Koenigs Theorem and Superintegrable Liouville Metrics

Koenigs Theorem and Superintegrable Liouville
Metrics

Galliano VALENT

G. Valent

Laboratoire de Physique Mathématique de Provence,
12 Rue Fabrot, 13100 Aix-en-Provence, France<EMAIL_ADDRESS>
Received January 26, 2023, in final form July 08, 2023; Published online July
19, 2023

In a first part, we give a new proof of Koenigs theorem and, in a second part,
we determine the local form of all the superintegrable Riemannian Liouville
metrics as well as their global geometries.

Koenigs metrics; Liouville metrics; superintegrable geodesic flows; two-
dimensional manifolds

32C05; 37E35; 37E99; 37J35

##  1 Introduction

The pioneering work of Koenigs [13] which led to a finite set of metrics with
non constant curvature and with a superintegrable (SI) geodesic flow was
rediscovered and generalized in [11] and [9] at the classical and at the
quantum levels, leading to the determination of its geodesics [1, 6, 17] and
allowed impressive three dimensional generalizations thanks to ideas from
conformal geometry [7]. Needless to say, this field is closely related to
superintegrability, a flourishing domain of research [16].

The original proof given by Koenigs is not so convincing and in [11] and [9]
some complex coordinates changes are even needed to recover the real forms of
the metrics.

The quest for a proof, in a more modern approach, was first given in [4]:
after having proved that SI metrics are metrics admitting three independent
projective fields the authors proceeded to give a complete classification and
description of these metrics with arbitrary signature.

Unaware of these results and starting from Matveev and Shevchishin framework
[15] the complete list of Riemannian Koenigs metrics was derived in [17,
Theorem 3]. According to the $y$ dependence of the quadratic integrals one has

hyperbolic: $\displaystyle g=\frac{(a\cos x+b)}{\sin^{2}x}\big{(}{\rm
d}x^{2}+{\rm d}y^{2}\big{)},$ trigonometric:
$\displaystyle\begin{cases}g_{0}=\big{(}a{\rm e}^{-x}+b{\rm
e}^{-2x}\big{)}\big{(}{\rm d}x^{2}+{\rm d}y^{2}\big{)},\vspace{1mm}\\\
g_{+}=\dfrac{a\sinh x+b}{\cosh^{2}x}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\vspace{1mm}\\\ g_{-}=\dfrac{a\cosh x+b}{\sinh^{2}x}\big{(}{\rm
d}x^{2}+{\rm d}y^{2}\big{)},\end{cases}$ affine: $\displaystyle
g=\frac{\big{(}a_{2}x^{2}+2a_{1}x+a_{0}\big{)}}{(a_{2}x+a_{1})^{2}}\big{(}{\rm
d}x^{2}+{\rm d}y^{2}\big{)}.$ (1.1)

The parameters which do appear must be restricted in order to avoid a constant
Gaussian curvature and to insure that the metric is indeed of Riemannian
signature.

Koenigs theorem stemmed from the following problem: start from a generic
Hamiltonian on some surface of revolution

$H=a^{2}(x)\big{(}P_{x}^{2}+P_{y}^{2}\big{)}$

find all the possible quadratic integrals

${\mathcal{S}}=A(x,y)P_{x}^{2}+B(x,y)P_{x}P_{y}+C(x,y)P_{y}^{2}.$

Our first aim will be to give a direct proof of Koenigs theorem facing up to
the PDEs which allow for the construction of ${\mathcal{S}}$.

On the way, we were also interested in another related problem: is it possible
to find all Riemannian Liouville metrics which are SI? Some partial answers
were already found in the literature [14] and not surprisingly this problem is
also related with Koenigs metrics.

Let us observe that this problem, in the pseudo-Riemannian setting, is much
richer than in the Riemannian case. It was first considered in [5] and
generalized in [2] where all the normal forms of the pseudo-Riemannian metrics
with a quadratic integral were determined, showing three possible classes
whereas for the Riemannian metrics there is a single class.

The content of this article is the following: in Section 2, we present our
direct proof of Koenigs theorem. In Section 3, we give the setting of SI
Liouville metrics and the PDE which determines these metrics. Then in Section
4, solving this PDE reveals a finite set of metrics. In Section 5, the global
structure of these metrics is discussed and, in Section 6, we interpret our
results in terms of coupling constant metamorphosis. Three appendices conclude
the paper: Appendix A which describes a non canonical metric on
${\mathbb{H}}^{2}$, Appendix B gives a summary of the various SI Liouville
metrics found and Appendix C which relates the metrics given in [4] and the
ones given by (1.1).

##  2 A direct proof of Koenigs theorem

The setting is the following: we have for Hamiltonian

$H=\Pi_{x}^{2}+a^{2}(x)P_{y}^{2},\qquad\Pi_{x}=a(x)P_{x},$

for which $\\{H,P_{y}\\}=0$ and we are looking for an extra quadratic integral
(which should be irreducible) having the most general form

${\mathcal{S}}=A(x,y)H+B(x,y)\Pi_{x}P_{y}+C(x,y)P_{y}^{2}.$

######  Remark 2.1.

We will exclude a constant Gaussian curvature
$R=aa^{\prime\prime}-(a^{\prime})^{2}$ since this implies the reducibility of
${\mathcal{S}}$.

######  Remark 2.2.

An additive constant either in $A$ or in $C$ is irrelevant.

Let us begin with

######  Lemma 2.3.

${\mathcal{S}}$ is an integral iff $(A,B,C)$ solve the following system of
PDEs:

$\\{H,{\mathcal{S}}\\}=0\quad\Longleftrightarrow\quad\begin{array}[]{@{}clcl}(1)\colon&\partial_{x}A=0,&\qquad(2)\colon&\partial_{x}B=-a\partial_{y}A,\\\\[17.07164pt]
(3)\colon&\partial_{x}C=-\partial_{y}(aB),&\qquad(4)\colon&\partial_{y}C=\partial_{x}(aB).\end{array}$

###### Proof.

It follows from an elementary computation of Poisson brackets. Let us observe
that the functions $aB$ and $C$ are harmonic conjugate. ∎

######  Lemma 2.4.

The previous system admits a solution iff

$\displaystyle\alpha(x)\partial_{y}^{3}A(y)+\lambda(x)\partial_{y}A(y)-\partial_{y}^{2}\beta(y)-\mu(x)\beta(y)=0,$
(2.1)

where

$\lambda=\frac{(\alpha\alpha^{\prime})^{\prime\prime}}{\alpha^{\prime}},\qquad\mu=\frac{\alpha^{\prime\prime\prime}}{\alpha^{\prime}},\qquad\alpha^{\prime}=a.$

###### Proof.

Relations $(1)$ and $(2)$ are easily integrated

$A=A(y),\qquad B(x,y)=-\alpha(x)\partial_{y}A+\beta(y).$

The remaining equations (3) and (4) become

$\partial_{x}C=\alpha\alpha^{\prime}\partial_{y}^{2}A-\alpha^{\prime}\partial_{y}\beta,\qquad\partial_{y}C=-(\alpha\alpha^{\prime})^{\prime}\partial_{y}A+\alpha^{\prime\prime}\beta.$

Their integrability condition gives relation (2.1). ∎

Let us prove:

######  Proposition 2.5.

The integrability equation (2.1) is equivalent to

$B=-\alpha\partial_{y}A,\qquad\partial_{y}^{3}A+s_{0}\partial_{y}A=0,\qquad\big{(}\alpha^{2}\big{)}^{\prime\prime\prime}-s_{0}\big{(}\alpha^{2}\big{)}^{\prime}=0,\qquad
s_{0}\in\\{0,\pm 1\\}.$

###### Proof.

Let us first deduce some necessary conditions. Applying $\partial_{x}$ to
(2.1) gives

$\displaystyle\alpha^{\prime}\partial_{y}^{3}A+\lambda^{\prime}\partial_{y}A-\mu^{\prime}\beta=0.$
(2.2)

Observing that $\mu^{\prime}$ cannot vanish, otherwise the Gaussian curvature
is constant, this relation determines $\beta(y)$ according to

$\beta=\frac{\alpha^{\prime}}{\mu^{\prime}}\partial_{y}^{3}A+\frac{\lambda^{\prime}}{\mu^{\prime}}\partial_{y}A.$

Applying on it $\partial_{x}$ leads to

$\displaystyle\left(\frac{\alpha^{\prime}}{\mu^{\prime}}\right)^{\prime}\partial_{y}^{3}A+\left(\frac{\lambda^{\prime}}{\mu^{\prime}}\right)^{\prime}\partial_{y}A=0,$

from which we deduce the separation relations

$\partial_{y}^{3}A+s_{0}\partial_{y}A=0,\qquad
s_{0}\in{\mathbb{R}},\qquad\left(\frac{\lambda^{\prime}-s_{0}\alpha^{\prime}}{\mu^{\prime}}\right)^{\prime}=0.$
(2.3)

By a scaling of $y$, we can restrict $s_{0}$ to be in $\\{0,\pm 1\\}.$

The relations obtained in (2.3) are necessary. To lift them up to sufficient
conditions we must discuss backwards successively the relations (2.1) and
(2.2).

Integration of the right-hand relation in (2.3) produces

$\lambda^{\prime}-s_{0}\alpha^{\prime}=l\mu^{\prime}\quad\Longrightarrow\quad\lambda-
s_{0}\alpha=l\mu+m,\qquad(l,m)\in{\mathbb{R}}^{2}.$ (2.4)

Since $\mu^{\prime}$ cannot vanish, relation (2.2) determines $\beta$:

$\beta=l\partial_{y}A\quad\Longrightarrow\quad B=-(\alpha-l)\partial_{y}A.$

Eventually relation (2.1) gives $m=-s_{0}l$. Hence the right-hand relation in
(2.4) leads to the following ODE

$\big{[}(\alpha-l)^{2}\big{]}^{\prime\prime\prime}-s_{0}\big{[}(\alpha-l)^{2}\big{]}^{\prime}=0,$

showing that we can take $l=0$ since $a=\alpha^{\prime}=(\alpha-l)^{\prime}$.

It follows that $\beta(y)=0$ and $B=-\alpha\partial_{y}A$, ending the proof. ∎

So we conclude, in agreement with [13] and [17]:

######  Theorem 2.6 (Koenigs).

Recalling that $H=(\alpha^{\prime})^{2}\big{(}P_{x}^{2}+P_{y}^{2}\big{)}$,
according to the values of $s_{0}$ we have:

$\displaystyle s_{0}=0,\qquad$ $\displaystyle
A=k_{1}y+k_{2}\frac{y^{2}}{2},\qquad$
$\displaystyle\alpha^{2}=a_{0}+a_{1}x+a_{2}x^{2},$ $\displaystyle
s_{0}=+1,\qquad$ $\displaystyle A=k_{1}\sin y+k_{2}\cos y,\qquad$
$\displaystyle\alpha^{2}=a_{0}+a_{1}\sinh x+a_{2}\cosh x,$ $\displaystyle
s_{0}=-1,\qquad$ $\displaystyle A=k_{1}\sinh y+k_{2}\cosh y,\qquad$
$\displaystyle\alpha^{2}=a_{0}+a_{1}\sin x+a_{2}\cos x,$

while ${\mathcal{S}}$ is given by

$\displaystyle s_{0}=0,\qquad$
$\displaystyle{\mathcal{S}}=AH-\frac{\big{(}\alpha^{2}\big{)}^{\prime}}{2}(\partial_{y}A)P_{x}P_{y}+\frac{1}{2}\big{(}\alpha^{2}\partial_{y}^{2}A-\big{(}\alpha^{2}\big{)}^{\prime\prime}A\big{)}P_{y}^{2},$
$\displaystyle s_{0}=\pm 1,\qquad$
$\displaystyle{\mathcal{S}}=AH-\frac{\big{(}\alpha^{2}\big{)}^{\prime}}{2}(\partial_{y}A)P_{x}P_{y}-\frac{s_{0}}{2}\big{(}\alpha^{2}-a_{0}\big{)}AP_{y}^{2}.$
(2.5)

###### Proof.

Proposition 2.5 gives

$aB=-\frac{\big{(}\alpha^{2}\big{)}^{\prime}}{2}\partial_{y}A$

and $C$ (see Lemma 2.3) has to be determined from

$\partial_{x}C=\frac{\alpha^{2}}{2}\partial^{2}_{y}A,\qquad\partial_{y}C=-\frac{1}{2}\big{(}\alpha^{2}\big{)}^{\prime\prime}\partial_{y}A.$

Elementary computations give $C$ in each case. ∎

######  Remark 2.7.

Let us notice the striking balance between trigonometric and hyperbolic lines
in the differential equations obtained in Proposition 2.5:

$\partial_{y}^{3}A+s_{0}\partial_{y}A=0,\qquad\big{(}\alpha^{2}\big{)}^{\prime\prime\prime}-s_{0}\big{(}\alpha^{2}\big{)}^{\prime}=0.$

######  Remark 2.8.

One can check that $aB$ and $C$ are indeed harmonic conjugate.

######  Remark 2.9.

From Theorem 2.6, we observe that we can define
${\mathcal{S}}=k_{1}{\mathcal{S}}_{1}+k_{2}{\mathcal{S}}_{2}$, where these
integrals are easily deduced from (2.5). It follows that the linear span of
the quadratic integrals is indeed four dimensional, with

$H,\quad P_{y}^{2},\quad{\mathcal{S}}_{1},\quad{\mathcal{S}}_{2}.$

######  Remark 2.10.

However a well known theorem (for a proof see [18, p. 461]) states that in
dimension 2 there may be at most three algebraically independent integrals
including the Hamiltonian. Indeed there is a quadratic relation between
${\mathcal{S}}_{1}$ and ${\mathcal{S}}_{2}$ given in [11] and [9] which does
reduce the number of algebraically independent integrals to three.

######  Remark 2.11.

The global properties of these SI metrics, their geodesics and their quantum
structure were discussed in [1] and in [17].

For further reference, let us mention three metrics of interest. The first one
is

$s_{0}=0,\qquad\alpha=2\sqrt{x}\colon\ g=x\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)}.$ (2.6)

A second one is

$s_{0}=0,\qquad\alpha=\sqrt{s+x^{2}}\colon\
g=\big{(}s+x^{2}\big{)}\frac{\big{(}{\rm d}x^{2}+{\rm d}y^{2}\big{)}}{x^{2}}.$
(2.7)

The last one is

$s_{0}=-1,\qquad\alpha=\frac{4}{s}\big{(}\sqrt{1+s{\rm e}^{x}}-1\big{)}\colon\
g=\big{(}s{\rm e}^{-x}+{\rm e}^{-2x}\big{)}\frac{\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)}}{4}.$

The change of coordinates $X={\rm e}^{-x/2}\cos(y/2)$ and $Y={\rm
e}^{-x/2}\sin(y/2)$ leads to

$g=\big{(}s+X^{2}+Y^{2}\big{)}\big{(}{\rm d}X^{2}+{\rm d}Y^{2}\big{)}.$ (2.8)

Let us consider now the second item: SI Liouville metrics.

##  3 The setting

The metric on a Liouville surface is defined locally as

$g=(f(x)+g(y))\big{(}{\rm d}x^{2}+{\rm d}y^{2}\big{)},$

and its geodesic flow, generated by the Hamiltonian

$H=\frac{P_{x}^{2}+P_{y}^{2}}{f(x)+g(y)},$

does exhibit the quadratic integral

$Q=\frac{g(y)P_{x}^{2}-f(x)P_{y}^{2}}{f(x)+g(y)}=P_{x}^{2}-f(x)H=g(y)H-P_{y}^{2},\qquad\\{H,Q\\}=0.$

To reach superintegrability, we need an extra integral which can be written
without loss of generality

${\mathcal{S}}=A(x,y)H+B(x,y)P_{x}P_{y}+C(x,y)P_{y}^{2}.$

######  Remark 3.1.

We will exclude from our analysis the metrics $g$ of constant curvature since
in this case all the integrals become reducible.

######  Remark 3.2.

Constants either in $A$ or in $C$ are irrelevant.

###  3.1 PDE system and a master equation

Let us start with

######  Proposition 3.3.

The conservation relation $\\{H,{\mathcal{S}}\\}=0$ is equivalent to the
following set of PDE:

  * $(1)$

$\partial_{x}A+\frac{\dot{g}}{2}B=0,$

  * $(2)$

$\partial_{y}A+(f+g)\partial_{x}B+\frac{f^{\prime}}{2}B+\dot{g}C=0,$

  * $(3)$

$\partial_{x}A+(f+g)(\partial_{y}B+\partial_{x}C)+\frac{\dot{g}}{2}B=0,$

  * $(4)$

$\partial_{y}A+(f+g)\partial_{y}C+\frac{f^{\prime}}{2}B+\dot{g}C=0.$

###### Proof.

Elementary calculation. ∎

###  3.2 Discussion

Computing relations (1)–(3) and (2)–(4) gives

$\partial_{x}C=-\partial_{y}B,\qquad\partial_{y}C=\partial_{x}B,$ (3.1)

implying that these two functions are harmonic conjugate while $A$ is
determined by

$\partial_{x}A=-\frac{\dot{g}}{2}B,\qquad\partial_{y}A=-f\partial_{y}C-\frac{f^{\prime}}{2}B-\partial_{y}(gC).$

So, defining $\ {\mathcal{A}}=A+gC,\ $ we end up with

$\begin{cases}\partial_{x}{\mathcal{A}}=-g\partial_{y}B-\dfrac{\dot{g}}{2}B,\vspace{1mm}\\\
\partial_{y}{\mathcal{A}}=-f\partial_{x}B-\dfrac{f^{\prime}}{2}B,\end{cases}\quad\Longrightarrow\quad{\mathcal{S}}={\mathcal{A}}H+BP_{x}P_{y}-CQ.$
(3.2)

The integrability condition of this system is therefore

$Bf^{\prime\prime}+3B^{\prime}f^{\prime}+2B^{\prime\prime}f=B\ddot{g}+3\dot{B}\dot{g}+2\ddot{B}g.$
(3.3)

This is the master equation to be solved.

Having observed in (3.1) that $B$ and $C$ are harmonic conjugate, let us
define

$(x+{\rm i}y)^{n}={\mathcal{R}}_{n}(x,y)+{\rm i}{\mathcal{I}}_{n}(x,y).$

Since equation (3.3) is linear, we can consider separately the following four
cases:

$\displaystyle{\rm I}\colon\hphantom{{\rm II}}\
B={\mathcal{R}}_{2n}(x,y),\qquad$ $\displaystyle{\rm II}\colon\ \
B={\mathcal{I}}_{2n+1}(x,y),$ $\displaystyle{\rm III}\colon\
B={\mathcal{R}}_{2n+1}(x,y),\qquad$ $\displaystyle{\rm IV}\colon\
B={\mathcal{I}}_{2n}(x,y).$

######  Remark 3.4.

Despite the usefulness of ${\mathcal{A}}$, and since $Q=g(y)H-P_{y}^{2}$, we
will write the integral ${\mathcal{S}}$ in its initial form

${\mathcal{S}}=A(x,y)H+B(x,y)P_{x}P_{y}+C(x,y)P_{y}^{2}.$

##  4 Discussion of the four cases

###  4.1 Case I

Let us prove

######  Proposition 4.1.

For $B$ given by

$B={\mathcal{R}}_{2n}(x,y)=\sum_{l=0}^{n}{2n\choose
2l}(-1)^{n-l}x^{2l}y^{2(n-l)},\qquad n\geq 0,$

the integrability constraint (3.3) never gives a Liouville metric except for
$n=0$ for which we have

$f(x)=\mu x^{2}+2a_{1}x+a_{0},\qquad g(y)=\mu y^{2}+2b_{1}y+b_{0}$ (4.1)

and

${\mathcal{A}}=-\mu xy-
b_{1}x-a_{1}y,\qquad{\mathcal{S}}={\mathcal{A}}H+P_{x}P_{y}.$ (4.2)

###### Proof.

For $n=0$, we have $B=1$ and $C=0$. Relation (3.3) becomes
$f^{\prime\prime}=\ddot{g}=2\mu$ with $\mu\in{\mathbb{R}}$ which gives (4.1).
Integrating (3.2) gives (4.2).

Let us consider $n\geq 1$. We will call $S_{1}$ (resp. $S_{2}$) the left-hand
side (resp. the right-hand side) of the relation (3.3). Positing
$C^{n}_{l}=(-1)^{n-l}{2n\choose 2l}$ and $\delta_{x}=x\partial_{x}$ and
$\delta_{y}=y\partial_{y}$, we have

$S_{1}=\sum_{l=0}^{n}C^{n}_{l}x^{2(l-1)}F_{l}(x)y^{2(n-l)},\qquad
F_{l}(x)=(\delta_{x}+2l-1)(\delta_{x}+4l)f(x),$

while

$S_{2}=\sum_{l=0}^{n}C^{n}_{l}x^{2l}y^{2(n-l-1)}G_{n-l}(y),\qquad
G_{\nu}(y)=(\delta_{y}+2\nu-1)(\delta_{y}+4\nu)g(y).$

The leading terms are

$S_{1}=C^{n}_{0}f^{\prime\prime}(x)y^{2n}+\cdots,\qquad
S_{2}=C^{n}_{n}x^{2n}\ddot{g}(y)+\cdots.$

Acting with $\partial_{x}^{2n}\partial_{y}^{2n}$ on the relation $S_{1}=S_{2}$
gives the separation relation

$\partial_{x}^{2(n+1)}f(x)=(-1)^{n}\partial_{y}^{2(n+1)}g(y),$

which leads to polynomials for $f$ and $g$

$f(x)=\sum_{k=0}^{2(n+1)}a_{k}x^{k},\qquad
g(y)=\sum_{k=0}^{2(n+1)}b_{k}y^{k},\qquad b_{2n+2}=(-1)^{n}a_{2n+2}.$

Now we can compute

$\displaystyle
S_{1}=\sum_{k=0}^{2(n+1)}\sum_{l=0}^{n}C^{n}_{l}a_{k}(k+2l-1)(k+4l)x^{k+2(l-1)}y^{2(n-l)},$
$\displaystyle
S_{2}=\sum_{k=0}^{2(n+1)}\sum_{l=0}^{n}C^{n}_{l}b_{k}(k+2(n-l)-1)(k+4(n-l))x^{2l}y^{k+2(n-l)}.$

The relation $S_{1}=S_{2}$ implies that $k$ can take only even values, giving

$\displaystyle\widetilde{S}_{1}=\frac{S_{1}}{2}=\sum_{k=0}^{n+1}\sum_{l=0}^{n}C^{n}_{l}a_{2k}(2k+2l-1)(k+2l)x^{2(k+l-1)}y^{2(n-l)},$
$\displaystyle\widetilde{S}_{2}=\frac{S_{2}}{2}=\sum_{k=0}^{n+1}\sum_{l=0}^{n}C^{n}_{l}b_{2k}(2(k+n-l)-1)(k+2(n-l))x^{2l}y^{2(k+n-l-1)}.$

The first sum can be written

$\widetilde{S}_{1}=a_{0}A_{0}+\sum_{k=1}^{n+1}\sum_{l=0}^{n}C^{n}_{l}a_{2k}(2k+2l-1)(k+2l)x^{2(k+l-1)}y^{2(n-l)},$
(4.3)

where

$A_{0}=\sum_{l=0}^{n-1}C^{n}_{l+1}(2l+1)(2l+2)x^{2l}y^{2(n-l-1)}.$

In (4.3), let us operate the change of summation index $l\to L=l+k-1$ giving

$\sum_{k=1}^{n+1}\sum_{L=k-1}^{n+k-1}C^{n}_{L-k+1}a_{2k}(2L+1)(2L-k+2)x^{2L}y^{2(k+n-L-1)}.$
(4.4)

Exchanging the summations in (4.4) leads to

$\displaystyle\widetilde{S}_{1}=a_{0}A_{0}$
$\displaystyle+\sum_{L=0}^{n}(2L+1)x^{2L}\sum_{k=1}^{L+1}C^{n}_{L-k+1}(2L-k+2)a_{2k}y^{2(k+n-L-1)}$
$\displaystyle{}+\sum_{L=n+1}^{2n}(2L+1)x^{2L}\sum_{k=L-n+1}^{n+1}C^{n}_{L-k+1}(2L-k+2)a_{2k}y^{2(k+n-L-1)}.$

Comparing with $\widetilde{S}_{2}$ shows that the last sum must vanish because
$L\geq n+1$ and this entails $a_{2s}=0$ for $s=2,3,\ldots,n+1$, reducing
$\widetilde{S}_{1}$ to the simple form

$\widetilde{S}_{1}=a_{0}A_{0}+a_{2}A_{2},\qquad
A_{2}=\sum_{l=0}^{n}C^{n}_{l}(2l+1)^{2}x^{2l}y^{2(n-l)}.$

Let us consider $\widetilde{S}_{2}$, which can be written

$\widetilde{S}_{2}=b_{0}B_{0}+b_{2}B_{2}+\sum_{l=0}^{n}C^{n}_{l}x^{2l}\sum_{k=2}^{n+1}b_{2k}(2(k+n-l)-1)(k+2(n-l))y^{2(n-l+k-1)},$

where

$\displaystyle
B_{0}=\sum_{l=0}^{n-1}C^{n}_{l}(2(n-l)-1)2(n-l)x^{2l}y^{2(n-l-1)},\qquad
B_{2}=\sum_{l=0}^{n}C^{n}_{l}(2(n-l)+1)^{2}x^{2l}y^{2(n-l)}.$

Comparing $\widetilde{S}_{2}$ with $\widetilde{S}_{1}$ implies $b_{2s}=0$ for
$s=2,3,\ldots,n+1$ and we are left with

$\widetilde{S}_{2}=b_{0}B_{0}+b_{2}B_{2}.$

The relation

$(2l+1)(2l+2)C^{n}_{l+1}=-(2(n-l)-1)(2n-2l)C^{n}_{l}\quad\Longrightarrow\quad
B_{0}=-A_{0},$

so we end up with

$\widetilde{S}_{1}-\widetilde{S}_{2}=(a_{0}+b_{0})A_{0}+a_{2}A_{2}-b_{2}B_{2}=0,$

or more explicitly

$\displaystyle\widetilde{S}_{1}-\widetilde{S}_{2}=(a_{0}+b_{0})\big{(}2C^{n}_{1}y^{2(n-1)}+\cdots+(2n-1)2nC^{n}_{n}x^{2(n-1)}\big{)}$
$\displaystyle\phantom{\widetilde{S}_{1}-\widetilde{S}_{2}=}{}+a_{2}\big{(}C^{n}_{0}y^{2n}+\cdots+(2n+1)^{2}C^{n}_{n}x^{2n}\big{)}-b_{2}\big{(}(2n+1)^{2}C^{n}_{0}y^{2n}+\cdots+C^{n}_{n}x^{2n}\big{)}=0.$

It follows that $a_{0}+b_{0}=0$ and $a_{2}=b_{2}=0$, hence $f(x)+g(y)\equiv
0$. ∎

Let us consider the second case.

###  4.2 Case II

Let us prove

######  Proposition 4.2.

For $B$ given by

$B={\mathcal{I}}_{2n+1}(x,y)=\sum_{l=0}^{n}{2n+1\choose
2l}(-1)^{n-l}x^{2l}y^{2(n-l)+1},\qquad n\geq 0,$

the integrability constraint (3.3) never gives a Liouville metric except if
$n=0$ for which we have

$f(x)=\mu x^{2}+2a_{1}x+a_{0},\qquad
g(y)=\frac{\mu}{4}y^{2}+\frac{b_{-2}}{y^{2}}+b_{0},$ (4.5)

and

${\mathcal{A}}=-\frac{\mu}{2}xy^{2}-\frac{a_{1}}{2}y^{2}-b_{0}x\quad\Longrightarrow\quad{\mathcal{S}}={\mathcal{A}}H+yP_{x}P_{y}+xQ.$
(4.6)

###### Proof.

For $n=0$, we have $B=y$ and $C=-x$. The relation (3.3) becomes

$f^{\prime\prime}=\ddot{g}+\frac{3}{y}\dot{g}=2\mu,\qquad\mu\in{\mathbb{R}},$

which implies (4.5). Integrating (3.2) gives (4.6).

Let us consider $n\geq 1$. Positing this time
$C^{n}_{l}=(-1)^{n-l}{2n+1\choose 2l},$ we have

$S_{1}=\sum_{l=0}^{n}C^{n}_{l}x^{2(l-1)}F_{l}(x)y^{2(n-l)+1},\qquad
F_{l}(x)=(\delta_{x}+2l-1)(\delta_{x}+4l)f(x),$

and

$S_{2}=\sum_{l=0}^{n-1}C^{n}_{l}x^{2l}y^{2(n-l)-1}G_{n-l}(y),\qquad
G_{\nu}(y)=(\delta_{y}+2\nu)(\delta_{y}+4\nu+2)g(y).$

The leading terms are

$\frac{S_{1}}{y}=C^{n}_{0}f^{\prime\prime}(x)y^{2n}+\cdots,\qquad\frac{S_{2}}{y}=C^{n}_{n}x^{2n}\left(\ddot{g}+\frac{3}{y}\dot{g}\right)+\cdots,$

so acting with $\partial_{x}^{2n}\partial_{y}^{2n}$ gives the separation
relation

$\partial_{x}^{2(n+1)}f(x)=(-1)^{n}(2n+1)\partial_{y}^{2n}\left(\ddot{g}+\frac{3}{y}\dot{g}\right),$

which are easily integrated into

$f(x)=\sum_{k=0}^{2(n+1)}a_{k}x^{k},\qquad g(y)=\sum_{k=-2,k\neq\pm
1}^{2(n+1)}b_{k}y^{k},\qquad b_{2(n+1)}=\frac{(-1)^{n}}{2(n+2)}a_{2(n+1)}.$

Similarly to Case I, we obtain

$\displaystyle\begin{split}&\widetilde{S}_{1}=\sum_{l=0}^{n}\sum_{k=0}^{n+1}C^{n}_{l}(2(k+l)-1)(k+2l)a_{2k}x^{2(k+l-1)}y^{2(n-l)+1},\\\
&\widetilde{S}_{2}=\sum_{l=0}^{n}\sum_{k=-1}^{n+1}C^{n}_{l}2(k+n-l)(k+2(n-l)+1)b_{2k}x^{2l}y^{2(k+n-l)-1}.\end{split}$

Following the same lines as in the proof of Proposition 2.5, we get

$\displaystyle\widetilde{S}_{1}=a_{0}A_{0}+a_{2}A_{2},$ $\displaystyle
A_{0}=\sum_{l=0}^{n-1}C^{n}_{l+1}(2l+1)(2l+2)x^{2l}y^{2(n-l)-1},\qquad
A_{2}=\sum_{l=0}^{n}C^{n}_{l}(2l+1)^{2}x^{2l}y^{2(n-l)+1}.$

Let us consider now $\widetilde{S}_{2}$. It can be expanded

$\displaystyle\widetilde{S}_{2}=b_{-2}B_{-2}+b_{0}B_{0}+b_{2}B_{2}+\sum_{l=0}^{n}\sum_{k=2}^{n+1}C^{n}_{l}2(k+n-l)(k+2(n-l)+1)b_{2k}x^{2l}y^{2(k+n-l)-1},$

where

$B_{-2}=\sum_{l=0}^{n}C^{n}_{l}2(n-l-1)2(n-l)x^{2l}y^{2(n-l)-3}.$

Comparing with $\widetilde{S}_{1}$ implies

$b_{-2}=0,\qquad b_{4}=b_{6}=\cdots=b_{2(n+1)}=0$

and we are left with

$\displaystyle\widetilde{S}_{2}=b_{0}B_{0}+b_{2}B_{2},$ $\displaystyle
B_{0}=\sum_{l=0}^{n-1}C^{n}_{l}2(n-l)(2(n-l)+1)x^{2l}y^{2(n-l)-1},$
$\displaystyle B_{2}=\sum_{l=0}^{n}C^{n}_{l}[2(n-l+1)]^{2}x^{2l}y^{2(n-l)+1}.$

Here also we have $B_{0}=-A_{0}$ and the relation
$\widetilde{S}_{1}=\widetilde{S}_{2}$ reduces to

$(a_{0}+b_{0})A_{0}+a_{2}A_{2}-b_{2}B_{2}=0.$

Since we have

$\displaystyle A_{0}$
$\displaystyle=2C^{n}_{1}y^{2n-1}+\cdots+(2n-1)2nC^{n}_{n}x^{2(n-1)}y,$
$\displaystyle A_{2}$
$\displaystyle=C^{n}_{0}y^{2n+1}+\cdots+(2n+1)^{2}C^{n}_{n}x^{2n}y,$
$\displaystyle B_{2}$
$\displaystyle=4(n+1)^{2}C^{n}_{0}y^{2n+1}+\cdots+4C^{n}_{n}x^{2n}y,$

it follows that for $n\geq 1$ we have $a_{0}+b_{0}=0$ and $a_{2}=b_{2}=0$
hence $f(x)+g(y)\equiv 0$. ∎

Let us consider the third case.

###  4.3 Case III

Let us prove

######  Proposition 4.3.

For $B$ given by

$B={\mathcal{R}}_{2n+1}(x,y)=\sum_{l=0}^{n}(-1)^{n-l}{2n+1\choose
2l+1}x^{2l+1}y^{2(n-l)},\qquad n\geq 0,$

the integrability constraint (3.3) never gives a Liouville metric except for
$n=0$ for which we have

$f(x)=\frac{\mu}{4}x^{2}+\frac{a_{-2}}{x^{2}}+a_{0},\qquad g(y)=\mu
y^{2}+2b_{1}y+b_{0},$ (4.7)

and

${\mathcal{A}}=-\frac{\mu}{2}x^{2}y-\frac{b_{1}}{2}x^{2}-a_{0}x\quad\Longrightarrow\quad{\mathcal{S}}={\mathcal{A}}H+xP_{x}P_{y}-yQ.$
(4.8)

###### Proof.

For $n=0$, we have $B=x$ and $C=y$. The relation (3.3) becomes

$f^{\prime\prime}+\frac{3}{x}f^{\prime}=\ddot{g}=2\mu,$

which implies (4.7). Integrating (3.2) gives (4.8).

Let us posit $C^{n}_{l}=(-1)^{n-l}{2n+1\choose 2l+1}$. This time we have

$S_{1}=\sum_{l=0}^{n}C^{n}_{l}x^{2l-1}F_{l}(x)y^{2(n-l)},\qquad
F_{l}(x)=(\delta_{x}+2l)(\delta_{x}+4l+2)f(x),$

and

$S_{2}=\sum_{l=0}^{n}C^{n}_{l}x^{2l+1}y^{2(n-l-1)}G_{n-l}(y),\qquad
G_{\nu}(y)=(\delta_{y}+2\nu-1)(\delta_{y}+4\nu)g(y).$

The leading terms are

$\frac{S_{1}}{x}=C^{n}_{0}\left(f^{\prime\prime}+\frac{3}{x}f^{\prime}\right)y^{2n}+\cdots,\qquad\frac{S_{2}}{x}=C^{n}_{n}x^{2n}\ddot{g}+\cdots,$

so acting with $\partial_{x}^{2n}\partial_{y}^{2n}$ we get the separation
relation

$\partial_{x}^{2n}\left(f^{\prime\prime}+\frac{3}{x}f^{\prime}\right)=\frac{(-1)^{n}}{2n+1}\partial_{y}^{2(n+1)}g,$

which implies

$f(x)=\sum_{k-2,k\neq\pm 1}^{2(n+1)}a_{k}x^{k},\qquad
g(y)=\sum_{k=0}^{2(n+1)}b_{k}y^{k},\qquad
b_{2(n+1)}=(-1)^{n}2(n+2)a_{2(n+1)}.$

As in the previous cases, we get

$\displaystyle\widetilde{S}_{1}$
$\displaystyle=\sum_{l=0}^{n}C^{n}_{l}y^{2(n-l)}\sum_{k=-1}^{n+1}2(k+l)(k+2l+1)a_{2k}x^{2(k+l)-1},$
$\displaystyle\widetilde{S}_{2}$
$\displaystyle=\sum_{l=0}^{n}C^{n}_{l}x^{2l+1}\sum_{k=0}^{n+1}(2(k+n-l)-1)(k+2(n-l))b_{2k}y^{2(k+n-l-1)}.$

By an analysis similar to the proof of Proposition 3.3, we have

$\widetilde{S}_{1}=a_{-2}A_{-2}+a_{0}A_{0}+a_{2}A_{2},$

with

$n=1\colon\ A_{-2}=0,\qquad n\geq 2\colon\
A_{-2}=2\sum_{l=0}^{n-2}C^{n}_{l+2}(2l+2)(2l+4)x^{2l+1}y^{2(n-l-2)}$

and

$A_{0}=2\sum_{l=0}^{n-1}C^{n}_{l+1}(2l+2)(2l+3)x^{2l+1}y^{2(n-l)-1},\qquad
A_{2}=2\sum_{l=0}^{n}C^{n}_{l}(2l+2)^{2}x^{2l+1}y^{2(n-l)}.$

Since the powers of $y$ appearing in $A_{-2}$ never appear in
$\widetilde{S}_{2}$ it follows that for $n\geq 2$ we have ${a_{-2}=0}$ and
$\widetilde{S}_{1}=a_{0}A_{0}+a_{2}A_{2}.$

By an argument similar to the proof of Proposition 2.5, we have

$\widetilde{S}_{2}=b_{0}B_{0}+b_{2}B_{2},$

where

$\displaystyle
B_{0}=\sum_{l=0}^{n-1}C^{n}_{l}(2(n-l)-1)2(n-l)x^{2l+1}y^{2(n-l-1)},$
$\displaystyle B_{2}=\sum_{l=0}^{n}C^{n}_{l}(2(n-l)+1)^{2}x^{2l+1}y^{2(n-l)}.$

Here also the relation $B_{0}=-A_{0}$ holds. So we have

$\widetilde{S}_{1}-\widetilde{S}_{2}=0\quad\Longleftrightarrow\quad(a_{0}+b_{0})A_{0}+a_{2}A_{2}-b_{2}B_{2}=0.$

The relations

$\displaystyle A_{0}$
$\displaystyle=12C^{n}_{1}xy^{2n-1}+\cdots+4n(2n+1)C^{n}_{n}x^{2n-1}y,$
$\displaystyle A_{2}$
$\displaystyle=8C^{n}_{0}xy^{2n}+\cdots+4(n+1)^{2}C^{n}_{n}x^{2n+1},$
$\displaystyle B_{2}$
$\displaystyle=(2n+1)^{2}C^{n}_{0}xy^{2n}+\cdots+C^{n}_{n}x^{2n+1}$

imply, for $n\geq 1$, that $a_{0}+b_{0}=0$ and $a_{2}=b_{2}=0$ hence
$f(x)+g(y)\equiv 0$. ∎

######  Remark 4.4.

One should observe that Case III is not really different from Case II since,
if we define

$f(x)=\mu x^{2}+2a_{1}x+a_{0},\qquad
g(y)=\frac{\mu}{4}y^{2}+\frac{b_{-2}}{y^{2}}+b_{0},$

the Hamiltonians are respectively

$H_{\rm II}=\frac{P_{x}^{2}+P_{y}^{2}}{f(x)+g(y)}\qquad\mbox{and}\qquad H_{\rm
III}=\frac{P_{x}^{2}+P_{y}^{2}}{g(x)+f(y)},$

and therefore they cannot be considered as different since they are related by
the substitution $x\leftrightarrow y$.

Let us consider the last case.

###  4.4 Case IV

Let us prove

######  Proposition 4.5.

For $B$ given by

$B={\mathcal{I}}_{2n}(x,y)=\sum_{l=0}^{n-1}(-1)^{n-l}{2n\choose
2l+1}x^{2l+1}y^{2(n-l)-1},\qquad n\geq 1,$

the integrability constraint (3.3) does not give a Liouville metric except for
$n=1$ and $n=2$. For $n=1$, we have

$f(x)=\mu x^{2}+\frac{a_{-2}}{x^{2}}+a_{0},\qquad g(y)=\mu
y^{2}+\frac{b_{-2}}{y^{2}}+b_{0},$ (4.9)

and

${\mathcal{A}}=-\mu
x^{2}y^{2}-\frac{b_{0}}{2}x^{2}-\frac{a_{0}}{2}y^{2},\qquad{\mathcal{S}}={\mathcal{A}}H+xyP_{x}P_{y}+\frac{1}{2}\big{(}x^{2}-y^{2}\big{)}Q,$
(4.10)

while for $n=2$ we have

$f(x)=\frac{a}{x^{2}},\qquad g(y)=\frac{b}{y^{2}},$

and

${\mathcal{S}}=4\big{(}ay^{2}-bx^{2}\big{)}H-4xy\big{(}x^{2}-y^{2}\big{)}P_{x}P_{y}-\big{(}x^{4}-6x^{2}y^{2}+y^{4}\big{)}Q.$

###### Proof.

For $n=1$, let us take $B=xy$ and $C=-\frac{1}{2}\big{(}x^{2}-y^{2}\big{)}$,
in which case the relation (3.3) becomes

$f^{\prime\prime}+\frac{3}{x}f^{\prime}=\ddot{g}+\frac{3}{y}\dot{g}=8\mu,$

leading to (4.9). Integrating (3.2) gives (4.10). Let us consider now $n\geq
2$. Positing this time $C^{n}_{l}=(-1)^{n-l}{2n\choose 2l+1}$, we have

$S_{1}=\sum_{l=0}^{n-1}C^{n}_{l}x^{2l-1}F_{l}(x)y^{2(n-l)-1},\qquad
F_{l}(x)=(\delta_{x}+2l)(\delta_{x}+4l+2)f(x),$

while

$S_{2}=\sum_{l=0}^{n-1}C^{n}_{l}x^{2l+1}y^{2(n-l-1)-1}G_{n-l}(y),\qquad
G_{\nu}(y)=(\delta_{y}+2(\nu-1))(\delta_{y}+4\nu-2)g(y).$

The higher order terms are given by

$S_{1}=C^{n}_{0}(xf^{\prime\prime}+3f^{\prime})y^{2n-1}+\cdots,\qquad
S_{2}=C^{n}_{n-1}x^{2n-1}(y\ddot{g}+3\dot{g})+\cdots.$

So, acting with $\partial_{x}^{2n-1}\partial_{y}^{2n-1}$ on the relation
$S_{1}=S_{2}$, gives the separation relation

$\partial_{x}^{2n-1}(xf^{\prime\prime}+3f^{\prime})=(-1)^{n-1}\partial_{y}^{2n-1}(y\ddot{g}+3\dot{g}),$

which implies

$f(x)=\sum_{k=-2,k\neq\pm 1}^{2n}a_{k}x^{k},\qquad g(y)=\sum_{k=-2,k\neq\pm
1}^{2n}b_{k}y^{k},\qquad b_{2n}=(-1)^{n-1}a_{2n}.$

It follows that

$\displaystyle\widetilde{S}_{1}$
$\displaystyle=\sum_{l=0}^{n-1}C^{n}_{l}y^{2(n-l)-1}\sum_{k=-1}^{n}2(k+l)(k+2l+1)a_{2k}x^{2(k+l)-1},$
$\displaystyle\widetilde{S}_{2}$
$\displaystyle=\sum_{l=0}^{n-1}C^{n}_{l}x^{2l+1}\sum_{k=-1}^{n}2(k+n-l-1)(k+2(n-l)-1)b_{2k}y^{2(k+n-l-1)-1}.$

As in the previous propositions, one can prove that

$\widetilde{S}_{1}=a_{-2}A_{-2}+a_{0}A_{0}+a_{2}A_{2},$ (4.11)

where

$n=2\colon\ A_{-2}=0,\qquad n\geq 3\colon\
A_{-2}=\sum_{l=0}^{n-3}C^{n}_{l+2}(2l+2)(2l+4)x^{2l+1}y^{2(n-l-2)-1},$

and

$\displaystyle
A_{0}=\sum_{l=0}^{n-2}C^{n}_{l+1}(2l+2)(2l+3)x^{2l+1}y^{2(n-l-1)-1},$
$\displaystyle A_{2}=\sum_{l=0}^{n-1}C^{n}_{l}(2l+2)^{2}x^{2l+1}y^{2(n-l)-1}.$

Similarly, one can show that $\widetilde{S}_{2}$ can be written

$\widetilde{S}_{2}=b_{2}B_{-2}+b_{0}B_{0}+b_{2}B_{2},$

where

$n=2\colon\ B_{-2}=0,\qquad n\geq 3\colon\
B_{-2}=\sum_{l=0}^{n-3}C^{n}_{l}2(n-l-2)2(n-l-1)x^{2l+1}y^{2(n-l-2)-1},$

and

$\displaystyle
B_{0}=\sum_{l=0}^{n-2}C^{n}_{l}2(n-l-1)(2(n-l)-1)x^{2l+1}y^{2(n-l-1)-1},$
$\displaystyle
B_{2}=\sum_{l=0}^{n-1}C^{n}_{l}(2(n-l))^{2}x^{2l+1}y^{2(n-l)-1}.$

Since the relation $B_{0}=-A_{0}$ remains valid the last step in the proof
requires solving

$\widetilde{S}_{1}=\widetilde{S}_{2}\quad\Longleftrightarrow\quad
a_{-2}A_{-2}-b_{-2}B_{-2}+(a_{0}+b_{0})A_{0}+a_{2}A_{2}-b_{2}B_{2}=0.$ (4.12)

At this stage we must discuss separately the cases $n=2$ and $n=3$ before the
general case $n\geq 4$.

####  4.4.1 First case: $\boldsymbol{n=2}$

In this case, we have $A_{-2}=B_{-2}=0$ and the previous relation reduces to

$-3(a_{0}+b_{0})xy+2a_{2}\big{(}xy^{3}-8x^{3}y\big{)}-2b_{2}\big{(}xy^{3}-x^{3}y\big{)}=0,$

implying $a_{2}=b_{2}=0$ and $a_{0}+b_{0}=0$ but no constraint on $a_{-2}=a$,
$b_{-2}=b$. So we have

$f(x)+g(y)=\frac{a}{x^{2}}+\frac{b}{y^{2}},$

and integrating (3.2) gives

${\mathcal{S}}=4\big{(}ay^{2}-bx^{2}\big{)}H-4xy\big{(}x^{2}-y^{2}\big{)}P_{x}P_{y}-\big{(}x^{4}-6x^{2}y^{2}+y^{4}\big{)}Q.$
(4.13)

####  4.4.2 Second case: $\boldsymbol{n=3}$

Comparing, in equation (4.12), the $y$ dependence of the various bivariate
polynomials shows that this equation reduces to

$a_{-2}A_{-2}-b_{-2}B_{-2}=0,\qquad(a_{0}+b_{0})A_{0}=0,\qquad
a_{2}A_{2}-b_{2}A_{2}=0.$

Taking into account that

$A_{-2}=B_{-2}=-48xy\quad\Longrightarrow\quad b_{-2}=a_{-2}=a.$

Since $A_{0}=30xy\big{(}3y^{2}-4x^{2}\big{)}$ it follows that $a_{0}+b_{0}=0$.

Observing that

$A_{2}=4\big{(}C^{3}_{0}xy^{3}+4C^{3}_{1}x^{3}y^{3}+9C^{3}_{2}x^{5}y\big{)},\qquad
B_{2}=4\big{(}9C^{3}_{0}xy^{3}+4C^{3}_{1}x^{3}y^{3}+C^{3}_{2}x^{5}y\big{)},$

implies $a_{2}=b_{2}=0$. Setting $a=1$, we have

$f(x)+g(y)=\frac{1}{x^{2}}+\frac{1}{y^{2}},$

and one can check that

$\displaystyle{\mathcal{S}}={}$
$\displaystyle-2\big{(}5x^{4}-6x^{2}y^{2}+5y^{4}\big{)}H-2xy\big{(}3x^{4}-10x^{2}y^{2}+3y^{4}\big{)}P_{x}P_{y}$
$\displaystyle-\big{(}x^{6}-15x^{4}y^{2}+15x^{2}y^{4}-y^{6}\big{)}Q.$ (4.14)

As shown in Appendix A, the corresponding Liouville metric is of constant
negative curvature and should not be considered any longer.

####  4.4.3 General case: $\boldsymbol{n\geq 4}$

This time we have

$\displaystyle
A_{-2}=4\big{(}2C^{n}_{2}xy^{2n-5}+\cdots+(n-2)(n-1)C^{n}_{n-1}x^{2n-5}y\big{)},$
$\displaystyle
B_{-2}=4\big{(}(n-2)(n-1)C^{n}_{0}xy^{2n-5}+\cdots+2C^{n}_{n-3}x^{2n-5}y\big{)},$

as well as

$A_{0}=2\big{(}3C^{n}_{1}xy^{2n-3}+\cdots+(n-1)(2n-1)C^{n}_{n-1}x^{2n-3}y\big{)}$

and

$\displaystyle
A_{2}=4\big{(}C^{n}_{0}xy^{2n-1}+\cdots+n^{2}C^{n}_{n-1}x^{2n-1}y\big{)},$
$\displaystyle
B_{2}=4\big{(}n^{2}C^{n}_{0}xy^{2n-1}+\cdots+C^{n}_{n-1}x^{2n-1}y\big{)}.$

It follows that $a_{0}+b_{0}=0$ and $a_{-2}=b_{-2}=a_{2}=b_{2}=0$ hence
$f(x)+g(y)\equiv 0$. ∎

##  5 Global structure

One should take care of the following point: in principle
$(H,Q,{\mathcal{S}})$ are independent integrals which ensure the
superintegrability, but for particular values of the parameters a Killing
vector may appear (let us call $K$ the corresponding linearly conserved
quantity) which may induce a reducibility either of $Q$ or of ${\mathcal{S}}$.
So we may have, in these particular cases, a SI system either of the form
$(H,Q,K)$ or $(H,{\mathcal{S}},K)$.

###  5.1 Case I

In Proposition 4.1, we got the Liouville metric

$\displaystyle f(x)=\mu x^{2}+a_{1}x+a_{0},\qquad g(y)=\mu
y^{2}+b_{1}y+b_{0},$ $\displaystyle{\mathcal{A}}=-\mu
xy-\frac{b_{1}}{2}x-\frac{a_{1}}{2}y,\qquad{\mathcal{S}}={\mathcal{A}}H+P_{x}P_{y}.$

If $\mu=0$ by a translation of $x$ and $y$, we can set $a_{0}=b_{0}=0$. Then,
by a linear change of coordinates and a scaling of the metric, one obtains

$g=x\big{(}{\rm d}x^{2}+{\rm d}y^{2}\big{)}\quad\Longrightarrow\quad
H=\frac{P_{x}^{2}+P_{y}^{2}}{x}.$ (5.1)

This geodesic flow is not globally defined because the conformal factor
vanishes for $x=0$ inducing a singularity in the Gaussian curvature $\
R=1/\big{(}2x^{3}\big{)}$.

This is one of the SI systems discovered by Koenigs [13], see (2.6). This case
is special since $P_{y}$ is now conserved and the SI system is

$H,\quad P_{y},\quad{\mathcal{S}}=-\frac{y}{2}H+P_{x}P_{y},$

since $Q=-P_{y}^{2}$ is reducible. It is well known [11] that there is another
quadratic integral

${\mathcal{T}}=-\frac{y^{2}}{2}H+2P_{y}(yP_{x}-xP_{y}),$

which is reducible since we have
$H{\mathcal{T}}=-2{\mathcal{S}}^{2}-2P_{y}^{4}$.

When $\mu\neq 0$ a translation of $x$ and $y$ allows $a_{1}=b_{1}=0$ and a
scaling of the metric allows $\mu=1$, leaving us with

$g=\big{(}s+x^{2}+y^{2}\big{)}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad(x,y)\in{\mathbb{R}}^{2},\qquad s=a_{0}+b_{0}.$ (5.2)

This metric is Koenigs [13], see (2.8). It was also considered by Matveev in
[14, p. 555]. It is globally defined iff $s>0$: since the conformal factor
never vanishes we have for manifold $M\cong{\mathbb{R}}^{2}$.

This metric was advocated in [14, p. 565] to be an example of a geodesic flow
with four independent integrals. Indeed there are 4 integrals:

$H,\qquad Q=(y^{2}+b_{0})H-P_{y}^{2},\qquad
K=xP_{y}-yP_{x},\qquad{\mathcal{S}}=-xyH+P_{x}P_{y},$

which are not functionally independent since we have

${\mathcal{S}}^{2}=-(Q+a_{0}H)(Q-b_{0}H)+HK^{2}.$

###  5.2 Case II

In Proposition 4.2, we got the Liouville metric with

$f(x)=\mu x^{2}+ax+a_{0},\qquad
g(y)=\frac{\mu}{4}y^{2}+\frac{b}{y^{2}}+b_{0},$

and

${\mathcal{A}}=-\frac{\mu}{2}xy^{2}-\frac{a}{4}y^{2}-b_{0}x\quad\Longrightarrow\quad{\mathcal{S}}={\mathcal{A}}H+yP_{x}P_{y}+xQ.$

If $\mu=0$ and $a\neq 0,\ b\neq 0$ a translation of $x$ reduces the metric to

$g=\big{(}b+(s+ax)y^{2}\big{)}\frac{\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)}}{y^{2}},\qquad x\in{\mathbb{R}},\quad y>0,$ (5.3)

which is never globally defined due to the zero of the conformal factor.

The three independent integrals are

$H=\frac{y^{2}}{b+(s+ax)y^{2}}\big{(}P_{x}^{2}+P_{y}^{2}\big{)},\qquad\widetilde{Q}=\frac{\big{(}b+sy^{2}\big{)}P_{x}^{2}-axy^{2}P_{y}^{2}}{b+(s+ax)y^{2}},$

and

${\mathcal{S}}=-\left(\frac{a}{4}y^{2}+sx\right)H+yP_{x}P_{y}+x\widetilde{Q}.$

A companion case is obtained via $(x\leftrightarrow y).$

For $a=0$, taking $b=s^{2}$ and scaling the metric gives

$g=\big{(}s+y^{2}\big{)}\frac{{\rm d}x^{2}+{\rm
d}y^{2}}{y^{2}}=\big{(}s+y^{2}\big{)}g_{0}\big{(}H^{2}\big{)},\qquad
x\in{\mathbb{R}},\quad y>0,$ (5.4)

we recover a second metric due to Koenigs, see (2.7), globally defined on
$M\cong{\mathbb{H}}^{2}$ iff $s>0$. Its three independent integrals are

$H=\frac{y^{2}}{s+y^{2}}\big{(}P_{x}^{2}+P_{y}^{2}\big{)},\qquad
P_{x},\qquad{\mathcal{S}}=-xH+P_{x}(xP_{x}+yP_{y}),$

since $Q=P_{x}^{2}-sH$ is reducible.

If $\mu\neq 0$ we can set $\mu=1$. For $a=b=0$, one gets

$g=\left(s+x^{2}+\frac{y^{2}}{4}\right)\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},$ (5.5)

a metric derived by Matveev [14], globally defined on $M\cong{\mathbb{R}}^{2}$
iff $s>0$. Its three independent integrals are

$H=\frac{P_{x}^{2}+P_{y}^{2}}{s+x^{2}+\frac{y^{2}}{4}},\qquad\widetilde{Q}=P_{x}^{2}-x^{2}H,\qquad{\mathcal{S}}=-x\left(s+\frac{y^{2}}{2}\right)H+yP_{x}P_{y}+x\widetilde{Q}.$

More generally, for $b\neq 0$, we have

$g=\Psi(x,y)g_{0}\big{(}H^{2}\big{)},\qquad\Psi(x,y)=b+y^{2}\left(s+x^{2}+\frac{y^{2}}{4}\right),$
(5.6)

globally defined on $M\cong{\mathbb{H}}^{2}$ iff $b>0$ and $s\geq 0$.

The three independent integrals are

$H=\frac{y^{2}}{\Psi}\big{(}P_{x}^{2}+P_{y}^{2}\big{)},\qquad\widetilde{Q}=P_{x}^{2}-x^{2}H,\qquad{\mathcal{S}}=-x\left(s+\frac{y^{2}}{2}\right)H+yP_{x}P_{y}+x\widetilde{Q}.$

Due to Remark 4.4, Case III does not produce new metrics and can be skipped.

###  5.3 Case IV

In Proposition 4.5, we got the Liouville metric

$f(x)=\mu x^{2}+\frac{a}{x^{2}}+a_{0},\qquad g(y)=\mu
y^{2}+\frac{b}{y^{2}}+b_{0},$

and

${\mathcal{A}}=-\mu
x^{2}y^{2}-\frac{b_{0}}{2}x^{2}-\frac{a_{0}}{2}y^{2},\qquad{\mathcal{S}}={\mathcal{A}}H+xyP_{x}P_{y}+\frac{1}{2}\big{(}x^{2}-y^{2}\big{)}Q.$
(5.7)

If $\mu=0$, the case where $a_{0}=b_{0}=0$ is special since the metric is

$g=\frac{bx^{2}+ay^{2}}{x^{2}y^{2}}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad a>0,\quad b>0,\quad a\neq b,$ (5.8)

encountered in Section 4.4.1 for $n=2$. It exhibits an extra Killing vector:
$x\partial_{x}+y\partial_{y}$.

We have for integrals

$H=\frac{x^{2}y^{2}}{bx^{2}+ay^{2}}\big{(}P_{x}^{2}+P_{y}^{2}\big{)},\qquad
Q=P_{x}^{2}-\frac{a}{x^{2}}H,\qquad K=xP_{x}+yP_{y}.$ (5.9)

However, there are two other quadratic integrals: the first one from (5.7) is
reducible

$2{\mathcal{S}}=2xyP_{x}P_{y}+\big{(}x^{2}-y^{2}\big{)}Q=K^{2}-(a+b)H.$

So we remain with 4 integrals: $H$, $Q$, $K$ and ${\mathcal{S}}$ given by
(4.13) which are not independent since one has the relation

$Q{\mathcal{S}}=-(a-b)^{2}H^{2}+2(a+b)HK^{2}-K^{4},$

and we remain with a SI geodesic flow with three independent integrals:
$(H,Q,K)$.

More generally, when $\mu\geq 0$ and $s=a_{0}+b_{0}\geq 0$ the metric is

$g=\left(\frac{bx^{2}+ay^{2}}{x^{2}+y^{2}}+s\frac{x^{2}y^{2}}{x^{2}+y^{2}}+\mu
x^{2}y^{2}\right)\frac{x^{2}+y^{2}}{x^{2}y^{2}}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad x>0,\quad y>0,$ (5.10)

and taking into account Proposition A.1, we have $g=\chi
g_{0}\big{(}H^{2}\big{)}.$

In terms of the global coordinates $\big{(}X^{1},X^{2},X^{3}\big{)}$, defined
in Appendix A, relation (A.1), we have

$x^{2}+y^{2}=\sqrt{\frac{X^{3}+X^{1}}{X^{3}-X^{1}}},\qquad
x^{2}-y^{2}=\frac{X^{2}}{X^{3}-X^{1}},$

leading to

$\frac{bx^{2}+ay^{2}}{x^{2}+y^{2}}=\frac{a+b}{2}-\frac{(a-b)}{2}\frac{X^{2}}{\sqrt{1+(X^{2})^{2}}}.$

Since $\frac{X^{2}}{\sqrt{1+(X^{2})^{2}}}\in(-1,+1)$ the sum of the first two
terms, for $a>0$ and $b>0$ is strictly positive, so if $(a>0$, $b>0)$ and
$(s\geq 0$, $\mu\geq 0)$ the non-vanishing of the conformal factor implies
$M\cong{\mathbb{H}}^{2}$.

The three independent integrals are

$H=\frac{x^{2}y^{2}}{x^{2}+y^{2}}\frac{\big{(}P_{x}^{2}+P_{y}^{2}\big{)}}{\chi},\qquad\widetilde{Q}=P_{x}^{2}-\left(\frac{a}{x^{2}}+\mu
x^{2}\right)H,$

and

${\mathcal{S}}=-x^{2}\left(\frac{s}{2}+\mu
y^{2}\right)H+xyP_{x}P_{y}+\frac{1}{2}\big{(}x^{2}-y^{2}\big{)}\widetilde{Q}.$

######  Remark 5.1.

The global structure meets the manifolds ${\mathbb{R}}^{2}$ and
${\mathbb{H}}^{2}$ but never ${\mathbb{S}}^{2}$. The explanation stems from a
theorem due to Kiyohara [12]: a SI geodesic flow of Hamiltonian $H$, globally
defined on ${\mathbb{S}}^{2}$, with two extra quadratic integrals implies that
its metric is of constant curvature hence cannot appear in our analysis. This
applies as well to Koenigs metrics which are never defined on
${\mathbb{S}}^{2}$.

Let us now relate our results, via coupling constant metamorphosis, to
previous work.

##  6 Coupling constant metamorphosis

The so-called “coupling constant metamorphosis” [8] or Stäckel transform [3]
establishes that if a system has for Hamiltonian

$H=P_{x}^{2}+P_{y}^{2}+V(x,y)$ (6.1)

and is quadratically SI then

$H^{\prime}=\frac{P_{x}^{2}+P_{y}^{2}}{V(x,y)}$

will be also quadratically SI. Since all the systems having the form (6.1)
were derived in [10] let us give their relation with our work.

In Case I (Section 5.1), we have

$V=\mu\big{(}x^{2}+y^{2}\big{)}+2(a_{1}x+b_{1}y)+s,\qquad s=a_{0}+b_{0},$

which is the case ${\bf E^{\prime}3}$, merely the case ${\bf E3}$ with a
translation of both variables.

In Case II (Section 5.2), the potential is

$V=\mu\Big{(}x^{2}+\frac{y^{2}}{4}\Big{)}+2a_{1}x+\frac{b}{y^{2}}+s,$

which is the case ${\bf E2}$. Case III is obtained by the permutation
$x\leftrightarrow y$.

In Case IV, the potential is

$V=\mu\big{(}x^{2}+y^{2}\big{)}+\frac{a}{x^{2}}+\frac{b}{y^{2}}+s,$

and this is the case ${\bf E1}$.

##  Appendix A Appendix

Let us prove:

######  Proposition A.1.

If one takes

$f(x)=\frac{1}{x^{2}},\qquad g(y)=\frac{1}{y^{2}}\quad\Longrightarrow\quad
g_{0}=\frac{\big{(}x^{2}+y^{2}\big{)}}{x^{2}y^{2}}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad x>0,\quad y>0,$

the Liouville metric is a non-canonical form of $g_{0}\big{(}H^{2}\big{)}$.

###### Proof.

The Gaussian curvature $R=-1$ implies a non-canonical metric of
${\mathbb{H}}^{2}$. There are three linear integrals

$K_{1}=xP_{x}+yP_{y},\qquad K_{2}=\frac{xP_{x}-yP_{y}}{x^{2}+y^{2}},\qquad
K_{3}=x\big{(}x^{2}-3y^{2}\big{)}P_{x}-y\big{(}y^{2}-3x^{2}\big{)}P_{y}.$

Within this geometry all the quadratic integrals are reducible:

$H=\frac{1}{4}\big{(}K_{1}^{2}-K_{2}K_{3}\big{)},\qquad Q=K_{1}K_{2},$

and despite the apparent complexity of ${\mathcal{S}}$, see (4.14), we have
merely $\ {\mathcal{S}}=-K_{3}^{2}$.

Defining the coordinates

$X^{1}=\frac{\big{(}x^{2}+y^{2}\big{)}^{2}-1}{4xy}\in{\mathbb{R}},\qquad
X^{2}=\frac{x^{2}-y^{2}}{2xy}\in{\mathbb{R}},\qquad
X^{3}=\frac{\big{(}x^{2}+y^{2}\big{)}^{2}+1}{4xy}\geq 1,$ (A.1)

one can check the relations

$\big{(}X^{1}\big{)}^{2}+\big{(}X^{2}\big{)}^{2}-\big{(}X^{3}\big{)}^{2}=-1$

and

$g_{0}\big{(}H^{2}\big{)}\equiv\big{(}{\rm d}X^{1}\big{)}^{2}+\big{(}{\rm
d}X^{2}\big{)}^{2}-\big{(}{\rm
d}X^{3}\big{)}^{2}=\frac{\big{(}x^{2}+y^{2}\big{)}}{x^{2}y^{2}}\big{(}{\rm
d}x^{2}+{\rm d}y^{2}\big{)}.$

##  Appendix B Appendix

The list of SI Liouville metrics is the following:

  1. 1.

The first Koenigs metric (5.1):

$g=x\big{(}{\rm d}x^{2}+{\rm d}y\big{)}^{2},\qquad x>0,\quad
y\in{\mathbb{R}},\quad M\cong\varnothing.$

  2. 2.

The second Koenigs metric (5.4):

$g=\big{(}s+x^{2}\big{)}\frac{{\rm d}x^{2}+{\rm
d}y^{2}}{x^{2}}=\big{(}s+x^{2}\big{)}g_{0}\big{(}H^{2}\big{)},\qquad x>0,\quad
y\in{\mathbb{R}},\qquad s>0\ \Longrightarrow\ M\cong{\mathbb{H}}^{2}.$

  3. 3.

The third Koenigs metric (5.2):

$g=\big{(}s+x^{2}+y^{2}\big{)}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad(x,y)\in{\mathbb{R}}^{2},\qquad
s>0\quad\Longrightarrow\quad M\cong{\mathbb{R}}^{2}.$

  4. 4.

The variant (5.3):

$g=\big{(}b+(s+ax)y^{2}\big{)}g_{0}\big{(}H^{2}\big{)},\qquad
x\in{\mathbb{R}},\quad y>0,\qquad M=\varnothing,$

and its companion via $(x\leftrightarrow y)$.

  5. 5.

Matveev metric (5.5):

$g=\left(s+x^{2}+\frac{y^{2}}{4}\right)\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad(x,y)\in{\mathbb{R}^{2}},\qquad
s>0\quad\Longrightarrow\quad M\cong{\mathbb{R}}^{2},$

and its companion via $(x\leftrightarrow y)$.

  6. 6.

The most general metric in Case II (see (5.6)):

$\displaystyle
g=\left[b+y^{2}\left(s+x^{2}+\frac{y^{2}}{4}\right)\right]g_{0}\big{(}H^{2}\big{)},\qquad
x\in{\mathbb{R}},\quad y>0,\qquad b>0,\quad s\geq 0$
$\displaystyle\Longrightarrow\quad M\cong{\mathbb{H}^{2}},$

and its companion via $(x\leftrightarrow y)$.

  7. 7.

The metric (5.8):

$\displaystyle g=\frac{bx^{2}+ay^{2}}{x^{2}y^{2}}\big{(}{\rm d}x^{2}+{\rm
d}y^{2}\big{)},\qquad a>0,\quad b>0,\quad a\neq b\quad\Longleftrightarrow\quad
M\cong{\mathbb{H}}^{2}.$

For $a=b$, it reduces to a non-canonical metric on ${\mathbb{H}}^{2}$, see
Appendix A.

  8. 8.

The most general metric in Case IV (Section 5.3) given by (5.10)

$g=\chi(x,y)g_{0}\big{(}H^{2}\big{)},\qquad\chi(x,y)=\frac{bx^{2}+ay^{2}}{x^{2}+y^{2}}+s\frac{x^{2}y^{2}}{x^{2}+y^{2}}+\mu
x^{2}y^{2},$

and if $a>0$, $b>0$, $a\neq b$, $s\geq 0$, $\mu\geq 0$, we have
$M\cong{\mathbb{H}}^{2}.$

The three independent integrals are given for each metric in Section 5.

##  Appendix C Appendix

The metrics derived in [4] are expressed in terms of algebraic functions while
Koenigs ones resort to trigonometric or hyperbolic functions. Let us put a
bridge, in the Riemannian case, between these metrics observing that a scaling
of the metric is irrelevant since this is tantamount to a scaling of the
Hamiltonian.

###  C.1 The affine case

The metric (a) given in [4], in which we substitute $u={\rm e}^{x}$, leads to

$u\big{(}{\rm d}u^{2}+{\rm d}y^{2}\big{)}.$ (C.1)

The metric (b) given in [4], in which we substitute $u=2\sqrt{{\rm
e}^{x}+\epsilon}$, becomes

$\big{(}u^{2}-4\epsilon\big{)}\frac{\big{(}{\rm d}u^{2}+{\rm
d}y^{2}\big{)}}{u^{2}}.$ (C.2)

Let us compare with the affine case in (1.1)

$g=\frac{\big{(}a_{2}x^{2}+2a_{1}x+a_{0}\big{)}}{(a_{2}x+a_{1})^{2}}\big{(}{\rm
d}x^{2}+{\rm d}y^{2}\big{)}.$

If $a_{2}=0$, we can set $a_{1}=1/2$ and taking $u=x+a_{0}$ we obtain (C.1).
If $a_{2}$ does not vanish, we take $a_{2}=1$. Defining $u=x+a_{1}$, we
recover (C.2) up to an overall scaling of $u$ and $y$.

What remains to be discussed in [4] is the metric (c) which can be written

$g=\frac{x}{\big{(}x^{2}+2ax+\epsilon\big{)}}\left(\frac{{\rm
d}x^{2}}{x^{2}\big{(}x^{2}+2ax+\epsilon\big{)}}+{\rm d}y^{2}\right),\qquad
a\in{\mathbb{R}},\qquad\epsilon=\pm 1.$

###  C.2 The hyperbolic case

If $\epsilon=-1$, let us define the coordinate change $x\to u$:

$u=\arctan\left(\frac{\sqrt{x^{2}+2ax-1}}{1-ax}\right),\qquad a<0,\qquad
x>x_{0}=\sqrt{a^{2}+1}+|a|.$

It maps $x\in(x_{0},+\infty)$ into $u\in(0,\arctan(1/|a|))$ and we have the
relations

$\frac{{\rm d}x^{2}}{x^{2}\big{(}x^{2}+2ax-1\big{)}}={\rm d}u^{2},\qquad
x=\frac{1}{\sqrt{a^{2}+1}\cos u-|a|},$

leading to

$g=\frac{1}{a^{2}+1}\frac{\big{(}\sqrt{a^{2}+1}\cos
u-|a|\big{)}}{\sin^{2}u}\big{(}{\rm d}u^{2}+{\rm d}y^{2}\big{)},$

which fits with the first metric in (1.1).

###  C.3 The trigonometric cases

For $\epsilon=+1$, we have two possible cases. The first one defines the
coordinate change

$u={\rm arctanh}\left(\frac{\sqrt{x^{2}+2ax+1}}{1+ax}\right),\qquad a>1,\quad
x>0,$

which maps $x\in(0,+\infty)$ into $u\in({\rm arctanh}(1/a),+\infty)$. We have
the relations

$\frac{{\rm d}x^{2}}{x^{2}\big{(}x^{2}+2ax+1\big{)}}={\rm d}u^{2},\qquad
x=\frac{1}{\sqrt{a^{2}-1}\cosh u-a},$

leading to

$g=\frac{1}{\big{(}a^{2}-1\big{)}}\frac{\sqrt{a^{2}-1}\cosh
u-a}{\sinh^{2}u}\big{(}{\rm d}u^{2}+{\rm d}y^{2}\big{)},$

which fits with the fourth metric in (1.1).

The second case is given by the coordinate change

$u={\rm arctanh}\left(\frac{1+ax}{\sqrt{x^{2}+2ax+1}}\right),\qquad
a\in(0,1),\quad x>0,$

which maps $x>0$ into $u\in({\rm arctanh}(a),+\infty)$. We have

$\frac{{\rm d}x^{2}}{x^{2}\big{(}x^{2}+2ax+1\big{)}}={\rm d}u^{2},\qquad
x=\frac{1}{\sqrt{1-a^{2}}\sinh u-a},$

leading to

$g=\frac{1}{1-a^{2}}\frac{\sqrt{1-a^{2}}\sinh u-a}{\cosh^{2}u}\big{(}{\rm
d}u^{2}+{\rm d}y^{2}\big{)},$

which fits with the third metric in (1.1).

Up to now $a\neq\pm 1$. For $a=1$, we have

$x=\frac{{\rm e}^{-u}}{{\rm e}^{-u}-1}\quad\Rightarrow\quad g=\big{(}{\rm
e}^{-u}-{\rm e}^{-2u}\big{)}\big{(}{\rm d}u^{2}+{\rm d}y^{2}\big{)}.$

For $a=-1$, we have two possible cases:

$\displaystyle x=\frac{{\rm e}^{u}}{{\rm e}^{u}-1}\quad\Rightarrow\quad
g=\big{(}{\rm e}^{2u}-{\rm e}^{u}\big{)}\big{(}{\rm d}u^{2}+{\rm
d}y^{2}\big{)},$ $\displaystyle x=\frac{{\rm e}^{u}}{{\rm
e}^{u}+1}\quad\Rightarrow\quad g=\big{(}{\rm e}^{u}+{\rm
e}^{2u}\big{)}\big{(}{\rm d}u^{2}+{\rm d}y^{2}\big{)}.$

Due to the freedom $u\to\pm u$, we are in agreement with the second case in
(1.1).

### Acknowledgements

The author is greatly indebted to the anonymous referees whose remarks allowed
useful corrections and put to light several references related with this work.

## References

  * [1] Ballesteros A., Enciso A., Herranz F.J., Ragnisco O., Riglioni D., Quantum mechanics on spaces of nonconstant curvature: the oscillator problem and superintegrability, Ann. Physics 326 (2011), 2053–2073, arXiv:1102.5494.
  * [2] Bolsinov A.V., Matveev V.S., Pucacco G., Normal forms for pseudo-Riemannian 2-dimensional metrics whose geodesic flows admit integrals quadratic in momenta, J. Geom. Phys. 59 (2009), 1048–1062, arXiv:0803.0289.
  * [3] Boyer C.P., Kalnins E.G., Miller Jr. W., Stäckel-equivalent integrable Hamiltonian systems, SIAM J. Math. Anal. 17 (1986), 778–797.
  * [4] Bryant R.L., Manno G., Matveev V.S., A solution of a problem of Sophus Lie: normal forms of two-dimensional metrics admitting two projective vector fields, Math. Ann. 340 (2008), 437–463, arXiv:0705.3592.
  * [5] Daskaloyannis C., Ypsilantis K., Unified treatment and classification of superintegrable systems with integrals quadratic in momenta on a two-dimensional manifold, J. Math. Phys. 47 (2006), 042904, 38 pages, arXiv:math-ph/0412055.
  * [6] Fordy A.P., First integrals from conformal symmetries: Darboux–Koenigs metrics and beyond, J. Geom. Phys. 145 (2019), 103475, 13 pages, arXiv:1804.06904.
  * [7] Fordy A.P., Huang Q., Generalised Darboux–Koenigs metrics and 3-dimensional superintegrable systems, SIGMA 15 (2019), 037, 30 pages, arXiv:1810.13368.
  * [8] Hietarinta J., Grammaticos B., Dorizzi B., Ramani A., Coupling-constant metamorphosis and duality between integrable Hamiltonian systems, Phys. Rev. Lett. 53 (1984), 1707–1710.
  * [9] Kalnins E.G., Kress J.M., Miller Jr. W., Winternitz P., Superintegrable systems in Darboux spaces, J. Math. Phys. 44 (2003), 5811–5848, arXiv:math-ph/0307039.
  * [10] Kalnins E.G., Kress J.M., Pogosyan G.S., Miller Jr. W., Completeness of superintegrability in two-dimensional constant-curvature spaces, J. Phys. A 34 (2001), 4705–4720, arXiv:math-ph/0102006.
  * [11] Kalnins E.G., Kress J.M., Winternitz P., Superintegrability in a two-dimensional space of nonconstant curvature, J. Math. Phys. 43 (2002), 970–983, arXiv:math-ph/0108015.
  * [12] Kiyohara K., Compact Liouville surfaces, J. Math. Soc. Japan 43 (1991), 555–591.
  * [13] Koenigs G., Sur les géodésiques à intégrales quadratiques, in Leçons sur la théorie générale des surfaces, Vol. 4, Editor J.G. Darboux, Chelsea Publishing, New York, 1972, 368–404.
  * [14] Matveev V.S., Lichnerowicz–Obata conjecture in dimension two, Comment. Math. Helv. 80 (2005), 541–570.
  * [15] Matveev V.S., Shevchishin V.V., Two-dimensional superintegrable metrics with one linear and one cubic integral, J. Geom. Phys. 61 (2011), 1353–1377, arXiv:1010.4699.
  * [16] Miller Jr. W., Post S., Winternitz P., Classical and quantum superintegrability with applications, J. Phys. A 46 (2013), 423001, 97 pages, arXiv:1309.2694.
  * [17] Valent G., Global structure and geodesics for Koenigs superintegrable systems, Regul. Chaotic Dyn. 21 (2016), 477–509, arXiv:1510.08379.
  * [18] Valent G., Duval C., Shevchishin V., Explicit metrics for a class of two-dimensional cubically superintegrable systems, J. Geom. Phys. 87 (2015), 461–481, arXiv:1403.0422.