# Quadratic, homogeneous and Kolmogorov vector fields on $S^{1}\times S^{2}$
and $S^{2}\times S^{1}$

Supriyo Jana  Department of Mathematics, Indian Institute of Technology
Madras, India<EMAIL_ADDRESS>and  Soumen Sarkar  Department of
Mathematics, Indian Institute of Technology Madras, India<EMAIL_ADDRESS>
###### Abstract.

In this paper, we consider the following two algebraic hypersurfaces

$S^{1}\times
S^{2}=\\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1=0;\,a>1\\}$

and

$S^{2}\times
S^{1}=\\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1=0;\,b>1\\}$

embedded in $\mathbb{R}^{4}$. We study polynomial vector fields in
$\mathbb{R}^{4}$ separately, having $S^{1}\times S^{2}$ and $S^{2}\times
S^{1}$ invariant by their flows. We characterize all linear, quadratic, cubic
Kolmogorov and homogeneous vector fields on $S^{1}\times S^{2}$ and
$S^{2}\times S^{1}$. We construct some first integrals of these vector fields
and find which of the vector fields are Hamiltonian. We give upper bounds for
the number of the invariant meridian and parallel hyperplanes of these vector
fields. In addition, we have shown that the upper bounds are sharp in many
cases.

###### Key words and phrases:

Polynomial vector fields, First integrals, Algebraic hypersurfaces, Quadratic
vector fields, Kolmogorov systems, Lotka-Volterra systems

###### 2020 Mathematics Subject Classification:

34A34, 34C14, 34C40, 37J35

##  1\. Introduction

Let $\mathbb{R}[x_{1},x_{2},x_{3},x_{4}]$ be the polynomial ring over real in
4 variables. Consider the following polynomial differential system in
$\mathbb{R}^{4}$

(1.1) $\frac{dx_{i}}{dt}=P_{i}(x_{1},x_{2},x_{3},x_{4})\,;i=1,...,4$

and its associated polynomial vector field

(1.2) $\chi=\sum\limits_{i=1}^{4}P_{i}\frac{\partial}{\partial x_{i}}$

where $P_{1},...,P_{4}\in\mathbb{R}[x_{1},x_{2},x_{3},x_{4}]$ of degrees
$n_{1},...,n_{4}$ respectively. For simplicity, we may write
$\chi:=(P_{1},...,P_{4})$. The number $n:=\max\\{n_{1},...,n_{4}\\}$ is called
the degree of the polynomial vector field (1.2). Polynomial vector fields of
degrees 1, 2 and 3 are called linear, quadratic and cubic vector fields,
respectively.

Polynomial vector fields on various algebraic manifolds embedded in either of
$\mathbb{R}^{2}$ and $\mathbb{R}^{3}$ are already well-studied, for instance,
see [14, 3, 10, 12, 18, 5]. However, only very few investigations are there in
$\mathbb{R}^{4}$ ([15, 11]). The advantage of having a polynomial vector field
on an algebraic manifold is that if a solution curve of a differential system
has an intersection with the manifold, then the whole solution curve lies on
that manifold.

Kolmogorov system in $\mathbb{R}^{4}_{+}$ has been studied in [15]. Then the
integrability of a class of Lotka-Volterra and Kolmogorov systems in
$\mathbb{R}^{n}$ have been explored in [13]. We note that a Lotka-Volterra
system (i.e., a degree two polynomial Kolmogorov system) in $\mathbb{R}^{N}$
describes the evolution of $N$ conflicting species in population biology. It
appears in many different areas like neural networks, laser physics, plasma
physics, etc.( see for instance [4, 8, 19]).

A first integral of the associated vector field of a differential system
reduces the dimension of the system by one, which makes the analysis of the
system easier. Therefore, finding first integrals is extremely important in
the qualitative theory of differential equations. However, the study of the
existence or non-existence of first integrals is a difficult problem in
general.If the vector field is Hamiltonian, then one easily gets at least one
first integral. In addition, if there is another independent first integral of
a Hamiltonian system then it is called integrabile in Liouville sense. A
Liouville integrable system satisfy the assumptions of the famous Liouville
theorem, see [1]. So, the corresponding system can be solved by quadratures,
i.e., by solving a finite number of algebraic equations and computing a finite
number of definite integrals. Also, Darboux theory of integrability [7,
Chapter 8] is very useful to compute first integrals.

We consider two algebraic hypersurfaces embedded in $\mathbb{R}^{4}$, the
products of spheres $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$, which are
manifolds of dimension 3. In [2], Benny and the second author proved that
product of two arbitrary spheres can be expressed as the zero set of a
polynomial. From [2], we found

(1.3) $S^{1}\times
S^{2}=\\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1=0;~{}a>1\\}$

and

(1.4) $S^{2}\times
S^{1}=\\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1=0;~{}b>1\\}.$

In this paper, we characterize all linear, quadratic, cubic Kolmogorov and
homogeneous vector fields on $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$. We
construct some first integrals of these vector fields and find which of the
vector fields are Hamiltonian. We give upper bounds for the number of the
invariant meridian and parallel hyperplanes of these vector fields.

We note that there is a diffeomorphism, possibly not polynomial, between
$S^{1}\times S^{2}$ and $S^{2}\times S^{1}$. However, the algebraic equations
in (1.3) and (1.4) are algebraically different. Therefore, we study them as
different hypersurfaces in $\mathbb{R}^{4}$.

This paper is organized in the following way. In Section 2, we present some
basic definitions and useful results on polynomial vector fields. We recall
the criteria to search for the invariant hypersurfaces of a vector field.

In Section 3, we characterize linear vector fields on $S^{1}\times S^{2}$. We
give a necessary and sufficient condition when a quadratic and a cubic
Kolmogorov vector field in $\mathbb{R}^{4}$ is a vector field on $S^{1}\times
S^{2}$, see Theorem 3.2 and 3.6 respectively. We show that there is no Lotka-
Volterra vector field on $S^{1}\times S^{2}$, see Corollary 3.4. We prove that
there is no Hamiltonian of a quadratic and cubic Kolmogorov vector field on
$S^{1}\times S^{2}$, see Theorem 3.5 and 3.7 respectively. Moreover, we
constructed rational first integral for them in those theorems. If
$\chi=(P_{1},P_{2},P_{3},P_{4})$ is a vector field in $\mathbb{R}^{4}$ with
each $P_{i}$ is a non-zero homogeneous polynomial, then we give some
sufficient as well as some necessary conditions for
${(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1}$ to be a first
integral of $\chi$, see Lemma 3.8 and Theorem 3.13 respectively. We
characterize Type-$n$ vector fields on $S^{1}\times S^{2}$, see Theorem 3.9
and study some properties of these vector fields.

In Section 4, we characterize linear vector fields on $S^{2}\times S^{1}$. We
exhibit a necessary and sufficient condition when a quadratic and a cubic
Kolmogorov vector field in $\mathbb{R}^{4}$ is a vector field on $S^{2}\times
S^{1}$, see Theorem 4.2 and 4.6 respectively. We prove that there is no Lotka-
Volterra vector field on $S^{1}\times S^{2}$; see Corollary 4.4. We show that
there is no Hamiltonian of a quadratic and cubic Kolmogorov vector field on
$S^{2}\times S^{1}$, see Theorem 4.5 and Theorem 4.7 respectively. Moreover,
we constructed rational first integral for them in those theorems. If
$\chi=(P_{1},P_{2},P_{3},P_{4})$ is a vector field in $\mathbb{R}^{4}$ with
each $P_{i}$ is a homogeneous polynomial, then we discuss some sufficient as
well as some necessary conditions for
${(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-a^{2})^{2}+x_{4}^{2}-1}$ is a first integral
of $\chi$, see Lemma 4.8 and Lemma 4.9 respectively. We characterize Pseudo
Type-$n$ vector field on $S^{2}\times S^{1}$, see Theorem 4.11 and study some
properties of these vector fields.

In Section 5, we investigate the maximum number of invariant meridian and
parallel hyperplanes of the vector fields on $S^{1}\times S^{2}$ and
$S^{2}\times S^{1}$. We show when these bounds can be reached. We conclude
this paper by establishing that there is no degree one polynomial
diffeomorphism between $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$, see
Theorem 6.1.

##  2\. Preliminaries

Let $f\in\mathbb{R}[x_{1},...,x_{4}]$ be a non-constant polynomial. The set
$\\{f=0\\}\subset\mathbb{R}^{4}$ is called an invariant algebraic hypersurface
of the vector field $\chi$ if there exists a polynomial
${K\in\mathbb{R}[x_{1},...,x_{4}]}$ such that

$\chi f=\sum\limits_{i=1}^{4}P_{i}\frac{\partial f}{\partial x_{i}}=Kf.$

Here, the polynomial $K$ is called the cofactor of $\chi$ for hypersurface
$\\{f=0\\}$.

One can observe that the gradient $(\frac{\partial f}{\partial
x_{1}},\cdots,\frac{\partial f}{\partial x_{4}})$ of $f$ is orthogonal to the
vector field $\chi=(P_{1},...,P_{4})$ at the points of the algebraic
hypersurface $\\{f=0\\}$. So the vector field $\chi$ is tangent to the
hypersurface $\\{f=0\\}$. Hence the hypersurface $\\{f=0\\}$ is formed by the
orbits of the vector field $\chi$. This justifies the name “invariant
algebraic hypersurface” given to $\\{f=0\\}$ because it is invariant under the
flow determined by $\chi$. Therefore, if a solution curve of system (1.1) has
a point on the algebraic hypersurface $f$, then the whole solution curve is
contained in $f$.

In this paper, we consider polynomial vector fields such that $S^{1}\times
S^{2}$ and $S^{2}\times S^{1}$ of (1.3) and (1.4) respectively are invariant
algebraic hypersurfaces of $\chi$. If $S^{1}\times S^{2}$ is invariant under
$\chi$, then $\chi$ is called a polynomial vector field on $S^{1}\times
S^{2}$. Similarly, we call $\chi$ a polynomial vector field on $S^{2}\times
S^{1}$ if $S^{2}\times S^{1}$ is invariant under $\chi$.

Such vector fields are called polynomial vector fields on $S^{1}\times S^{2}$
and polynomial vector fields on $S^{2}\times S^{1}$ respectively.

One of the best tools in order to search for invariant algebraic hypersurfaces
is the following. Let $W$ be a vector subspace of
$\mathbb{R}[x_{1},...,x_{4}]$ generated by the independent polynomials
$v_{1},...,v_{l}$, i.e., $W=\langle v_{1},...,v_{\ell}\rangle$. The extactic
polynomial of $\chi$ associated with $W$ is the polynomial

$\mathcal{E}_{W}(\chi)=\begin{vmatrix}v_{1}&\dots&v_{\ell}\\\
\chi(v_{1})&\dots&\chi(v_{\ell})\\\ \dots&\dots&\dots\\\
\chi^{{\ell}-1}(v_{1})&\dots&\chi^{{\ell}-1}(v_{\ell})\end{vmatrix}$

where $\chi^{j}(v_{i})=\chi^{j-1}(\chi(v_{i}))$ for all $i,j$. From the
properties of determinant, it follows that the definition of the extactic
polynomial is independent of the chosen basis of $W$. The proof of the
following is similar to the proof of [9, Proposition 1].

######  Proposition 2.1.

Let $\chi$ be a polynomial vector field in $\mathbb{R}^{n}$ and $W$ a finite
dimensional vector subspace of $\mathbb{R}[x_{1},x_{2},\cdots,x_{n}]$ with
$\dim(W)>1$. Then, an invariant algebraic hypersurface given by $\\{f=0\\}$
for the vector field $\chi$, with $f\in W$, is a factor of
$\mathcal{E}_{W}(\chi)$.

The multiplicity of an invariant algebraic hypersurface $f=0$ with $f\in W$ is
the largest positive integer $k$ such that $f^{k}$ divides the polynomial
$\mathcal{E}_{W}(\chi)$ when $\mathcal{E}_{W}(\chi)\neq 0$, otherwise the
multiplicity is infinite. For more details on the multiplicity, see [6, 17].

######  Definition 2.2.

Let $U$ be an open subset of $\mathbb{R}^{4}$. A non-constant analytic map
$H\colon U\to\mathbb{R}$ is called a first integral of the vector field (1.2)
on $U$ if $H$ is constant on all solution curves of system (1.1) contained in
$U$ ; i.e. $H(x_{1}(t),x_{2}(t),x_{3}(t),x_{4}(t))=$ constant for all values
of $t$ for which the solution $(x_{1}(t),x_{2}(t),x_{3}(t),x_{4}(t))$ is
defined and contained in $U$.

Note that $H$ is a first integral of the vector field (1.2) on $U$ if and only
if $\chi H=0$ on $U$.

######  Definition 2.3.

Given a non-constant analytic function $H\colon\mathbb{R}^{4}\to\mathbb{R}$,
the vector field

$\chi=-\frac{\partial H}{\partial x_{2}}\frac{\partial}{\partial
x_{1}}+\frac{\partial H}{\partial x_{1}}\frac{\partial}{\partial
x_{2}}-\frac{\partial H}{\partial x_{4}}\frac{\partial}{\partial
x_{3}}+\frac{\partial H}{\partial x_{3}}\frac{\partial}{\partial x_{4}}$

is called a Hamiltonian vector field in $\mathbb{R}^{4}$ with Hamiltonian $H$.

So, if the system (1.1) is written by

$P_{1}=-\frac{\partial H}{\partial x_{2}},P_{2}=\frac{\partial H}{\partial
x_{1}},P_{3}=-\frac{\partial H}{\partial x_{4}},P_{4}=\frac{\partial
H}{\partial x_{3}}$

then the vector field (1.2) is Hamiltonian. Observe that a Hamiltonian $H$ of
a Hamiltonian vector field is a first integral of that vector field.

######  Definition 2.4.

A Hamiltonian system in $\mathbb{R}^{4}$ with Hamiltonian $H$ is called
integrable in the Liouville sense in $\mathbb{R}^{4}$ if the system has a
first integral $F$, which is independent with the Hamiltonian $H$.

Note that the Poisson bracket between $H$ and $F$ is zero if and only if $F$
is a first integral of the Hamiltonian system in $\mathbb{R}^{4}$ with
Hamiltonian $H$. See details in [1].

######  Definition 2.5.

The system (1.1) is called a polynomial Kolmogorov system in $\mathbb{R}^{4}$
when $P_{i}=x_{i}\psi_{i}$ and its associated vector field is called
polynomial Kolmogorov vector field where
$\psi_{i}\in\mathbb{R}[x_{1},...,x_{4}]$ for $1\leq i\leq 4$.

In addition, if $\psi_{i}$ is a linear polynomial for each $i$, then the
polynomial Kolmogorov system is called the Lotka-Volterra system in
$\mathbb{R}^{4}$, and its associated vector field is called Lotka-Volterra
vector field.

######  Definition 2.6.

Suppose $\chi=(P_{1},...,P_{4})$ is a polynomial vector field in
$\mathbb{R}^{4}$. We say $\chi$ is Type-$n$ vector field when
$P_{1},...,P_{4}$ are homogeneous polynomials of degree $n$ and $\chi$ is
Pseudo Type-$n$ vector field when $P_{1},P_{2},P_{3}$ are homogeneous
polynomials of degree $n$ but $P_{4}=0$.

##  3\. Polynomial vector fields on $S^{1}\times S^{2}$

In this section, we study and characterize polynomial vector fields on
$S^{1}\times S^{2}$. Recall the presentation of $S^{1}\times
S^{2}\subset\mathbb{R}^{4}$ from (1.3). A vector field
$\chi=(P_{1},...,P_{4})$ in $\mathbb{R}^{4}$ is a vector field on $S^{1}\times
S^{2}$ if

(3.1)
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=K((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1)$

for some $K\in\mathbb{R}[x_{1},...,x_{4}]$.

######  Lemma 3.1.

Let $Q_{1},Q_{2},R_{1},R_{2}\in\mathbb{R}[x_{1},...,x_{4}]$ be such that
${Q_{1}R_{1}+Q_{2}R_{2}}$ is the zero polynomial and $\gcd(R_{1},R_{2})=1$.
Then $Q_{1}=AR_{2},Q_{2}=-AR_{1}$ for some polynomial
${A\in\mathbb{R}[x_{1},...,x_{4}].}$

###### Proof.

$Q_{1}R_{1}+Q_{2}R_{2}=0$ implies that $Q_{1}R_{1}=-Q_{2}R_{2}.$ Hence $R_{2}$
divides $Q_{1}$ and $R_{1}$ divides $Q_{2}$ since $\gcd(R_{1},R_{2})=1.$
Assuming $Q_{1}=AR_{2}$ and $Q_{2}=A^{{}^{\prime}}R_{1}$, we get
$(A+A^{{}^{\prime}})R_{1}R_{2}=0$. This implies that $A^{{}^{\prime}}=-A$. ∎

######  Theorem 3.2.

Let $\chi=(P_{1},...,P_{4})$ be a quadratic vector field in $\mathbb{R}^{4}$.
Then $\chi$ is a quadratic vector field on $S^{1}\times S^{2}$ if and only if

(3.2) $\begin{split}P_{1}&=\frac{1}{4}Kx_{1}+fx_{2},\\\
P_{2}&=\frac{1}{4}Kx_{2}-fx_{1},\\\
P_{3}&=\frac{k_{3}}{2}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}+a^{4}-1)+gx_{4},\quad\text{and}\\\
P_{4}&=\frac{k_{4}}{2}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}+a^{4}-1)-gx_{3}\end{split}$

where $K=k_{3}x_{3}+k_{4}x_{4}$ for some $k_{3},k_{4}\in\mathbb{R}$ and $f,g$
are linear polynomials. Moreover, $K$ is the cofactor of $\chi$ for
$S^{1}\times S^{2}$.

###### Proof.

Suppose $\chi$ is a quadratic vector field on $S^{1}\times S^{2}$. So (3.1)
gives the following.

(3.3)
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(\sum_{i=1}^{2}P_{i}x_{i})+2(\sum_{i=3}^{4}P_{i}x_{i})=(\sum\limits_{j=1}^{4}k_{j}x_{j})((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1),$

since there is no constant term on the left side. We write

(3.4) $P_{i}=P_{i}^{(2)}+P_{i}^{(1)}+P_{i}^{(0)}$

where $P_{i}^{(j)}$ is the degree $j$ homogeneous part of the polynomial
$P_{i}$. Comparing the degree 5 terms in (3.3), we get

$4(x_{1}^{2}+x_{2}^{2})(\sum_{i=1}^{2}P_{i}^{(2)}x_{i})=(\sum\limits_{j=1}^{4}k_{j}x_{j})(x_{1}^{2}+x_{2}^{2})^{2}.$

Hence

(3.5)
$\sum_{i=1}^{2}P_{i}^{(2)}x_{i}=\frac{1}{4}(\sum\limits_{j=1}^{4}k_{j}x_{j})(x_{1}^{2}+x_{2}^{2}).$

Also comparing the degree $4$ terms in (3.3), we get

$4(x_{1}^{2}+x_{2}^{2})(\sum_{i=1}^{2}P_{i}^{(1)}x_{i})=0.$

Hence $\sum\limits_{i=1}^{2}P_{i}^{(1)}x_{i}=0$. Then, by Lemma 3.1, we obtain
$P_{1}^{(1)}=Ax_{2}$ and $P_{2}^{(1)}=-Ax_{1}$ for some $A\in\mathbb{R}$.

In particular, if $x_{3}=x_{4}=0$, then (3.3) gives the following.

(3.6)
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(\sum_{i=1}^{2}P_{i}(x_{1},x_{2},0,0)x_{i})=(k_{1}x_{1}+k_{2}x_{2})((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}-1).$

Comparing the degree 3 terms in (3.6), we get

$4(x_{1}^{2}+x_{2}^{2})(\sum_{i=1}^{2}P_{i}^{(0)}(x_{1},x_{2},0,0)x_{i})-4a^{2}(\sum_{i=1}^{2}P_{i}^{(2)}(x_{1},x_{2},0,0)x_{i})=-2a^{2}(\sum_{i=1}^{2}k_{i}x_{i})(x_{1}^{2}+x_{2}^{2}).$

Then, for $x_{3}=0=x_{4}$, the above equation together with (3.5) implies the
following.

(3.7) $4(\sum_{i=1}^{2}P_{i}^{(0)}x_{i})=-a^{2}(k_{1}x_{1}+k_{2}x_{2}).$

Also comparing the degree 1 terms in (3.6), we get

(3.8)
$-4a^{2}(\sum_{i=1}^{2}P_{i}^{(0)}x_{i})=(a^{4}-1)(k_{1}x_{1}+k_{2}x_{2}).$

Now, from (3.7) and (3.8), we obtain $k_{1}=k_{2}=0$. Hence
$P_{1}^{(0)}=P_{2}^{(0)}=0$. Therefore, from (3.5), we get the following.

(3.9)
$\sum_{i=1}^{2}P_{i}^{(2)}x_{i}=\frac{1}{4}(k_{3}x_{3}+k_{4}x_{4})(x_{1}^{2}+x_{2}^{2}).$

Taking $K:=k_{3}x_{3}+k_{4}x_{4}$, the above equation implies that

$(P_{1}^{(2)}-\frac{1}{4}Kx_{1})x_{1}+(P_{2}^{(2)}-\frac{1}{4}Kx_{2})x_{2}=0.$

By Lemma 3.1,

$P_{1}^{(2)}=\frac{1}{4}Kx_{1}+Bx_{2}\text{ and
}P_{2}^{(2)}=\frac{1}{4}Kx_{2}-Bx_{1}$

where $B$ is a linear homogeneous polynomial. Therefore, we can get $P_{1}$
and $P_{2}$ since all the homogeneous parts are obtained for both the
polynomials.

Let us now find $P_{3}$ and $P_{4}$. Comparing the degree 1 terms in (3.3), we
get

$P_{3}^{(0)}=\frac{a^{4}-1}{2}k_{3},P_{4}^{(0)}=\frac{a^{4}-1}{2}k_{4}.$

Comparing the degree 2 terms in (3.3), we get
$P_{3}^{(1)}x_{3}+P_{4}^{(1)}x_{4}=0$ which follows that
$P_{3}^{(1)}=Cx_{4},P_{4}^{(1)}=-Cx_{3}$ for some $C\in\mathbb{R}$ using Lemma
3.1. Now comparing the degree 3 terms in (3.3), we get

$-4a^{2}(\sum_{i=1}^{2}P_{i}^{(2)}x_{i})+2\sum_{i=3}^{4}P_{i}^{(2)}x_{i}=(k_{3}x_{3}+k_{4}x_{4})(-2a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}).$

Then using (3.9) in the above equation, we have

$\sum_{i=3}^{4}P_{i}^{(2)}x_{i}=\frac{1}{2}(k_{3}x_{3}+k_{4}x_{4})(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}).$

This implies

$(P_{3}^{(2)}-\frac{1}{2}k_{3}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}))x_{3}+(P_{4}^{(2)}-\frac{1}{2}k_{4}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}))x_{4}=0.$

Therefore, by Lemma 3.1,

$P_{3}^{(2)}=\frac{1}{2}k_{3}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2})+Dx_{4}\quad\text{and}\quad
P_{4}^{(2)}=\frac{1}{2}k_{4}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2})-Dx_{3}$

where $D\in\mathbb{R}[x_{1},...,x_{4}]$ is a linear homogeneous polynomial.
Hence, $P_{3},P_{4}$ are completely determined.

If $P_{1},\ldots,P_{4}$ are given by (3.2), then they satisfy (3.1). Thus, the
converse part is true. ∎

######  Corollary 3.3.

$\chi=(P_{1},...,P_{4})$ is a linear vector field on $S^{1}\times S^{2}$ if
and only if there exists $\alpha,\beta\in\mathbb{R}$ such that

$P_{1}=\alpha x_{2},P_{2}=-\alpha x_{1},P_{3}=\beta x_{4},P_{4}=-\beta x_{3}.$

Observe that if $P_{1},...,P_{4}\neq 0$ then a linear vector field
$\chi=(P_{1},...,P_{4})$ on $S^{1}\times S^{2}$ is a Type-1 vector field.

###### Proof.

The vector field (3.2) on $S^{1}\times S^{2}$ is linear if and only if $K=0$
and $f,g$ are constants. Hence the result follows. ∎

######  Corollary 3.4.

There is no Lotka-Volterra vector field on $S^{1}\times S^{2}$.

###### Proof.

Suppose $\chi=(P_{1},...,P_{4})$ defines a Lotka-Volterra vector field on
$S^{1}\times S^{2}$. Then $x_{i}$ divides $P_{i}$ for $i\in\\{1,...,4\\}$.
Since $\chi$ is a quadratic vector field, it will be of the form (3.2). Now
$x_{1}$ divides $P_{1}$ and $x_{2}$ divides $P_{2}$ implies $x_{1}$ divides
$f$ and $x_{2}$ divides $f$ respectively. So $x_{1}x_{2}$ divides $f$.
Therefore, $f$ must be zero since it is linear.

Again, $x_{3}$ divides $P_{3}$ implies $k_{3}=0$ and $x_{3}$ divides $g$.
Similarly, $x_{4}$ divides $P_{4}$ implies $k_{4}=0$ and $x_{4}$ divides $g$.
So $x_{3}x_{4}$ divides $g$. Therefore, $g$ must be zero since it is linear.
Hence $P_{1}=\cdots=P_{4}=0$. ∎

######  Theorem 3.5.

There is no Hamiltonian of a quadratic vector field on $S^{1}\times S^{2}$.
However, every quadratic vector field on $S^{1}\times S^{2}$ has a rational
first integral.

###### Proof.

Any quadratic vector field on $S^{1}\times S^{2}$ can be described by (3.2).
Suppose $H$ is a Hamiltonian of the quadratic vector field (3.2) on
$S^{1}\times S^{2}$, where $f=f_{0}+\sum\limits_{i=1}^{4}f_{i}x_{i}$ and
${g=g_{0}+\sum\limits_{i=1}^{4}g_{i}x_{i}}$; $f_{i},g_{i}\in\mathbb{R}$. Since
${-\frac{\partial H}{\partial
x_{2}}=P_{1}=\frac{1}{4}Kx_{1}+(f_{0}+\sum\limits_{i=1}^{4}f_{i}x_{i})x_{2}}$,
we get

$H=-\frac{1}{4}Kx_{1}x_{2}-\frac{1}{2}(f_{0}+f_{1}x_{1}+f_{3}x_{3}+f_{4}x_{4})x_{2}^{2}-\frac{1}{3}f_{2}x_{2}^{3}+H_{1}(x_{1},x_{3},x_{4}),$

where $H_{1}$ is a function independent of $x_{2}$. Now, from $\frac{\partial
H}{\partial x_{1}}=P_{2},$ we get

$-\frac{1}{4}Kx_{2}-\frac{1}{2}f_{1}x_{2}^{2}+\frac{\partial H_{1}}{\partial
x_{1}}=\frac{1}{4}Kx_{2}-f_{0}x_{1}-(\sum_{i=1}^{4}f_{i}x_{i})x_{1}.$

Thus, $(-\frac{1}{2}K+f_{1}x_{2}+f_{2}x_{1})x_{2}=\frac{\partial
H_{1}}{\partial x_{1}}+(f_{0}+f_{3}x_{3}+f_{4}x_{4})x_{1}$. Therefore,

(3.10) $K=f_{1}=f_{2}=0~{}\mbox{and}~{}\frac{\partial H_{1}}{\partial
x_{1}}=-(f_{0}+f_{3}x_{3}+f_{4}x_{4})x_{1},$

since $H_{1}$ is independent of $x_{2}$ and $K=k_{3}x_{3}+k_{4}x_{4}$. Hence,

$H_{1}=-\frac{1}{2}(f_{0}+f_{3}x_{3}+f_{4}x_{4})x_{1}^{2}+H_{2}(x_{3},x_{4}),$

where $H_{2}$ is a function of $x_{3},x_{4}$ only. Therefore,
$H=-\frac{1}{2}(f_{0}+f_{3}x_{3}+f_{4}x_{4})(x_{1}^{2}+x_{2}^{2})+H_{2}$. Note
that $P_{3}=gx_{4},P_{4}=-gx_{3}$ since $K=0$. Now, $-\frac{\partial
H}{\partial x_{4}}=P_{3}$ implies that

$\frac{1}{2}f_{4}(x_{1}^{2}+x_{2}^{2})-\frac{\partial H_{2}}{\partial
x_{4}}=(g_{0}+\sum_{i=1}^{4}g_{i}x_{i})x_{4}.$

Thus,

$\frac{1}{2}f_{4}(x_{1}^{2}+x_{2}^{2})-(g_{1}x_{1}+g_{2}x_{2})x_{4}=\frac{\partial
H_{2}}{\partial x_{4}}+(g_{0}+g_{3}x_{3}+g_{4}x_{4})x_{4}$

Therefore,

(3.11) $f_{4}=g_{1}=g_{2}=0~{}\quad\mbox{and }\quad\frac{\partial
H_{2}}{\partial x_{4}}=-(g_{0}+g_{3}x_{3}+g_{4}x_{4})x_{4},$

since $H_{2}$ is a funcion of $x_{3},x_{4}$ only. Hence,

$H_{2}=-\frac{1}{2}(g_{0}+g_{3}x_{3})x_{4}^{2}-\frac{1}{3}g_{4}x_{4}^{3}+H_{3}(x_{3}),$

where $H_{3}$ is a function of $x_{3}$ only. Hence

$H=-\frac{1}{2}(f_{0}+f_{3}x_{3})(x_{1}^{2}+x_{2}^{2})-\frac{1}{2}(g_{0}+g_{3}x_{3})x_{4}^{2}-\frac{1}{3}g_{4}x_{4}^{3}+H_{3}.$

Again, $\frac{\partial H}{\partial x_{3}}=P_{4}$ implies that
$-\frac{1}{2}f_{3}(x_{1}^{2}+x_{2}^{2})-\frac{1}{2}g_{3}x_{4}^{2}+H^{{}^{\prime}}_{3}=-(g_{0}+g_{3}x_{3}+g_{4}x_{4}).$
Hence

(3.12) $f_{3}=g_{3}=g_{4}=0,$

since $H_{3}$ is a function of $x_{3}$ only. Combining (3.10), (3.11) and
(3.12), we obtain $f_{i}=g_{i}=0$ for $i\in\\{1,...,4\\}$. Hence the
polynomial vector field becomes linear, which contradicts the assumption. So,
there is no Hamiltonian of a quadratic vector field on $S^{1}\times S^{2}$.

Notice that the hypersurfaces given by
$(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1=0$ and
$x_{1}^{2}+x_{2}^{2}=0$ are two invariant algebraic hypersurfaces of the
vector field (3.2) with cofactor $K$ and $\frac{1}{2}K$ respectively.
Therefore, by Darboux Integrability Theory (Theorem 5, [16]), the function
${((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1)(x_{1}^{2}+x_{2}^{2})^{-2}}$
is a rational first integral of the vector field (3.2). ∎

######  Theorem 3.6.

Let $\chi=(P_{1},...,P_{4})$ be a cubic Kolmogorov vector field in
$\mathbb{R}^{4}$. Then $\chi$ is a vector field on $S^{1}\times S^{2}$ if and
only if

(3.13) $\begin{split}P_{1}&=x_{1}(\frac{1}{4}K+\alpha x_{2}^{2}),\\\
P_{2}&=x_{2}(\frac{1}{4}K-\alpha x_{1}^{2}),\\\
P_{3}&=x_{3}(\frac{k_{33}}{2}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}+a^{4}-1)+\beta
x_{4}^{2}),\quad\text{and}\\\
P_{4}&=x_{4}(\frac{k_{44}}{2}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}+a^{4}-1)-\beta
x_{3}^{2})\end{split}$

where $K=k_{33}x_{3}^{2}+k_{44}x_{4}^{2}$ with $k_{33},k_{44}\in\mathbb{R}$.
Moreover, $K$ is the cofactor of $\chi$ for $S^{1}\times S^{2}$.

###### Proof.

Consider a cubic Kolmogorov vector field $\chi=(P_{1},...,P_{4})$ in
$\mathbb{R}^{4}$ with

$P_{i}=x_{i}\psi_{i}$

where $\psi_{i}\in\mathbb{R}[x_{1},...,x_{4}]$ is a quadratic polynomial for
$i=1,...,4$. We write

$\psi_{i}=\psi_{i}^{(2)}+\psi_{i}^{(1)}+\psi_{i}^{(0)}$

as in (3.4).

Suppose $\chi$ is a vector field on $S^{1}\times S^{2}$. Then it must satisfy

(3.14)
$4(\sum_{i=1}^{2}x_{i}^{2}-a^{2})(\sum_{i=1}^{2}\psi_{i}x_{i}^{2})+2(\sum_{i=3}^{4}\psi_{i}x_{i}^{2})=(\sum\limits_{1\leq
i\leq j\leq
4}k_{ij}x_{i}x_{j})((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1),$

since there is no constant and degree 1 term on the left side. Comparing the
degree 6 terms in (3.14), we get

$4(x_{1}^{2}+x_{2}^{2})(\sum_{i=1}^{2}\psi_{i}^{(2)}x_{i}^{2})=(\sum\limits_{1\leq
i\leq j\leq 4}k_{ij}x_{i}x_{j})(x_{1}^{2}+x_{2}^{2})^{2}.$

Hence

(3.15) $\sum_{i=1}^{2}\psi_{i}^{(2)}x_{i}^{2}=\frac{1}{4}(\sum\limits_{1\leq
i\leq j\leq 4}k_{ij}x_{i}x_{j})(x_{1}^{2}+x_{2}^{2}).$

Also comparing the degree $5$ terms in (3.14), we obtain

$4(x_{1}^{2}+x_{2}^{2})(\sum_{i=1}^{2}\psi_{i}^{(1)}x_{i}^{2})=0.$

Hence $\sum\limits_{i=1}^{2}\psi_{i}^{(1)}x_{i}^{2}=0$. This implies
$x_{1}^{2}$ divides $\psi_{2}^{(1)}$ and $x_{2}^{2}$ divides $\psi_{1}^{(1)}$.
Therefore, $\psi_{1}^{(1)},\psi_{2}^{(2)}$ must be zero since they are linear
polynomials.

In particular, if $x_{3}=x_{4}=0$, then (3.14) gives the following.

(3.16)
$4(\sum_{i=1}^{2}x_{i}^{2}-a^{2})(\sum_{i=1}^{2}\psi_{i}(x_{1},x_{2},0,0)x_{i}^{2})=(\sum\limits_{1\leq
i\leq j\leq 2}k_{ij}x_{i}x_{j})((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}-1).$

Comparing the degree 4 terms in (3.16), we get

$4(\sum_{i=1}^{2}x_{i}^{2})(\sum_{i=1}^{2}\psi_{i}^{(0)}(x_{1},x_{2},0,0)x_{i}^{2})-4a^{2}(\sum_{i=1}^{2}\psi_{i}^{(2)}(x_{1},x_{2},0,0)x_{i}^{2})=-2a^{2}(\sum\limits_{1\leq
i\leq j\leq}k_{ij}x_{i}x_{j})(\sum_{i=1}^{2}x_{i}^{2}).$

Then, for $x_{3}=0=x_{4}$, the above equation together with (3.15) implies the
following.

(3.17) $4(\sum_{i=1}^{2}\psi_{i}^{(0)}x_{i}^{2})=-a^{2}(\sum\limits_{1\leq
i\leq j\leq 2}k_{ij}x_{i}x_{j}).$

Also, comparing the degree 2 terms in (3.16), we get

(3.18)
$-4a^{2}(\sum_{i=1}^{2}\psi_{i}^{(0)}x_{i}^{2})=(a^{4}-1)(\sum\limits_{1\leq
i\leq j\leq 2}k_{ij}x_{i}x_{j}).$

From (3.17) and (3.18), we obtain $k_{ij}=0$ for $1\leq i\leq j\leq 2$. Hence
$\psi_{1}^{(0)}=\psi_{2}^{(0)}=0$. Therefore, from (3.15), we get

(3.19)
$\sum_{i=1}^{2}\psi_{i}^{(2)}x_{i}^{2}=\frac{1}{4}(\sum\limits_{\begin{subarray}{c}1\leq
i\leq 4,\,3\leq j\leq 4\\\ i\leq
j\end{subarray}}k_{ij}x_{i}x_{j})(x_{1}^{2}+x_{2}^{2}).$

Taking $K:=\sum\limits_{\begin{subarray}{c}1\leq i\leq 4,\,3\leq j\leq 4\\\
i\leq j\end{subarray}}k_{ij}x_{i}x_{j}$, the equation (3.19) implies that

$(\psi_{1}^{(2)}-\frac{1}{4}K)x_{1}^{2}+(\psi_{2}^{(2)}-\frac{1}{4})x_{2}^{2}=0.$

Hence, by Lemma 3.1, $\psi_{1}^{(2)}=\frac{1}{4}K+\alpha x_{2}^{2}$ and
$\psi_{2}^{(2)}=\frac{1}{4}K-\alpha x_{1}^{2}$ for some $\alpha\in\mathbb{R}$.
So, we have found all the homogeneous components of $\psi_{1}$ and $\psi_{2}$.

Let us now find $\psi_{3}$ and $\psi_{4}$. Comparing the coefficients of the
degree 2 monomials in (3.14), we obtain

$\psi_{3}^{(0)}=\frac{a^{4}-1}{2}k_{33},\quad\psi_{4}^{(0)}=\frac{a^{4}-1}{2}k_{44}\text{
and }k_{ij}=0\text{ for }1\leq i\leq 4,3\leq j\leq 4,i<j.$

Hence $K=k_{33}x_{3}^{2}+k_{44}x_{4}^{2}.$ Comparing the degree 3 terms in
(3.14), we get $\sum\limits_{i=3}^{4}\psi_{i}^{(1)}x_{i}^{2}=0$ which follows
that ${\psi_{3}^{(1)}=\psi_{4}^{(1)}=0}$, since these are linear polynomials.
Now comparing the degree 4 terms in (3.14), we get

$-4a^{2}(\sum_{i=1}^{2}\psi_{i}^{(2)}x_{i}^{2})+2(\sum_{i=3}^{4}\psi_{i}^{(2)}x_{i}^{2})=(k_{33}x_{3}^{2}+k_{44}x_{4}^{2})(-2a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}).$

Then using (3.19) in the above equation, we have

$\sum_{i=3}^{4}\psi_{i}^{(2)}x_{i}^{2}=\frac{1}{2}(k_{33}x_{3}^{2}+k_{44}x_{4}^{2})(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}).$

This implies

$(\psi_{3}^{(2)}-\frac{1}{2}k_{33}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}))x_{3}^{2}+(\psi_{4}^{(2)}-\frac{1}{2}k_{44}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}))x_{4}^{2}=0.$

Therefore, by Lemma 3.1,

$\psi_{3}^{(2)}=\frac{1}{2}k_{33}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2})+\beta
x_{4}^{2}\quad\text{and}\quad\psi_{4}^{(2)}=\frac{1}{2}k_{44}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2})-\beta
x_{3}^{2}$

for some $\beta\in\mathbb{R}$. Hence, $\psi_{3},\psi_{4}$ are completely
determined.

If $P_{1},\ldots,P_{4}$ are given by (3.13), then they satisfy (3.1). Thus,
the converse part is true. ∎

######  Theorem 3.7.

There is no Hamiltonian of a cubic Kolmogorov vector field on $S^{1}\times
S^{2}$. However, every cubic Kolmogorov vector field on $S^{1}\times S^{2}$
has a rational first integral.

###### Proof.

Any cubic Kolmogorov vector field on $S^{1}\times S^{2}$ can be described by
(3.13). Suppose $H$ is a Hamiltonian of the vector field (3.13). Then

$H=x_{1}x_{2}(\sum\limits_{i+j=0}^{2}a_{ij}x_{1}^{i}x_{2}^{j})+x_{3}x_{4}(\sum\limits_{i+j=0}^{2}b_{ij}x_{3}^{i}x_{4}^{j})+c_{0}x_{1}x_{2}x_{3}x_{4}$

by [15, Theorem 1]. Since $-\frac{\partial H}{\partial x_{2}}=P_{1}$, we get

$-x_{1}(\sum\limits_{i+j=0}^{2}a_{ij}x_{1}^{i}x_{2}^{j})-x_{1}x_{2}(a_{01}+a_{11}x_{1}+2a_{02}x_{2})-c_{0}x_{1}x_{3}x_{4}=x_{1}(\frac{1}{4}K+\alpha
x_{2}^{2}).$

Thus, $K=a_{00}=a_{10}=a_{01}=a_{11}=a_{20}=c_{0}=0$. Hence,
$H=a_{02}x_{1}x_{2}^{3}+x_{3}x_{4}(\sum\limits_{i+j=0}^{2}b_{ij}x_{3}^{i}x_{4}^{j})$.
Again, $-\frac{\partial H}{\partial x_{4}}=P_{3}$ implies

$-x_{3}(\sum\limits_{i+j=0}^{2}b_{ij}x_{3}^{i}x_{4}^{j})-x_{3}x_{4}(b_{01}+b_{11}x_{3}+2b_{02}x_{4})=\beta
x_{3}x_{4}^{2}.$

Then $b_{00}=b_{10}=b_{01}=b_{11}=b_{20}=0$. Hence,
$H=a_{02}x_{1}x_{2}^{3}+b_{02}x_{3}x_{4}^{3}$. Now from $\frac{\partial
H}{\partial x_{1}}=P_{2}=-\alpha x_{1}^{2}x_{2}$ and $\frac{\partial
H}{\partial x_{3}}=P_{4}=-\beta x_{3}^{2}x_{4}$, we obtain that $H=0$, which
is absurd. Hence, a cubic Kolmogorov vector field on $S^{1}\times S^{2}$
cannot have any Hamiltonian.

Let $\chi=(P_{1},...,P_{4})$ be a vector field of the form (3.13). Then the
hypesurfaces given by
$(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1=0$ and
$x_{1}^{2}+x_{2}^{2}=0$ are invariant algebraic hypersurfaces of $\chi$ with
cofactors $K$ and $\frac{1}{2}K$ respectively. Hence by Darboux Integrability
Theory ([16, Theorem 5]), the function
$((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1)(x_{1}^{2}+x_{2}^{2})^{-2}$
is a rational first integral of $\chi$. ∎

######  Lemma 3.8.

Suppose $\chi=(P_{1},...,P_{4})$ is a polynomial vector field on $S^{1}\times
S^{2}$ where $P_{i}$ are homogeneous polynomials with
$\deg(P_{1})=\deg(P_{2})=m$ and $\deg(P_{3})=\deg(P_{4})=n$. If $m-1\leq n\leq
m+3$ then $G=(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1$ is a first
integral of $\chi$.

###### Proof.

Each monomial in the expression of $\chi G$ has degree either $m+3$, $m+1$ or
$n+1$. If $\chi$ is a vector field on $S^{1}\times S^{2}$ then $\chi
G=KG=K((x_{1}^{2}+x_{2}^{2})^{2}-2a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2})+K(a^{4}-1)$
for some $K\in\mathbb{R}[x_{1},...,x_{4}]$.

If $m-1\leq n\leq m+2$ then $\chi G$ has degree at most $m+3$. So $K$ has the
degree at most $m-1$. Hence $K(a^{4}-1)$ has degree at most $m-1$. However,
the degree of each monomial in the expression of $\chi G$ is at least $m$.
Hence $K$ must be zero.

If $n=m+3$ then $\chi G$ has degree $m+4$. So $K$ has degree $m$. Hence
$K(a^{4}-1)$ has degree $m$. However, the degree of each monomial in the
expression of $\chi G$ is at least $m+1$. Hence $K$ must be zero. Therefore,
$G$ must be a first integral of $\chi$, since in each case above $K=0$. ∎

######  Theorem 3.9.

Let $\chi=(P_{1},...,P_{4})$ be a Type-$n$ vector field in $\mathbb{R}^{4}$.
Then $\chi$ is a Type-$n$ vector field on $S^{1}\times S^{2}$ if and only if
there exists $A,B\in\mathbb{R}[x_{1},...,x_{4}]$ such that

(3.20) $P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=Bx_{4},P_{4}=-Bx_{3}.$

###### Proof.

By Lemma 3.8, any Type-$n$ vector field satisfies the following.

$\begin{split}&4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=0\\\
\implies&4(x_{1}^{2}+x_{2}^{2})(P_{1}x_{1}+P_{2}x_{2})-4a^{2}(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=0.\end{split}$

Comparing the same degree parts in the above equation, we get
$4(x_{1}^{2}+x_{2}^{2})(P_{1}x_{1}+P_{2}x_{2})=0$ and
$-4a^{2}(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=0$. Then,
$P_{1}x_{1}+P_{2}x_{2}=0$ since $x_{1}^{2}+x_{2}^{2}\neq 0$. Hence
$P_{3}x_{3}+P_{4}x_{4}=0$. Now by Lemma 3.1,
$P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=Bx_{4}$ and $P_{4}=-Bx_{3}$ for some
$A,B\in\mathbb{R}[x_{1},...,x_{4}]$.

If $P_{1},...,P_{4}$ are given by (3.20), then they satisfy (3.1). So, the
converse part follows. ∎

######  Remark 3.10.

Similar to the above proof, using the argument of separating the same degree
terms of $\chi G$, it can be proved that if $\chi=(P_{1},...,P_{4})$ is a
polynomial vector field in $\mathbb{R}^{4}$ satisfying the hypothesis of Lemma
3.8, it will be a vector field on $S^{1}\times S^{2}$ if and only if there
exists $A,B\in\mathbb{R}[x_{1},...,x_{4}]$ such that

$P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=Bx_{4},P_{4}=-Bx_{3}.$

######  Proposition 3.11.

There exists no Pseudo Type-$n$ vector field on $S^{1}\times S^{2}$.

###### Proof.

Let $\chi=(P_{1},P_{2},P_{3},0)$ be a Pseudo Type-$n$ vector field on
$S^{1}\times S^{2}$. Thus,

$4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2P_{3}x_{3}=K((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1)$

for some $K\in\mathbb{R}[x_{1},...,x_{4}]$ with $\deg K\leq n-1$. So, $K=0$,
since the degree of each term on the left side is greater than $n$. Then,
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2P_{3}x_{3}=0$. Thus,
$x_{1}^{2}+x_{2}^{2}-a^{2}$ is a factor of $P_{3}$. This is not possible, as
$P_{3}$ is homogeneous and $a\neq 0$.

∎

######  Proposition 3.12.

Any Type-$1$ vector field on $S^{1}\times S^{2}$ is Hamiltonian.

###### Proof.

Suppose $\chi$ is a Type-1 vector field on $S^{1}\times S^{2}$. Then, by
Theorem (3.9),

$P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=Bx_{4},P_{4}=-Bx_{3}\quad\text{for some
}A,B\in\mathbb{R}.$

One can check that
$H=-\frac{A}{2}(x_{1}^{2}+x_{2}^{2})-\frac{B}{2}(x_{3}^{2}+x_{4}^{2})$ is a
Hamiltonian for $\chi$. ∎

######  Theorem 3.13.

Let $P_{i}\in\mathbb{R}[x_{1},...,x_{4}]$ be non-zero homogeneous polynomials
such that ${\deg(P_{i})=m_{i}}$ for $1\leq i\leq 4$. Assume that
$G=(x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1$ is a first integral
of $\chi=(P_{1},...,P_{4})$. Then the following holds.

  1. (a)

If $m_{1}\neq m_{2}$ then $\max(m_{1},m_{2})=\min(m_{1},m_{2})+2$ and

$\\{m_{3},m_{4}\\}=\\{\max(m_{1},m_{2})+2,\min(m_{1},m_{2})\\}$.

  2. (b)

If $m_{1}=m_{2}$ then either $m_{3}=m_{4}$ or
$\\{m_{3},m_{4}\\}=\\{m_{1},m_{1}+2\\}$

###### Proof.

Since $G$ is a first integral of $\chi$, $\chi G=0$, i.e.,

(3.21)
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=0.$

  1. (a)

Since $m_{1}\neq m_{2}$, $P_{1}x_{1}+P_{2}x_{2}$ is a non-zero polynomial.
Assume that $m_{1}<m_{2}$. So either of $m_{3},m_{4}$ should be $m_{2}+2$.
Without loss of generality, suppose $m_{3}=m_{2}+2$. We rewrite $\chi G$ by
separating the same degree homogeneous parts as follows.

$\chi
G=(4(x_{1}^{2}+x_{2}^{2})P_{2}x_{2}+2P_{3}x_{3})+4(x_{1}^{2}+x_{2}^{2})P_{1}x_{1}-4a^{2}P_{1}x_{1}-4a^{2}P_{2}x_{2}+2P_{4}x_{4}.$

Notice that

$\begin{split}\deg(4(x_{1}^{2}+x_{2}^{2})P_{2}x_{2}+2P_{3}x_{3})=m_{2}+3,\deg(((x_{1}^{2}+x_{2}^{2})P_{1}x_{1})=m_{1}+3,\deg(P_{1}x_{1})=m_{1}+1,\\\
\deg(P_{2}x_{2})=m_{2}+1\text{ and }\deg(P_{4}x_{4})=m_{4}+1.\end{split}$

In the next step, we will match the degrees of the terms in $\chi G=0$ under
certain conditions.

     * •

$m_{4}=m_{2}+2$: Then there are no term of degree $m_{1}+1$ in $\chi G$ except
$P_{1}x_{1}$. Hence $P_{1}=0$, which is not possible.

     * •

$m_{4}>m_{2}+2$: Then every term of $\chi G$ has strictly lower degree than
$P_{4}x_{4}$. Hence $P_{4}=0$, which is not possible.

     * •

$m_{4}<m_{2}+2:$ Then $m_{4}\in\\{m_{1},m_{2},m_{1}+2\\}$. If $m_{4}=m_{1}$
then $m_{1}+3$ must be $m_{2}+1$, i.e., $m_{2}=m_{1}+2$. If $m_{4}=m_{2}$ or
$m_{1}+2$, then $P_{1}x_{1}=0$ since $P_{1}x_{1}$ will be the only lowest
degree term in $\chi G$. Hence $P_{1}=0$, which is not possible.

So, in this case, $m_{2}=m_{1}+2,m_{3}=m_{2}+2,m_{4}=m_{1}$.

  2. (b)

Now consider $m_{1}=m_{2}$. If $P_{1}x_{1}+P_{2}x_{2}=0$ then
$P_{3}y+P_{4}z=0$ from (3.21), which implies $m_{3}=m_{4}$.

Suppose $P_{1}x_{1}+P_{2}x_{2}\neq 0$. We rewrite $\chi G$ by separating same
degree homogeneous parts as follows.

$\chi
G=4(x_{1}^{2}+x_{2}^{2})(P_{1}x_{1}+P_{2}x_{2})-4a^{2}(P_{1}x_{1}+P_{2}x_{2})+2P_{3}x_{3}+2P_{4}x_{4}.$

Notice that

$\begin{split}\deg(4(x_{1}^{2}+x_{2}^{2})(P_{1}x_{1}+P_{2}x_{2})=m_{1}+3,\deg(-4a^{2}(P_{1}x_{1}+P_{2}x_{2}))=m_{1}+1,\\\
\deg(2P_{3}x_{3})=m_{3}+1,\deg(2P_{4}x_{4})=m_{4}+1.\end{split}$

Observe that $m_{1}+3$ must be either $m_{3}+1$ or $m_{4}+1$. Otherwise, from
$\chi G=0$, we obtain $P_{1}x_{1}+P_{2}x_{2}=0$, which contradicts our
assumption. Without loss of generality, suppose $m_{1}+3=m_{3}+1$. Then
$m_{1}+1=m_{4}+1$, since $P_{1}x_{1}+P_{2}x_{2}$ and $P_{4}$ are nonzero.

∎

######  Theorem 3.14.

A Type-$n$ vector field on $S^{1}\times S^{2}$ has at least two independent
first integrals.

###### Proof.

Suppose $f$ is a first integral of the Type-$n$ vector field (3.20) on
$S^{1}\times S^{2}$. Then $\sum\limits_{i=1}^{4}P_{i}\frac{\partial
f}{\partial x_{i}}=0$. This implies

$Ax_{2}\frac{\partial f}{\partial x_{1}}-Ax_{1}\frac{\partial f}{\partial
x_{2}}+Bx_{4}\frac{\partial f}{\partial x_{3}}-Bx_{3}\frac{\partial
f}{\partial x_{4}}=0.$

By [20, Theorem 3(Page 53)], solutions of the above can be given in terms of
the solutions of the equations

$\frac{dx_{1}}{Ax_{2}}=\frac{dx_{2}}{-Ax_{1}}=\frac{dx_{3}}{Bx_{4}}=\frac{dx_{4}}{-Bx_{3}}=\frac{df}{0}=\frac{x_{1}dx_{1}+x_{2}dx_{2}}{0}=\frac{x_{3}dx_{3}+x_{4}dx_{4}}{0}.$

Hence, we obtain two independent first integrals $x_{1}^{2}+x_{2}^{2}$ and
$x_{3}^{2}+x_{4}^{2}$. ∎

######  Corollary 3.15.

Any Type-$n$ Hamiltonian vector field on $S^{1}\times S^{2}$ is integrable in
the Liouville sense.

##  4\. Polynomial vector fields on $S^{2}\times S^{1}$

In this section, we study and characterize polynomial vector fields on
$S^{2}\times S^{1}$. Recall the presentation of $S^{2}\times
S^{1}\subset\mathbb{R}^{4}$ from (1.4). A vector field
$\chi=(P_{1},...,P_{4})$ in $\mathbb{R}^{4}$ is a vector field on $S^{2}\times
S^{1}$ if

(4.1)
$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})(P_{1}x_{1}+P_{2}x_{2}+P_{3}x_{3})+2P_{4}x_{4}=K((x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1)$

for some $K\in\mathbb{R}[x_{1},...,x_{4}]$.

######  Lemma 4.1.

Let $n\in\mathbb{N}$ and $Q_{1},Q_{2},Q_{3}\in\mathbb{R}[x_{1},...,x_{4}]$ be
non-zero polynomials such that $Q_{1}x_{i}^{n}+Q_{2}x_{j}^{n}+Q_{3}x_{k}^{n}$
is the zero polynomial for distinct $i,j,k\in\\{1,...,4\\}$. Then
$Q_{1}=Ax_{j}^{n}+Bx_{k}^{n},Q_{2}=-Ax_{i}^{n}+Cx_{k}^{n}$ and
$Q_{3}=-Bx_{i}^{n}-Cx_{j}^{n}$ for some polynomials $A,B,C.$

###### Proof.

The equation $Q_{1}x_{i}^{n}+Q_{2}x_{j}^{n}+Q_{3}x_{k}^{n}=0$ implies that
$Q_{1}=\frac{Q_{2}x_{j}^{n}+Q_{3}x_{k}^{n}}{x_{i}^{n}}$. Hence we get
$Q_{1}=Ax_{j}^{n}+Bx_{k}^{n}$ for some polynomials $A$ and $B$ since
$x_{i},x_{j}$ and $x_{k}$ are mutually prime and any monomial in
$Q_{2}x_{j}^{n}+Q_{3}x_{k}^{n}$ has power of both $x_{j}$ and $x_{k}$ not less
than $n$. Similarly, $Q_{2}=A^{{}^{\prime}}x_{i}^{n}+B^{{}^{\prime}}x_{k}^{n}$
and $Q_{3}=A^{{}^{\prime\prime}}x_{i}^{n}+B^{{}^{\prime\prime}}x_{j}^{n}$.

Now
$Q_{3}x_{k}^{n}=-Q_{1}x_{i}^{n}-Q_{2}x_{j}^{n}=-Ax_{i}^{n}x_{j}^{n}-Bx_{i}^{n}x_{k}^{n}-A^{{}^{\prime}}x_{i}^{n}x_{j}^{n}-B^{{}^{\prime}}x_{j}^{n}x_{k}^{n}$.
This implies $x_{k}^{n}$ divides $A+A^{{}^{\prime}}$ and hence
$A+A^{{}^{\prime}}=Lx_{k}^{n}$ for some polynomial
$L\in\mathbb{R}[x_{1},...,x_{4}]$. So,
$Q_{2}=-Ax_{i}^{n}+(Lx_{i}^{n}+B^{{}^{\prime}})x_{k}^{n}$.

Again
$Q_{2}x_{j}^{n}=-Q_{1}x_{i}^{n}-Q_{3}x_{k}^{n}=-Ax_{i}^{n}x_{j}^{n}-Bx_{i}^{n}x_{k}^{n}-A^{{}^{\prime\prime}}x_{i}^{n}x_{k}^{n}-B^{{}^{\prime\prime}}x_{j}^{n}x_{k}^{n}$.
This implies that $x_{j}^{n}$ divides $A^{{}^{\prime\prime}}+B$ and hence
$A^{{}^{\prime\prime}}+B=L^{{}^{\prime}}x_{j}^{n}$ for some polynomial
$L^{{}^{\prime}}$. Hence, we have
${Q_{3}=-Bx_{i}^{n}+(L^{{}^{\prime}}x_{i}^{n}+B^{{}^{\prime\prime}})x_{j}^{n}}$.
Substituting these values of $Q_{1},Q_{2}$ and $Q_{3}$ in the equation
${Q_{1}x_{i}^{n}+Q_{2}x_{j}^{n}+Q_{3}x_{k}^{n}=0}$, we get

$\begin{split}&(Ax_{j}^{n}+Bx_{k}^{n})x_{i}^{n}+(-Ax_{i}^{n}+(x_{i}^{n}L+B^{\prime})x_{k}^{n})x_{j}^{n}+(-Bx_{i}^{n}+(x_{i}^{n}L^{\prime}+B^{\prime\prime})x_{j}^{n})x_{k}^{n}=0\\\
\implies&L^{{}^{\prime}}x_{i}^{n}+B^{{}^{\prime\prime}}=-(Lx_{i}^{n}+B^{{}^{\prime}})=0.\end{split}$

Hence, assuming $C:=Lx_{i}^{n}+B^{{}^{\prime}}$, we get our result. ∎

######  Theorem 4.2.

Let $\chi=(P_{1},...,P_{4})$ be a quadratic vector field in $\mathbb{R}^{4}$.
Then $\chi$ is a quadratic vector field on $S^{2}\times S^{1}$ if and only if

(4.2) $\begin{split}P_{1}&=\frac{c}{4}x_{1}x_{4}+fx_{2}+gx_{3},\\\
P_{2}&=\frac{c}{4}x_{2}x_{4}-fx_{1}+hx_{3},\\\
P_{3}&=\frac{c}{4}x_{3}x_{4}-gx_{1}-hx_{2},\text{ and}\\\
P_{4}&=\frac{c}{2}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}+b^{4}-1)\end{split}$

where $f,g,h$ are linear polynomials and $c\in\mathbb{R}$. Moreover,
$K=cx_{4}$ is the cofactor of $\chi$ for $S^{2}\times S^{1}$.

###### Proof.

Suppose $\chi$ is a quadratic vector field on $S^{2}\times S^{1}$. So it must
satisfy

(4.3)
$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})(\sum_{i=1}^{3}P_{i}x_{i})+2P_{4}x_{4}=(\sum\limits_{j=1}^{4}k_{j}x_{j})((x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1)$

since there is no constant term on the left side. We write

$P_{i}=P_{i}^{(2)}+P_{i}^{(1)}+P_{i}^{(0)}$

as in (3.4). Comparing degree 5 terms in (4.3), we get

$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})(\sum_{i=1}^{3}P_{i}^{(2)}x_{i})=(\sum\limits_{j=1}^{4}k_{j}x_{j})(x_{1}^{2}+x_{2}^{2}+x_{3})^{2}.$

Hence

(4.4)
$\sum_{i=1}^{3}P_{i}^{(2)}x_{i}=\frac{1}{4}(\sum\limits_{j=1}^{4}k_{j}x_{j})(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}).$

Also comparing degree $4$ terms in (4.3), we get

$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})(\sum_{i=1}^{3}P_{i}^{(1)}x_{i})=0.$

Hence $\sum_{i=1}^{3}P_{i}^{(1)}x_{i}=0$. Then, by Lemma 4.1,
$P_{1}^{(1)}=\alpha x_{2}+\beta x_{3},P_{2}^{(1)}=-\alpha x_{1}+\gamma x_{3}$
and $P_{3}^{(1)}=-\beta x_{1}-\gamma x_{2}$ for some
$\alpha,\beta,\gamma\in\mathbb{R}$.

In particular, if $x_{4}=0$, then (4.3) gives the following.

(4.5)
$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})(\sum_{i=1}^{3}P_{i}(x_{1},x_{2},x_{3},0)x_{i})=(\sum_{j=1}^{3}k_{j}x_{j})((x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}-1).$

Comparing degree 3 terms in (4.5),

$4(\sum_{i=1}^{3}x_{i}^{2})(\sum_{i=1}^{3}P_{i}^{(0)}(x_{1},x_{2},x_{3},0)x_{i})-4b^{2}(\sum_{i=1}^{3}P_{i}^{(2)}(x_{1},x_{2},x_{3},0)x_{i})=-2b^{2}(\sum_{j=1}^{3}k_{j}x_{j})(\sum_{i=1}^{3}x_{i}^{2}).$

Then, for $x_{4}=0$, the above equation together with (4.4) gives the
following.

(4.6) $4(\sum_{i=1}^{3}P_{i}^{(0)}x_{i})=-b^{2}(\sum_{j=1}^{3}k_{j}x_{j}).$

Also comparing degree 1 terms in (4.5), we get

(4.7)
$-4b^{2}(\sum_{i=1}^{3}P_{i}^{(0)}x_{i})=(b^{4}-1)(\sum_{j=1}^{3}k_{j}x_{j}).$

Now, from (4.6) and (4.7), we obtain $k_{1}=k_{2}=k_{3}=0$. Hence
$P_{1}^{(0)}=P_{2}^{(0)}=P_{3}^{(0)}=0$. Therefore, from (4.4) we get the
following.

(4.8)
$\sum_{i=1}^{3}P_{i}^{(2)}x_{i}=\frac{1}{4}k_{4}x_{4}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}).$

Then, by Lemma 4.1,
$P_{1}^{(2)}=\frac{1}{4}k_{4}x_{1}x_{4}+Ax_{2}+Bx_{3},P_{2}^{(2)}=\frac{1}{4}k_{4}x_{2}x_{4}-Ax_{1}+Cx_{3}$
and ${P_{3}^{(2)}=\frac{1}{4}k_{4}x_{3}x_{4}-Bx_{1}-Cx_{2}}$ where $A,B,C$ are
linear homogeneous polynomials. So we have found all the homogeneous parts of
$P_{1},P_{2}$ and $P_{3}$.

Let us now find $P_{4}$. Comparing degree 1 terms in (4.3), we get

$P_{4}^{(0)}=\frac{b^{4}-1}{2}k_{4}.$

Comparing degree 2 terms in (4.3), we get $P_{4}^{(1)}x_{4}=0$ which follows
that $P_{4}^{(1)}=0$. Now, comparing degree 3 terms in (4.3), we get

$-4b^{2}(\sum_{i=1}^{3}P_{i}^{(2)}x_{i})+2P_{4}^{(2)}x_{4}=k_{4}x_{4}(-2b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}).$

Then using (4.8) in the above equation, we obtain

$P_{4}^{(2)}x_{4}=\frac{1}{2}k_{4}x_{4}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}).$

Hence
$P_{4}^{(2)}=\frac{k_{4}}{2}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}).$
So, we have found all the homogeneous components of $P_{4}$. Renaming $k_{4}$
as $c$, we get our result.

If $P_{1},...,P_{4}$ are given by (4.2), then they satisfy (4.1). Thus, the
converse part is true. ∎

######  Corollary 4.3.

$\chi=(P_{1},...,P_{4})$ is a linear vector field on $S^{2}\times S^{1}$ if
and only if there exists $\alpha,\beta,\gamma\in\mathbb{R}$ such that

$P_{1}=\alpha x_{2}+\beta x_{3},P_{2}=-\alpha x_{1}+\gamma x_{3},P_{3}=-\beta
x_{1}-\gamma x_{2},\text{ and }P_{4}=0.$

Observe that if $P_{1},P_{2},P_{3}\neq 0$ then a linear vector field
$\chi=(P_{1},...,P_{4})$ on $S^{2}\times S^{1}$ is a Pseudo Type-1 vector
field.

###### Proof.

The vector field (4.2) on $S^{2}\times S^{1}$ is a linear if and only if $c=0$
and $f,g,h$ are constants. Hence the result follows. ∎

######  Corollary 4.4.

There is no Lotka-Volterra vector field on $S^{2}\times S^{1}$.

###### Proof.

Suppose $\chi=(P_{1},...,P_{4})$ defines a Lotka-Volterra vector field on
$S^{2}\times S^{1}$. Then $x_{i}$ divides $P_{i}$ for $i\in\\{1,...,4\\}$.
Since $\chi$ is a quadratic vector field, it will be of the form (4.2).
$x_{4}$ divides $P_{4}$ only when $c=0$. In that case,
$P_{4}=0,P_{1}=fx_{2}+gx_{3},P_{2}=-fx_{1}+hx_{3}$ and $P_{3}=-gx_{1}-hx_{2}$.

Suppose $P_{i}=\ell_{i}x_{i}$ for $i\in\\{1,2,3\\}$. Then
$\sum\limits_{i=1}^{3}P_{i}x_{i}=0=\sum\limits_{i=1}^{3}\ell_{i}x_{i}^{2}.$
Since $\ell_{i}$ are linear polynomials, $\ell_{i}=0$ for each
$i\in\\{1,2,3\\}$. Hence, $P_{1}=P_{2}=P_{3}=0$. ∎

######  Theorem 4.5.

There is no Hamiltonian of a quadratic vector field on $S^{2}\times S^{1}$.
However, every quadratic vector field on $S^{2}\times S^{1}$ has a rational
first integral.

###### Proof.

Any quadratic vector field on $S^{2}\times S^{1}$ can be described by (4.2).
Suppose $H$ is a Hamiltonian of the quadratic vector field (4.2) on
$S^{2}\times S^{1}$, where $f=f_{0}+\sum\limits_{i=1}^{4}f_{i}x_{i},\break
g=g_{0}+\sum\limits_{i=1}^{4}g_{i}x_{i}$ and
$h=h_{0}+\sum\limits_{i=1}^{4}h_{i}x_{i};f_{i},g_{i},h_{i}\in\mathbb{R}$.

Since $\frac{\partial H}{\partial
x_{3}}=P_{4}=\frac{c}{2}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}+b^{4}-1)$,
we get

$H=\frac{c}{2}(-b^{2}(x_{1}^{2}+x_{2}^{2})+x_{4}^{2}+b^{4}-1)x_{3}-\frac{b^{2}c}{6}x_{3}^{3}+H_{1}(x_{1},x_{2},x_{4})$

where $H_{1}$ is a function independent of $x_{3}$. Now, from $-\frac{\partial
H}{\partial x_{4}}=P_{3}$, we get

$-cx_{3}x_{4}-\frac{\partial H_{1}}{\partial
x_{4}}=\frac{c}{4}x_{3}x_{4}-gx_{1}-hx_{2}.$

Thus,

(4.9) $c=g_{3}=h_{3}=0\quad\text{and}$

since $H_{1}$ is a function independent of $x_{3}$. So, $\frac{\partial
H_{1}}{\partial x_{4}}=(\sum\limits_{\begin{subarray}{c}i=0,i\neq
3\end{subarray}}^{4}g_{i}x_{i})x_{1}+(\sum\limits_{\begin{subarray}{c}i=0,i\neq
3\end{subarray}}^{4}h_{i}x_{i})x_{2}$ and hence

$H_{1}=(g_{0}+g_{1}x_{1}+g_{2}x_{2}+\frac{1}{2}g_{4}x_{4})x_{1}x_{4}+(h_{0}+h_{1}x_{1}+h_{2}x_{2}+\frac{1}{2}h_{4}x_{4})x_{2}x_{4}+H_{2}(x_{1},x_{2})$

where $H_{2}$ is a function of $x_{1}$ and $x_{2}$ only. Notice that $H=H_{1}$
since $c=0$. As $\frac{\partial H}{\partial x_{1}}=P_{2}$, we get

$(g_{0}+2g_{1}x_{1}+g_{2}x_{2}+\frac{1}{2}g_{4}x_{4})x_{4}+h_{1}x_{2}x_{4}+\frac{\partial
H_{2}}{\partial x_{1}}=-fx_{1}+hx_{3}.$

Thus,

$(g_{0}+\frac{1}{2}g_{4}x_{4})x_{4}+(2g_{1}+f_{4})x_{1}x_{4}+(g_{2}+h_{1})x_{2}x_{4}+(f_{3}-h_{1})x_{1}x_{3}-(\sum\limits_{\begin{subarray}{c}i=0,i\neq
1,3\end{subarray}}^{4}h_{i}x_{i})x_{3}=-\frac{\partial H_{2}}{\partial
x_{1}}-(f_{0}+f_{1}x_{1}+f_{2}x_{2})x_{1}$

Then

(4.10) $g_{0}=g_{4}=h_{0}=h_{2}=h_{4}=0,f_{4}=-2g_{1},f_{3}=-g_{2}=h_{1},$

since $H_{2}$ is a function of $x_{1},x_{2}$ only. Therefore,

$H_{2}=-\frac{1}{2}(f_{0}+f_{2}x_{2})x_{1}^{2}-\frac{1}{3}f_{1}x_{1}^{3}+H_{3}(x_{2})$

where $H_{3}$ is a function of $x_{2}$ only. Hence,

$H=H_{1}=-\frac{1}{2}(f_{0}+f_{2}x_{2})x_{1}^{2}-\frac{1}{3}f_{1}x_{1}^{3}+g_{1}x_{1}^{2}x_{4}+H_{3}(x_{2}).$

Again, $-\frac{\partial H}{\partial x_{2}}=P_{1}$ gives
$\frac{1}{2}f_{2}x_{1}^{2}-H_{3}^{{}^{\prime}}=fx_{2}+gx_{3}$. Thus

(4.11) $f_{1}=f_{2}=f_{4}=g_{1}=0,$

since $H_{3}$ is a function of $x_{2}$ only. Combining (4.9), (4.10) and
(4.11), we obtain ${f_{3}=-g_{2}=h_{1}}$ and other coefficients of $f,g$ and
$h$ are zero except $f_{0}$. Hence, $f=f_{0}+h_{1}x_{3},$ $g=-h_{1}x_{2},$
$h=h_{1}x_{1}$. For these $f,g$ and $h$ the vector field (4.2) becomes a
linear, which contradicts the assumption. So, there is no Hamiltonian of the
quadratic vector field (4.2).

The hypersurfaces given by $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=0$ and
${(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1=0}$ are two invariant
algebraic hypersurfaces of the vector field (4.2) with cofactors
$\frac{c}{2}x_{4}$ and $cx_{4}$ respectively. By Darboux Integrability Theory
(Theorem 5, [16]), the rational function
$(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{-2}((x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1)$
is a first integral of the vector field. ∎

######  Theorem 4.6.

Let $\chi=(P_{1},...,P_{4})$ be a cubic Kolmogorov vector field in
$\mathbb{R}^{4}$. Then $\chi$ is a vector field on $S^{2}\times S^{1}$ if and
only if

(4.12) $\begin{split}P_{1}&=x_{1}(\frac{c}{4}x_{4}^{2}+\alpha x_{2}^{2}+\beta
x_{3}^{2}),\\\ P_{2}&=x_{2}(\frac{c}{4}x_{4}^{2}-\alpha x_{1}^{2}+\gamma
x_{3}^{2}),\\\ P_{3}&=x_{3}(\frac{c}{4}x_{4}^{2}-\beta x_{1}^{2}-\gamma
x_{2}^{2}),\text{ and}\\\
P_{4}&=\frac{1}{2}cx_{4}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}+b^{4}-1)\end{split}$

where $c,\alpha,\beta,\gamma\in\mathbb{R}$. Moreover, $K=cx_{4}^{2}$ is the
cofactor of $\chi$ for $S^{2}\times S^{1}$.

###### Proof.

Consider a cubic Kolmogorov vector field $\chi=(P_{1},...,P_{4})$ in
$\mathbb{R}^{4}$ with

$P_{i}=x_{i}\psi_{i}$

for $i=1,...,4$. We write

$\psi_{i}=\psi_{i}^{(2)}+\psi_{i}^{(1)}+\psi_{i}^{(0)}$

as in (3.4).

Suppose $\chi$ is a vector field on $S^{2}\times S^{1}$. So it must satisfy

(4.13)
$4(\sum_{i=1}^{3}x_{i}^{2}-b^{2})(\sum_{i=1}^{3}\psi_{i}x_{i}^{2})+2\psi_{4}x_{4}^{2}=(\sum_{1\leq
i\leq j\leq
4}k_{ij}x_{i}x_{j})((\sum_{i=1}^{3}x_{i}^{2}-b^{2})^{2}+x_{4}^{2}-1)$

since there is no constant term on the left side. Comparing degree 6 terms in
(4.13), we get

$4(\sum_{i=1}^{3}x_{i}^{2})(\sum_{i=1}^{3}\psi_{i}^{(2)}x_{i}^{2})=(\sum_{1\leq
i\leq j\leq 4}k_{ij}x_{i}x_{j})(\sum_{i=1}^{3}x_{i}^{2})^{2}.$

Hence

(4.14) $\sum_{i=1}^{3}\psi_{1}^{(2)}x_{i}^{2}=\frac{1}{4}(\sum_{1\leq i\leq
j\leq 4}k_{ij}x_{i}x_{j})(\sum_{i=1}^{3}x_{i}^{2}).$

Also comparing degree $5$ terms in (4.13), we get

$4(\sum_{i=1}^{3}x_{i}^{2})(\sum_{i=1}^{3}\psi_{i}^{(1)}x_{i}^{2})=0.$

Hence $\sum_{i=1}^{3}\psi_{i}^{(1)}x_{i}^{2}=0$. Then
$\psi_{1}^{(1)},\psi_{2}^{(1)},\psi_{3}^{(1)}$ are zero since they are degree
1 polynomials.

In particular, if $x_{4}=0$ then (4.13) gives the following.

(4.15)
$4(\sum_{i=1}^{3}x_{i}^{2}-b^{2})(\sum_{i=1}^{3}\psi_{i}(x_{1},x_{2},x_{3},0)x_{i}^{2})=(\sum_{1\leq
i\leq j\leq 3}k_{ij}x_{i}x_{j})((\sum_{i=1}^{3}x_{i}^{2}-b^{2})^{2}-1).$

Comparing degree 4 terms in (4.15)

$4(\sum_{i=1}^{3}x_{i}^{2})(\sum_{i=1}^{3}\psi_{i}^{(0)}(x_{1},x_{2},x_{3},0)x_{i}^{2})-4b^{2}(\sum_{i=1}^{3}\psi_{i}^{(2)}(x_{1},x_{2},x_{3},0)x_{i}^{2})=-2b^{2}(\sum_{1\leq
i\leq j\leq 3}k_{ij}x_{i}x_{j})(\sum_{i=1}^{3}x_{i}^{2}).$

Then, for $x_{4}=0$, the above equation together with (4.14) gives the
following.

(4.16) $4(\sum_{i=1}^{3}\psi_{i}^{(0)}x_{i}^{2})=-b^{2}(\sum_{1\leq i\leq
j\leq 3}k_{ij}x_{i}x_{j}).$

Also comparing degree 2 terms in (4.15), we get

(4.17) $-4b^{2}(\sum_{i=1}^{3}\psi_{i}^{(0)}x_{i}^{2})=(b^{4}-1)(\sum_{1\leq
i\leq j\leq 3}k_{ij}x_{i}x_{j}).$

Now, from (4.16) and (4.17), we obtain $k_{ij}=0$ for $1\leq i\leq j\leq 3$.
Hence we get, $\psi_{1}^{(0)}=\psi_{2}^{(0)}=\psi_{3}^{(0)}=0$. Therefore,
from (4.14) we get the following.

(4.18)
$\sum_{i=1}^{3}\psi_{i}^{(2)}x_{i}^{2}=\frac{1}{4}(\sum_{i=1}^{4}k_{i4}x_{i}x_{4})(\sum_{i=1}^{3}x_{i}^{2}).$

Taking $K:=\sum_{i=1}^{4}k_{i4}x_{i}x_{4}$, we get

$(\psi_{1}^{(2)}-\frac{1}{4}K)x_{1}^{2}+(\psi_{2}^{(2)}-\frac{1}{4}K)x_{2}^{+}(\psi_{3}^{(2)}-\frac{1}{4}K)x_{3}^{2}=0.$

Then, by Lemma 4.1,

$\psi_{1}^{(2)}=\frac{1}{4}K+\alpha x_{2}^{2}+\beta
x_{3}^{2},\psi_{2}^{(2)}=\frac{1}{4}K-\alpha x_{1}^{2}+\gamma
x_{3}^{2},\psi_{3}^{(2)}=\frac{1}{4}K-\beta x_{1}^{2}-\gamma x_{2}^{2}$

for some $\alpha,\beta,\gamma\in\mathbb{R}$. So we have found all the
homogeneous parts of $\psi_{1}$, $\psi_{2}$ and $\psi_{3}$. Now, let us find
$\psi_{4}$.

Comparing the degree 2 terms in (4.13), we get

$\psi_{4}^{(0)}=\frac{b^{4}-1}{2}k_{44}\text{ and }k_{i4}=0\text{ for
}i=1,2,3.$

Hence $K=k_{44}x_{4}^{2}$. Comparing degree 3 terms in (4.13), we get
$\psi_{4}^{(1)}x_{4}^{2}=0$ which follows that $\psi_{4}^{(1)}=0$. Now
comparing degree 4 terms in (4.13), we get

$-4b^{2}(\sum_{i=1}^{3}\psi_{i}^{(2)}x_{i}^{2})+2\psi_{4}^{(2)}x_{4}^{2}=k_{44}x_{4}^{2}(-2b^{2}(\sum_{i=1}^{3}x_{i}^{2})+x_{4}^{2}).$

Then using (4.18) in the above equation, we obtain

$\psi_{4}^{(2)}x_{4}^{2}=\frac{1}{2}k_{44}x_{4}^{2}(-b^{2}(\sum_{i=1}^{3}x_{i}^{2})+x_{4}^{2}).$

Hence
$\psi_{4}^{(2)}=\frac{1}{2}k_{44}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2})$.
So we have found all homogeneous components of $\psi_{4}$. Renaming $k_{44}$
as $c$, we get our result.

If $P_{1},...,P_{4}$ are given by (4.12), then they satisfy (4.1). Thus, the
converse part is true. ∎

######  Theorem 4.7.

There is no Hamiltonian of a cubic Kolmogorov vector field on $S^{2}\times
S^{1}$. However, every cubic Kolmogorov vector field on $S^{2}\times S^{1}$
has a rational first integral.

###### Proof.

Any cubic Kolmogorov vector field on $S^{2}\times S^{1}$ can be described by
(4.12). Suppose $H$ is a Hamiltonian of the vector field (4.12). Then

$H=x_{1}x_{2}(\sum\limits_{i+j=0}^{2}a_{ij}x_{1}^{i}x_{2}^{j})+x_{3}x_{4}(\sum\limits_{i+j=0}^{2}b_{ij}x_{3}^{i}x_{4}^{j})+c_{0}x_{1}x_{2}x_{3}x_{4}$

by [15, Theorem 1]. Since $\frac{\partial H}{\partial x_{3}}=P_{4}$, we get

$x_{4}(\sum\limits_{i+j=0}^{2}b_{ij}x_{3}^{i}x_{4}^{j})+x_{3}x_{4}(a_{10}+a_{11}x_{4}+2a_{20}x_{3})+c_{0}x_{1}x_{2}x_{4}=\frac{1}{2}cx_{4}(-b^{2}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})+x_{4}^{2}+b^{4}-1).$

Thus, $c=c_{0}=0$ and $b_{ij}=0$ for $0\leq i+j\leq 2$. Hence,
$H=x_{1}x_{2}(\sum\limits_{i+j=0}^{2}a_{ij}x_{1}^{i}x_{2}^{j})$. Again,
$-\frac{\partial H}{\partial x_{2}}=P_{1}$ implies that

$-x_{1}(\sum\limits_{i+j=0}^{2}a_{ij}x_{1}^{i}x_{2}^{j})-x_{1}x_{2}(a_{01}+a_{11}x_{1}+2a_{02}x_{2})=x_{1}(\alpha
x_{2}^{2}+\beta x_{3}^{2}).$

Then, $a_{00}=a_{10}=a_{01}=a_{11}=a_{20}=\beta=0$ and $-3a_{02}=\alpha$.
Hence, $H=a_{02}x_{1}x_{2}^{3}$. Now, $\frac{\partial H}{\partial
x_{1}}=P_{2}$ implies that

$a_{02}x_{2}^{3}=x_{2}(-\alpha x_{1}^{2}+\gamma x_{3}^{2}).$

Thus, $a_{02}=0$ and this gives $H=0$, which is absurd. Hence, a cubic
Kolmogorov vector field on $S^{2}\times S^{1}$ cannot have any Hamiltonian.

Let $\chi=(P_{1},...,P_{4})$ be a vector field of the form (4.12). Then the
hypersurfaces given by $x_{1}=0,$ ${x_{2}=0,x_{3}=0,}$ and
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=0$ are invariant hypersurfaces $\chi$ with
cofactors ${\frac{c}{4}x_{4}^{2}+\alpha x_{2}^{2}+\beta x_{3}^{2},}$
$\frac{c}{4}x_{4}^{2}-\alpha x_{1}^{2}+\gamma
x_{3}^{2},\frac{c}{4}x_{4}^{2}-\beta x_{1}^{2}-\gamma x_{2}^{2},$ and
$\frac{c}{2}x_{4}^{2}$ respectively. Notice that

$\gamma(\frac{c}{4}x_{4}^{2}+\alpha x_{2}^{2}+\beta
x_{3}^{2})-\beta(\frac{c}{4}x_{4}^{2}-\alpha x_{1}^{2}+\gamma
x_{3}^{2})+\alpha(\frac{c}{4}x_{4}^{2}-\beta x_{1}^{2}-\gamma
x_{2}^{2})-\frac{\gamma-\beta+\alpha}{2}\frac{c}{2}x_{4}^{2}=0.$

Hence by Darboux Integrability Theory (Theorem 5, [16]), the function

$f=x_{1}^{\gamma}x_{2}^{-\beta}x_{3}^{\alpha}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{-\frac{\gamma-\beta+\alpha}{2}}$

is a rational first integral of $\chi$. ∎

######  Lemma 4.8.

Suppose $P_{1},P_{2},P_{3},P_{4}\in\mathbb{R}[x_{1},...,x_{4}]$ are non-zero
homogeneous polynomials with $\deg(P_{i})=m$ for $1\leq i\leq 3$ and
$\deg(P_{4})=n$. If $m-1\leq n\leq m+3$ then
${G=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1}$ is a first
integral of $\chi=(P_{1},...,P_{4})$.

###### Proof.

The arguments are similar to the proof of Lemma 3.8. ∎

######  Lemma 4.9.

Let $P_{i}\in\mathbb{R}[x_{1},...,x_{4}]$ is a homogeneous polynomial for
$i=1,...,4$ such that
${G=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1}$ is a first
integral of $\chi=(P_{1},...,P_{4})$. Then $P_{4}=0$.

###### Proof.

Since $G$ is a first integral of $\chi$, we get the following.

$4(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})(\sum\limits_{i=1}^{3}P_{i}x_{i})+2P_{4}x_{4}=0.$

Hence $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2}$ divides $P_{4}$. This is possible
only when $P_{4}=0$ since $P_{4}$ is a homogeneous polynomial. ∎

######  Corollary 4.10.

There is no Type-$n$ vector field on $S^{2}\times S^{1}$ for $n\in\mathbb{N}$.

###### Proof.

If $\chi$ is a Type-$n$ vector field on $S^{2}\times S^{1}$ then by Lemma 4.8,
$G=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})^{2}+x_{4}^{2}-1$ is a first integral
of $\chi$. Again from Lemma 4.9, if $G$ is a first integral of $\chi$, then
$P_{4}$ must be zero, which contradicts the assumption. Hence there is no
Type-$n$ vector field on $S^{2}\times S^{1}$. ∎

######  Theorem 4.11.

Suppose $\chi=(P_{1},...,P_{4})$ is a Pseudo Type-$n$ vector field in
$\mathbb{R}^{4}$. Then $\chi$ is a Pseudo Type-$n$ vector field on
$S^{2}\times S^{1}$ if and only if there exists
$A,B,C\in\mathbb{R}[x_{1},...,x_{4}]$ such that

(4.19)
$P_{1}=Ax_{2}+Bx_{3},P_{2}=-Ax_{1}+Cx_{3},P_{3}=-Bx_{1}-Cx_{2},P_{4}=0.$

###### Proof.

The vector field $\chi$ satisfies (4.1) and also $P_{4}=0$ since $\chi$ is
Pseudo Type-$n$. Hence, we get $P_{1}x_{1}+P_{2}x_{2}+P_{3}x_{3}=0$. Now,
using Lemma 4.1, we get the stated form of a Pseudo Type-$n$ vector field on
$S^{2}\times S^{1}$.

If $P_{1},...,P_{4}$ are given by (4.19), then they satisfy (4.1). Thus, the
converse part is true. ∎

######  Theorem 4.12.

There is no Hamiltonian of a Pseudo Type-$n$ vector field on $S^{2}\times
S^{1}$. However, each Pseudo Type-$n$ vector field on $S^{2}\times S^{1}$ has
atleast two independent first integrals.

###### Proof.

Suppose $\chi$ is the Pseudo Type-$n$ vector field (4.19). Assume that $\chi$
is a Hamiltonian vector field with Hamiltonian $H$. $H$ is independent of
$x_{3}$ since $\frac{\partial H}{\partial x_{3}}=P_{4}=0$. Hence,

$-\frac{\partial H}{\partial x_{2}}=P_{1}=Ax_{2}+Bx_{3}\mbox{ and
}\frac{\partial H}{\partial x_{1}}=P_{2}=-Ax_{1}+Cx_{3}.$

Now, we get that $-x_{1}\frac{\partial H}{\partial x_{2}}+x_{2}\frac{\partial
H}{\partial x_{1}}=(Bx_{1}+Cx_{2})x_{3}$. Then $Bx_{1}+Cx_{2}=0$ since $H$ is
independent of $x_{3}$. Hence, $P_{3}=-(Bx_{1}+Cx_{2})=0$, which contradicts
that $\chi$ is a Pseudo Type-$n$ vector field. Hence, $\chi$ is not
Hamiltonian.

Suppose $f$ is a first integral of the Pseudo Type-$n$ vector field (4.19).
Then $\sum\limits_{i=1}^{3}P_{i}\frac{\partial f}{\partial x_{i}}=0$. This
implies

$(Ax_{2}+Bx_{3})\frac{\partial f}{\partial
x_{1}}+(-Ax_{1}+Cx_{3})\frac{\partial f}{\partial
x_{2}}+(-Bx_{1}-Cx_{2})\frac{\partial f}{\partial x_{3}}=0.$

By [20, Theorem 3(Page 53)], solutions of the above can be given in terms of
the solutions of the equations

$\frac{dx_{1}}{Ax_{2}+Bx_{3}}=\frac{dx_{2}}{-Ax_{1}+Cx_{3}}=\frac{dx_{3}}{-Bx_{1}-Cx_{2}}=\frac{dx_{4}}{0}=\frac{df}{0}=\frac{x_{1}dx_{1}+x_{2}dx_{2}+x_{3}dx_{3}}{0}=\frac{dx_{4}}{0}.$

Hence, we obtain two independent first integrals,
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ and $x_{4}$. ∎

######  Corollary 4.13.

Any Pseudo Type-$n$ Hamiltonian system on $S^{2}\times S^{1}$ is integrable in
the Liouville sense.

##  5\. Invariant meridian and parallel hyperplanes

This section introduces meridian and parallel hyperplanes in $\mathbb{R}^{4}$.
Then we compute an upper bounds for a degree $n$ vector field on $S^{1}\times
S^{2}$ and $S^{2}\times S^{1}$. We also show when these bounds can be
achieved. We define,

  1. (1)

a hyperplane of the form $a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}=0$ as a meridian
hyperplane, and

  2. (2)

a hyperplane of the form $x_{4}=k$ as a parallel hyperplane

where $a_{1},a_{2},a_{3},k\in\mathbb{R}$.

######  Theorem 5.1.

Suppose $\chi$ is a vector field on $S^{1}\times S^{2}$.

  1. (i)

If $\chi$ is quadratic, then the number of invariant hyperplanes of the form
${a_{1}x_{1}+a_{2}x_{2}=0}$ is either 0, 1, or infinite, and the number of
invariant hyperplanes of the form $a_{3}x_{3}+a_{4}x_{4}=0$ is either 0 or 1.

  2. (ii)

Let $\chi$ be a cubic Kolmogorov vector field such that it has finitely many
invariant hyperplanes of the form $a_{1}x_{1}+a_{2}x_{2}=0$ and
$a_{3}x_{3}+a_{4}x_{4}=0$. Then only such invariant hyperplanes are
$x_{1}=0,x_{2}=0,x_{3}=0$, and $x_{4}=0$.

  3. (iii)

Let $\chi$ be the Type-$n$ vector field (3.20). Hyperplane
$a_{1}x_{1}+a_{2}x_{2}=0$ is invariant if and only if $a_{1}x_{1}+a_{2}x_{2}$
divides $A$. Also, the hyperplane $a_{3}x_{3}+a_{4}x_{4}=0$ is invariant if
and only if $a_{3}x_{3}+a_{4}x_{4}$ divides $B$. Hence the sharp bound for the
number of such invariant hyperplanes of $\chi$ is $n-1$.

###### Proof.

  1. (i)

Suppose $\chi$ is of the form (3.2). Then

$\mathcal{E}_{\langle x_{1},x_{2}\rangle}(\chi)=-f(x_{1}^{2}+x_{2}^{2})\text{
and }$ $\mathcal{E}_{\langle
x_{3},x_{4}\rangle}(\chi)=\frac{1}{2}(-a^{2}(x_{1}^{2}+x_{2}^{2})+x_{3}^{2}+x_{4}^{2}+a^{4}-1)(k_{4}x_{3}-k_{3}x_{4})-g(x_{3}^{2}+x_{4}^{2}).$

From Proposition 2.1, if $a_{1}x_{1}+a_{2}x_{2}=0$ is invariant under $\chi$
then $a_{1}x_{1}+a_{2}x_{2}$ must divide $\mathcal{E}_{\langle
x_{1},x_{2}\rangle}(\chi)$, which implies that $a_{1}x_{1}+a_{2}x_{2}$ divides
$f$. Hence if $f$ has no divisor of the form $a_{1}x_{1}+a_{2}x_{2}$, then
$\chi$ has no such invariant hyperplane. If $f$ is a non-zero polynomial
having a divisor of the form $a_{1}x_{1}+a_{2}x_{2}$, then $f$ has exactly one
such invariant hyperplane, namely $f=0$. If $f$ is zero polynomial, then one
can check that any hyperplane of the form $a_{1}x_{1}+a_{2}x_{2}=0$ is
invariant under $\chi$.

Suppose $a_{3}x_{3}+a_{4}x_{4}=0$ is invariant under $\chi$. Then by
Proposition 2.1, the polynomial $a_{3}x_{3}+a_{4}x_{4}$ must divide
$\mathcal{E}_{\langle x_{3},x_{4}\rangle}(\chi)$, say $\mathcal{E}_{\langle
x_{3},x_{4}\rangle}(\chi)=K^{{}^{\prime}}(a_{3}x_{3}+a_{4}x_{4})$. Suppose $c$
is the constant term of $K^{{}^{\prime}}$. So comparing degree 1 terms in the
previous equation, we get $\frac{a^{4}-1}{2}k_{4}=ca_{3}$ and
$-\frac{a^{4}-1}{2}k_{3}=ca_{4}$. If $c\neq 0$ then $k_{4}x_{3}-k_{3}x_{4}=0$
is invariant since
$a_{3}x_{3}+a_{4}x_{4}=\frac{a^{4}-1}{2c}(k_{4}x_{3}-k_{3}x_{4})$. But
$\chi(k_{4}x_{3}-k_{3}x_{4})=g(k_{3}x_{3}+k_{4}x_{4})$ and
$k_{4}x_{3}-k_{3}x_{4}$ doesnot divide $k_{3}x_{3}+k_{4}x_{4}$, hence
$k_{4}x_{3}-k_{3}x_{4}$ is a divisor of $g$. So $a_{3}x_{3}+a_{4}x_{4}$ is a
divisor of $g$.

If $c=0$ then $k_{3}=k_{4}=0$, which implies that $a_{3}x_{3}+a_{4}x_{4}$
divides $g$ since ${a_{3}x_{3}+a_{4}x_{4}}$ divides $\mathcal{E}_{\langle
x_{3},x_{4}\rangle}(\chi)$. Hence, the number of invariant hyperplanes of the
form ${a_{3}x_{3}+a_{4}x_{4}=0}$ of $\chi$ is either 0 or 1, depending on the
divisor of the linear polynomial $g$.

  2. (ii)

This follows immediately from Proposition 2.1.

  3. (iii)

We can check that if $a_{1}x_{1}+a_{2}x_{2}$ is a factor of $A$ then
$a_{1}x_{1}+a_{2}x_{2}=0$ is invariant hyperplane of $\chi$. Similarly, if
$a_{3}x_{3}+a_{4}x_{4}$ is a factor of $B$ then $a_{3}x_{3}+a_{4}x_{4}=0$ is
invariant hyperplane of $\chi$.

Converse parts follow immediately from Proposition 2.1.

∎

One can see that for a degree $n$ vector field $\chi$, the extactic polynomial
$\mathcal{E}_{\langle x_{1},x_{2},x_{3}\rangle}(\chi)$ has degree at most
$3n$. Hence the number of invariant meridian hyperplanes of $\chi$ is at most
$3n$. But in some special cases, we could reduce the bound.

######  Theorem 5.2.

Suppose $\chi$ is a Type-$n$ vector field on $S^{1}\times S^{2}$. Assume that
$\chi$ has finitely many invariant meridian hyperplanes, taking into account
their multiplicities. Then the maximum number of invariant meridian
hyperplanes is $3n-2$. Also, this bound is sharp.

###### Proof.

The vector field $\chi=(P_{1},...,P_{4})$ is of the form (3.20). Now

$\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=\begin{vmatrix}x_{1}&x_{2}&x_{3}\\\
Ax_{2}&-Ax_{1}&Bx_{4}\\\ \chi(Ax_{2})&\chi(-Ax_{1})&\chi(Bx_{4})\end{vmatrix}$

Suppose $C_{i}$ is the $i$-th column of the above determinant. Then performing
the column operation $C_{1}=x_{1}C_{1}+x_{2}C_{2}$, we get

$\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=\begin{vmatrix}x_{1}^{2}+x_{2}^{2}&x_{2}&x_{3}\\\
0&-Ax_{1}&Bx_{4}\\\
x_{1}\chi(Ax_{2})+x_{2}\chi(-Ax_{1})&\chi(-Ax_{1})&\chi(Bx_{4})\end{vmatrix}$

Note that
$x_{1}\chi(Ax_{2})+x_{2}\chi(-Ax_{1})=x_{1}(\chi(A)x_{2}+AP_{2})-x_{2}(\chi(A)x_{1}+AP_{1})=-A^{2}(x_{1}^{2}+x_{2}^{2})$.
Hence, $x_{1}^{2}+x_{2}^{2}$ divides $\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)$. But $x_{1}^{2}+x_{2}^{2}$ has no linear
factor over $\mathbb{R}$. So $\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)$ can have maximum $3n-2$ divisors of the form
$a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$ since the extactic polynomial has degree at
most $3n$. Hence, by Proposition 2.1, $\chi$ can have at most $3n-2$ invariant
hyperplanes(counting multiplicity) of the form
$a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}=0$.

To show that the bound can be obtained, consider the following polynomials.

$P_{1}=x_{1}^{n-1}x_{2}x_{3},P_{2}=-x_{1}^{n}x_{3},P_{3}=cx_{1}^{n-1}x_{3}x_{4},P_{4}=-cx_{1}^{n-1}x_{3}^{2},$

where $c\in\mathbb{R}\backslash\\{\pm 1\\},n\geq 2$. Then by Theorem 3.9,
$\chi=(P_{1},...,P_{4})$ is a Type-$n+1$ vector field on $S^{1}\times S^{2}$.
One can compute $\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=(x_{1}^{2}+x_{2}^{2})(1-c^{2})x_{1}^{3n-3}x_{3}^{4}$.
Hence, $x_{1}=0$ and $x_{3}=0$ are the only invariant hyperplanes of $\chi$
with multiplicities $3n-3$ and $4$, respectively. ∎

######  Theorem 5.3.

Suppose $\chi=(P_{1},P_{2},P_{3},0)$ is a Pseudo Type-$n$ vector field on
$S^{2}\times S^{1}$ and $\chi$ has finitely many invariant meridian
hyperplanes, taking into account their multiplicities. If
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ divides $P_{1},P_{2},P_{3}$ then the maximum
number of invariant meridian hyperplanes is $3n-6$.

###### Proof.

The vector field $\chi$ is of the form (4.19). Now

$\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=\begin{vmatrix}x_{1}&x_{2}&x_{3}\\\
P_{1}&P_{2}&P_{3}\\\ \chi(P_{1})&\chi(P_{2})&\chi(P_{3})\end{vmatrix}$

Suppose $C_{i}$ is the $i$-th column of the above determinant. Then performing
the column operation $C_{1}=x_{1}C_{1}+x_{2}C_{2}+x_{3}C_{3}$, we get

$\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=\begin{vmatrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}&x_{2}&x_{3}\\\
x_{1}P_{1}+x_{2}P_{2}+x_{3}P_{3}&x_{2}P_{2}&x_{3}P_{3}\\\
x_{1}\chi(P_{1})+x_{2}\chi(P_{2})+x_{3}\chi(P_{3})&x_{2}\chi(P_{2})&x_{3}\chi(P_{3})\end{vmatrix}$

We get $x_{1}P_{1}+x_{2}P_{2}+x_{3}P_{3}=0$ from (4.19). Also

$\begin{split}&x_{1}\chi(P_{1})+x_{2}\chi(P_{2})+x_{3}\chi(P_{3})\\\
=&x_{1}\chi(Ax_{2}+Bx_{3})+x_{2}\chi(-Ax_{1}+Cx_{3})+x_{3}\chi(-Bx_{1}-Cx_{2})\\\
=&x_{1}(\chi(A)x_{2}+AP_{2}+\chi(B)x_{3}+BP_{3})+x_{2}(-\chi(A)x_{1}-AP_{1}+\chi(C)x_{3}+CP_{3})\\\
&+x_{3}(-\chi(B)x_{1}-BP_{1}-\chi(C)x_{2}-CP_{2})\\\
=&-P_{1}(Ax_{2}+Bx_{3})-P_{2}(-Ax_{1}+Cx_{3})-P_{3}(-Bx_{1}-Cx_{2})\\\
=&-P_{1}^{2}-P_{2}^{2}-P_{3}^{2}.\end{split}$

Hence

$\mathcal{E}_{\langle
x_{1},x_{2},x_{3}\rangle}(\chi)=\begin{vmatrix}x_{1}^{2}+x_{2}^{2}+x_{3}^{2}&x_{2}&x_{3}\\\
0&x_{2}P_{2}&x_{3}P_{3}\\\
-P_{1}^{2}-P_{2}^{2}-P_{3}^{2}&x_{2}\chi(P_{2})&x_{3}\chi(P_{3})\end{vmatrix}.$

Observe that if $\sum\limits_{i=1}^{3}x_{i}^{2}$ divides $P_{j}$, say
$P_{j}=L(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})$ then

$\chi(P_{j})=\chi(L(\sum\limits_{i=1}^{3}x_{i}^{2}))=\chi(L)(\sum\limits_{i=1}^{3}x_{i}^{2})+2L(\sum\limits_{i=1}^{3}x_{i}P_{i})=\chi(L)(\sum\limits_{i=1}^{3}x_{i}^{2}).$

Hence, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ divides $P_{1},P_{2},P_{3}$ implies
that $(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{3}$ is a divisor of
$\mathcal{E}_{\langle x_{1},x_{2},x_{3}\rangle}(\chi)$ since we can take out
$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ from the first column, second row and third
row.

Now since $\mathcal{E}_{\langle x_{1},x_{2},x_{3}\rangle}(\chi)$ has degree
$3n$ and $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ has no divisor of the form
$a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}$, the result follows. ∎

######  Example 5.4.

Consider the Pseudo Type-1 vector field $\chi$ of the form (4.19) on
$S^{2}\times S^{1}$. Now,
$\chi(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3})=(-a_{2}A-a_{3}B)x_{1}+(a_{1}A-a_{3}C)x_{2}+(a_{1}B+a_{2}C)x_{3}$.
If $\chi$ has invariant hyperplane of the form
$a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}=0$ then there exists a constant $K$ such
that
$\chi(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3})=K(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3})$,
i.e.,

$a_{1}(Ax_{2}+Bx_{3})+a_{2}(-Ax_{1}+Cx_{3})+a_{3}(-Bx_{1}-Cx_{2})=K(a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}).$

Hence comparing coefficients both sides, we obtain $-a_{2}A-a_{3}B=Ka_{1}$,
${a_{1}A-a_{3}C=Ka_{2}}$, $a_{1}B+a_{2}C=Ka_{3}$. Now

$a_{1}(-a_{2}A-a_{3}B)+a_{2}(a_{1}A-a_{3}C)+a_{3}(a_{1}B+a_{2}C)=0=K(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}).$

So $K=0$ since atleast one of $a_{1},a_{2},a_{3}$ is non-zero. This implies
$a_{2}=-a_{3}\frac{B}{A}$ and ${a_{1}=a_{3}\frac{C}{A}}$ (assuming $A\neq 0$).
Hence $\chi$ has exactly one invariant meridian hyperplane, namely the
hyperplane $Cx_{1}-Bx_{2}+Ax_{3}=0$.

A Type-$n$ vector field on $S^{1}\times S^{2}$ has no invariant parallel
hyperplane. However, every parallel hyperplane is invariant under a Pseudo
Type-$n$ vector field on $S^{2}\times S^{1}$.

######  Theorem 5.5.

Let $\chi$ be a degree $n$ polynomial vector field on $S^{1}\times S^{2}$.
Assume that $\chi$ has finitely many invariant parallel hyperplanes, taking
into account their multiplicities. Then the maximum number of invariant
parallel hyperplanes is $(n-1)$. Also, this bound can be reached.

###### Proof.

$\mathcal{E}_{\langle 1,x_{4}\rangle}=P_{4}$. By Proposition 2.1, $\chi$ can
have maximum $n$ invariant parallel hyperplanes. Suppose $\chi$ has $n$
invariant parallel hyperplanes(counting multiplicities). Then
$P_{4}=c\prod\limits_{i=1}^{n}(x_{4}-k_{i})$($c\neq 0)$ for some
$k_{i}\in\mathbb{R}$. As $\chi$ is a vector field on $S^{1}\times S^{2}$, it
satisfies (3.1) i.e.,

(5.1)
$4(x_{1}^{2}+x_{2}^{2}-a^{2})(P_{1}x_{1}+P_{2}x_{2})+2(P_{3}x_{3}+P_{4}x_{4})=K((x_{1}^{2}+x_{2}^{2}-a^{2})^{2}+x_{3}^{2}+x_{4}^{2}-1).$

Equation (5.1) is satisfied for each $(x_{1},...,x_{4})\in\mathbb{R}$, in
particular for $x_{1}=x_{2}=x_{3}=0$. So we get

$2cx_{4}\prod_{i=1}^{n}(x_{4}-k_{i})=K(0,0,0,x_{4})(x_{4}^{2}+a^{4}-1).$

Note that $K(0,0,0,x_{4})\neq 0$ as $c\neq 0$. Also,
$x_{4}\prod\limits_{i=1}^{n}(x_{4}-k_{i})$ is a factor of $K(0,0,0,x_{4})$
since $(x_{4}^{2}+a^{4}-1)$ has no real linear factor. But $K(0,0,0,x_{4})$
has degree at most $n-1$, which arises a contradiction. Thus $\chi$ can have
at most $n-1$ invariant parallel hyperplanes.

To show that the above bound is reached, consider the following polynomials.

$P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=x_{4}\prod_{i=1}^{n-1}(x_{4}-k_{i}),P_{4}=-x_{3}\prod_{i=1}^{n-1}(x_{4}-k_{i});$

where $A$ is a polynomial of degree $n-1$ and $k_{i}$ are distinct real
numbers. Then $P_{i}$’s satisfy (3.1) and thus $\chi=(P_{1},...,P_{4})$ is a
vector field on $S^{1}\times S^{2}$. Notice that each parallel hyperplane
$x_{4}-k_{i}=0$ is invariant for $i\in\\{1,...,n-1\\}$. ∎

Theorem 4.2 tells that a quadratic vector field on $S^{2}\times S^{1}$ has
either 0 or infinitely many (only when $c=0)$ invariant parallel hyperplanes.

######  Theorem 5.6.

Let $\chi$ be a degree $n\geq 3$ polynomial vector field on $S^{2}\times
S^{1}$. Assume that $\chi$ has finitely many invariant parallel hyperplanes,
taking into account their multiplicities. Then the maximum number of invariant
parallel hyperplanes is $(n-1)$. Also, this bound can be reached.

###### Proof.

The arguments to the first part are similar to the proof of the first part of
Theorem 5.5.

Consider the following polynomials to prove the second part.

$P_{1}=P_{2}=P_{3}=\frac{1}{4}x_{4}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}-b^{2})\prod\limits_{i=1}^{n-3}(x_{4}-k_{i}),P_{4}=\frac{1}{2}(x_{1}+x_{2}+x_{3})(x_{4}^{2}-1)\prod\limits_{i=1}^{n-3}(x_{4}-k_{i})$

where $k_{i}\neq\pm 1$ are distinct real numbers. Then $P_{i}$’s satisfy (4.1)
and thus $\chi=(P_{1},...,P_{4})$ is a vector field on $S^{2}\times S^{1}$.
Notice that $x_{4}-1=0,x_{4}+1=0$ and $x_{4}-k_{i}=0$ are invariant parallel
hyperplanes of $\chi$ for $i=1,...,n-3$. This completes the proof. ∎

##  6\. An Application

We recall the algebraic representation of $S^{1}\times S^{2}$ and $S^{2}\times
S^{1}$ from (1.3) and (1.4) respectively. In this section, we study if there
are polynomial diffeomorphisms between $S^{1}\times S^{2}$ and $S^{2}\times
S^{1}$. Note that the spaces $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$ are
diffeomorphic via a polynomial map with each component linear polynomial when
both the spaces are embedded in $\mathbb{R}^{5}$. But this is not true when we
consider $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$ as hyperplanes embedded
in $\mathbb{R}^{4}$.

Let $F\colon\mathbb{R}^{4}\to\mathbb{R}^{4}$ be a polynomial map. If
$F|_{S^{1}\times S^{2}}\colon S^{1}\times S^{2}\to S^{2}\times S^{1}$ then
$F|_{S^{1}\times S^{2}}$ is called a polynomial map from $S^{1}\times S^{2}$
to $S^{2}\times S^{1}$. If $F|_{S^{1}\times S^{2}}$ is a diffeomorphism, then
it is called a polynomial diffeomorphism from $S^{1}\times S^{2}$ to
$S^{2}\times S^{1}$.

######  Theorem 6.1.

There does not exist any polynomial diffeomorphism from $S^{1}\times S^{2}$
onto $S^{2}\times S^{1}$ with each component as a linear polynomial.

###### Proof.

Let $F:S^{1}\times S^{2}\to S^{2}\times S^{1}$ be a polynomial diffeomorphism
where $F=(F_{1},...,F_{4})$ with each $F_{i}$ linear. Any $F_{i}$ cannot be
constant. Otherwise, $F$ will not be surjective.

Consider a Type-1 vector field $\chi=(P_{1},...,P_{4})$ on $S^{1}\times
S^{2}$. Thus, by Corollary 3.3,

$P_{1}=Ax_{2},P_{2}=-Ax_{1},P_{3}=Bx_{4},P_{4}=-Bx_{3}$

where $A$ and $B$ are non-zero real numbers. Next, we compute the push-forward
vector field $F_{*}\chi$ on $S^{2}\times S^{1}$. We consider the Jacobian
matrix of $F$.

$DF(x)=\begin{pmatrix}\frac{\partial F_{1}}{\partial
x_{1}}(x)&.&.&\frac{\partial F_{1}}{\partial x_{4}}(x)\\\ .&.&.&.\\\
.&.&.&.\\\ \frac{\partial F_{4}}{\partial x_{1}}(x)&.&.&\frac{\partial
F_{4}}{\partial x_{4}}(x)\end{pmatrix}.$

Observe that $D:=DF(x)$ is a constant matrix, and no row of $D$ is zero since
$F_{i}$ is non-constant for $i=1,...,4$.

Let $D=DF(x)=(d_{ij})_{4\times 4}$ and ${\bf y}\in S^{2}\times S^{1}$ such
that $F^{-1}({\bf y})=(x_{1},...,x_{4})$. Note that $F^{-1}=(G_{1},...,G_{4})$
with each $G_{i}$ is non-constant linear. Therefore

$F_{*}\chi({\bf y})=D(Ax_{2},-Ax_{1},Bx_{4},-Bx_{3})^{t}=D(AG_{2}({\bf
y}),-AG_{1}{(\bf y)},BG_{4}({\bf y}),-BG_{3}({\bf y}))^{t}.$

Hence, it is a linear vector field on $S^{2}\times S^{1}$. Thus, by Corollary
4.3, the fourth component of this vector field is zero. In particular, this is
true for $A=0$, $B=1$, and $A=1,B=0$. So,

$d_{43}G_{4}({\bf y})-d_{44}G_{3}({\bf y})=0,\quad\mbox{and}\quad
d_{41}G_{2}({\bf y})-d_{42}G_{1}({\bf y})=0$

for any ${\bf y}\in S^{2}\times S^{1}$. Since $(d_{41},d_{42},d_{43},d_{44})$
is non-zero and each $G_{i}$ is non-constant, we have the following cases.

Suppose, if $d_{41}\neq 0$, then $d_{42}\neq 0$ and $G_{2}({\bf
y})=\frac{d_{42}}{d_{41}}G_{1}({\bf y})$. Since $F$ is a diffeomorphism, there
exist ${\bf y}_{1},{\bf y}_{2}\in S^{2}\times S^{1}$ such that

$(1,a,0,0)=(G_{1}({\bf y}_{1}),G_{2}({\bf y}_{1}),G_{3}({\bf
y}_{1}),G_{4}({\bf y}_{1})),~{}(a,1,0,0)=(G_{1}({\bf y}_{2}),G_{2}({\bf
y}_{2}),G_{3}({\bf y}_{2}),G_{4}({\bf y}_{2})).$

Thus

$1=G_{2}({\bf y}_{2})=\frac{d_{42}}{d_{41}}a=\frac{G_{2}({\bf
y}_{1})}{G_{1}({\bf y}_{1})}a=a\times a=a^{2}.$

This contradicts $a>1$ which is needed to show that the set in (1.3) is
manifold.

If $d_{41}=0$, then $d_{42}=0$, otherwise $G_{1}=0$. So assume, $d_{43}\neq
0$. Then $d_{44}\neq 0$ and $G_{4}({\bf y})=\frac{d_{43}}{d_{44}}G_{3}({\bf
y})$. Therefore, $G_{3}({\bf y})=0$ for some ${\bf y}$ if and only if
$G_{4}({\bf y})=0$ for those ${\bf y}$. However,
$(a,2^{-\frac{1}{4}},2^{-\frac{1}{2}},0)\in S^{1}\times S^{2}$. Thus there
exists ${\bf y}\in S^{2}\times S^{1}$ such that

$(a,2^{-\frac{1}{4}},2^{-\frac{1}{2}},0)=(G_{1}({\bf y}),G_{2}({\bf
y}),G_{3}({\bf y}),G_{4}({\bf y})).$

Hence, we arrived at a contradiction. Therefore, the claim is proved. ∎

######  Remark 6.2.

The above result justifies why a separate discussion is needed to study
polynomial vector fields on $S^{1}\times S^{2}$ and $S^{2}\times S^{1}$.

Acknowledgment. The authors thank Joji Benny for the helpful discussion. The
first author is supported by Prime Minister’s Research Fellowship, Government
of India. The second author thanks ICSR of IIT Madras for SEED grant from
April 2021 to March 2024.

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