Generalised Darboux–Koenigs Metrics and 3-Dimensional Superintegrable Systems

Generalised Darboux–Koenigs Metrics
and 3-Dimensional Superintegrable Systems

Allan P. FORDY † and Qing HUANG ‡

A.P. Fordy and Q. Huang

† School of Mathematics, University of Leeds, Leeds LS2 9JT, UK
<EMAIL_ADDRESS>
‡ School of Mathematics, Northwest University, Xi’an 710069, People’s Republic
of China<EMAIL_ADDRESS>
Received November 01, 2018, in final form April 16, 2019; Published online May
05, 2019

The Darboux–Koenigs metrics in 2D are an important class of conformally flat,
non-constant curvature metrics with a single Killing vector and a pair of
quadratic Killing tensors. In [arXiv:1804.06904] it was shown how to derive
these by using the conformal symmetries of the 2D Euclidean metric. In this
paper we consider the conformal symmetries of the 3D Euclidean metric and
similarly derive a large family of conformally flat metrics possessing between
1 and 3 Killing vectors (and therefore not constant curvature), together with
a number of quadratic Killing tensors. We refer to these as generalised
Darboux–Koenigs metrics. We thus construct multi-parameter families of super-
integrable systems in 3 degrees of freedom. Restricting the parameters
increases the isometry algebra, which enables us to fully determine the
Poisson algebra of first integrals. This larger algebra of isometries is then
used to reduce from 3 to 2 degrees of freedom, obtaining Darboux–Koenigs
kinetic energies with potential functions, which are specific cases of the
known super-integrable potentials.

Darboux–Koenigs metrics; Hamiltonian system; super-integrability; Poisson
algebra; conformal algebra

17B6; 37J15; 37J35; 70G45; 70G65; 70H06

##  1 Introduction

In recent years there has been a burst of activity in the identification and
classification of super-integrable systems, both classical and quantum (see
the review [14] and references therein). Most of the interest is in
Hamiltonians which are in “natural form” (the sum of kinetic and potential
energies), with the kinetic energy being quadratic in momenta. A non-
degenerate kinetic energy is associated with (pseudo-)Riemannian metric, so
the leading order term in any integral defines a Killing tensor for this
metric, and itself commutes with the kinetic energy.

For a manifold with coordinates $(q_{1},\dots,q_{n})$, metric coefficients
$g_{ij}$, with inverse $g^{ij}$, the geodesic equations are Hamiltonian, with
kinetic energy

$\displaystyle
H=\frac{1}{2}\sum_{i,j=1}^{n}g^{ij}p_{i}p_{j},\qquad\mbox{where}\quad
p_{i}=\sum_{k}g_{ik}\dot{q}_{k}.$ (1.1)

For a metric with isometries, the infinitesimal generators (Killing vectors)
give rise to first integrals, which are linear in momenta (Noether constants).

The most common examples found in the literature have kinetic energies which
are associated with flat or constant curvature metrics, so finding the leading
order parts of higher order integrals (the Killing tensors) is
straightforward, since, as is well known [8], they are all built as polynomial
expressions in the first order integrals (Killing vectors). Indeed, in this
case, (1.1) is actually the second order Casimir function of the symmetry
algebra. However, for the non-constant curvature case, there is no algorithmic
way known for building Killing tensors. In [6] a new method was proposed for
the “next” class of metrics, namely the conformally flat case.

Conformally flat metrics possess an algebra of conformal symmetries,
corresponding to the particular flat metric to which they are related. In 2
dimensions, as is well known, the conformal algebra is infinite. For $n\geq 3$
this algebra is finite and has maximal dimension $\frac{1}{2}(n+1)(n+2)$,
which is achieved for conformally flat spaces (which includes flat and
constant curvature spaces). Any two conformally equivalent metrics have the
same conformal algebra, so we can describe this in terms of the corresponding
flat metric. In flat spaces, the infinitesimal generators consist of $n$
translations, $\frac{1}{2}n(n-1)$ rotations, 1 scaling and $n$ inversions,
totalling $\frac{1}{2}(n+1)(n+2)$. This algebra is isomorphic to
$\mathfrak{so}(n+1,1)$ (see [3, Vol. 1, p. 143]). Whilst the conformal algebra
in two-dimensional spaces is infinite, there still exists the 6-dimensional
subalgebra described above (with $n=2$).

The true symmetries (isometries) of a metric form a subalgebra of the
conformal algebra. The maximum dimension of the algebra of isometries is
$\frac{1}{2}n(n+1)$, which is realised when the space is flat or constant
curvature. To discuss conformally flat metrics, we wish to avoid the flat and
constant curvature cases, so should not have “too many isometries”. In 2
dimensions we cannot have more than one Killing vector, since, by a theorem of
Darboux and Koenigs, if such a metric possesses at least two Killing vectors,
then it possesses three and the space has constant curvature (see [9, 10]).
Killing vectors correspond to first integrals which are linear in momenta.
Metrics found by Koenigs (see [9, 10, 11]) are characterised by having one
linear and two quadratic first integrals. In modern day jargon, these are
super-integrable systems, defined on spaces which are conformally flat, but
not constant curvature. In [6] these systems were rederived by a new approach
which builds higher order integrals as polynomial expressions in conformal
symmetries, together with an important non-constant multiple of the
Hamiltonian itself (see Section 3 below). The Darboux–Koenigs metrics have
been further generalised in 2 dimensions, by considering Hamiltonians with one
linear and one or more higher order first integrals (see [13] for the cubic
case and [15] for integrals of any integer order).

In this paper, we consider the generalisation to 3 degrees of freedom. There
exist $N$-dimensional generalisations of Darboux–Koenigs metrics, both in the
classical and quantum domain [1, 2]. In these papers the approach is more
geometric and the connection to the 2-dimensional Darboux–Koenigs metrics is
different from the one presented below.

In Section 2 we describe the structure of the 10-dimensional conformal algebra
in this case, and show that it has a particular decomposition introduced in
[7]. We show that this algebra has 4 involutive automorphisms, which play an
important role in our calculations. In Section 3 we describe the method
introduced in [6] for the construction of (in this paper) quadratic first
integrals out of conformal symmetries. We initially only assume the existence
of one Killing vector $K$, but in 3 dimensions we can have up to $4$ Killing
vectors without forcing constant curvature (although the examples in this
paper have at most 3). The calculations give rise to metrics which depend upon
a number of parameters, and degeneration of these parameters leads to metrics
with additional Killing vectors (“over-degeneration” leading to flat or
constant curvature metrics). The significance of the additional Killing
vectors is that we can use these to reduce from 3 to 2 degrees of freedom,
using the “Kaluza–Klein reduction” approach described in [5]. This process
leads to Hamiltonians in 2 degrees of freedom with one linear and two
quadratic first integrals, which are Darboux–Koenigs kinetic energies with the
addition of potential functions, which are compared with those of [9, 10].
With enough Killing vectors, we can reduce in different ways, leading to the
same Darboux–Koenigs kinetic energy, but different potentials. Within the
3-dimensional framework we give point transformations between these.

Another consequence of having more Killing vectors is that the Poisson algebra
of first integrals is simplified, so that we can find the full set of
relations. The involutive automorphisms play a crucial role here, since each
algebra is invariant with respect to one of these.

On the other hand, an example treated in Section 9.2 has only one Killing
vector, so reduction to 2 dimensions results in a metric with no Killing
vectors, but with higher order first integrals, thus leading to a maximally
super-integrable system which falls out of the Darboux–Koenigs and later
classifications.

The results on Poisson algebras are given in Sections 4 to 6, whilst those on
reductions can be found in Sections 7 to 10.

##  2 The 3D Euclidean metric and its conformal algebra

Consider metrics which are conformally related to the standard Euclidean
metric in 3 dimensions, with Cartesian coordinates $(q_{1},q_{2},q_{3})$. The
corresponding kinetic energy (1.1) takes the form

$\displaystyle
H=\varphi(q_{1},q_{2},q_{3})\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}.$
(2.1)

A conformal invariant $X$, linear in momenta, will satisfy
$\\{X,H\\}=\lambda(X)H$, for some function $\lambda(X)$. The conformal
invariants form a Poisson algebra, which we call the conformal algebra. For
special cases of $\varphi(q_{1},q_{2},q_{3})$ there will be a subalgebra for
which $\\{X,H\\}=0$, thus forming true invariants of $H$. These correspond to
infinitesimal isometries (Killing vectors) of the metric. Constant curvature
metrics possess $\frac{1}{2}n(n+1)=6$ Killing vectors (when $n=3$).

As discussed in the introduction, the conformal algebra of this 3D metric has
dimension $\frac{1}{2}(n+1)(n+2)=10$ (when $n=3$). A convenient basis is as
follows

$\displaystyle e_{1}=p_{1},\qquad h_{1}=-2(q_{1}p_{1}+q_{2}p_{2}+q_{3}p_{3}),$
$\displaystyle
f_{1}=\big{(}q_{2}^{2}+q_{3}^{2}-q_{1}^{2}\big{)}p_{1}-2q_{1}q_{2}p_{2}-2q_{1}q_{3}p_{3},$
(2.2a) $\displaystyle e_{2}=p_{2},\qquad h_{2}=2(q_{1}p_{2}-q_{2}p_{1}),$
$\displaystyle
f_{2}=-4q_{1}q_{2}p_{1}-2\big{(}q_{2}^{2}-q_{1}^{2}-q_{3}^{2}\big{)}p_{2}-4q_{2}q_{3}p_{3},$
(2.2b) $\displaystyle e_{3}=p_{3},\qquad h_{3}=-2q_{3}p_{1}+2q_{1}p_{3},$
$\displaystyle
f_{3}=-4q_{1}q_{3}p_{1}-4q_{2}q_{3}p_{2}-2\big{(}q_{3}^{2}-q_{1}^{2}-q_{2}^{2}\big{)}p_{3},$
(2.2c) $\displaystyle h_{4}=4q_{3}p_{2}-4q_{2}p_{3}.$ (2.2d)

The Poisson relations of the ten elements in the conformal algebra (2.2) are
given in Table 1.

Table 1: The 10-dimensional conformal algebra (2.2). | $e_{1}$ | $h_{1}$ | $f_{1}$ | $e_{2}$ | $h_{2}$ | $f_{2}$ | $e_{3}$ | $h_{3}$ | $f_{3}$ | $h_{4}$
---|---|---|---|---|---|---|---|---|---|---
$e_{1}$ | 0 | $2e_{1}$ | $-h_{1}$ | 0 | $-2e_{2}$ | $-2h_{2}$ | 0 | $-2e_{3}$ | $-2h_{3}$ | 0
$h_{1}$ |  | 0 | $2f_{1}$ | $-2e_{2}$ | 0 | $2f_{2}$ | $-2e_{3}$ | 0 | $2f_{3}$ | 0
$f_{1}$ |  |  | 0 | $-h_{2}$ | $-f_{2}$ | 0 | $-h_{3}$ | $-f_{3}$ | 0 | 0
$e_{2}$ |  |  |  | 0 | $2e_{1}$ | $-2h_{1}$ | 0 | 0 | $h_{4}$ | $4e_{3}$
$h_{2}$ |  |  |  |  | 0 | $-4f_{1}$ | 0 | $-h_{4}$ | 0 | $4h_{3}$
$f_{2}$ |  |  |  |  |  | 0 | $h_{4}$ | 0 | 0 | $4f_{3}$
$e_{3}$ |  |  |  |  |  |  | 0 | $2e_{1}$ | $-2h_{1}$ | $-4e_{2}$
$h_{3}$ |  |  |  |  |  |  |  | 0 | $-4f_{1}$ | $-4h_{2}$
$f_{3}$ |  |  |  |  |  |  |  |  | 0 | $-4f_{2}$
$h_{4}$ |  |  |  |  |  |  |  |  |  | 0

Note that this is an example of the conformal algebra given in [7, Table 3]
corresponding to the case $a_{2}=a_{3}=2$, $a_{4}=0$. The subalgebra
$\mathfrak{g}_{1}$, with basis (2.2a) is just a copy of $\mathfrak{sl}(2)$. We
then make the vector space decomposition of the full algebra $\mathfrak{g}$
into invariant subspaces under the action of $\mathfrak{g}_{1}$:

$\displaystyle\mathfrak{g}=\mathfrak{g}_{1}+\mathfrak{g}_{2}+\mathfrak{g}_{3}+\mathfrak{g}_{4}.$

The basis elements for $\mathfrak{g}_{i}$ have the same subscript and are
given in the rows of (2.2).

The algebra (2.2) possesses a number of involutive automorphisms:

$\displaystyle\iota_{12}\colon\
(q_{1},q_{2},q_{3})\mapsto(q_{2},q_{1},q_{3}),\qquad\iota_{13}\colon\
(q_{1},q_{2},q_{3})\mapsto(q_{3},q_{2},q_{1}),$
$\displaystyle\iota_{23}\colon\
(q_{1},q_{2},q_{3})\mapsto(q_{1},q_{3},q_{2}),$
$\displaystyle\iota_{ef}\colon\
(q_{1},q_{2},q_{3})\mapsto\left(-\frac{q_{1}}{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}},-\frac{q_{2}}{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}},-\frac{q_{3}}{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}\right),$

whose actions are given in Table 2.

Table 2: The involutions of the conformal algebra (2.2). | $e_{1}$ | $h_{1}$ | $f_{1}$ | $e_{2}$ | $h_{2}$ | $f_{2}$ | $e_{3}$ | $h_{3}$ | $f_{3}$ | $h_{4}$
---|---|---|---|---|---|---|---|---|---|---
$\iota_{12}$ | $e_{2}$ | $h_{1}$ | $\frac{1}{2}f_{2}$ | $e_{1}$ | $-h_{2}$ | $2f_{1}$ | $e_{3}$ | $-\frac{1}{2}h_{4}$ | $f_{3}$ | $-2h_{3}$
$\iota_{13}$ | $e_{3}$ | $h_{1}$ | $\frac{1}{2}f_{3}$ | $e_{2}$ | $\frac{1}{2}h_{4}$ | $f_{2}$ | $e_{1}$ | $-h_{3}$ | $2f_{1}$ | $2h_{2}$
$\iota_{23}$ | $e_{1}$ | $h_{1}$ | $f_{1}$ | $e_{3}$ | $h_{3}$ | $f_{3}$ | $e_{2}$ | $h_{2}$ | $f_{2}$ | $-h_{4}$
$\iota_{ef}$ | $-f_{1}$ | $-h_{1}$ | $-e_{1}$ | $-\frac{1}{2}f_{2}$ | $h_{2}$ | $-2e_{2}$ | $-\frac{1}{2}f_{3}$ | $h_{3}$ | $-2e_{3}$ | $h_{4}$

###  2.1 Different choices for the decomposition of
$\boldsymbol{\mathfrak{g}}$

The first block of Table 2 shows different copies of $\mathfrak{sl}(2)$, which
we might have chosen as our $\mathfrak{g}_{1}$. The remaining blocks are just
the corresponding invariant subspaces.

In [7] we showed that the $6$-dimensional algebra
$\mathfrak{g}_{1}+\mathfrak{g}_{2}$ (which here we call $\mathfrak{g}^{(0)}$)
has two quadratic Casimirs

$\displaystyle{\cal
C}_{2}^{(0)}=4e_{1}f_{1}+h_{1}^{2}+2e_{2}f_{2}-h_{2}^{2},\qquad{\cal
C}_{4}^{(0)}=h_{1}h_{2}+e_{1}f_{2}-2e_{2}f_{1},$ (2.3a) the second of which is
derived from a quartic Casimir, which happens to be a perfect square in this
case. In the $6\times 6$ matrix representation (the adjoint representation),
this is a multiple of the identity matrix, but in our Poisson representation,
it vanishes identically, so the 6-dimensional Poisson algebra has a quadratic
constraint.

Under the action of the involutions, this $6$-dimensional algebra is
transformed into the first 6 elements found to the right of $\iota_{ij}$ in
Table 2 (here labelled $\mathfrak{g}^{(ij)}$), and the Casimirs take the form

$\displaystyle{\cal C}_{2}^{(12)}={\cal C}_{2}^{(0)},\qquad{\cal
C}_{4}^{(12)}=-{\cal C}_{4}^{(0)},$ (2.3b) $\displaystyle{\cal
C}_{2}^{(13)}=2e_{2}f_{2}+2e_{3}f_{3}+h_{1}^{2}-\tfrac{1}{4}h_{4}^{2},\qquad{\cal
C}_{4}^{(13)}=e_{3}f_{2}-e_{2}f_{3}+\tfrac{1}{2}h_{1}h_{4},$ (2.3c)
$\displaystyle{\cal
C}_{2}^{(23)}=4e_{1}f_{1}+h_{1}^{2}+2e_{3}f_{3}-h_{3}^{2},\qquad{\cal
C}_{4}^{(23)}=h_{1}h_{3}+e_{1}f_{3}-2e_{3}f_{1},$ (2.3d) $\displaystyle{\cal
C}_{2}^{(ef)}={\cal C}_{2}^{(0)},\qquad{\cal C}_{4}^{(ef)}=-{\cal
C}_{4}^{(0)}.$ (2.3e)

Each of the Casimirs ${\cal C}_{2}^{(0)}$ and ${\cal C}_{2}^{(ij)}$ represent
spaces of constant curvature, whilst each of ${\cal C}_{4}^{(0)}$ and ${\cal
C}_{4}^{(ij)}$ identically vanishes, so is a quadratic constraint on the whole
$10$-dimensional conformal algebra.

There are 2 more $6$-dimensional algebras:

$\displaystyle\mathfrak{g}^{(he)}=\langle
h_{2},h_{3},h_{4},e_{1},e_{2},e_{3}\rangle,\qquad\mathfrak{g}^{(hf)}=\langle
h_{2},h_{3},h_{4},f_{1},f_{2},f_{3}\rangle,$ (2.3f)

with Casimirs

$\displaystyle{\cal C}_{2}^{(he)}=e_{1}^{2}+e_{2}^{2}+e_{3}^{2},\qquad{\cal
C}_{4}^{(he)}=e_{1}h_{4}+2e_{2}h_{3}-2e_{3}h_{2},$ (2.3g) $\displaystyle{\cal
C}_{2}^{(hf)}=f_{1}^{2}+\tfrac{1}{4}\big{(}f_{2}^{2}+f_{3}^{2}\big{)},\qquad{\cal
C}_{4}^{(hf)}=f_{3}h_{2}-f_{2}h_{3}-f_{1}h_{4},$ (2.3h)

which are related through the involution $\iota_{ef}$. Again, ${\cal
C}_{4}^{(he)}$ and ${\cal C}_{4}^{(hf)}$ vanish identically.

##  3 Geodesic flows in 3D with linear and quadratic integrals

In this and later sections we seek Hamiltonian functions (2.1) which possess
up to 4 first order invariants (Killing vectors), together with a number of
quadratic integrals. In 3 dimensions, if we have 5 isometries, then we must
have 6, so the space has constant curvature. We already mentioned that in 2
dimensions, Darboux and Koenigs proved that we can only have 1 isometry (i.e.,
2 implied 3). This “gap phenomenon” occurs in all dimensions and is the
subject of [12].

Initially we only assume the existence of one Killing vector $K$, the
calculations giving rise to metrics which depend upon a number of parameters.
Degeneration of these parameters leads to metrics with additional Killing
vectors, with “over-degeneration” leading to flat or constant curvature
metrics. We use the method introduced in [6] to construct quadratic invariants
out of conformal invariants.

A quadratic conformal invariant is any expression of the form

$\displaystyle
F=\sum_{i,j=1}^{10}\beta_{ij}X_{i}X_{j}+\psi(q_{1},q_{2},q_{3})H,$ (3.1)

where $\beta_{ij}$ is a symmetric matrix of (constant) coefficients, $X_{i}$
are linear conformal invariants, and $\psi(q_{1},q_{2},q_{3})$ is an arbitrary
function, which satisfies

$\displaystyle\\{F,H\\}=\left(\sum_{i=1}^{3}\mu_{i}(q_{1},q_{2},q_{3})p_{i}\right)H,$

where $\mu_{i}(q_{1},q_{2},q_{3})$ are some functions.

We can ask whether there is a choice of $\beta_{ij}$ and
$\psi(q_{1},q_{2},q_{3})$ for which $\mu_{i}(q_{1},q_{2},q_{3})\equiv 0$, in
which case $F$ is a quadratic invariant. In fact, we have more structure. If
both $K$ and $F$ are invariants, then so are $\\{F,K\\}$, $\\{\\{F,K\\},K\\}$,
etc.

###  3.1 Using the involutions

It follows from the involutions of Table 2, that we can limit our choice of a
single first order invariant to just 3 elements of our algebra: $e_{1},h_{1}$
and $h_{4}$. Any other element is equivalent to one of these. Having fixed
this invariant we choose a quadratic conformal invariant (3.1) with specific
form

$\displaystyle F=X_{i}X_{j}+\psi(q_{1},q_{2},q_{3})H,$

where $X_{i}$ and $X_{j}$ are linear conformal invariants. At this stage we
still have some equivalent choices from involutions which leave our choice of
Killing vector invariant. For example, $e_{1}$ is fixed under the action of
$\iota_{23}$, while $h_{1}$ is fixed (up to sign) by them all and $h_{4}$ is
fixed (up to sign) by $\iota_{23}$ and $\iota_{ef}$.

###  3.2 Reductions to two dimensions

In Sections 4, 5 and 6 we present a number of systems in 3 degrees of freedom,
which have at least one first integral of degree one in momenta, as well as 3
independent quadratic integrals. These systems depend upon a number of
parameters and, in full generality, the Poisson algebras are complicated, but
by restricting parameters we can obtain systems with larger isometry algebras
(first order integrals) and this enables us to calculate the full set of
Poisson relations. The involutions play an important role in this calculation.

There is clearly an analogy with the Darboux–Koenigs systems in 2 degrees of
freedom, but we show that there is a much deeper connection. The increased
isometry algebra, obtained by restricting parameters, can be used to reduce
the system from 3 to 2 degrees of freedom. After this process, we find that
each of our systems reduces to one particular Darboux–Koenigs system, but with
the addition of a potential. We compare the resulting potentials with those
presented in [9, 10].

We use the particular method of reduction introduced in [5], referred to as
the “Kaluza–Klein reduction”, since it is essentially the reverse procedure to
the Kaluza–Klein extension. By adapting coordinates to a linear first
integral, we can reduce from 3 to 2 degrees of freedom. In principle, the
lower-dimensional system would possess vector potential terms, but in the
examples of this paper (and of [5]) these can be removed by gauge
transformation. Reduction to 2 degrees of freedom is not enough. We would like
our system to be at least completely integrable and, preferably, maximally
super-integrable. This requires the existence of integrals of the original
system, which commute with the linear integral. We would like to choose our
coordinate system so that the reduced kinetic energy is explicitly conformally
related to the standard Cartesian kinetic energy. We would like to be able to
construct any symmetries and a 6-dimensional subalgebra of conformal
symmetries of the corresponding $2$-dimensional metric from invariant
combinations of the conformal algebra (2.2). All of these things will be done
for the systems discussed in this paper.

Each additional Killing vector allows us to reduce to 2 dimensions in a
different way. From a given starting point, we always arrive at the same
Darboux–Koenigs metric, but the specific reduction gives us a particular
potential function. This allows us to relate different 2-dimensional
potentials by a “rotation” in 3 space.

##  4 Systems with $\boldsymbol{K=e_{1}}$

In the case $K=e_{1}$, we consider the Hamiltonian (2.1) of the form

$\displaystyle
H=\varphi(q_{2},q_{3})\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$ (4.1)

and exploit the following chains of length 3, found in the algebra:

$\displaystyle\\{\cdot,e_{1}\\}\colon\ f_{1}\mapsto h_{1}\mapsto-2e_{1}\mapsto
0\qquad\mbox{and}\qquad\\{\cdot,e_{1}\\}\colon\ f_{k}\mapsto 2h_{k}\mapsto
4e_{k}\mapsto 0,\qquad k=2,3.$

We use this to build 3 functions which satisfy $\\{\cdot,e_{1}\\}\colon
F_{1}\mapsto F_{2}\mapsto F_{3}\mapsto 0$. If we can find $\varphi$ and
$F_{1}$, such that $\\{F_{1},H\\}=0$, then, by the Jacobi identity, we
automatically have $\\{F_{2},H\\}=\\{F_{3},H\\}=0$.

###  4.1 $\boldsymbol{F_{1}=e_{2}f_{2}+\psi(q_{1},q_{2},q_{3})H}$

We consider a metric (4.1) with the chain of functions generated by
$F_{1}=e_{2}f_{2}+\psi(q_{1},q_{2},q_{3})H$, under the action of $e_{1}$:
$F_{2}=\frac{1}{2}\\{F_{1},e_{1}\\}$, $F_{3}=\frac{1}{2}\\{F_{2},e_{1}\\}$,
namely,

$\displaystyle F_{1}=e_{2}f_{2}+\psi H,\qquad
F_{2}=e_{2}h_{2}+\tfrac{1}{2}\psi_{q_{1}}H,\qquad
F_{3}=e_{2}^{2}+\tfrac{1}{4}\psi_{q_{1}q_{1}}H.$ (4.2)

The conditions $\\{F_{1},H\\}=\\{F_{3},e_{1}\\}=0$ lead to

$\displaystyle\varphi=\frac{q_{2}^{2}q_{3}^{2}}{\alpha q_{2}^{2}+\beta
q_{3}^{2}+\gamma
q_{2}^{2}q_{3}^{2}},\qquad\psi=-2\beta\frac{q_{1}^{2}+q_{3}^{2}}{q_{2}^{2}}+2\gamma
q_{2}^{2}.$ (4.3)

Since $H$, $e_{1}$, $F_{1}$, $F_{2}$, $F_{3}$ are independent first integrals
of $H$, this Hamiltonian is maximally super-integrable. We also have that $H$,
$e_{1}$, $F_{3}$ are in involution.

######  Remark 4.1 (the Stäckel transform).

The Hamiltonian (4.1), with $\varphi$ given by (4.3) can be regarded as the
Stäckel transform of the Euclidean kinetic energy with potential
$V=\varphi^{-1}$. Such potentials were classified in [4] and this example
corresponds to a reduction of $V_{[3,1,1]}$, given in [4, Section 5], by
equating $(q_{1},q_{2},q_{3})=(x,y,z)$ and
$(\alpha,\beta,\gamma)=(a_{3},a_{2},a_{5})$.

Notice that the resulting metric is invariant under the involution
$\iota_{23}$ (extended to include $\alpha\leftrightarrow\beta$), so we also
have the integrals $G_{i}=\iota_{23}(F_{i})$. These are not, of course,
functionally independent, but could be useful when considering the Poisson
algebra with the involution $\iota_{23}$. In fact, $G_{3}$ satisfies the
simple relation

$\displaystyle G_{3}+F_{3}=\gamma H-e_{1}^{2}.$

The Poisson algebra for the general case seems to be quite complicated.
Restricting the parameters introduces additional isometries. Setting any two
of the parameters to zero reduces the metric to a flat or constant curvature
case. However,

  1. 1.

Setting just $\alpha=0$, we have the isometry algebra $\langle
e_{1},e_{3},h_{3}\rangle$, satisfying Euclidean relations.

  2. 2.

Setting just $\beta=0$, we have the isometry algebra $\langle
e_{1},e_{2},h_{2}\rangle$, satisfying Euclidean relations. This case is
related to that of $\alpha=0$ by the action of the involution $\iota_{23}$.

  3. 3.

Setting just $\gamma=0$, we have the isometry algebra $\langle
e_{1},h_{1},f_{1}\rangle$, satisfying the relations of $\mathfrak{sl}(2)$.

####  4.1.1 The Poisson algebra of integrals when $\boldsymbol{\alpha=0}$

Now we have

$\displaystyle\varphi=\frac{q_{2}^{2}}{\beta+\gamma
q_{2}^{2}}\qquad\mbox{and}\qquad\psi=-2\beta\frac{q_{1}^{2}+q_{3}^{2}}{q_{2}^{2}}+2\gamma
q_{2}^{2},$ (4.4a) so we have lost the involutive symmetry (so no integrals
$G_{i}$), but gained additional Killing vectors, with isometry algebra
$\langle e_{1},e_{3},h_{3}\rangle$, satisfying the Euclidean relations
$\displaystyle\\{e_{1},e_{3}\\}=0,\qquad\\{e_{1},h_{3}\\}=-2e_{3},\qquad\\{e_{3},h_{3}\\}=2e_{1}.$
(4.4b)

If we define $K_{1}=e_{1}$, $K_{2}=e_{3}$, $K_{3}=h_{3}$, then
$\\{K_{i},F_{j}\\}$ are easily determined, requiring only the introduction of
one additional element $F_{4}=e_{2}h_{4}-\frac{4\beta q_{3}}{q_{2}^{2}}H$,
when calculating $\\{K_{2},F_{1}\\}$ and $\\{K_{3},F_{2}\\}$. The formula for
$F_{4}$ is easily determined from Table 1, by considering
$\\{e_{3},e_{2}f_{2}\\}=-e_{2}h_{4}$. The cubic expressions
$\\{F_{i},F_{j}\\}$ can all be written in terms of our linear and quadratic
functions. The full set of Poisson relations is given in Table 3. Adding $H$,
we have an 8-dimensional algebra obeying constraints:

$\displaystyle I_{1}=K_{1}^{2}+K_{2}^{2}+F_{3}-\gamma H=0,\qquad
I_{2}=K_{1}F_{4}-2K_{2}F_{2}+2K_{3}F_{3}=0,$ $\displaystyle
I_{3}=4F_{2}^{2}-8F_{1}F_{3}+F_{4}^{2}-16\beta\gamma H^{2}=0.$ (4.5)

######  Remark 4.2.

Notice that the first two expressions are related to ${\cal C}_{2}^{(he)}$ and
${\cal C}_{4}^{(he)}$ respectively (see (2.3g)). We could, in fact, use
$I_{1}=0$ to eliminate $F_{3}$ and just write a $6\times 6$ table, but $F_{3}$
was an integral part of the definition of this case.

Table 3: The Poisson algebra of first integrals when $\alpha=0$.

| $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$
---|---|---|---|---|---|---|---
$K_{1}$ | $0$ | $0$ | $-2K_{2}$ | $-2F_{2}$ | $-2F_{3}$ | $0$ | $0$
$K_{2}$ |  | $0$ | $2K_{1}$ | $-F_{4}$ | $0$ | $0$ | $-4F_{3}$
$K_{3}$ |  |  | $0$ | $0$ | $F_{4}$ | $0$ | $-4F_{2}$
$F_{1}$ |  |  |  | $0$ | $-2(2K_{1}F_{1}+K_{3}F_{4})$ | $-2(2K_{1}F_{2}+K_{2}F_{4})$ | $8(K_{3}F_{2}-K_{2}F_{1})$
$F_{2}$ |  |  |  |  | $0$ | $-4K_{1}F_{3}$ | $2(2K_{2}F_{2}+2K_{3}F_{3}-K_{1}F_{4})$
$F_{3}$ |  |  |  |  |  | $0$ | $8K_{2}F_{3}$
$F_{4}$ |  |  |  |  |  |  | $0$

Since $H$ is invariant under the action of $\iota_{13}$, the Poisson algebra
has this symmetry. The action of $\iota_{13}$ is summarised in Table 4.

Table 4: The action of $\iota_{13}$ on $H$ and its integrals. | $H$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$
---|---|---|---|---|---|---|---|---
$\iota_{13}$ | $H$ | $K_{2}$ | $K_{1}$ | $-K_{3}$ | $F_{1}$ | $\frac{1}{2}F_{4}$ | $F_{3}$ | $2F_{2}$

Several of the entries in Table 3 are related through this involution. We also
have that the constraints (4.5) transform as

$\displaystyle\iota_{13}\colon\
(I_{1},I_{2},I_{3})\mapsto(I_{1},-I_{2},I_{3}).$

####  4.1.2 The Poisson algebra of integrals when $\boldsymbol{\gamma=0}$

When $\gamma=0$, the Hamiltonian $H$ is invariant under the action of
$\iota_{ef}$ (as is the $3$-dimensional isometry algebra). We define
$K_{1}=e_{1}$, $K_{2}=h_{1}$, $K_{3}=f_{1}$, with $F_{1}$, $F_{2}$, $F_{3}$
given by (4.2). The action of $\iota_{ef}$ on $F_{2}$ and $F_{3}$ gives us two
more quadratic elements, which close the algebra:

$\displaystyle F_{4}=f_{2}h_{2}-\frac{4\beta
q_{1}\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}}{q_{2}^{2}}H,\qquad
F_{5}=f_{2}^{2}-\frac{4\beta\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}^{2}}{q_{2}^{2}}H.$
(4.6)

The action of $\iota_{ef}$ is summarised in Table 5.

Table 5: The action of $\iota_{ef}$ on $H$ and its integrals. | $H$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$ | $F_{5}$
---|---|---|---|---|---|---|---|---|---
$\iota_{ef}$ | $H$ | $-K_{3}$ | $-K_{2}$ | $-K_{1}$ | $F_{1}$ | $-\frac{1}{2}F_{4}$ | $\frac{1}{4}F_{5}$ | $-2F_{2}$ | $4F_{3}$

A consequence of this is that we can deduce the entries for
$\\{K_{i},F_{4}\\}$ and $\\{K_{i},F_{5}\\}$ (in Table 6) from those of
$\\{K_{j},F_{2}\\}$ and $\\{K_{j},F_{3}\\}$. The entries $\\{F_{3},F_{4}\\}$
and $\\{F_{2},F_{5}\\}$ are similarly related. Some of the longer entries are
labelled $P_{ij}$ and listed below.

Table 6: The Poisson algebra of first integrals when $\gamma=0$.

| $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$ | $F_{5}$
---|---|---|---|---|---|---|---|---
$K_{1}$ | 0 | $2K_{1}$ | $-K_{2}$ | $-2F_{2}$ | $-2F_{3}$ | $0$ | $-6F_{1}-8{\cal C}_{K}+4(2\alpha+\beta)H$ | $-4F_{4}$
$K_{2}$ |  | 0 | $2K_{3}$ | $0$ | $-2F_{2}$ | $-4F_{3}$ | $2F_{4}$ | $4F_{5}$
$K_{3}$ |  |  | 0 | $-F_{4}$ | $-3F_{1}-4{\cal C}_{K}+2(2\alpha+\beta)H$ | $-2F_{2}$ | $-F_{5}$ | 0
$F_{1}$ |  |  |  | 0 | $8(K_{3}F_{3}+\beta K_{1}H)$ | $4K_{2}F_{3}$ | $4(K_{1}F_{5}+4\beta K_{3}H)$ | $-4K_{2}F_{5}$
$F_{2}$ |  |  |  |  | 0 | $-4K_{1}F_{3}$ | $P_{24}$ | $P_{25}$
$F_{3}$ |  |  |  |  |  | 0 | $P_{34}$ | $P_{35}$
$F_{4}$ |  |  |  |  |  |  | 0 | $-8K_{3}F_{5}$
$F_{5}$ |  |  |  |  |  |  |  | 0

The longer entries are given by

$\displaystyle P_{24}=8K_{2}(-F_{1}-2{\cal C}_{K}+(2\alpha+\beta)H),\qquad
P_{25}=-4(K_{2}F_{4}+2K_{3}F_{1}+4\beta K_{3}H),$ $\displaystyle
P_{34}=-4(K_{2}F_{2}-K_{1}F_{1}-2\beta K_{1}H),\qquad
P_{35}=-8K_{2}(F_{1}+2\beta H),$

where ${\cal C}_{K}=K_{1}K_{3}+\frac{1}{4}K_{2}^{2}$ is the Casimir function
of the isometry algebra.

Adding $H$, we have a 9-dimensional algebra obeying the constraints:

$\displaystyle I_{1}=K_{1}F_{1}+K_{2}F_{2}-2K_{3}F_{3}-2\beta K_{1}H=0,$
$\displaystyle I_{2}=K_{1}F_{5}+K_{2}F_{4}-2K_{3}F_{1}+4\beta K_{3}H=0,$
$\displaystyle
I_{3}=F_{1}F_{2}-F_{3}F_{4}+2K_{2}(K_{1}F_{1}+K_{2}F_{2}-2K_{3}F_{3})-2\beta
HF_{2}=0,$ $\displaystyle
I_{4}=F_{1}F_{4}-F_{2}F_{5}+2K_{2}(K_{1}F_{5}+K_{2}F_{4}-2K_{3}F_{1})-2\beta
HF_{4}=0.$

######  Remark 4.3.

Notice that the first two expressions are related to the Casimir ${\cal
C}_{4}^{(0)}$, of the algebra $\mathfrak{g}^{(0)}$ of Section 2.1. Under the
action of $\iota_{ef}$, they obey

$\displaystyle(I_{1},I_{2},I_{3},I_{4})\mapsto\big{(}\tfrac{1}{2}I_{2},2I_{1},-\tfrac{1}{2}I_{4},-2I_{3}\big{)}.$

###  4.2 $\boldsymbol{F_{1}=e_{1}h_{2}+\psi_{1}(q_{1},q_{2},q_{3})H}$,
$\boldsymbol{G_{1}=e_{1}h_{3}+\psi_{2}(q_{1},q_{2},q_{3})H}$

We again take the Hamiltonian (4.1), with first integral $K=e_{1}$, and
consider the chains of functions generated by

$\displaystyle F_{1}=e_{1}h_{2}+\psi_{1}(q_{1},q_{2},q_{3})H,\qquad$
$\displaystyle
F_{2}=\tfrac{1}{2}\\{F_{1},e_{1}\\}=e_{1}e_{2}+\tfrac{1}{2}\psi_{1\,q_{1}}H,$
(4.7a) $\displaystyle G_{1}=e_{1}h_{3}+\psi_{2}(q_{1},q_{2},q_{3})H,\qquad$
$\displaystyle
G_{2}=\tfrac{1}{2}\\{G_{1},e_{1}\\}=e_{1}e_{3}+\tfrac{1}{2}\psi_{2\,q_{1}}H,$
(4.7b) related through the involution $\iota_{23}$. The conditions
$\\{F_{1},H\\}=0$ and $\\{G_{1},H\\}=0$ give the functions
$\displaystyle\varphi=\frac{1}{\alpha q_{2}+\beta
q_{3}+\gamma},\qquad\psi_{1}=-\frac{\alpha}{2}q_{1}^{2},\qquad\psi_{2}=-\frac{\beta}{2}q_{1}^{2},$
(4.7c)

so $\\{F_{2},e_{1}\\}=-\frac{\alpha}{2}H$ and
$\\{G_{2},e_{1}\\}=-\frac{\beta}{2}H$. The integrals $H$, $F_{1}$, $F_{2}$,
$G_{1}$, $G_{2}$ are functionally independent.

######  Remark 4.4 (the Stäckel transform).

Again, the Hamiltonian (4.1), but with $\varphi$ given by (4.7c) can be
regarded as the Stäckel transform of the Euclidean kinetic energy with
potential $V=\varphi^{-1}$. In the classification of [4], this example
corresponds to a reduction of $V_{[0]}$, given in [4, Section 5], by equating
$(q_{1},q_{2},q_{3})=(z,x,y)$ and $(\alpha,\beta,\gamma)=(a_{1},a_{2},a_{5})$.

This solution has a $3$-dimensional isometry algebra

$\displaystyle K_{1}=e_{1},\qquad K_{2}=\beta e_{2}-\alpha e_{3},\qquad
K_{3}=\beta h_{2}-\alpha h_{3},$ (4.8a) satisfying
$\displaystyle\\{K_{1},K_{2}\\}=0,\qquad\\{K_{1},K_{3}\\}=-2K_{2},\qquad\\{K_{2},K_{3}\\}=2\big{(}\alpha^{2}+\beta^{2}\big{)}K_{1},$
(4.8b) with Casimir $\displaystyle{\cal
C}_{K}=\big{(}\alpha^{2}+\beta^{2}\big{)}K_{1}^{2}+K_{2}^{2}.$ (4.8c)

We then find that

$\displaystyle\alpha G_{1}-\beta
F_{1}+K_{1}K_{3}=0,\qquad\mbox{and}\qquad\alpha G_{2}-\beta
F_{2}+K_{1}K_{2}=0,$

with the integrals $H$, $K_{1}$, $K_{2}$, $K_{3}$, $F_{1}$ being functionally
independent.

Table 7: The Poisson algebra of first integrals of (4.1), with (4.7c).

| $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$ | $F_{5}$ | $F_{6}$
---|---|---|---|---|---|---|---|---|---
$K_{1}$ | $0$ | $0$ | $-2K_{2}$ | $-2F_{2}$ | $\frac{1}{2}\alpha H$ | $-2F_{4}$ | $0$ | $0$ | $0$
$K_{2}$ |  | $0$ | $2\delta K_{1}$ | $2\beta K_{1}^{2}$ | $0$ | $-4\delta F_{2}+8\beta K_{1}K_{2}$ | $\alpha\delta H$ | $-\alpha\beta H$ | $\bar{\alpha}(\delta F_{4}-2\beta K_{2}^{2})$
$K_{3}$ |  |  | $0$ | $F_{3}$ | $F_{4}\\!-\\!2\beta K_{1}^{2}$ | $4\beta K_{1}K_{3}-8\delta F_{1}-4\alpha F_{6}$ | $-4\delta F_{2}-4\beta K_{1}K_{2}$ | $4(\beta F_{2}\\!+\\!K_{1}K_{2})$ | $\bar{\alpha}(\delta F_{3}-2\beta K_{2}K_{3})$
$F_{1}$ |  |  |  | $0$ | $-2K_{1}F_{5}$ | $4K_{3}F_{5}-8\beta K_{1}F_{1}$ | $-4K_{1}(K_{1}K_{2}\\!+\\!\beta F_{2})$ | $8K_{1}F_{2}$ | $\bar{\alpha}K_{1}(\beta F_{3}-2K_{2}K_{3})$
$F_{2}$ |  |  |  |  | $0$ | $4(K_{1}^{2}K_{2}\\!-\\!\beta K_{1}F_{2}\\!+\\!K_{2}F_{5})$ | $\alpha\beta K_{1}H$ | $-2\alpha K_{1}H$ | $\bar{\alpha}K_{1}(\beta F_{4}-2K_{2}^{2})$
$F_{3}$ |  |  |  |  |  | $0$ | $P_{34}$ | $P_{35}$ | $P_{36}$
$F_{4}$ |  |  |  |  |  |  | $0$ | $-2\alpha K_{2}H$ | $4\bar{\alpha}K_{2}(\delta(K_{1}^{2}-F_{5})-K_{2}^{2})$
$F_{5}$ |  |  |  |  |  |  |  | $0$ | $2\bar{\alpha}K_{2}(F_{4}\\!+\\!2\beta F_{5}\\!-\\!2\beta K_{1}^{2})$
$F_{6}$ |  |  |  |  |  |  |  |  | $0$

We can use $\langle K_{1},K_{2},K_{3},F_{1},F_{2}\rangle$ to generate a
Poisson algebra. The action of $K_{i}$ on $F_{i}$ and the Poisson bracket
$\\{F_{1},F_{2}\\}$ require the introduction of four more quadratic elements:

$\displaystyle F_{3}=2K_{2}h_{2}-\alpha K_{1}h_{4}-2\alpha q_{1}(\beta
q_{2}-\alpha q_{3})H,\qquad F_{4}=2e_{2}K_{2}-\alpha(\beta q_{2}-\alpha
q_{3})H,$ $\displaystyle F_{5}=e_{1}^{2}-e_{2}^{2}+\alpha q_{2}H,\qquad
F_{6}=K_{2}h_{4}+(\beta q_{2}-\alpha q_{3})^{2}H,$

satisfying the relations given in Table 7. In this table we have
$\delta=\alpha^{2}+\beta^{2}$, $\bar{\alpha}=\frac{2}{\alpha}$ and

$\displaystyle P_{34}=4K_{1}\big{(}\beta
F_{4}-2\alpha^{2}K_{1}^{2}-6K_{2}^{2}+2\alpha^{2}\gamma H\big{)},$
$\displaystyle P_{35}=8K_{1}\big{(}3F_{4}+\beta F_{5}-\beta
K_{1}^{2}\big{)}-24K_{2}F_{2}-8\alpha K_{3}H,$ $\displaystyle
P_{36}=\frac{4}{\alpha}K_{3}\big{(}\beta
F_{4}-2\alpha^{2}K_{1}^{2}-4K_{2}^{2}+2\alpha^{2}\gamma
H\big{)}+\frac{4\beta}{\alpha}K_{2}F_{3}.$

This 10-dimensional algebra (including $H$ itself) is constrained by the
following relations

$\displaystyle I_{1}=K_{1}F_{3}+2K_{3}F_{2}-4K_{2}F_{1}=0,\qquad I_{2}=2\alpha
K_{1}F_{6}+K_{2}F_{3}-K_{3}F_{4}=0,$ $\displaystyle I_{3}=\alpha
K_{3}H+4K_{2}F_{2}-2K_{1}F_{4}=0,\qquad
I_{4}=8K_{1}K_{2}F_{5}-8K_{1}^{3}K_{2}+4F_{2}F_{4}+\alpha F_{3}H=0,$
$\displaystyle
I_{5}=\big{(}2\alpha^{2}+\beta^{2}\big{)}K_{1}^{2}+K_{2}^{2}-\big{(}\alpha^{2}+\beta^{2}\big{)}F_{5}-\beta
F_{4}-\alpha^{2}\gamma H=0.$

##  5 Systems with $\boldsymbol{K=h_{1}}$

In the case $K=h_{1}$, we consider the Hamiltonian (2.1) of the form

$\displaystyle
H=\varphi(q_{1},q_{2},q_{3})\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}=q_{1}^{2}\Phi\left(\frac{q_{2}}{q_{1}},\frac{q_{3}}{q_{1}}\right)\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$
(5.1)

which is the most general form commuting with $h_{1}$. This is also generally
invariant with respect to the involution $\iota_{ef}$, so we choose our
functions $F_{i}$ to have this invariance, noting that
$\iota_{ef}\colon(e_{2},h_{2},f_{2})\mapsto\big{(}{-}\frac{1}{2}f_{2},h_{2},-2e_{2}\big{)}$.

###  5.1 $\boldsymbol{F_{1}=e_{2}h_{2}+\psi_{1}(q_{1},q_{2},q_{3})H}$,
$\boldsymbol{F_{2}=f_{2}h_{2}+\psi_{2}(q_{1},q_{2},q_{3})H}$,
$\boldsymbol{F_{3}=h_{2}^{2}+\psi_{3}(q_{1},q_{2},q_{3})H}$

With the Hamiltonian (5.1), with first integral $K=h_{1}$, we consider
functions of the form

$\displaystyle F_{1}=e_{2}h_{2}+\psi_{1}(q_{1},q_{2},q_{3})H,\qquad
F_{2}=f_{2}h_{2}+\psi_{2}(q_{1},q_{2},q_{3})H,$ $\displaystyle
F_{3}=h_{2}^{2}+\psi_{3}(q_{1},q_{2},q_{3})H,$

and require the conditions $\\{F_{i},H\\}=0$ to find

$\displaystyle\varphi=\frac{q_{2}^{2}q_{3}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}}{\big{(}\alpha
q_{2}^{2}+\beta q_{3}^{2}\big{)}\sqrt{q_{1}^{2}+q_{2}^{2}}+\gamma
q_{1}q_{3}^{2}},\qquad\psi_{1}=-\frac{\gamma\big{(}2q_{1}^{2}+q_{2}^{2}\big{)}+2\beta
q_{1}\sqrt{q_{1}^{2}+q_{2}^{2}}}{q_{2}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}},$ (5.2)
$\displaystyle\psi_{2}=2\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}\psi_{1},\qquad\psi_{3}=-\frac{4q_{1}\big{(}\beta
q_{1}+\gamma\sqrt{q_{1}^{2}+q_{2}^{2}}\big{)}}{q_{2}^{2}}.$

If we extend $\iota_{ef}$ to act on the parameters:
$(\alpha,\beta,\gamma)\mapsto(\alpha,\beta,-\gamma)$ then

$\displaystyle(H,F_{1},F_{2},F_{3})\mapsto\big{(}H,-\tfrac{1}{2}F_{2},-2F_{1},F_{3}\big{)}.$

######  Remark 5.1 (the Stäckel transform).

Again, the Hamiltonian (5.1), with $\varphi$ given by (5.2) can be regarded as
the Stäckel transform of the Euclidean kinetic energy with potential
$V=\varphi^{-1}$. In the classification of [4], this example corresponds to a
reduction of system iii, given in [4, Section 7], by equating
$(q_{1},q_{2},q_{3})=(z,y,x)$ and $(\alpha,\beta,\gamma)=(a_{3},a_{4},a_{5})$.

The integrals $H$, $K$, $F_{1}$, $F_{2}$, $F_{3}$ are functionally
independent. The simpler Poisson relations are

$\displaystyle\\{F_{1},K\\}=2F_{1},\qquad\\{F_{2},K\\}=-2F_{2},\qquad\\{F_{3},K\\}=0,$
$\displaystyle\\{F_{1},F_{2}\\}=4K(2\beta H-F_{3}).$ (5.3)

Again, we can restrict the parameters to increase the algebra of isometries:

  1. 1.

By setting $\alpha=0$, we have the isometry algebra $\langle
e_{3},h_{1},f_{3}\rangle$, satisfying the relations of $\mathfrak{sl}(2)$.

  2. 2.

By setting $\beta=0$, we just have the single isometry $h_{1}$.

  3. 3.

By setting $\gamma=0$, the metric reduces to that of Section 4.1 (with
$\gamma=0$) so has the isometry algebra $\langle e_{1},h_{1},f_{1}\rangle$.

####  5.1.1 The Poisson algebra of integrals when $\boldsymbol{\alpha=0}$

Now we have

$\displaystyle\varphi=\frac{q_{2}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}}{\beta\sqrt{q_{1}^{2}+q_{2}^{2}}+\gamma
q_{1}},$

with $\psi_{i}$ as before, and gain additional Killing vectors, with isometry
algebra $\langle h_{1},e_{3},f_{3}\rangle$.

Table 8: The Poisson algebra of first integrals when $\alpha=0$. | $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$
---|---|---|---|---|---|---|---
$K_{1}$ | $0$ | $-2K_{2}$ | $2K_{3}$ | $-2F_{1}$ | $2F_{2}$ | $0$ | $0$
$K_{2}$ |  | $0$ | $-2K_{1}$ | $0$ | $F_{4}$ | $0$ | $4F_{1}$
$K_{3}$ |  |  | $0$ | $F_{4}$ | $0$ | $0$ | $4F_{2}$
$F_{1}$ |  |  |  | $0$ | $4K_{1}(2\beta H-F_{3})$ | $2(K_{2}F_{4}-2K_{1}F_{1})$ | $8K_{2}(2\beta H-F_{3})$
$F_{2}$ |  |  |  |  | $0$ | $2(K_{3}F_{4}+2K_{1}F_{2})$ | $8K_{3}(2\beta H-F_{3})$
$F_{3}$ |  |  |  |  |  | $0$ | $-8(K_{2}F_{2}+K_{3}F_{1})$
$F_{4}$ |  |  |  |  |  |  | $0$

If we define $K_{1}=h_{1}$, $K_{2}=e_{3}$, $K_{3}=f_{3}$, then
$\\{K_{i},F_{j}\\}$ are easily determined, requiring only the introduction of
one additional element $F_{4}=-h_{2}h_{4}-4q_{3}\psi_{1}H$, when calculating
$\\{K_{2},F_{2}\\}$ and $\\{K_{3},F_{1}\\}$. The formula for $F_{4}$ is easily
determined from Table 1, by considering $\\{e_{3},f_{2}h_{2}\\}=-h_{2}h_{4}$.
The cubic expressions $\\{F_{i},F_{j}\\}$ can all be written in terms of our
linear and quadratic functions. The full set of Poisson relations is given in
Table 8. Since $H$ is invariant under the action of $\iota_{ef}$, the Poisson
algebra has this symmetry. The action of $\iota_{ef}$ is summarised in Table
9. Several of the entries in Table 8 are related through this involution.

Table 9: The action of $\iota_{ef}$ on $H$ and its integrals. | $H$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$
---|---|---|---|---|---|---|---|---
$\iota_{ef}$ | $H$ | $-K_{1}$ | $-\frac{1}{2}K_{3}$ | $-2K_{2}$ | $-\frac{1}{2}F_{2}$ | $-2F_{1}$ | $F_{3}$ | $F_{4}$

Adding $H$, we have an 8-dimensional algebra obeying the constraints:

$\displaystyle I_{1}=K_{1}^{2}+2K_{2}K_{3}+F_{3}-4\beta H=0,\qquad
I_{2}=K_{1}F_{4}-2K_{2}F_{2}+2K_{3}F_{1}=0.$

######  Remark 5.2.

Both of these are Casimirs of the algebra, with the first extending the
Casimir of the isometry algebra and the second being related to ${\cal
C}_{4}^{(13)}$ of (2.3c).

##  6 Systems with $\boldsymbol{K=h_{4}}$

In the case $K=h_{4}$, we consider the Hamiltonian (2.1) of the form

$\displaystyle
H=\varphi(q_{1},q_{2},q_{3})\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}=\Phi\big{(}q_{1},q_{2}^{2}+q_{3}^{2}\big{)}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$
(6.1)

which is the most general form commuting with $h_{4}$. This is also generally
invariant with respect to the involution $\iota_{23}$, so we choose our
functions $F_{i}$ to have this invariance.

###  6.1 $\boldsymbol{F_{1}=e_{1}^{2}+\psi_{1}(q_{1},q_{2},q_{3})H}$,
$\boldsymbol{F_{2}=e_{1}e_{2}+\psi_{2}(q_{1},q_{2},q_{3})H}$,
$\boldsymbol{F_{3}=e_{1}e_{3}+\psi_{3}(q_{1},q_{2},q_{3})H}$

With the Hamiltonian (6.1), with first integral $K=h_{4}$, we consider
functions of the form

$\displaystyle F_{1}=e_{1}^{2}+\psi_{1}(q_{1},q_{2},q_{3})H,\quad
F_{2}=e_{1}e_{2}+\psi_{2}(q_{1},q_{2},q_{3})H,$ $\displaystyle
F_{3}=e_{1}e_{3}+\psi_{3}(q_{1},q_{2},q_{3})H,$

and require the conditions $\\{F_{i},H\\}=0$ to find

$\displaystyle
H=\varphi(q_{1},q_{2},q_{3})\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}=\frac{1}{\alpha\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}+\beta
q_{1}+\gamma}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$ (6.2)
$\displaystyle\psi_{1}=-\alpha q_{1}^{2}-\beta
q_{1},\qquad\psi_{2}=-\left(\alpha
q_{1}+\frac{\beta}{2}\right)q_{2},\qquad\psi_{3}=-\left(\alpha
q_{1}+\frac{\beta}{2}\right)q_{3},$

which clearly transform as $(H,F_{1},F_{2},F_{3})\mapsto(H,F_{1},F_{3},F_{2})$
under $\iota_{23}$.

######  Remark 6.1 (the Stäckel transform).

The Hamiltonian (6.2), can be regarded as the Stäckel transform of the
Euclidean kinetic energy with potential $V=\varphi^{-1}$. In the
classification of [4], this example corresponds to a reduction of $V_{[0]}$,
given in [4, Section 5], by equating $(q_{1},q_{2},q_{3})=(x,y,z)$ and
$(\alpha,\beta,\gamma)=(a_{4},a_{1},a_{5})$.

This $H$ has first order symmetries

$\displaystyle K_{1}=h_{4},\qquad K_{2}=\alpha h_{2}+\beta e_{2},\qquad
K_{3}=\alpha h_{3}+\beta e_{3},$ (6.3a) satisfying
$\displaystyle\\{K_{1},K_{2}\\}=-4K_{3},\qquad\\{K_{1},K_{3}\\}=4K_{2},\qquad\\{K_{3},K_{2}\\}=\alpha^{2}K_{1}.$
(6.3b) The action of $K_{i}$ on $F_{2}$ and $F_{3}$ introduces three new
quadratic elements to our algebra: $\displaystyle F_{4}=e_{2}^{2}-\alpha
q_{2}^{2}H,\qquad F_{5}=e_{2}e_{3}-\alpha q_{2}q_{3}H,\qquad
F_{6}=e_{3}^{2}-\alpha q_{3}^{2}H,$ (6.3c)

giving 9 first integrals, which satisfies the Poisson relations of Table 10.
Adding $H$ gives us a 10-dimensional algebra obeying the constraints:

$\displaystyle I_{1}=\alpha K_{1}F_{2}-2K_{2}F_{5}+2K_{3}F_{4}=0,\qquad
I_{2}=\alpha K_{1}F_{3}-2K_{2}F_{6}+2K_{3}F_{5}=0,$ $\displaystyle
I_{3}=4\alpha K_{1}F_{1}-\beta^{2}K_{1}H-8K_{2}F_{3}+8K_{3}F_{2}=0,$
$\displaystyle I_{4}=8F_{3}F_{4}-8F_{2}F_{5}+K_{1}K_{2}H=0,\qquad
I_{5}=8F_{3}F_{5}-8F_{2}F_{6}+K_{1}K_{3}H=0.$

######  Remark 6.2.

The first three of these are related to the Casimir ${\cal C}_{4}^{(he)}$ of
(2.3g). The elements $K_{i}$ and $F_{i}$ are built from the algebra
$\mathfrak{g}^{(he)}$ of Section 2.1. We also have a linear relation
$F_{1}+F_{4}+F_{6}=\gamma H$, which could be used to eliminate $F_{6}$, but
this would complicate Table 11.

Table 10: The Poisson algebra of first integrals of (6.2).

| $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$ | $F_{5}$ | $F_{6}$
---|---|---|---|---|---|---|---|---|---
$K_{1}$ | $0$ | $-4K_{3}$ | $4K_{2}$ | $0$ | $-4F_{3}$ | $4F_{2}$ | $-8F_{5}$ | $4(F_{4}-F_{6})$ | $8F_{5}$
$K_{2}$ |  | $0$ | $-\alpha^{2}K_{1}$ | $4\alpha F_{2}$ | $2\alpha(F_{4}-F_{1})+\frac{1}{2}\beta^{2}H$ | $2\alpha F_{5}$ | $-4\alpha F_{2}$ | $-2\alpha F_{3}$ | $0$
$K_{3}$ |  |  | $0$ | $4\alpha F_{3}$ | $2\alpha F_{5}$ | $2\alpha(F_{6}-F_{1})+\frac{1}{2}\beta^{2}H$ | $0$ | $-2\alpha F_{2}$ | $-4\alpha F_{3}$
$F_{1}$ |  |  |  | $0$ | $-K_{2}H$ | $-K_{3}H$ | $0$ | $0$ | $0$
$F_{2}$ |  |  |  |  | $0$ | $\frac{1}{4}\alpha K_{1}H$ | $-K_{2}H$ | $-\frac{1}{2}K_{3}H$ | $0$
$F_{3}$ |  |  |  |  |  | $0$ | $0$ | $-\frac{1}{2}K_{2}H$ | $-K_{3}H$
$F_{4}$ |  |  |  |  |  |  | $0$ | $\frac{1}{2}\alpha K_{1}H$ | $0$
$F_{5}$ |  |  |  |  |  |  |  | $0$ | $\frac{1}{2}\alpha K_{1}H$
$F_{6}$ |  |  |  |  |  |  |  |  | $0$

The Hamiltonian (6.2) is invariant under the action of $\iota_{23}$, as is
this algebra (see Table 11).

Table 11: The action of $\iota_{23}$ on $H$ and its integrals. | $H$ | $K_{1}$ | $K_{2}$ | $K_{3}$ | $F_{1}$ | $F_{2}$ | $F_{3}$ | $F_{4}$ | $F_{5}$ | $F_{6}$
---|---|---|---|---|---|---|---|---|---|---
$\iota_{23}$ | $H$ | $-K_{1}$ | $K_{3}$ | $K_{2}$ | $F_{1}$ | $F_{3}$ | $F_{2}$ | $F_{6}$ | $F_{5}$ | $F_{4}$

Several of the entries of Table 10 (such as $\\{K_{2},F_{2}\\}$ and
$\\{K_{3},F_{3}\\}$) are related through this involution.

##  7 Reductions of the system of Section 4.1

We now consider reductions of the Hamiltonian

$\displaystyle H=\frac{q_{2}^{2}q_{3}^{2}}{\alpha q_{2}^{2}+\beta
q_{3}^{2}+\gamma
q_{2}^{2}q_{3}^{2}}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$

associated with the restrictions of Sections 4.1.1 and 4.1.2. We adapt
coordinates to the appropriate invariant, rendering it as $P_{3}$. The
2-dimensional reduction is then $P_{3}=\mu$, but we keep the label $P_{3}$ so
that we can later transform (in the 3-dimensional context) between these
reductions.

In this first section on reductions we give more detailed calculations. The
same methods are used in later sections, but we omit the details.

###  7.1 Reductions when $\boldsymbol{\alpha=0}$

For the case with (4.4a) and isometry algebra (4.4b) we use respectively
$K_{2}=e_{3}$ and $K_{3}=h_{3}$ to reduce to 2 dimensions.

####  7.1.1 Reduction using the isometry $\boldsymbol{K_{2}=e_{3}}$

Our original coordinate system $(q_{1},q_{2},q_{3})$ is already adapted to
$e_{3}=p_{3}=P_{3}$. Since $\\{F_{2},e_{3}\\}=\\{F_{3},e_{3}\\}=0$, we can
reduce these integrals in this case, with

$\displaystyle H=\frac{q_{2}^{2}}{\beta+\gamma
q_{2}^{2}}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+\frac{q_{2}^{2}p_{3}^{2}}{\beta+\gamma
q_{2}^{2}},\qquad F_{2}=e_{2}h_{2}-\frac{2\beta q_{1}}{q_{2}^{2}}H,\qquad
F_{3}=e_{2}^{2}-\frac{\beta}{q_{2}^{2}}H,$ (7.1)

still satisfying $\\{H,e_{1}\\}=0$, $\\{F_{2},e_{1}\\}=2F_{3}$,
$\\{F_{3},e_{1}\\}=0$. The system (7.1) is just the Darboux–Koenigs system of
type $D_{2}$ with the potential of “Type D” in [9].

By considering linear and quadratic conformal elements (in 3D) that Poisson
commute with $e_{3}$, we can reduce the $10$-dimensional conformal algebra to
the required $6$-dimensional algebra. For example

$\displaystyle e_{1},\ e_{2},\ h_{2},\ e_{1}h_{1}-e_{3}h_{3},\
h_{3}^{2}-4e_{1}f_{1},\ \mbox{and}\ h_{4}^{2}-8e_{2}f_{2},$

give us the standard 6-dimensional subalgebra of conformal symmetries of 2D
space:

$\displaystyle T_{1}=p_{1},\qquad T_{2}=p_{2},\qquad
T_{3}=q_{2}p_{1}-q_{1}p_{2},\qquad T_{4}=q_{1}p_{1}+q_{2}p_{2},$
$\displaystyle
T_{5}=\big{(}q_{1}^{2}-q_{2}^{2}\big{)}p_{1}+2q_{1}q_{2}p_{2},\qquad
T_{6}=2q_{1}q_{2}p_{1}+\big{(}q_{2}^{2}-q_{1}^{2}\big{)}p_{2}.$

######  Remark 7.1 (the reduction of quadratic elements).

In the case of quadratic expressions we set $p_{3}=0$, with only the parts
involving $p_{1}^{2}$, $p_{1}p_{2}$, $p_{2}^{2}$ representing a quadratic
conformal element of the 2-dimensional kinetic energy. From the 3 linear and 3
quadratic expressions, we can deduce 6 linear elements. For example,

$\displaystyle e_{1}=p_{1},\qquad
e_{1}h_{1}-e_{3}h_{3}=-2p_{1}(q_{1}p_{1}+q_{2}p_{2})-2q_{1}p_{3}^{2}$

give us $T_{1}$ and $T_{4}$.

The functions $F_{2}$ and $F_{3}$ in (7.1) can be written in terms of these,
with $T_{1}$ corresponding to the single Killing vector:

$\displaystyle F_{2}=-2T_{2}T_{3}-2\beta\frac{q_{1}}{q_{2}^{2}}H,\qquad
F_{3}=T_{2}^{2}-\frac{\beta}{q_{2}^{2}}H,$

satisfying

$\displaystyle\\{T_{1},F_{2}\\}=-2F_{3},\qquad\\{T_{1},F_{3}\\}=0,\qquad\\{F_{2},F_{3}\\}=4T_{1}\big{(}T_{1}^{2}+\mu^{2}-\gamma
H\big{)}.$

####  7.1.2 Reduction using the isometry $\boldsymbol{K_{3}=h_{3}}$

We adapt coordinates to $h_{3}$:

$\displaystyle Q_{1}=\sqrt{q_{1}^{2}+q_{3}^{2}},\qquad Q_{2}=q_{2},\qquad
Q_{3}=-\tfrac{1}{2}\arctan\left(\frac{q_{1}}{q_{3}}\right)\quad\Rightarrow\quad
h_{3}=P_{3},$ (7.2)

and

$\displaystyle H=\frac{Q_{2}^{2}}{\beta+\gamma
Q_{2}^{2}}\left(P_{1}^{2}+P_{2}^{2}+\frac{P_{3}^{2}}{4Q_{1}^{2}}\right),$
(7.3)

which is just the Darboux–Koenigs system of type $D_{2}$ with the potential of
“Type B” (parameter $b_{3}$) in [9].

By considering linear and quadratic conformal elements (in 3D) that Poisson
commute with $h_{3}$, we can reduce the $10$-dimensional conformal algebra to
the required $6$-dimensional algebra. For example

$\displaystyle h_{1},\ e_{2},\ f_{2},\ e_{1}^{2}+e_{3}^{2},\
4f_{1}^{2}+f_{3}^{2},\ \mbox{and}\ 2e_{1}h_{2}+e_{3}h_{4},$

give us (setting $P_{3}=0$) the standard conformal algebra for 2D space:

$\displaystyle T_{1}=P_{1},\qquad T_{2}=P_{2},\qquad
T_{3}=Q_{2}P_{1}-Q_{1}P_{2},\qquad T_{4}=Q_{1}P_{1}+Q_{2}P_{2},$
$\displaystyle
T_{5}=\big{(}Q_{1}^{2}-Q_{2}^{2}\big{)}P_{1}+2Q_{1}Q_{2}P_{2},\qquad
T_{6}=2Q_{1}Q_{2}P_{1}+\big{(}Q_{2}^{2}-Q_{1}^{2}\big{)}P_{2}.$

Furthermore, since $e_{1}$ and $e_{3}$ are first integrals, so is

$\displaystyle
J_{2}=e_{1}^{2}+e_{3}^{2}=T_{1}^{2}+\frac{P_{3}^{2}}{4Q_{1}^{2}}.$

The other independent quadratic integral is $F_{1}$, which, in these
coordinates, takes the form

$\displaystyle
J_{1}=-2T_{1}T_{5}+\frac{Q_{2}^{2}-Q_{1}^{2}}{2Q_{1}^{2}}P_{3}^{2}-2\big{(}\beta-\gamma
Q_{1}^{2}\big{)}H.$

From the definition of the constraint $I_{1}$ in (4.5), the integral
$F_{3}=-K_{1}^{2}-K_{2}^{2}+\gamma H=-J_{2}+\gamma H$, so is not independent.

######  Remark 7.2 (transformation of potentials).

Notice that the change of coordinates (7.2) transforms the Hamiltonian of
(7.1) to that of (7.3), preserving the form of the 2-dimensional metric
$D_{2}$, but changing the potential.

######  Remark 7.3 (reduction with $\beta=0$).

To get this reduction, we can apply the involution $\iota_{23}$ to the case
$\alpha=0$.

###  7.2 Reductions when $\boldsymbol{\gamma=0}$

Now we have

$\displaystyle\varphi=\frac{q_{2}^{2}q_{3}^{2}}{\alpha q_{2}^{2}+\beta
q_{3}^{2}}\qquad\mbox{and}\qquad\psi=-2\beta\frac{q_{1}^{2}+q_{3}^{2}}{q_{2}^{2}},$

so still have the involutive symmetry $\iota_{23}$ (extended to include
$\alpha\leftrightarrow\beta$). We have also gained additional Killing vectors,
giving us the isometry algebra $\mathfrak{g}_{1}=\langle
e_{1},h_{1},f_{1}\rangle$. We can therefore reduce with respect to either
$h_{1}$ or $f_{1}$.

####  7.2.1 Reduction using the isometry $\boldsymbol{K_{2}=h_{1}}$

We now want to reduce our Hamiltonian (4.1) to the submanifold $h_{1}=\mu$. To
do this, we choose canonical coordinates $Q_{i},P_{i}$ so that $h_{1}=P_{3}$,
which makes $H$ independent of $Q_{3}$. It is easy to see that

$\displaystyle\left\\{\frac{q_{1}}{q_{3}},h_{1}\right\\}=0,\qquad\left\\{\frac{q_{2}}{q_{3}},h_{1}\right\\}=0,\qquad\left\\{-\tfrac{1}{2}\log
q_{3},h_{1}\right\\}=1,$

so clearly we should choose $Q_{3}=-\frac{1}{2}\log q_{3}$ and then some
convenient functions of $z_{1}=\frac{q_{1}}{q_{3}}$,
$z_{2}=\frac{q_{2}}{q_{3}}$ for $Q_{1}$, $Q_{2}$. To obtain a diagonal metric
on our reduced space we need the orthogonality relation
$\sum\limits_{i,j=1}^{3}g^{ij}\partial_{i}Q_{1}\partial_{j}Q_{2}=0$. We choose
$Q_{1}=\frac{q_{1}}{q_{3}}$ and
$Q_{2}=s\big{(}\frac{q_{1}}{q_{3}},\frac{q_{2}}{q_{3}}\big{)}$, leading to

$\displaystyle\big{(}\big{(}z_{1}^{2}+1\big{)}\partial_{z_{1}}+z_{1}z_{2}\partial_{z_{2}}\big{)}s(z_{1},z_{2})=0\quad\Rightarrow\quad
s(z_{1},z_{2})=f\left(\frac{z_{2}}{\sqrt{z_{1}^{2}+1}}\right).$

With the point transformation

$\displaystyle Q_{1}=\frac{q_{1}}{q_{3}},\qquad
Q_{2}=\frac{\sqrt{q_{1}^{2}+q_{3}^{2}}}{q_{2}},\qquad Q_{3}=-\tfrac{1}{2}\log
q_{3},$ (7.4a) we find $\displaystyle
H=\sum_{i,j=1}^{3}\hat{g}^{ij}P_{i}P_{j},\qquad\hat{g}^{ij}=\frac{1}{\delta}\left(\begin{matrix}\big{(}1+Q_{1}^{2}\big{)}^{2}&0&\frac{1}{2}Q_{1}\big{(}1+Q_{1}^{2}\big{)}\\\\[2.84526pt]
0&Q_{2}^{2}\big{(}1+Q_{2}^{2}\big{)}&-\frac{1}{2}Q_{2}\\\\[2.84526pt]
\frac{1}{2}Q_{1}\big{(}1+Q_{1}^{2}\big{)}&-\frac{1}{2}Q_{2}&\frac{1}{4}\big{(}1+Q_{1}^{2}\big{)}\end{matrix}\right),$
(7.4b)

where $\delta=\alpha\big{(}1+Q_{1}^{2}\big{)}+\beta Q_{2}^{2}$.

#### The resulting 2D system

By setting $P_{3}=\mu$, we can consider our Hamiltonian as $2$-dimensional,
with (upper index) metric $\hat{g}^{ij}$, given by the $2\times 2$ diagonal
matrix in (7.4b). However, since we wish to consider this as an embedding into
3 dimensions, it is better to leave $P_{3}$ in the formulae, remembering that
we can always set $P_{3}=\mu$ if we wish to consider the $2$-dimensional
reduction. The quadratic terms $P_{i}P_{3}$ are incorporated in the formulae
by “completing the square”, writing the Hamiltonian as one with electro-
magnetic, vector potential:

$\displaystyle
H=\sum_{i=1}^{2}\hat{g}^{ii}(P_{i}-A_{i}P_{3})^{2}+h(Q_{1},Q_{2})P_{3}^{2},$
(7.5)

where

$\displaystyle A_{1}=\frac{-Q_{1}}{2\big{(}1+Q_{1}^{2}\big{)}},\qquad
A_{2}=\frac{1}{2Q_{2}\big{(}1+Q_{2}^{2}\big{)}},\qquad
h=\frac{Q_{2}^{2}}{4\big{(}1+Q_{1}^{2}\big{)}\big{(}\alpha\big{(}1+Q_{1}^{2}\big{)}+\beta
Q_{2}^{2}\big{)}}.$

Since, for these particular $A_{i}$ we have
$\\{P_{1}-A_{1}P_{3},P_{2}-A_{2}P_{3}\\}=(\partial_{1}A_{2}-\partial_{2}A_{1})P_{3}=0$,
we may redefine the momenta (a gauge transformation) by
$\hat{P}_{i}=P_{i}-A_{i}P_{3}$, so that (7.5) takes the form
$H=\sum\limits_{i=1}^{2}\hat{g}^{ii}\hat{P}_{i}^{2}+\hat{h}(Q_{1},Q_{2})P_{3}^{2}$.
To incorporate this into the canonical transformation, we must adjust $Q_{3}$:

$\displaystyle\hat{Q}_{3}=Q_{3}+M_{1}(Q_{1})+M_{2}(Q_{2})\qquad\mbox{so
that}\qquad\big{\\{}\hat{Q}_{3},P_{i}-A_{i}(Q_{i})P_{3}\big{\\}}=0,$

leading to $M_{1}=-\frac{1}{4}\log\big{(}1+Q_{1}^{2}\big{)}$,
$M_{2}=\frac{1}{4}\log\big{(}\frac{Q_{2}^{2}}{1+Q_{2}^{2}}\big{)}$, giving the
point transformation (dropping the hats):

$\displaystyle Q_{1}=\frac{q_{1}}{q_{3}},\qquad
Q_{2}=\frac{\sqrt{q_{1}^{2}+q_{3}^{2}}}{q_{2}},\qquad
Q_{3}=-\tfrac{1}{4}\log\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}.$

In the original $3$-dimensional setting we had the isometry algebra
$\mathfrak{g}_{1}=\langle e_{1},h_{1},f_{1}\rangle$. Since neither of $e_{1}$,
$f_{1}$ commute with $h_{1}$, we cannot reduce these to the $2$-dimensional
setting. However, since $\\{e_{1}f_{1},h_{1}\\}=0$, we can reduce this
quadratic element, which takes the form

$\displaystyle
e_{1}f_{1}=T_{1}^{2}+\left(P_{1}+\frac{Q_{1}Q_{2}P_{2}}{1+Q_{1}^{2}}\right)Q_{1}P_{3},\qquad
T_{1}=\frac{1}{Q_{2}}\sqrt{\frac{1+Q_{2}^{2}}{1+Q_{1}^{2}}}\big{(}\big{(}1+Q_{1}^{2}\big{)}P_{1}+Q_{1}Q_{2}P_{2}\big{)},\\!\\!\\!$

where $T_{1}$ is a first order, first integral of the kinetic energy
$H_{0}=\sum\limits_{i=1}^{2}\hat{g}^{ii}P_{i}^{2}$.

#### Adapting coordinates to $\boldsymbol{T_{1}}$

By noting that

$\displaystyle y_{1}=\frac{Q_{2}}{\sqrt{1+Q_{1}^{2}}},\qquad
y_{2}=\log\left(\frac{Q_{1}Q_{2}}{\sqrt{1+Q_{1}^{2}}}+\sqrt{1+Q_{2}^{2}}\right)$

satisfy $\\{y_{1},T_{1}\\}=0$, $\\{y_{2},T_{1}\\}=1$, we may choose
$(y_{1},y_{2}+B(y_{1}))$ as new coordinates, with $B(y_{1})$ chosen so that
the resulting metric is diagonal. In this way we find

$\displaystyle u=\frac{Q_{2}}{\sqrt{1+Q_{1}^{2}}},\qquad
v=\log\left(\frac{Q_{1}Q_{2}+\sqrt{\big{(}1+Q_{1}^{2}\big{)}\big{(}1+Q_{2}^{2}\big{)}}}{\sqrt{1+Q_{1}^{2}+Q_{2}^{2}}}\right),$

in which coordinates $T_{1}$ and $H$ take the form

$\displaystyle T_{1}=p_{v},\qquad H=\frac{1}{\alpha+\beta
u^{2}}\left(u^{2}\big{(}1+u^{2}\big{)}p_{u}^{2}+\frac{u^{2}}{1+u^{2}}\,p_{v}^{2}+\frac{u^{2}\operatorname{sech}^{2}v\,p_{w}^{2}}{4\big{(}1+u^{2}\big{)}}\right).$

The final step is to put $H_{0}$ into conformal form
$H_{0}=\bar{\varphi}(\bar{u})\big{(}p_{\bar{u}}^{2}+p_{\bar{v}}^{2}\big{)}$,
in which case, the Laplace–Beltrami operator takes a very simple form:
$L_{b}=\varphi(\bar{u},\bar{v})\big{(}\partial_{\bar{u}}^{2}\pm\partial_{\bar{v}}^{2}\big{)}$,
with no first order terms. As a consequence, the coordinates $\bar{u}$ and
$\bar{v}$ satisfy $L_{b}\bar{u}=L_{b}\bar{v}=0$. Solutions of these equations
are

$\displaystyle\bar{u}=\arctan(u),\qquad\bar{v}=v,$

which, dropping bars, gives

$\displaystyle T_{1}=p_{v},\qquad
H=\frac{\sin^{2}(2u)}{2(\alpha+\beta)+2(\alpha-\beta)\cos(2u)}\big{(}p_{u}^{2}+p_{v}^{2}+\tfrac{1}{4}\operatorname{sech}^{2}v\,p_{w}^{2}\big{)},$
(7.6)

which has the $D_{4}$ kinetic energy, with potential (equivalent to the
“$b_{3}$” part of Case B in [9]).

We can see that $h_{1}$ commutes with the 6 functions

$\displaystyle e_{1}f_{1},\ e_{2}f_{2},\ e_{3}f_{3},\ h_{2},\ h_{3},\
\mbox{and}\ h_{4},$

which correspond to the $6$-dimensional conformal algebra of our 2D kinetic
energy $H_{0}$. In the present coordinates, these take the form

$\displaystyle T_{1}=p_{v},\qquad\\!T_{2}=p_{u}\cosh v\sin u+p_{v}\sinh v\cos
u,\qquad\\!T_{3}=p_{u}\cosh v\cos u-p_{v}\sinh v\sin u,$ $\displaystyle
T_{4}=p_{u}\sinh v\sin u+p_{v}\cosh v\cos u,\qquad\\!T_{5}=p_{u}\sinh v\cos
u-p_{v}\cosh v\sin u,\qquad\\!T_{6}=p_{u}.$

We have seen that both $J_{1}=F_{1}$ and $J_{2}=e_{1}f_{1}$ commute with
$h_{1}$ (in the 3D space), so can be reduced to the $2$-dimensional setting.
We therefore have that

$\displaystyle J_{1}=2T_{2}^{2}+\beta(\cos 2u-\cosh
2v)\operatorname{sech}^{2}u\,H-\tfrac{1}{4}\cos^{2}u\operatorname{sech}^{2}v\,p_{w}^{2},$
$\displaystyle J_{2}=T_{1}^{2}-\tfrac{1}{4}\tanh^{2}vp_{w}^{2},$

commute with $H$.

####  7.2.2 Reduction using the isometry $\boldsymbol{K_{3}=f_{1}}$

We now want to reduce our Hamiltonian (4.1) to the submanifold $f_{1}=\mu$, so
choose canonical coordinates $Q_{i}$, $P_{i}$ so that $f_{1}=P_{3}$, which
makes $H$ independent of $Q_{3}$. It is easy to see that we should choose
$Q_{3}=\frac{q_{1}}{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}$ and then some convenient
functions of $z_{1}=\frac{q_{3}}{q_{2}}$ and
$z_{2}=\frac{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}{q_{2}}$ for $Q_{1}$ and $Q_{2}$.
To obtain a diagonal metric on our reduced space we need the orthogonality
relation $\sum\limits_{i,j=1}^{3}g^{ij}\partial_{i}Q_{1}\partial_{j}Q_{2}=0$,
which leads us to

$\displaystyle Q_{1}=\frac{q_{3}}{q_{2}},\qquad
Q_{2}=\frac{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}{\sqrt{q_{2}^{2}+q_{3}^{2}}},\qquad
Q_{3}=\frac{q_{1}}{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}},$

and

$\displaystyle H=\frac{Q_{1}^{2}}{\alpha+\beta
Q_{1}^{2}}\left(\big{(}Q_{1}^{2}+1\big{)}P_{1}^{2}+\frac{Q_{2}^{2}}{Q_{1}^{2}+1}P_{2}^{2}+\frac{P_{3}^{2}}{Q_{2}^{2}\big{(}Q_{1}^{2}+1\big{)}}\right).$

From the Casimir ${\cal C}=e_{1}f_{1}+\frac{1}{4}h_{1}^{2}$, of our isometry
algebra, we obtain

$\displaystyle{\cal C}=T_{1}^{2}+\frac{P_{3}^{2}}{Q_{2}^{2}},$ (7.7)

where $T_{1}=Q_{2}P_{2}$ corresponds to a Killing vector on the reduced space.

#### Adapting coordinates to $\boldsymbol{T_{1}}$

We have

$\displaystyle\\{Q_{1},T_{1}\\}=\\{Q_{3},T_{1}\\}=0\qquad\mbox{and}\qquad\\{\log(Q_{2}),T_{1}\\}=1,$

so must build coordinates out of these, such that the kinetic energy $H_{0}$
is transformed into conformal form
$H_{0}=\varphi(u)\big{(}p_{u}^{2}+p_{v}^{2}\big{)}$. This is achieved by the
choice

$\displaystyle u=\arctan(Q_{1}),\qquad v=\log(Q_{2}),\qquad
w=Q_{3}\qquad\Rightarrow\qquad T_{1}=p_{v}$

and

$\displaystyle
H=\frac{\sin^{2}(2u)}{2(\alpha+\beta)+2(\alpha-\beta)\cos(2u)}\big{(}p_{u}^{2}+p_{v}^{2}+{\rm
e}^{-2v}p_{w}^{2}\big{)},$ (7.8)

which has the $D_{4}$ kinetic energy, with potential (equivalent to the
“$a_{1}$” part of Case A in [9]).

#### The 6-dimensional conformal algebra

We can identify 3 linear and 3 quadratic conformal elements in the 3D space,
which commute with $f_{1}$, so can be reduced to the 2-dimensional setting:

$\displaystyle e_{1}f_{1}+\tfrac{1}{4}h_{1}^{2},\ f_{2},\ f_{3},\ h_{4},\
e_{2}f_{2}-\tfrac{1}{2}h_{2}^{2},\ e_{3}f_{3}+\tfrac{1}{2}h_{3}^{2},$

leading to

$\displaystyle T_{1}=p_{v},\qquad T_{2}={\rm e}^{v}(\sin up_{u}+\cos
up_{v}),\qquad T_{3}={\rm e}^{v}(\cos up_{u}-\sin up_{v}),$ $\displaystyle
T_{4}=p_{u},\qquad T_{5}={\rm e}^{-v}(\sin up_{u}-\cos up_{v}),\qquad
T_{6}={\rm e}^{-v}(\cos up_{u}+\sin up_{v}),$

with $T_{1}$ being the sole isometry.

The Hamiltonian (7.8) has 2 independent quadratic integrals, derived from the
Casimir (7.7) and the former integral $F_{5}$ (see (4.6)), which can be
written in terms of the above conformal elements:

$\displaystyle J_{1}=T_{1}^{2}+{\rm e}^{-2v}\,p_{w}^{2},\qquad
J_{2}=4T_{2}^{2}-4\beta{\rm e}^{2v}\sec^{2}u\,H.$

####  7.2.3 The relationship between these reductions

If we consider the $h_{1}$ reduction to have coordinates
$(\bar{u},\bar{v},\bar{w})$, then combining the transformations, we obtain

$\displaystyle u=\bar{u},\qquad v=2\bar{w}+\log(\cosh\bar{v}),\qquad w={\rm
e}^{2\bar{w}}\tanh\bar{v},$

and the Hamiltonian (7.8) is transformed onto that of (7.6) (with bars). This
transformation preserves the form of the 2-dimensional kinetic energy, but
changes the potential, thus relating two of the potentials obtained in [9].

##  8 Reductions of the system of Section 4.2

We saw in Section 4.2 that the Hamiltonian

$\displaystyle H=\frac{1}{\alpha q_{2}+\beta
q_{3}+\gamma}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}$

admitted the $3$-dimensional isometry algebra (4.8a), with Poisson relations
(4.8b). In this section we reduce this Hamiltonian to a 2D space in two
different ways, using the isometries $K_{2}$ and $K_{3}$ respectively.

###  8.1 Reduction using the isometry $\boldsymbol{K_{2}=\beta e_{2}-\alpha
e_{3}}$

We see that

$\displaystyle\\{\alpha q_{2}+\beta
q_{3},K_{2}\\}=\\{q_{1},K_{2}\\}=0\qquad\mbox{and}\qquad\left\\{\frac{q_{2}}{\beta},K_{2}\right\\}=1,$

and we need to choose new coordinates $(Q_{1},Q_{2},Q_{3})$, such that
$K_{2}=P_{3}$ (so the new $H$ is independent of $Q_{3}$) and $H$ takes the
form

$\displaystyle
H=\tilde{\varphi}(Q_{1},Q_{2})\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\tilde{V}(Q_{1},Q_{2})P_{3}^{2},$

which can be achieved in stages as described in Section 7 or by the single
canonical transformation generated by

$\displaystyle S=\frac{(\alpha q_{2}+\beta
q_{3}+\gamma)}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{3}}}\
P_{1}+\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{6}}q_{1}P_{2}+\left(\frac{q_{2}}{\beta}-\frac{\alpha(\alpha
q_{2}+\beta
q_{3}+\gamma)}{\beta\big{(}\alpha^{2}+\beta^{2}\big{)}}\right)P_{3},$ (8.1)

which transforms the Hamiltonian to

$\displaystyle
H=\frac{1}{Q_{1}}\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\frac{P_{3}^{2}}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{4}{3}}Q_{1}},$
(8.2)

so the entire Hamiltonian (not just the kinetic energy) has inherited the
symmetry $K_{1}=\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{6}}P_{2}$, which
is a result of the property $\\{K_{1},K_{2}\\}=0$.

Since $P_{3}$ is a first integral, we can reduce to the $2$-dimensional space
with $P_{3}=\mu$. Again, we can derive the $6$-dimensional conformal algebra
from our original $10$-dimensional algebra (2.2), by considering linear and
quadratic conformal elements which commute with $K_{2}$. The 3 linear elements

$\displaystyle\alpha e_{2}+\beta e_{3},\ e_{1},\ \alpha h_{2}+\beta
h_{3}\qquad\text{give us}\qquad T_{1}=P_{1},\qquad T_{2}=P_{2},\qquad
T_{3}=Q_{2}P_{1}-Q_{1}P_{2},$

while the 3 quadratic elements

$\displaystyle\alpha K_{1}h_{1}+h_{3}K_{2},\qquad 8(\alpha e_{2}+\beta
e_{3})(\alpha f_{2}+\beta
f_{3})-\big{(}\alpha^{2}+\beta^{2}\big{)}h_{4}^{2},\qquad
4\big{(}\alpha^{2}+\beta^{2}\big{)}e_{1}f_{1}-K_{3}^{2},$

factorise (when $P_{3}=0$) to give the 3 remaining conformal elements:

$\displaystyle T_{4}=Q_{1}P_{1}+Q_{2}P_{2},\qquad
T_{5}=\big{(}Q_{1}^{2}-Q_{2}^{2}\big{)}P_{1}+2Q_{1}Q_{2}P_{2},$ $\displaystyle
T_{6}=2Q_{1}Q_{2}P_{1}+\big{(}Q_{2}^{2}-Q_{1}^{2}\big{)}P_{2}.$

The $2$-dimensional Hamiltonian (8.2) (with $P_{3}=\mu$) is the $D_{1}$
kinetic energy, with potential (equivalent to the “$b_{2}$” part of Case 1 in
[10]).

####  8.1.1 The quadratic integrals

We consider the 2 functions (4.7a) and see that $\\{F_{2},K_{2}\\}=0$, but
$\\{F_{1},K_{2}\\}\neq 0$. However, it is easy to see that
$\big{\\{}\big{(}\alpha^{2}+\beta^{2}\big{)}F_{1}-\beta
K_{1}K_{3},K_{2}\big{\\}}=\\{K_{1},K_{2}\\}=0$. Noting that

$\displaystyle\big{(}\alpha^{2}+\beta^{2}\big{)}F_{1}-\beta K_{1}K_{3}=\alpha
K_{1}(\alpha h_{2}+\beta
h_{3})-\tfrac{1}{2}\alpha\big{(}\alpha^{2}+\beta^{2}\big{)}q_{1}^{2}H,$

and that $\alpha h_{2}+\beta h_{3}\mapsto
2\gamma\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{6}}T_{2}+2\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{2}}T_{3}$
(under the action of the transformation generated by (8.1)) this integral
reduces to

$\displaystyle
J_{1}=T_{2}T_{3}-\tfrac{1}{4}Q_{2}^{2}H=P_{2}(Q_{2}P_{1}-Q_{1}P_{2})-\tfrac{1}{4}Q_{2}^{2}H.$

The function $F_{2}$ is more straightforward and reduces to

$\displaystyle J_{2}=T_{1}T_{2}-\frac{Q_{2}}{2}H=P_{1}P_{2}-\frac{Q_{2}}{2}H.$

In both these formulae, $H$ is the full (8.2) (with potential) and
$\\{J_{i},H\\}=0$, for each $i$, with $\\{J_{1},P_{2}\\}=4J_{2}$.

###  8.2 Reduction using the isometry $\boldsymbol{K_{3}=\beta h_{2}-\alpha
h_{3}}$

We see that

$\displaystyle\\{\alpha q_{2}+\beta
q_{3},K_{3}\\}=\big{\\{}q_{1}^{2}+q_{2}^{2}+q_{3}^{2},K_{3}\big{\\}}=0\qquad\mbox{and}$
$\displaystyle\left\\{\frac{1}{2\sqrt{\alpha^{2}+\beta^{2}}}\tan^{-1}\left(\frac{\beta
q_{2}-\alpha
q_{3}}{q_{1}\sqrt{\alpha^{2}+\beta^{2}}}\right),K_{3}\right\\}=1,$

and we again need to choose new coordinates $(Q_{1},Q_{2},Q_{3})$, such that
$K_{2}=P_{3}$ (so the new $H$ is independent of $Q_{3}$) and $H$ takes the
form

$\displaystyle
H=\tilde{\varphi}(Q_{1},Q_{2})\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\tilde{V}(Q_{1},Q_{2})P_{3}^{2},$

which can be achieved in stages as described in Section 7 or by the single
canonical transformation generated by

$\displaystyle S=\frac{\alpha q_{2}+\beta
q_{3}+\gamma}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{3}}}P_{1}+\frac{\big{(}\alpha^{2}+\beta^{2}\big{)}q_{1}^{2}+(\beta
q_{2}-\alpha
q_{3})^{2}}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{1}{3}}}P_{2}$
$\displaystyle\hphantom{S=}{}+\frac{1}{2\sqrt{\alpha^{2}+\beta^{2}}}\tan^{-1}\left(\frac{\beta
q_{2}-\alpha q_{3}}{q_{1}\sqrt{\alpha^{2}+\beta^{2}}}\right)P_{3},$

which transforms the Hamiltonian to

$\displaystyle
H=\frac{1}{Q_{1}}\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\frac{P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{1}Q_{2}^{2}},$
(8.3)

so this Hamiltonian has not inherited the symmetry $K_{1}$, but the kinetic
energy has acquired the symmetry $P_{2}$, which is the reduction of $K_{1}$ to
the 2D space, corresponding to the level set $P_{3}=\mu$.

Again, we can derive the $6$-dimensional conformal algebra from our original
$10$-dimensional algebra (2.2), by considering linear and quadratic conformal
elements which commute with $K_{3}$. The 3 linear elements

$\displaystyle\alpha e_{2}+\beta e_{3},\ h_{1},\ \alpha f_{2}+\beta
f_{3}\qquad\text{give us}$ $\displaystyle T_{1}=P_{1},\qquad
T_{4}=Q_{1}P_{1}+Q_{2}P_{2},\qquad
T_{5}=\big{(}Q_{1}^{2}-Q_{2}^{2}\big{)}P_{1}+2Q_{1}Q_{2}P_{2},$

while the 3 quadratic elements

$\displaystyle 4\big{(}h_{2}^{2}+h_{3}^{2}\big{)}+h_{4}^{2},\qquad
K_{2}^{2}+\big{(}\alpha^{2}+\beta^{2}\big{)}K_{1}^{2},\qquad
4\big{(}\alpha^{2}+\beta^{2}\big{)}f_{1}^{2}(\beta f_{2}-\alpha f_{3})^{2},$

simplify (when $P_{3}=0$) to give the 3 remaining conformal elements:

$\displaystyle T_{2}=P_{2},\qquad T_{3}=Q_{2}P_{1}-Q_{1}P_{2},\qquad
T_{6}=2Q_{1}Q_{2}P_{1}+\big{(}Q_{2}^{2}-Q_{1}^{2}\big{)}P_{2}.$

The $2$-dimensional Hamiltonian (8.3) (with $P_{3}=\mu$) is the $D_{1}$
kinetic energy, with potential (equivalent to the “$b_{3}$” part of Case 1 in
[10]).

###  8.3 The relationship between these reductions

For this section we change notation for the reduction of Section 8.1, writing
the 3D coordinates as $(u,v,w)$:

$\displaystyle
H_{u}=\frac{1}{u}\big{(}p_{u}^{2}+p_{v}^{2}\big{)}+\frac{p_{w}^{2}}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{4}{3}}u},\qquad
J_{1u}=p_{v}(vp_{u}-up_{v})-\tfrac{1}{4}v^{2}H_{u},$ $\displaystyle
J_{2u}=p_{u}p_{v}-\frac{v}{2}H_{u}.$ (8.4)

Composing the two point transformations
$(u,v,w)\mapsto(q_{1},q_{2},q_{3})\mapsto(Q_{1},Q_{2},Q_{3})$, we obtain

$\displaystyle u=Q_{1},\qquad
v=Q_{2}\cos\big{(}2\sqrt{\alpha^{2}+\beta^{2}}Q_{3}\big{)},$ $\displaystyle
w=\frac{1}{\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{2}{3}}}Q_{2}\sin\big{(}2\sqrt{\alpha^{2}+\beta^{2}}Q_{3}\big{)}-\frac{\alpha}{\beta\big{(}\alpha^{2}+\beta^{2}\big{)}}.$

Under the corresponding canonical transformation of the full $6$-dimensional
phase space, the Hamiltonian $H_{u}$ is transformed to $H_{Q}$, given by
(8.3). This means that it preserves the form of the kinetic energy (associated
with $D_{1}$) and changes the “$b_{2}$” part of the potential of Case 1 in
[10] to the “$b_{3}$” part.

Any other integral $F_{u}$ is transformed to

$\displaystyle
F_{Q}=F_{Q}^{(0)}+F_{Q}^{(c)}\cos(kQ_{3})+F_{Q}^{(s)}\sin(kQ_{3}),\qquad\mbox{for
some}\quad k.$

In the full $6$-dimensional phase space, $\\{H_{Q},F_{Q}\\}=0$ (noting that
the $P_{3}^{2}$ term in $H_{Q}$ will interact with $Q_{3}$), but in the
$2$-dimensional reduction (with $P_{3}=\mu$), we have
$\big{\\{}H_{Q},F_{Q}^{(0)}\big{\\}}=0$.

Carrying this out for $J_{1u}$, $J_{2u}$, we find

$\displaystyle
J_{1Q}^{(0)}=P_{2}(Q_{2}P_{1}-Q_{1}P_{2})-\frac{Q_{1}P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{2}^{2}}-\tfrac{1}{4}Q_{2}^{2}H_{Q},\qquad
J_{2Q}^{(0)}=0.$

The function $J_{2u}^{2}$ would lead to something nontrivial, but it would be
quartic. However, we can use the Casimir (4.8c) of the original isometry
algebra:

$\displaystyle{\cal
C}_{K}=\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{4}{3}}p_{v}^{2}+p_{w}^{2}=\big{(}\alpha^{2}+\beta^{2}\big{)}^{\frac{4}{3}}\left(P_{2}^{2}+\frac{P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{2}^{2}}\right).$

We have thus found that the system (8.4) (together with combinations of its
linear integrals) can be transformed onto the system

$\displaystyle
H_{Q}=\frac{1}{Q_{1}}\,\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\frac{P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{1}Q_{2}^{2}},$
$\displaystyle
R_{1Q}=P_{2}(Q_{2}P_{1}-Q_{1}P_{2})-\frac{Q_{1}P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{2}^{2}}-\frac{1}{4}Q_{2}^{2}H_{Q},$
$\displaystyle
R_{2Q}=P_{2}^{2}+\frac{P_{3}^{2}}{4\big{(}\alpha^{2}+\beta^{2}\big{)}Q_{2}^{2}},$

corresponding to two particular parametric reductions of the potential of Case
1 in [10].

##  9 Reductions of the system of Section 5.1

We now consider reductions of the Hamiltonian

$\displaystyle H=\frac{q_{2}^{2}q_{3}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}}{(\alpha
q_{2}^{2}+\beta q_{3}^{2})\sqrt{q_{1}^{2}+q_{2}^{2}}+\gamma
q_{1}q_{3}^{2}}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}.$

We found that we had three cases, corresponding to $\alpha=0$ or $\beta=0$ or
$\gamma=0$. However, $\gamma=0$ is just the Hamiltonian discussed in Sections
4.1.2 and 7.2, so we are left with just the first two. Our first example is
the case with $\alpha=0$, which has a $3$-dimensional isometry algebra and
reduces to a Hamiltonian with Darboux–Koenigs metric $D_{4}$. However, in
Section 9.2, we consider the case $\beta=0$, which has only one linear
integral $h_{1}$, with the consequence that the reduced metric has no Killing
vectors at all. Nevertheless, the reduced Hamiltonian commutes with a pair of
functions, one quadratic and the other quartic, so is a maximally super-
integrable system which is outside the Darboux–Koenigs classification.

###  9.1 The case when $\boldsymbol{\alpha=0}$

We now consider the Hamiltonian

$\displaystyle
H=\frac{q_{2}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}}{\beta\sqrt{q_{1}^{2}+q_{2}^{2}}+\gamma
q_{1}}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$

which has the isometry algebra $K_{1}=h_{1}$, $K_{2}=e_{3}$, $K_{3}=f_{3}$,
satisfying

$\displaystyle\\{K_{1},K_{2}\\}=-2K_{2},\qquad\\{K_{1},K_{3}\\}=2K_{3},\qquad\\{K_{2},K_{3}\\}=-2K_{1}.$

We can therefore reduce with respect to either $K_{1}$ or $K_{3}$. Reduction
with respect to $K_{2}$ would be equivalent to that with respect to $K_{3}$
through the action of the involution $\iota_{ef}$.

####  9.1.1 Reduction using the isometry $\boldsymbol{K_{1}}$

Since $K_{1}=h_{1}$, we can again use the transformation (7.4a):

$\displaystyle Q_{1}=\frac{q_{1}}{q_{3}},\qquad
Q_{2}=\frac{\sqrt{q_{1}^{2}+q_{3}^{2}}}{q_{2}},\qquad Q_{3}=-\tfrac{1}{2}\log
q_{3},$

this time giving

$\displaystyle
H=\frac{Q_{1}^{2}}{\beta\big{(}1+Q_{1}^{2}\big{)}+\gamma\sqrt{1+Q_{1}^{2}}}\left(\big{(}1+Q_{1}^{2}\big{)}^{2}\left(P_{1}+\frac{Q_{1}P_{3}}{2\big{(}1+Q_{1}^{2}\big{)}}\right)^{2}\right.$
$\displaystyle\left.\hphantom{H=}{}+\big{(}1+Q_{2}^{2}\big{)}\left(P_{2}+\frac{Q_{2}P_{3}}{2\big{(}1+Q_{2}^{2}\big{)}}\right)^{2}+\frac{P_{3}^{2}}{4(1+Q_{2}^{2})}\right).$

Here we have already “completed the square” to give the vector potential

$\displaystyle(A_{1},A_{2})=\left(-\frac{Q_{1}}{2\big{(}1+Q_{1}^{2}\big{)}},-\frac{Q_{2}}{2\big{(}1+Q_{2}^{2}\big{)}}\right),$

satisfying

$\displaystyle\\{P_{1}-A_{1}P_{3},P_{2}-A_{2}P_{3}\\}=(\partial_{1}A_{2}-\partial_{2}A_{1})P_{3}=0,$

so we can use these to define new momenta
$\big{(}\tilde{P}_{1},\tilde{P}_{2}\big{)}$, but must adjust the definition of
$Q_{3}$ in order to keep the transformation canonical:

$\displaystyle\\{Q_{3}+M_{1}(Q_{1})+M_{2}(Q_{2}),P_{i}-A_{i}P_{3}\\}=0\qquad\Rightarrow\qquad
M_{1}^{\prime}(Q_{1})=A_{1},\qquad M_{2}^{\prime}(Q_{2})=A_{2},$

leading to the point transformation

$\displaystyle Q_{1}=\frac{q_{1}}{q_{3}},\qquad
Q_{2}=\frac{\sqrt{q_{1}^{2}+q_{3}^{2}}}{q_{2}},\qquad
Q_{3}=-\tfrac{1}{4}\log\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)},$

now giving

$\displaystyle
H=\frac{Q_{1}^{2}}{\beta\big{(}1+Q_{1}^{2}\big{)}+\gamma\sqrt{1+Q_{1}^{2}}}\left(\big{(}1+Q_{1}^{2}\big{)}^{2}P_{1}^{2}+\big{(}1+Q_{2}^{2}\big{)}P_{2}^{2}+\frac{P_{3}^{2}}{4\big{(}1+Q_{2}^{2}\big{)}}\right).$

Since $\\{K_{2}K_{3},K_{1}\\}=0$, this quantity defines a Killing vector for
the 2-dimensional metric, associated with this Hamiltonian:

$\displaystyle
K_{2}K_{3}=2\left(T_{1}^{2}-\frac{Q_{2}^{2}P_{3}^{2}}{4\big{(}1+Q_{2}^{2}\big{)}}\right),\quad\mbox{with}\qquad
T_{1}=\sqrt{\big{(}1+Q_{2}^{2}\big{)}}P_{2},$

where $T_{1}$ is a first order, first integral of the 2D kinetic energy
$H_{0}$.

We choose coordinates $(u,v,w)$ so that $T_{1}=p_{v}$ and
$H_{0}=\tilde{\varphi}(u)\big{(}p_{u}^{2}+p_{v}^{2}\big{)}$:

$\displaystyle Q_{1}=\tan u,\qquad Q_{2}=\sinh v,\qquad Q_{3}=w,$

giving

$\displaystyle H=\frac{\sin^{2}u}{\beta+\gamma\cos
u}\big{(}p_{u}^{2}+p_{v}^{2}+\tfrac{1}{4}\operatorname{sech}^{2}v\,p_{w}^{2}\big{)},$
(9.1)

which has the $D_{4}$ kinetic energy, with potential (equivalent to the
“$b_{3}$” part of Case B in [9]).

#### The conformal algebra and quadratic integrals

We can see that $K_{1}=h_{1}$ commutes with the 6 functions

$\displaystyle
e_{3}f_{3},\;\;\;e_{1}f_{1},\;\;\;e_{2}f_{2},\;\;\;h_{2},\;\;\;h_{3},\;\;\;\mbox{and}\;\;\;h_{4},$

which correspond to the $6$-dimensional conformal algebra of our 2D kinetic
energy $H_{0}$. In the present coordinates, these take the form

$\displaystyle T_{1}=p_{v},\qquad\\!T_{2}=p_{u}\cosh v\sin u+p_{v}\sinh v\cos
u,\qquad\\!T_{3}=p_{u}\cosh v\cos u-p_{v}\sinh v\sin u,$ $\displaystyle
T_{4}=p_{u},\qquad\\!T_{5}=p_{u}\sinh v\sin u+p_{v}\cosh v\cos
u,\qquad\\!T_{6}=p_{u}\sinh v\cos u-p_{v}\cosh v\sin u.$

The element $T_{1}$ is an invariant of the kinetic energy, but the full
expression for $K_{2}K_{3}=e_{3}f_{3}$ is a quadratic integral for $H$:

$\displaystyle
J_{1}=\tfrac{1}{2}e_{3}f_{3}=T_{1}^{2}-\tfrac{1}{4}\tanh^{2}v\,p_{w}^{2}.$

We also have that $\\{F_{3},K_{1}\\}=\\{F_{4},K_{1}\\}=0$ (see Table 8), but
$F_{3}=4J_{1}-4\beta H+p_{w}^{2}$, so is not independent. However $F_{4}$ is
independent and takes the form

$\displaystyle J_{2}=F_{4}=-8T_{4}T_{6}+2(4\beta\cos u+\gamma(3+\cos
2u))\csc^{2}u\sinh v\,H,$

satisfying

$\displaystyle\\{J_{1},J_{2}\\}=16T_{1}T_{3}T_{4}-4(4\beta\cos u+\gamma(3+\cos
2u))\cosh v\csc^{2}u\,T_{1}H$
$\displaystyle\hphantom{\\{J_{1},J_{2}\\}=}{}-4\operatorname{sech}v\tanh v\sin
u\,T_{4}p_{w}^{2}.$

####  9.1.2 Reduction using the isometry $\boldsymbol{K_{3}=f_{3}}$

With $K_{3}=f_{3}$, it is again possible to choose coordinates such that
$K_{3}=P_{3}$, and to set the coefficient of $P_{1}P_{2}$ to be zero:

$\displaystyle Q_{1}=\frac{q_{2}}{q_{1}},\qquad
Q_{2}=\frac{q_{1}^{2}+q_{2}^{2}+q_{3}^{2}}{\sqrt{q_{1}^{2}+q_{2}^{2}}},\qquad
Q_{3}=\frac{q_{3}}{2\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}},$

giving

$\displaystyle
H=\frac{Q_{1}^{2}}{\beta\big{(}1+Q_{1}^{2}\big{)}+\gamma\sqrt{1+Q_{1}^{2}}}\left(\big{(}1+Q_{1}^{2}\big{)}^{2}P_{1}^{2}+Q_{2}^{2}P_{2}^{2}+\frac{P_{3}^{2}}{4Q_{2}^{2}}\right).$

A further change of coordinates

$\displaystyle Q_{1}=\tan u,\qquad Q_{2}={\rm e}^{v},\qquad Q_{3}=w,$

gives

$\displaystyle H=\frac{\sin^{2}u}{\beta+\gamma\cos
u}\big{(}p_{u}^{2}+p_{v}^{2}+\tfrac{1}{4}{\rm e}^{-2v}p_{w}^{2}\big{)},$ (9.2)

which has the $D_{4}$ kinetic energy, with potential (equivalent to the
“$a_{1}$” part of Case A in [9]).

#### The conformal algebra and quadratic integrals

We can see that $K_{3}=f_{3}$ commutes with the 6 functions

$\displaystyle 2e_{3}f_{3}+h_{1}^{2},\ 4e_{1}f_{1}-h_{3}^{2},\
8e_{2}f_{2}-h_{4}^{2},\ h_{2},\ f_{1},\ \mbox{and}\ f_{2},$

which correspond to the $6$-dimensional conformal algebra of our 2D kinetic
energy $H_{0}$. In the present coordinates, these take the form

$\displaystyle T_{1}=p_{v},\qquad T_{2}={\rm e}^{v}(p_{u}\sin u+p_{v}\cos
u),\qquad T_{3}={\rm e}^{v}(p_{u}\cos u-p_{v}\sin u),$ $\displaystyle
T_{4}=p_{u},\qquad T_{5}={\rm e}^{-v}(p_{u}\sin u-p_{v}\cos u),\qquad
T_{6}={\rm e}^{-v}(p_{u}\cos u+p_{v}\sin u).$

The element $T_{1}$ is an invariant of the kinetic energy, but the full
expression for $2K_{2}K_{3}-K_{1}^{2}=2e_{3}f_{3}+h_{1}^{2}$ (the Casimir of
the isometry algebra) is a quadratic integral for $H$:

$\displaystyle
J_{1}=\tfrac{1}{4}\big{(}2K_{2}K_{3}-K_{1}^{2}\big{)}=T_{1}^{2}+\tfrac{1}{4}{\rm
e}^{-2v}p_{w}^{2}.$

We also have that $\\{F_{2},K_{3}\\}=\\{F_{3},K_{3}\\}=0$ (see Table 8), but
$F_{3}=4(J_{1}-\beta H)$, so is not independent. However $F_{2}$ is
independent and takes the form

$\displaystyle J_{2}=F_{2}=-4T_{3}T_{4}-(4\beta\cos u+\gamma(3+\cos
2u))\csc^{2}u\,{\rm e}^{v}H.$

####  9.1.3 The relationship between these reductions

If we consider the $K_{1}$ reduction to have coordinates
$(\bar{u},\bar{v},\bar{w})$, then combining the transformations, we obtain

$\displaystyle u=\bar{u},\qquad v=-2\bar{w}+\log(\cosh\bar{v}),\qquad
w=\tfrac{1}{2}{\rm e}^{2\bar{w}}\tanh\bar{v},$

and the Hamiltonian (9.2) is transformed onto that of (9.1) (with bars). This
transformation preserves the form of the 2-dimensional kinetic energy, but
changes the potential, thus relating two of the potentials obtained in [9].

###  9.2 The case when $\boldsymbol{\beta=0}$

We now consider the Hamiltonian

$\displaystyle H=\frac{q_{2}^{2}q_{3}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}}{\alpha
q_{2}^{2}\sqrt{q_{1}^{2}+q_{2}^{2}}+\gamma
q_{1}q_{3}^{2}}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$ (9.3)

which has the single isometry $K_{1}=h_{1}$. We therefore reduce with respect
to this by making an appropriate choice of coordinates $Q_{i}$, which would
generally give “vector potential” terms, which again turn out to be gauge
equivalent to zero. We can again incorporate this gauge transformation into
the definition of $Q_{3}$, to obtain

$\displaystyle Q_{1}=\frac{q_{2}}{q_{1}},\qquad
Q_{2}=\frac{q_{3}}{\sqrt{q_{1}^{2}+q_{2}^{2}}},\qquad
Q_{3}=-\tfrac{1}{4}\log\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)},$

giving

$\displaystyle H=\frac{Q_{1}^{2}Q_{2}^{2}}{\alpha Q_{1}^{2}+\gamma
Q_{2}^{2}\sqrt{Q_{1}^{2}+1}}\left(\big{(}Q_{1}^{2}+1\big{)}^{2}P_{1}^{2}+\big{(}Q_{2}^{2}+1\big{)}P_{2}^{2}+\frac{P_{3}^{2}}{4\big{(}Q_{2}^{2}+1\big{)}}\right).$

The final transformation puts the kinetic energy into the form
$H_{0}=\tilde{\varphi}(u,v)\big{(}p_{u}^{2}+p_{v}^{2}\big{)}$:

$\displaystyle Q_{1}=\tan u,\qquad Q_{2}=\sinh v,\qquad Q_{3}=w,$

giving

$\displaystyle
H=\frac{\sin^{2}{u}\sinh^{2}{v}}{\alpha\sin^{2}{u}+\gamma\cos{u}\,\sinh^{2}{v}}\,\left(p_{u}^{2}+p_{v}^{2}+\frac{1}{4}\operatorname{sech}^{2}v\,p_{w}^{2}\right).$
(9.4)

The corresponding metric has no Killing vectors, so we cannot reduce further.
However, we can construct the $6$-dimensional conformal algebra from the
linear and quadratic invariants of $h_{1}$:

$\displaystyle h_{2},\ h_{3},\ h_{4},\ e_{1}f_{1},\ e_{2}f_{2},\ e_{3}f_{3},$

giving

$\displaystyle T_{1}=p_{u},\qquad
T_{2}=\sin{u}\sinh{v}p_{u}+\cos{u}\cosh{v}p_{v},\qquad
T_{3}=\cos{u}\sinh{v}p_{u}-\sin{u}\cosh{v}p_{v},$ $\displaystyle
T_{4}=\sin{u}\cosh{v}p_{u}+\cos{u}\sinh{v}p_{v},\qquad
T_{5}=\cos{u}\cosh{v}p_{u}-\sin{u}\sinh{v}p_{v},\qquad T_{6}=p_{v}.$

We see from (5.3) that $h_{1}$ commutes with $F_{3}$ and $F_{1}F_{2}$, which
therefore give us a quadratic and a quartic integral for $H$, which can be
written in terms of the conformal algebra:

$\displaystyle J_{1}=T_{1}^{2}-\gamma\cot u\operatorname{cosec}u\,H,$
$\displaystyle J_{2}=\big{(}2T_{1}T_{5}-\gamma\cosh
v\operatorname{cosec}^{2}u\big{(}1+\cos^{2}u\big{)}H\big{)}^{2}-T_{1}^{2}p_{w}^{2}\operatorname{sech}^{2}v\sin^{2}u.$

We thus find that the Hamiltonian (9.4) is maximally super-integrable, but
that it falls out of the Darboux–Koenigs classification, since it has no first
order integral, together with one quadratic and one quartic integral.

##  10 Reductions of the system of Section 6.1

We now consider the Hamiltonian

$\displaystyle
H=\frac{1}{\alpha\big{(}q_{1}^{2}+q_{2}^{2}+q_{3}^{2}\big{)}+\beta
q_{1}+\gamma}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)},$

with symmetry algebra $\langle K_{1},K_{2},K_{3}\rangle$, defined by (6.3a).

We can therefore reduce with respect to either $K_{1}$ or $K_{2}$. Reduction
with respect to $K_{3}$ would be equivalent to that with respect to $K_{2}$
through the action of the involution $\iota_{23}$.

###  10.1 Reduction using the isometry $\boldsymbol{K_{1}=h_{4}}$

To transform $K_{1}=h_{4}$ to $P_{3}$ and to eliminate the $P_{1}P_{2}$ term,
we choose coordinates

$\displaystyle Q_{1}=q_{1},\qquad Q_{2}=\sqrt{q_{2}^{2}+q_{3}^{2}},\qquad
Q_{3}=\tfrac{1}{4}\arctan\left(\frac{q_{2}}{q_{3}}\right),$

giving

$\displaystyle H=\frac{1}{\alpha\big{(}Q_{1}^{2}+Q_{2}^{2}\big{)}+\beta
Q_{1}+\gamma}\left(P_{1}^{2}+P_{2}^{2}+\frac{P_{3}^{2}}{16Q_{2}^{2}}\right).$
(10.1)

Since $\big{\\{}K_{2}^{2}+K_{3}^{2},K_{1}\big{\\}}=0$, we use this to build an
invariant of (10.1)

$\displaystyle K_{2}^{2}+K_{3}^{2}=4\left(T_{1}^{2}+\frac{(2\alpha
Q_{1}+\beta)^{2}}{16Q_{2}^{2}}P_{3}^{2}\right),\qquad T_{1}=2\alpha
Q_{2}P_{1}-(2\alpha Q_{1}+\beta)P_{2},$

is a first order integral of the kinetic energy.

The change of coordinates

$\displaystyle\bar{u}=\alpha\big{(}Q_{1}^{2}+Q_{2}^{2}\big{)}+\beta
Q_{1},\qquad\bar{v}=\frac{1}{2\alpha}\arctan{\frac{2\alpha
Q_{1}+\beta}{2\alpha Q_{2}}},\qquad\bar{w}=Q_{3},$

gives

$\displaystyle T_{1}=p_{\bar{v}},\qquad
H=\frac{1}{\bar{u}+\gamma}\left(\big{(}4\alpha\bar{u}+\beta^{2}\big{)}p_{\bar{u}}^{2}+\frac{1}{4\alpha\bar{u}+\beta^{2}}p_{\bar{v}}^{2}+\frac{\alpha^{2}p_{\bar{w}}^{2}}{4\big{(}4\alpha\bar{u}+\beta^{2}\big{)}\cos^{2}{(2\alpha\bar{v})}}\right).$

With the final step in the transformation,

$\displaystyle u=\ln\left(\bar{u}+\frac{\beta^{2}}{4\alpha}\right),\qquad
v=4\alpha\bar{v},\qquad w=\bar{w},$

we obtain

$\displaystyle T_{1}=4\alpha p_{v},\qquad
H=\frac{16\alpha^{2}\mathrm{e}^{-u}}{4\alpha\mathrm{e}^{u}+4\alpha\gamma-\beta^{2}}\left(p_{u}^{2}+p_{v}^{2}+\frac{p_{w}^{2}}{64\cos^{2}{\frac{v}{2}}}\right),$

which has the $D_{3}$ kinetic energy, with potential (equivalent to the
“$b_{1}$” part of Case B in [9]).

####  10.1.1 The conformal algebra and quadratic integrals

We can see that $K_{1}=h_{4}$ commutes with the 6 functions

$\displaystyle K_{2}^{2}+K_{3}^{2},\ e_{2}^{2}+e_{3}^{2},\
f_{2}^{2}+f_{3}^{2},\ e_{1},\ h_{1},\ \mbox{and}\ f_{1},$

which correspond to the $6$-dimensional conformal algebra of our 2D kinetic
energy $H_{0}$. In the present coordinates, these take the form

$\displaystyle T_{1}=p_{v},\qquad T_{2}={\rm
e}^{-\frac{1}{2}u}\left(p_{u}\sin\frac{v}{2}+p_{v}\cos\frac{v}{2}\right),\qquad
T_{3}={\rm
e}^{-\frac{1}{2}u}\left(p_{u}\cos\frac{v}{2}-p_{v}\sin\frac{v}{2}\right),$
$\displaystyle T_{4}=p_{u},\qquad T_{5}={\rm
e}^{\frac{1}{2}u}\left(p_{u}\sin\frac{v}{2}-p_{v}\cos\frac{v}{2}\right),\qquad
T_{6}={\rm
e}^{\frac{1}{2}u}\left(p_{u}\cos\frac{v}{2}+p_{v}\sin\frac{v}{2}\right).$

The element $T_{1}$ is an invariant of the kinetic energy, but the full
expression for $K_{2}^{2}+K_{3}^{2}$ gives a quadratic integral for $H$:

$\displaystyle
J_{1}=T_{1}^{2}+\frac{1}{64}\operatorname{sech}^{2}\frac{v}{2}\,p_{w}^{2}.$

We also have that $\\{F_{1},K_{1}\\}=0$ (see Table 10), giving us

$\displaystyle J_{2}=T_{2}^{2}+\left(\beta^{2}-4\alpha{\rm
e}^{u}\sin^{2}\frac{v}{2}\right)H.$

###  10.2 Reduction using the isometry $\boldsymbol{K_{2}=\alpha h_{2}+\beta
e_{2}}$

To transform $K_{2}=\alpha h_{2}+\beta e_{2}$ to $P_{3}$ and to eliminate the
$P_{1}P_{2}$ term, we choose coordinates

$\displaystyle Q_{1}=q_{3},\qquad Q_{2}=\sqrt{\alpha q_{1}^{2}+\alpha
q_{2}^{2}+\beta q_{1}},\qquad
Q_{3}=-\frac{1}{2\alpha}\arctan\left(\frac{2\alpha q_{1}+\beta}{2\alpha
q_{2}}\right),$

giving

$\displaystyle H=\frac{1}{\alpha
Q_{1}^{2}+Q_{2}^{2}+\gamma}\left(P_{1}^{2}+\frac{4\alpha
Q_{2}^{2}+\beta^{2}}{4Q_{2}^{2}}P_{2}^{2}+\frac{P_{3}^{2}}{4\alpha
Q_{2}^{2}+\beta^{2}}\right).$ (10.2)

Since
$\big{\\{}K_{2}^{2}+K_{3}^{2}+\frac{\alpha^{2}}{4}K_{1}^{2},K_{2}\big{\\}}=0$,
we use this to build an invariant of (10.2)

$\displaystyle
K_{2}^{2}+K_{3}^{2}+\frac{\alpha^{2}}{4}K_{1}^{2}=4\left(T_{1}^{2}+\frac{\big{(}4\alpha^{2}Q_{1}^{2}+4\alpha
Q_{2}^{2}+\beta^{2}\big{)}}{4\alpha Q_{2}^{2}+\beta^{2}}P_{3}^{2}\right),$

where $T_{1}=\frac{\sqrt{4\alpha
Q_{2}^{2}+\beta^{2}}}{Q_{2}}(Q_{2}P_{1}-\alpha Q_{1}P_{2})$ is a first order
integral of the kinetic energy.

The change of coordinates,

$\displaystyle\bar{u}=\alpha
Q_{1}^{2}+Q_{2}^{2},\qquad\bar{v}=\frac{1}{2\alpha}\arctan{\frac{2\alpha
Q_{1}}{\sqrt{4\alpha Q_{2}^{2}+\beta^{2}}}},\qquad\bar{w}=Q_{3},$

gives

$\displaystyle T_{1}=p_{\bar{v}},\qquad
H=\frac{1}{\bar{u}+\gamma}\left(\big{(}4\alpha\bar{u}+\beta^{2}\big{)}p_{\bar{u}}^{2}+\frac{1}{4\alpha\bar{u}+\beta^{2}}p_{\bar{v}}^{2}+\frac{p_{\bar{w}}^{2}}{\big{(}4\alpha\bar{u}+\beta^{2}\big{)}\cos^{2}{(2\alpha\bar{v})}}\right).$

With the final step in the transformation,

$\displaystyle u=\ln\left(\bar{u}+\frac{\beta^{2}}{4\alpha}\right),\qquad
v=4\alpha\bar{v},\qquad w=\bar{w},$

we obtain

$\displaystyle T_{1}=4\alpha p_{v},\qquad
H=\frac{16\alpha^{2}\mathrm{e}^{-u}}{4\alpha\mathrm{e}^{u}+4\alpha\gamma-\beta^{2}}\left(p_{u}^{2}+p_{v}^{2}+\frac{p_{w}^{2}}{16\alpha^{2}\cos^{2}{\frac{v}{2}}}\right),$

which has the $D_{3}$ kinetic energy, with the same potential (equivalent to
the “$b_{1}$” part of Case B in [9]).

##  11 Conclusions

The purpose of this paper was to extend the method developed in [6] from 2 to
3 degrees of freedom. The original motivation for [6] was to understand
Darboux–Koenigs super-integrable systems, which are characterised as having
exactly one first order integral and 2 second order integrals. These are
therefore associated with non-constant curvature metrics, which, in 2
dimensions, are necessarily conformally flat. The method of [6] exploited the
conformal flatness by building first integrals from conformal symmetries.

In the current paper we considered 3-dimensional conformally flat metrics and
built integrals from the corresponding conformal algebra (2.2). We initially
assumed one first order integral $K$ and three quadratic ones $F_{i}$, giving
us five integrals in all. A further requirement was involutivity of three of
these and independence of them all, to give us a super-integrable system.
These requirements are the obvious generalisation of the Darboux–Koenigs
systems to 3 degrees of freedom.

The classification of such systems is a difficult task. As can be seen in the
examples, we could start with a different collection of quadratic integrals to
arrive at the same Hamiltonian (in particular, for the cases with larger
symmetry algebras). However, the structure of the conformal algebra (2.2),
with its involutive automorphisms, enabled us to limit the starting choice of
Killing vector to either $e_{1}$, $h_{1}$ or $h_{4}$, thus reducing the
overall number of choices. We did not tackle the classification problem, but
presented a number of interesting, multi-parameter examples, which, upon
restriction, led to systems with a higher degree of symmetry, which were then
able to be reduced to 2 degrees of freedom. Almost all our examples, upon
reduction, had one linear and two quadratic first integrals, so were
inevitably in the Darboux–Koenigs classification. However, the third dimension
is “remembered” through the addition of a potential function, all of which
were just special cases of those classified in [9, 10]. Furthermore, when
there existed two inequivalent symmetries (with respect to involutions of the
system under study) we were able to reduce in two different ways, leading to
the introduction of two different potentials, and the two potentials could be
related through a transformation in the 3-dimensional space. This means that
the potentials in the classification of [9, 10] are not fully independent, if
we allow transformations which take us out of the 2-dimensional space.

We see from the results of Section 7, that different parameter restrictions of
the 3-parameter Hamiltonian with conformal factor (4.3) can reduce to
different Darboux–Koenigs cases. This means that the Hamiltonian (4.1), with
$\varphi$ given by (4.3), simultaneously contains both Darboux–Koenigs cases
$D_{2}$ and $D_{4}$. Furthermore, since there are only 4 metrics in the
Darboux–Koenigs classification, we inevitably arrive at the same metric, from
two or more different 3-dimensional metrics (see the examples of Sections 7.2
and 9.1).

Whilst most of our examples (before reduction) possessed a 3-dimensional
isometry algebra, the Hamiltonian (9.3) had only one first order integral, so
the reduced metric had no Killing vectors at all. Nevertheless, we were able
to show that the reduced Hamiltonian was maximally super-integrable,
possessing quadratic and quartic integrals, thus falling out of the
Darboux–Koenigs class of metric.

This paper represents an important first step in our investigations into the
use of the conformal algebra to build higher order first integrals. Clearly we
need a better understanding of the underlying structure in order to tackle the
important open problem of the classification of such higher-dimensional
superintegrable systems and their reductions.

Reducing from 3 to 2 dimensions, we acquired 1 parameter potentials. It is
likely that if we start from a 4-dimensional system, with a large enough
isometry algebra, we could produce a 2-parameter restriction of the potentials
given in [9, 10]. Perhaps the full potential could be recovered by going to
high enough dimensions.

### Acknowledgements

This work was supported by NSFC (Grant No. 11871396) and NSF of Shaanxi
Province of China (Grant No. 2018JM1005). APF thanks Boris Kruglikov for
useful discussions on the “gap problem”. We thank the referees and an editor
for their useful remarks.

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