# Integrable systems with linear periodic integral
for the Lie algebra $\mathrm{e}(3)$

I. K. Kozlov and A. A. Oshemkov No Affiliation, E-mail:
<EMAIL_ADDRESS>of Mechanics and Mathematics, Moscow State
University, Moscow, 119991 Russia, E-mail<EMAIL_ADDRESS>
###### Abstract

Integrable systems with a linear periodic integral for the Lie algebra
$\mathrm{e}(3)$ are considered. One investigates singulariries of the
Liouville foliation, bifurcation diagram of the momentum mapping,
transformations of Liouville tori, topology of isoenergy surfaces and other
topological properties of such systems.

Keywords and phrases: Integrable Hamiltonian system, periodic integral,
bifurcation diagram, momentum mapping, Liouville tori

##  1 Introduction

In this paper we study some topological properties of integrable Hamiltonian
systems with an $S^{1}$-symmetry given by the Euler equations for the Lie
algebra $\mathrm{e}(3)$. Probably, the most well-known example of such a
system is the classical Lagrange top. Roughly speaking, we consider a
‘‘generalized’’ Lagrange top which Hamiltonian has an arbitrary potential
function and linear terms in momenta, but possesses the same $S^{1}$-symmetry.

We are interested in local and global topological properties of the Liouville
foliation defined by the system under consideration, namely, the structure of
bifurcation diagram and transformations of Liouville tori for critical values
of the momentum mapping, non-degeneracy of equilibria and other singular
points, the topology of isoenergy surfaces.

Note that there is a number of integrable systems with periodic linear
integral which are well known in mechanics and mathematical physics, which
phase topology were studied by various authors. In particular, there are
Lagrange and Kirchhoff integrable cases in rigid body dynamics (for the
description of their topology see [1, 2, 3]), the integrable case of Leggett
equations describing dynamics of spin in the superfluid 3He (the bifurcation
diagram and Fomenko invariants for this system are described in [6]), the
integrable case of the motion of heavy ellipsoid on a smooth horizontal plane
(topological invariants for this system were found in [7]).

Topological properties of all these systems are quite similar because of an
$S^{1}$-symmetry which imposes strong restrictions on the structure of their
singularities. Therefore, they can be studied under a uniform scheme. In this
paper we perform such an investigation for an example of Hamiltonian
possessing a periodic linear integral on $\mathrm{e}(3)^{*}$. Note that the
problem of topological investigation of integrable systems with S1-action is
discussed in paper [4], which contains a list of various open problems in the
theory of integrable systems.

Apart from the systems on $\mathrm{e}(3)^{*}$ considered in this paper there
are other integrable systems with $S^{1}$-symmetry, which were also studied by
various authors. For instance, natural mechanical systems on surfaces of
revolution homeomorphic to the sphere were studied recently in [5] (see also
[1]). Another example is the classical Euler case in the rigid body dynamics,
where the $S^{1}$-action is given not by a linear, but by a quadratic
integral. The results obtained in this paper show in particular that there are
some differences between the topological properties of the systems under
consideration and other cases with an $S^{1}$-symmetry (for example, the one
investigated in [5] or the Euler case).

The article is organized as follows. In Section 2 we describe the systems
under consideration. We start the analysis with the study of non-deneracy and
types of singular points of rank 0 in Section 3 (Corollary 1 ). In Section 4
we find singular points of rank 1 (Theorem 3) and describe the bifurcation
diagrams of the system (Theorems 4 and 5). In Section 5 we determine types of
non-degenerate points of rank 1 (Theorem 6) and specify the corresponding
Liouville tori bifurcations (Theorem 7). Finally, in Section 6 we list all
possible isoenergy surfaces for the system (Theorem 8).

##  2 Description of the system

Let us recall that the Lie–Poisson bracket for the Lie algebra $\mathrm{e}(3)$
is given by the formulas

$\\{S_{i},S_{j}\\}=\varepsilon_{ijk}S_{k},\quad\\{S_{i},R_{j}\\}=\varepsilon_{ijk}R_{k},\quad\\{R_{i},R_{j}\\}=0,$
(1)

where $S_{1},S_{2},S_{3},R_{1},R_{2},R_{3}$ are linear coordinates on the dual
space $\mathrm{e}(3)^{*}$ for the Lie algebra $\mathrm{e}(3)$. We will use the
notation $\textbf{S}=(S_{1},S_{2},S_{3})$ and $\textbf{R}=(R_{1},R_{2},R_{3})$
and also $\langle{\cdot}{,}{\cdot}\rangle$ and ${\,}\times{\,}$ for the scalar
and vector product of 3-dimensional vectors.

A Hamiltonian system with Hamiltonian $H$ is given by the Euler equations

$\dot{x}_{i}=\\{x_{i},H\\},$

which for the Lie algebra $\mathrm{e}(3)$ take the form

$\dot{\textbf{S}}=\frac{\partial
H}{\partial\textbf{S}}\times\textbf{S}+\frac{\partial
H}{\partial\textbf{R}}\times\textbf{R},\qquad\dot{\textbf{R}}=\frac{\partial
H}{\partial\textbf{S}}\times\textbf{R}.$

Bracket (1) has two Casimir functions:

$F_{1}=\langle\textbf{R},\textbf{R}\rangle,\qquad
F_{2}=\langle\textbf{S},\textbf{R}\rangle.$

Their regular common level surfaces

$M^{4}_{a,g}=\\{(\textbf{S},\textbf{R})\mid\,F_{1}(\textbf{S},\textbf{R})=a,\,F_{2}(\textbf{S},\textbf{R})=g,\\},\qquad
a>0,$ (2)

are the sympectic leaves of bracket (1) and are the orbits of the coadjoint
repsresentation for the Lie algebra $\mathrm{e}(3)$. We are interested in
integrable Hamiltonian systems on the orbits $M^{4}_{a,g}$ for which some
linear function on $\mathrm{e}(3)^{*}$ is a first integral defining an
$S^{1}$-action.

Let us describe several examples of such systems from mechanics and
mathematical physics, which are integrable cases of the Euler equations for
the Lie algebra $\mathrm{e}(3)$ with Hamiltonian $H$ and integral $K$ (an
explanation of physical sense for parameters and variables of these systems
can be found in [1, 2, 6, 7]).

1) The Lagrange case. This is a symmetric top with two equal moments of
inertia which center of gravity lies on the symmetry axis:

$H=\frac{S_{1}^{2}}{A}+\frac{S_{2}^{2}}{A}+\frac{S_{3}^{2}}{B}-pR_{3},\quad
K=S_{3},\quad\text{where }A,B,p=\mathrm{const}.$

2) The Kirchhoff case. This system describes the motion of a dynamically
symmetric rigid body in an ideal fluid:

$\displaystyle H$
$\displaystyle=AS_{1}^{2}+AS_{2}^{2}+aS_{3}^{2}+2(BS_{1}R_{1}+2BS_{2}R_{2}+bS_{3}R_{3})+$
$\displaystyle+CR_{1}^{2}+CR_{2}^{2}+cR_{3}^{2},\quad K=S_{3},\quad\text{where
}A,a,B,b,C,c=\mathrm{const}.$

3) The following integrable case for the Leggett system describing the
dynamics of spin in the superfluid 3He:

$H=S_{1}^{2}+S_{2}^{2}+S_{3}^{2}-\gamma S_{3}-R_{3}^{2},\quad
K=S_{3},\quad\text{where }\gamma=\mathrm{const}.$

4) Integrable system describing the motion of a dynamically and geometrically
symmetric heavy ellipsoid on a smooth horizontal plane:

$\displaystyle H$
$\displaystyle=\frac{S_{1}^{2}+S_{2}^{2}+A(S_{1}R_{1}+S_{2}R_{2})^{2}}{2b(1+A(R_{1}^{2}+R_{2}^{2}))}+\frac{S_{3}^{2}}{2J}+\sqrt{1+cR_{3}^{2}}+sR_{3},$
$\displaystyle K$ $\displaystyle=S_{3}\quad\text{where
}A=\frac{cR_{3}^{2}}{1+cR_{3}^{2}},\quad b,c,J,s=\mathrm{const}.$

In all these examples the additional integral is the function $S_{3}$ on
$\mathrm{e}(3)^{*}$. Let us explain that this is a general case if we require
that the integral is linear and periodic.

######  Assertion 1.

Let $K$ be a linear functions on $\mathrm{e}(3)^{*}$ which Hamiltonian flow
$\operatorname{sgrad}K$ defined by bracket (1) is periodic. Then there is a
linear change of variables preserving the bracket (1) taking the function $K$
to $cS_{3}$, where $c$ is some constant.

###### Proof.

Let
$K=\alpha_{1}S_{1}+\alpha_{2}S_{2}+\alpha_{3}S_{3}+\beta_{1}R_{1}+\beta_{2}R_{2}+\beta_{3}R_{3}$.
For an arbitrary orthogonal matrix $A$ the transformation
$\Phi_{A}:(\textbf{S},\textbf{R})\to(A\textbf{S},A\textbf{R})$ preserves
bracket (1). If $\alpha_{1}=\alpha_{2}=\alpha_{3}=0$, then we can choose a
matrix $A$ such that $\Phi_{A}$ takes the function $K$ to $\lambda R_{3}$,
where $\lambda=\mathrm{const}$. It is clear that the Hamiltonian flow of the
function $\lambda R_{3}$ is not periodic, since the trajectories of the field
$\operatorname{sgrad}R_{3}=(-R_{2},R_{1},0,0,0,0)$ are straight lines in
$\mathrm{e}(3)^{*}$.

If there are non-zero $\alpha_{i}$, then applying an appropriate
transformation $\Phi_{A}$ we can transform $K$ to a function of the form
$cS_{3}+\beta^{\prime}_{1}R_{1}+\beta^{\prime}_{2}R_{2}+\beta^{\prime}_{3}R_{3}$.
It is easy to check that for any vector v the transformations
$\Psi_{\textbf{v}}:(\textbf{S},\textbf{R})\to(\textbf{S}+\textbf{v}\times\textbf{R},\textbf{R})$
also preserve bracket (1). This allows one to transform the function $K$ to
the form $cS_{3}+\lambda R_{3}$, where $c\neq 0$.

Now consider the function $K=S_{3}+\lambda R_{3}$ and determine for which
$\lambda$ the Hamiltonian flow of $K$ is periodic. Integral trajectories for
the field $\operatorname{sgrad}K=(-S_{2}-\lambda R_{2},S_{1}+\lambda
R_{1},0,-R_{2},R_{1},0)$ can be explicitly written:

$\displaystyle\gamma(t)=($ $\displaystyle(s_{1}{-}\lambda r_{2}t)\cos
t{-}(s_{2}{+}\lambda r_{1}t)\sin t,\,(s_{2}{+}\lambda r_{1}t)\cos
t{+}(s_{1}{-}\lambda r_{2}t)\sin t,$ $\displaystyle s_{3},\,r_{1}\cos
t-r_{2}\sin t,\,r_{2}\cos t+r_{1}\sin t,\,r_{3}),$

where $s_{1},s_{2},s_{3},r_{1},r_{2},r_{3}$ are constants. It is clear from
this formula that the trajectories are periodic only for $\lambda=0$. ∎

######  Remark 1.

It is well known that an action of any compact group can be linearized at a
fixed point and that for an action of the circle $S^{1}$ the corresponding
tangent space can be represented as a sum of invariant two-dimensional
subspaces. Thus among all linear functions on $\mathrm{e}(3)^{*}$ the periodic
integrals are distiguished by the property that their linearization at any
singular point is a unitary operator with respect to a complex structure on
the tangent space. It also follows that up to the choice of the coordinate
system and multipltication by a constant any periodic linear integral on
$\mathrm{e}(3)^{*}$ is $S_{3}$.

Further we will consider Hamiltonian systems for the Lie algebra
$\mathrm{e}(3)$ which possess the first integral $K=S_{3}$ and which
Hamiltonian $H$ is quadratic in $S$, i.e.,

$H=A_{1}S_{1}^{2}+A_{2}S_{2}^{2}+A_{3}S_{3}^{2}+f_{1}(\textbf{R})S_{1}+f_{2}(\textbf{R})S_{2}+f_{3}(\textbf{R})S_{3}+f_{4}(\textbf{R}),$
(3)

where $A_{1},A_{2},A_{3}$ are arbitrary positive constants and
$f_{1},f_{2},f_{3},f_{4}$ are smooth functions of $R_{1},R_{2},R_{3}$.

First of all, let us rewrite Hamiltonian (3) in a more convient way using its
commutativity with the function $S_{3}$.

######  Assertion 2.

Up to multiplication by a constant any Hamiltonian of the form (3) commuting
with the function $K=S_{3}$ has the form

$\displaystyle H$
$\displaystyle=\frac{1}{2}\Bigl{(}S_{1}^{2}+S_{2}^{2}+\frac{S_{3}^{2}}{\beta}\Bigr{)}+g_{1}(\textbf{R}^{2},R_{3})(S_{1}R_{2}-S_{2}R_{1})+$
(4)
$\displaystyle+g_{2}(\textbf{R}^{2},R_{3})\langle\textbf{S},\textbf{R}\rangle+g_{3}(\textbf{R}^{2},R_{3})S_{3}+V(\textbf{R}^{2},R_{3}),$

where $\beta>0$ and the functions $g_{1},g_{2},g_{3},V$ depend only on
$\textbf{R}^{2}$ and $R_{3}$ and are smooth if $\textbf{R}^{2}\neq 0$.

###### Proof.

The Hamiltonian vector field for the function $K$ is equal to

$\operatorname{sgrad}K=-R_{2}\frac{\partial}{\partial
R_{1}}+R_{1}\frac{\partial}{\partial R_{2}}-S_{2}\frac{\partial}{\partial
S_{1}}+S_{1}\frac{\partial}{\partial S_{2}}.$

Since $\\{H,K\\}=(\operatorname{sgrad}K)H=0$, we get

$\begin{gathered}(\operatorname{sgrad}K)H=2(A_{2}-A_{1})S_{1}S_{2}+\\\
+\Bigl{(}{-}R_{2}\frac{\partial f_{1}}{\partial R_{1}}{+}R_{1}\frac{\partial
f_{1}}{\partial
R_{2}}{+}f_{2}(\textbf{R})\Bigr{)}S_{1}+\Bigl{(}{-}R_{2}\frac{\partial
f_{2}}{\partial R_{1}}{+}R_{1}\frac{\partial f_{2}}{\partial
R_{2}}{-}f_{1}(\textbf{R})\Bigr{)}S_{2}+\\\ +\Bigl{(}-R_{2}\frac{\partial
f_{3}}{\partial R_{1}}+R_{1}\frac{\partial f_{3}}{\partial
R_{2}}\Bigr{)}S_{3}+\Bigl{(}-R_{2}\frac{\partial f_{4}}{\partial
R_{1}}+R_{1}\frac{\partial f_{4}}{\partial R_{2}}\Bigr{)}=0.\end{gathered}$

Hence, $A_{1}=A_{2}$ (multiplying by a constant we can make both these
constants equal to $\frac{1}{2}$) and the four expressions in the brackets are
equal to zero.

In polar coordinates $(\rho,\varphi)$ on the plane $(R_{1},R_{2})$ the vector
field $\frac{\partial}{\partial\varphi}$ is exactly
$-R_{2}\frac{\partial}{\partial R_{1}}+R_{1}\frac{\partial}{\partial R_{2}}$.
Therefore,

$\frac{\partial f_{3}}{\partial\varphi}=0,\quad\frac{\partial
f_{4}}{\partial\varphi}=0,\qquad\frac{\partial
f_{1}}{\partial\varphi}=-f_{2},\quad\frac{\partial
f_{2}}{\partial\varphi}=f_{1}.$

The first two of these equations imply that $f_{3}$ and $f_{4}$ depend only on
$\rho$ and $R_{3}$ or, equivalently,
$f_{3}(\textbf{R})=g_{3}(\textbf{R}^{2},R_{3})$ and
$f_{4}(\textbf{R})=V(\textbf{R}^{2},R_{3})$. The latter two equations can be
cosidered as a system of ODE with parameters $\rho$ and $R_{3}$. Solving it,
we obtain

$\displaystyle f_{1}$
$\displaystyle=f_{11}(\rho,R_{3})\cos\varphi+f_{12}(\rho,R_{3})\sin\varphi=\frac{f_{11}(\rho,R_{3})}{\rho}R_{1}+\frac{f_{12}(\rho,R_{3})}{\rho}R_{2},$
$\displaystyle f_{2}$
$\displaystyle=-f_{12}(\rho,R_{3})\cos\varphi{+}f_{11}(\rho,R_{3})\sin\varphi=\frac{-f_{12}(\rho,R_{3})}{\rho}R_{1}{+}\frac{f_{11}(\rho,R_{3})}{\rho}R_{2}.$

Since $\rho=\sqrt{R_{1}^{2}+R_{2}^{2}}$ we get the desired form for the
Hamiltonian $H$. ∎

##  3 Singularities of rank 0

It turns out that equilibria points for a Hamiltonian system on
$\mathrm{e}(3)^{*}$ possessing a linear periodic integral $K$ are exactly the
points where $\operatorname{sgrad}K=0$. This gives the following simple
description for singularities of rank 0 of such integrable Hamiltonian systems
(not necessarily with Hamiltonian of the form (3)).

######  Theorem 1.

The set of singular points of rank $0$ for an integrable Hamiltonian system on
$\mathrm{e}(3)^{*}$ with arbitrary Hamiltonian $H$ possessing the integral
$K=S_{3}$ is the two-dimensional subspace

$\\{(0,0,S_{3},0,0,R_{3})\\}$ (5)

in $\mathrm{e}(3)^{*}$. In particular, for each orbit $M^{4}_{a,g}$ there are
precisely two singular points of rank $0$:

$\Bigl{(}0,0,\pm\frac{g}{\sqrt{a}},0,0,\pm\sqrt{a}\Bigr{)}.$

###### Proof.

The Hamiltonian vector field of a function $f$ on $\mathrm{e}(3)^{*}$ has the
form

$\operatorname{sgrad}f=\Bigl{(}\frac{\partial
f}{\partial\textbf{S}}\times\textbf{S}+\frac{\partial
f}{\partial\textbf{R}}\times\textbf{R},\frac{\partial
f}{\partial\textbf{S}}\times\textbf{R}\Bigr{)},$ (6)

and for the function $K=S_{3}$ we have
$\operatorname{sgrad}K=(-S_{2},S_{1},0,-R_{2},R_{1},0)$. Therefore,
$\operatorname{sgrad}K=0$ exactly at points (5). Thus, points other than (5)
can not be singular points of rank $0$.

Let us prove that $\operatorname{sgrad}H$ vanishes at points (5). The
functions $H$ and $K$ commute with respect to bracket (1), i.e.,
$d_{y}H(\operatorname{sgrad}_{y}K)=0$ for any point $y\in\mathrm{e}(3)^{*}$
(the index $y$ in $d_{y}f$ or $\operatorname{sgrad}_{y}f$ denotes the point at
which the differential or, respectively, skew-gradient of the function $f$ is
taken). Taking the differential of the function
$d_{y}H(\operatorname{sgrad}_{y}K)$ at any point $y=(0,0,S_{3},0,0,R_{3})$, we
get

$A_{K}^{*}(d_{y}H)=0,$ (7)

where $A_{K}$ is the linearization operator for the vector field
$\operatorname{sgrad}K$ at the point $y$, since $\operatorname{sgrad}_{y}K=0$.
The matrix of the operator $A_{K}:\mathrm{e}(3)^{*}\to\mathrm{e}(3)^{*}$ has
the form

$\begin{pmatrix}0&-1&0&0&0&0\\\ 1&0&0&0&0&0\\\ 0&0&0&0&0&0\\\ 0&0&0&0&-1&0\\\
0&0&0&1&0&0\\\ 0&0&0&0&0&0\end{pmatrix}$

and therefore condition (7) implies that $\frac{\partial H}{\partial
S_{1}}=\frac{\partial H}{\partial S_{2}}=\frac{\partial H}{\partial
R_{1}}=\frac{\partial H}{\partial R_{2}}=0$ at any point
$y=(0,0,S_{3},0,0,R_{3})$. Hence $\operatorname{sgrad}H$ vanishes at points
(5), since at a point $y=(0,0,S_{3},0,0,R_{3})$ formula (6) becomes

$\operatorname{sgrad}_{y}f=\Bigl{(}S_{3}\frac{\partial f}{\partial
S_{2}}+R_{3}\frac{\partial f}{\partial R_{2}},-S_{3}\frac{\partial f}{\partial
S_{1}}-R_{3}\frac{\partial f}{\partial R_{1}},0,R_{3}\frac{\partial
f}{\partial S_{2}},-R_{3}\frac{\partial f}{\partial S_{1}},0\Bigr{)}.$

Theorem 1 is proved. ∎

Now, let us state when these zero-rank points are non-degenerate and determine
their type (for more information about non-degeneracy of singular points of a
momentum mapping see [1]).

######  Theorem 2.

For an integrable Hamiltonian system on $\mathrm{e}(3)^{*}$ with arbitrary
Hamiltonian $H$ possessing the integral $K=S_{3}$, the singular point of rank
$0$

$P_{\pm}=\Bigl{(}0,0,\pm\frac{g}{\sqrt{a}},0,0,\pm\sqrt{a}\Bigr{)}$

on the orbit $M^{4}_{a,g}$ is non-degenerate iff $q\neq 0$, where

$\displaystyle q$ $\displaystyle=p^{2}+R_{3}^{2}(H_{11}H_{22}-|H_{12}|^{2}),$
(8) $\displaystyle p$
$\displaystyle=\frac{g}{2R_{3}}\frac{\partial^{2}H}{\partial
S_{1}^{2}}+R_{3}\frac{\partial^{2}H}{\partial S_{1}\partial
R_{1}}-\frac{\partial H}{\partial S_{3}},$ (9)

and

$\begin{gathered}H_{11}=\frac{\partial^{2}H}{\partial S_{1}^{2}},\qquad
H_{12}=\Bigl{(}\frac{\partial^{2}H}{\partial S_{1}\partial
R_{1}}-\frac{1}{R_{3}}\frac{\partial H}{\partial
S_{3}}\Bigr{)}+i\frac{\partial^{2}H}{\partial S_{2}\partial R_{1}},\\\
H_{22}=\frac{\partial^{2}H}{\partial
R_{1}^{2}}+\frac{g}{R_{3}^{3}}\frac{\partial H}{\partial
S_{3}}-\frac{1}{R_{3}}\frac{\partial H}{\partial R_{3}}.\end{gathered}$

Also, if the point $P_{\pm}$ is non-degenerate, then its type is

  1. 1.

center-center if $q>0$,

  2. 2.

focus-focus if $q<0$.

Theorem 2 holds for any Hamiltonian $H$ that commutes (and is functionally
independent) with $K=S_{3}$. For the Hamiltonian $H$ quadratic in S the
condition of non-degeneracy and types of singular points of rank 0 are as
follows.

######  Corollary 1.

For Hamiltonian $\eqref{Eq:H}$ the type of singular points of rank $0$ is
completely determined as in Theorem 2 by

$q=\frac{g^{2}}{4R_{3}^{2}}-R_{3}^{2}g_{1}^{2}(a,R_{3})+gR_{3}\frac{\partial
g_{2}}{\partial R_{3}}(a,R_{3})-g\frac{\partial g_{3}}{\partial
R_{3}}(a,R_{3})-R_{3}\frac{\partial V}{\partial R_{3}}(a,R_{3}).$

###### Proof.

Calculating all expressions from Theorem 2, we have

$\begin{gathered}H_{11}=1,\qquad H_{12}=-\frac{1}{R_{3}}\Bigl{(}\frac{g}{\beta
R_{3}}+g_{3}(a,R_{3})\Bigr{)}-ig_{1}(a,R_{3}),\\\
H_{22}=\frac{g}{R^{3}_{3}}\Bigl{(}\frac{g}{\beta
R_{3}}+g_{3}(a,R_{3})\Bigr{)}-\\\ -\frac{1}{R_{3}}\Bigl{(}g\frac{\partial
g_{2}}{\partial R_{3}}(a,R_{3})+\frac{g}{R_{3}}\frac{\partial g_{3}}{\partial
R_{3}}(a,R_{3})+\frac{\partial V}{\partial
R_{3}}(a,R_{3})\Bigr{)},\end{gathered}$ (10)

and

$p=\frac{g}{R_{3}}\Bigl{(}\frac{1}{2}-\frac{1}{\beta}\Bigr{)}-g_{3}(a,R_{3}).$

Substituting them into (8), one obtains the required formula for $q$. ∎

In order to prove Theorem 2 we use the following criteria of non-degeneracy
(see [1]), which can be regarded as a definition.

######  Definition 1.

A point $P$ of rank $0$ for an integrable Hamiltonian system with Hamiltonian
$H$ and integral $K$ on a symplectic manifold $M^{4}$ is non-degenerate iff
the following two conditions hold:

  * •

the linearizations $A_{H}$ and $A_{K}$ of the Hamiltonian vector fields
$\operatorname{sgrad}H$ and $\operatorname{sgrad}K$ at the point $P$ are
linear independent,

  * •

there exists a linear combination $\lambda A_{H}+\mu A_{K}$ with four
different non-zero eigenvalues.

Let us study the spectrum of linearization of $\operatorname{sgrad}H$ at the
points of rank 0. Taking functions $S_{1},S_{2},R_{1},R_{2}$ as local
coordinates in a neighbourhood of $0$-rank point $P_{\pm}$ on an orbit
$M^{4}_{a,g}$ we have

$R_{3}=\pm\sqrt{a-R_{1}^{2}-R_{2}^{2}},\qquad
S_{3}=\frac{1}{R_{3}}(g-S_{1}R_{1}-S_{2}R_{2}).$

Denote by $\widehat{H}(S_{1},S_{2},R_{1},R_{2})$ the restriction of the
fucntion $H$ onto $M^{4}_{a,g}$.

######  Lemma 1.

For any function $H$ commuting with $K=S_{3}$ the spectrum of the
linearization operator
$A_{\widehat{H}}=\mathrm{Lin}(\operatorname{sgrad}\widehat{H})$ at the
singular points $P_{\pm}$ of rank $0$ has the form
$\sigma(A_{\widehat{H}})=\\{\pm i(p+\sqrt{q}),\pm i(p-\sqrt{q})\\}$, where $p$
and $q$ are given by (9) and (8).

###### Proof.

In the coordinates $S_{1},S_{2},R_{1},R_{2}$ the Poisson bracket on the
symplectic leaf $M^{4}_{a,g}$ has the form

$\mathcal{A}=\left(\begin{matrix}0&S_{3}&0&R_{3}\\\ -S_{3}&0&-R_{3}&0\\\
0&R_{3}&0&0\\\ -R_{3}&0&0&0\end{matrix}\right).$

It is easy to check that the linearization of $\operatorname{sgrad}K$ defines
a complex structure on the tangent space:

$A_{\widehat{K}}=\mathrm{Lin}(\operatorname{sgrad}\widehat{K})=\left(\begin{matrix}0&-1&0&0\\\
1&0&0&0\\\ 0&0&0&-1\\\ 0&0&1&0\end{matrix}\right).$ (11)

Since $[A_{\widehat{H}},A_{\widehat{K}}]=0$, the operator $A_{\widehat{H}}$
can be complexified. The matrix of the Poisson structure can also be
complexified, i.e., we can identify $(2\times 2)$-blocks
$\left(\begin{smallmatrix}\alpha&-\beta\\\
\beta&\alpha\end{smallmatrix}\right)$ in matrices with complex numbers
$\alpha+i\beta$. Thus, in the complex coordinates $S_{1}+iS_{2},R_{1}+iR_{2}$
the matrix $\mathcal{A}$ of the Poisson structure has the form

$\mathcal{A}=\begin{pmatrix}-iS_{3}&-iR_{3}\\\ -iR_{3}&0\end{pmatrix}.$

On a symplectic manifold we have
$A_{\widehat{H}}=\mathcal{A}\,d^{2}\widehat{H}$, and therefore
$d^{2}\widehat{H}$ can also be complexified. By direct calculation we get

$d^{2}\widehat{H}=\begin{pmatrix}H_{11}&H_{12}\\\
\,\overline{\hbox{\vphantom{|}$\\!H\\!$}}\,_{12}&H_{22}\end{pmatrix},$

where $H_{lj}$ are given by formulas (10). The imaginary parts of $H_{11}$ and
$H_{22}$ vanish because $H$ commutes with $K$.

Using the fact that if $\mu_{1},\mu_{2}$ are eigenvalues of a matrix $(A+iB)$
for real matrices $A,B$, then the matrix $\left(\begin{smallmatrix}A&B\\\
-B&A\end{smallmatrix}\right)$ has the eigenvalues
$\mu_{1},\mu_{2},\overline{\mu}_{1},\overline{\mu}_{2}$, we obtain that the
specturm of the (real) operator $A_{\widehat{H}}$ is given by the equation

$\mu^{2}-i(S_{3}H_{11}+R_{3}H_{12}+R_{3}\,\overline{\hbox{\vphantom{|}$\\!H\\!$}}\,_{12})\mu+R_{3}^{2}(H_{11}H_{22}-|H_{12}|^{2})=0,$

which solutions give the desired spectrum. Lemma 1 is proved. ∎

######  Remark 2.

It is clear from (11) that for the integral $K=S_{3}$ the spectrum of the
corresponding operator $A_{\widehat{K}}$ is
$\sigma(A_{\widehat{K}})=\\{i,-i,i,-i\\}$. This doesn’t immediately prove non-
deneracy of points but shows that non-degenerate points can be only of center-
center or focus-focus type.

###### Proof of Theorem 2.

Using Lemma 1 and Definition 1 of non-degeneracy we get the condition of the
theorem in all cases except for $q=0$ or $p^{2}=q$.

If $q=0$, then the spectra of $A_{\widehat{H}}$ and $A_{\widehat{K}}$ are
proportional, thus the point is degenerate (this is precisely the moment when
the image of a focus-focus point meets an arc of the bifurcation diagram while
transforming into a center-center point).

If $p^{2}=q$, then the point is non-degenerate, and one should just take
another linear combination with different eigenvalues (such a linear
combination exists since the spectra of $A_{\widehat{H}}$ and
$A_{\widehat{K}}$ are non-proportional). ∎

##  4 Bifurcation diagrams

In order to construct the bifurcation diagram let us describe all critical
points of the momentum mapping. The singular points of rank $0$ are found in
Section 3. Thus, it remains to describe only singular points of rank $1$. The
next two lemmas show that we can use some convenient coordinates for
investigating them.

######  Lemma 2.

For a Hamiltonian system with Hamiltonian $H$ of the form (4) and integral
$K=S_{3}$, the subspace $\\{(\textbf{S},\textbf{R})\mid R_{1}=R_{2}=0\\}$ in
$\mathrm{e}(3)^{*}$ does not contain points of rank $1$.

###### Proof.

Since we know all singular points of rank 0 (they are points with
$R_{1}=R_{2}=S_{1}=S_{2}=0$; see Theorem 1), it suffices to prove that if
$y=(S_{1},S_{2},S_{3},0,0,R_{3})\in\mathrm{e}(3)^{*}$ is a singular point,
then its coordinates $S_{1}$ and $S_{2}$ vanish. Suppose that this is not the
case. Then $\operatorname{sgrad}_{y}K=(-S_{2},S_{1},0,0,0,0)\neq 0$ and,
therefore, $\operatorname{sgrad}_{y}H=\lambda\operatorname{sgrad}_{y}K$ for a
certain $\lambda$. Hence, by formula (6) (taking into account that $R_{3}\neq
0$), we have $\frac{\partial H}{\partial S_{1}}=\frac{\partial H}{\partial
S_{2}}=0$ at the point $y$. But for a Hamiltonian of the form (4) this is
possible only if $S_{1}=S_{2}=0$ for the point $y$. ∎

Now, since we can assume that $R_{1}^{2}+R_{2}^{2}\neq 0$, we choose new
coordinates on the remaining set of points
$U=\mathbb{R}^{6}(\textbf{S},\textbf{R})\setminus\\{R_{1}=R_{2}=0\\}$. Note
that the set $U$ is homeomorphic to $\mathbb{R}^{5}\times S^{1}$.

######  Lemma 3.

Formulas

$\begin{gathered}S_{1}=\frac{(g-kx)\cos\varphi+m\sin\varphi}{\sqrt{a-x^{2}}},\quad
S_{2}=\frac{(g-kx)\sin\varphi-m\cos\varphi}{\sqrt{a-x^{2}}},\\\ S_{3}=k,\quad
R_{1}=\sqrt{a-x^{2}}\,\cos\varphi,\quad
R_{2}=\sqrt{a-x^{2}}\,\sin\varphi,\quad R_{3}=x\end{gathered}$ (12)

define regular coordinates $(x,m,\varphi,k,a,g)$ on the set $U$, where
$x^{2}<a$ and $\varphi$ is an angular coordinate, i.e., is defined modulo
$2\pi$.

The inverse change of variables on the set $U$, i.e., the expression of
$(x,m,\varphi,k,a,g)$ through $(\textbf{S},\textbf{R})$ is as follows:

$\begin{gathered}x=R_{3},\quad
m=M(\textbf{S},\textbf{R})=S_{1}R_{2}-S_{2}R_{1},\quad\varphi=\arg(R_{1}+iR_{2}),\\\
k=S_{3},\quad
a=F_{1}(\textbf{S},\textbf{R})=\langle\textbf{R},\textbf{R}\rangle,\quad
g=F_{2}(\textbf{S},\textbf{R})=\langle\textbf{S},\textbf{R}\rangle.\end{gathered}$

###### Proof.

By direct calculation, it is easy to check that given formulas define a
bijection and that the Jacobian does not vanish on $U$:

$\det\frac{\partial(x,m,\varphi,k,a,g)}{\partial(S_{1},S_{2},S_{3},R_{1},R_{2},R_{3})}=2(R_{1}^{2}+R_{2}^{2})\neq
0.$

∎

Substituting expressions (12) into (4), we obtain that the Hamiltonian in the
coordinates $(x,m,\varphi,k,a,g)$ on the set $U$ has the form

$H=\frac{(g{-}kx)^{2}{+}m^{2}}{2(a-x^{2})}+\frac{k^{2}}{2\beta}+g_{1}(a,x)m+g_{2}(a,x)g+g_{3}(a,x)k+V(a,x).$
(13)

Futher we will often write $g_{1},g_{2},g_{3},V$ without arguments assuming
that they are functions of $a$ and $x$.

The next statement describes the set of singular points of rank 1.

######  Theorem 3.

The set of all singular points of rank $1$ for the system with Hamiltonian (4)
and integral $K=S_{3}$ on $\mathrm{e}(3)^{*}$ is given by the following two
equations in the coordinates $(x,m,\varphi,k,a,g)$:

$\displaystyle m=-(a-x^{2})g_{1},$ (14)
$\displaystyle\frac{(kx{-}g)(ka{-}gx)}{(a-x^{2})^{2}}+xg_{1}^{2}-(a{-}x^{2})g_{1}\frac{\partial
g_{1}}{\partial x}+g\frac{\partial g_{2}}{\partial x}+k\frac{\partial
g_{3}}{\partial x}+\frac{\partial V}{\partial x}=0.$ (15)

###### Proof.

Calculating the matrix of the Poisson bracket in the coordinates
$(x,m,\varphi,k,a,g)$, one obtains

$\begin{pmatrix}0&a-x^{2}&0&0&0&0\\\ x^{2}-a&0&0&0&0&0\\\ 0&0&0&1&0&0\\\
0&0&-1&0&0&0\\\ 0&0&0&0&0&0\\\ 0&0&0&0&0&0\end{pmatrix}.$

Therefore, in these coordinates the skew-gradients of $H$ and $K$ are

$\displaystyle\operatorname{sgrad}H$
$\displaystyle=\Bigl{(}(a-x^{2})\frac{\partial H}{\partial
m},(x^{2}-a)\frac{\partial H}{\partial x},\frac{\partial H}{\partial
k},0,0,0\Bigr{)},$ (16) $\displaystyle\operatorname{sgrad}K$
$\displaystyle=(0,0,1,0,0,0).$

Here we take into account that $\frac{\partial
H}{\partial\varphi}=\\{H,K\\}\equiv 0$.

Thus the condition of linear dependence of $\operatorname{sgrad}H$ and
$\operatorname{sgrad}K$ at a point $y\in\mathrm{e}(3)^{*}$

$\operatorname{sgrad}H=\lambda\operatorname{sgrad}K$

is equivalent to the conditions

$\frac{\partial H}{\partial m}=0,\qquad\frac{\partial H}{\partial
x}=0,\qquad\frac{\partial H}{\partial k}=\lambda$ (17)

at the point $y$. Differentiating Hamiltonian (13) with respect to $m$ and
$x$, we see that $\frac{\partial H}{\partial m}=0$ is equivalent to (14) and
$\frac{\partial H}{\partial x}=0$ is equivalent to (15) after the substitution
of $m$ from (14). ∎

######  Corollary 2.

On each orbit $M^{4}_{a,g}$ the set of singular points of rank $1$ form a one-
parameter family of critical circles, which is parametrized by points $(k,x)$
of curves defined by equation (15). For each point $(k,x)$ satisfying (15) the
corresponding critical circle in $M^{4}_{a,g}$ is given by the formulas

$\begin{gathered}S_{1}{=}\frac{(g{-}kx)\cos\varphi{-}(a{-}x^{2})g_{1}\sin\varphi}{\sqrt{a-x^{2}}},\quad
S_{2}{=}\frac{(g{-}kx)\sin\varphi{+}(a{-}x^{2})g_{1}\cos\varphi}{\sqrt{a-x^{2}}},\\\
S_{3}=k,\quad R_{1}=\sqrt{a-x^{2}}\,\cos\varphi,\quad
R_{2}=\sqrt{a-x^{2}}\,\sin\varphi,\quad R_{3}=x,\end{gathered}$

where $\varphi$ is a parameter on the circle.

###### Proof.

As it is shown in the proof of Theorem 3,
$\operatorname{sgrad}K=\frac{\partial}{\partial\varphi}$ in the coordinates
$(x,m,\varphi,k,a,g)$. Therefore, each critical circle is a coordinate line of
the coordinate $\varphi$. Substituting (14) into expressions (12), we obtain
the required formulas. ∎

Now we can describe the bifurcation diagram. For each pair of parameters
$a,g$, where $a>0$, consider the function

$W_{a,g}(k,x)=\frac{(g-kx)^{2}}{2(a-x^{2})}+\frac{k^{2}}{2\beta}-\frac{g^{2}_{1}}{2}(a-x^{2})+g_{2}g+g_{3}k+V,$
(18)

which is an analogue of a reduced potential. Recall that $g_{1},g_{2},g_{3},V$
are functions of $a$ and $x$.

######  Theorem 4.

The bifurcation diagram of the integrable Hamiltonian system with Hamiltonian
$\eqref{Eq:H}$ and the integral $K=S_{3}$ on orbit $\eqref{Eq:LevelSurf}$
consists of the following subsets on the plane $\mathbb{R}^{2}(h,k)$:

$1)$ two points $Z_{\pm}$ (they can coinside if $g=0$) with coordinates

$h=\frac{g^{2}}{2\beta
a}+g\,g_{2}(a,\pm\sqrt{a})\pm\frac{g}{\sqrt{a}}\,g_{3}(a,\pm\sqrt{a})+V(a,\pm\sqrt{a}),\quad
k=\pm\frac{g}{\sqrt{a}},$

which are the images of two singular points of rank $0$;

$2)$ the points $(h(x),k(x))$ which are the images of singular points of rank
$1$ and are parametrized by the parameter $x$, where the function $k(x)$ is
implicitly defined by the quadratic (or linear) equation $\frac{\partial
W_{a,g}}{\partial x}(k,x)=0$, and $h(x)=W_{a,g}(k(x),x)$.

###### Proof.

The first statement immediately follows from Theorem 1 describing singular
points of rank 0. Similarly, the second one follows from Theorem 3 describing
singular points of rank 1 by taking into account expression (13) for the
Hamiltonian $H$ and definition (18) of the function $W_{a,g}$. ∎

######  Remark 3.

For each fixed $a,g$ the equations from Theorem 4

$h=W_{a,g}(k,x),\qquad\frac{\partial W_{a,g}}{\partial x}(k,x)=0$ (19)

describing the image of the set of singular points of rank $1$ belonging to
the orbit $M^{4}_{a,g}$ are exactly the equations for the envelope of the
family of parabolas

$h=\Bigl{(}\frac{x^{2}}{2(a-x^{2})}+\frac{1}{2\beta}\Bigr{)}k^{2}+B_{a,g}(x)k+C_{a,g}(x)$

on the plane $\mathbb{R}^{2}(h,k)$ depending on the parameter $x$, where

$\displaystyle B_{a,g}(x)$ $\displaystyle=g_{3}(a,x)-\frac{gx}{a-x^{2}},$ (20)
$\displaystyle C_{a,g}(x)$
$\displaystyle=\frac{g^{2}}{2(a-x^{2})}-\frac{g^{2}_{1}(a,x)}{2}(a-x^{2})+g_{2}(a,x)g+V(a,x)$

(see formula (18)). In other words, the bifurcation diagram (without points
$Z_{\pm}$) can be regarded as the envelope of this family of parabolas.

The bifurcation diagram $\Sigma$ is the union of $\Sigma_{0}=\\{Z_{\pm}\\}$
and $\Sigma_{1}$ which consists of the images of singular points of rank $1$.
Let us rewrite conditions (19) describing $\Sigma_{1}$ in a more explicit
parametric form.

The relation $\frac{\partial W_{a,g}}{\partial x}(k,x)=0$ from Theorem 4 is
exactly equation (15). In notation (20) it can be written as

$\frac{ax}{(a-x^{2})^{2}}k^{2}+B^{\prime}_{a,g}(x)k+C^{\prime}_{a,g}(x)=0,$
(21)

where

$\displaystyle B^{\prime}_{a,g}(x)$ $\displaystyle=\frac{\partial
g_{3}}{\partial x}-\frac{g(a+x^{2})}{(a-x^{2})^{2}},$ $\displaystyle
C^{\prime}_{a,g}(x)$
$\displaystyle=\frac{g^{2}x}{(a-x^{2})^{2}}+xg_{1}^{2}-(a-x^{2})g_{1}\frac{\partial
g_{1}}{\partial x}+g\frac{\partial g_{2}}{\partial x}+\frac{\partial
V}{\partial x}.$

Equation (21) is quadratic with respect to $k$ for $x\neq 0$ (it is reduced to
linear equation for $x=0$). Its discriminant equals

$\displaystyle D_{a,g}(x)$
$\displaystyle=(B^{\prime}_{a,g}(x))^{2}-\frac{4ax}{(a{-}x^{2})^{2}}C^{\prime}_{a,g}(x)=\frac{1}{(a{-}x^{2})^{2}}\Bigl{(}g-(a{+}x^{2})\frac{\partial
g_{3}}{\partial x}\Bigr{)}^{2}-$
$\displaystyle-\frac{4ax}{(a-x^{2})^{2}}\Bigl{(}xg_{1}^{2}-(a-x^{2})g_{1}\frac{\partial
g_{1}}{\partial x}+g\frac{\partial g_{2}}{\partial x}+x\Bigl{(}\frac{\partial
g_{3}}{\partial x}\Bigr{)}^{2}+\frac{\partial V}{\partial x}\Bigr{)}.$

In order to describe a parametrization of bifurcational curves consider the
set

$\Theta_{a,g}=\\{x\in\mathbb{R}\mid x^{2}<a,\,\,x\neq 0,\,\,D_{a,g}(x)\geq
0\\}.$

Each its (arcwise) connected component is an interval, which is either non-
degenerate (i.e., has a non-zero length) or degenerate (i.e., is a point).
Denote the set of all non-degenerate intervals by $\mathcal{I}_{a,g}$ and
denote the set of degenerate intervals by $\Theta^{0}_{a,g}$. Clearly,
$\Theta_{a,g}\setminus\Theta^{0}_{a,g}=\bigcup_{I\in\mathcal{I}_{a,g}}I$.

Since $\Theta_{a,g}$ is, evidently, a closed subset of
$(-\sqrt{a},0)\cup(0,\sqrt{a})$, intervals from $\mathcal{I}_{a,g}$ contain
their endpoints except for the case when an endpoint is $\pm\sqrt{a}$ or $0$.

Thus, the set $\Sigma_{1}$ in the plane $\mathbb{R}^{2}(h,k)$ contains curves
defined on intervals from $\mathcal{I}_{a,g}$, ‘‘separate’’ points
corresponding to points from $\Theta^{0}_{a,g}$, and, possibly, something else
corresponding to $x=0$. An explicite description of $\Sigma_{1}$ is given in
the following statement.

######  Theorem 5.

The set $\Sigma_{1}$ for the integrable Hamiltonian system with Hamiltonian
$\eqref{Eq:H}$ and the integral $K=S_{3}$ on orbit $\eqref{Eq:LevelSurf}$ is
the union of the following parametric curves and points on the plane
$\mathbb{R}^{2}(h,k)$:

$1)$ the pairs of curves $(h_{\pm}(x),k_{\pm}(x))$, $x\in I$, for each
$I\in\mathcal{I}_{a,g}$, where

$\displaystyle
h_{\pm}(x)=\frac{(g{-}k_{\pm}(x)x)^{2}}{2(a-x^{2})}{+}\frac{k_{\pm}^{2}(x)}{2\beta}\,{-}\,\frac{(a{-}x^{2})g^{2}_{1}}{2}+g_{2}g+g_{3}k_{\pm}(x)+V,\\!\\!\\!$
(22) $\displaystyle
k_{\pm}(x)=\frac{g(a+x^{2})}{2ax}-\frac{(a-x^{2})^{2}}{2ax}\frac{\partial
g_{3}}{\partial x}\pm\frac{(a-x^{2})}{2ax}\times\\!\\!\\!$
$\displaystyle\\!\\!\\!\\!\times$
$\displaystyle\sqrt{\\!\\!\Bigl{(}\\!g{-}(a{+}x^{2})\frac{\partial
g_{3}}{\partial
x}\\!\Bigr{)}^{2}\\!\\!\\!{-}4ax\Bigl{(}\\!xg_{1}^{2}{-}(a{-}x^{2})g_{1}\frac{\partial
g_{1}}{\partial x}{+}g\frac{\partial g_{2}}{\partial
x}{+}x\Bigl{(}\\!\frac{\partial g_{3}}{\partial
x}\\!\Bigr{)}^{2}\\!\\!{+}\frac{\partial V}{\partial x}\\!\Bigr{)}};\\!\\!\\!$

$2)$ the points $(h(x_{0}),k(x_{0}))$ for each $x_{0}\in\Theta^{0}_{a,g}$,
where

$\displaystyle h$
$\displaystyle(x_{0})=\frac{(g{-}k(x_{0})x_{0})^{2}}{2(a-x_{0}^{2})}+\frac{k^{2}(x_{0})}{2\beta}-\frac{(a{-}x_{0}^{2})g^{2}_{1}}{2}+g_{2}g+g_{3}k(x_{0})+V,$
$\displaystyle k$
$\displaystyle(x_{0})=\frac{g(a+x_{0}^{2})}{2ax_{0}}-\frac{(a-x_{0}^{2})^{2}}{2ax_{0}}\frac{\partial
g_{3}}{\partial x}(a,x_{0}),$

and $g_{1},g_{2},g_{3},V$ in these formulas mean the values of the
corresponding functions at the point $(a,x_{0})$;

$3)$ for the orbits $M^{4}_{a,g}$, where $g\neq a\frac{\partial
g_{3}}{\partial x}(a,0)$, the point $(h_{0},k_{0})$, where

$\displaystyle h_{0}$
$\displaystyle=\frac{g^{2}}{2a}+\frac{k_{0}^{2}}{2\beta}-\frac{ag_{1}^{2}(a,0)}{2}+g_{2}(a,0)g+g_{3}(a,0)k_{0}+V(a,0),$
$\displaystyle k_{0}$ $\displaystyle=\frac{ag_{1}(a,0)\frac{\partial
g_{1}}{\partial x}(a,0)-g\frac{\partial g_{2}}{\partial x}(a,0)-\frac{\partial
V}{\partial x}(a,0)}{\frac{\partial g_{3}}{\partial x}(a,0)-\frac{g}{a}};$

$4)$ for the orbits $M^{4}_{a,g}$, where $g=a\frac{\partial g_{3}}{\partial
x}(a,0)$ and $a$ satisfies the relation

$ag_{1}(a,0)\frac{\partial g_{1}}{\partial x}(a,0)-a\frac{\partial
g_{3}}{\partial x}(a,0)\frac{\partial g_{2}}{\partial x}(a,0)-\frac{\partial
V}{\partial x}(a,0)=0,$

the parabola

$h=\frac{k^{2}}{2\beta}{+}g_{3}(a,0)k{+}\frac{a}{2}\Bigl{(}\\!\frac{\partial
g_{3}}{\partial
x}(a,0)\\!\Bigr{)}^{2}\\!\\!{-}\frac{a}{2}g_{1}(a,0){+}a\frac{\partial
g_{3}}{\partial x}(a,0)g_{2}(a,0){+}V(a,0).$

###### Proof.

All formulas in cases 1)–4) follow from equations (19) and expression (18).
The cases 1) and 2) correspond to solutions of quadratic equation (21) for
each parameters $x$ from $\Theta_{a,g}$, but in the case 2), when
$x\in\Theta^{0}_{a,g}$, the corresponding discriminant $D_{a,g}(x)$ vanishes,
since $D_{a,g}$ is a continuous function on $(-\sqrt{a},\sqrt{a})$.

The case 3) corresponds to $x=0$ in equation (21). If
$B^{\prime}_{a,g}(0)=\frac{\partial g_{3}}{\partial x}(a,0)-\frac{g}{a}\neq
0$, then $-C^{\prime}_{a,g}(0)/B^{\prime}_{a,g}(0)$ is the unique solution
$k_{0}$ of linear equation (21) for $x=0$, and we obtain the point
$(h_{0},k_{0})$ in the case 3). Note that if $B^{\prime}_{a,g}(0)\neq 0$, then
the discriminant $D_{a,g}(x)$ is positive on some interval
$(-\varepsilon,\varepsilon)$ and there are two bifurcational curves (22)
defined on $(-\varepsilon,0)$ and $(0,\varepsilon)$ which tend to the point
$(h_{0},k_{0})$ as $x\to 0$ and form one smooth bifurcational curve glued from
two curves at this point.

The case 4) also corresponds to $x=0$, but the conditions on $g$ and $a$ in
the case 4) are equivalent to the conditions
$B^{\prime}_{a,g}(0)=C^{\prime}_{a,g}(0)=0$, which imply that an arbitrary $k$
is a solution of (21) for $x=0$. Thus, we obtain the required parabola in the
case 4). ∎

Note that for arbitrary functions $g_{1},g_{2},g_{3},V$ the behavior of
bifurcational curves described in Theorem 5 by explicit formulas can be fairly
complicated. They can have many cusps, intersect one another or coincide on
some their arcs. Some general properties concerning the behavior of
bifurcational curves are described in the following statement.

######  Corollary 3.

$1)$ If $J\subset\Theta_{a,g}$ is an open interval such that $D_{a,g}|_{J}>0$,
then the bifurcational curve $(h_{\pm}(x),k_{\pm}(x))$ defined on $J$ by
formulas (22) is a smooth parametric curve which is regular for all $x$, where
$\frac{dk_{\pm}}{dx}(x)\neq 0$.

$2)$ Exactly two arcs of the bifurcational curves described in the items $1)$
and $4)$ of Theorem 5 tend to infinity such that
$h(k)\sim\frac{k^{2}}{2\beta}$ (one arc for $k\to+\infty$ and one arc for
$k\to-\infty$). For the curves defined by formulas (22) these arcs correspond
to $x\to 0$.

$3)$ For each singular point $P_{\pm}$ of rank $0$ which is of center-center
type (by Theorem 2 there can be $0$, $1$, or $2$ such points) there are
exactly two arcs of the bifurcational curves described by formulas (22) which
tend to the corresponding point $Z_{\pm}$ described in Theorem 4 as
$x\to\pm\sqrt{a}$.

###### Proof.

Since $h=h_{\pm}(x)$, $k=k_{\pm}(x)$ satisfy equations (19), we have

$\frac{dh_{\pm}}{dx}(x)=\frac{\partial W_{a,g}}{\partial
k}(k_{\pm}(x),x)\frac{dk_{\pm}}{dx}(x).$

Therefore, the parametric curve (22) is regular iff
$\frac{dk_{\pm}}{dx}(x)\neq 0$ and can have singularities (for example, cusps)
only at points, where $\frac{dk_{\pm}}{dx}=0$.

Items 2) and 3) follow from formulas (22) by investigating the behavior of the
parametric curves $(h_{\pm}(x),k_{\pm}(x))$ as $x$ tends to $0$ or
$\pm\sqrt{a}$. Note that $D_{a,g}$ is positive in a neighborhood of the points
$\pm\sqrt{a}$ iff $q$ from Corollary 1 is positive for $R_{3}=\pm\sqrt{a}$. ∎

##  5 Liouville tori bifurcations

All basic definitions and facts about Liouville tori bifurcations can be found
in [1].

######  Theorem 6.

A singular point of rank $1$ (described in Theorem 3 and Corollary 2) is non-
degenerate iff $\frac{\partial^{2}W_{a,g}(k,x)}{\partial x^{2}}\neq 0$, where
$W_{a,g}(k,x)$ is given by (18). Moreover,

  * •

if $\frac{\partial^{2}W_{a,g}(k,x)}{\partial x^{2}}>0$, then the type of the
point is elliptic;

  * •

if $\frac{\partial^{2}W_{a,g}(k,x)}{\partial x^{2}}<0$, then the type of the
point is hyperbolic.

The non-degeneracy and the type of a singular point $y$ of rank $1$ are
completely determined by the spectrum of linearization of the Hamiltonian
vector field which is a (non-trivial) linear combination of
$\operatorname{sgrad}H$ and $\operatorname{sgrad}K$ vanishing at $y$. Thus,
Theorem 6 follows from the following statement.

######  Lemma 4.

Each point $y$ of rank $1$ (described in Theorem 3 and Corollary 2) is a
singular point for the vector field $\operatorname{sgrad}F_{y}$, where
$F_{y}=H-\lambda K$ and $\lambda=\frac{\partial H}{\partial k}\big{|}_{y}$.
The spectrum of the linearization
$A_{F_{y}}=\mathrm{Lin}(\operatorname{sgrad}F_{y})$ at the point $y$ consists
of $4$ zeroes and

$\mu_{\pm}=\pm i\sqrt{\frac{\partial^{2}W_{a,g}(k,x)}{\partial x^{2}}}.$

###### Proof.

The proof is by direct calculation. The Hamiltonian vector fields
$\operatorname{sgrad}H$ and $\operatorname{sgrad}K$ in the coordinates
$(x,m,\varphi,k,a,g)$ from Lemma 3 are given by (16), and at a point
$y\in\mathrm{e}(3)^{*}$ of rank 1 conditions (17) are fulfilled. Hence for the
function $F_{y}=H-\lambda K$, where $\lambda=\frac{\partial H}{\partial
k}\big{|}_{y}$, we have $\operatorname{sgrad}_{y}F_{y}=0$, and therefore the
linearization $A_{F_{y}}$ of the field

$\operatorname{sgrad}F_{y}=\Bigl{(}(a-x^{2})\frac{\partial H}{\partial
m},-(a-x^{2})\frac{\partial H}{\partial x},\frac{\partial H}{\partial
k}-\lambda,0,0,0\Bigr{)}$

at the point $y$ is well-defined. Taking into account conditions (17), we get
the following equation for the spectrum of $A_{F_{y}}$:

$\det(A_{F_{y}}-\mu\,\mathrm{Id})=\mu^{4}(a-x^{2})^{2}\det\left(\begin{matrix}\frac{\partial^{2}H}{\partial
m\partial x}-\mu&\frac{\partial^{2}H}{\partial m^{2}}\\\
-\frac{\partial^{2}H}{\partial x^{2}}&-\frac{\partial^{2}H}{\partial x\partial
m}-\mu\end{matrix}\right)=0.$

Thus the non-zero eigenvalues of $A_{F_{y}}$ are

$\mu_{\pm}=\pm\sqrt{\Bigl{(}\frac{\partial^{2}H}{\partial x\partial
m}\Bigr{)}^{2}-\frac{\partial^{2}H}{\partial
x^{2}}\frac{\partial^{2}H}{\partial m^{2}}}.$ (23)

For the function $H$ given by (13) we have

$\begin{gathered}\frac{\partial^{2}H}{\partial
m^{2}}=\frac{1}{a-x^{2}},\qquad\frac{\partial^{2}H}{\partial x\partial
m}=\frac{\partial g_{1}}{\partial x}(a,x)+\frac{2mx}{(a-x^{2})^{2}},\\\
\frac{\partial^{2}H}{\partial
x^{2}}=\frac{(g^{2}+ak^{2}+m^{2})(a+3x^{2})-2gkx(x^{2}+3a)}{(a-x^{2})^{3}}+\\\
+m\frac{\partial^{2}g_{1}}{\partial
x^{2}}(a,x)+g\frac{\partial^{2}g_{2}}{\partial
x^{2}}(a,x)+k\frac{\partial^{2}g_{3}}{\partial
x^{2}}(a,x)+\frac{\partial^{2}V}{\partial x^{2}}(a,x).\\\ \end{gathered}$ (24)

Since, by Theorem 3, at a singular point we have $m=-(a-x^{2})g_{1}(a,x)$,
equalities (24) can be rewritten as

$\begin{gathered}\frac{\partial^{2}H}{\partial
m^{2}}=\frac{1}{a-x^{2}},\qquad\frac{\partial^{2}H}{\partial x\partial
m}=\frac{\partial g_{1}}{\partial x}(a,x)-\frac{2xg_{1}(a,x)}{a-x^{2}},\\\
\frac{\partial^{2}H}{\partial x^{2}}=\frac{\partial^{2}W_{a,g}(k,x)}{\partial
x^{2}}+(a-x^{2})\Bigl{(}\frac{\partial g_{1}}{\partial
x}(a,x)-\frac{2xg_{1}(a,x)}{a-x^{2}}\Bigr{)}^{2},\\\ \end{gathered}$ (25)

where $W_{a,g}(k,x)$ is given by (18). Substituting expressions (25) into
formula (23) we get the desired expression for $\mu_{\pm}$.

Lemma 4 and, consequently, Theorem 6 are proved. ∎

######  Theorem 7.

The only possible non-degenerate Liouville tori bifurcations for the isoenergy
surfaces $Q^{3}$ of the integrable Hamiltonian system with Hamiltonian
$\eqref{Eq:H}$ and the integral $K=S_{3}$ on orbit $\eqref{Eq:LevelSurf}$ are
the so-called $A$ and $V_{k}$ bifurcations. In particular, if there is only
one singular circle in a fiber, then the bifurcation is either $A$ or $B$.

###### Proof.

There is only one elliptic bifurcation (of type $A$), thus we consider
hyperbolic bifurcations. Since all critical points of rank $1$ satisfy the
condition $R_{1}^{2}+R_{2}^{2}\not=0$, we can work in the coordinates
$(x,m,\varphi,k,a,g)$.

Consider the inverse image of a point $(h_{0},k_{0})$ under the momentum
mapping $M^{4}_{a,g}\to\mathbb{R}^{2}(h,k)$. Then $\varphi$ is arbitrary and
$m$ is given by

$\ \frac{(m+(a-x^{2})g_{1}(a,x))^{2}}{2(a-x^{2})}=h_{0}-W_{a,g}(k_{0},x),$
(26)

where $x$ satisfies the condition $h_{0}\geq W_{a,g}(k_{0},x)$.

Thus any connected component of a singular fiber for a non-degenerate
singularity is a product of $S^{1}$ and a wedge sum of $k$ circles as in
Figure 1.

Figure 1: Atom $V_{k}$.

More precisely, the set in the plane $(m,x)$ given by equation (26) is
homeomorphic to the union of circles that are joined at the points
$h_{0}=W_{a,g}(k_{0},x)$.

Since the singularity is non-degenerate, this is precisely the bifurcation for
the $V_{k}$ atom. Theorem 7 is proved. ∎

##  6 Isoenergy surfaces

For a Hamiltonian function $H$ on $\mathrm{e}(3)^{*}$ which is a positive
definite quadratic form in S, the topology of isoenergy surfaces is completely
determined by their projections on the Poisson shere (for details see [1]). By
Theorem 2, the projection is invariant under rotation around the $R_{3}$-axis.
As a direct consequence we get the following statement.

######  Theorem 8.

Any isoenergy surface $Q^{3}$ of the integrable Hamiltonian system with
Hamiltonian $\eqref{Eq:H}$ and the integral $K=S_{3}$ on orbit
$\eqref{Eq:LevelSurf}$ is either $\mathbb{RP}^{3}$ or a disjoint union of $k$
products $S^{1}\times S^{2}$ and not more than two spheres $S^{3}$.

###### Proof.

If the projection of $Q^{3}$ on the Poisson sphere is surjective, then
$Q^{3}=\mathbb{RP}^{3}$. Otherwise the image of the projection is the unioun
of $l$ rings and not more than two disks with centers in the poles
$\textbf{R}=(0,0,R_{3})$. Each ring corresponds to $S^{1}\times S^{2}$ and
each disk to $S^{3}$. ∎

### ACKNOWLEDGMENTS

This work was supported by the Russian Science Foundation, project no.
17-11-01303.

## References

  * [1] A.V. Bolsinov and A.T. Fomenko, Integrable Hamiltonian Systems: Geometry, Topology, Classification (CRC, Boca Raton, FL, 2004).
  * [2] A.V. Borisov and I.S. Mamaev, Rigid body dynamics (NIC Regular Chaotic Dynamics,Moscow, Izhevsk, 2001) [in Russian].
  * [3] A. A. Oshemkov, ‘‘Fomenko invariants for the main integrable cases of the rigid body motion equations’’, in Topological Classification of Integrable Systems, Adv. Sov. Math. 6, 67146 (1991).
  * [4] A. V. Bolsinov, A. M. Izosimov, A. Yu. Konjaev, and A. A. Oshemkov, ‘‘Algebra and topology of integrable systems. Research problems’’, Tr. Sem. Vektor. Tenzor. Anal. 28, 119191 (2012).
  * [5] E.O. Kantonistova, ‘‘Topological classification of integrable Hamiltonian systems in a potential field on surfaces of revolution’’, Mathematics, 207, 358399 (2016).
  * [6] B.S. Kruglikov, Topological classification of Leggett systems in an integrable case for 3He-A, Uspekhi Mat. Nauk, 46: 179181 (1991).
  * [7] M.Yu. Ivochkin, Mat. Sb., 199:6 (2008), 85104 Topological analysis of the motion of an ellipsoid on a smooth plane, Mat. Sb., 199, 871890 (2008).