Geometrical Aspects of the Hamiltonization Problem
of Dynamical Systems Geometrical Aspects of the Hamiltonization Problem of
Dynamical Systems

Misael Avendaño-Camacho ${}^{1^{\dagger}}$, Claudio César García-Mendoza
${}^{2^{\dagger}}$, José Crispín Ruíz-Pantaleón ${}^{3^{\ddagger}}$ and
Eduardo Velasco-Barreras ${}^{4^{\dagger}\S}$ Avendaño-Camacho, García-
Mendoza, Ruíz-Pantaleón and Velasco-Barreras

† Departamento de Matemáticas, Universidad de Sonora, México ‡ Instituto de
Matemáticas, Universidad Nacional Autónoma de México, México § Instituto de
Matemática e Estatística, Universidade Federal Fluminense 1
<EMAIL_ADDRESS>2<EMAIL_ADDRESS>3<EMAIL_ADDRESS>4
<EMAIL_ADDRESS>
Some geometric solutions to the problem of endowing a dynamical system with a
Hamiltonian formulation are presented within the class of Poisson structures.
We address this problem on orientable manifolds, and by using decomposable
Poisson structures. In the first case, the existence of a Hamiltonian
formulation is ensured under the vanishing of some topological obstructions,
improving a result of Gao. In the second case, we apply a variant of the
Hojman construction to solve the problem for vector fields admitting a
transversally invariant metric and, in particular, infinitesimal generators of
proper actions. Finally, we also consider the hamiltonization problem for Lie
group actions, and give solutions in the particular case in which the acting
Lie group is a low-dimensional torus.

Hamiltonian formulation, Poisson manifold, first integral, unimodularity,
transversally invariant metric, symmetry

37J06, 37J39, 53D17, 37C86, 70G45, 37C79

##  1 Introduction

The hamiltonization problem on a smooth manifold is the question of whether a
dynamical system admits a Hamiltonian formulation. More precisely, given a
vector field $X$ on a smooth manifold $M$, the problem consists in finding a
scalar function $h$ and a Poisson structure $\pi$ on $M$ such that

$X=\pi(\mathrm{d}h,\cdot).$ (1.1)

In this case, the vector field $X$ is said to be Hamiltonian for the Poisson
manifold $(M,\pi)$.

The problem of finding a Poisson structure with respect to which a flow of
ODEs, or a vector field, is Hamiltonian has received considerable attention
from several decades ago. At the earlier stages, this problem concerned to the
existence of a Hamiltonian formulation of a set of ODEs and their explicit
formulation, mainly in two and three dimensions [3, 10, 11, 12, 23]. Some
geometric approaches to the hamiltonization problem can be found in the works
of Whittaker [23], Perlick [19], S. Hojman [14, 15] and R. Alvarado Flores et.
al. [1]. A solution to this problem is given by the Lie-Königs theorem [23],
which requires the solvability of the ODEs. V. Perlick gives a generalization
of the Lie–Konigs theorem in a more global setting, but only when the original
system is even–dimensional, so that a time–dependent symplectic structure
exists [19]. S. Hojman solves the hamiltonization problem of a vector field by
constructing a Poisson structure, provided that it admits an infinitesimal
symmetry and a first integral [15]. A geometric reformulation of Hojman
results is presented in [1], where the hamiltonization problem is solved for a
vector field being in a two–dimensional Lie subalgebra and also admitting a
first integral. In recent works, the hamiltonization problem has also been
considered, both theoretically [17] and in applied contexts [2].

In this paper, we present a geometric approach to the hamiltonization problem.
In particular, we give solutions to the hamiltonization problem for vector
fields on orientable manifolds with sufficient first integrals, and also for
vector fields admitting first integrals that are not necessarily regular. In
the latter case, this improves the Hojman construction and the approach of
[1]. Furthermore, we provide solutions to the hamiltonization problem for Lie
group actions in some particular cases. Here, we have coined the term
_hamiltonizable_ to refer to a vector field for which it is possible to solve
the hamiltonization problem.

Our first result deals with vector fields on orientable manifolds admitting
the maximum number of functionally independent first integrals. In Theorem
3.1, we prove, under these assumptions, that the vector field is
hamiltonizable for as many Poisson structures as first integrals. This is a
generalization of the results presented in [10]. Moreover, we also improve the
previous result by relaxing the hypothesis on the number of first integrals:
if a vector field $X$ on an $m$–dimensional manifold admits $m-2$ first
integrals and an invariant volume form, then $X$ is hamiltonizable under some
suitable topological condition on the common level sets of the first integrals
(Theorem 3.3). This theorem in the three–dimensional case recovers the
hamiltonization criteria given by Gao in [9] for Lotka–Volterra systems; but
the topological hypotheses of our theorem were obviated there. These
hypotheses are necessary in general as we exhibit it in Example 3.10. Theorem
3.17 deals with the hamiltonization problem for vector fields tangent to a
regular oriented foliation of dimension $r$: if a vector field preserves the
leaf–wise volume and belongs to the kernel of a nowhere vanishing leaf–wise
closed $(r-2)$–form, then the hamiltonization problem reduces to the
triviality of the foliated de Rham cohomology in degree one for the foliation
integrating the kernel of the closed form. On the other hand, by considering a
correspondence between multivector fields and differential forms on orientable
manifolds, we have formulated a criterion of hamiltonization by unimodular
Poisson structures in terms of the existence of integrating factors of
primitives (Theorem 3.13). In this case, the given vector field is a modular
vector field.

We also present various results on the hamiltonization problem where
orientability is no longer required. Instead, we aim for decomposable Poisson
structures. More precisely, we state conditions under which a given vector
field $X$ with first integral $h$ on a manifold $M$ is hamiltonizable by a
Poisson structure of the form $\pi=Y\wedge X$, for some vector field $Y$ on
$M$. In particular, Theorem 4.2 recovers the results of [1] for the case when
the first integral $h$ is regular. We also present geometric setups in which
Theorem 4.2 is applicable: in Theorem 5.1 we have shown that a vector field is
hamiltonizable under suitable conditions involving transversally invariant
Riemannian metrics. At the same time, we apply Theorem 5.1 to the case of
proper actions of 1–dimensional Lie groups. It turns out that if the
infinitesimal generator is complete, then it is hamiltonizable if and only if
the orbit space is not compact (Proposition 5.5). Moreover, in Proposition
5.12 we adapt Theorem 5.1 to obtain a hamiltonization criteria for
infinitesimal generators of proper actions of general Lie groups.

Finally, we consider the hamiltonization problem of Lie group actions: given a
Lie group action on a manifold $M$, whether a Poisson structure on $M$ exists
with respect to which the action is Hamiltonian. We want to comment that we
have not found in the literature any formulation nor results concerning to the
hamiltonization problem of Lie group actions. For the abelian Lie group
$\mathbb{T}^{k}$, we state some appropriate hypothesis under which there
exists a bivector field $\pi$ on $M$ such that the infinitesimal generators
are of the form (1.1). If $\pi$ is Poisson, then the $\mathbb{T}^{k}$–action
is Hamiltonian. In Theorem 6.12 we provide sufficient conditions to solve the
hamiltonization problem for 2–dimensional torus actions. The one–dimensional
case is the content of Proposition 5.7.

##  2 Preliminaries

A _Poisson structure_ on a smooth manifold $M$ is a bivector field
$\pi\in\Gamma(\wedge^{2}TM)$ satisfying the Jacobi identity

$[\pi,\pi]=0.$

Here, the bracket $[-,-]$ stands for the Schouten–Nijenhuis bracket for
multivector fields [7, Section 1.8]. The pair $(M,\pi)$ is called a _Poisson
manifold_.

The _rank_ of a Poisson structure $\pi$ at a point $x\in M$ is defined by
$\operatorname{rank}\pi_{x}:=\dim\pi^{\sharp}(T^{\ast}_{x}M)$, where the
vector bundle map $\pi^{\sharp}:T^{*}M\to TM$ is given by the usual
contraction $\pi^{\sharp}\alpha:=\mathbf{i}_{\alpha}\pi$, for all $\alpha\in
T^{\ast}M$. A _singular point_ $x$ of $\pi$ is characterized by the condition
that the rank of $\pi$ is not constant around $x$. Otherwise, the point $x$ is
said to be a _regular point_ of $\pi$.

The image $\pi^{\sharp}(T^{\ast}M)$ is an integrable distribution on $M$,
called the _characteristic distribution_ of $\pi$, and each integral
submanifold carries a symplectic structure canonically induced by the Poisson
structure $\pi$ on $M$. Consequently, the integral symplectic submanifolds
define a smooth symplectic foliation of $M$, which may be singular in general.

A vector field $X$ on a Poisson manifold $(M,\pi)$ is _tangent_ to the
symplectic foliation if $X_{x}\in\pi^{\sharp}(T^{*}_{x}M)$, for all $x\in M$.
Note that for each $\alpha\in\Gamma(T^{*}M)$ the vector field
$X=\pi^{\sharp}\alpha$ is tangent to the symplectic foliation. In particular,
if $\alpha=\mathrm{d}{h}$, for some $h\in C^{\infty}(M)$, then $X$ is called
_Hamiltonian vector field_. In this case, the function $h$ is said to be a
_Hamiltonian function_ for $X$. We denote by $\mathfrak{ham}(M,\pi)$ the Lie
algebra of all Hamiltonian vector fields on $M$.

The _Casimir functions_ of $\pi$ are the Hamiltonian functions $c\in
C^{\infty}(M)$ of the zero vector field, that is,
$\pi^{\sharp}\mathrm{d}{c}=0$. These functions are constant along the leaves
of the symplectic foliation of $M$. Thus, every tangent vector field is also
tangent to the level sets of every Casimir function. In particular, this is
true for Hamiltonian vector fields.

The infinitesimal Poisson automorphisms of $\pi$, or _Poisson vector fields_ ,
for short, are the vector fields $X\in\Gamma(TM)$ such that
$\mathrm{L}_{X}\pi=0$. By $\mathfrak{poiss}(M,\pi)$ we denote the Lie algebra
of Poisson vector fields, and the Lie subalgebra of tangent Poisson vector
fields by $\mathfrak{poiss}_{\mathrm{tan}}(M,\pi)$. We recall that the Lie
algebra $\mathfrak{ham}(M,\pi)$ is an ideal of both $\mathfrak{poiss}(M,\pi)$
and $\mathfrak{poiss}_{\mathrm{tan}}(M,\pi)$.

Now, we observe that if $\pi$ is a Poisson structure and $f$ is an arbitrary
function on $M$, then $f\pi$ is not a Poisson structure, in general. We say
that $\pi$ is _conformally invariant_ if $f\pi$ is again a Poisson structure
on $M$ for any $f\in C^{\infty}(M)$. A simple computation shows that

$[f\pi,f\pi]=-2f\,\pi^{\sharp}\mathrm{d}{f}\wedge\pi.$

######  Lemma 2.1.

Every Poisson structure with _rank at most two_ is conformally invariant.

This follows from the fact that the 3–vector field
$\pi^{\sharp}\mathrm{d}{f}\wedge\pi$ is a section of the bundle
$\wedge^{3}(\pi^{\sharp}(T^{*}M))$, which is zero due to our rank hypothesis.
Finally, note that for every Poisson structure $\pi$, the bivector field
$f\pi$ is Poisson if $f$ is a Casimir function of $\pi$.

#### Tangential Poisson Cohomology.

Associated with a Poisson structure $\pi$ on $M$, there is a coboundary
operator
$\mathrm{d}_{\pi}:\Gamma(\wedge^{\bullet}TM)\to\Gamma(\wedge^{\bullet+1}TM)$
given by $\mathrm{d}_{\pi}A:=[\pi,A]$, for $A\in\Gamma(\wedge^{\bullet}T{M})$
[18]. The pair $(\Gamma(\wedge^{\bullet}TM),\mathrm{d}_{\pi})$ is the
_Lichnerowicz–Poisson_ complex. Its cohomology, denoted by
$H^{\bullet}(M,\pi)$, is called the _Poisson cohomology_ of the Poisson
manifold $(M,\pi)$.

By definition, $H^{0}(M,\pi)$ consists of the Casimir functions of $(M,\pi)$.
The 1–coboundaries are the Hamiltonian vector fields, and the 1–cocycles are
the Poisson vector fields. Hence, the cohomology in degree one is the Lie
algebra

$H^{1}(M,\pi)={\mathfrak{poiss}(M,\pi)}\,\big{/}\,{\mathfrak{ham}(M,\pi)}.$

The _tangential Poisson cohomology_ in degree one is the Lie subalgebra
$H^{1}_{\mathrm{tan}}(M,\pi)\subseteq H^{1}(M,\pi)$ consisting of the Poisson
cohomology classes with tangent representatives. More precisely,

$H^{1}_{\mathrm{tan}}(M,\pi):={\mathfrak{poiss}_{\mathrm{tan}}(M,\pi)}\,\big{/}\,{\mathfrak{ham}(M,\pi)}.$

Now, suppose that $(M,\pi)$ is a _regular_ Poisson manifold, and let
$\mathcal{S}$ be the symplectic foliation of $M$. We recall that the _foliated
de Rham complex_
$(\Gamma(\wedge^{\bullet}T^{*}\mathcal{S}),\mathrm{d}_{\mathcal{S}})$ of
$\mathcal{S}$ consists of the graded algebra
$\Gamma(\wedge^{\bullet}T^{*}\mathcal{S})$ of foliated 1–forms endowed with
the foliated exterior derivative $\mathrm{d}_{\mathcal{S}}$. Then, we have the
following fact [21, Chapter 5].

######  Proposition 2.2.

The tangential Poisson cohomology in degree one of a regular Poisson manifold
is isomorphic to the foliated de Rham cohomology of $\mathcal{S}$ and, in
particular, independent of the leaf–wise symplectic form on $\mathcal{S}$.

In the case when the foliation $\mathcal{S}$ is _simple_ , that is, given by
the fibers of a submersion, we have the following criterion for the vanishing
of the tangential Poisson cohomology in degree one.

######  Proposition 2.3.

Let $(M,\pi)$ be a regular Poisson manifold such that the symplectic foliation
$\mathcal{S}$ is given by the fibers of a submersion. If the fibers are
connected and simply connected, then $H^{1}_{\mathrm{tan}}(M,\pi)=0$.

This is consequence of Proposition 2.2 and of the fact that the topology of
the fibers imply the triviality of the foliated de Rham cohomology of
$\mathcal{S}$ in degree one [5, Prop. 7.4].

#### Orientable Poisson Manifolds.

Recall that on each oriented $m$–dimensional manifold $M$, equipped with a
volume form $\Omega$, there exists a one–to–one correspondence between
$(m-2)$–differential forms and bivector fields given by the following formula:

$\mathbf{i}_{\pi}\Omega=\varrho,\qquad\varrho\in\Gamma(\wedge^{m-2}T^{\ast}M),\
\pi\in\Gamma(\wedge^{2}TM).$ (2.1)

Here, the interior product of multivector fields and differential forms is
defined by the rule $\mathbf{i}_{A\wedge
B}=\mathbf{i}_{A}\circ\mathbf{i}_{B}$, for any
$A,B\in\Gamma(\wedge^{\bullet}TM)$.

Consider the $(m-2)$–differential form
$\varrho=\frac{1}{2}\sum_{i,j}\varrho_{ij}\mathrm{d}{x^{1}}\wedge\cdots\wedge\widehat{\mathrm{d}{x^{i}}}\wedge\cdots\wedge\widehat{\mathrm{d}{x^{j}}}\wedge\cdots\wedge\mathrm{d}{x^{m}}$,
where the 1–forms with $\ \widehat{}\ $ are omitted. We define the _rank_ of
$\varrho$ at a point $x\in M$ as the rank of the matrix $[\varrho_{ij}(x)]$.
Here, the set $\\{x^{1},\ldots,x^{m}\\}$ is a local coordinate system on $M$.
We also recall that an orientable Poisson manifold is said to be _unimodular_
if admits a volume form invariant under the all Hamiltonian flows [22].

######  Lemma 2.4.

If $\varrho$ is a closed $(m-2)$–differential form of rank at most two, then
the bivector field $\pi$ defined by the formula (2.1) is a unimodular Poisson
structure on $M$ of rank at most two.

######  Lemma 2.5.

Let $\varrho$ be a $(m-2)$–differential form, and suppose that the bivector
field $\pi$ defined by the formula (2.1) is a Poisson structure on $M$. Then,
a vector field $X$ on $M$ is a Hamiltonian vector field, with respect to
$\pi$, if and only if there exists $h\in C^{\infty}(M)$ such that

$\mathbf{i}_{X}\Omega=\mathrm{d}{h}\wedge\varrho.$ (2.2)

In particular, a function $K\in C^{\infty}(M)$ is Casimir for $\pi$ if and
only if $\mathrm{d}{K}\wedge\varrho=0$.

Finally, one can verify that the Poisson bivector field given by (2.1) is
unimodular if and only if $\varrho$ admits an (non–zero) integrating factor
$a\in C^{\infty}(M)$, $\mathrm{d}(a\varrho)=0$.

##  3 Hamiltonization problem on orientable manifolds

We begin by observing that every vector field on an orientable smooth
$m$–dimensional manifold admitting $m-1$ independent first integrals is
hamiltonizable.

######  Theorem 3.1.

Let $X$ be a vector field on an orientable $m$–dimensional manifold $M$.
Suppose that we have $m-1$ independent first integrals
$h_{1},\ldots,h_{m-1}\in C^{\infty}(M)$ of $X$. Then, there exist unique
Poisson structures $\pi_{1},\ldots,\pi_{m-1}$ of rank at most two on $M$ such
that

$\pi_{i}^{\sharp}\mathrm{d}h_{j}=\delta_{ij}X,\quad\text{for all}\quad
i,j=1,\ldots,m-1.$

Moreover, we have that $[\pi_{i},\pi_{j}]=0$, for all $i$, and $j$.

######  Proof 3.2.

Fix a volume form $\Omega$ on $M$, and define
$\psi_{i}\in\Gamma(\wedge^{2}TM)$ by

$\mathbf{i}_{\psi_{i}}\Omega:=\mathrm{d}h_{1}\wedge\cdots\wedge\widehat{\mathrm{d}h_{i}}\wedge\cdots\wedge\mathrm{d}h_{m-1}.$

Since the $(m-2)$–form on the right–hand side is closed and of rank two, it
follows from Lemma 2.4 that $\psi_{i}$ is Poisson of rank two, and hence, is
conformally invariant due to Lemma 2.1. Now, let us show that
$X\wedge\psi_{i}^{\sharp}\mathrm{d}h_{i}=0$. Indeed, by using that each
$h_{i}$ is a first integral of $X$, we have

$\mathbf{i}_{X\wedge\psi_{i}^{\sharp}\mathrm{d}h_{i}}\Omega=\mathbf{i}_{X}(\mathrm{d}h_{i}\wedge\mathbf{i}_{\psi_{i}}\Omega)=(-1)^{i-1}\mathbf{i}_{X}(\mathrm{d}h_{1}\wedge\cdots\wedge\mathrm{d}h_{m-1})=0.$

Since $\psi_{i}^{\sharp}\mathrm{d}h_{i}$ is nowhere vanishing, then
$X=f_{i}\psi_{i}^{\sharp}\mathrm{d}h_{i}$, for some $f_{i}\in C^{\infty}(M)$.
Moreover, by the conformal invariance of $\psi_{i}$, the bivector field
$\pi_{i}:=f_{i}\psi_{i}$ is Poisson of rank at most two. Then, we have that
$\pi_{i}^{\sharp}\mathrm{d}h_{i}=X$, and $\pi_{i}^{\sharp}\mathrm{d}h_{j}=0$
for $i\neq j$. Finally, to prove that $[\pi_{i},\pi_{j}]=0$, note that
$\mathbf{i}_{\pi_{i}}\Omega=f_{i}\mathbf{i}_{\psi_{i}}\Omega$, from where the
relations $\mathbf{i}_{\pi_{j}}\mathbf{i}_{\pi_{i}}\Omega=0$, and
$\mathbf{i}_{\pi_{j}}\mathrm{d}\mathbf{i}_{\pi_{i}}\Omega=0$ follow for every
$i$ and $j$. Therefore, $\mathbf{i}_{[\pi_{i},\pi_{j}]}\Omega=0$.

As an improvement of the previous result, we proceed to show that we can relax
the hypothesis on the number of first integrals: in the case when we have
$(m-2)$ independent first integrals of a given vector field $X$, the existence
of an $X$–invariant volume form is equivalent to the existence of a Poisson
structure such that $X$ is an infinitesimal automorphism. Hence, the
hamiltonization problem of $X$ turns into providing conditions for the
triviality of its Poisson cohomology.

######  Theorem 3.3.

Let $X$ be a vector field on an orientable $m$–dimensional manifold $M$, and
$c_{1},\ldots,c_{m-2}\in C^{\infty}(M)$ independent first integrals of $X$.
Suppose that

  1. 1.

there exists an $X$–invariant volume form on $M$,

  2. 2.

the level sets of
$\boldsymbol{c}:=(c_{1},\ldots,c_{m-2}):M\to\mathbb{R}^{m-2}$ are connected
and simply connected.

Then, the vector field $X$ is Hamiltonian with respect to a unimodular Poisson
structure of rank two admitting $c_{1},\ldots,c_{m-2}$ as Casimir functions.

In order to prove the proposition above, we need the following result.

######  Lemma 3.4.

Let $c_{1},\ldots,c_{m-2}\in C^{\infty}(M)$ be independent functions on an
orientable manifold $M$ of dimension $m$. There exists a one–to–one
correspondence between volume forms on $M$ and unimodular Poisson structures
of rank two on $M$ admitting $c_{1},\ldots,c_{m-2}$ as Casimir functions.

######  Proof 3.5.

Let us show that the correspondence between volume forms $\Omega$ on $M$ and
Poisson structures $\pi$ of rank two such that each $c_{i}$ is Casimir is
given by the relation

$\mathbf{i}_{\pi}\Omega=\mathrm{d}c_{1}\wedge\cdots\wedge\mathrm{d}c_{m-2}.$
(3.1)

Indeed, the correspondence $\Omega\mapsto\pi$ follows from Lemmas 2.4 and 2.5.
Conversely, suppose we are given a Poisson structure $\pi$ of rank two with
prescribed Casimir functions $c_{1},\ldots,c_{m-2}$, and fix a volume form
$\Omega_{0}$. By Lemma 2.5, we have that
$\mathbf{i}_{\pi}\Omega_{0}\wedge\mathrm{d}c_{i}=0$, for all $i=1,\ldots,m-2$.
Then, it follows from the independence of the $c_{i}$s that
$\mathbf{i}_{\pi}\Omega_{0}=f\mathrm{d}c_{1}\wedge\cdots\wedge\mathrm{d}c_{m-2}$
for a certain nowhere vanishing $f\in C^{\infty}(M)$. By setting
$\Omega:=\tfrac{1}{f}\Omega_{0}$, the relation (3.1) follows.

######  Remark 3.6.

The Poisson structures defined as in the formula (3.1) are called
_Flashcka–Ratiu Poisson structures_ [4].

######  Proof 3.7 (Proof of Theorem 3.3).

Let $\Omega$ be a volume form such that $\mathrm{div}_{\Omega}(X)=0$. By Lemma
3.4, the bivector field $\pi$ defined by (3.1) is a unimodular Poisson
structure of rank two such that $c_{1},\ldots,c_{m-2}$ are Casimir functions.
So, the regular foliation consisting of the level sets of
$\boldsymbol{c}:M\to\mathbb{R}^{m-2}$ is the symplectic foliation of $\pi$. By
the $X$-invariance of $\Omega$, and since $c_{1},\ldots,c_{m-2}$ are first
integrals of $X$, we get that $X$ is a tangent Poisson vector field of $\pi$.
Hence, the Poisson cohomology class of $X$ lies in the tangential Poisson
cohomology in degree one,

$[X]\in
H^{1}_{\mathrm{tan}}(M,\pi)={\mathfrak{poiss}_{\mathrm{tan}}(M,\pi)}\,\big{/}\,{\mathfrak{ham}(M,\pi)}.$

Since the fibers of $\boldsymbol{c}$ are connected and simply connected, we
have that $H^{1}_{\mathrm{tan}}(M,\pi)=0$, due to Proposition 2.3. Therefore,
the cohomology class of $X$ is trivial, so $X\in\mathfrak{ham}(M,\pi)$.

Consider the setting of Theorem 3.3. Observe that on the open domain where the
vector field $X$ is non–vanishing, the Hamiltonian function of $X$ and
$c_{1},\ldots,c_{m-2}$ define a maximal set of independent first integrals.
This prevents a chaotic behavior of the dynamical system associated with $X$.

######  Example 3.8.

Here we present families of linear vector fields which admit hamiltonization.

  1. 1.

Consider the linear vector field $X(x):=Ax\cdot\frac{\partial}{\partial x}$ on
$\mathbb{R}^{n}_{x}$, associated with an $n\times n$ matrix $A$ such that
$\operatorname{tr}A=0$, and $\operatorname{rank}A\leq 2$. By the first
condition, the canonical volume form on $\mathbb{R}^{n}_{x}$ is $X$–invariant.
The second one implies that there exist (linear) independent first integrals
$c_{1}(x):=v_{1}\cdot x,\ldots,c_{n-2}(x):=v_{n-2}\cdot x$ of $X$, associated
with some independent $v_{1},\ldots,v_{n-2}\in\ker{A^{\top}}$. Hence, by
Theorem 3.3, the vector field $X$ is hamiltonizable on $\mathbb{R}^{n}_{x}$.
Furthermore, the (constant) Poisson structure in (3.1) is given by

$\pi=\sum_{1\leq i<j\leq
n}(-1)^{i+j}\,\det{P_{[i,j]}}\frac{\partial}{\partial{x^{i}}}\wedge\frac{\partial}{\partial{x^{j}}},$
(3.2)

where $P_{[i,j]}$ denotes the $(n-2)\times(n-2)$ submatrix of $P=(v_{1}\cdots
v_{n-2})$ without the rows $i$ and $j$. The symplectic foliation consists of
the 2–dimensional planes given as the common level sets of the $c_{i}$s
equipped with the constant symplectic structure induced by the restriction of
$\omega=\omega_{ij}\mathrm{d}{x^{i}}\wedge\mathrm{d}{x^{j}}$, where

$\omega_{ij}=(-1)^{i+j+1}\,\frac{\det{P_{[i,j]}}}{|\pi|^{2}},\qquad|\pi|^{2}:=\sum_{1\leq
i<j\leq n}\big{(}\det{P_{[i,j]}}\,\big{)}^{2};\qquad 1\leq i<j\leq n.$ (3.3)

In particular, a Hamiltonian for $X$ is given by
$h(x)=\frac{1}{2}x^{\top}(WA)x$, where $W=[\omega_{ij}]_{n\times n}$.

  2. 2.

Let $X=Ax\cdot\frac{\partial}{\partial x}$ be a linear vector field on
$\mathbb{R}^{n}_{x}$, associated with a matrix $A$ such that
$\operatorname{tr}A=0$, and $\operatorname{rank}A=3$. Suppose that the
non–zero eigenvalues $\lambda_{1},\lambda_{1},\lambda_{3}$ of $A^{\top}$ are
all real and distinct. Let $v_{1},v_{2},v_{3}$ be eigenvectors of
$\lambda_{1},\lambda_{2},\lambda_{3}$, respectively; and $v_{4},\ldots,v_{n}$
a basis of $\ker{A}^{\top}$. As in above, the canonical volume form on
$\mathbb{R}^{n}_{x}$ is $X$–invariant, and the linear functions
$c_{4}(x):=v_{4}\cdot x,\ldots,c_{n}(x):=v_{n}\cdot x$ are independent first
integrals of $X$. Moreover, the cubic function $c(x):=(v_{1}\cdot
x)(v_{2}\cdot x)(v_{3}\cdot x)$ is also a first integral of $X$. This follows
from $\mathrm{L}_{X}(v_{i}\cdot x)=\lambda_{i}v_{i}\cdot x$, and
$\operatorname{tr}A=0$. Hence, by Theorem 3.3, the vector field $X$ is
hamiltonizable on the open set complementary to the union of the 2–planes
generated by pairs of $v_{1},v_{2},v_{3}$. The (quadratic) Poisson structure
and the leaf–wise symplectic form are given by (3.2) and (3.3), respectively,
with $P=(u(x)\,\ v_{4}\,\cdots\,v_{n})$, and
$u(x)=\sum_{\mathrm{cyclic}}(v_{1}\cdot x)(v_{2}\cdot x)\,v_{3}$. A
Hamiltonian function for $X$ can be given analogously as in above.

  3. 3.

Fix $a,v\in\mathbb{R}^{3}$, such that $\|a\|>\|v\|$, and $a\cdot v=0$.
Consider the linear vector field on $\mathbb{R}^{4}_{(x,y)}$

$X(x,y):=(a\times x+yv)\cdot\tfrac{\partial}{\partial x}+(v\cdot
x)\tfrac{\partial}{\partial y},\qquad x\in\mathbb{R}^{3},y\in\mathbb{R}.$

We show that $X$ can be hamiltonized in two different ways by using the
Theorem 3.3. First, note that the quadratic function
$q:\mathbb{R}^{4}_{(x,y)}\setminus\\{0\\}\to\mathbb{R}_{s}$,
$q(x,y):=\tfrac{1}{2}(x\cdot x-y^{2})$, is a first integral of $X$ whose level
sets $q^{-1}(s)$ are diffeomorphic to $\mathbb{S}^{2}\times\mathbb{R}$ if
$s\geq 0$, and to $\mathbb{S}^{0}\times\mathbb{R}^{3}$ if $s<0$. On the other
hand, our hypotheses on $a$ and $v$ imply that there exists
$w\in\mathbb{R}^{3}$, with $\|w\|<1$, such that $w\times a=v$. Thus, the
linear function $\ell(x,y):=w\cdot x-y$ is also a first integral of $X$.
Finally, set $U:=\mathbb{R}^{4}_{(x,y)}\setminus\\{(x,y)\mid x-yw=0\\}$. It
follows from the property $\|w\|<1$ that the level sets of
$\boldsymbol{c}:=(q,\ell):U\to\mathbb{R}^{2}$ are diffeomorphic to
$\mathbb{S}^{2}$, which is connected and simply connected. Since the Euclidean
volume on $\mathbb{R}^{4}$ is $X$–invariant, Theorem 3.3 implies that $X$ is a
Hamiltonian vector field on $U$ with respect to a linear Poisson structure. On
the other hand, observe that the linear function $\varphi(x,y):=a\cdot x$ is
also a first integral of $X$. By restricting to the level sets of
$(\ell,\varphi):\mathbb{R}^{4}\to\mathbb{R}^{2}$, we get from the first incise
that $X$ is hamiltonian with respect to a constant Poisson structure, and with
hamiltonian function $h=\lambda(x\cdot x-y^{2})$, for some constant factor
$\lambda$ (see also [16, Theorem 2]).

$\triangleleft$

A remarkable difference between the Poisson structures obtained in Theorem 3.1
and Theorem 3.3 is that the former encodes the zeroes of the vector field,
while latter is always regular, so that the zeroes of the vector field are
encoded in the Hamiltonian function. The regularity property in Theorem 3.3,
which is due to the independence of the first integrals, is in general a
necessary condition, as the following example illustrates.

######  Example 3.9.

Consider the functions $f,c:\mathbb{R}^{3}\to\mathbb{R}$ given by
$f=\tfrac{1}{2}(x^{2}+y^{2}+z^{2})$, and $c=2f^{2}$. The fibers of $c$ are
connected and simply connected since they consist of 2–spheres and a single
point. For the Euclidean volume form $\Omega$ on $\mathbb{R}^{3}$, the
bivector field $\pi$ defined by $\mathbf{i}_{\pi}\Omega=\mathrm{d}c$ is a
Poisson structure with Casimir function $c$. Now, consider the vector field
$X=-y\tfrac{\partial}{\partial x}+x\tfrac{\partial}{\partial y}$, for which
$\Omega$ is invariant, and $c$ is a first integral. Then, by Theorem 3.3, the
vector field $X$ is Hamiltonian for $\pi$ on the open dense set
$U:=\mathbb{R}^{3}\setminus\\{0\\}$ where $\mathrm{d}c$ is non–vanishing.
However, _the vector field $X$ is not Hamiltonian for $\pi$ on the whole
$\mathbb{R}^{3}$_. In fact, no 1–form $\alpha$ satisfies
$\pi^{\sharp}\alpha=X$ on $\mathbb{R}^{3}$. To see this, suppose such an
$\alpha=\alpha_{1}\mathrm{d}{x}+\alpha_{2}\mathrm{d}{y}+\alpha_{3}\mathrm{d}{z}$
exists. By restricting to $U$, we get

$\alpha_{1}|_{U}=\tfrac{xz}{2f^{2}}+\lambda
xf,\qquad\alpha_{2}|_{U}=\tfrac{yz}{2f^{2}}+\lambda
yf,\qquad\alpha_{3}|_{U}=\tfrac{-y^{2}-x^{2}}{2f^{2}}+\lambda zf,$

for some $\lambda\in C^{\infty}(U)$. The substitution $\lambda
f=-\tfrac{z}{2f^{2}}+\mu$, for $\mu\in C^{\infty}(U)$, yields

$\alpha_{1}|_{U}=x\mu,\qquad\alpha_{2}|_{U}=y\mu,\qquad\alpha_{3}|_{U}=\tfrac{1}{f}+z\mu.$

From the first equation, it follows that $\alpha_{1}$ vanishes on the
$yz$–plane, so there exists a global $\rho\in C^{\infty}(\mathbb{R}^{3})$ such
that $\alpha_{1}=x\rho$. From here, we get that $\rho|_{U}=\mu$. By the third
equation, we get $\tfrac{1}{f}=(z\rho-\alpha_{3})|_{U}$, which is impossible,
because $\tfrac{1}{f}$ cannot be smoothly extended to $\mathbb{R}^{3}$.
$\triangleleft$

We observe that in the 3–dimensional case, Theorem 3.3 recovers the
hamiltonization criteria of [9, First theorem], where the topological
hypotheses on the level sets defined by the first integrals were obviated.

Let us explain why the topological condition 2 for the fibers in Theorem 3.3
is necessary in general. Suppose we are given a submersion
$\boldsymbol{c}=(c_{1},\ldots,c_{m-2}):M\to\mathbb{R}^{m-2}$ and a fiber–wise
1–form $\alpha$ along the fibers of $\boldsymbol{c}$ satisfying the following
properties:

  1. (a)

$\alpha$ is nowhere vanishing;

  2. (b)

$\alpha$ is fiber-wise closed;

  3. (c)

for every nowhere vanishing $f\in C^{\infty}(M)$, the 1–form $f\alpha$ is not
fiber–wise exact.

In this setting, for any volume form $\Omega$, the vector field $X$ defined by

$\mathbf{i}_{X}\Omega=\alpha\wedge\mathrm{d}c_{1}\wedge\ldots\wedge\mathrm{d}c_{m-2},$

has $\Omega$ as invariant volume form, but $X$ cannot be Hamiltonian for a
Poisson structure $\pi$ with Casimir functions $c_{1},\ldots,c_{m-2}$. In
fact, by Lemma 3.4, any two of such Poisson structures differ by
multiplication of nowhere vanishing functions. Therefore, if $\beta$ is a
fiber–wise 1–form such that $\pi^{\sharp}\beta=X$, then $\beta=f\alpha$ for
some nowhere vanishing $f\in C^{\infty}(M)$. So, by property (c) on $\alpha$,
the vector field $X$ is not Hamiltonian for $\pi$. We illustrate this
situation in dimension three with a couple of examples.

######  Example 3.10.

Consider the vector field $X=\tfrac{\partial}{\partial z}$ on
$M=\mathbb{R}^{3}\setminus\\{z\text{--axis}\\}$. Clearly,
$c=\frac{1}{2}(x^{2}+y^{2})$ is a first integral for $X$, and the Euclidean
volume form on $\mathbb{R}^{3}$ is $X$–invariant. Observe that every nowhere
vanishing Poisson structure $\pi$ with Casimir function $c$ is of the form

$\pi=\frac{1}{f}\left(x\tfrac{\partial}{\partial y}-y\tfrac{\partial}{\partial
x}\right)\wedge\tfrac{\partial}{\partial z},$

for a nowhere vanishing function $f\in C^{\infty}(M)$. Moreover, the vector
field $X$ is an infinitesimal Poisson automorphism for $\pi$ if and only if
$\frac{\partial f}{\partial z}=0$, i.e., $f=f(x,y)$. Note that a 1–form
$\alpha\in\Gamma(T^{*}M)$ satisfies $\pi^{\sharp}\alpha=X$ if and only if

$\alpha=f\,\frac{x\mathrm{d}y-y\mathrm{d}x}{x^{2}+y^{2}}+g\mathrm{d}c,$

for some $g\in C^{\infty}(M)$. We claim that there is no $\alpha$ of the above
form which is exact. Indeed, let $\gamma$ be the unitary circle on the
$xy$–plane centered at the origin. Then,

$\int_{\gamma}\alpha=\int_{\gamma}f\neq 0$

because $f$ is nowhere vanishing, and $\gamma$ is connected. In particular,
$\alpha$ is not exact. Therefore, $X$ is not a hamiltonian vector field for
Poisson structures with Casimir $c$. $\triangleleft$

Although the previous example exhibits the necessity of condition 2 in Theorem
3.3, the vector field $\tfrac{\partial}{\partial z}$ obviously admits a
maximal number of independent first integrals, and hence, is hamiltonizable.
In the following example, we present a vector field that admits an invariant
volume form, but only a first integral.

######  Example 3.11.

Fix $\lambda\in\mathbb{R}$. On $\mathbb{R}^{3}$, the vector field

$X_{\lambda}=(2xz+\lambda y)\tfrac{\partial}{\partial x}+(2yz-\lambda
x)\tfrac{\partial}{\partial y}+(1-x^{2}-y^{2}+z^{2})\tfrac{\partial}{\partial
z}$ (3.4)

has the first integral $c=\frac{x^{2}+y^{2}}{(x^{2}+y^{2}+z^{2}+1)^{2}}$, and
admits the invariant volume form

$\Omega=\frac{\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z}{(x^{2}+y^{2}+z^{2}+1)^{2}}.$

Moreover, the level surface $T_{a}:=c^{-1}(a)$ is a torus, for $a>1/4$. Note
that if $\lambda$ is rational, then the orbits of $X_{\lambda}$ along $T_{a}$
are closed paths, and consequently the vector field $X_{\lambda}$ admits a
second first integral independent with $c$. By Theorem 3.1, $X_{\lambda}$ is
Hamiltonizable. If, instead, $\lambda$ is irrational, then the orbits of
$X_{\lambda}$ are dense curves on $T_{a}$. This implies that any first
integral of $X_{\lambda}$ is constant on $T_{a}$, and hence cannot be
independent with $c$. In particular, it cannot be hamiltonized as in Theorem
3.3. $\triangleleft$

######  Remark 3.12.

Even though the vector field $X_{\lambda}$ in (3.4) does not admit a second
first integral for $\lambda\in\mathbb{R}\setminus\mathbb{Q}$, one can
construct a Poisson structure with respect to which $X_{\lambda}$ is
Hamiltonian (see Example 5.14).

A well–known fact about the hamiltonization problem, which can be seen as a
trivial instance of Theorem 3.3, is that a vector field $X$ on a simply
connected 2–dimensional manifold is hamiltonizable if and only if admits an
invariant volume form $\Omega$ (we recall that every simply connected manifold
is orientable). In this case, the Poisson structure $\pi$ is defined by
$\pi^{\sharp}=(\Omega^{\flat})^{-1}$, and a hamiltonian function is a
primitive of $\mathbf{i}_{X}\Omega$. Motivated by this situation, we formulate
the following.

######  Theorem 3.13.

Let $X$ be a vector field on an orientable $m$–dimensional manifold $M$.
Suppose that there exists a volume form $\Omega$ such that

$\mathbf{i}_{X}\Omega=\mathrm{d}\varrho,$ (3.5)

for some $\varrho\in\Gamma(\wedge^{m-2}T^{*}M)$ of rank at most two. If
$\varrho$ admits an integrating factor $a\in C^{\infty}(M)$, in the sense that
$\mathrm{d}(a\varrho)=0$, then $X$ is hamiltonizable on the open set
$U:=\\{a\neq 0\\}$ by a unimodular Poisson structure of rank at most two.

######  Proof 3.14.

First note from Lemma 2.4 that the bivector field $\pi$ defined by
$\mathbf{i}_{\pi}\Omega=a\varrho$ is a unimodular Poisson structure of rank at
most two on $M$. Now, from the integrating factor property, the function
$h=\frac{1}{a}$ satisfies
$\mathbf{i}_{X}\Omega=\mathrm{d}{h}\wedge(a\varrho)$. Hence, by Lemma 2.5, the
vector field $X$ is Hamiltonian for $\pi$ on $U$, with Hamiltonian function
$h$.

######  Remark 3.15.

Observe that Theorem 3.13 is a hamiltonization criteria via
“unimodularization”: the Poisson structure $\pi$ is unimodular, and $X$ is its
_modular vector field_ with respect to $\Omega$ [22].

Note that a necessary condition for (3.5) is that $\mathrm{div}_{\Omega}{X}=0$
holds on $M$. This is sufficient if the de Rham cohomology of $M$ in degree
$m-1$ is trivial.

######  Corollary 3.16.

Let $M$ be a orientable $m$–dimensional manifold with
$H^{m-1}_{\mathrm{dR}}(M)=0$, and $X$ a vector field on $M$ admitting an
invariant volume form $\Omega$. Then, the condition (3.5) holds for some
$\varrho$. Furthermore, if $\varrho$ admits a nowhere vanishing integrating
factor, then $X$ is hamiltonizable on $M$ in the following cases:

  1. (a)

The manifold $M$ is 3–dimensional.

  2. (b)

The manifold $M$ is 4–dimensional, and $\varrho\wedge\varrho=0$.

This last corollary follows from Theorem 3.13, since the items (a) and (b)
ensure that $\varrho$ has rank at most two.

###  3.1 Orientability of foliations

In this subsection we present a generalization of Theorem 3.3 in the framework
of orientable foliations. Recall that a regular foliation $\mathcal{F}$ is
said to be orientable if there exists a nowhere vanishing element
$\eta\in\Gamma(\wedge^{\mathrm{top}}T^{\ast}\mathcal{F})$, called a leaf-wise
volume form of $\mathcal{F}$.

######  Theorem 3.17.

Let $\mathcal{F}$ be an oriented regular foliation on $M$ of dimension $r$,
and $X\in\Gamma(TM)$ a vector field tangent to $\mathcal{F}$ preserving the
leaf–wise volume. Under the following conditions, the vector field $X$ is
hamiltonizable by a unimodular Poisson structure of rank two on $M$:

  1. 1.

There exists a nowhere vanishing, leaf–wise closed, and locally decomposable
leaf–wise $(r-2)$–form $\beta\in\Gamma(\wedge^{r-2}T^{*}\mathcal{F})$ such
that $\mathbf{i}_{X}\beta=0$.

  2. 2.

The foliated de Rham cohomology of the foliation $\mathcal{F}_{\beta}$
integrating $\ker\beta$ is trivial in degree one,
$H^{1}_{\mathrm{dR}}(M,\mathcal{F}_{\beta})=0$.

######  Proof 3.18.

Let $\eta\in\Gamma(\wedge^{r}T^{*}\mathcal{F})$ be an $X$–invariant leaf–wise
volume form on $\mathcal{F}$. Define the bivector field
$\pi\in\Gamma(\wedge^{2}T\mathcal{F})$ by

$\mathbf{i}_{\pi}\eta=\beta.$

Note that $\beta$ is of rank at most two since it is locally decomposable. By
the fact that $\beta$ is leaf–wise closed, we get from Lemma 2.4 that $\pi$ is
Poisson. Moreover, the symplectic foliation of $\pi$ is precisely
$\mathcal{F}_{\beta}$. On the other hand, since $\mathbf{i}_{X}\beta=0$, we
get that $X$ is tangent to $\mathcal{F}_{\beta}$ and that
$\mathrm{L}_{X}\beta=0$. From here and the $X$–invariance of $\eta$, we get
that $X$ is an infinitesimal Poisson automorphism for $\pi$. Finally, taking
into account that $H^{1}_{\mathrm{dR}}(M,\mathcal{F}_{\beta})=0$, we conclude
from Proposition 2.2 that $X$ is Hamiltonian.

######  Corollary 3.19.

Let $X$ be a volume–preserving vector field on an oriented $m$–dimensional
manifold $M$. Then, the vector field $X$ is hamiltonizable by a unimodular
Poisson structure of rank two on $M$ if:

  1. 1.

There exist independent 1–forms
$\alpha_{1},\ldots,\alpha_{k}\in\Gamma(T^{*}M)$ satisfying the integrability
condition

$\mathrm{d}\alpha_{i}\wedge\alpha_{1}\wedge\cdots\wedge\widehat{\alpha}_{i}\wedge\cdots\wedge\alpha_{k}=0,\quad\text{for
all}\ \ i=1,\ldots,k;$ (3.6)

and such that $X$ is tangent to the foliation $\mathcal{F}_{\alpha}$
integrating $\bigcap_{i=1}^{k}\ker\alpha_{i}$.

  2. 2.

There exists a nowhere vanishing, leaf–wise closed, and locally decomposable
leaf–wise $(m-k-2)$–form
$\beta\in\Gamma(\wedge^{m-k-2}T^{*}\mathcal{F}_{\alpha})$ such that
$\mathbf{i}_{X}\beta=0$.

  3. 3.

The foliated de Rham cohomology of the foliation $\mathcal{F}_{\beta}$
integrating $\ker\beta$ is trivial in degree one,
$H^{1}_{\mathrm{dR}}(M,\mathcal{F}_{\beta})=0$.

######  Proof 3.20.

For $\alpha:=\alpha_{1}\wedge\cdots\wedge\alpha_{k}$, the bivector field $\pi$
defined by

$\mathbf{i}_{\pi}\Omega=\alpha\wedge\beta$

is Poisson, and $X\in\mathfrak{ham}(M,\pi)$.

As a generalization of Theorem 3.3, the results presented in Theorem 3.17 and
Corollary 3.19 allow us to hamiltonize vector fields without having explicit
first integrals. Indeed, we can recover Theorem 3.3 by setting $\mathcal{F}$
as the trivial foliation and
$\beta=\mathrm{d}c_{1}\wedge\cdots\wedge\mathrm{d}c_{m-2}$, where
$c_{1},\ldots,c_{m-2}\in C^{\infty}(M)$ are independent first integrals of the
given vector field $X$.

######  Example 3.21.

Consider the manifold
$M=N\times\big{(}\mathbb{R}^{3}\setminus\\{y_{3}\text{--axis}\\}\big{)}$
endowed with the oriented regular foliation $(\mathcal{F},\eta)$ given by

$\mathcal{F}:=\bigcup_{x\in
N}\\{x\\}\times\big{(}\mathbb{R}^{3}\setminus\\{y_{3}\text{--axis}\\}\big{)},\qquad\eta:=\frac{e^{a(x)}}{y_{1}^{2}+y_{2}^{2}}\,\mathrm{d}{y_{1}}\wedge\mathrm{d}{y_{2}}\wedge\mathrm{d}{y_{3}}\big{|}_{\mathbb{R}^{3}\setminus\\{y_{3}\text{--axis}\\}};$

and the $\mathcal{F}$–tangent, and $\eta$–preserving vector field on $M$

$X=f_{1}(x)\,y_{3}\left(y_{1}\frac{\partial}{\partial
y_{1}}+y_{2}\frac{\partial}{\partial
y_{2}}\right)+f_{2}(x,y)\frac{\partial}{\partial y_{3}}.$

Here, $a,f_{1}\in C^{\infty}(N)$, and $f_{2}\in C^{\infty}(M)$ with
$\frac{\partial f_{2}}{\partial y_{3}}=0$. Note that the nowhere vanishing,
and leaf–wise closed differential 1–form

$\beta=e^{b(x)}\,\frac{y_{2}\mathrm{d}y_{1}-y_{1}\mathrm{d}y_{2}}{y_{1}^{2}+y_{2}^{2}},\qquad
b\in C^{\infty}(N);$

is such that $\mathbf{i}_{X}\beta=0$. Moreover, the kernel of $\beta$
integrates to the subfoliation $\mathcal{F}_{\beta}$ of $\mathcal{F}$ given by
the fibers of the submersion

$p:M\to N\times\mathbb{S}^{1},\qquad
p(x;y):=\left(x;\tfrac{y_{1}}{\sqrt{y_{1}^{2}+y_{2}^{2}}},\tfrac{y_{2}}{\sqrt{y_{1}^{2}+y_{2}^{2}}}\right).$

Since the leaves of $\mathcal{F}_{\beta}$ are connected and simply connected,
we have that $H^{1}_{\mathrm{dR}}(M,\mathcal{F}_{\beta})=0$ (see [5, Prop.
7.4]). Therefore, by the proof of Theorem 3.17, the vector field $X$ is
Hamiltonian on $M$ with respect to the Poisson structure

$\pi_{\eta,\beta}=-e^{b(x)}\left(y_{1}\frac{\partial}{\partial
y_{1}}+y_{2}\frac{\partial}{\partial
y_{2}}\right)\wedge\frac{\partial}{\partial y_{3}}.$

Furthermore, in the case when $f_{2}$ depends radially on the variables
$y_{1},y_{2}$, that is, $f_{2}(x,y)=g(x;y_{1}^{2}+y_{2}^{2})$ for some $g\in
C^{\infty}(N\times\mathbb{R})$, a Hamiltonian function for $X$ is given by

$h(x,y)=-\frac{e^{-b(x)}}{2}\left[G\big{(}x,\ln(y_{1}^{2}+y_{2}^{2})\big{)}+y_{3}^{2}f_{1}(x)\right],$

where $G\in C^{\infty}(N\times\mathbb{R}_{t})$ is such that $\frac{\partial
G}{\partial t}=g(x,e^{t})$. $\triangleleft$

##  4 Hamiltonization through decomposable Poisson structures

In this section, we give conditions under which a vector field $X$ on a smooth
manifold $M$ is hamiltonizable via a decomposable Poisson structure $\pi$.
More precisely, we look after Poisson structures of the form $\pi=Y\wedge X$,
for a vector field $Y\in\Gamma(TM)$.

Consider the open set $U\subseteq M$ where the vector field $X$ is
non–vanishing. Denote by $\mathcal{F}_{X}$ the 1–dimensional foliation of $U$
given by the orbits of $X$, and let $\nu_{X}:=TU/T\mathcal{F}_{X}$ be its
normal bundle. If $X$ is hamiltonian with respect to a Poisson structure $\pi$
of rank at most two, then we have $\pi\wedge
X\in\Gamma(\wedge^{3}(\pi^{\sharp}(T^{*}M)))=\\{0\\}$. This and the nowhere
vanishing property of $X$ on $U$ imply that there exists a vector field
$Y\in\Gamma(TU)$ such that $\pi=Y\wedge X$ on $U$. Moreover, it is clear that
$\pi|_{U}$ only depends of the class $s:=[Y]\in\Gamma(\nu_{X})$ rather than on
$Y$ itself. Finally, if $h\in C^{\infty}(M)$ is a hamiltonian of $X$, then
$\mathrm{d}h(Y)=1$ on $U$.

The properties discussed in the previous paragraph motivate the notion of
normal class relative to a vector field $X$, which encodes the freedom of the
choice of the vector field $Y$ such that $\pi=Y\wedge X$.

######  Definition 4.1.

Let $X$ be a vector field on $M$. A normal class relative to $X$, or simply a
_normal class_ , is an equivalence class of vector fields on $M$ with respect
to the relation

$Y_{1}\sim_{X}Y_{2}\text{ if and only if }(Y_{2}-Y_{1})\wedge X=0.$

Moreover, a normal class $[Y]$ is said to be

  * •

_invariant_ if $[X,Y]\wedge X\wedge Y=0$;

  * •

_normalized_ with respect to a first integral $h$ of $X$ (or, _$h$
–normalized_) if $\mathrm{d}h(Y)X=X$.

It is straightforward to verify that, for any normal class, the properties of
invariance and of $h$–normalization are independent of the choice of a
representative. Furthermore,

######  Theorem 4.2.

A vector field $X$ on $M$ is _hamiltonizable_ by a Poisson structure of the
form $\pi=Y\wedge X$ if and only if $X$ admits an invariant normal class that
is normalized with respect to a first integral of $X$ that is regular on
$\mathrm{supp}(X)$.

In other words, the hamiltonization problem by decomposable Poisson structures
translates into the construction of normalized invariant normal classes. As a
first step of the proof of this, observe that one can associate to each normal
class $s=[Y]$ relative to a vector field $X$ on $M$ a well–defined bivector
field $\pi_{s}$ given by $\pi_{s}:=Y\wedge X$. Furthermore:

######  Lemma 4.3.

The assignment

$s\longmapsto\pi_{s}$

is a one–to–one correspondence between normal classes relative to $X$ and
bivector fields decomposable by $X$. Moreover,

  1. (i)

the class $s$ is invariant if and only if $\pi_{s}$ is Poisson;

  2. (ii)

the class $s$ is $h$–normalized if and only if
$\pi_{s}^{\sharp}\mathrm{d}h=X$.

######  Proof 4.4.

By definition of normal class, the assignment $s\mapsto\pi_{s}$ is injective.
For the surjectivity, simply note that the decomposable bivector $\pi=Y\wedge
X$ is the image of the normal class defined by $Y$. On the other hand, if $Y$
is a representative of the normal class $s$, then
$[\pi_{s},\pi_{s}]=2[X,Y]\wedge X\wedge Y$, so the Poisson property for
$\pi_{s}$ is equivalent to the invariance of $s$. Finally, if $h$ is a first
integral of $X$, then we have that
$\pi_{s}^{\sharp}\mathrm{d}h=\mathrm{d}h(Y)X$, which implies that $s$ is
$h$–normalized if and only if $\pi_{s}^{\sharp}\mathrm{d}h=X$.

Note that Lemma 4.3 implies every but one assertion in Theorem 4.2. To
complete the proof, it is left to show that the first integrals normalizing
some normal class of a vector field $X$ are those that are regular on the
support of $X$.

######  Lemma 4.5.

Let $X$ be a vector field with first integral $h$. Then, an $h$–normalized
class relative to $X$ exists if and only if $h$ is regular on
$\mathrm{supp}(X)$. In this case, if $U$ and $V$ are the respective open sets
where $X$ and $\mathrm{d}h$ are non–vanishing, then:

  1. 1.

Every vector field $Y_{0}$ on $V$ satisfying $\mathrm{d}h|_{U}(Y_{0})=1$
determines a unique $h$–normalized normal class $s_{0}$ on $M$.

  2. 2.

If, in addition, the identity $[X,Y_{0}]\wedge X\wedge Y_{0}=0$ holds on $U$,
then $s_{0}$ is invariant on $M$.

######  Proof 4.6.

If $Y$ represents an $h$–normalized normal class, then $\mathrm{d}h(Y)=1$ on
$U$, which implies that $\mathrm{d}h(Y)=1$ holds at
$\overline{U}=\mathrm{supp}(X)$. Therefore, $\mathrm{d}h$ does not vanish on
the support of $X$. Now, suppose that $h$ is regular on $\mathrm{supp}(X)$.
Since $\overline{U}\subseteq V$, we can find disjoint open sets $W$ and
$W^{\prime}$ satisfying $\overline{U}\subseteq W$, and $M\setminus V\subseteq
W^{\prime}$, together with a smooth function $\mu\in C^{\infty}(M)$ such that
$\mu|_{\overline{U}}=1$, and $\mu|_{M\setminus W}=0$. Given any
$Y_{0}\in\Gamma(TV)$ such that $\mathrm{d}h|_{U}(Y_{0})=1$, a smooth vector
field $Y$ on $M$ is well defined by $Y:=\mu Y_{0}$ on $V$, and $Y:=0$ on
$W^{\prime}$. Since $Y$ agrees with $Y_{0}$ on $U$, we have
$\mathrm{d}h(Y)X=X$. In other words, the normal class $s_{0}:=[Y]$ is
$h$–normalized. Finally, since $X(\mu)=0$ and $X$ vanishes outside $U$, the
invariance of $s_{0}$ follows from $[X,Y_{0}]\wedge X\wedge Y_{0}=0$ on $U$.

Finally, we describe a reformulation of the invariance property for normalized
classes.

######  Lemma 4.7.

Let $X$ be a vector field on $M$ with first integral $h$. Let also $s$ be an
$h$–normalized normal class, and $Y$ a representative of $s$. The following
assertions are equivalent:

  1. 1.

The class $s$ is invariant.

  2. 2.

The identity $[X,Y]\wedge X=0$ holds on $M$.

  3. 3.

There exists $a\in C^{\infty}(U)$ such that $[X,Y]=aX$ holds on $U$.

Here, we denote by $U$ the open subset of $M$ where $X$ is non–vanishing.

######  Proof 4.8.

Since, by definition, $X$ is nowhere vanishing on $U$, it readily follows that
2 and 3 are equivalent. Moreover, it is clear that 2 implies 1. So, it is left
to show that 1 implies 2. Observe that

$\displaystyle 0$ $\displaystyle=\mathbf{i}_{\mathrm{d}h}([X,Y]\wedge X\wedge
Y)$ $\displaystyle=\mathrm{d}h([X,Y])X\wedge Y-\mathrm{d}h(X)[X,Y]\wedge
Y+[X,Y]\wedge(\mathrm{d}h(Y)X)$ $\displaystyle=\mathrm{d}h([X,Y])X\wedge
Y+[X,Y]\wedge X,$

where in the last step, we have applied that $\mathrm{d}h(X)=0$, and
$\mathrm{d}h(Y)X=X$. Since $\mathrm{d}h([X,Y])X\wedge Y=0$ trivially holds on
$M\setminus U$, it suffices to show that $\mathrm{d}h([X,Y])=0$ on $U$. From
$\mathrm{d}h(Y)X=X$, we get that $\mathrm{d}h(Y)=1$ on $U$. Moreover, by the
Koszul’s formula,

$\mathrm{d}h([X,Y])=\mathrm{L}_{X}(\mathrm{d}h(Y))-\mathrm{L}_{Y}(\mathrm{d}h(X))-\mathrm{d}^{2}h(X,Y).$

Since $\mathrm{d}h(X)$, and $\mathrm{d}h(Y)$ are constant on $U$, we conclude
that $\mathrm{d}h([X,Y])=0$ on $U$, as desired.

######  Remark 4.9.

There is a cohomological approach to the proof of Lemma 4.7, where the
non–trivial part is to show that $[X,Y]\wedge X\wedge Y=0$ implies
$[X,Y]\wedge X=0$. Indeed, set $\pi:=Y\wedge X$. Since $\mathrm{d}h(Y)X=X$,
and $[X,Y]\wedge X\wedge Y=0$, we get from Theorem 4.2 that $\pi$ is Poisson
and $X$ is Hamiltonian. In particular, $X$ is an infinitesimal Poisson
automorphism for $\pi$, which leads to $[X,Y]\wedge X=0$.

We left for the next sections the discussion of further applications of
Theorem 4.2 to the problem of hamiltonization by means of possibly non–regular
first integrals.

###  4.1 The case of regular first integrals

In this part we review under the light of our previous development some known
hamiltonization criteria that require the existence of _regular_ first
integrals, that is, without critical points. This requirement, however,
imposes strong topological conditions on the manifold: it excludes, for
instance, a global hamiltonization of vector fields on compact manifolds.

As a consequence of Theorem 4.2, we recover the following hamiltonization
criteria [1, 15], which implicitly require the regularity of the first
integral.

######  Corollary 4.10.

Let $X$ be a vector field admitting a regular first integral $h$.

  1. (a)

If a vector field $Y$ is such that $\mathrm{d}h(Y)=1$, and $[X,Y]=aX$ for some
$a\in C^{\infty}(M)$, then $\pi:=Y\wedge X$ is Poisson, and satisfies
$\pi^{\sharp}\mathrm{d}h=X$.

  2. (b)

If a vector field $Z$ is such that $\mathrm{d}h(Z)$ is nowhere vanishing, and
$[X,Z]=pX+qZ$ for some $p,q\in C^{\infty}(M)$, then
$\pi:=\tfrac{1}{\mathrm{d}h(Z)}Z\wedge X$ is Poisson and satisfies
$\pi^{\sharp}\mathrm{d}h=X$.

######  Proof 4.11.

For item $(b)$, it is straightforward to verify that the normal class of
$Y:=\tfrac{1}{\mathrm{d}h(Z)}Z$ is invariant, and $h$–normalized. Hence, by
Theorem 4.2, the bivector field $\pi:=Y\wedge
X=\tfrac{1}{\mathrm{d}h(Z)}Z\wedge X$ is Poisson and satisfies
$\pi^{\sharp}\mathrm{d}h=X$. Moreover, item $(a)$ readily follows from $(b)$.

######  Remark 4.12.

The hamiltonization criteria in $(a)$ of Corollary 4.10 appears for the first
time in [15]. An intrinsic formulation of $(a)$ is found in [1], where
statement $(b)$ is presented as a generalization of $(a)$. However, we remark
that $(b)$ _is not_ a generalization but a reformulation of $(a)$, since it
does not provide new solutions to the hamiltonization problem at all: every
vector field $X$ satisfying $(b)$ also satisfies $(a)$. To see this, set
$Y:=\tfrac{1}{\mathrm{d}h(Z)}Z$. Then, note that $\mathrm{d}h(Y)=1$, and
$[X,Y]=aX$ for $a:=\tfrac{p}{\mathrm{d}h(Z)}$.

######  Example 4.13.

On $M=\\{x\in\mathbb{R}^{n}\mid x_{i}\neq 0,\ i=1,\ldots,n\\}$, $n\geq 2$,
consider the vector field

$X=\sum_{i=1}^{n}x_{i}F_{i}\frac{\partial}{\partial x_{i}},$

where $F_{1},\ldots,F_{n}$ are homogeneous functions of degree $r$ on $M$ such
that $F_{1}+\cdots+F_{n}=0$. Note that the function

$h(x)=x_{1}\cdots x_{n}$

is a regular first integral of $X$ on $M$. On the other hand, since $X$ is an
homogeneous vector field of degree $r+1$, we have that $[X,E]=-rX$, where
$E=x_{1}\frac{\partial}{\partial x_{1}}+\cdots+x_{n}\frac{\partial}{\partial
x_{n}}$ is the Euler vector field. Furthermore, the function
$\mathrm{d}h(E)=nh$ is nowhere vanishing on $M$. By Corollary 4.10, the
bivector field

$\pi=\frac{1}{nh}\sum_{1\leq i<j\leq
n}x_{i}x_{j}(F_{j}-F_{i})\frac{\partial}{\partial
x_{i}}\wedge\frac{\partial}{\partial x_{j}},$

is Poisson and satisfies $\pi^{\sharp}\mathrm{d}h=X$. Therefore, $X$ is
hamiltonizable on $M$. $\triangleleft$

######  Example 4.14.

Let $X=Ax\cdot\frac{\partial}{\partial x}$ be a linear vector field on
$\mathbb{R}^{n}_{x}$, associated with a real matrix $A$, and $E$ the Euler
vector field on $\mathbb{R}^{n}_{x}$. Observe that $[X,E]=0$.

  * •

First, suppose that $A^{\top}$ admits two distinct real eigenvalues
$\lambda_{1},\lambda_{2}$, with eigenvectors $v_{1},v_{2}$, respectively. Fix
real numbers $r_{1},r_{2}$ satisfying $\lambda_{1}r_{1}+\lambda_{2}r_{2}=0$,
and $r_{1}+r_{2}\neq 0$. Then, on the open set $U=\\{x\in\mathbb{R}^{n}\mid
x\cdot v_{1}\neq 0,\ x\cdot v_{2}\neq 0\\}$, the function $h(x)=|v_{1}\cdot
x|^{r_{1}}|v_{2}\cdot x|^{r_{2}}$ is a regular first integral of $X$.
Moreover, $\mathrm{d}h(E)=(r_{1}+r_{2})h$ is nowhere vanishing on $U$. Then,
by Corollary 4.10, the vector field $X$ is Hamiltonian on $U$ with respect to
$\pi_{h}=\frac{1}{h}E\wedge X$, with hamiltonian function $h$.

  * •

On the other hand, if the kernel of $A^{\top}$ is non–trivial, then for every
non–zero $v\in\ker{A^{\top}}$ the linear function $f(x)=v\cdot x$ is a regular
first integral of $X$. Moreover, $\mathrm{d}f(E)=f$ is nowhere vanishing on
$W=\\{x\in\mathbb{R}^{n}\ |\ x\cdot v\neq 0\\}$. By Corollary 4.10, the vector
field $X$ is Hamiltonian on $W$ with respect to $\pi_{f}=\frac{1}{f}E\wedge
X$, with hamiltonian function $f$.

$\triangleleft$

##  5 Hamiltonization via transversally invariant metrics

In this part, we work with the notion of transversally invariant Riemannian
metric (for a more general view of this notion and its geometric applications,
see [6, 20]). We derive some results on hamiltonization of infinitesimal
generators of proper actions. In particular, we consider the case of periodic
flow vector fields.

Let $(M,g)$ be a Riemannian manifold, and $X$ a vector field on $M$. First, we
recall that the metric $g$ is said to be _$X$ –invariant_ if
$\mathrm{L}_{X}g=0$. A weaker notion is that of transversal invariance: let
$O$ be the orbit of $X$ through a point $m\in M$. Then, the metric
$\eta\in\Gamma(T^{*}M\otimes T^{*}M)$ dual to $g$ induces a metric $\eta_{O}$
on the annihilator subbundle $TO^{\circ}$. We say that $g$ is _transversally
invariant_ if, for each $m\in M$, the map
$(\mathrm{d}_{m}\mathrm{Fl}^{t}_{X})^{*}:T_{\mathrm{Fl}^{t}_{X}(m)}O^{\circ}\to
T_{m}O^{\circ}$ induced by the flow $\mathrm{Fl}^{t}_{X}$ of $X$ is an
isometry, wherever is well defined. This is equivalent to require that the map
$\overline{\mathrm{d}_{m}\mathrm{Fl}^{t}_{X}}:T_{m}M/T_{m}O\to
T_{\mathrm{Fl}^{t}_{X}(m)}M/T_{\mathrm{Fl}^{t}_{X}(m)}O$ is an isometry.
Infinitesimally, this is just $\mathrm{L}_{X|_{O}}\eta_{O}=0$ for every orbit
$O$ of $X$. It is clear that every $X$–invariant metric is transversally
invariant, but the converse is not true.

######  Theorem 5.1.

Let $X$ be a vector field on the manifold $M$ that admits

  1. 1.

a first integral $h\in C^{\infty}(M)$ that is regular on $\mathrm{supp}(X)$,
and

  2. 2.

a transversally invariant metric $g\in\Gamma(TM\otimes TM)$.

Then, the vector field $X$ is hamiltonizable on $M$.

######  Proof 5.2.

Let $U$ and $V$ be the respective open sets where $X$ and $\mathrm{d}h$ are
nowhere vanishing. Define $Y_{0}\in\Gamma(TV)$ by

$Y_{0}:=\frac{\eta^{\sharp}\mathrm{d}h}{\eta(\mathrm{d}h,\mathrm{d}h)},$

where $\eta\in\Gamma(T^{*}M\otimes T^{*}M)$ is the dual metric of $g$. Since
$h$ is a first integral of $X$, we have $\mathrm{d}h\in\Gamma(TO^{\circ})$ for
each orbit $O$ of $X$. From here, the invariance of $h$, and the transversal
invariance of $g$, we have $[X,Y_{0}]=0$. On the other hand, since $h$ is
regular on $\mathrm{supp}(X)$, and $\mathrm{d}h(Y_{0})=1$ on $V$, we have from
Lemma 4.5 that $Y_{0}$ induces a unique invariant and $h$–normalized normal
class on $M$. By Theorem 4.2, $X$ is hamiltonizable on $M$.

###  5.1 The case of proper actions of 1–dimensional Lie groups

We now apply Theorem 5.1 to the case when the vector field is an infinitesimal
generator of a proper action, namely, if it is complete and the action induced
by its flow is proper. In this case, we can benefit from several properties,
which we now recall.

First, we take into account the well–known fact on the existence of invariant
metrics for proper Lie group actions [8, Proposition 2.5.2].

######  Proposition 5.3.

Every proper Lie group action admits an invariant Riemannian metric.

On the other hand, recall that every function on a compact manifold always has
a critical point. Furthermore, the converse is also true, in the following
sense [13, Theorem 4.8]:

######  Proposition 5.4.

Every non–compact manifold admits a smooth function with no critical points.

As a consequence of these facts, and of Theorem 5.1, we have:

######  Proposition 5.5.

Let $X$ be a complete vector field on $M$ such that its flow induces a proper
$\mathbb{R}$–action. Then, the following assertions are equivalent:

  1. 1.

The vector field $X$ is hamiltonizable on $M$.

  2. 2.

The orbit space $N:=M/\mathbb{R}$ is non–compact.

######  Proof 5.6.

Note that proper actions do not admit compact orbits unless the acting Lie
group is compact. Thus, the proper $\mathbb{R}$–action given by the flow of
$X$ is also free. So, the vector field $X$ is nowhere vanishing, and the orbit
space $N$ is a smooth manifold. Therefore, we have a one–to–one correspondence
between first integrals $h\in C^{\infty}(M)$ of $X$, and functions $f\in
C^{\infty}(N)$. Now, suppose that there exist $\pi$ and $h$ such that
$\pi^{\sharp}\mathrm{d}h=X$. Then, $h$ is a first integral of $X$ with no
critical points, and the corresponding $f\in C^{\infty}(N)$ has no critical
points. Thus, $N$ cannot be compact. Conversely, if $N$ is non–compact, then
there exists $f\in C^{\infty}(N)$ with no critical points, and the induced
first integral $h\in C^{\infty}(M)$ of $X$ is regular. Since the
$\mathbb{R}$–action by the flow of $X$ is proper, there exists an invariant
metric. By applying Theorem 5.1, the result follows.

Now, we consider the case of vector fields of periodic flow, that is, when the
flow induces an $\mathbb{S}^{1}$–action. Although the following fact readily
follows from Theorem 5.1, it is also convenient to give a more constructive
proof of it.

######  Proposition 5.7.

Every vector field $X$ on $M$ of periodic flow admitting a first integral $h$
that is regular on $\mathrm{supp}(X)$ is hamiltonizable.

######  Proof 5.8.

By Lemma 4.5, there exists $Y_{0}\in\Gamma(TM)$ such that
$\mathrm{d}h(Y_{0})X=X$. Now, consider the average $Y:=\langle Y_{0}\rangle$
of $Y_{0}$ under the $\mathbb{S}^{1}$–action induced by the flow of $X$. Then,
we have $[X,Y]=fX$, for some $f\in C^{\infty}(M)$. Moreover, by the
$\mathbb{S}^{1}$–invariance of $h$ and $X$, and the properties of the
averaging operator, we get that $\mathrm{d}h(Y)X=X$. In other words, the
normal class induced by $Y$ is invariant and $h$–normalized. By Theorem 4.2,
the vector field $X$ is hamiltonizable on $M$.

######  Corollary 5.9.

Let $X$ be a vector field on $M$ of periodic flow with no critical points.
Then, $X$ is globally hamiltonizable on $M$ if and only if $M$ is non–compact.

######  Proof 5.10.

If $M$ is compact, then any function $h\in C^{\infty}(M)$ has a critical
point. Since $X$ has no critical points, the equation
$\pi^{\sharp}\mathrm{d}h=X$ cannot globally hold on $M$. Conversely, if $M$ is
non–compact, then the orbit space of the vector field $N:=M/\mathbb{S}^{1}$ is
also non–compact. So, there exists $f\in C^{\infty}(N)$ with no critical
points. In particular, if $p:M\to N$ denotes the canonical projection to the
orbit space, then $h:=p^{*}f$ is a regular first integral of $X$. By
Proposition 5.7, the result follows.

We now illustrate with an example our hamiltonization criteria for periodic
vector fields.

######  Example 5.11.

Consider the torus $\mathbb{T}^{2}$ with natural coordinates
$(\varphi_{1},\varphi_{2})$, $\varphi_{i}\in\mathbb{R}/2\pi\mathbb{Z}$. Recall
that, for coprime integers $m$ and $n$, the vector field

$\Upsilon=m\frac{\partial}{\partial\varphi_{1}}+n\frac{\partial}{\partial\varphi_{2}}$

is the infinitesimal generator of an $\mathbb{S}^{1}$–action on
$\mathbb{T}^{2}$. Now, fix a $2\pi$–periodic $G\in C^{\infty}(\mathbb{R})$,
and define

$X:=\varpi\Upsilon,$

with $\varpi(\varphi_{1},\varphi_{2}):=G(n\varphi_{1}-m\varphi_{2})$. On the
other hand, the function
$h(\varphi_{1},\varphi_{2}):=F(n\varphi_{1}-m\varphi_{2})$, for some
$2\pi$–periodic $F\in C^{\infty}(\mathbb{R})$, is a first integral of $X$
(indeed, every first integral of $\Upsilon$ is of this form). Moreover, for
integers $r,s$ satisfying $nr-ms=1$, the vector field

$Y=\frac{r}{F^{\prime}(n\varphi_{1}-m\varphi_{2})}\frac{\partial}{\partial\varphi_{1}}+\frac{s}{F^{\prime}(n\varphi_{1}-m\varphi_{2})}\frac{\partial}{\partial\varphi_{2}}$

defines an invariant and $h$–normalized normal class relative to $X$ on its
domain. Therefore, $X$ is a Hamiltonian vector field with respect to the
Poisson bivector field

$\pi:=Y\wedge
X=\frac{\varpi(\varphi_{1},\varphi_{2})}{F^{\prime}(n\varphi_{1}-m\varphi_{2})}\frac{\partial}{\partial\varphi_{1}}\wedge\frac{\partial}{\partial\varphi_{2}},$

and Hamiltonian function $h$. In particular, if
$\int_{0}^{2\pi}G(t)\mathrm{d}t=0$, then we can take $F$ as an antiderivative
of $G$. In this case, $X$ is Hamiltonian respect to the canonical Poisson
vector field
$\pi=\frac{\partial}{\partial\varphi_{1}}\wedge\frac{\partial}{\partial\varphi_{2}}$
on the whole $\mathbb{T}^{2}$. $\triangleleft$

###  5.2 Hamiltonization of infinitesimal generators

Here, we adapt the result of Theorem 5.1 to obtain a hamiltonization criteria
for infinitesimal generators of proper actions.

######  Proposition 5.12.

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, acting properly on
$M$. For each $\xi\in\mathfrak{g}$, let $\xi_{M}$ be the infinitesimal
generator of the $G$–action. Suppose that there exists a $G$–invariant
function $h\in C^{\infty}(M)$ with no critical points. Then, there exists a
linear map $\xi\mapsto\pi_{\xi}$ from the Lie algebra $\mathfrak{g}$ of $G$ to
a vector space of Poisson structures of rank at most two such that
$\pi_{\xi}^{\sharp}\mathrm{d}h=\xi_{M}$.

######  Proof 5.13.

Since $G$ acts properly on $M$, there exists an invariant riemannian metric on
$M$. Let the dual metric be denoted by $\eta\in\Gamma(TM\otimes TM)$, and set
$Y:=\tfrac{\eta^{\sharp}\mathrm{d}h}{\eta(\mathrm{d}h,\mathrm{d}h)}$. Clearly,
$\mathrm{d}h(Y)=1$. On the other hand, since $h$ and $\eta$ are invariant, we
get that $Y$ is invariant. In other words, for each $\xi\in\mathfrak{g}$, we
have $[\xi_{M},Y]=0$, and $\mathrm{d}h(\xi_{M})=0$. By setting
$\pi_{\xi}:=Y\wedge\xi_{M}$, the result follows.

We now illustrate the previous ideas to the vector fields of Example 3.11.

######  Example 5.14.

Consider the vector fields on $\mathbb{R}^{3}$

$\displaystyle X_{1}$ $\displaystyle=2xz\tfrac{\partial}{\partial
x}+2yz\tfrac{\partial}{\partial
y}+(1-x^{2}-y^{2}+z^{2})\tfrac{\partial}{\partial z},$ $\displaystyle X_{2}$
$\displaystyle=y\tfrac{\partial}{\partial x}-x\tfrac{\partial}{\partial y}.$

Note that $X_{1}$ and $X_{2}$ commute and have periodic flow. Therefore, the
flows of $X_{1}$ and $X_{2}$ induce a $\mathbb{T}^{2}$–action on
$\mathbb{R}^{3}$, which is free on

$M=\mathbb{R}^{3}\setminus(\\{z-\text{axis}\\}\cup\\{x^{2}+y^{2}=1,z=0\\}).$

Indeed, $X_{1}$ and $X_{2}$ have as common first integral
$h=\frac{x^{2}+y^{2}}{(x^{2}+y^{2}+z^{2}+1)^{2}}$, whose level sets on $M$ are
the orbits of the $\mathbb{T}^{2}$–action. Now, let $Y_{0}$ any vector field
on $M$ such that $\mathrm{d}h(Y_{0})=1$. Since $h$ is
$\mathbb{T}^{2}$–invariant, the averaged vector field $Y:=\langle
Y_{0}\rangle^{\mathbb{T}^{2}}$ is invariant and also satisfies
$\mathrm{d}h(Y)=1$. Note that for each $\lambda\in\mathbb{R}$, the Poisson
structure

$\pi_{\lambda}:=Y\wedge X_{\lambda},\qquad\text{where
}X_{\lambda}:=X_{1}+\lambda X_{2},$

is such that $\pi_{\lambda}^{\sharp}\mathrm{d}h=X_{\lambda}$. In other words,
the vector field $X_{\lambda}$ is hamiltonizable by Proposition 5.12,
regardless of the value of $\lambda$. Moreover, recall from Example 3.11 that
if $\lambda$ is irrational, then the orbits of $X_{\lambda}$ are dense on the
level sets of $h$, which are compact. Thus, its flow does not define a proper
action, and therefore, the vector field $X_{\lambda}$ cannot be hamiltonized
by the result of Proposition 5.5. $\triangleleft$

##  6 Hamiltonization of toric actions

In this section we address the question on hamiltonization of Lie group
actions for the particular case in which the acting Lie group is a torus. We
adapt some of the ideas developed in Section 4 to provide a Poisson structure
so that a given toric action is Hamiltonian with momentum map.

Let $X_{1},\ldots,X_{k}\in\Gamma(TM)$ be infinitesimal generators of an action
of $\mathbb{T}^{k}$ on $M$. Suppose also that we are given
$\mathbb{T}^{k}$–invariant functions $h_{1},\ldots,h_{k}\in C^{\infty}(M)$
satisfying the following condition: there exist
$Y_{1},\ldots,Y_{k}\in\Gamma(TM)$ such that

$\sum_{j=1}^{k}\mathrm{d}h_{i}(Y_{j})X_{j}=X_{i},\text{ for every
}i\in\\{1,\ldots,k\\}.$ (6.1)

######  Remark 6.1.

In the case $k=1$, condition (6.1) means that $h_{1}$ is a first integral of
$X_{1}$ that is regular on $\mathrm{supp}(X_{1})$, due to Lemma 4.5.

Let also be $\mathbb{R}^{k}=\mathrm{Lie}(\mathbb{T}^{k})$ be the abelian Lie
algebra of $\mathbb{T}^{k}$. For each
$\xi=(\xi_{1},\ldots,\xi_{k})\in\mathbb{R}^{k}$, consider the infinitesimal
generator $\xi_{M}:=\sum_{i=1}^{k}\xi_{i}X_{i}$, and $h_{\xi}\in
C^{\infty}(M)$ given by $h_{\xi}:=\sum_{i=1}^{k}\xi_{i}h_{i}$.

Now, define $\pi\in\Gamma(\wedge^{2}TM)$ by

$\pi:=\sum_{j=1}^{k}Y_{j}\wedge X_{j}.$ (6.2)

######  Lemma 6.2.

For each $\xi\in\mathbb{R}^{k}$, the identity
$\pi^{\sharp}\mathrm{d}h_{\xi}=\xi_{M}$ holds. In particular, in the case when
$\pi$ is Poisson, the $\mathbb{T}^{k}$–action on $M$ is hamiltonian with
momentum map $\mathbb{J}:M\to(\mathbb{R}^{k})^{*}$ given by
$\mathbb{J}(x)(\xi):=h_{\xi}(x)$.

######  Proof 6.3.

Since $h_{i}$ is a first integral of $X_{i}$, we have from the condition (6.1)
that
$\pi^{\sharp}\mathrm{d}h_{i}=\sum_{j=1}^{k}\mathrm{d}h_{i}(Y_{j})X_{j}=X_{i}$.
Therefore,
$\pi^{\sharp}\mathrm{d}h_{\xi}=\sum_{i=1}^{k}\xi_{i}\pi^{\sharp}\mathrm{d}h_{i}=\sum_{i=1}^{k}\xi_{i}X_{i}=\xi_{M}$.

Now, let us study conditions under which $\pi$ is a Poisson structure. First,
observe that, since the action is abelian, the infinitesimal generators
$X_{i}$ commute, $[X_{i},X_{j}]=0$. Moreover, by averaging with respect to the
$\mathbb{T}^{k}$–action, and taking into account the invariance of each
$h_{i}$ and $X_{i}$, one can assume that the vector fields $Y_{i}$ given in
the condition (6.1) are also invariant, $[X_{i},Y_{j}]=0$. Therefore,

$[\pi,\pi]=\sum_{1\leq i,j\leq k}[Y_{i},Y_{j}]\wedge X_{i}\wedge X_{j}.$

######  Lemma 6.4.

Let $U_{ij}\subseteq M$ be the open set on which $X_{i}\wedge X_{j}\neq 0$.
If, for every $i,j\in\\{1,\ldots,k\\}$, one has
$[Y_{i},Y_{j}]|_{U_{ij}}=a_{ij}X_{i}+b_{ij}X_{j}$, for some $a_{ij},b_{ij}\in
C^{\infty}(U_{ij})$, then $\pi$ is a Poisson structure on $M$.

###  6.1 Actions of 2–dimensional torus

Recall that the Poisson property for $\pi$ in (6.2) holds in particular for
$k=1$: this is the content of Proposition 5.7. Now, we focus on the case of an
action of a 2–dimensional torus on $M$, with infinitesimal generators
$X_{1},X_{2}$, and invariant functions $h_{1},h_{2}$ satisfying the condition

$\mathrm{d}h_{i}(Y_{1})X_{1}+\mathrm{d}h_{i}(Y_{2})X_{2}=X_{i},\quad i=1,2,$
(6.3)

for some $\mathbb{T}^{2}$–invariant $Y_{1},Y_{2}\in\Gamma(TM)$. We then have
that the bivector field

$\pi=Y_{1}\wedge X_{1}+Y_{2}\wedge X_{2}$ (6.4)

satisfies $[\pi,\pi]=[Y_{1},Y_{2}]\wedge X_{1}\wedge X_{2}$.

We denote by $U$ the open set in which the vector fields $X_{1}$ and $X_{2}$
are independent.

######  Lemma 6.5.

For $i=1,2$, we have that $\mathrm{d}h_{i}[Y_{1},Y_{2}]|_{U}=0$. In
particular, $\mathbf{i}_{\mathrm{d}h_{i}}[\pi,\pi]=0$.

######  Proof 6.6.

From the condition (6.3), we have that
$\mathrm{L}_{Y_{i}}h_{j}=\mathrm{d}h_{j}(Y_{i})=\delta_{ij}$ holds on $U$.
Then,
$\mathrm{d}h_{i}[Y_{1},Y_{2}]=\mathrm{L}_{Y_{1}}\mathrm{L}_{Y_{2}}h_{i}-\mathrm{L}_{Y_{2}}\mathrm{L}_{Y_{1}}h_{i}=0$
on $U$.

######  Proposition 6.7.

Suppose that $\operatorname{dim}{M}=4$. Then, the bivector field $\pi$ in
(6.4) is Poisson on $M$, and has rank four on $U$.

######  Proof 6.8.

The condition (6.3) implies that $h_{1}$ and $h_{2}$ are functionally
independent on $U$, so their common level sets induce a 2–dimensional
foliation $\mathcal{L}$ of $U$. Moreover, by the invariance of $h_{1}$ and
$h_{2}$, the independent vector fields $X_{1}$ and $X_{2}$ generate the
tangent bundle of $\mathcal{L}$. On the other hand, by Lemma 6.5, the vector
field $[Y_{1},Y_{2}]$ is tangent to $\mathcal{L}$ on $U$. In particular, the
hypothesis in Lemma 6.4 is satisfied, proving that $\pi$ is Poisson on $M$.
Finally, from the condition (6.3), the vector fields $Y_{1}$ and $Y_{2}$ are
independent, and normal to $\mathcal{L}$ on $U$. Therefore, the vector fields
$Y_{1},Y_{2},X_{1},X_{2}$ are independent, and $\pi$ has maximal rank on $U$.

In general, the orbits of an action are contained in the level sets of
invariant functions. In Proposition 6.7, the dimension hypothesis allows to
describe the orbits as level sets of invariant functions. This motives the
consideration of the following setting, in which the dimension hypothesis can
be relaxed: the orbits and the level sets agree along the leaves of a
foliation of the manifold, with an additional transversality condition.

Denote by $\mathcal{L}$ the foliation of $M$ given by the common level sets of
$h_{1}$ and $h_{2}$, and by $\mathcal{O}$ the foliation of $M$ by the orbits
of the $\mathbb{T}^{2}$–action.

######  Definition 6.9.

A foliation $\mathcal{S}$ of $M$ is said to be _compatible_ with $\mathcal{L}$
and $\mathcal{O}$ if the following conditions are satisfied:

  1. 1.

The foliations $\mathcal{S}$ and $\mathcal{L}$ are transversal,
$T\mathcal{S}+T\mathcal{L}=TM$.

  2. 2.

The leaves of the foliation $\mathcal{S}$ are invariant, $T\mathcal{O}\subset
T\mathcal{S}$.

  3. 3.

On the open set $U$ in which $X_{1}\wedge X_{2}\neq 0$, one has
$T\mathcal{O}=T\mathcal{S}\cap T\mathcal{L}$.

Observe that, in the context of Proposition 6.7, such a compatible foliation
exists: it consists of the connected components of $M$. More generally, we
have:

######  Proposition 6.10.

The bivector field $\pi$ in (6.4) is Poisson on $M$ if and only if there
exists a foliation $\mathcal{S}$ on $M$ that is compatible with $\mathcal{L}$
and $\mathcal{O}$ on $U$, and $Y_{1},Y_{2}$ are tangent to $\mathcal{S}|_{U}$.

######  Proof 6.11.

By the condition (6.3), the invariant functions $h_{1}$ and $h_{2}$ are
independent on $U$, as well as $Y_{1}$ and $Y_{2}$. In particular, the
foliation $\mathcal{L}$ is regular on $U$. Now, suppose that there exists a
foliation $\mathcal{S}$ on $M$ such that is compatible with $\mathcal{L}$ and
$\mathcal{O}$ on $U$, and $Y_{1},Y_{2}\in\Gamma(T\mathcal{S}|_{U})$. Then,
$[Y_{1},Y_{2}]|_{U}\in\Gamma(T\mathcal{S})$. On the other hand, by Lemma 6.5,
we have that $\mathrm{d}h_{i}[Y_{1},Y_{2}]|_{U}=0$, so
$[Y_{1},Y_{2}]|_{U}\in\Gamma(T\mathcal{L})$. By the compatibility of
$\mathcal{S}$, this implies that $[Y_{1},Y_{2}]|_{U}\in\Gamma(T\mathcal{O})$.
Finally, since $X_{1}$, and $X_{2}$ generate $T\mathcal{O}$, we have that
$[Y_{1},Y_{2}]\wedge X_{1}\wedge X_{2}|_{U}=0$. From here, and the fact that
$X_{1}\wedge X_{2}=0$ outside of $U$, we conclude that $\pi$ is Poisson on
$M$. Conversely, suppose that $\pi$ is Poisson. Let us show that the
symplectic foliation $\mathcal{S}$ of $\pi$ is compatible with $\mathcal{L}$
and $\mathcal{O}$ on $U$, and satisfies
$Y_{1},Y_{2}\in\Gamma(T\mathcal{S}|_{U})$. By (6.3), we have
$\mathrm{d}h_{j}(X_{i})=0$, and $\mathrm{d}h_{j}(Y_{i})=\delta_{ji}$ on $U$,
so the vector fields $Y_{1},Y_{2},X_{1},X_{2}$ are independent on $U$. Thus,
there exist $\alpha_{1},\alpha_{2}\in\Gamma(T^{*}U)$ such that
$\alpha_{i}(X_{j})=-\delta_{ij}$, and $\alpha_{i}(Y_{j})=0$. Then,
$\pi^{\sharp}\alpha_{i}=Y_{i}$, so $Y_{1}$, and $Y_{2}$ are tangent to
$\mathcal{S}$ on $U$. From here, we also have that $\mathcal{S}$ and
$\mathcal{L}$ are transversal on $U$:

$T\mathcal{S}|_{U}+T\mathcal{L}|_{U}\supseteq\mathrm{span}\\{Y_{1}|_{U},Y_{2}|_{U}\\}+T\mathcal{L}|_{U}=TM|_{U}$

Finally, if $v\in(T\mathcal{S}\cap T\mathcal{L})|_{U}$, then
$v=a_{1}X_{1}+a_{2}X_{2}+b_{1}Y_{2}+b_{2}Y_{2}$, with
$b_{i}=\mathrm{d}h_{i}(v)=0$. So, $v\in T\mathcal{O}$, showing that
$T\mathcal{S}\cap T\mathcal{L}=T\mathcal{O}$.

We now give sufficient conditions under which the vector fields $Y_{1}$ and
$Y_{2}$ in (6.3) can be chosen as sections of a given compatible foliation.

######  Lemma 6.12.

Let $V$ be a $\mathbb{T}^{2}$–saturated open set satisfying
$\mathrm{supp}(X_{1})\cup\mathrm{supp}(X_{2})\subseteq V$ and
$\mathrm{d}h_{1}\wedge\mathrm{d}h_{2}\neq 0$ on $V$. If $\mathcal{S}$ is a
compatible regular foliation on $V$, then there exist
$\mathbb{T}^{2}$–invariant vector fields $Y_{1},Y_{2}\in\Gamma(TM)$ tangent to
$\mathcal{S}$ on $V$ and satisfying (6.3).

######  Proof 6.13.

Since $\mathrm{d}h_{1}$ and $\mathrm{d}h_{2}$ are invariant and independent in
the saturated open set $V$, there exist invariant
$\widetilde{Y}_{1},\widetilde{Y}_{2}\in\Gamma(TV)$ satisfying
$\mathrm{d}h_{i}(\widetilde{Y}_{j})=\delta_{ij}$. Moreover, recall that
$\mathcal{S}$ and $\mathcal{L}$ are regular and transversal on $V$, so there
exist $Y^{\prime}_{1},Y^{\prime}_{2}\in\Gamma(T\mathcal{S})$, and
$Z^{\prime}_{1},Z^{\prime}_{2}\in\Gamma(T\mathcal{L})$ such that
$\widetilde{Y}_{j}=Y^{\prime}_{j}+Z^{\prime}_{j}$. Furthermore, since
$T\mathcal{O}\subseteq T\mathcal{S}\cap T\mathcal{L}$, we have that the action
restricts to the leaves of these foliations. So, by averaging and taking into
account the invariance of $\widetilde{Y}_{i}$, we may assume that
$Y^{\prime}_{j},Z^{\prime}_{j}$ are invariant. Since
$\mathrm{d}h_{i}(Z^{\prime}_{j})=0$, we have
$\mathrm{d}h_{i}(Y^{\prime}_{j})=\delta_{ij}$. On the other hand, observe that
the closed sets $C:=\mathrm{supp}(X_{1})\cup\mathrm{supp}(X_{2})$ and
$M\setminus V$ are disjoint. Thus, there exist disjoint open sets $V^{\prime}$
and $W^{\prime}$ such that $C\subseteq W^{\prime}$ and $M\setminus V\subseteq
V^{\prime}$. Let $\mu\in C^{\infty}(M)$ be such that $\mu|_{C}=1$ and
$\mu|_{\overline{V^{\prime}}}=0$. Since $X_{1},X_{2}$ are trivial outside $C$,
and $\mu$ is constant on $C$, we have that $\mu$ is invariant. Then, there
exist smooth, global, and invariant vector fields $Y_{1},Y_{2}\in\Gamma(TM)$,
well–defined by $Y_{j}:=\mu Y^{\prime}_{j}$ on $V$, and $Y_{j}=0$ on
$V^{\prime}$. It is left to show that $Y_{1},Y_{2}$ satisfy the condition
(6.3). First, observe that the condition trivially holds on $M\setminus C$. On
the other hand, since $Y_{j}$ agrees with $Y^{\prime}_{j}$ on $C\subseteq V$,
we get that $\mathrm{d}h_{i}(Y_{j})=\delta_{ij}$ on $C$. Therefore, the
condition is satisfied.

######  Theorem 6.14.

Let $X_{1}$ and $X_{2}$ be infinitesimal generators of a
$\mathbb{T}^{2}$–action on the manifold $M$, and $h_{1},h_{2}\in
C^{\infty}(M)$ $\mathbb{T}^{2}$–invariant functions independent on a saturated
open neighborhood $V$ of $\mathrm{supp}(X_{1})\cup\mathrm{supp}(X_{2})$.
Suppose that there exists a compatible regular foliation $\mathcal{S}$ on $V$,
in the sense of Definition 6.9. Then, there exists a Poisson structure $\pi$
on $M$ such that the $\mathbb{T}^{2}$–action is Hamiltonian on $(M,\pi)$ with
momentum map $\mathbb{J}:M\to(\mathbb{R}^{2})^{*}$ given by
$\mathbb{J}(x)(\xi):=\xi_{1}h_{1}(x)+\xi_{2}h_{2}(x)$, $\xi\in\mathbb{R}^{2}$.

######  Proof 6.15.

Let $U$ be the open set in which $X_{1}\wedge X_{2}\neq 0$. By Lemma 6.12,
there exist invariant vector fields $Y_{1},Y_{2}$ on $M$ tangent to
$\mathcal{S}$ on $V$ satisfying (6.3). Since $\mathcal{S}$ is compatible on
$U$, the bivector field $\pi$ in (6.4) is Poisson, by Proposition 6.10.
Finally, the fact that $\mathbb{J}$ is a momentum map for the
$\mathbb{T}^{2}$–action follows from Lemma 6.2

######  Remark 6.16.

We believe that Lemma 6.12, and hence Theorem 6.14, are still true in the more
general case when the invariant functions $h_{1},h_{2}$ satisfy the condition
(6.3), not only when $\mathrm{supp}(X_{1})\cup\mathrm{supp}(X_{2})\subseteq
V$. However, establishing such fact requires a more profound analysis of the
condition (6.3), perhaps by providing an intrinsic formulation of it.

#### ACKNOWLEDGMENTS.

This research was partially supported by the Mexican National Council of
Science and Technology (CONACYT) under the grant CB2015 no. 258302 and the
University of Sonora (UNISON) under the project no. USO315007338. J.C.R.P.
thanks CONACyT for a postdoctoral fellowship held during the production of
this work. E.V.B. was supported by FAPERJ grants E-26/202.411/2019 and
E-26/202.412/2019.

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