_1^2 + (\sin^2 y_1 + m^2 \cos^2 y_1) \, dy_2^2
        + 2m \cos y_1 \, dy_2 dy_3 + dy_3^2 \eqno(126)
$$
is similar to the first, one must necessarily have $n^2=m^2$.

Adopting for this the same method which has served us in
the analogous cases, we observe that the group $\Gamma_4$
of motions of the space (126)   is generated by the four
infinitesimal transformations:
\begin{eqnarray}
&& Y_1 f = \frac{\partial f}{\partial y_2}\ ,\nonumber\\ && Y_2 f = \cos y_2
\frac{\partial f}{\partial y_1}
      - \cot y_1\sin y_2 \frac{\partial f}{\partial y_2}
      + \frac{m\sin y_2}{\sin y_1} \frac{\partial f}{\partial y_3}\ ,
\nonumber\\
&&
Y_3 f = -\sin y_2 \frac{\partial f}{\partial y_1}
      - \cot y_1\cos y_2 \frac{\partial f}{\partial y_2}
      + \frac{m\cos y_2}{\sin y_1} \frac{\partial f}{\partial y_3}\ ,
\nonumber\\ && Y_4 f = \frac{\partial f}{\partial y_3}\ , \nonumber
\end{eqnarray}
with the composition:
\begin{eqnarray}
&&
 [Y_1,Y_2]f = Y_3 f\ ,\
 [Y_2,Y_3]f = Y_1 f\ ,\
 [Y_3,Y_1]f = Y_2 f\ ,
\nonumber\\
&&
 [Y_1,Y_4]f
 =[Y_2,Y_4]f
 =[Y_3,Y_4]f = 0\ .
\nonumber
\end{eqnarray}

First we must determine if the group $G_4$ of the first space is similar to the
$\Gamma_4$ of the second. Assuming that the equations of transformation change
$X_1 f$, $X_2 f$, $X_3 f$, $X_4 f$, respectively into $\bar Y_1 f$, $\bar Y_2
f$, $\bar Y_3 f$, $\bar Y_4 f$, the $\bar Y f$ must be combinations of the $Y f
$ and have their same composition. {}From this it follows that, since $Y_4 f $
is the only infinitesimal transformation of $\Gamma_4 $ which commutes with
every other, $ \bar Y_4 f$ will not differ from it other than by a constant
factor $ a$, while $\bar Y_1 f$, $\bar Y_2 f$, $\bar Y_3 f$, belonging to the
derived group, will not involve $Y_4 f $ and one will have\footnote{The
original paper had $X_2$ instead of $Y_2$ in the second equation, an obvious
typo [Editor].}
\begin{eqnarray}
&&
\bar Y_1 f = c_{11} Y_1 f + c_{12} Y_2 f + c_{13} Y_3 f \ ,\nonumber\\
&&
\bar Y_2 f = c_{21} Y_1 f + c_{22} Y_2 f + c_{23} Y_3 f \ ,\nonumber\\
&&
  \bar Y_3 f = c_{31} Y_1 f + c_{32} Y_2 f + c_{33} Y_3 f \ ,\
  \bar Y_4 f = a Y_4 f\ .\nonumber
\end{eqnarray}

The composition equations
$$
 [\bar Y_1,\bar Y_2]f = \bar Y_3 f\ ,\
 [\bar Y_2,\bar Y_3]f = \bar Y_1 f\ ,\
 [\bar Y_3,\bar Y_1]f = \bar Y_2 f
$$
show that the nine constants $c_{ik} $ are the coefficients
of an orthogonal matrix of determinant $= +1$.  Now among
$X_1 f$,
$X_2 f$,
$X_3 f$,
$X_4 f$,
holds the unique relation
$$
  X_4 f = \frac{1}{n} \left\{
       \cos x_1 \, X_1 f +\sin x_1 \sin x_2 \, X_2 f
             + \sin x_1 \cos x_2 \, X_3 f \right\}
$$
and similarly among the $Y_i f $  the other
$$
  Y_4 f =  \frac{1}{m} \left\{
       \cos y_1 \, Y_1 f +\sin y_1 \, \sin y_2 Y_2 f
             + \sin y_1 \cos y_2 \, Y_3 f \right\}\ .
$$

Expressing the $x$ in terms of the $y$,
we have
$$
  \bar Y_4 f = \frac{1}{n} \left\{
       \cos x_1 \, \bar Y_1 f +\sin x_1 \sin x_2 \, \bar Y_2 f
             + \sin x_1 \cos x_2 \, \bar Y_3 f \right\}
$$
or equivalently
\begin{eqnarray}
&&\kern-1.3cm
  Y_4 = \frac{1}{an} \left\{
(c_{11} \cos x_1 + c_{21} \sin x_1 \sin x_2
        + c_{31} \sin x_1 \cos x_2) Y_1 f \right.
\nonumber\\ &&\kern-1.3cm\qquad
+(c_{12} \cos x_1 + c_{22} \sin x_1 \sin x_2
        + c_{32} \sin x_1 \cos x_2) Y_2 f
\nonumber\\ &&\kern-1.3cm\qquad
\left.
+(c_{13} \cos x_1 + c_{23} \sin x_1 \sin x_2
        + c_{33} \sin x_1 \cos x_2) Y_3 f \right\}
\ .
\nonumber
\end{eqnarray}

Comparing this with ($\gamma$) leads to the three equations
\begin{eqnarray}
&& (an/m) \cos y_1 = c_{11} \cos x_1 + c_{21} \sin x_1 \sin x_2
        + c_{31} \sin x_1 \cos x_2\ ,
\nonumber\\ && (an/m) \sin y_1 \sin y_2 = c_{12} \cos x_1 + c_{22} \sin x_1
\sin x_2
        + c_{32} \sin x_1 \cos x_2\ ,
\nonumber\\ && (an/m) \sin y_1 \cos y_2 = c_{13} \cos x_1 + c_{23} \sin x_1
\sin x_2
        + c_{33} \sin x_1 \cos x_2\ .
\nonumber
\end{eqnarray}

The compatibility of these three equations in $y_1 $, $y_2 $ gives, according
to the general theory, the necessary and sufficient condition for the
similarity of the two groups $G_4 $, $\Gamma_4 $. This condition is found
immediately by squaring and summing the above three equations, which gives
$a^2n^2/m^2=1 $. It suffices therefore to take $a=\pm m/n$ in order that
corresponding equations of transformation of $G_4 $ into $\Gamma_4 $ exist.
Equations (127) show that one has $\partial y_1/\partial x_3= \partial
y_2/\partial x_3=0$.

For the rest,
expressing the fact that the $X_i f $ are transformed
respectively into the $\bar Y_i f $, we can find all the values of
the first partial derivatives of the $y $ with respect to the $x$.
It suffices for us to note here, in addition to the
two above, the following
\begin{eqnarray}
\kern-0.8cm
\frac{\partial y_2}{\partial x_2}
  &=& c_{11} - c_{12} \cot y_1 \sin y_2 - c_{13} \cot y_1 \cos y_2
\ ,\seteqno{128a}\\
\kern-0.8cm
\cos x_2 \frac{\partial y_1}{\partial x_1}
  &=& c_{22} \cos y_2 - c_{23} \sin y_2
\nonumber\\
&&\kern-0.8cm\qquad
+ \cot x_1 \sin x_2
         (c_{12} \cos y_2 - c_{13} \sin y_2)
\ ,\seteqno{128b}\\
\kern-0.8cm
\frac{\partial y_3}{\partial x_3}
  &=& a
\ .\seteqno{128c}
\end{eqnarray}

Assuming now that the two line elements are transformable
one into the other, except for a constant factor, we utilize
as in \S19 the Christoffel formula
$$
\frac{\partial^2 y_\nu}{\partial x_r \partial x_s}
  + \sum_{i,k} \christop{\nu}{ik}_y
      \frac{\partial y_i}{\partial x_r}  \frac{\partial y_k}{\partial x_s}
  = \sum_\mu \christop{\mu}{rs}_x \frac{\partial y_\nu}{\partial x_\mu}\ ,
$$
setting $\nu=1 $, $r=2$, $s=3$
and substituting for the Christoffel symbols
their actual values, we obtain
$$
\sin y_1 \frac{\partial y_2}{\partial x_2}
= \frac{n}{am}\sin x_1 \frac{\partial y_1}{\partial x_1} \ ,
$$
or equivalently by (128)
\begin{eqnarray}
&&
\cos x_2 ( c_{11} \sin y_1 - c_{12}\cos y_1 \sin y_2
              - c_{13} \cos y_1 \cos y_2)
\nonumber\\
&&
  = \frac{n}{am} \left\{\sin x_1 (c_{22} \cos y_2 - c_{23} \sin y_2 )
\right.
\nonumber\\
&&\qquad
\left.
       + \cos x_1 \sin x_1 (c_{12} \cos y_2 - c_{13} \sin y_2 )\right\}
\ .
\nonumber
\end{eqnarray}

Multiplying this last equation by $a^2n^2/m^2 \, \sin y_1=\sin y_1 $,
noting (127) one obtains the equation
\begin{eqnarray}
&&\kern-0.7cm
c_{11} \cos x_2\left\{ 1-(c_{11}\cos x_1 + c_{21}\sin x_1 \sin x_2
               + c_{31} \sin x_1 \cos x_2)^2 \right\}
\nonumber\\
&&\kern-0.7cm
 - c_{12} \cos x_2 (c_{11}\cos x_1 + c_{21}\sin x_1 \sin x_2
               + c_{31} \sin x_1 \cos x_2 )
\nonumber\\ &&\kern-0.7cm \qquad \times
           (c_{12} \cos x_1 + c_{22} \sin x_1 \sin x_2
                  + c_{32} \sin x_1 \cos x_2 )
\nonumber\\
&&\kern-0.7cm
 - c_{13} \cos x_2 (c_{11}\cos x_1 + c_{21}\sin x_1 \sin x_2
               + c_{31} \sin x_1 \cos x_2  )
\nonumber\\ &&\kern-0.7cm \qquad \times
                   (c_{13} \cos x_1 + c_{23} \sin x_1 \sin x_2
        + c_{33} \sin x_1 \cos x_2 )
\nonumber\\
&&\kern-0.7cm
=\frac{n^2}{m^2} \left\{ c_{22} \sin x_1 (c_{13} \cos x_1
           + c_{23} \sin x_1 \sin x_2 + c_{33} \sin x_1 \cos x_2)
 \right.
\nonumber\\
&&\kern-0.7cm
   + c_{12} \cos x_1 \sin x_2 (c_{13} \cos x_1 + c_{23} \sin x_1 \sin x_2
        + c_{33} \sin x_1 \cos x_2)
\nonumber\\
&&\kern-0.7cm
- c_{23} \sin x_1 (c_{12} \cos x_1 + c_{22} \sin x_1 \sin x_2
                  + c_{32} \sin x_1 \cos x_2 )
\nonumber\\
&&\kern-0.7cm
-\left.
 c_{13} \cos x_1 \sin x_2 (c_{12} \cos x_1 + c_{22} \sin x_1 \sin x_2
                  + c_{32} \sin x_1 \cos x_2 )
 \right\}\ ,
\nonumber
\end{eqnarray}
which must prove to be an identity in $x_1,x_2 $.
Setting $x_2=0 $ in this equation we find
$$
\frac{n^2}{m^2}( c_{11}\sin^2 x_1 - c_{31} \sin x_1 \cos x_1)
= c_{11} \sin^2 x_1 - c_{31} \sin x_1 \cos x_1
$$
so $n^2=m^2$,
unless one has $c_{11}=0 $, $c_{31}=0$
so that also
$c_{22}=0$, $c_{23}=0$, $c_{21}=\pm1 $.

Introducing these values of $c$ into the above identity
leads to:
$(n^2/m^2-1)c_{21}=0 $
and so again $n^2=m^2 $, Q.E.D.

\typeout{**** Begin section 36}

\section{%36.
The impossibility of other spaces with continuous groups of motions.}

In the previous sections we have exhausted the study
of the 3-dimensional spaces which admit intransitive
groups of motion or transitive 3-parameter groups.
And now we show that with this we have also determined all
the possible spaces which admit continuous groups of motions.

Therefore, since the group of motions of a space cannot
have more than 6 parameters, it will clearly suffice to
show that a (transitive) group of motions with 6, 5, or 4
parameters necessarily contains some {\it real\/}
3-parameter subgroup.

If we treat a $G_6$ this is clear since then the motions
which leave a point of the space fixed form precisely a
real subgroup with $6-3=3 $ parameters.\footnote{%(34)
S. Lie-F.  Engel, Vol.~I, p.~204.}

If the complete group of motions is a $G_4$ we easily
find the same thing recalling that the derived group of
a $G_4$ possesses at most 3 parameters and therefore,
in any case, there exist real 3-parameter subgroups
in $G_4$.
And indeed if the $G_4 $ is generated by the four infinitesimal
transformations
$X_1 f$,
$X_2 f$,
$X_3 f$,
$X_4 f$,
and the derived group is the identity,
or $(X_1 f) $
or $(X_1 f,X_2 f) $
or even $(X_1 f,X_2 f,X_3 f) $,
then $(X_1 f,X_2 f,X_3 f) $
will always be a real 3-parameter subgroup.

It remains to show the same property for a $G_5 $.
In this transitive group those motions which leave an
arbitrary point of the space fixed form a real $G_2$
and we propose to establish that such a $G_2 $ would
necessarily be contained in a real subgroup $G_3$ of the $G_5 $.
Lie\footnote{%(35)
S. Lie-F.  Engel, Vol.~I, pp.592--593.}
shows that indeed every $G_2 $ in a group with $r\geq3 $ parameters
is contained in at least one subgroup $G_3 $;
however,
it could easily happen that in the general case
these subgroups $G_3$ are only complex.
But if we apply the same derivation given
by Lie ({\it ibid.\/}) we see that our assertion will be
proved when it is shown that if
$$
  E_1 f = \sum_{i,k} a_{ik} x_k \frac{\partial f}{\partial x_i}\ ,\
  E_2 f = \sum_{i,k} b_{ik} x_k \frac{\partial f}{\partial x_i}
$$
are two linear homogeneous transformations in three
variables $x_1,x_2,x_3 $
such that one has
$ [E_1,E_2]f =k E_1 f $ ($k$ constant)
and one interprets $x_1,x_2,x_3 $
as homogeneous coordinates of a point in a
plane, then there will be at least one {\it real\/} point
that will remain fixed by both transformations (fixed point).
It is known that to find the fixed points with respect
to the $E_1 f $ one has the system of equations
\begin{eqnarray}
&& a_{11} x_1 + a_{12} x_2 + a_{13} x_3 = \rho x_1\ ,\nonumber\\
&& a_{21} x_1 + a_{22} x_2 + a_{23} x_3 = \rho x_2\ ,\nonumber\\
&& a_{31} x_1 + a_{32} x_2 + a_{33} x_3 = \rho x_3\nonumber
\end{eqnarray}
and since the cubic equation with real coefficients
$$
 \left|\begin{array}{lll}
       a_{11}-\rho & a_{12} & a_{13}\\
       a_{21}& a_{22} -\rho & a_{23}\\
       a_{31}-\rho & a_{32} & a_{33} -\rho \end{array}\right| =0
$$
has at least one real root, there will certainly be at
least one real fixed point with respect to $E_1 f $.
If there exists for  $E_1 $ an {\it isolated\/} real fixed
point, then since by assumption
$[E_1,E_2]f =k E_1 f $,
it  will be fixed with respect to $E_2 f $.\footnote{%(36).
See  S. Lie-F.  Engel, Vol.~1, p.~507, Theor.~104.}
So it will suffice to consider the case in which $E_1 f $
has no real isolated fixed points.  This happens only
when the above cubic equation has a single root, which
furthermore makes all the second order minors of the same determinant
zero.\footnote{%(37)
See the precise discussion in S. Lie-G.  Scheffers, pp.510-511.} Then all the
fixed points are distributed over a (real) line and if we assume this line as
the side $x_3=0 $ of the fundamental triangle, we give to $E_1 f$, as is
immediately seen, the form\footnote{The original paper had $\partial f/\partial
x_3$ in the second term, an obvious typo [Editor].} $$
 E_1 f = \rho \left(
  x_1 \frac{\partial f}{\partial x_1}
+ x_2 \frac{\partial f}{\partial x_2}
+ x_3 \frac{\partial f}{\partial x_3} \right)
+ \alpha x_3 \frac{\partial f}{\partial x_1}
+ \beta  x_3 \frac{\partial f}{\partial x_2}
\ .
$$

If one had\footnote{The original paper had $\alpha_3 = \beta_3 = 0$ here, an
obvious typo [Editor].} $\alpha=\beta=0 $,   $E_1 f $ would leave every point
fixed and a real fixed point of $E_2 f $  would satisfy the required condition.
If $\beta\neq0 $, changing $x_1 $ into $x_1+h x_2 $, we can make $\alpha=0 $
and we will thus have $$
 E_1 f = \rho \left(
  x_1 \frac{\partial f}{\partial x_1}
+ x_2 \frac{\partial f}{\partial x_2}
+ x_3 \frac{\partial f}{\partial x_3} \right)
+ \beta  x_3 \frac{\partial f}{\partial x_2}
 \ .
$$

If $E_1 f,E_2 f $ were in the involution
relation\footnote{In Italian: ``relazione involutoria" [Translator].}
$[E_1,E_2]f =0 $,
the above considerations are already sufficient to demonstrate
the assertion, since in the most unfavorable case
where neither $E_1 f $ nor $E_2 f $ possess a real isolated
fixed point,
the meeting point of the two lines of the respective
invariant points would satisfy the desired condition.

Therefore we assume in $$ [E_1,E_2]f  = k E_1 f \eqno(a) $$ that $k\neq0 $. One
then has\footnote{The original paper had $\partial f/\partial x_1$ instead of
$\partial f/\partial x_3$ in the last term [Editor].}
\begin{eqnarray}
&&\kern-1.4cm
 E_2 f
= (a_1 x_1 + a_2 x_2 + a_3 x_3)
             \frac{\partial f}{\partial x_1}
+ (b_1 x_1 + b_2 x_2 + b_3 x_3)
             \frac{\partial f}{\partial x_2}
\nonumber\\ &&\kern-1.4cm \qquad \qquad
+ (c_1 x_1 + c_2 x_2 + c_3 x_3)
             \frac{\partial f}{\partial x_3}
\ .\nonumber
\end{eqnarray}

The condition $(a)$ gives
$k\rho=0 $, $a_2=c_1=c_2=0$, $b_2=c_3+k$,
so that $\rho=0 $ and we can make
%%
\begin{eqnarray}
&&\kern-1cm
 E_1 f= x_3 \frac{\partial f}{\partial x_2}\ ,\nonumber\\
&&\kern-1cm
 E_2 f= (a_1 x_1 + a_3 x_3) \frac{\partial f}{\partial x_1}
      + (b_1 x_1 + b_2 x_2) \frac{\partial f}{\partial x_2}
      +  c_3 x_3 \frac{\partial f}{\partial x_3}\ .
\nonumber
\end{eqnarray}

The real point of coordinates $(0,1,0)$ remains fixed by both
transformations.

\typeout{**** Begin section 37}

\section{%37.
The impossibility of groups $G_5$ of motions.}

By what we have shown in the previous section,
there does not exist any space which has a $G_5$ for the
complete group of motions.  From this it follows that
if a space should admit a subgroup $G_5 $ of motions,
also admitting a $G_6 $, it would be of constant curvature.
But we can easily go farther and show that the groups $G_6 $
of motions of the spaces of constant curvature do not
contain a real subgroups of 5 parameters, namely:

{\it
There does not exist any 3-dimensional space whose group
of motions contains a real 5-parameter subgroup.}

Assuming the existence of such a $G_5$ of motions, its subgroup $G_2$ which
leaves any point $P$ whatsoever of the space fixed is contained, by the
previous section, in a real $G_3$.  This $G_3$ would necessarily be transitive
since otherwise with motions of $G_3$ one could transport $P$ anywhere, but
every point would remain fixed by a double infinity of motions of the $G_3$
which is absurd. The group $G_3$ being transitive, we can apply the methods
developed in \S\S 5--11 and therefore give the line element of the space of
constant curvature one of the following 6 forms:
\begin{eqnarray}
&&\kern-1cm
ds^2= dx_1^2 + dx_2^2 + dx_3^2\ ,\
K=0\ ,\seteqno{$\alpha$1}\\
&&\kern-1cm
ds^2= dx_1^2 + e^{2x_1} (dx_2^2 + dx_3^2)\ ,\
K=-1\ ,\seteqno{$\alpha$2}\\
&&\kern-1cm
ds^2= dx_1^2 + x_1^2(dx_2^2 + \sin^2 x_2 \, dx_3^2)\ ,\
K=0\ ,\seteqno{$\beta$1}\\
&&\kern-1cm
ds^2= dx_1^2 + \sin^2 x_1 (dx_2^2 + \sin^2 x_2 \, dx_3^2)\ ,\
K=1\ ,\seteqno{$\beta$2}\\
&&\kern-1cm
ds^2= dx_1^2 + \sinh^2 x_1(dx_2^2 + \sin^2 x_2 \, dx_3^2)\ ,\
K=-1\ ,\seteqno{$\beta$3}\\
&&\kern-1cm
ds^2= dx_1^2 + \cosh^2 x_1(dx_2^2 + \sin^2 x_2 \, dx_3^2)\ ,\
K=-1\ ,\seteqno{$\gamma$}
\end{eqnarray}
which is adapted to the subgroup $G_3$ of rotations about a
point\footnote{%(38)
As is seen, in the space of zero curvature (Euclidean),
we have two different forms for the line element,
one ($\alpha$) corresponding to the case of a center of
rotation at infinity,
the second ($\beta$) to the case of the center of rotation at a
finite distance.
For the pseudospherical spaces ($K = -1$) we have three
distinct forms ($\alpha$),  ($\beta$), ($\gamma$),
according to whether the center of rotation is at infinity or
a finite distance, or is ideal, and finally for the space of Riemann
($K = +1$) only one form.
These geometric circumstances are well known from the theory
of spaces of constant curvature.}
generated in the respective cases
($\alpha$),  ($\beta$), ($\gamma$)
by the three infinitesimal transformations designated in \S5 by
($\alpha^\ast$),  ($\beta^\ast$), ($\gamma^\ast$).
For each of these forms we have to determine, by integrating
the fundamental equations (A), the form of the complete group
$G_6$ of motions and see if there exists a subgroup $G_5$ of the $G_6$
containing the $G_3$.  The answer being negative, the stated
property will be established.

Here I limit myself to carrying out the calculations for one case.
We choose, for example, the (parabolic) form
$$
ds^2= dx_1^2 + e^{2x_1}(dx_2^2 + dx_3^2)
$$
of the line element of the pseudospherical spaces.
Integrating the equations of \S7 we easily find that
the complete group $G_6$ of motions is generated by the 6
infinitesimal transformations:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2}\ ,\
X_2 f = \frac{\partial f}{\partial x_3}\ ,\
X_3 f = x_3 \frac{\partial f}{\partial x_2}
          - x_2 \frac{\partial f}{\partial x_3}\ ,
\nonumber\\
&&
X_4 f
= x_2 \frac{\partial f}{\partial x_1}
 + \frac12 (e^{-2x_1} + x_3^2-x_2^2) \frac{\partial f}{\partial x_2}
 - x_2 x_3 \frac{\partial f}{\partial x_3}\ ,
\nonumber\\
&&
X_5 f
= x_3 \frac{\partial f}{\partial x_1}
- x_2 x_3\frac{\partial f}{\partial x_2}
 + \frac12 (e^{-2x_1} + x_2^2-x_3^2) \frac{\partial f}{\partial x_3}
 \ ,
\nonumber\\
&&
X_6 f
= \frac{\partial f}{\partial x_1}
- x_2 \frac{\partial f}{\partial x_2}
- x_3 \frac{\partial f}{\partial x_3}
\ .
\nonumber
\end{eqnarray}

We now write the related composition equations:
\begin{eqnarray}
&&\kern-1cm
[X_1,X_2]f =0\ ,\
[X_1,X_3]f =-X_2 f \ ,\
[X_1,X_4]f =X_6 f \ ,\nonumber\\ &&\kern-1cm\qquad
[X_1,X_5]f =-X_3 f\ ,
[X_1,X_6]f =-X_1 f \ ,\nonumber\\
&&\kern-1cm
[X_2,X_3]f =X_1 f\ ,\
[X_2,X_4]f =X_3 f\ ,\
[X_2,X_5]f =X_6 f\ ,\nonumber\\ &&\kern-1cm\qquad
[X_2,X_6]f =-X_2 f\ ,\nonumber\\
&&\kern-1cm
[X_3,X_4]f =X_5 f\ ,\
[X_3,X_5]f =-X_4 f\ ,\
[X_3,X_6]f =0\ ,\nonumber\\
&&\kern-1cm
[X_4,X_5]f = 0\ ,\
[X_4,X_6]f =X_4 f\ ,\nonumber\\
&&\kern-1cm
[X_5,X_6]f =X_5 f\ ,\nonumber
\end{eqnarray}
the inspection of which would suffice to show us that
there does not exist in the $G_6$ any real $G_5$ containing
the subgroup
$G_3=(X_1 f,X_2 f,X_3 f)$.

In fact let $Y f $ be an infinitesimal transformation of $G_5$
that does not belong to $G_3$; we can set
$Y f=a X_4 f+b X_5 f +c X_6 f $
with $a,b,c $ constants.
In $G_5$ there will therefore also exist the three infinitesimal
transformations
\begin{eqnarray}
&&
 [X_1,Y]f  = a X_6 f - b X_3 f - c X_1 f\ ,
\nonumber\\
&&
 [X_2,Y]f  = a X_3 f + b X_6 f - c X_2 f\ ,
\nonumber\\
&&
 [X_3,Y]f  = a X_5 f - b X_4 f \ ,
\nonumber
\end{eqnarray}
and so also $a X_6 f $, $b X_6 f $, and hence in any case $X_6 f$ since if
$a=b=0 $, $Y f $ reduces to $X_6 f$. Now the four transformations $X_1 f$, $X_2
f$, $X_3 f$, $X_6 f$, of $G_5$ actually generate a $G_4$ and if by $Z f =aX_4 f
+ bX_5 f$ we indicate the last infinitesimal transformation, then $[X_3,Z]f =
aX_5 f - bX_4 f $ must be a combination of $X_1 f$, $X_2 f$, $X_3 f$, $X_6 f$,
$Z f$ and so differs from $Z f $ only by a constant factor $\rho$. Therefore
one will have $a=\rho b $, $b=-\rho a$ from which $\rho^2+1=0 $ and so $Z f=X_4
f + i X_5 f $, which gives only a complex $G_5$. Demonstrations completely
analogous, as the reader can verify, are valid in all the other cases.

\typeout{**** Begin section 38}

\section{%38.
Summarized table of the line elements.}

It will be useful to summarize the results obtained by gathering
together in a table the various types to which we have reduced,
in the course of this study, the line elements of all possible
spaces which admit continous groups of motions.

We divide these spaces into six categories according to the type
of their complete group $G$ of motions.  We assign a space to the
category A) when its group of motions is a $G_1$,
to B) when it is a $G_2$,
to C) when it is an {\it intransitive\/} $G_3$.
The other two categories D) , E) contain the spaces whose group
of motions is {\it transitive\/}, D) those with a $G_3$, E) those
with a $G_4$.
Finally the sixth category F) will include the spaces of constant
curvature which admit a group $G_6$ of motions.
In the same table we also give the infinitesimal transformation
generators and their composition.

\typeout{**** Begin section 38 tables}

\subsection*{Category A}
\subsubsection*{Groups $G_1$}

\beqfl
ds^2 = \Sigma \, a_{ik} \, dx_i dx_k
\eeqfl
with coefficients $a_{ik}$ independent of $x_1$

\medskip group:
$$
  X_1 f = \frac{\partial f}{\partial x_1}
$$

\subsection*{Category B}
\subsubsection*{Groups $G_2$}

\beqfl
ds^2 = dx_1^2 + \alpha \, dx_2^2 + 2\beta \, dx_2 dx_3 + \gamma \, dx_3^2
\eeqfl
with $\alpha,\beta,\gamma$ functions only of $x_1$

\medskip group:
$$
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3}
$$

composition:
$$
[X_1, X_2] =  0
$$

\beqfl
ds^2 = dx_1^2 + \alpha \, dx_2^2 + 2(\beta-\alpha x_2) \, dx_2 dx_3
       + (\alpha x_2^2 - 2\beta x_2 + \gamma) \, dx_3^2
\eeqfl
with $\alpha,\beta,\gamma$ functions of $x_1$

\medskip group:
$$
X_1 f = \frac{\partial f}{\partial x_3} \ ,\
X_2 f = e^{x_3} \frac{\partial f}{\partial x_2}
$$

composition:
$$
 [X_1, X_2] f =  X_2 f
$$

\subsection*{Category C}
\subsubsection*{Groups $G_3$ intransitive}

\beqfl \mbox{\boldmath  $\alpha$)}\qquad ds^2 = dx_1^2 + \varphi^2(x_1) (dx_2^2
+ dx_3^2) \eeqfl

with $\varphi(x_1)$ an arbitrary function of $x_1$

\medskip group:
$$
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
X_3 f = x_3 \frac{\partial f}{\partial x_2}
       - x_2 \frac{\partial f}{\partial x_3}
$$

 composition:
$$
 [X_1, X_2] f =  0\ ,\
 [X_1, X_3] f =  - X_2 f\ ,\
 [X_2, X_3] f =  X_1 f
$$

\beqfl \mbox{\boldmath  $\beta$)} \qquad ds^2 = dx_1^2 + \varphi^2(x_1) (dx_2^2
+ \sin^2 x_2 \, dx_3)^2 \eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_3} \ ,\
X_2 f = \sin x_3 \frac{\partial f}{\partial x_2}
        +\cot x_2 \cos x_3 \frac{\partial f}{\partial x_3}\ ,
\nonumber\\ &&
\hbox{\phantom{$X_1 f = \frac{\partial f}{\partial x_3} \ ,\ $}}\
X_3 f = \cos x_3 \frac{\partial f}{\partial x_2}
        -\cot x_2 \sin x_3 \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2] f =  X_3 f\ ,\
 [X_2, X_3] f =  X_1 f\ ,\
 [X_3, X_1] f =  X_2 f
$$

\beqfl \mbox{\boldmath  $\gamma$)} \qquad ds^2 = dx_1^2 + \varphi^2(x_1) \,
(dx_2^2 + e^{2 x_2} \, dx_3)^2 \eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_3} \ ,\
X_2 f = \frac{\partial f}{\partial x_2}
        - x_3 \frac{\partial f}{\partial x_3} \ ,
\nonumber\\ &&
\hbox{\phantom{$X_1 f = \frac{\partial f}{\partial x_3} \ ,\ $}}\
X_3 f = x_3 \frac{\partial f}{\partial x_2}
       +\frac12(e^{-2 x_2} - x_3^2) \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2] f =  -X_1 f\ ,\
 [X_1, X_3] f =  X_2 f\ ,\
 [X_2, X_3] f =  -X_3 f
$$

\subsection*{Category D}
\subsubsection*{Groups $G_3$ transitive}

\paragraph*{\it Type IV}

\beqfl
ds^2 = dx_1^2
   + e^{x_1} [dx_2^2 + 2x_1 \, dx_2 dx_3 +(x_1^2+n^2) \, dx_3)^2]
\eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = 2\frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
\nonumber\\ &&
X_3 f = -2 \frac{\partial f}{\partial x_1}
    + (x_2 + 2 x_3) \frac{\partial f}{\partial x_2}
       + x_3 \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2]   =  0\ ,\
 [X_1, X_3] f =  X_1 f\ ,\
 [X_2, X_3] f =  X_1 f + X_2 f
$$

\paragraph*{\it Type VI}

\beqfl
ds^2 = dx_1^2 + e^{2x_1} \, dx_2^2 + 2n e^{(h+1)x_1} \, dx_2 dx_3
        + e^{2h x_1} \, dx_3^2
\eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
\nonumber\\ &&
X_3 f = - \frac{\partial f}{\partial x_1}
    + x_2 \frac{\partial f}{\partial x_2}
    + h x_3 \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2] =  0\ ,\
 [X_1, X_3] f =  X_1 f\ ,\
 [X_2, X_3] f =  h X_2 f
$$

\paragraph*{\it Type VII}

\beqfl
ds^2 = dx_1^2 + e^{-h x_1} \left\{ (n+\cos v x_1) \, dx_2^2
  + (h \cos v x_1 + v \sin x_1 + hn) \, dx_2 dx_3 \right.
\eeqfl\beqfl \hskip65pt\hfill\left.\textstyle
  + \left( \frac{2-v^2}{2} \cos v x_1 + \frac{hv}{2} \sin v x_1 +n
   \right) \, dx_3^2 \right\}
\eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
\nonumber\\ &&
X_3 f =  \frac{\partial f}{\partial x_1}
    - x_3 \frac{\partial f}{\partial x_2}
    + (x_2+ h x_3) \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2] =  0\ ,\
 [X_1, X_3] f =  X_2 f\ ,\
 [X_2, X_3] f =  -X_1 f + h X_2 f
$$

\paragraph*{\it Type VIII\footnote{Bianchi's $Q^{IV}$ was
replaced by the more familiar $Q^{(4)}$ [Editor].}}

\beqfl
ds^2 = \frac{Q^{(4)}(x_1)}{24} \, dx_1^2 + Q(x_1) \, dx_2^2
  + \left(Q(x_1) x_2^2 - \frac{Q'(x_1)}{2} x_2 +\frac{Q''(x_1)}{2}
         -\frac{h}{2} \right)   \, dx_3^2
\eeqfl\beqfl \hskip25pt
+ 2\left( \frac{Q''(x_1)}{12} + h\right) \, dx_1 dx_2
+ 2\left\{ \frac{Q'''(x_1)}{24}
       - \left( \frac{Q''(x_1)}{12} + h \right) x_2 \right\} \, dx_1 dx_3
\eeqfl\beqfl \hskip25pt
+ 2\left( \frac{Q'(x_1)}{4} - Q(x_1) x_2 \right) \, dx_2 dx_3\ ,
\eeqfl

with $Q(x_1)$ a fourth degree polynomial in $x_1$ with its first
\par
coefficient  positive (or zero), and $h$ a constant

\medskip group:
\begin{eqnarray}
&&
X_1 f = e^{-x_3} \frac{\partial f}{\partial x_1}
    - x_2^2 e^{-x_3}  \frac{\partial f}{\partial x_2}
    - 2 x_2 e^{-x_3} \frac{\partial f}{\partial x_3}\ ;
\nonumber\\
&&
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
X_3 f = \frac{\partial f}{\partial x_2}
\nonumber
\end{eqnarray}

composition:
$$
 [X_1, X_2] f =  X_1 f\ ,\
 [X_1, X_3] f =  2 X_2 f\ ,\
 [X_2, X_3] f =   X_3 f
$$


\paragraph*{\it Type IX\footnote{The original
paper had $x_3/2$ instead of $2x_3$, which was incorrect. Also, the second term
in $a_{12}$ had the coefficient 1/2 instead of 2, corrected here after the {\it
Opere} [Editor].}}

\beqfl
ds^2 = \Sigma_{i,k} \, a_{ik} \, dx_i dx_k
\eeqfl
\begin{eqnarray}
&& a_{11} = 2 e \cos 2 x_3 + 2 f \sin 2 x_3 + (a^2+d^2)/2 \ ,\nonumber\\ &&
a_{22} = 2 \sin x_1 \cos x_1 (b \sin x_3 - c \cos x_3)
         - a_{11} \sin^2 x_1 + a^2 + d \sin^2 x_1
\ ,\nonumber\\ &&
a_{33} = a^2\ ,\
a_{13} = b \cos x_3 + c \sin x_3
\ ,\nonumber\\ &&
a_{12} =  \cos x_1 (b \cos x_3 + c \sin x_3)
         + 2 \sin x_1 (e \sin 2x_3 - f \cos 2x_3)
\ ,\nonumber\\ &&
a_{23} = a^2 \cos x_1 + \sin x_1 (b \sin x_3 - c \cos x_3)
\nonumber
\end{eqnarray}

group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \cos x_2 \frac{\partial f}{\partial x_1}
        -\cot x_1 \sin x_2\frac{\partial f}{\partial x_2}
        +\frac{\sin x_2}{\sin x_1} \frac{\partial f}{\partial x_3}
\ ,\
\nonumber\\
&&
X_3 f = -\sin x_2 \frac{\partial f}{\partial x_1}
        -\cot x_1 \cos x_2 \frac{\partial f}{\partial x_2}
        +\frac{\cos x_2}{\sin x_1} \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition\footnote{The original paper had $Xf$ in the second commutator on
the r.h.s., an obvious typo [Editor].}: $$
 [X_1, X_2] f =  X_3 f\ ,\
 [X_2, X_3] f =  X_1 f\ ,\
 [X_3, X_1] f =  X_2 f
$$

\subsection*{Category E}
\subsubsection*{Groups $G_4$
\protect\footnote{Simply transitive subgroup Bianchi types added in brackets by
translator for clarity.}}

\paragraph*{a) \it [Type II]}

\beqfl
ds^2 = dx_1^2 + dx_2^2 + 2x_1 \, dx_2 dx_3 + (x_1^2+1) \, dx_3^2
\eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
X_3 f = -\frac{\partial f}{\partial x_1}
        + x_3 \frac{\partial f}{\partial x_2} \ ,
\nonumber\\ &&
X_4 f = x_3 \frac{\partial f}{\partial x_1}
    + \frac12(x_1^2- x_3^2)  \frac{\partial f}{\partial x_2}
    -  x_1  \frac{\partial f}{\partial x_3}\
\nonumber
\end{eqnarray}

composition:
\begin{eqnarray}
&&
 [X_1, X_2] =  [X_1, X_3] = [X_1, X_4] = 0\ ,
\nonumber\\ &&
 [X_2, X_3] f = X_1 f\ ,\
 [X_2, X_4] f =  -X_3 f\ ,\
 [X_3, X_4] f =  X_2 f
\nonumber
\end{eqnarray}

\paragraph*{b) \it [Types III, VIII]}

\beqfl
ds^2 = dx_1^2 + e^{2x_1} \, dx_2^2 + 2n e^{x_1} \, dx_2 dx_3 +  dx_3^2
\eeqfl

\medskip group:
\begin{eqnarray}
&&
X_1 f = \frac{\partial f}{\partial x_2} \ ,\
X_2 f = \frac{\partial f}{\partial x_3} \ ,\
X_3 f = \frac{\partial f}{\partial x_1}
        - x_2 \frac{\partial f}{\partial x_2} \ ,
\nonumber\\ &&
X_4 f = x_2 \frac{\partial f}{\partial x_1}
    + \frac12\left( \frac{e^{-2 x_1}}{1-n^2} - x_2^2\right)
             \frac{\partial f}{\partial x_2}
    -  \frac{n e^{-x_1}}{1-n^2} \frac{\partial f}{\partial x_3}
\nonumber
\end{eqnarray}

composition:
\begin{eqnarray}
&&
 [X_1, X_2] = 0\ ,\
 [X_1, X_3] f = -X_1 f\ ,\
 [X_1, X_4] f = X_3 f\ ,
\nonumber\\ &&
 [X_2, X_3] = 0\ ,\
 [X_2, X_4] = 0\ ,\
 [X_3, X_4] f =  -X_4 f
\nonumber
\end{eqnarray}

\paragraph*{c) \it [Type IX]}

\null\ \newline
\beqfl
ds^2 = dx_1^2 + (\sin^2 x_1 + n^2 \cos^2 x_1) \, dx_2^2
    + 2n \cos x_1 \, dx_2 dx_3 +  dx_3^2
\eeqfl

\medskip group:
\begin{eqnarray}
&& X_1 f = \frac{\partial f}{\partial x_2} \ ,\ X_2 f = \cos x_2 \frac{\partial
f}{\partial x_1}
       - \cot x_1 \sin x_2 \frac{\partial f}{\partial x_2}
       +\frac{n \sin x_2}{\sin x_1} \frac{\partial f}{\partial x_3}
 \ ,\nonumber\\ &&
X_3 f = -\sin x_2 \frac{\partial f}{\partial x_1}
       - \cot x_1 \cos x_2 \frac{\partial f}{\partial x_2}
       + \frac{n \cos x_2}{\sin x_1} \frac{\partial f}{\partial x_3} \ ,\
X_4 f = \frac{\partial f}{\partial x_3} \nonumber
\end{eqnarray}

composition:
\begin{eqnarray}
&&
 [X_1, X_2] f = X_3 f\ ,\
 [X_2, X_3] f = X_1 f\ ,\
 [X_3, X_1] f = X_2 f\ ,
\nonumber\\ &&
 [X_1, X_4] = [X_2, X_4] = [X_3, X_4] =  0
\nonumber
\end{eqnarray}

\subsection*{Category F}
\subsubsection*{Groups $G_6$ --- spaces of constant curvature}

\typeout{**** Begin section 39}

\section{%39
Conclusion.}

Having classified all possible types of spaces which admit a continuous
group of motions, it remains only that we say how, given the line element
of a space, one can verify whether that same space admits a continuous
group of motions, and if so, how the equations are found which reduce the
line element to one of the typical forms of our table.

For this purpose it is enough to recall the equations (A) \S1 which are
precisely according to Lie, the equations of definition\footnote{%39
S. Lie-F. Engel, Vol 1, \S 50, p.186.} of the group.
With only algebraic operations and differentiation one evaluate the
number $r$ of parameters of the group and decides on its transitivity
or intransitivity,\footnote{%40
S. Lie-F. Engel, Vol 1, p.217.}
so that one sees immediately to which of our categories the given space
belongs.

The integration of the fundamental equations (A) then gives us the actual
form of the infinitesimal transformations of the group and this makes the
composition evident to us, after which one will decide immediately to which
type in the category the space belongs since one will clearly find in the
table one and only one group which offers the same or an equivalent
composition.

Then one tries to identify the two groups, namely to assign the values of
the constants which enter in the group of canonical form and to calculate
the equations of transformation.  To this task one responds perfectly
applying the general criteria for the similarity of groups given in the
work of Lie.

\subsection*{NOTE}

After the editing of the present {\it Memoria\/} Professor Ricci brought
my attention to a {\it Nota\/} of Professor Levi-Civita, where by chance
particular 3-dimensional spaces with 3 or 4-parameter groups of
motions are already given (see T. Levi-Civita, {\it Sul moto di un corpo
rigido attorno ad un punto fisso\/} [On the motion of a rigid body around
a fixed point], Rendiconti della Reale Accademia dei Lincei (5), 5 (2nd sem.~1896),
3--9; 122--123).

\footnote{The original paper has a ``correzione" here that corrects a sentence
at the end of sec. 21. In the translation, the appropriate correction was made
where it belongs [Editor].}

\end{document}
