Toward Classification of 2nd Order Superintegrable Systems

Toward Classification of 2nd Order Superintegrable
Systems in 3-Dimensional Conformally Flat Spaces
with Functionally Linearly Dependent Symmetry
Operators

Bjorn K. BERNTSON †, Ernest G. KALNINS ‡ and Willard MILLER Jr. §

B.K. Berntson, E.G. Kalnins and W. Miller Jr.

† Department of Mathematics, KTH Royal Institute of Technology, Stockholm,
Sweden<EMAIL_ADDRESS>
‡ Department of Mathematics, University of Waikato, Hamilton, New Zealand
<EMAIL_ADDRESS>
§ School of Mathematics, University of Minnesota, Minneapolis, Minnesota, USA
<EMAIL_ADDRESS>http://www-users.math.umn.edu/~mille003/

Received April 07, 2020, in final form December 09, 2020; Published online
December 16, 2020

We make significant progress toward the classification of 2nd order
superintegrable systems on 3-dimensional conformally flat space that have
functionally linearly dependent (FLD) symmetry generators, with special
emphasis on complex Euclidean space. The symmetries for these systems are
linearly dependent only when the coefficients are allowed to depend on the
spatial coordinates. The Calogero–Moser system with 3 bodies on a line and
2-parameter rational potential is the best known example of an FLD
superintegrable system. We work out the structure theory for these FLD systems
on 3D conformally flat space and show, for example, that they always admit a
1st order symmetry. A partial classification of FLD systems on complex 3D
Euclidean space is given. This is part of a project to classify all 3D 2nd
order superintegrable systems on conformally flat spaces.

superintegrable systems; Calogero 3 body system; functional linear dependence

20C35; 35B06; 70H20; 81Q80; 81R12

##  1 Introduction

There is a hierarchy of 2nd order classical and quantum superintegrable
systems in 3-dimensional conformally flat spaces, ranging from the most
tractable at the top, nondegenerate (i.e., 4-parameter) potentials with 6
linearly independent symmetries, all of which have been classified, followed
by semidegenerate (i.e., 3-parameter) potentials on which much progress has
been made, to the least tractable (1-parameter) for classification at the
bottom. By definition the 2 classes at the top admit 5 functionally linearly
independent symmetry operators, i.e., they are not only linearly independent
in the usual sense but also if the coefficients are allowed to depend on the
spatial variables. However, there exist 2nd order superintegrable systems with
at least 5 functionally linearly dependent symmetry operators and 2-parameter
potentials; such systems have never been classified. We initiate the study of
such systems by developing their structure theory on conformally flat spaces
and performing a partial classification of these systems in constant curvature
spaces.

We recall some basic facts and results about conformally flat superintegrable
systems. An $n$-dimensional complex Riemannian space is conformally flat if
and only if it admits a set of local coordinates $\\{x_{1},\dots,x_{n}\\}$
such that the contravariant metric tensor takes the form
$g^{ij}=\delta^{ij}/\lambda({\bf x})$ [10, 14]. A classical superintegrable
system ${\mathcal{H}}=\sum_{ij}g^{ij}p_{i}p_{j}+V({\bf x})$ on the phase space
of this manifold is one that admits $2n-1$ functionally independent
generalized symmetries (or constants of the motion) ${\mathcal{S}}_{k}$ for
$k=1,\dots,2n-1$ with ${\mathcal{S}}_{1}={\mathcal{H}}$ where the
${\mathcal{S}}_{k}$ are polynomials in the momenta $p_{j}$. It is easy to see
that $2n-1$ is the maximum possible number of functionally independent
symmetries and, locally, such (in general nonpolynomial) symmetries always
exist. The system is second order maximal superintegrable if the $2n-1$
functionally independent symmetries can be chosen to be quadratic in the
momenta. (Second order superintegrable systems, though complicated, are
tractable because standard orthogonal separation of variables techniques are
associated with second order symmetries, and these techniques can be brought
to bear.)

For a classical 3D system in a conformally flat space (note that all 2D spaces
are conformally flat) we can always choose local coordinates $\\{x,y,z\\}$,
not unique, such that the Hamiltonian takes the form
${\mathcal{H}}=\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}/{\lambda(x,y,z)}+V(x,y,z).$
This system is second order superintegrable with semidegenerate potential
$V=V(x,y,z;\alpha,\beta,\gamma)=\alpha V^{\alpha}({\bf x})+\beta
V^{\beta}({\bf x})+\gamma V^{\gamma}({\bf x})$ if it admits 5 functionally
independent quadratic constants of the motion, i.e., generalized symmetries,

${\mathcal{S}}_{k}=\sum_{i,j}a^{ij}_{k}p_{i}p_{j}+W_{k}(x,y,z;\alpha,\beta,\gamma)={\mathcal{S}}_{k}^{0}+W_{k},\qquad
k=1,\dots,5.$

Here the functions $V^{\alpha}$, $V^{\beta}$, $V^{\gamma}$ are independent of
the parameters $\alpha$, $\beta$, $\gamma$, the set
$\big{\\{}V^{\alpha},V^{\beta},V^{\gamma}\big{\\}}$ must have linearly
independent gradients, and we ignore the additive constant. We call this a
3-parameter potential.

In some cases the system may also have a 6th symmetry ${\mathcal{S}}_{6}$,
(but no more) such the set
$\big{\\{}{\mathcal{S}}_{k}^{0}\,|\,k=1,\dots,6\big{\\}}$ is functionally
linearly independent and this implies that the potential depends on 4
parameters [10]. Furthermore the classification theory requires that the 5, 6
constants of the motion be functionally linearly independent, i.e., the
equation

$\sum_{k=1}^{5,6}f_{k}({\bf x})\mathcal{S}_{k}^{0}({\bf x})=0$ (1.1)

is satisfied if and only if $f_{k}({\bf x})=0$ for all $k$. If equation (1.1)
is satisfied for functions $f_{k}({\bf x})$ not identically $0$, the set of
constants of the motion are functionally linearly dependent (FLD).

For 2nd order superintegrable systems in 3 dimensions that are functionally
linearly independent, the systems that admit 6 linearly independent second
order constants of the motion (the maximum possible) have all been classified
[6, 13] and there has been considerable progress on the remaining 5 linear
independent case [7, 12]. However, little has been done to classify
superintegrable systems in 3 dimensions that are FLD. The best known such
system is the rational 3-body Calogero–Moser system on the line with
2-parameter potential. To the best of our knowledge there are no 2nd order FLD
superintegrable systems with trigonometric, hyperbolic, or elliptic
potentials. In this paper we derive structure results for all 2nd order
superintegrable FLD systems with $r\geq 5$ linearly independent second order
symmetries on conformally flat real or complex spaces that have potentials
that depend on 2 functionally independent variables (the maximum possible),
and such that the FLD equation $\sum_{k}f_{k}({\bf x})\mathcal{S}_{k}^{0}({\bf
x})=0$ is satisfied with at most 5 nonzero terms $f_{k}({\bf x})$. (For the
analogous 2nd order 2-dimensional FLD systems the answer is known: there is
only one such family of systems [9].)

The paper is organized as follows: in Section 2 we present the Calogero–Moser
system and a system on 3-dimensional Minkowski space as examples. In Section 3
we present structure results for all FLD systems on conformally flat spaces.
The most important result is that all such systems admit a 1st order constant
of the motion. In Section 4 we work out a partial classification of all
3-dimensional second order superintegrable FLD systems in flat space with
2-parameter potentials, such that the FLD equation $\sum_{k}f_{k}({\bf
x})\mathcal{S}_{k}^{0}({\bf x})=0$ is satisfied with at most 5 nonzero terms
$f_{k}({\bf x})$, (including the structure of the symmetry algebras for most
of these systems). In Section 5 we summarize the corresponding result for
3-dimensional FLD systems on the complex 3-sphere. In Section 6 we present
some conclusions and a brief discussion of related properties of these
systems. Here all of our systems are classical. However the quantum analogs
follow easily by symmetrization of the symmetry operators and there is a 1-1
matching of the Hamiltonians modulo the scalar curvature [11]. In particular
the Euclidean space Hamiltonians are identical.

##  2 Examples

###  2.1 An FLD example: the rational Calogero–Moser system
with 2-parameter potential

This potential takes the form [1, 4, 5, 8, 16, 17, 18, 19, 20]

$V=\frac{\alpha}{(x-y)^{2}}+\frac{\beta}{(y-z)^{2}}+\frac{\gamma}{(z-x)^{2}}.$
(2.1)

Let us consider the system of symmetries defining the system with potential
$V$. A basis for the space of symmetries is (using $J_{12}=xp_{2}-yp_{1}$,
$J_{23}=yp_{3}-zp_{2}$, $J_{13}=xp_{3}-zp_{1}$)

$\displaystyle{\mathcal{S}}_{1}={\mathcal{H}}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+V,\qquad{\mathcal{S}}_{2}=(p_{1}+p_{2}+p_{3})^{2},\qquad{\mathcal{S}}_{3}=J_{12}^{2}+J_{23}^{2}+J_{13}^{2}+W_{3},$
$\displaystyle{\mathcal{S}}_{4}=p_{1}(J_{13}-J_{12})+p_{2}(J_{12}-J_{23})+p_{3}(J_{23}-J_{13})+W_{4},$
$\displaystyle{\mathcal{S}}_{5}=J_{12}J_{13}+J_{23}J_{12}+J_{13}J_{23}+W_{5},$

where the potential terms $W_{i}$ contain the parameters $\alpha$, $\beta$,
$\gamma$. In this case, the Bertrand–Darboux equations [10, 11] for each
symmetry ${\mathcal{S}}_{k}=\sum_{ij}a^{ij}_{k}p_{i}p_{j}+W_{k}$ of
${\mathcal{H}}$ are

$\displaystyle
V_{x}+V_{y}+V_{z}=0,\qquad(x-y)V_{xy}+(z-y)V_{yz}-V_{x}+2V_{y}-V_{z}=0,$
$\displaystyle(x-z)V_{xz}+(y-z)V_{yz}-V_{x}-V_{y}+2V_{z}=0,$ (2.2)

and their differential consequences.

We say that this is a (functionally independent) 2-parameter potential. A
2-parameter potential is one that can be expressed in the form
$V=a_{1}f(x,y,z)+a_{2}g(x,y,z)$ where $a_{1}$, $a_{2}$ are arbitrary
parameters, $f$ and $g$ are independent of these parameters, and the set
$\\{f,g\\}$ is linearly independent. (Here we are ignoring the trivially
additive parameter in the potential.) Functional independence for the
potential is the additional requirement that the set $\\{f,g\\}$ is
functionally independent. Functional dependence means essentially that the
system could be recast as 1-parameter.

What is important to notice here is the occurrence of the first order
condition $V_{x}+V_{y}+V_{z}=0$ for the potential as a consequence of the
Bertrand–Darboux equations. Thus the potential is a function of only two
functionally independent variables, impossible for nondegenerate potentials.

We observe the FLD relation

$(x+y+z)^{2}{{\mathcal{S}}}_{1}^{0}-\big{(}x^{2}+y^{2}+z^{2}\big{)}{{\mathcal{S}}}_{2}^{0}+2{{\mathcal{S}}}_{3}^{0}-2(x+y+z){{\mathcal{S}}}_{4}^{0}-2{{\mathcal{S}}}_{5}^{0}=0$

obeyed by the purely quadratic terms in the symmetries, i.e., we have set
${\mathcal{S}}_{i}={{\mathcal{S}}}_{i}^{0}+W_{i}$. We show below, in Theorem
3.1, that the existence of such an FLD relation implies the existence of a
first order condition in the Bertrand–Darboux equations (2.2). Furthermore,
the 5 quadratic symmetries are functionally dependent:

${\mathcal{H}}({\mathcal{S}}_{3}-{\mathcal{S}}_{5})-\frac{{\mathcal{S}}_{4}^{2}}{2}-\frac{{\mathcal{S}}_{2}{\mathcal{S}}_{3}}{2}+\frac{(\alpha+\beta+\gamma)}{2}{\mathcal{H}}-\frac{(\alpha+\beta+\gamma)}{2}{\mathcal{S}}_{2}=0.$

Hence the system defined by (2.1) is minimally superintegrable with 4
functionally independent symmetries. We show below, in Corollary 3.2, that
this is a generic feature of FLD systems with exactly 5 linearly independent
generators.

###  2.2 A Minkowski space FLD example

Here

${\mathcal{H}}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\alpha(x-z)+\beta(y+{\rm
i}z)+\gamma(y+{\rm i}z)^{2},$ (2.3)

which admits the 1st order symmetry

${\mathcal{J}}=p_{1}-{\rm i}p_{2}+p_{3}$

and the 2nd order symmetries [7]

$\displaystyle{\mathcal{S}}_{1}={\mathcal{H}}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\alpha(x-z)+\beta(y+{\rm
i}z)+\gamma(y+{\rm i}z)^{2},$
$\displaystyle{\mathcal{S}}_{2}={\mathcal{J}}^{2},\qquad{\mathcal{S}}_{3}=p_{1}^{2}+\alpha
x,\qquad{\mathcal{S}}_{4}=(-{\rm i}p_{2}+p_{3})p_{1}+(p_{3}-{\rm
i}p_{2})^{2}+\tfrac{\alpha}{2}({\rm i}y-x-z),$
$\displaystyle{\mathcal{S}}_{5}=(p_{1}-{\rm i}p_{2}+p_{3})({\rm
i}J_{12}-J_{13})-\tfrac{{\rm i}}{2}\alpha yz-\tfrac{{\rm i}}{2}\alpha
xy+\tfrac{1}{4}\alpha x^{2}+\tfrac{1}{2}\alpha xz-\tfrac{1}{4}\alpha
y^{2}+\tfrac{1}{4}\alpha z^{2}.$

The 5 generators are linearly independent and satisfy the FLD relation

$({\rm i}y-z){\mathcal{S}}_{2}^{0}+(-{\rm
i}y+x+z){{\mathcal{S}}}_{4}^{0}+{{\mathcal{S}}}_{5}^{0}=0,$

where as before $\mathcal{S}_{k}^{0}$ is the quadratic momentum part of the
symmetry $\mathcal{S}_{k}$.

##  3 Some theory

Functional linear dependence of a functionally independent maximal set of
symmetries is hard to achieve. We recall the following result where the system
need not be superintegrable [11]:

######  Theorem 3.1.

Let the linearly independent set
$\\{{\mathcal{H}}={\mathcal{S}}_{1},{{\mathcal{S}}}_{2},\dots,{\mathcal{S}}_{t}\\}$,
$(t>2)$ be a functionally linearly dependent basis of $2$nd order symmetries
for the system
${\mathcal{H}}=\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{3}\big{)}/\lambda({\bf
x})+V={\mathcal{H}}^{0}+V$ with nontrivial potential $V$, i.e., there is a
relation $\sum_{k}f_{k}({\bf x}){{\mathcal{S}}}_{k}^{0}\equiv 0$ in an open
set, where not all $f_{k}({\bf x})$ are constants, and no such relation holds
for the $f_{k}({\bf x})$ all constant, except if the constants are all zero.
$($Here ${\mathcal{S}}_{i}={\mathcal{S}}_{i}^{0}+W_{i}$ where the $W_{i}$ are
the potential terms.$)$ Then the potential must satisfy a first order relation
$AV_{x}+BV_{y}+CV_{z}=0$ where not all of the functions $A$, $B$, $C$ vanish.

###### Proof.

By relabeling, we can express one of the quadratic parts of the constants of
the motion ${{\mathcal{S}}}_{0}^{0}$ as a linear combination of a linearly
independent subset

$\big{\\{}{{\mathcal{S}}}_{1}^{0},\dots,{{\mathcal{S}}}_{r}^{0},\,1\leq r\leq
t-1\big{\\}},$

i.e.,

${\mathcal{S}}_{0}^{0}=\sum_{\ell=1}^{r}f_{\ell}(x,y,z){{\mathcal{S}}}_{\ell}^{0}.$

Taking the Poisson bracket of both sides of this equation with
$\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{3}\big{)}/\lambda$ and using the fact that
each of the ${{\mathcal{S}}}_{h}$ is a constant of the motion, we obtain the
identity

$\sum_{\ell=1}^{r}\sum_{i,j=1}^{3}(\partial_{x_{k}}f_{\ell})a_{\ell}^{ij}p_{i}p_{j}p_{k}=0,$

where $(x,y,z)\equiv(x_{1},x_{2},x_{3})$. It is straightforward to check that
this identity can be satisfied if and only if the functions

$c_{k}^{ij}=\sum_{\ell=1}^{r}(\partial_{x_{k}}f_{\ell})a_{\ell}^{ij},\qquad
1\leq i,j,k\leq 3$

satisfy the equations

$c_{i}^{ii}=0,\qquad c_{j}^{ii}+2c_{i}^{ij}=0,\quad i\neq j,\qquad
c_{3}^{12}+c_{1}^{23}+c_{2}^{31}=0.$ (3.1)

Note that $c_{k}^{ij}=c_{k}^{ji}$. Corresponding to each of the basis
symmetries ${\mathcal{S}}_{\ell}$ there is a linear set $C_{\ell}=0$ of
Bertrand–Darboux equations [11]. A straightforward substitution into the
identity $C_{0}-\sum_{\ell=1}^{r}f_{\ell}({\bf x})C_{\ell}=0$ yields the
relation

$\left(\begin{matrix}c^{12}_{1}-c^{11}_{2}\\\ c^{31}_{1}-c^{11}_{3}\\\
c^{31}_{2}-c^{21}_{3}\end{matrix}\right)V_{1}+\left(\begin{matrix}c^{22}_{1}-c^{21}_{2}\\\
c^{32}_{1}-c^{12}_{3}\\\
c^{32}_{2}-c^{22}_{3}\end{matrix}\right)V_{2}+\left(\begin{matrix}c^{32}_{1}-c^{31}_{2}\\\
c^{33}_{1}-c^{13}_{3}\\\ c^{33}_{2}-c^{23}_{3}\end{matrix}\right)V_{3}=0.$
(3.2)

These first order differential equations for the potential cannot all vanish
identically. Indeed if they did all vanish then we would have the conditions

$\displaystyle c^{12}_{1}=c^{11}_{2},\qquad c^{31}_{1}=c^{11}_{3},\qquad
c^{31}_{2}=c^{21}_{3},\qquad c^{22}_{1}=c^{21}_{2},\qquad
c^{32}_{1}=c^{12}_{3},$ $\displaystyle c^{32}_{2}=c^{22}_{3},\qquad
c^{32}_{1}=c^{31}_{2},\qquad c^{33}_{1}=c^{13}_{3},\qquad
c^{33}_{2}=c^{23}_{3}.$

These conditions, together with conditions (3.1) show that $c_{i}^{jk}=0$ for
all $i$, $j$, $k$. Thus we have
$\sum_{\ell=1}^{r}(\partial_{x_{k}}f_{\ell})a_{\ell}^{ij}=0$, $1\leq i,j,k\leq
3$. Since the set
$\big{\\{}{{\mathcal{S}}}_{1}^{0},\dots,{{\mathcal{S}}}_{r}^{0}\big{\\}}$, is
functionally linearly independent, we have $\partial_{x_{k}}f_{\ell}\equiv 0$
for $1\leq k\leq 3$, $1\leq\ell\leq r$. Hence the $f_{\ell}$ are constants,
which means that
${{\mathcal{S}}}_{0}^{0}-\sum_{\ell=1}^{r}f_{\ell}{{\mathcal{S}}}_{\ell}^{0}=0$.
Thus the set
$\big{\\{}{\mathcal{S}}_{0}^{0},\dots,{\mathcal{S}}_{4}^{0}\big{\\}}$ is
linearly dependent. This is a contradiction! ∎

This shows that the potential function for any system, superintegrable or not,
with a basis of symmetries that is functionally linearly dependent must
satisfy at least one nontrivial first order partial differential equation
$AV_{x}+BV_{y}+CV_{z}=0$ where the functions $A$, $B$, $C$ are parameter-free.
This means that all such potentials depend on either one or two functionally
independent coordinates.

######  Corollary 3.2.

Suppose the system has exactly $5$ linearly independent generators
$\\{{\mathcal{S}}_{1}={\mathcal{H}},\dots,\allowbreak{\mathcal{S}}_{5}\\}$ and
is a functionally linearly dependent basis of $2$nd order symmetries for the
Hamiltonian
${\mathcal{H}}=\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{3}\big{)}/\lambda({\bf
x})+V={\mathcal{H}}^{0}+V$ with $3$-parameter potential. Then this set of $5$
generators must be functionally dependent.

###### Proof.

Suppose the set is functionally independent. Then from [12, equation (2)] at
any fixed point there is a potential for any prescribed values of $V$,
$V_{x}$, $V_{y}$, $V_{z}$. However, since the system is FLD the potentials
must satisfy $A(x,y,z)V_{x}+B(x,y,z)V_{y}+C(x,y,z)V_{z}=0$ for $A$, $B$, $C$
not all zero, so the possible derivatives of $V$ are not independent.
Contradiction! ∎

Thus for systems with exactly 5 linearly independent symmetries at most 4 of
the 5 FLD generators can form a functionally independent set. However we shall
show that there are FLD systems with 2-parameter potentials that admit $>5$
linearly independent and 5 functionally independent 2nd order symmetries in
which case Corollary 3.2 does not apply.

######  Lemma 3.3.

Equations (3.1) imply

$\partial_{x_{i}}\big{(}c^{ij}_{i}-c^{ii}_{j}\big{)}=0,\qquad\partial_{x_{i}}\big{(}c^{ij}_{k}-c^{ik}_{j}\big{)}=0.$

A new result is

######  Theorem 3.4.

Under the hypotheses of Theorem 3.1 there exists a $1$st order Killing vector
$\mathcal{J}$ for $\mathcal{H}$, i.e.,
$\\{{\mathcal{J}},{\mathcal{H}\\}=\\{\mathcal{J}},V\\}=0$, of the form

${\mathcal{J}}=a_{1}p_{1}+a_{2}p_{2}+a_{3}p_{3}+a_{4}(xp_{2}-yp_{1})+a_{5}(yp_{3}-zp_{2})+a_{6}(zp_{1}-xp_{z})$

for some constants $a_{j}$, not all zero.

###### Proof.

Let

${\mathcal{J}}=\big{(}c_{1}^{12}-c_{2}^{11}\big{)}p_{1}+\big{(}c_{1}^{22}-c_{2}^{12}\big{)}p_{2}+\big{(}c_{1}^{23}-c_{2}^{13}\big{)}p_{3}={\mathcal{J}}^{x}p_{1}+{\mathcal{J}}^{y}p_{2}+{\mathcal{J}}^{z}p_{3},$

so that the first of equations (3.2) is $\\{{\mathcal{J}},V\\}=0$. From
equations (3.1) and Lemma 3.3 we can verify that

$\displaystyle\big{\\{}{\mathcal{J}},{\mathcal{H}}^{0}\big{\\}}=-\left[\frac{\big{(}c^{12}_{1}-c_{2}^{11}\big{)}\lambda_{1}}{\lambda}+\frac{\big{(}c^{22}_{1}-c_{2}^{12}\big{)}\lambda_{2}}{\lambda}+\frac{\big{(}c^{23}_{1}-c_{2}^{13}\big{)}\lambda_{3}}{\lambda}\right]{\mathcal{H}}^{0}$
$\displaystyle\hphantom{\big{\\{}{\mathcal{J}},{\mathcal{H}}^{0}\big{\\}}}{}=-\frac{1}{\lambda}\big{[}3c_{1}^{12}\lambda_{1}-3c_{2}^{12}\lambda_{2}+\big{(}c_{1}^{23}-c_{2}^{13}\big{)}\lambda_{3}\big{]}{\mathcal{H}}^{0},$

so either ${\mathcal{J}}=0$ or ${\mathcal{J}}$ is a conformal symmetry of
${\mathcal{H}}^{0}$. However, from Lemma 3.3 we see that

$\partial_{x}{\mathcal{J}}^{x}=\partial_{y}{\mathcal{J}}^{y}=\partial_{z}{\mathcal{J}}^{z}=0.$
(3.3)

The first order conformal symmetries of ${\mathcal{H}}^{0}$ are the same as
for the case $\lambda=1$, and the only such symmetries that satisfy the
requirements (3.3) are linear combinations of $p_{1}$, $p_{2}$, $p_{3}$ and

$J_{12}=xp_{2}-yp_{1},\qquad J_{23}=yp_{3}-zp_{2},\qquad
J_{13}=xp_{3}-zp_{1},$

and these would be actual symmetries of ${\mathcal{H}}^{0}$ (true conformal
symmetries such as $xp_{1}+yp_{2}+zp_{3}$ fail this test). Thus either
${\mathcal{J}}$ vanishes or it is a 1st order symmetry of $\mathcal{H}$.

Analogous constructions and conclusions can be obtained for the 2nd and 3rd of
equations (3.2). However, at least one of these equations is nonzero. ∎

Since any Euclidean coordinate transformation applied to the Hamiltonian
$\mathcal{H}$ takes it into one of similar form

${\tilde{\mathcal{H}}}=\frac{{\tilde{p}}^{2}_{1}+{\tilde{p}}^{2}_{2}+{\tilde{p}}^{2}_{3}}{\tilde{\lambda}}+{\tilde{V}},$

without loss of generality, we can assume that, up to conjugacy, $\mathcal{J}$
takes one of the five canonical forms:

$\displaystyle p_{1},\quad p_{1}+{\rm i}p_{2},\quad
xp_{2}-yp_{1},\quad(xp_{2}-yp_{1})+{\rm i}(yp_{3}-zp_{2}),$
$\displaystyle(xp_{2}-yp_{1})+{\rm i}(yp_{3}-zp_{2})+p_{3}+{\rm i}p_{1}.$
(3.4)

With the same assumptions for FLD systems as in Theorem 3.1, let
${\mathcal{O}}_{{\mathcal{H}}^{0}}(r)$ be the set of all subsets
${\mathcal{B}}$ of
$\\{{\mathcal{S}}_{1}={\mathcal{H}},{{\mathcal{S}}}_{2},\dots,{\mathcal{S}}_{t}\\}$
with $r+1$ elements such that, after relabeling, there is an FLD relation

$\hat{\mathcal{S}}_{0}^{0}=\sum_{\ell=1}^{r}\hat{f}_{\ell}(x,y,z)\hat{\mathcal{S}}_{\ell}^{0},$
(3.5)

and such that

  1. 1)

${\mathcal{H}}^{0},{\mathcal{J}}^{2}\in\operatorname{span}{\mathcal{B}}$,

  2. 2)

$\operatorname{span}\mathcal{B}\subseteq\operatorname{span}\operatorname{Ad}_{\mathcal{J}}\mathcal{B}$,

  3. 3)

$\mathcal{H}$ admits a 2-parameter potential.

In this paper we find all superintegrable Hamiltonians $\mathcal{H}$ on
constant curvature spaces for which
${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)\neq\varnothing$. Note that if
$\mathcal{H}$ admits exactly 5 linearly independent symmetries, all cases are
included in ${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$. If $\mathcal{H}$ admits
more than 5 linearly independent 2nd order symmetries we have no proof of
completeness but we have not as yet found a verifiable counterexample.

##  4 Euclidean space

We first study the possible FLD 2nd order superintegrable systems in 3D
complex Euclidean space. Complex metrics were commonly used in the 19th
century. Of particular interest for superintegrability and separation of
variables are the paper [15] and the book [3]. Bôcher was the first president
of the American Mathematical Society. The metrics are defined as usual as are
the curvature conditions but all the variables are complex. Thus a space is
conformally flat if the metric can be expressed as $\lambda(x,y,z)\big{(}{\rm
d}x^{2}+{\rm d}y^{2}+{\rm d}z^{2}\big{)}$ for complex variables $x$, $y$, $z$.
The advantage is that one complex system can describe several real forms of
this system by specializing the coordinates. For example the complex metric
${\rm d}x^{2}+{\rm d}y^{2}+{\rm d}z^{2}$ is Euclidean for $x$, $y$, $z$ real
and Minkowski space for $z={\rm i}w$ for $x$, $y$, $w$ real. In this paper,
potentials $V(x,y,z)$ that are real for $x$, $y$, $z$ real live on Euclidean
space and potentials that are real for $x$, $y$, $w$ real live on Minkowski
space. Every potential belongs to one of these classes. Similar remarks are
true for the complex 3-sphere, with real forms the real 3-sphere, and the
3-hyperboloids of one and two sheets.

By relabeling, we can express one of the quadratic parts of the constants of
the motion ${{\mathcal{S}}}_{0}^{0}$ as a linear combination of the quadratic
parts of the remaining $r$ generators through (3.5). Without loss of
generality we can reduce to the case where the expansion (3.5) is unique. The
generators
${{\mathcal{S}}_{0}^{0}},{{\mathcal{S}}_{1}^{0}},{{\mathcal{S}}_{2}^{0}},\dots,{{\mathcal{S}}_{r}^{0}}$
are polynomials in $x$, $y$, $z$ of order at most 2 and are linearly
independent. Thus we can solve for the expansion coefficients in the form
$f_{\ell}(x,y,z)=s_{\ell}(x,y,z)/s_{0}(x,y,z)$, $\ell=1,\dots,4$ where
$s_{0},s_{1},\dots,s_{r}$ are polynomials in $x$, $y$, $z$ of order at most 2.
It follows that

$\sum_{a_{1},a_{2},a_{3}}A(a_{1},a_{2},a_{3})x^{a_{1}}y^{a_{2}}z^{a_{3}}\equiv
s_{0}{{\mathcal{S}}_{0}^{0}}-\sum_{\ell=1}^{4}s_{\ell}{{\mathcal{S}}_{\ell}^{0}}=0,$
(4.1)

where each coefficient $A(a_{1},a_{2},a_{3})$ must vanish. In particular, the
sum of all terms homogeneous of degree $n$ must vanish for each $n=0,\dots,4$:

$B(n)\equiv\sum_{a_{1}+a_{2}+a_{3}=n}A(a_{1},a_{2},a_{3})x^{a_{1}}y^{a_{2}}z^{a_{3}}=0.$

Each of the generators ${{\mathcal{S}}_{r}^{0}}$ is a linear combination of
terms $J_{ij}J_{k\ell}$, (order 2), $J_{ij}p_{k}$, (order 1) and $p_{i}p_{j}$,
(order 0).

Since we have assumed that the expansion (3.5) is unique, there must be only a
single term $B(N)$ that is not identically zero and each
${{\mathcal{S}}_{\ell}^{0}}$ is homogeneous of degree $0$, $1$, or $2$. Thus
each $s_{\ell}$ must be homogeneous of degree $b$ and each
${{\mathcal{S}}_{\ell}^{0}}$ must be homogeneous of degree $c=0,1,2$ where
$b+c=N$. This greatly restricts the possibilities for (4.1).

###  4.1 Classification criteria

In the subsequent five subsections we obtain all FLD-superintegrable bases
${\mathcal{B}}$ on 3D complex Euclidean space that belong to the class
${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$. Each such basis is associated with a
Hamiltonian ${\mathcal{H}}={\mathcal{H}}^{0}+V$ with a two-parameter potential
$V$. We emphasize that ${\mathcal{B}}$ does not necessarily contain all the
(momentum parts of) symmetries of ${\mathcal{H}}$. We compute $V$ as the
general solution of the Bertrand–Darboux equations associated with
${\mathcal{B}}$. However, the Hamiltonians ${\mathcal{H}}={\mathcal{H}}^{0}+V$
obtained in this way may admit additional symmetries not obtained from
${\mathcal{B}}$. Additionally, we remark that a particular solution $V_{{\rm
p}}$ of the general solution $V$ may correspond to a Hamiltonian with more
symmetries than the Hamiltonian with $V$. We make no attempt to classify these
special cases.

The classification is performed modulo complex Euclidean transformations: by
the discussion in Section 3, the Hamiltonian $\mathcal{H}$ must admit one of
the first order symmetries in (3.4). Starting from each of the symmetries in
(3.4), which we denote by $\mathcal{J}$, we use the following algorithm to
identify FLD-superintegrable systems.

  1. 1.

We compute the action of $\operatorname{Ad}_{\mathcal{J}}$ on a basis of
second order symmetries of $\mathcal{H}^{0}$. We use this to construct a
generalized eigenbasis (with respect to $\operatorname{Ad}_{\mathcal{J}}$) of
such possible second order symmetries.

  2. 2.

We then consider 5-element subsets ${\mathcal{B}}$ of this basis and verify
that ${\mathcal{B}}\in{\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$.

  3. 3.

For each possible action of $\operatorname{Ad}_{\mathcal{J}}$ on
$\mathcal{B}$, we identify all possibilities where 1) the elements of
$\mathcal{B}$ are homogeneous in the spatial variables, in accordance with the
discussion in the previous subsection, 2) the elements of $\mathcal{B}$ are
FLD, 3) the elements $\mathcal{H}^{0}$ and $\mathcal{J}^{2}$ are contained in
$\operatorname{span}\mathcal{B}$.

  4. 4.

For each basis satisfying the criteria in the previous step, we use the
Bertrand–Darboux equations to compute the corresponding potential. We require
that the potential be 2-parameter functionally independent. In this case
$\mathcal{B}\in\mathcal{O}_{\mathcal{H}^{0}}(4)$. We verify that $\mathcal{H}$
is superintegrable: it must admit at least 4 functionally independent
symmetries. The final list of the potentials defining such systems is given in
Table 1.

In the case of $\mathcal{J}=p_{1}$, the space of quadratic forms in
$\\{p_{1},p_{2},p_{3},J_{12},J_{13},J_{23}\\}$, modulo the relation
$\mathbf{p}\cdot(\mathbf{p}\times\mathbf{x})=0$, provides a generalized
eigenbasis of order two symmetries with respect to
$\operatorname{Ad}_{\mathcal{J}}$. Hence we provide details of steps 2-4 of
the algorithm above and also show that our examples in Section 2 are contained
in this case.

The computations involved in the cases of the remaining forms in (3.4) are
lengthier and we provide only the essential details. In all cases we supply
the potentials and the algebra generated by the FLD symmetries.

###  4.2 First case: $\boldsymbol{{\mathcal{J}}=p_{1}}$

Here the centralizer of $\mathcal{J}$ is the group generated by translation in
$y$, $z$ and rotation about the $x$-axis. We can use this freedom to simplify
the computation. Since $p_{1}$ is a symmetry the potential must be of the form
$V(y,z)$. Any degree two symmetry can be written as a quadratic form in
$\\{p_{1},p_{2},p_{3},J_{12},J_{13},J_{23}\\}$. Due to the triple product
identity $\mathbf{p}\cdot(\mathbf{p}\times\mathbf{x})=0$, the space of such
quadratic forms has dimension $21-1=20$.

To be concrete, we write a general symmetry as

$\mathcal{S}=RQR^{\rm T}+F_{0}(x,y,z),$ (4.2)

where

$R=(p_{1},p_{2},p_{3},J_{12},J_{13},J_{23})$ (4.3)

and

$Q=\frac{1}{2}\left(\begin{matrix}2a_{1}&a_{2}&a_{3}&a_{4}&a_{5}&a_{6}\\\
a_{2}&2a_{7}&a_{8}&a_{9}&a_{10}&a_{11}\\\
a_{3}&a_{8}&2a_{12}&a_{13}&a_{14}&a_{15}\\\
a_{4}&a_{9}&a_{13}&2a_{16}&a_{17}&a_{18}\\\
a_{5}&a_{10}&a_{14}&a_{17}&2a_{19}&a_{20}\\\
a_{6}&a_{11}&a_{15}&a_{18}&a_{20}&2a_{21}\end{matrix}\right).$

(To get a true basis of second order symmetries of $\mathcal{H}^{0}$, we set
one of $a_{6}$, $a_{10}$, $a_{13}$ to zero.)

We use the fact that the adjoint action
${\mathcal{S}}\to\\{p_{1},{\mathcal{S}}\\}\equiv\operatorname{Ad}_{p_{1}}{\mathcal{S}}$
will map the 5-dimensional space of a solution set into itself. Since this
action is essentially differentiation with respect to $x$, it is clear that
$\operatorname{Ad}_{p_{1}}^{3}=0$, so the generalized eigenvalues of
$\operatorname{Ad}_{p_{1}}$ must all be $0$. Thus the possible Jordan
canonical forms for the operator $\operatorname{Ad}_{p_{1}}$ on a generalized
eigenbasis of solutions ${\mathcal{S}}$ are

$\displaystyle\left(\begin{matrix}0&1&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&1&0&0&0\\\
0&0&1&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&1\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&1&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&1&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right),$
$\displaystyle\left(\begin{matrix}0&1&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right).$ (4.4)

We get 5 different forms depending on the smallest integer $k$ such that
$\operatorname{Ad}_{p_{1}}^{k}=0$. We will consider each of these 5 forms in
turn to determine its implications for the generalized eigenbasis of solutions
${\mathcal{S}}$.

####  4.2.1 Form (4.4a)

We first look at the possibilities for form (4.4a). In this case
$\operatorname{Ad}_{p_{1}}^{2}\neq 0$ so that part of the eigenbasis must be
$\\{{\mathcal{L}},{\mathcal{L}}_{1},{\mathcal{L}}_{2}\\}$, symmetries that
generate a chain of length 3.

The action of $\operatorname{Ad}_{p_{1}}$ is nontrivial on only two of
components of $R$ in (4.3):

$\operatorname{Ad}_{p_{1}}J_{12}=-p_{2},\qquad\operatorname{Ad}_{p_{1}}J_{13}=-p_{3}.$
(4.5)

The action of $\operatorname{Ad}_{p_{1}}$ on any monomial in $\mathcal{S}$ can
then be determined from (4.5) and the Leibniz property. We find that

$\displaystyle\mathcal{L}=a_{16}J_{12}^{2}+a_{17}J_{12}J_{13}+a_{19}J_{13}^{2}+W$
(4.6)

(where here and below, the $a_{ij}$ are assumed to be arbitrary parameters) is
the most general homogeneous solution of $\operatorname{Ad}_{p_{1}}^{3}=0$.
Starting from $\mathcal{L}$, a chain is generated with

$\displaystyle\mathcal{L}_{1}=\operatorname{Ad}_{p_{1}}\mathcal{L}=-2a_{16}p_{2}J_{12}-a_{17}(p_{2}J_{13}+p_{3}J_{12})-2a_{19}p_{3}J_{13}+W_{1},$
$\displaystyle\mathcal{L}_{2}=\operatorname{Ad}_{p_{1}}\mathcal{L}_{1}=2a_{16}p_{2}^{2}+2a_{17}p_{2}p_{3}+2a_{19}p_{3}^{2}+W_{2}.$
(4.7)

where we omit the expressions for the functions $W$, $W_{1}$, $W_{2}$. In
addition there must be 2 eigenfunctions of $\operatorname{Ad}_{p_{1}}$ with
eigenvalue $0$ and independent of ${\mathcal{L}}_{2}$.

The symmetries that are annihilated by $\operatorname{Ad}_{p_{1}}$ take the
form

$\displaystyle{\mathcal{K}}=b_{1}p_{1}^{2}+b_{2}p_{1}p_{2}+b_{3}p_{1}p_{3}+b_{6}p_{1}J_{23}+b_{7}p_{2}^{2}+b_{8}p_{2}p_{3}+b_{11}p_{2}J_{23}$
$\displaystyle\hphantom{{\mathcal{K}}=}{}+b_{12}p_{3}^{2}+b_{15}p_{3}J_{23}+b_{21}J_{23}^{2}+U,$
(4.8)

where the $b_{j}$, analogous to $a_{j}$, are constants to be determined, and
$U$ is the potential.

${\mathcal{L}}$ is homogeneous of order 2 in the variables $x$, $y$, $z$. We
consider cases for the form of $\mathcal{L}_{2}$. A very special case is that
where, by a rotation if necessary, ${\mathcal{L}}_{2}$ takes the form where
$a_{16}=a_{19}\neq 0$, $a_{17}=0$. Thus we have
${\mathcal{L}}_{2}^{0}=2a_{16}({\mathcal{H}}^{0}-{\mathcal{J}}^{2})$. Always
${\mathcal{H}}$ can be assumed to be a basis symmetry, so to achieve form
(4.4a) we have to select a symmetry $\mathcal{K}$ that is linearly independent
of the 4 forms already exhibited.

If we choose $\mathcal{K}$ of order 2 in the spatial variables, so
${\mathcal{K}}=b_{21}J_{23}^{2}$ it is straightforward to show that
${\mathcal{B}}=\mathcal{L},\mathcal{L}_{1},\mathcal{L}_{2},\mathcal{K},\mathcal{H}$
is an FLD basis. The Bertrand–Darboux equations for $V(y,z)$ and the
potentials associated with these symmetries are obtained from requiring

$\\{\mathcal{H},\mathcal{L}\\}=\\{\mathcal{H},\mathcal{L}_{1}\\}=\\{\mathcal{H},\mathcal{L}_{2}\\}=\\{\mathcal{H},\mathcal{K}\\}=0.$

We consider the equations for $V(y,z)$ and $W(x,y,z)$ arising (as coefficients
of $p_{1}$, $p_{2}$, $p_{3}$) from $\\{\mathcal{H},\mathcal{L}\\}$:

$\displaystyle a_{16}xzV_{z}+a_{16}xyV_{y}+W_{x}=0,\qquad
a_{16}x^{2}V_{y}-W_{y}=0,\qquad a_{16}x^{2}V_{z}-W_{z}=0.$

The second and third equations are satisfied when
$W(x,y,z)=a_{16}V(y,z)+W_{00}(x)$, where $W_{00}$ is at this point arbitrary.
Upon substituting this form for $W$ into the first equation, we observe that
we must have $W_{00}(x)=c_{1}x^{2}+c_{2}$, for some constants $c_{1}$,
$c_{2}$, to obtain a well-defined equation for $V(y,z)$. The general solution
of the first equation is then

$V(y,z)=\frac{F(z/y)}{y^{2}}$ (4.9)

for $F$ an arbitrary function (up to an additive constant, $-c_{1}$, which we
set to zero without loss of generality). The Jacobi identity guarantees that
this potential is compatible with the symmetries $\mathcal{L}_{1}$,
$\mathcal{L}_{2}$. We can verify compatibility with $\mathcal{K}$ directly: a
function $U$ of $x$, $y$, $z$ can be found so that
$\\{\mathcal{H},\mathcal{K}\\}=0$.

The Calogero potential (2.1) belongs to the class (4.9). Indeed, under the
Jacobi transformation

$x=\tfrac{1}{\sqrt{3}}(r_{1}+r_{2}+r_{3}),\qquad
y=\tfrac{1}{\sqrt{2}}(r_{2}-r_{1}),\qquad
z=\tfrac{1}{\sqrt{6}}(2r_{3}-r_{2}-r_{1}),$ (4.10)

we obtain the Calogero potential (2.1) in variables $r_{1}$, $r_{2}$, $r_{3}$
by choosing

$F(w)=\frac{\beta}{2(1-\sqrt{3}w)^{2}}+\frac{\gamma}{2(1+\sqrt{3}w)^{2}}+\frac{\alpha}{2}.$

If we choose $\mathcal{K}$ of order 1, so that
${\mathcal{K}}=b_{11}p_{2}J_{23}+b_{15}p_{3}J_{23}+U$ where
$|b_{11}|+|b_{15}|>0$, we can verify that the symmetries
${\mathcal{B}}=\\{\mathcal{L},\mathcal{L}_{1},\mathcal{L}_{2},\mathcal{K},\mathcal{H}\\}$
is an FLD basis and solve the Bertrand–Darboux equations to obtain

$V(y,z)=\frac{F_{\rm p}\big{(}\frac{y}{z}\big{)}}{y^{2}},\qquad F_{\rm
p}(t)\equiv\frac{\beta_{1}}{(t+q)^{2}}+\frac{\beta_{2}(qt-1)}{(t+q)^{2}\sqrt{1+t^{2}}},$
(4.11)

where $\beta_{1}$, $\beta_{2}$ are arbitrary parameters and $q\equiv
b_{11}/b_{15}$. Similarly, applying the Jacobi transformation (4.10) to (4.11)
we can obtain a solution adapted to translation invariance.

If we choose $\mathcal{K}$ of order 0, there is no 3-parameter solution for
the potential. The other possibilities for $\mathcal{L}$ of order 2 are that
1) ${\mathcal{L}}_{2}$ can be transformed so that $a_{17}=a_{19}=0$ and the
one chains are $\mathcal{H}$ and $p_{1}^{2}$, in which case there is no
3-parameter potential, and 2) ${\mathcal{L}}_{2}$ can be transformed so that
$a_{17}=2ia_{16}$, $a_{19}=-a_{16}$ and the one chains are $\mathcal{H}$ and
$p_{1}^{2}$, which is not FLD.

####  4.2.2 Form (4.4b)

Here there is one chain of length 3 and one chain of length 2. The general
form for the chain of length 3 is (4.6)–(4.7) again. The general form for a
chain of length 2 is

$\displaystyle\mathcal{L}_{1}^{\prime}=b_{9}p_{2}J_{12}+b_{10}p_{2}J_{13}+b_{13}p_{3}J_{12}+b_{14}p_{3}J_{13}+b_{18}J_{12}J_{23}+b_{20}J_{13}J_{23}+W_{3},$
$\displaystyle\mathcal{L}_{2}^{\prime}=\operatorname{Ad}_{p_{1}}\mathcal{L}_{1}^{\prime}=-b_{9}p_{2}^{2}-b_{10}p_{2}p_{3}-b_{13}p_{2}p_{3}-b_{14}p_{3}^{2}-b_{18}p_{2}J_{23}-b_{20}p_{3}J_{23}+W_{4},$
(4.12)

where $W_{3}$ and $W_{4}$ are potentials that will play no role in our
analysis. We consider three cases based on the order (in the spatial
variables) of $\mathcal{L}_{1}^{\prime}$

Case: ${\mathcal{L}}^{\prime}_{1}$ of order 2. In this case we have
$b_{9}=b_{10}=b_{13}=b_{14}=0$ so that so that ${\mathcal{L}}_{1}^{\prime}$
takes the form
${\mathcal{L}}_{1}^{\prime}=b_{18}J_{12}J_{23}+b_{20}J_{13}J_{23}+W_{3}$ and
${\mathcal{L}}_{2}^{\prime}$ takes the form
$-b_{18}p_{2}J_{23}-b_{20}p_{3}J_{23}+W_{4}$. Since both $\mathcal{H}$ and
$p_{1}^{2}$ are of order 0, and since they both must be included in form
(4.4b), this case cannot occur.

Case: ${\mathcal{L}}^{\prime}_{1}$ of order 1. In this case we have
$b_{18}=b_{20}=0$ and
$\mathcal{L}_{1}^{\prime}=b_{9}p_{2}J_{12}+b_{10}p_{2}J_{13}+b_{13}p_{3}J_{12}+b_{14}p_{3}J_{13}+W_{3}$,
$\mathcal{L}_{2}^{\prime}=-b_{9}p_{2}^{2}-b_{10}p_{2}p_{3}-b_{13}p_{2}p_{3}-b_{14}p_{3}^{2}$.
However, there is no choice of the surviving parameters $a_{j}$ and $b_{j}$ so
that $\mathcal{H}$ or $p_{1}^{2}$ is contained in
$\operatorname{span}\\{\mathcal{L}_{2},\mathcal{L}_{2}^{\prime}\\}$ and this
case cannot occur.

Case: ${\mathcal{L}}_{1}^{\prime}$ of order 0. This case cannot occur since
${\mathcal{L}}_{1}^{\prime}$ vanishes.

Thus we conclude that form (4.4b) does not occur.

####  4.2.3 Form (4.4c)

Now we have 2 chains of length 2 and one of length 1. The general form for a
chain of length 2 is (4.12). We use the convention that the first chain of
length two, $\\{\mathcal{L}_{1},\mathcal{L}_{2}\\}$, has parameters $a_{j}$
and the second chain of length two,
$\\{\mathcal{L}_{1}^{\prime},\mathcal{L}_{2}^{\prime}\\}$, has parameters
$b_{j}$.

The general form for a chain of length 1 is (4.8).

It is not possible for both ${\mathcal{L}}_{1}$ and
${\mathcal{L}}_{1}^{\prime}$ to be of order 2 since then there would only be
one symmetry of order 0, not enough to contain both ${\mathcal{H}}$ and
$p_{1}^{2}$. We perform case-based analysis on the allowable cases.

Case: ${\mathcal{L}}_{1}$ of order 2, ${\mathcal{L}}^{\prime}_{1}$ of order 1.
This implies that ${\mathcal{K}}$ must be of order 0, so that ${\mathcal{H}}$
and $p_{1}^{2}$ can be contained in the spanning set. We consider the symmetry
$\mathcal{L}_{2}=-a_{18}p_{2}J_{23}-a_{20}p_{3}J_{23}$.

By rotation of coordinates about the $z$-axis we can achieve one of the forms
$a_{20}\neq 0$, $a_{18}=0$ or $a_{20}\neq 0$, $a_{18}=-{\rm i}a_{20}$. For the
second form the basis is not FLD, so can be ruled out. For the first form the
basis is FLD but fails the requirement of yielding a 2-parameter potential
depending on 2 functionally independent coordinates.

Case: Both ${\mathcal{L}}_{1}$ and ${\mathcal{L}}^{\prime}_{1}$ are of order
1. Then, since $p_{1}^{2}$ and $\mathcal{H}$ are always basis vectors, the
remaining basis symmetry $\mathcal{K}$ must be of order 0. It can be chosen as
either $p_{2}^{2}$ or $(p_{2}+{\rm i}p_{3})^{2}$. In the 1st case we determine
all possible choices of basis vectors such that the set is FLD. There are only
4 general cases and we verify that none of them define a superintegrable
system, i.e., yields a 2-parameter potential. In the 2nd case there are 9
possible FLD families, but they all fail the symmetry test.

####  4.2.4 Form (4.4d)

Here we have 1 chain of length 2 and 3 chains of length 1.

The general form for a chain of length 2 is (4.12) while the general form for
a chain of length 1 is (4.8).

There are 2 basic cases: 1) ${\mathcal{L}}_{1}^{\prime}$ is of order 2,
${\mathcal{L}}_{2}^{\prime}$ is of order 1 and ${\mathcal{K}}$ is of orders,
2, 1, or 0; 2) ${\mathcal{L}}_{1}^{\prime}$ is of order 1,
${\mathcal{L}}_{2}^{\prime}$ is of order 0 and ${\mathcal{K}}$ is of orders,
2, 1, or 0. We check all of the possibilities and find the Hamiltonian
$\mathcal{H}=p_{x}^{2}+p_{y}^{2}+p_{z}^{2}+V(y,z)$, with

$V(y,z)=b(z-{\rm i}y)+F(z+{\rm i}y),$ (4.13)

where $b$ is a free constant and $F$ is an arbitrary function. The
corresponding FLD basis is
${\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},({\rm
i}p_{2}+p_{3})^{2},p_{1}({\rm i}p_{2}+p_{3}),(z+{\rm
i}y)^{2}p_{1}^{2},p_{1}({\rm i}J_{12}+J_{13})\big{\\}}$. The Minkowski example
in Section 2 is a special case of potential (4.13). Indeed, under the complex
orthogonal change of coordinates

$x=-2{\rm i}r_{1},\qquad y=\tfrac{1}{2}(r_{1}+r_{2}-(1-{\rm i})r_{3}),\qquad
z=\tfrac{{\rm i}}{2}(r_{1}-r_{2}-(1-{\rm i})r_{3})$

the potential (4.13) becomes that in (2.3) when we choose $F(w)=\beta w+\gamma
w^{2}$ and $b=\alpha$.

A special case of (4.13) with increased symmetry is

$V(y,z)=b_{1}(z-{\rm i}y)+b_{2}(z+{\rm i}y)^{2}.$ (4.14)

Another case is

$V(y,z)=b_{1}z+\frac{b_{2}}{y^{2}}.$ (4.15)

A third FLD basis is
$\big{\\{}{\mathcal{H}}^{0},p_{1}^{2},p_{2}^{2},p_{1}p_{2},p_{1}J_{12}\big{\\}}$
with corresponding potential

$V(y,z)=by+F(z),$ (4.16)

where $F$ is an arbitrary function, and $b$ is an arbitrary constant. A
special case with increased symmetry is

$V(y,z)=b_{1}y+b_{2}z.$ (4.17)

######  Remark 4.1.

The symmetry algebras of the Hamiltonians corresponding the potentials (4.14)
and (4.17) are omitted below due to their complexity.

####  4.2.5 Form (4.4e)

Here we have 5 chains of length 1. The possibilities are 1) 1 symmetry of
order 2, 2 symmetries of order 1 and 2 symmetries of order 0; 2) 1 symmetry of
order 2, 1 symmetry of order 1 and 3 symmetries of order 0; 3) 2 symmetries of
order 1 and 3 symmetries of order 0; 4) 1 symmetry of order 1 and 4 symmetries
of order 0. In all cases the systems are FLD but they do not admit 2-parameter
functionally independent potentials.

####  4.2.6 Structure algebras

For the generalized Calogero system (4.9) a basis for the generators is

$\displaystyle\mathcal{J}=p_{1},\qquad{\mathcal{S}}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{F\big{(}\frac{y}{z}\big{)}}{y^{2}},\qquad{\mathcal{S}}_{2}=p_{1}^{2},$
$\displaystyle{\mathcal{S}}_{3}=\frac{1}{2}J_{23}^{2}+\frac{F\big{(}\frac{y}{z}\big{)}y^{2}+F\big{(}\frac{y}{z}\big{)}z^{2}}{2y^{2}},\qquad{\mathcal{S}}_{4}=\frac{1}{2}\big{(}J_{12}^{2}+J_{13}^{2}\big{)}+\frac{x^{2}F\big{(}\frac{y}{z}\big{)}}{2y^{2}},$
$\displaystyle{\mathcal{S}}_{5}=p_{2}J_{12}+p_{3}J_{13}+\frac{xF\big{(}\frac{y}{z}\big{)}}{y^{2}}.$

The nonzero commutators of the generators are

$\displaystyle\\{\mathcal{J},{\mathcal{S}}_{4}\\}=-{\mathcal{S}}_{5},\qquad\\{\mathcal{J},{\mathcal{S}}_{5}\\}={\mathcal{J}}^{2}-{\mathcal{H}},\qquad\\{{\mathcal{S}}_{4},{\mathcal{S}}_{5}\\}=-2{\mathcal{J}}{\mathcal{S}}_{3}-2{\mathcal{J}}{\mathcal{S}}_{4},$
(4.18)

and the functional relationship is

$x^{2}{\mathcal{S}}_{1}^{0}-\big{(}x^{2}+y^{2}+z^{2}\big{)}{\mathcal{S}}_{2}^{0}+2{\mathcal{S}}_{4}^{0}-2x{\mathcal{S}}_{5}^{0}=0.$
(4.19)

Note that both $\mathcal{H}$ and ${\mathcal{S}}_{3}$ lie in the center of this
algebra.

For the system (4.11) a basis for the 1st and 2nd order generators is

$\displaystyle{\mathcal{J}}=p_{1},\qquad{\mathcal{S}}_{1}={\mathcal{H}},\qquad{\mathcal{S}}_{2}={\mathcal{J}}^{2},\qquad{\mathcal{S}}_{3}=\frac{1}{2}J_{23}^{2}+\frac{F_{\rm
p}\big{(}\frac{y}{z}\big{)}\big{(}y^{2}+z^{2}\big{)}}{2y^{2}},$
$\displaystyle{\mathcal{S}}_{4}=\frac{1}{2}\big{(}J_{12}^{2}+J_{13}^{2}\big{)}+\frac{x^{2}F_{\rm
p}\big{(}\frac{y}{z}\big{)}}{2y^{2}},\qquad{\mathcal{S}}_{5}=p_{2}J_{12}+p_{3}J_{13}+\frac{xF_{\rm
p}\big{(}\frac{y}{x}\big{)}}{y^{2}},$
$\displaystyle{\mathcal{S}}_{6}=(qJ_{12}+J_{13})J_{23}+W_{6},\qquad{\mathcal{S}}_{7}=(qp_{2}+p_{3})J_{23}+W_{7},$

where we omit the complicated forms of the potentials $W_{6}$, $W_{7}$. Since
the potential-free parts of the generators satisfy (4.19) the set of
generators is FLD. The subset
$\\{{\mathcal{J}},{\mathcal{H}},{\mathcal{S}}_{1},\dots,{\mathcal{S}}_{5}\\}$
generates a closed quadratic algebra with nonzero relations (4.18). However,
if any linear combination of ${\mathcal{S}}_{6}$, ${\mathcal{S}}_{7}$ is added
to the generators, a new 3rd order symmetry is produced that is not a
polynomial in the generators, so the resulting algebra doesn’t close at second
order.

######  Remark 4.2.

The set of symmetries $\\{\mathcal{S}_{1},\dots,\mathcal{S}_{7}\\}$ contains 5
independent symmetries. However, the set of FLD symmetries
$\\{{\mathcal{L}},{\mathcal{L}}_{1},{\mathcal{L}}_{2},{\mathcal{H}},{\mathcal{K}}\\}$,
equivalent to
$\\{\mathcal{S}_{1},{\mathcal{S}}_{2},{\mathcal{S}}_{4},{\mathcal{S}}_{5},{\mathcal{S}}_{7}\\}$
via a general linear transformation, contains only 4 independent symmetries
(as is the maximum possible by Corollary 3.2). The symmetries
$\mathcal{S}_{3}$, $\mathcal{S}_{6}$ are obtained in addition to the FLD
symmetries by seeking all 2nd order, linearly independent symmetries of the
potential (4.11).

For the generalized Minkowski system (4.13) it is convenient to pass from the
original variables $\\{x,y,z\\}$ to new variables $\\{X,Y,Z\\}$ where $X=x$,
$Y=z-{\rm i}y$, $Z=z+{\rm i}y$. The Hamiltonian then can be written as
${\mathcal{H}}=p_{X}^{2}+4p_{Y}p_{Z}+bY+F(Z)$. The generating symmetries are

$\displaystyle{\mathcal{J}}=p_{X},\qquad\mathcal{S}_{1}={\mathcal{H}}=p_{X}^{2}+4p_{Y}p_{Z}+bY+F(Z),\qquad\mathcal{S}_{2}=\mathcal{J}^{2},$
$\displaystyle{\mathcal{S}}_{3}=Zp_{X}^{2}-2Xp_{X}p_{Y}-\tfrac{1}{2}bX^{2},\qquad{\mathcal{S}}_{4}=p_{X}p_{Y}+\tfrac{1}{2}bX,\qquad{\mathcal{S}}_{5}=p_{Y}^{2}+\tfrac{1}{2}bZ,$

and the nonzero structure relations are

$\\{{\mathcal{J}},{\mathcal{S}}_{3}\\}=2{\mathcal{S}}_{4},\qquad\\{{\mathcal{J}},{\mathcal{S}}_{4}\\}=-\tfrac{b}{2},\qquad\\{{\mathcal{S}}_{3},{\mathcal{S}}_{4}\\}=-2{\mathcal{J}}{\mathcal{S}}_{5},$

with ${\mathcal{H}}$ in the center of the algebra. The potential-free parts of
the generators satisfy
$-z{\mathcal{J}}^{2}+{\mathcal{S}}_{3}^{0}+2x{\mathcal{S}}_{4}^{0}=0$, so the
system is FLD.

For the system (4.15) the generating symmetries are

$\displaystyle{\mathcal{J}}=p_{1},\qquad{\mathcal{S}}_{1}={\mathcal{H}}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+b_{1}z+\frac{b_{2}}{y^{2}},\qquad{\mathcal{S}}_{2}={\mathcal{J}}^{2},$
$\displaystyle{\mathcal{S}}_{3}=p_{1}J_{13}+\frac{b_{1}x^{2}}{4},\qquad{\mathcal{S}}_{4}=p_{1}p_{3}+\frac{b_{1}x}{2},\qquad{\mathcal{S}}_{5}=J_{12}^{2}+\frac{b_{2}x^{2}}{y^{2}},$
$\displaystyle{\mathcal{S}}_{6}=2p_{2}J_{12}+\frac{2b_{2}x}{y^{2}},\qquad{\mathcal{S}}_{7}=2p_{2}^{2}+\frac{2b_{2}}{y^{2}},\qquad{\mathcal{S}}_{8}=p_{2}J_{23}+\frac{b_{1}y^{2}}{4}-\frac{b_{2}z}{y^{2}}.$

Since the potential-free parts of the generators satisfy
$z{\mathcal{S}}_{2}^{0}+{\mathcal{S}}_{3}^{0}-x{\mathcal{S}}_{4}=0$, the set
of generators is FLD. The subset
$\\{{\mathcal{J}},{\mathcal{S}}_{1},{\mathcal{S}}_{2},{\mathcal{S}}_{3},{\mathcal{S}}_{4},{\mathcal{S}}_{7}\\}$
generates a closed quadratic algebra with nonzero relations:

$\displaystyle\\{{\mathcal{J}},{\mathcal{S}}_{3}\\}=-{\mathcal{S}}_{4},\qquad\\{{\mathcal{J}},{\mathcal{S}}_{4}\\}=-\tfrac{b_{1}}{2},\qquad\\{{\mathcal{S}}_{2},{\mathcal{S}}_{3}\\}=-2{\mathcal{J}}{\mathcal{S}}_{4},$
$\displaystyle\\{{\mathcal{S}}_{2},{\mathcal{S}}_{4}\\}=-b_{1}{\mathcal{J}},\qquad\\{{\mathcal{S}}_{3},{\mathcal{S}}_{4}\\}={\mathcal{J}}\left({\mathcal{S}}_{1}-\tfrac{1}{2}{\mathcal{S}}_{7}-2{\mathcal{S}}_{2}\right).$

However, if any linear combination of ${\mathcal{S}}_{5}$,
${\mathcal{S}}_{6}$, ${\mathcal{S}}_{8}$ is added to the generators, a new 3rd
order symmetry is produced that is not a polynomial in the generators, so the
resulting algebra doesn’t close at second order.

For the system (4.16) the generating symmetries are

$\displaystyle{\mathcal{J}}=p_{1},\qquad{\mathcal{S}}_{1}={\mathcal{H}}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+by+F(z),\qquad{\mathcal{S}}_{2}=p_{1}^{2},$
$\displaystyle{\mathcal{S}}_{3}=-yp_{1}^{2}+xp_{1}p_{2}+\frac{bx^{2}}{4},\qquad{\mathcal{S}}_{4}=p_{1}p_{2}+\frac{bx}{2},\qquad{\mathcal{S}}_{5}=p_{2}^{2}+by,$

and the nonzero structure relations are

$\displaystyle\\{\mathcal{J},\mathcal{S}_{3}\\}=-{\mathcal{S}}_{4},\qquad\\{\mathcal{J},\mathcal{S}_{3}\\}=-\tfrac{b}{2},\qquad\\{\mathcal{S}_{2},\mathcal{S}_{3}\\}=-2\mathcal{J}{\mathcal{S}}_{4},\qquad\\{\mathcal{S}_{2},\mathcal{S}_{4}\\}=-b\mathcal{J},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-\mathcal{J}(\mathcal{S}_{2}-\mathcal{S}_{4}),\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-2\mathcal{J}{\mathcal{S}}_{4},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-b\mathcal{J}.$

The potential-free parts of the generators satisfy
$y\mathcal{J}^{2}+x\mathcal{S}_{3}^{0}-\mathcal{S}_{4}^{0}=0$, so the system
is FLD.

###  4.3 Second case: $\boldsymbol{{\mathcal{J}}=p_{1}+{\rm i}p_{2}}$

We introduce appropriate new coordinates $\\{\eta,\xi,z\\}$ where
$x=\frac{1}{2}(\xi+\eta)$, $y=\frac{{\rm i}}{2}(\eta-\xi)$, $z=z$. In the new
coordinates the 1st order symmetries for the potential-free case are:

$\displaystyle p_{1}+{\rm i}p_{2}=2p_{\eta}={\mathcal{J}},\qquad p_{2}={\rm
i}(p_{\xi}-p_{\eta}),\qquad J_{12}={\rm i}(\xi p_{\xi}-\eta p_{\eta}),$
$\displaystyle J_{13}=\tfrac{1}{2}(\eta+\xi)p_{z}-z(p_{\eta}+p_{\xi}),\qquad
J_{23}=\tfrac{{\rm i}}{2}(\xi-\eta)p_{z}+{\rm i}z(p_{\eta}-p_{\xi}).$

In this case $\operatorname{Ad}_{p_{1}+{\rm i}p_{2}}^{3}=0$. For convenience
we prefer to work with $\tilde{J}=p_{\eta}=(p_{1}+{\rm i}p_{2})/2$. The
canonical forms associated with $\operatorname{Ad}_{\tilde{J}}^{3}=0$ are
again (4.4).

######  Remark 4.3.

A basis of second order symmetries in this case is again given by (4.2). The
formulas for the momentum parts of the symmetry operators appearing below are
most naturally expressed in terms of
$\\{p_{1},p_{2},p_{3},J_{12},J_{13},J_{23}\\}$, as before. However, the
potentials we obtain are most naturally expressed in terms of the new
coordinates $\\{\eta,\xi,z\\}$. We take this approach below and in Sections
4.4 and 4.5.

$\operatorname{Ad}_{\tilde{J}}$ has nontrivial action on three components of
$R$ in (4.3):

$\operatorname{Ad}_{\tilde{J}}J_{12}={\rm i}(p_{1}+{\rm i}p_{2})/2={\rm
i}p_{\eta},\qquad\operatorname{Ad}_{\tilde{J}}J_{13}=-p_{3}/2,\qquad\operatorname{Ad}_{\tilde{J}}J_{23}=-{\rm
i}p_{3}/2.$

From here we can construct a convenient generalized eigenbasis for the
20-dimensional space of symmetries:

$\displaystyle L_{1}=-\tfrac{1}{2}J_{12}^{2},\qquad L_{2}=\tfrac{{\rm
i}}{2}J_{12}(J_{13}-{\rm i}J_{23}),\qquad L_{3}=2J_{13}^{2},$ $\displaystyle
M_{1}=-\tfrac{{\rm i}}{2}(p_{1}+{\rm i}p_{2})J_{12},\qquad
M_{2}=-\tfrac{1}{4}(p_{1}+{\rm i}p_{2})(J_{13}-{\rm i}J_{23})-\tfrac{{\rm
i}}{2}p_{3}J_{12},$ $\displaystyle M_{3}=-2p_{3}J_{13},\qquad
M_{4}=J_{13}^{2}+J_{23}^{2},\qquad M_{5}={\rm i}J_{12}(J_{13}+{\rm i}J_{23}),$
$\displaystyle M_{6}={\rm i}p_{2}(J_{13}-{\rm i}J_{23}),\qquad M_{7}=-2{\rm
i}(p_{1}-{\rm i}p_{2})J_{12}-p_{3}(J_{13}-{\rm i}J_{23}),$ $\displaystyle
N_{1}=\tilde{J}^{2}=\frac{1}{4}(p_{1}+{\rm i}p_{2})^{2},\\!\qquad
N_{2}=\tfrac{1}{2}(p_{1}+{\rm i}p_{2})p_{3},\\!\qquad
N_{3}=p_{3}^{2},\\!\qquad N_{4}=-p_{3}(J_{13}+{\rm i}J_{23}),$ $\displaystyle
N_{5}=-\tfrac{1}{2}(p_{1}+{\rm i}p_{2})(J_{13}+{\rm i}J_{23}),\qquad
N_{6}=-{\rm i}p_{2}p_{3},\qquad
N_{7}=\mathcal{H}^{0}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2},$ $\displaystyle
N_{8}=(J_{13}+{\rm i}J_{23})^{2},\qquad N_{9}=-\tfrac{1}{2}(p_{1}-{\rm
i}p_{2})(J_{13}+{\rm i}J_{23}),\qquad N_{10}=\tfrac{1}{4}(p_{1}-{\rm
i}p_{2})^{2},$

where the 3-chains and 2-chains are $\\{L_{1},M_{1},N_{1}\\}$,
$\\{L_{2},M_{2},N_{2}\\}$, $\\{L_{3},M_{3},N_{3}\\}$, $\\{M_{4},N_{4}\\}$,
$\\{M_{5},N_{5}\\}$, $\\{M_{6},N_{6}\\}$, and $\\{M_{7},N_{7}\\}$. $N_{8}$,
$N_{9}$, and $N_{10}$ are 1-chains.

####  4.3.1 Form (4.4a)

Here we have a 3-chain and two 1-chains, one of which must be
$\mathcal{H}^{0}$. There are two cases to consider. Either the terminal
element of the three chain or the second 1-chain must be
$N_{1}=\tilde{\mathcal{J}}^{2}$.

In the first case, the 3-chain is
$\\{L_{1}+\beta_{1}M_{4}+\beta_{2}M_{5}+\gamma
N_{8},M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5},N_{1}\\}$ (where here and below
Greek letters with subscripts are arbitrary parameters analogous to the
$a_{ij}$ in the previous subsection; they are fixed by requiring certain
combinations of them are FLD) and the 1-chain is one of
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$,
$\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$, or $N_{8}$ (in which case we can
take $\gamma=0$ by a canonical form-preserving change of basis). The first
1-chain possibility is FLD when 1) $\beta_{1}=-1/4$,
$\mu_{2}=\mu_{3}=\mu_{4}=0$, or 2) $\beta_{1}=0$, $\gamma=\beta_{2}^{2}/2$,
$\mu_{1}=2\beta_{2}(2\mu_{2}-2\beta_{2}^{2}-1)$, $\mu_{3}=-2\beta_{2}$,
$\mu_{4}=1$ or 3) $\beta_{1}=\mu_{3}=\mu_{4}=0$, $\mu_{1}=4\beta_{2}\mu_{2}$,
or 4) $\beta_{1}=\beta_{2}=\mu_{1}=\mu_{3}=\gamma\mu_{4}=0$. The third subcase
with $\gamma=\beta_{2}^{2}/2$ and the fourth subcase with $\gamma=\mu_{4}=0$
lead to the admissible potentials

$V(\xi,z)=\frac{b}{\xi^{2}}+F(q\xi+z)$ (4.20)

and

$V(\xi,z)=\frac{b}{\xi^{2}}+F(z),$ (4.21)

respectively. Note that (4.21) is special case of (4.20) with increased
symmetry.

The second 1-chain possibility is FLD when $\mu_{1}=\mu_{3}=0$ and
$\beta_{1}=-1/4$ but does not lead to an admissible potential.

The third 1-chain possibility is FLD when $\beta_{1}=-1/2$ and $\beta_{2}=0$,
leading to the admissible potential

$\displaystyle V(\xi,z)=\frac{F(z/\xi)}{\xi^{2}}.$ (4.22)

In the second case, the 3-chain is
$\\{\alpha_{1}L_{1}+\alpha_{2}L_{2}+\alpha_{3}L_{3}+\beta_{1}M_{4}+\beta_{2}M_{5}+\gamma
N_{8},\alpha_{1}M_{1}+\alpha_{2}M_{2}+\alpha_{3}M_{3}+\beta_{1}N_{4}+\beta_{2}N_{5},\alpha_{1}N_{1}+\alpha_{2}N_{2}+\alpha_{3}N_{3}\\}$.
This case is not FLD for any choice of parameters.

####  4.3.2 Form (4.4b)

Here we have one 3-chain and one 2-chain. The 3-chain must be
$\big{\\{}L_{1}+\beta_{1}M_{4}+\beta_{2}M_{5}+\gamma
N_{8},M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5},N_{1}=\tilde{\mathcal{J}}^{2}\big{\\}}$
and the 2-chain must be
$\big{\\{}M_{7}+\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9},N_{7}=\mathcal{H}^{0}\big{\\}}$.
The symmetries are not FLD for any choice of parameters.

####  4.3.3 Form (4.4c)

Here we have two 2-chains and a single 1-chain. There are three cases to
consider: the terminal elements of the 2-chains are $\tilde{\mathcal{J}}^{2}$
and $\mathcal{H}^{0}$, one 2-chain terminates in $\tilde{\mathcal{J}}^{2}$ and
the 1-chain is $\mathcal{H}^{0}$, one 2-chain terminates with
$\mathcal{H}^{0}$ and the 1-chain is $\tilde{\mathcal{J}}^{2}$.

In the first case, the 2-chains are
$\\{M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},N_{1}\\}$ and
$\\{M_{7}+\gamma_{1}N_{4}+\gamma_{2}N_{5}+\gamma_{3}N_{9},N_{7}\\}$ and the
1-chain is one of $N_{8}$, $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$,
$\mu_{1}N_{2}+\mu_{3}N_{6}+\mu_{4}N_{10}$. For the first choice of the
1-chain, the symmetries are FLD when $\beta_{1}=-1/2,$
$\beta_{2}=\beta_{3}=0$, but this does not lead to an admissible potential.
The second 1-chain possibility is FLD when $\beta_{1}=-1/4$,
$\beta_{3}=\mu_{1}=\mu_{3}=0$, but this does not lead to an admissible
potential. For the third 1-chain possibility, the symmetries are FLD when
either $\beta_{1}=\beta_{3}=\mu_{3}=\mu_{4}=0$, $\mu_{1}=4\beta_{2}\mu_{2}$ or
$\beta_{1}=-1/4,$ $\beta_{3}=\mu_{2}=\mu_{3}=\mu_{4}=0$, but neither
corresponds to an admissible potential.

In the second case, one $2$-chain is
$\\{M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},N_{1}\\}$ and the
second $2$-chain is either
$\\{\gamma_{1}M_{1}+\gamma_{2}M_{2}+\gamma_{3}M_{3}+\gamma_{4}M_{6}+\gamma_{5}M_{7}+\delta_{1}N_{4}+\delta_{2}N_{5}+\delta_{3}N_{9},\gamma_{1}N_{1}+\gamma_{2}N_{2}+\gamma_{3}N_{3}+\gamma_{4}N_{6}+\gamma_{5}N_{7}\\}$
(we can take $\gamma_{1}=0$ by a canonical form-preserving change of basis) or
$\\{\gamma_{1}M_{4}+\gamma_{2}M_{5}+\delta
N_{8},\gamma_{1}N_{4}+\gamma_{2}N_{5}\\}$. To simplify the analysis, we
observe that the symmetry $M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9}$
leads to an inadmissible potential unless $\beta_{3}=0$; similarly, if
$\gamma_{1}N_{1}+\gamma_{2}N_{2}+\gamma_{3}N_{3}+\gamma_{4}N_{6}+\gamma_{5}N_{7}$
is a symmetry of an admissible potential we must have $\gamma_{4}=0$. For the
first choice of the second 2-chain, we find three sets of FLD symmetries:
$\beta_{1}=\beta_{3}=\gamma_{1}=\gamma_{4}=0$,
$\gamma_{2}=4\beta_{2}\gamma_{3}$; $\beta_{1}=-1/4$,
$\beta_{3}=\gamma_{1}=\gamma_{3}=\gamma_{4}=0$; and
$\beta_{3}=\gamma_{1}=\gamma_{3}=\gamma_{4}=\gamma_{5}=0$,
$\gamma_{2}=2\delta_{3}$, but none of these lead to admissible potentials. The
second choice of the second 2-chain leads to an FLD basis when
$\beta_{1}=-1/4$, $\beta_{3}=\gamma_{1}=0$, but this does not lead to an
admissible potential.

In the third case, one $2$-chain is
$\\{M_{7}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},N_{7}\\}$ and the
second $2$-chain is either
$\\{\gamma_{1}M_{1}+\gamma_{2}M_{2}+\gamma_{3}M_{3}+\gamma_{4}M_{6}+\gamma_{5}M_{7}+\delta_{1}N_{4}+\delta_{2}N_{5}+\delta_{3}N_{9},\gamma_{1}N_{1}+\gamma_{2}N_{2}+\gamma_{3}N_{3}+\gamma_{4}N_{6}+\gamma_{5}N_{7}\\}$
(we can take $\gamma_{5}=0$ by a canonical form-preserving change of basis) or
$\\{\gamma_{1}M_{4}+\gamma_{2}M_{5}+\delta
N_{8},\gamma_{1}N_{4}+\gamma_{2}N_{5}\\}$. Using the requirement
$\gamma_{4}=0$ from the second case, we find that the first choice for the
second 2-chain does not yield an FLD basis for any choice of parameters. The
second choice for the second 2-chain also does not lead to an FLD basis for
any choice of parameters.

####  4.3.4 Form (4.4d)

Here we have a 2-chain and three 1-chains. There are again three cases to
consider: $\tilde{\mathcal{J}}^{2}$ and $\mathcal{H}^{0}$ are 1-chains,
$\tilde{\mathcal{J}}^{2}$ is the terminal element of a 2-chain and $H$ is a
1-chain, and $\mathcal{H}^{0}$ is the terminal element of a $2$-chain and
$\tilde{\mathcal{J}}^{2}$ is a 1-chain.

In the first case, the 2-chain is either
$\\{\alpha_{1}M_{1}+\alpha_{2}M_{2}+\alpha_{3}M_{3}+\alpha_{4}M_{6}+\alpha_{5}M_{7}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},\alpha_{1}N_{1}+\alpha_{2}N_{2}+\alpha_{3}N_{3}+\alpha_{4}N_{6}+\alpha_{5}N_{7}\\}$
or $\\{\alpha_{1}M_{4}+\alpha_{2}M_{5}+\beta
N_{8},\alpha_{1}N_{4}+\alpha_{2}N_{5}\\}$ and the final 1-chain is one of
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$,
$\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$, $N_{8}$. To simplify the analysis,
it is sometimes useful to find conditions under which the nontrivial 1-chains
are compatible (both correspond to the same admissible potential) before
searching for FLD systems. For the first choice of the 2-chain where the final
1-chain is order-0, we have the conditions
$\alpha_{2}=-2\alpha_{3}\mu_{3}/\mu_{4}$ and
$\mu_{2}=\big{(}\mu_{3}^{3}-2\mu_{1}\mu_{4}^{2}+2\mu_{3}\mu_{4}^{2}\big{)}/4\mu_{3}\mu_{4}$
when $\mu_{4}\neq 0$ (we must also assume $\alpha_{3}\mu_{3}\neq 0$ to avoid
linear dependence), but this does not lead to an FLD system with admissible
potential. When $\mu_{4}=0$, the 1-chains are incompatible. For the first
choice of the 2-chain where the final 1-chain is order-1, we have the
compatibility conditions $\alpha_{3}=0$ or
$\alpha_{2}=2\alpha_{3}\mu_{2}/\mu_{1}$ ($\mu_{1}\neq 0$); the first of these
leads to an FLD system ($\alpha_{3}=\alpha_{4}=\alpha_{5}=\mu_{1}=\mu_{3}=0$,
$\beta_{3}=3\alpha_{2}/2$) with admissible potential

$V(\xi,z)=\frac{b_{1}}{\xi^{2}}+b_{2}(q\xi+z)$ (4.23)

and an FLD system ($\alpha_{3}=\alpha_{4}=\alpha_{5}=\mu_{1}=\mu_{3}=0$,
$\beta_{3}=-5\alpha_{2}/2$) with admissible potential

$V(\xi,z)=\frac{b_{1}}{\xi^{2/3}}+\frac{b_{2}(q\xi+z)}{\xi^{1/3}}.$ (4.24)

In the first choice of the 2-chain where the final 1-chain is order-2, the
symmetries are FLD when $\alpha_{3}=\alpha_{4}=\alpha_{5}=0$ and
$\beta_{3}=\alpha_{2}/2$, but this does not lead to an admissible potential.
For the second choice of the 2-chain where the final 1-chain is order-0, the
symmetries are FLD when $\alpha_{1}=\mu_{2}=\mu_{3}=\mu_{4}=0$, but this does
not lead to an admissible potential. For the second choice of the 2-chain
where the final 1-chain is order-1, imposing $\mu_{3}=0$ we find that the
symmetries are not FLD for any choice of parameters. For the second choice of
the 2-chain where the final 1-chain is order-2, the symmetries are not FLD for
any choice of parameters.

In the second case, the 2-chain is
$\\{M_{1}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},N_{1}\\}$ and there are
five subcases for the two remaining 1-chains: one order-2 and one order-1
1-chain, one order-2 and one order-0 1-chain, two order-1 1-chains, one
order-1 and one order-0 1-chain, and two order-0 1-chains.

In the first subcase, the symmetries are FLD when $\beta_{1}=-1/4$ and
$\beta_{3}=\mu_{1}=\mu_{3}=0$, but this does not lead to an admissible
potential.

In the second subcase, the symmetries are FLD when either $\beta_{1}=-1/2$,
$\beta_{2}=\beta_{3}=0$ or $\beta_{3}=\mu_{3}=\mu_{4}=0$. From here we obtain
three admissible potentials. When $\beta_{1}=-1/2$, $\beta_{2}=\beta_{3}=0$,
$\mu_{3},\mu_{4}\neq 0$ and
$\mu_{2}=\big{(}\mu_{3}^{3}-2\mu_{1}\mu_{4}^{2}+2\mu_{3}\mu_{4}^{2}\big{)}/4\mu_{3}\mu_{4}$,
we have the potential

$V(\xi,z)=b_{1}\xi+\frac{b_{2}}{(q\xi+z)^{2}},$ (4.25)

when $\beta_{2}=(\mu_{1}+2\beta_{1}\mu_{1})/4\mu_{2}$,
$\beta_{3}=\mu_{3}=\mu_{4}=0$, we have the potential

$V(\xi,z)=\frac{b}{(q\xi+z)^{2}}+F(\xi),$ (4.26)

and when $\beta_{3}=\mu_{2}=\mu_{3}=\mu_{4}=0$, we have the potential

$V(\xi,z)=\frac{bz}{\xi^{3}}+F(\xi).$ (4.27)

In the third subcase, we recall that $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$
only leads to an admissible potential when $\mu_{3}=0$. Then, by a canonical
form-preserving change of basis, we see that $N_{4}$ and $N_{5}$ must be
independent symmetries. The symmetries are FLD when $\beta_{3}=0$ and lead to
an admissible potential

$V(\xi,z)=\frac{bz}{\xi^{3/2}}+F(\xi).$ (4.28)

In the fourth subcase, we write $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$ and
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the order-1 and
order-2 1-chains, respectively. The symmetries are FLD when $\beta_{1}=-1/4$,
$\beta_{3}=\mu_{1}=\mu_{3}=0$ or $\beta_{3}=\nu_{3}=\nu_{4}=0$. There are two
resulting FLD systems with admissible potentials:
$\beta_{2}=(\mu_{2}+2\beta_{1}\mu_{2})/2\mu_{1}$,
$\nu_{1}=2\mu_{2}\nu_{2}/\mu_{1}$, $\beta_{3}=\mu_{3}=\nu_{3}=\nu_{4}=0$, we
obtain a potential equivalent to (4.26) and
$\beta_{3}=\mu_{1}=\mu_{3}=\nu_{3}=\nu_{2}=\nu_{4}=0$ with

$V(\xi,z)=bz\xi^{a}+F(\xi).$ (4.29)

In the fifth subcase, we
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$ and
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the two order-0
1-chains. Assume first that $\mu_{4}$ and $\nu_{4}$ are not both zero. Without
loss of generality we assume $\mu_{4}\neq 0$, so we can take $\nu_{4}=0$ by a
canonical form-preserving change of basis. It is then required that
$\nu_{3}=0$ if we are to have an admissible potential. The 1-chains are
incompatible unless $\mu_{3}=-\nu_{1}\mu_{4}/2\nu_{2}$. When additionally
$\nu_{1}=4\beta_{2}\nu_{2}$, $\beta_{1}=\beta_{3}=\nu_{3}=\nu_{4}=0$, we find
an FLD system with admissible potential

$V(\xi,z)=b\xi+F(q\xi+z).$ (4.30)

If $\mu_{4}=\nu_{4}=0$, we must also have $\mu_{3}=\nu_{3}=0$ and we can
consider $N_{2}$ and $N_{3}$ as independent symmetries. The symmetries are FLD
when $\beta_{3}=0$; when additionally $\beta_{1}=-1/10$, we find the
admissible potential

$V(\xi,z)=b\xi z+F(\xi),$ (4.31)

and when additionally $\beta_{1}=0$, we find the admissible potential

$V(\xi,z)=bz+F(\xi).$ (4.32)

In the third case, the 2-chain is
$\\{M_{7}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9},N_{1}\\}$ and there are
five subcases for the two remaining 1-chains: one order-2 and one order-1
1-chain, one order-2 and one order-0 1-chain, two order-1 1-chains, one
order-1 and one order-0 1-chain, and two order-0 1-chains. The first three
subcases are not FLD for any choice of parameters. In the fourth subcase, we
write $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$ and
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the order-1 and
order-2 1-chains, respectively. The symmetries are FLD when
$\mu_{1}=\mu_{3}=\nu_{2}=\nu_{3}=\nu_{4}=0$, but this does not lead to an
admissible potential. In the fifth subcase, we write
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$ and
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the two 1-chains.
Compatibility of these 1-chains requires $\mu_{3}=\mu_{4}=\nu_{3}=\nu_{4}=0$
and we may take $N_{2}$ and $N_{3}$ as independent symmetries. However, the
simultaneous admissible potential of $N_{2}$ and $N_{3}$ is incompatible with
the $M_{7}+\beta_{1}N_{4}+\beta_{2}N_{5}+\beta_{3}N_{9}$ for all choices of
parameters.

####  4.3.5 Form (4.4e)

Here we have five 1-chains, two of which must be $\mathcal{H}^{0}$ and
$\tilde{\mathcal{J}}^{2}$. There are seven cases for the three additional
1-chains:

  1. 1)

one order-2 1-chain and two order-1 1-chains,

  2. 2)

one order-2, one order-1, and one order-0 1-chain,

  3. 3)

one order-2 1-chain and two order-0 1-chains,

  4. 4)

three order-1 1-chains,

  5. 5)

two order-1 1-chains and one order-0 1-chain,

  6. 6)

one order-1 and two order-0 1-chains,

  7. 7)

three order-0 1-chains.

In the first case, we write $N_{8}$, $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$
and $\nu_{1}N_{4}+\nu_{2}N_{5}+\nu_{3}N_{9}$ for the three 1-chains. The
potential is admissible only if $\mu_{3}=\nu_{3}=0$, so we may take $N_{4}$
and $N_{5}$ as independent symmetries. The symmetries are incompatible (do not
have a simultaneous admissible potential).

In the second case, we write $N_{8}$, $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$
and $\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the three
1-chains. The symmetries are FLD when $\mu_{3}=\nu_{3}=\nu_{4}=0$. When also
$\mu_{2}=\nu_{1}=0$, we find the potential

$V(\xi,z)=\frac{b}{z^{2}}+F(\xi);$ (4.33)

when also $\nu_{2}=\mu_{1}\nu_{1}/2\mu_{2}$, we find the admissible potential

$V(\xi,z)=\frac{b_{1}\xi^{2}+b_{2}z(\mu_{1}z+\mu_{2}\xi)}{\xi^{2}(2\mu_{1}z+\mu_{2}\xi)^{2}}+F(\xi),$
(4.34)

which contains (4.33) as a special case.

In the third case, we write $N_{8}$,
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$ and
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$ for the 1-chains. We
first assume that one of $\mu_{4}$, $\nu_{4}$ is nonzero. Without loss of
generality we take $\mu_{4}\neq 0$ so that we may take $\nu_{4}=0$ by a
canonical form-preserving change of basis. We can only have an admissible
potential if also $\nu_{3}=0$. The symmetries are not FLD for any choice of
the remaining parameters. We then consider the case where
$\mu_{3}=\mu_{4}=\nu_{3}=\nu_{4}=0$. We can then take $N_{2}$ and $N_{3}$ as
independent symmetries, but the symmetries are not FLD.

In the fourth case, we can make a canonical-form preserving change of basis
and consider $N_{4}$, $N_{5}$ and $N_{9}$ as independent symmetries. These
symmetries are incompatible (in particular, $N_{9}$ does not produce an
admissible symmetry).

The fifth case is similar to the first case: we may take $N_{4}$ and $N_{5}$
as independent symmetries. We write
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$ for the remaining
nontrivial 1-chain. The symmetries are FLD when $\mu_{3}=\mu_{4}=0$; when also
$\mu_{2}=0$, we find the admissible potential (4.28).

In the sixth case, we write $\mu_{1}N_{4}+\mu_{2}N_{5}+\mu_{3}N_{9}$,
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$, and
$\sigma_{1}N_{2}+\sigma_{2}N_{3}+\sigma_{3}N_{6}+\sigma_{4}N_{10}$ for the
three 1-chains. This case is similar to the third case: the two subcases
reduce to $\nu_{4}\neq 0$, $\sigma_{3}=\sigma_{4}=0$ and
$\nu_{2}=\nu_{3}=\nu_{4}=\sigma_{1}=\sigma_{3}=\sigma_{4}=0$. The first
subcase is FLD when also $\mu_{1}=\sigma_{2}=0$, but we do not get an
admissible potential. The second subcase is FLD and when also $\mu_{1}=0$, we
find the admissible potential (4.33).

In the seventh case, we write
$\mu_{1}N_{2}+\mu_{2}N_{3}+\mu_{3}N_{6}+\mu_{4}N_{10}$,
$\nu_{1}N_{2}+\nu_{2}N_{3}+\nu_{3}N_{6}+\nu_{4}N_{10}$, and
$\sigma_{1}N_{2}+\sigma_{2}N_{3}+\sigma_{3}N_{6}+\sigma_{4}N_{10}$ for the
three 1-chains. We assume that at least one of $\mu_{4}$, $\nu_{4}$,
$\sigma_{4}$ is nonzero. Without loss of generality, we take $\mu_{4}\neq 0$
so we can make a canonical form-preserving change of basis and take
$\nu_{4}=\sigma_{4}=0$. The second and third symmetries will only have an
admissible potential if also $\nu_{3}=\sigma_{3}=0$, so we may also take
$\nu_{2}=\sigma_{1}=0$: $N_{2}$ and $N_{3}$ are independent symmetries. The
symmetries are incompatible unless $\mu_{4}=0$, a contradiction. We next
assume $\mu_{4}=\nu_{4}=\sigma_{4}=0$. Then we may consider $N_{2}$, $N_{3}$,
and $N_{6}$ as independent symmetries. These symmetries are incompatible.

####  4.3.6 Structure algebras

For the potential (4.20), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{b}{\xi^{2}}+F(q\xi+z),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{1}+qM_{5}+\frac{q^{2}}{2}N_{8}+\frac{b(2qz-\eta)}{2\xi},\qquad\mathcal{S}_{4}=M_{1}+qN_{5}+\frac{b}{2\xi},$
$\displaystyle\mathcal{S}_{5}=N_{3}+4qN_{2}+F(qz+\xi).$

They satisfy
$4(2qz-\eta)\tilde{\mathcal{J}}^{2}+\xi\mathcal{S}_{1}^{0}-4\mathcal{S}_{4}^{0}-\xi\mathcal{S}_{5}^{0}=0$
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\mathcal{S}_{4},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{3}-q^{2}b\tilde{\mathcal{J}},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=8q^{2}\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=8q^{2}\tilde{\mathcal{J}}^{3}.$

For the potential (4.21), the symmetries and their FLD relation and algebra
are obtained from that of (4.20) in the limit $q\to 0$.

For the potential (4.22), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{F(z/\xi)}{\xi^{2}},\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{1}-\frac{1}{2}M_{4}-\frac{\big{(}\xi\eta+z^{2}\big{)}F(z/\xi)}{2\xi^{2}},\qquad\mathcal{S}_{4}=M_{1}-\frac{1}{2}N_{4}+\frac{F(z/\xi)}{2\xi},$
$\displaystyle\mathcal{S}_{5}=N_{8}+F(z/\xi).$

They satisfy
$4\big{(}\xi\eta+z^{2}\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{4}^{0}-\mathcal{S}_{5}^{0}=0$,
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\mathcal{S}_{4},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{3}.$

For the potential (4.23), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{b_{1}}{\xi^{2}}+b_{2}(q\xi+z),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{1}+qM_{5}+\frac{q^{2}}{2}N_{8}+\frac{b_{1}(2qz-\eta)}{2\xi},\qquad\mathcal{S}_{4}=M_{1}+\frac{b_{1}}{2\xi}+\frac{qb_{2}\xi^{2}}{8},$
$\displaystyle\mathcal{S}_{5}=M_{2}-qN_{4}+\frac{3}{2}N_{9}+\frac{b_{1}z}{\xi^{2}}+\frac{b_{2}(2qz-\eta)\xi}{4},\qquad\mathcal{S}_{6}=N_{2}+\frac{b_{2}\xi}{4},$
$\displaystyle\mathcal{S}_{7}=N_{3}+b_{2}z,\qquad\mathcal{S}_{8}=N_{5}-\frac{b_{2}\xi^{2}}{8}.$

They satisfy
$4\eta\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{1}^{0}+4\mathcal{S}_{4}^{0}+\xi\mathcal{S}_{7}^{0}=2z\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{6}^{0}-\mathcal{S}_{8}^{0}=0$.
The subset
$\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{4},\mathcal{S}_{6},\mathcal{S}_{7},\mathcal{S}_{8}\big{\\}}$
generates a closed quadratic algebra with nonzero relations

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{6},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{8}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{8},$
$\displaystyle\\{\mathcal{S}_{6},\mathcal{S}_{7}\\}=-b_{2}\tilde{\mathcal{J}},\qquad\\{\mathcal{S}_{6},\mathcal{S}_{8}\\}=-2\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{7},\mathcal{S}_{8}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{6}.$

However, if any linear combination of $\mathcal{S}_{3}$, $\mathcal{S}_{5}$ is
added to the generators, a new 3rd order symmetry is produced that is not a
polynomial in the generators, so the resulting algebra doesn’t close at second
order.

For the potential (4.24), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{b_{1}}{\xi^{2/3}}+\frac{b_{2}(q\xi+z)}{\xi^{4/3}},\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}+2N_{4}-\frac{b_{1}\xi^{1/3}}{2}-\frac{b_{2}(q\xi+16z)}{8\xi^{4/3}},$
$\displaystyle\mathcal{S}_{4}=M_{2}-qN_{4}-\frac{5}{2}N_{9}-\frac{b_{1}z}{\xi^{2/3}}+\frac{b_{2}\big{(}2q\xi
z+3\xi\eta-4z^{2}\big{)}}{4\xi^{4/3}},$
$\displaystyle\mathcal{S}_{5}=N_{2}-\frac{3b_{2}}{4\xi^{1/3}},\qquad\mathcal{S}_{6}=N_{5}-\frac{3b_{2}\xi^{2/3}}{8}.$

They satisfy
$2z\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{5}^{0}-\mathcal{S}_{6}^{0}=0$. The
subset
$\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{3},\mathcal{S}_{5},\mathcal{S}_{6}\big{\\}}$
generates a closed quadratic algebra with nonzero relations

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=3\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{6}\\}=-6\tilde{\mathcal{J}}\mathcal{S}_{6},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-2\tilde{\mathcal{J}}^{3}.$

However, if $\mathcal{S}_{4}$ is added to the generators, a new 3rd order
symmetry is produced that is not a polynomial in the generators, so the
resulting algebra doesn’t close at second order.

For the potential (4.25), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+b_{1}\xi+\frac{b_{2}}{(q\xi+z)^{2}},$
$\displaystyle\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},\qquad\mathcal{S}_{3}=M_{1}+qN_{5}+\frac{b_{1}\xi^{2}}{8},$
$\displaystyle\mathcal{S}_{4}=M_{3}+4qM_{2}+2q(1-2q^{2})N_{5}+\frac{b_{1}z(z+2q\xi)}{2}+\frac{b_{2}\big{(}2qz+2q^{2}\xi-(\xi+\eta)\big{)}}{(q\xi+z)^{2}},$
$\displaystyle\mathcal{S}_{5}=N_{3}+4qN_{2}+\frac{b_{2}}{(q\xi+z)^{2}},\qquad\mathcal{S}_{6}=N_{4}+2qN_{5}-\frac{b_{2}\xi}{(q\xi+z)^{2}},\qquad\mathcal{S}_{7}=N_{8}+\frac{b_{2}\xi^{2}}{(q\xi+z)^{2}},$
$\displaystyle\mathcal{S}_{8}=N_{10}-2qN_{6}-2q(1+2q^{2})N_{2}-\frac{b_{1}(2qz-\eta)}{2}-\frac{b_{2}q^{2}}{(q\xi+z)^{2}}.$

They satisfy

$\displaystyle
4(2qz-\eta)\tilde{\mathcal{J}}^{2}+\xi\mathcal{S}_{1}^{0}-4\mathcal{S}_{3}^{0}-\xi\mathcal{S}_{5}^{0}=4\big{(}\xi\eta+z^{2}\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{3}^{0}-2\xi\mathcal{S}_{6}^{0}-\mathcal{S}_{7}^{0}=0.$

The subset
$\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{3},\mathcal{S}_{5},\mathcal{S}_{6},\mathcal{S}_{7}\big{\\}}$
generates a closed quadratic algebra with nonzero relations

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=8q^{2}\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{6}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{6},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{7}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{7},$
$\displaystyle\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{5}-16q^{2}\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{7}\\}=-8\tilde{\mathcal{J}}\mathcal{S}_{6},\qquad\\{\mathcal{S}_{6},\mathcal{S}_{7}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{7}.$

However, if any linear combination of $\mathcal{S}_{4}$, $\mathcal{S}_{8}$ is
added to the generators, a new 3rd order symmetry is produced that is not a
polynomial in the generators, so the resulting algebra doesn’t close at second
order.

For the potential (4.26), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{b}{(q\xi+z)^{2}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}+qN_{5}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\qquad\mathcal{S}_{4}=N_{3}+4qN_{2}+\frac{b}{(q\xi+z)^{2}},$
$\displaystyle\mathcal{S}_{5}=N_{4}+2qN_{5}-\frac{b\xi}{(q\xi+z)^{2}},\qquad\mathcal{S}_{6}=N_{8}+\frac{b\xi^{2}}{(q\xi+z)^{2}}.$

They satisfy

$4(2qz-\eta)\tilde{\mathcal{J}}^{2}+\xi\mathcal{S}_{1}^{0}-4\mathcal{S}_{3}^{0}-\xi\mathcal{S}_{4}^{0}=4\big{(}z^{2}+\xi\eta\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{3}^{0}-2\xi\mathcal{S}_{5}^{0}-\mathcal{S}_{6}^{0}=0$

and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=8q^{2}\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{6}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{6},$
$\displaystyle\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{4}-16q^{2}\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-8\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{6}.$

For the potential (4.27), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{bz}{\xi^{3}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}-\frac{1}{2}N_{4}+\frac{bz}{2\xi^{2}}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\qquad\mathcal{S}_{4}=N_{2}-\frac{b}{8\xi^{2}},$
$\displaystyle\mathcal{S}_{5}=N_{5}+\frac{b}{4\xi},\qquad\mathcal{S}_{6}=N_{8}+\frac{bz}{\xi}.$

They satisfy
$4\big{(}\xi\eta+z^{2}\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{3}^{0}-\mathcal{S}_{6}^{0}=2z\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{4}^{0}-\mathcal{S}_{5}^{0}=0$
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{5},$
$\displaystyle\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=b\tilde{\mathcal{J}}.$

The case of the potential (4.28) is treated as a special case of (4.29) (with
$a=-3/2$) below.

We consider the potential (4.29):

$V(\xi,z)=bz\xi^{a}+F(\xi),\qquad a\neq-2,-3/2,-1;$

we cover these exclusions as special cases below. Under our assumptions we
have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+bz\xi^{a}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}-\frac{a}{2(2a+3)}N_{4}+\frac{2abz\xi^{a+1}}{4(2a+3)}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,$
$\displaystyle\mathcal{S}_{4}=N_{2}+\frac{b\xi^{a+1}}{4(1+a)},\qquad\mathcal{S}_{5}=N_{5}-\frac{b\xi^{a+2}}{4(a+2)}.$

They satisfy

$2z\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{4}^{0}-\mathcal{S}_{5}^{0}=0$
(4.35)

and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-\frac{3(a+1)}{2a+3}\tilde{\mathcal{J}}\mathcal{S}_{4},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\frac{3(a+2)}{2a+3}\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-2\tilde{\mathcal{J}}^{3}.$

In the case $a=-2$ we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{bz}{\xi^{2}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}-N_{4}+\frac{bz}{\xi}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\quad\mathcal{S}_{4}=N_{2}-\frac{b}{4\xi},\qquad\mathcal{S}_{5}=N_{5}-\frac{b\log\xi}{4}.$

They satisfy (4.35) and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-3\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\frac{3b}{4}\tilde{\mathcal{J}},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-2\tilde{\mathcal{J}}^{3}.$

In the case $a=-3/2$ we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{bz}{\xi^{3/2}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=N_{4}-\frac{bz}{\xi^{1/2}},\qquad\mathcal{S}_{4}=N_{2}-\frac{b}{2\xi^{1/2}},\qquad\mathcal{S}_{5}=N_{5}-\frac{b\xi^{1/2}}{2}.$

They satisfy (4.35) and their nonzero commutators are

$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=2\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-2\tilde{\mathcal{J}}^{3}.$

In the case $a=-1$ we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{bz}{\xi}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}-\frac{1}{2}N_{4}-\frac{bz}{2}+\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\quad\mathcal{S}_{4}=N_{2}+\frac{b\log\xi}{4},\qquad\mathcal{S}_{5}=N_{5}-\frac{b\xi}{4}.$

They satisfy (4.35) and their nonzero commutators are

$\displaystyle\\{\tilde{\mathcal{J}},\mathcal{S}_{3}\\}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-\tfrac{3b}{4}\tilde{\mathcal{J}},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-3\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-2\tilde{\mathcal{J}}^{3}.$

For the potential (4.30), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+b\xi+F(q\xi+z),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}+qN_{5}+\frac{b\xi^{2}}{8},\qquad\mathcal{S}_{4}=4qN_{2}+N_{3}+F(q\xi+z),$
$\displaystyle\mathcal{S}_{5}=2q(1+2q^{2})N_{2}+2qN_{6}-N_{10}-\frac{b\eta}{2}-qbz+q^{2}F(q\xi+z).$

They satisfy

$4(2qz-\eta)\tilde{\mathcal{J}}^{2}+\xi\mathcal{S}_{1}^{0}-4\mathcal{S}_{3}^{0}-\xi\mathcal{S}_{4}^{0}=0$

and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{5}\big{\\}}=\tfrac{b}{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=8q^{2}\tilde{\mathcal{J}}^{3},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=8q^{4}\tilde{\mathcal{J}}^{3}+q^{2}\tilde{\mathcal{J}}\mathcal{S}_{2}-3q^{2}\tilde{\mathcal{J}}\mathcal{S}_{4}+2\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-4q^{2}b\tilde{\mathcal{J}}.$

The case of the potential (4.31) is obtained exactly as a special case of
(4.29) (with $a=1$) above.

For the potential (4.32) (a special case of (4.29) with $a=0$, but with an
additional symmetry), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+bz+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\qquad\mathcal{S}_{4}=N_{2}+\frac{b\xi}{4},\qquad\mathcal{S}_{5}=N_{3}-bz,\qquad\mathcal{S}_{6}=N_{5}-\frac{b\xi^{2}}{8}.$

They satisfy
$4\eta\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{1}^{0}+4\mathcal{S}_{3}^{0}+\xi\mathcal{S}_{5}^{0}=2z\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{4}^{0}-\mathcal{S}_{6}^{0}=0$
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{6}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{6},$
$\displaystyle\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-b\tilde{\mathcal{J}},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{4}.$

For the potential (4.33), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{b}{z^{2}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}+\frac{1}{4}\int\xi
F^{\prime}(\xi)\,\mathrm{d}\xi,\qquad\mathcal{S}_{4}=N_{2}+\frac{b}{z^{2}},\qquad\mathcal{S}_{5}=N_{4}-\frac{b\xi}{z^{2}},\qquad\mathcal{S}_{6}=N_{8}+\frac{b\xi^{2}}{z^{2}}.$

They satisfy

$4\big{(}\xi\eta+z^{2}\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{3}^{0}-2\xi\mathcal{S}_{5}^{0}-\mathcal{S}_{6}^{0}=4\tilde{\mathcal{J}}^{2}-\xi\mathcal{S}_{1}^{0}+4\mathcal{S}_{3}^{0}+\xi\mathcal{S}_{4}^{0}=0$

and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{6}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{6},$
$\displaystyle\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-4\tilde{\mathcal{J}},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-8\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{6}.$

For the potential (4.34), we consider two cases. In the first case,
$\mu_{2}=0$ and (4.34) reduces to (4.33) after a redefinition of $F(\xi)$. In
the second case, we take $\mu_{2}\neq 0$, so we define $q=\mu_{1}/\mu_{2}$ so
that (4.34) reduces to

$V(\xi,z)=\frac{bz(\xi+qz)}{\xi^{2}(\xi+2qz)^{2}}+F(\xi)$

after a redefinition of $F(\xi)$ and introduction of a new free parameter $b$.
For this potential we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=(p_{1}+{\rm
i}p_{2})/2,\qquad\mathcal{S}_{1}=\mathcal{H}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}+\frac{bz(\xi+qz)}{\xi^{2}(\xi+2qz)^{2}}+F(\xi),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{1}-\frac{1}{2}N_{4}+\frac{bz(\xi+qz)}{2\xi(\xi+2qz)^{2}}+\frac{1}{4}\int\xi
F(\xi)\,\mathrm{d}\xi,\qquad\mathcal{S}_{4}=N_{2}+\frac{q}{2}N_{3}-\frac{b}{8(\xi+2qz)^{2}},$
$\displaystyle\mathcal{S}_{5}=N_{5}+qN_{4}+\frac{b\xi}{4(\xi+2qz)^{2}},\qquad\mathcal{S}_{6}=N_{8}-\frac{b\xi^{2}}{4q(\xi+2qz)^{2}}.$

They satisfy

$4(q\eta-z)\tilde{\mathcal{J}}^{2}-q\xi\mathcal{S}_{1}^{0}+4q\mathcal{S}_{3}^{0}+2\xi\mathcal{S}_{4}^{0}+2\mathcal{S}_{5}^{0}=4\big{(}\xi\eta+z^{2}\big{)}\tilde{\mathcal{J}}^{2}-\xi^{2}\mathcal{S}_{1}^{0}+4\xi\mathcal{S}_{3}^{0}-\mathcal{S}_{6}^{0}=0$

and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\tilde{\mathcal{J}}^{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-\tilde{\mathcal{J}}\mathcal{S}_{5},$
$\displaystyle\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=-4q\tilde{\mathcal{J}}\mathcal{S}_{4}-2\tilde{\mathcal{J}}^{3},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{6}\\}=-4\tilde{\mathcal{J}}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{5},\mathcal{S}_{6}\\}=-4q\tilde{\mathcal{J}}\mathcal{S}_{6}.$

###  4.4 Third case: $\boldsymbol{{\mathcal{J}}=xp_{2}-yp_{1}}$

Here the centralizer of $\mathcal{J}$ is the group generated by translation in
$z$ and rotations about the $z$-axis. We can use this freedom to simplify the
computation. Since ${\mathcal{J}}$ is a symmetry the potential must be of the
form $V\big{(}x^{2}+y^{2},z\big{)}$. A basis for symmetries is again given by
(4.2), but as in the previous section, we will construct a more convenient
basis by consider the action of $\operatorname{Ad}_{J_{12}}$.

In addition we obtain a series of equations for the first derivatives
$\partial_{x}F_{0}$, $\partial_{y}F_{0}$, $\partial_{z}F_{0}$, which lead to
Bertrand–Darboux equations for $V\big{(}x^{2}+y^{2},z\big{)}$. At the end we
have to find 5 linearly independent solutions for $\mathcal{S}$ and verify
that they admit one functionally linearly dependent solution.

The adjoint action
${\mathcal{S}}\to\\{J_{12},{\mathcal{S}}\\}\equiv\operatorname{Ad}_{J_{12}}{\mathcal{S}}$
will map the 5-dimensional space of a solution set into itself. This action
preserves the order of symmetry operators that are homogeneous in Cartesian
coordinates. However, it is also convenient to introduce cylindrical
coordinates $\\{r,\theta,z\\}$ where $x=r\cos(\theta)$, $y=r\sin(\theta)$,
$z=z$, and

$\displaystyle p_{1}=p_{r}\cos(\theta)-p_{\theta}\sin(\theta)/r,\qquad
p_{2}=p_{r}\sin(\theta)+p_{\theta}\cos(\theta)/r,\qquad p_{3}=p_{z},$
$\displaystyle J_{12}=p_{\theta},\qquad
J_{13}=(rp_{z}-zp_{r})\cos\theta+\frac{z\sin\theta\,p_{\theta}}{r},\qquad
J_{23}=(rp_{z}-zp_{r})\sin\theta-\frac{z\cos\theta\,p_{\theta}}{r}.$

On the components of $R$ in (4.3), $\mathcal{J}=J_{12}$ has the following
nontrivial actions:

$\displaystyle\operatorname{Ad}_{J_{12}}p_{1}=p_{2},\qquad\operatorname{Ad}_{J_{12}}p_{2}=-p_{1},\qquad\operatorname{Ad}_{J_{12}}J_{13}=J_{23},\qquad\operatorname{Ad}_{J_{12}}J_{23}=-J_{13}.$

We can use these to construct a basis consisting of eigenvectors of
$\operatorname{Ad}_{J_{12}}$. We label the eigenvectors to take advantage of
their transformation under rotation: eigenvectors with subscripts $\pm 2$,
$\pm 1$, and $0$ indicate corresponding eigenvalues of $\pm 2{\rm i}$,
$\pm{\rm i}$, and 0, respectively (the second subscript, when applicable,
distinguishes between multiple eigenvectors of the same order with the same
eigenvalue). A complex eigenbasis for the 6-dimensional space of symmetries of
order 2 is

$\displaystyle L_{0,1}=J_{12}^{2},\qquad L_{0,2}=J_{13}^{2}+J_{23}^{2},\qquad
L_{+1}=-\tfrac{1}{2}J_{12}(J_{13}-{\rm i}J_{23}),$ $\displaystyle
L_{-1}=\tfrac{1}{2}J_{12}(J_{13}+{\rm i}J_{23}),\qquad L_{-2}=-\tfrac{{\rm
i}}{8}(J_{13}+{\rm i}J_{23})^{2},\qquad L_{+2}=\tfrac{{\rm i}}{8}(J_{13}-{\rm
i}J_{23})^{2}.$

A complex eigenbasis for the 8-dimensional space of symmetries of order 1 is

$\displaystyle M_{0,1}=-p_{3}J_{12},\qquad
M_{0,2}=-p_{1}J_{13}-p_{2}J_{23},\qquad
M_{+1,2}=-\tfrac{1}{2}p_{3}(J_{13}+{\rm i}J_{23}),$ $\displaystyle
M_{+1,1}=(p_{1}+{\rm i}p_{2})J_{12},\qquad
M_{-1,2}=\tfrac{1}{2}p_{3}(J_{13}-{\rm i}J_{23}),\qquad
M_{-1,1}=-\tfrac{1}{2}J_{12}(p_{1}-{\rm i}p_{2}),$ $\displaystyle
M_{+2}=-\tfrac{1}{4}(p_{1}-{\rm i}p_{2})(J_{13}-{\rm i}J_{23}),\qquad
M_{-2}=\tfrac{1}{4}(p_{1}+{\rm i}p_{2})(J_{13}+{\rm i}J_{23}).$

A complex eigenbasis for the 6-dimensional space of symmetries of order 0 is

$\displaystyle N_{0,1}=p_{3}^{2},\qquad
N_{0,2}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2},\qquad N_{-2}=-\tfrac{1}{4}(p_{1}+{\rm
i}p_{2})^{2},$ $\displaystyle N_{+2}=\tfrac{1}{4}(p_{1}-{\rm
i}p_{2})^{2},\qquad N_{-1}=-\tfrac{1}{2}(p_{1}+{\rm i}p_{2})p_{3},\qquad
N_{+1}=\tfrac{1}{2}(p_{1}-{\rm i}p_{2})p_{3}.$

Because $\mathcal{J}$ and $\mathcal{H}$ must be basis vectors, it follows that
the possible actions of $\operatorname{Ad}_{p_{\theta}}$ on an eigenbasis are
described by the canonical forms

$\displaystyle\left(\begin{matrix}\lambda_{1}&0&0&0&0\\\
0&\lambda_{2}&0&0&0\\\ 0&0&\lambda_{3}&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}\lambda_{1}&0&0&0&0\\\
0&\lambda_{2}&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),$
$\displaystyle\left(\begin{matrix}\lambda_{1}&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right),$ (4.36)

where $\lambda_{j}=\pm{\rm i},\pm 2{\rm i}$. Note that there are a large
number of cases to consider. The matrices in (4.36) are all diagonal and each
contains at least 2 zeros on the diagonal because $\mathcal{J}$ and
$\mathcal{H}$ must always be included as eigenfunctions. The remaining
eigenfunctions correspond to $3$, $2$, $1$ or 0 eigenvalues $\lambda_{j}$. All
possible choices have to be considered from the eigenfunctions listed above.

####  4.4.1 Form (4.36a)

Since the eigenvalues for real Euclidean space must occur in complex-conjugate
pairs, a system of this form is only possible for Minkowski space. We examine
all such cases and find numerous FLD systems, but none are 2-parameter
functionally independent superintegrable.

####  4.4.2 Form (4.36b)

We find the following FLD bases and potentials (in each case $F$ is an
arbitrary function of its argument and $b$ is an arbitrary parameter)

$\displaystyle{\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},L_{+1},L_{-1},L_{0,2}\big{\\}},\qquad
V(r,z)=F\big{(}r^{2}+z^{2}\big{)}+\frac{bz}{r\big{(}r^{2}+z^{2}\big{)}},$
(4.37)
$\displaystyle{\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},L_{+2},L_{-2},L_{0,2}\big{\\}},\qquad
V(r,z)=F\big{(}r^{2}+z^{2}\big{)}+\frac{b}{z^{2}},$ (4.38)
$\displaystyle{\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},N_{+2},N_{-2},N_{0,1}\big{\\}},\qquad
V(r,x)=br^{2}+F(z),$ (4.39)

and

${\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},M_{+1,1},M_{-1,1},N_{0,1}\big{\\}},\qquad
V(r,z)=\frac{b}{r}+F(z),$ (4.40)

In addition, there is the FLD basis and potential

${\mathcal{B}}=\big{\\{}{\mathcal{H}}^{0},{\mathcal{J}}^{2},N_{+2},N_{-2},N_{0,1}\big{\\}},\qquad
V(r,z)=b_{1}\big{(}4r^{2}+z^{2}+2qz\big{)}+\frac{b_{2}}{(z+q)^{2}},$ (4.41)

which is 2-parameter superintegrable.

####  4.4.3 Form (4.36c)

Since the eigenvalues for real Euclidean space must occur in complex-conjugate
pairs, a system of this form is only possible for Minkowski space. We examine
all such systems and find that none are FLD.

####  4.4.4 Form (4.36d)

Checking over all possibilities for systems with this eigenvalue form, we find
that none are FLD.

####  4.4.5 Symmetry algebras

For the potential (4.37), we have the symmetries

$\displaystyle\mathcal{J}=J_{12},\qquad\mathcal{S}_{1}=\mathcal{H}=N_{0,2}+F\big{(}r^{2}+z^{2}\big{)}+\frac{bz}{r\big{(}r^{2}+z^{2}\big{)}},\qquad\mathcal{S}_{2}=\mathcal{J}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{0,2}+\frac{bz}{r},\qquad\mathcal{S}_{4}=L_{-1}-\frac{{\rm
i}b{\rm e}^{-{\rm i}\theta}}{4},\qquad\mathcal{S}_{5}=L_{+1}-\frac{{\rm
i}b{\rm e}^{{\rm i}\theta}}{4}.$

They satisfy ${\rm i}z{\rm e}^{{\rm i}\theta}\mathcal{J}^{2}+r{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{4}^{0}+r\mathcal{S}_{5}^{0}=0$ and their nonzero
commutators are

$\displaystyle\\{\mathcal{J},\mathcal{S}_{4}\\}={\rm
i}\mathcal{S}_{4},\qquad\\{\mathcal{J},\mathcal{S}_{5}\\}=-{\rm
i}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2{\rm
i}\mathcal{J}\mathcal{S}_{4},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=2{\rm
i}\mathcal{J}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=\tfrac{{\rm
i}}{2}\mathcal{J}\mathcal{S}_{3}-\tfrac{{\rm i}}{2}\mathcal{J}^{3}.$

For the potential (4.38), we have the symmetries

$\displaystyle\mathcal{J}=J_{12},\qquad\mathcal{S}_{1}=\mathcal{H}=N_{0,2}+F\big{(}r^{2}+z^{2}\big{)}+\frac{b}{z^{2}},\qquad\mathcal{S}_{2}=\mathcal{J}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{0,2}+\frac{br^{2}}{z^{2}},\qquad\mathcal{S}_{4}=L_{+2}-\frac{{\rm
i}br^{2}{\rm e}^{2{\rm
i}\theta}}{8z^{2}},\qquad\mathcal{S}_{5}=L_{-2}-\frac{{\rm i}br^{2}{\rm
e}^{-2{\rm i}\theta}}{8z^{2}}.$

They satisfy $2{\rm i}z^{2}{\rm e}^{2{\rm i}\theta}\mathcal{J}^{2}-{\rm
i}r^{2}{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{3}^{0}-4r^{2}\mathcal{S}_{4}^{0}-4r^{2}{\rm e}^{4{\rm
i}\theta}\mathcal{S}_{5}^{0}=0$ and their nonzero commutators are

$\displaystyle\\{\mathcal{J},\mathcal{S}_{4}\\}=-2{\rm
i}\mathcal{S}_{4},\qquad\\{\mathcal{J},\mathcal{S}_{5}\\}=2{\rm
i}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-4{\rm
i}\mathcal{J}\mathcal{S}_{4},$
$\displaystyle\\{\mathcal{S}_{3},\mathcal{S}_{5}\\}=-4{\rm
i}\mathcal{J}\mathcal{S}_{5},\qquad\\{\mathcal{S}_{4},\mathcal{S}_{5}\\}=\tfrac{{\rm
i}}{8}\mathcal{J}\mathcal{S}_{3}+\tfrac{b}{4}\mathcal{J}^{3}.$

For the potential (4.39), we have the symmetries

$\displaystyle\mathcal{J}=J_{12},\qquad\mathcal{S}_{1}=\mathcal{H}=N_{0,2}+br^{2}+G(z),\qquad\mathcal{S}_{2}=\mathcal{J}^{2},$
$\displaystyle\mathcal{S}_{3}=N_{+2}-\frac{br^{2}{\rm e}^{2{\rm
i}\theta}}{4},\qquad\mathcal{S}_{4}=N_{-2}+\frac{br^{2}{\rm e}^{-2{\rm
i}\theta}}{4},\qquad\mathcal{S}_{5}=N_{0,1}+G(z).$

They satisfy $2{\rm e}^{2{\rm i}\theta}\mathcal{J}^{2}-r^{2}{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{2}^{0}-2r^{2}\mathcal{S}_{3}^{2}r^{2}{\rm e}^{4{\rm
i}\theta}\mathcal{S}_{4}^{0}+r^{2}{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{5}^{0}=0$ and their nonzero commutators are

$\displaystyle\\{\mathcal{J},\mathcal{S}_{3}\\}=-2{\rm
i}\mathcal{S}_{3},\qquad\\{\mathcal{J},\mathcal{S}_{4}\\}=2{\rm
i}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}={\rm
i}b\mathcal{J}.$

For the potential (4.40), we have the symmetries

$\displaystyle\mathcal{J}=J_{12},\qquad\mathcal{S}_{1}=\mathcal{H}=N_{0,2}+F(z)+\frac{b}{r},\qquad\mathcal{S}_{2}=\mathcal{J}^{2}$
$\displaystyle\mathcal{S}_{3}=M_{+1,1}+\frac{{\rm i}b{\rm e}^{{\rm
i}\theta}}{4},\qquad\mathcal{S}_{4}=M_{-1,1}+\frac{{\rm i}b{\rm e}^{-{\rm
i}\theta}}{4},\qquad\mathcal{S}_{5}=N_{0,1}+F(z).$

They satisfy ${\rm i}{\rm e}^{{\rm
i}\theta}\mathcal{J}^{2}-r\mathcal{S}_{3}^{0}-r{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{4}^{0}=0$ and their nonzero commutators are

$\displaystyle\\{\mathcal{J},\mathcal{S}_{3}\\}=-{\rm
i}\mathcal{S}_{3},\qquad\\{\mathcal{J},\mathcal{S}_{4}\\}={\rm
i}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=\tfrac{{\rm
i}}{2}\mathcal{J}(\mathcal{S}_{5}-\mathcal{S}_{2}).$

For the potential (4.41), we have the symmetries

$\displaystyle\mathcal{J}=J_{12},\qquad\mathcal{S}_{1}=\mathcal{H}=N_{0,2}+b_{1}\big{(}4r^{2}+z^{2}+2qz\big{)}+\frac{b_{2}}{(z+q)^{2}},\qquad\mathcal{S}_{2}=\mathcal{J}^{2},$
$\displaystyle\mathcal{S}_{3}=N_{+2}-b_{1}r^{2}{\rm e}^{2{\rm
i}\theta},\qquad\mathcal{S}_{4}=N_{-2}+b_{1}r^{2}{\rm e}^{-2{\rm
i}\theta},\qquad\mathcal{S}_{5}=N_{0,1}+b_{1}z(z+2q)+\frac{b_{2}}{(z+q)^{2}},$
$\displaystyle\mathcal{S}_{6}=M_{+1,2}-qN_{+1}+\frac{r{\rm e}^{{\rm
i}\theta}(b_{1}(z+q)^{4}-b_{2})}{2(z+q)^{2}},$
$\displaystyle\mathcal{S}_{7}=M_{-1,2}-qN_{-1}-\frac{r{\rm e}^{-{\rm
i}\theta}(b_{1}(z+q)^{4}-b_{2})}{2(z+q)^{2}}.$

They satisfy $\mathcal{J}^{2}-r^{2}\mathcal{S}_{1}^{0}-2r^{2}{\rm e}^{-2{\rm
i}\theta}\mathcal{S}_{5}+2r^{2}{\rm e}^{2{\rm
i}\theta}\mathcal{S}_{6}+r^{2}\mathcal{S}_{7}=0$. The subset
$\\{\mathcal{J},\mathcal{S}_{1},\mathcal{S}_{2},\mathcal{S}_{5},\mathcal{S}_{6},\mathcal{S}_{7}\\}$
generates a closed quadratic algebra with nonzero relations

$\displaystyle\\{\mathcal{J},\mathcal{S}_{3}\\}=-2{\rm
i}\mathcal{S}_{3},\qquad\\{\mathcal{J},\mathcal{S}_{4}\\}=2{\rm
i}\mathcal{S}_{4},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=4{\rm
i}b_{1}\mathcal{J}.$

However, if any linear combination of ${\mathcal{S}}_{3}$, ${\mathcal{S}}_{4}$
is added to the generators, a new 3rd order symmetry is produced that is not a
polynomial in the generators, so the resulting algebra doesn’t close at second
order.

###  4.5 Fourth case: $\boldsymbol{{\mathcal{J}}=J_{12}+{\rm i}J_{23}}$

This case is similar to the second case, treated in Section 4.3. We make the
change of variables

$\displaystyle x=-\rho\big{[}{\rm e}^{-\theta}+{\rm
e}^{\theta}\big{(}1/4-r^{2}\big{)}\big{]},\qquad y=-\rho r\exp(\theta),\qquad
z={\rm i}\rho\big{[}{\rm e}^{-\theta}-{\rm
e}^{\theta}\big{(}1/4+r^{2}\big{)}\big{]},$

so that $p_{r}=2(J_{12}+{\rm i}J_{23})$ and

$\displaystyle p_{1}=\frac{{\rm
e}^{2\theta}\big{(}4r^{2}-1\big{)}(p_{\theta}-\rho
p_{\rho})-4(p_{\theta}-2rp_{r}+\rho p_{\rho})}{4\rho{\rm e}^{\theta}},\qquad
p_{2}=\frac{r{\rm e}^{2\theta}(\rho p_{\rho}-p_{\theta}-p_{r})}{\rho{\rm
e}^{\theta}},$ $\displaystyle p_{3}={\rm i}\frac{{\rm
e}^{2\theta}(4r^{2}+1)(p_{\theta}-\rho p_{\rho})+4(p_{\theta}-2rp_{r}+\rho
p_{\rho})}{4\rho{\rm e}^{\theta}},$ $\displaystyle
J_{12}=-2rp_{\theta}+\big{(}\tfrac{1}{4}+r^{2}+{\rm
e}^{-2\theta}\big{)}p_{r},\qquad J_{13}={\rm i}(rp_{r}-p_{\theta}),$
$\displaystyle J_{23}={\rm i}\big{(}r^{2}+{\rm e}^{-2\theta}\big{)}p_{r}-2{\rm
i}rp_{\theta}-\tfrac{{\rm i}}{4}p_{r}.$

Similarly to Section 4.3, we prefer to work with $\tilde{J}=p_{r}$. The action
of $\operatorname{Ad}_{\tilde{J}}$ on the elements of $R$ in (4.3) is

$\displaystyle\operatorname{Ad}_{\tilde{J}}p_{1}=2p_{2},\qquad\operatorname{Ad}_{\tilde{J}}p_{2}=-2(p_{1}-{\rm
i}p_{3}),\qquad\operatorname{Ad}_{\tilde{J}}p_{3}=-2{\rm i}p_{2},$
$\displaystyle\operatorname{Ad}_{\tilde{J}}J_{12}=2{\rm
i}J_{13},\qquad\operatorname{Ad}_{\tilde{J}}J_{13}=-2{\rm i}(J_{12}+{\rm
i}J_{23}),\qquad\operatorname{Ad}_{\tilde{J}}J_{23}=-2J_{13}.$

From here we can construct a convenient generalized eigenbasis of symmetries.

A basis for the six-dimensional space of order-two symmetries is

$\displaystyle L_{1}=\tfrac{1}{24}J_{12}^{2},\qquad L_{2}=\tfrac{{\rm
i}}{6}J_{12}J_{23},\qquad L_{3}=\tfrac{1}{3}\big{(}J_{12}^{2}-J_{13}^{2}+{\rm
i}J_{12}J_{23}\big{)},$ $\displaystyle L_{4}=2{\rm i}(J_{12}+{\rm
i}J_{23})J_{13},\qquad L_{5}=4(J_{12}+{\rm i}J_{23})^{2},\qquad
L_{6}=J_{12}^{2}+J_{13}^{2}+J_{23}^{2}.$

Here, $\\{L_{1},L_{2},L_{3},L_{4},L_{5}\\}$ form a chain and
$\\{L_{5},L_{6}\\}\subset\ker\operatorname{Ad}_{p_{r}}$.

A basis for the eight-dimensional space of order-one symmetries is

$\displaystyle M_{1}=\tfrac{1}{24}p_{1}J_{12},\qquad
M_{2}=\tfrac{1}{12}(p_{2}J_{12}+{\rm i}p_{1}J_{13}),\qquad M_{3}=\tfrac{{\rm
i}}{6}(p_{1}J_{23}+2p_{2}J_{13}+p_{3}J_{12}),$ $\displaystyle M_{4}=-{\rm
i}p_{1}J_{13}+p_{2}(J_{12}+{\rm i}J_{23})-p_{3}J_{13},\qquad
M_{5}=-4(p_{1}-{\rm i}p_{3})(J_{12}+{\rm i}J_{23}),$ $\displaystyle
M_{6}=\tfrac{1}{2}(p_{2}J_{12}-{\rm i}p_{1}J_{13}),\qquad
M_{7}=-2p_{1}J_{12}-{\rm i}p_{1}J_{23}+{\rm i}p_{3}J_{12},$ $\displaystyle
M_{8}=-2({\rm i}p_{1}+p_{3})J_{13}-2p_{2}(J_{12}+{\rm i}J_{23}).$

Here, we have two separate chains: $\\{M_{1},M_{2},M_{3},M_{4},M_{5}\\}$ and
$\\{M_{6},M_{7},M_{8}\\}$:
$\\{M_{5},M_{8}\\}\subset\ker\operatorname{Ad}_{p_{r}}$.

A basis for the six-dimensional space of order-zero symmetries is

$\displaystyle N_{1}=\tfrac{1}{24}p_{1}^{2},\qquad
N_{2}=\tfrac{1}{6}p_{1}p_{2},\qquad
N_{3}=-\tfrac{1}{3}\big{(}p_{1}^{2}-p_{2}^{2}+{\rm i}p_{1}p_{3}\big{)},$
$\displaystyle N_{4}=-2(p_{1}-{\rm i}p_{3})p_{2},\qquad N_{5}=4(p_{1}-{\rm
i}p_{3})^{2},\qquad N_{6}=\mathcal{H}^{0}=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}.$

Here, $\\{N_{1},N_{2},N_{3},N_{4},N_{5}\\}$ form a chain and
$\\{N_{5},N_{6}\\}\subset\ker\operatorname{Ad}_{\tilde{J}}$.

The possible canonical forms are

$\displaystyle\allowdisplaybreaks\left(\begin{matrix}0&1&0&0&0\\\ 0&0&1&0&0\\\
0&0&0&1&0\\\ 0&0&0&0&1\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&1&0&0&0\\\
0&0&1&0&0\\\ 0&0&0&1&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right),$ (4.42)
$\displaystyle\left(\begin{matrix}0&1&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&1\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&1&0&0&0\\\
0&0&1&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right),$ (4.43)
$\displaystyle\left(\begin{matrix}0&1&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&1&0\\\
0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&1&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{matrix}\right),\qquad\left(\begin{matrix}0&0&0&0&0\\\
0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\end{matrix}\right).$ (4.44)

####  4.5.1 Form (4.42a)

Here we have a 5-chain. This form does not occur because it cannot not contain
$\mathcal{H}$ (moreover, $\mathcal{H}$ and $p_{r}^{2}$ cannot be in the same
chain).

####  4.5.2 Form (4.42b)

This form can (and must) contain both $\tilde{\mathcal{J}}^{2}$ and
$\mathcal{H}^{0}$. Because $\mathcal{H}^{0}$ is not in a nontrivial chain, the
basis must be

$\big{\\{}\mathcal{H}^{0},L_{2}+\beta L_{3}+\gamma
L_{4}+\delta_{1}L_{5}+\delta_{2}L_{6},L_{3}+\beta L_{4}+\gamma
L_{5},L_{4}+\beta L_{5},L_{5}=p_{r}^{2}\big{\\}},$

but we can take $\beta=\gamma=\delta_{1}=0$ by a canonical form-preserving
change of basis. The chain $\\{L_{2}+\delta_{2}L_{6},L_{3},L_{4},L_{5}\\}$ is
FLD but does not correspond to an admissible potential.

####  4.5.3 Form (4.43c)

Here we have a 3-chain and a 2-chain. This form does not occur because it
cannot not contain $\mathcal{H}$.

####  4.5.4 Form (4.43d)

Here we have a 3-chain and two 1-chains. One of the 1-chains is
$\mathcal{H}^{0}$. First suppose the second one-chain is $p_{r}^{2}$. Using
canonical form-preserving changes of basis when necessary, the possible
3-chains are equivalent to one of $\\{N_{3},N_{4},N_{5}\\}$,

$\displaystyle\\{\alpha_{1}M_{3}+M_{6}+\beta_{1}M_{4}+\gamma_{1}M_{5},\alpha_{1}M_{4}+M_{7}+\beta_{1}M_{5},\alpha_{1}M_{5}+M_{8}\\},$
$\displaystyle\\{M_{3}+\alpha_{2}M_{6}+\beta_{2}M_{7}+\gamma_{2}M_{8},M_{4}+\alpha_{2}M_{7}+\beta_{2}M_{8},M_{5}+\alpha_{2}M_{8}\\}.$

The first case is FLD and provides the admissible potential

$\displaystyle V(\rho,\theta)=b\rho^{2}+F\big{(}\rho{\rm e}^{\theta}\big{)}.$
(4.45)

The second case is not FLD. The third case is FLD when $\alpha_{2}=0$ and
$\beta_{2}=\pm 1/2$ but these cases do not provide 2-parameter potentials.

If $\tilde{\mathcal{J}}^{2}$ is not one of the 1-chains, our basis must
contain (after a canonical form-preserving change of basis)
$\\{L_{3}+\gamma_{2}L_{6},L_{4},L_{5}\\}$. It is left to chose a second
1-chain, for which there are three possibilities: $L_{6}$ (in which case we
can take $\gamma=0$ by a canonical form-preserving change of basis), $\mu
M_{5}+\nu M_{8}$, and $N_{5}$. The first possibility gives an FLD basis and
has the admissible potential

$V(\rho,\theta)=\frac{b{\rm e}^{-2\theta}}{\rho^{2}}+F(\rho).$ (4.46)

The second and third possibilities are FLD when $\gamma=1/3$, but neither
leads to a 2-parameter potential.

####  4.5.5 Form (4.44e)

Here we have two 2-chains and a 1-chain, which must be $\mathcal{H}^{0}$. One
of the 2-chains must be $\\{L_{4}+\mu L_{6},L_{5}\\}$. The possibilities for
the other 2-chain are (after canonical form-preserving changes of basis)
$\\{\alpha M_{4}+M_{7}+\gamma M_{5},\alpha M_{5}+M_{8}\\}$ or $\\{M_{4}+\beta
M_{7}+\delta M_{8},M_{5}+\beta M_{8}\\}$. Only the latter (together with
$L_{5}$) is FLD when $\alpha=1$, $\beta=\gamma=0$, and $\delta=-1/2$ but does
not yield an admissible potential.

####  4.5.6 Form (4.44f)

Here we have a 2-chain and three 1-chains, one of which must be
$\mathcal{H}^{0}$. We first assume that the 2-chain is $\\{L_{4}+\mu
L_{6},L_{5}\\}$. There are then four ways to choose the remaining two one
chains: $\\{N_{5},\alpha M_{5}+\beta M_{8}\\}$, $\\{\alpha M_{5}+\beta
M_{8},L_{6}\\}$ (in which case we take $\mu=0$), $\\{N_{5},L_{6}\\}$ (again,
$\mu=0$), or $\\{M_{5},M_{8}\\}$. The first case is FLD when $\alpha=0$ but
does yield an admissible potential. The second, third, and fourth cases are
not FLD.

If the 2-chain is not $\\{L_{4}+\mu L_{6},L_{5}\\}$, one of the 1-chains must
be $L_{5}=\tilde{\mathcal{J}}^{2}$. Then we have one 1-chain ($N_{5}$,
$L_{6}$, or $\mu M_{5}+\nu M_{8}$) and one 2-chain ($\\{N_{4},N_{5}\\}$,
$\\{\alpha M_{4}+M_{7}+\gamma M_{5},\alpha M_{5}+M_{8}\\}$, or $\\{M_{4}+\beta
M_{7}+\delta M_{8},M_{5}+\beta M_{8}\\}$) to choose. There are several FLD
bases but only one leads to an admissible potential

$\displaystyle V(\rho,\theta)=\frac{b{\rm
e}^{-3\theta}}{\rho}+F\big{(}\rho{\rm e}^{\theta}\big{)}.$ (4.47)

####  4.5.7 Form (4.44g)

This case consists of five 1-chains, two of which must be $\mathcal{H}^{0}$
and $\tilde{\mathcal{J}}^{2}$. There are therefore three subcases to consider:
the remaining symmetries are either $\\{L_{6},\alpha M_{5}+\beta
M_{8},N_{5}\\}$, $\\{L_{6},M_{5},M_{8}\\}$, or $\\{M_{5},M_{8},N_{5}\\}$. The
first and third cases are FLD in certain cases but the corresponding
potentials do not have 2 independent parameters.

####  4.5.8 Structure algebras

For the potential (4.45), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=2(J_{12}+{\rm
i}J_{23}),\qquad\mathcal{S}_{1}=\mathcal{H}=N_{6}+b\rho^{2}+F\big{(}\rho{\rm
e}^{\theta}\big{)},\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=L_{5}+\frac{-\big{[}4+{\rm
e}^{2\theta}\big{(}1-12r^{2}\big{)}\big{]}b\rho^{2}-4F\big{(}\rho{\rm
e}^{\theta}\big{)}}{24},\qquad\mathcal{S}_{4}=N_{4}-b\rho^{2}r{\rm
e}^{2\theta},$ $\displaystyle\mathcal{S}_{5}=N_{5}+b\rho^{2}{\rm
e}^{2\theta}.$

They satisfy $\rho^{2}{\rm
e}^{2\theta}\big{(}4{\mathcal{S}}_{1}^{0}+24{\mathcal{S}}_{3}^{0}+24{\mathcal{S}}_{4}^{0}+\big{(}1+12r^{2}\big{)}{\mathcal{S}}_{5}^{0}\big{)}-12\tilde{\mathcal{J}}^{2}=0$
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\mathcal{S}_{4},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=\mathcal{S}_{5},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=b\tilde{\mathcal{J}}.$

For the potential (4.46), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=2(J_{12}+{\rm
i}J_{23}),\qquad\mathcal{S}_{1}=\mathcal{H}=N_{6}+\frac{b{\rm
e}^{-2\theta}}{\rho^{2}}+F(\rho),\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2}+v_{0},$
$\displaystyle\mathcal{S}_{3}=L_{3}+\frac{br^{2}}{2}-\frac{b{\rm
e}^{-2\theta}}{3},\qquad\mathcal{S}_{4}=L_{4}-br,\qquad\mathcal{S}_{5}=L_{6}+b{\rm
e}^{-2\theta}.$

They satisfy $\big{(}1+12{\rm
e}^{-2\theta}-12r^{2}\big{)}\tilde{\mathcal{J}}^{2}-12{\mathcal{S}}_{3}^{0}-12r{\mathcal{S}}_{4}^{0}-4{\mathcal{S}}_{5}^{0}=0$
and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\mathcal{S}_{4},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=\mathcal{S}_{2},\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-2\tilde{\mathcal{J}}\mathcal{S}_{3}+\tfrac{1}{3}\tilde{\mathcal{J}}\mathcal{S}_{5}+\tfrac{1}{6}\tilde{\mathcal{J}}^{3}+\tfrac{b}{12}\tilde{\mathcal{J}}.$

For the potential (4.47), we have the symmetries

$\displaystyle\tilde{\mathcal{J}}=2(J_{12}+{\rm
i}J_{23}),\qquad\mathcal{S}_{1}=\mathcal{H}=N_{6}+\frac{b{\rm
e}^{-3\theta}}{\rho}+F\big{(}{\rm
e}^{\theta}\rho\big{)},\qquad\mathcal{S}_{2}=\tilde{\mathcal{J}}^{2},$
$\displaystyle\mathcal{S}_{3}=M_{4}-\frac{1}{2}M_{8}-\frac{b\rho
r^{2}}{2\rho},\qquad\mathcal{S}_{4}=M_{5}+br,\qquad\mathcal{S}_{5}=N_{5}-\frac{2b{\rm
e}^{-\theta}}{\rho}.$

They satisfy $\tilde{\mathcal{J}}^{2}+\rho{\rm
e}^{\theta}\hat{\mathcal{S}}_{3}^{0}+\rho r{\rm
e}^{\theta}\hat{\mathcal{S}}_{4}^{0}=0$, and their nonzero commutators are

$\displaystyle\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{3}\big{\\}}=\mathcal{S}_{4},\qquad\big{\\{}\tilde{\mathcal{J}},\mathcal{S}_{4}\big{\\}}=-b,\qquad\\{\mathcal{S}_{3},\mathcal{S}_{4}\\}=-\tfrac{1}{2}\tilde{\mathcal{J}}\mathcal{S}_{4}.$

###  4.6 Fifth case: $\boldsymbol{{\mathcal{J}}=-{\rm i}J_{12}+J_{23}-{\rm
i}p_{1}+p_{3}}$

This case does not occur for complex Euclidean systems since the symmetry
$\mathcal{J}$ is not homogeneous.

###  4.7 Additional comments

We note that for all of the systems classified we can find a complete integral
for the Hamilton–Jacobi equation. For example, the system (4.29) with $a=-3$,
has the Hamilton–Jacobi equation

$\displaystyle 4\frac{\partial S}{\partial\xi}\frac{\partial
S}{\partial\eta}+\left(\frac{\partial S}{\partial
z}\right)^{2}+\frac{bz}{\xi^{3}}+F(\xi)=E.$

For this equation, we find the complete integral

$\displaystyle
S(\xi,\eta,z)=\frac{b^{2}}{768c_{1}^{3}\xi^{3}}+\frac{b\big{(}2c_{1}^{2}z+c_{2}\xi\big{)}}{16c_{1}^{3}\xi^{2}}+c_{1}\eta+\frac{c_{1}^{2}(4c_{2}z+E\xi)-c_{2}^{2}\xi}{4c_{1}^{3}}-\frac{1}{4c_{1}}\int
F(\xi)\,\mathrm{d}\xi,$

where $c_{1}$, $c_{2}$ are arbitrary constants and another constant arises
from the indefinite integral of $F$. The corresponding Schrödinger equation

$\displaystyle\left(4\partial_{\xi}\partial_{\eta}+\partial_{z}^{2}+\frac{bz}{\xi^{3}}+F(\xi)\right)\psi=E\psi$

has the solution

$\psi(\xi,\eta,z)=\exp S(\xi,\eta,z).$

The symmetry algebras for these FLD superintegrable systems don’t always
close. However the symmetries always provide some information about the
classical trajectories of solutions of the Hamilton–Jacobi equation. If a
superintegrable system is functionally independent the trajectories are
uniquely determined, However, if the system is FLD then we can solve for one
of the constants of the motion in terms of the others. Thus a 2-parameter
manifold can be computed from the symmetries such that the trajectories of
solutions of the Hamilton–Jacobi equation must lie on this manifold.

Table 1: Summary of the FLD systems. (4.9) | $V(y,z)=\frac{F(z/y)}{y^{2}}$
---|---
(4.11) | $V(y,z)=\frac{\beta_{1}}{y^{2}}+\frac{1}{(b_{11}y+b_{15}z)^{2}}\left(\beta_{3}+\frac{(-b_{11}z+b_{15}y)\beta_{2}}{\sqrt{y^{2}+z^{2}}}+\frac{(2y^{2}-z^{2})\beta_{1}b_{15}^{2}-2\beta_{1}b_{11}b_{15}yz}{y^{2}}\right)$
(4.13) | $V(y,z)=b(z-{\rm i}y)+F(z+{\rm i}y)$
(4.14) | $V(y,z)=b_{1}(z-{\rm i}y)+b_{2}(z+{\rm i}y)^{2}$
(4.15) | $V(y,z)=b_{1}z+\frac{b_{2}}{y^{2}}$
(4.16) | $V(y,z)=cy+F(z)$
(4.17) | $V(y,z)=c_{1}y+c_{2}z$
(4.20) | $V(\xi,z)=\frac{b}{\xi^{2}}+F(q\xi+z)$
(4.21) | $V(\xi,z)=\frac{b}{\xi^{2}}+F(z)$
(4.22) | $V(\xi,z)=\frac{F(z/\xi)}{\xi^{2}}$
(4.23) | $V(\xi,z)=\frac{b}{\xi^{2}}+b_{2}(q\xi+z)$
(4.24) | $V(\xi,z)=\frac{b_{1}}{\xi^{2/3}}+\frac{b_{2}(q\xi+z)}{\xi^{4/3}}$
(4.25) | $V(\xi,z)=b_{1}\xi+\frac{b_{2}}{(q\xi+z)^{2}}$
(4.26) | $V(\xi,z)=\frac{b}{(q\xi+z)^{2}}+F(\xi)$
(4.27) | $V(\xi,z)=\frac{bz}{\xi^{3}}+F(\xi)$
(4.28) | $V(\xi,z)=\frac{bz}{\xi^{3/2}}+F(\xi)$
(4.29) | $V(\xi,z)=bz\xi^{a}+F(\xi)$
(4.30) | $V(\xi,z)=b\xi+F(q\xi+z)$
(4.31) | $V(\xi,z)=b\xi z+F(\xi)$
(4.32) | $V(\xi,z)=bz+F(\xi)$
(4.33) | $V(\xi,z)=\frac{b}{z^{2}}+F(\xi)$
(4.34) | $V(\xi,z)=\frac{b_{1}\xi^{2}+b_{2}z(\mu_{1}z+\mu_{2}\xi)}{\xi^{2}(2\mu_{1}z+\mu_{2}\xi)^{2}}+F(\xi)$
(4.37) | $V(r,z)=F\big{(}r^{2}+z^{2}\big{)}+\frac{bz}{r(r^{2}+z^{2})}$
(4.38) | $V(r,z)=F\big{(}r^{2}+z^{2}\big{)}+\frac{b}{z^{2}}$
(4.39) | $V(r,x)=br^{2}+F(z)$
(4.40) | $V(r,z)=\frac{b}{r}+F(z)$
(4.41) | $V(r,z)=\frac{c_{1}(4b_{5}r^{2}+b_{5}z^{2}+2b_{3}z)}{4a_{1}b_{5}}-\frac{c_{2}}{(2b_{5}(a_{1}+2k_{1})(b_{5}z+b_{3})^{2}}$
(4.45) | $V(\rho,\theta)=b\rho^{2}+F\big{(}\rho{\rm e}^{\theta}\big{)}$
(4.46) | $V(\rho,\theta)=\frac{b{\rm e}^{-2\theta}}{\rho^{2}}+F(\rho)$
(4.47) | $V(\rho,\theta)=\frac{b{\rm e}^{-3\theta}}{\rho}+F\big{(}\rho{\rm e}^{\theta}\big{)}$

##  5 The complex 3-sphere

We choose a standardized Cartesian-like coordinate system $\\{x,y,z\\}$ on the
3-sphere such that the Hamiltonian is

${\mathcal{H}}=\left(1+\frac{r^{2}}{4}\right)^{2}\big{(}p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\big{)}+V,$

where $r^{2}=x^{2}+y^{2}+z^{2}$. These coordinates can be related to the
standard realization of the sphere via complex coordinates ${\bf
s}=(s_{1},s_{2},s_{3},s_{4})$ such that $\sum_{j=1}^{4}s_{j}^{2}=1$ and ${\rm
d}s^{2}=\sum_{j}{\rm d}s_{j}^{2}$ via

$s_{1}=\frac{4x}{4+r^{2}},\qquad s_{2}=\frac{4y}{4+r^{2}},\qquad
s_{3}=\frac{4z}{4+r^{2}},\qquad s_{4}=\frac{4-r^{2}}{4+r^{2}}$

with inverse $x=2s_{1}/(1+s_{4})$, $y=2s_{2}/(1+s_{4})$, $z=2s_{3}/(1+s_{4})$.
A basis of Killing vectors for the zero potential system is $J_{h}$, $K_{h}$,
$h=1,2,3$ where

$\displaystyle J_{23}=yp_{z}-zp_{y},\qquad J_{31}=zp_{x}-xp_{z},\qquad
J_{12}=xp_{y}-yp_{x},$ $\displaystyle
K_{1}=\left(1+\frac{x^{2}-y^{2}-z^{2}}{4}\right)p_{x}+\frac{xy}{2}p_{y}+\frac{xz}{2}p_{z},$
$\displaystyle
K_{2}=\left(1+\frac{y^{2}-x^{2}-z^{2}}{4}\right)p_{y}+\frac{xy}{2}p_{x}+\frac{yz}{2}p_{z},$
$\displaystyle
K_{3}=\left(1+\frac{z^{2}-x^{2}-y^{2}}{4}\right)p_{z}+\frac{xz}{2}p_{x}+\frac{yz}{2}p_{y}.$

The relation between this basis and the standard basis of rotation generators
on the sphere $I_{\ell m}=s_{\ell}p_{m}-s_{m}p_{\ell}=-I_{m\ell}$ is

$J_{23}=I_{23},\qquad J_{13}=I_{13},\qquad J_{12}=I_{12},\qquad
K_{1}=I_{41},\qquad K_{2}=I_{42},\qquad K_{3}=I_{43}.$

To solve the classification problem for Hamiltonians in the class
${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$ on the complex 3-sphere we can use
methods analogous to those for Euclidean space. From result (3.4) applied to
the 3-sphere we see that, up to conjugacy, there are just 2 cases to consider:
${\mathcal{J}}=J_{12}$ and ${\mathcal{J}}=J_{12}+{\rm i}J_{23}$. The details
are complicated but we find that there are no class
${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$ FLD superintegrable systems on the
complex 3-sphere in this class. To save space we do not provide the details
here. They can be found in the online paper [2].

##  6 Conclusions

This paper is part of a program to classify all 2nd order superintegrable
classical and quantum systems on 3-dimensional conformally flat complex
manifolds. We have worked out the basic structure theory for certain FLD-
superintegrable systems on these manifolds and classified all such systems on
constant curvature spaces that are in the class
${\mathcal{O}}_{{\mathcal{H}}^{0}}(4)$. There turn out to be no such systems
on the complex 3-sphere [2]. The remaining systems to classify are highly
degenerate, admitting at least 6 linearly independent symmetries and as yet we
have found no verifiable examples. For complex Euclidean space we list the
2-parameter potentials in Table 1. In most of the cases the potential depends
on at least one arbitrary function. The key to the classification is a proof
that all such systems admit a 1st order symmetry.

### Acknowledgements

We thank the referees for suggestions that improved our presentation, in
particular one referee for pointing out a gap in our original classification.
B.K.B. acknowledges support from the Göran Gustafsson Foundation. W.M. was
partially supported by a grant from the Simons Foundation (# 412351 to Willard
Miller, Jr.).

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