# Invariant Hyperplane sections of vector fields on the product of spheres

Joji Benny  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.

Let $S_{p,q}$ be the hypersurface in $\mathbb{R}^{n},$ where $n=p+q+1,$
defined by the following:

$\displaystyle
S_{p,q}:=\left\\{(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}~{}~{}\big{|}~{}~{}\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)^{2}+\sum_{j=p+2}^{n}x_{j}^{2}=1\right\\}$

where $a>1$. We show that $S_{p,q}$ is homeomorphic to the product
$S^{p}\times S^{q}$. We classify all degree one and two polynomial vector
fields on $S_{p,q}$. We consider the polynomial vector field
$\mathcal{X}=(R_{1},...,R_{p+1},R_{p+2},...,R_{n})$ in $\mathbb{R}^{p+q+1}$
which keeps $S_{p,q}$ invariant. Then we study the number of certain invariant
algebraic subsets of $S_{p,q}$ for the vector field $\mathcal{X}$ if either
$p>1$ or $q>1$.

###### Key words and phrases:

polynomial vector fields, invariant meridians, invariant parallels, extactic
polynomial, invariant hypersurfaces

###### 2020 Mathematics Subject Classification:

34A34, 34C14, 34C40, 34C45, 58J90

##  1\. Introduction

Let $R_{1},\ldots,R_{n}$ be polynomials in $\mathbb{R}[x_{1},\ldots,x_{n}]$.
Then the following system of differential equations

(1.1) $\frac{dx_{i}}{dt}=R_{i}(x_{1},\ldots,x_{n})$

for $i=1,\ldots,n$ is called a polynomial differential system in
$\mathbb{R}^{n}$. The differential operator

(1.2) $\mathcal{X}=\sum_{i=1}^{n}R_{i}\frac{\partial}{\partial x_{i}}$

is called the vector field associated with the system (1.1). The degree of the
polynomial vector field in (1.2) is defined to be
$\max\\{\deg(R_{i})|i=1,\ldots,n\\}$.

When $n=2$ in (1.1), this differential system has been studied since 1900
possibly because of the second part of the Hilbert sixteenth problem (see [6]
and some references therein).

An invariant algebraic set for (1.2) is a subset $A\subset\mathbb{R}^{n}$ such
that $A$ is the zero set of some
$f(x_{1},x_{2},\ldots,x_{n})\in\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]$ and
$\mathcal{X}f=Kf$ for some $K\in\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]$. Here
the polynomial $K$ is called the cofactor of $f$.

In the case of the torus $S^{1}\times S^{1}$, the maximum number of invariant
meridians and parallels are studied in [10] and [13]. Bounds on the number of
invariant hyperplanes and co-dimension one spheres for polynomial vector
fields in $\mathbb{R}^{n}$ are obtained in [9] and [2] respectively. In
relation to the second part of Hilbert’s sixteenth problem, the maximum number
of algebraic limit cycles of a polynomial vector field in $\mathbb{R}^{2}$ as
a function of its degree has been studied in [11], [12], and [16]. The maximum
number of straight lines that are invariant for a vector field in the real
plane as a function of its degree has been studied in [1] and [15].

Inspired by the above works, we introduce an algebraic representation of
$S^{p}\times S^{q}$ for any positive integers $p,q$ and study the number of
invariant algebraic sets of a vector field on $S^{p}\times S^{q}$. We are
primarily interested in the algebraic sets obtained by the intersection of
$S^{p}\times S^{q}$ with hyperplanes. We classify all degree one and two
polynomial vector fields on $S_{p,q}$. We obtain an upper bound of the number
of possible invariant algebraic sets that are intersections of $S^{p}\times
S^{q}$ with hyperplanes. Therefore, many other types of invariant
hypersurfaces in $S^{p}\times S^{q}$ remain to be explored.

The paper is organized as follows. In Section 2, we recall the definition of
the extactic algebraic polynomial associated to a vector subspace of the ring
of polynomials and the given vector field. We also state some basic properties
of the extactic polynomials.

In Section 3, we show that $S_{p,q}$ with the subspace topology is
homeomorphic to $S^{p}\times S^{q}$. Then we classify all degree one and two
polynomial vector fields on $S_{p,q}$.

In Section 4, we define meridians and parallels on $S_{p,q}$ analogously to
their definitions given in [10] for $S^{1}\times S^{1}$. We prove that
meridians and parallels are connected if $p>1$ and $q>1$. We give an upper
bound for the number of invariant meridians and parallels in Theorem 4.3. We
compute an upper bound for the number of invariant meridians of degree one
vector fields on $S_{p,q}$. The bound for the degree one vector fields on
$S_{p,q}$ is attained. We demonstrate that the bounds in Theorem 4.3 are close
to being tight for the cases $p=2,3$ and $\deg(\mathcal{X})\geq 4$.

In Section 5, we show that the maximum number of invariant meridians on
$S_{1,q}(\cong S^{1}\times S^{q})$ is $2(m-1)$ where $m$ is the degree of
$\mathcal{X}$ on $S_{1,q}$ and that this bound can be reached. We also discuss
invariant parallels on $S_{1,q}$ and derive a bound on the number of invariant
parallels. We show that there exists a vector field on $S_{1,q}$ with rational
first integral.

In Section 6, we show that the maximum number of invariant parallels for
$\mathcal{X}$ on $S_{p,1}:=S^{p}\times S^{1}$ is $2(m-2)$ where
$m=\deg\mathcal{X}$. We also demonstrate that this bound can be attained. We
also make a remark on an upper bound for the number of invariant meridians for
$\mathcal{X}$ on $S_{p,1}$ and exhibit a vector field on $S_{p,1}$ with
rational first integral.

##  2\. Invariant Algebraic Sets and Extactic polynomials

In this section, we recall the concept of invariant algebraic sets and
extactic polynomials for polynomial vector fields on $\mathbb{R}^{n}$
following [10]. Then we discuss some basic properties of extactic polynomials.

Let $S$ be a hypersurface in $\mathbb{R}^{n}$ defined by the zeroes of a non
constant polynomial $h\in\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]$. We say that a
vector field $\mathcal{X}$ of the form (1.2) is defined on $S$ if
$(R_{1},R_{2},\ldots,R_{n})\cdot\nabla h=0$ for all points on the hypersurface
$S$. This is equivalent to saying that $\mathcal{X}h=Kh$ for some polynomial
$K\in\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]$. Because
$h(a_{1},a_{2},...,a_{n})=0$ for all points $(a_{1},a_{2},...,a_{n})$ on the
hypersurface S. The hypersurface $S$ is called a regular hypersurface if
$\nabla h\neq 0$ for all points on $S$. This hypersurface is called
irreducible if $h$ is irreducible. The degree of an irreducible hypersurface
$S$ is defined to be the degree of $h$.

In order to study invariant algebraic sets on an algebraic hypersurface
$S\subset\mathbb{R}^{n}$, one may use the idea of extactic algebraic
polynomial. We briefly recall this concept following [9]. Let $W$ be a
$k$-dimensional vector subspace of $\mathbb{R}[x_{1},x_{2},x_{3},\dots,x_{n}]$
with basis $\\{v_{1},\dots,v_{k}\\}$. Then the extactic algebraic polynomial
of the vector field $\mathcal{X}$ associated to $W$ is given by

$\mathcal{E}_{W}(\mathcal{X})=\mathsf{det}\begin{pmatrix}v_{1}&v_{2}&\cdots&v_{k}\\\
\mathcal{X}(v_{1})&\mathcal{X}(v_{2})&\cdots&\mathcal{X}(v_{k})\\\
\vdots&\vdots&\ddots&\vdots\\\
\mathcal{X}^{k-1}(v_{1})&\mathcal{X}^{k-1}(v_{2})&\cdots&\mathcal{X}^{k-1}(v_{k})\end{pmatrix},$

where $\mathcal{X}^{j}(v_{i})=\mathcal{X}^{j-1}(\mathcal{X}(v_{i}))$ for
$i\geq 2$. We note that the definition of the extactic algebraic set is
independent of the choice of basis of $W$, see Section 2 of [10]. In this
paper we will work with $W$ mostly of the form
$\\{x_{1},x_{2},\ldots,x_{n}\\}$ or $\\{1,x_{1},x_{2},\ldots,x_{n}\\}.$

We recall the definition of the algebraic multiplicity of an irreducible
algebraic set given by an irreducible polynomial $f=0$ from [14].

######  Definition 2.1.

The hypersurface given by $f=0$ with $f\in W$ has algebraic multiplicity, or
simply, multiplicity $m$ for $\mathcal{X}$ if
$\mathcal{E}_{W}(\mathcal{X})\neq 0$ and $(f)^{m}$ divides
$\mathcal{E}_{W}(\mathcal{X})$ and for $m^{\prime}>m$, $(f)^{m^{\prime}}$ is
not a factor of $\mathcal{E}_{W}(\mathcal{X})$. It has no defined algebraic
multiplicity if $\mathcal{E}_{W}(\mathcal{X})=0$.

We note that, in this paper, we are interested in hypersurfaces with finite
multiplicity which is accounted when we count the number of invariant
algebraic hypersurfaces.

We shall use the following proposition whose proof can be found in [9,
Proposition 1].

######  Proposition 2.2.

Let $\mathcal{X}$ be a polynomial vector field on $\mathbb{R}^{n}$ and $W$ a
finite dimensional vector sub-space of $\mathbb{R}[x_{1},x_{2},\dots,x_{n}]$
with $\dim(W)>1$. If $\\{f=0\\}$ is an invariant algebraic set for the vector
field $\mathcal{X}$ and $f\in W,$ then $f$ is a factor of
$\mathcal{E}_{W}(\mathcal{X}).$

We recall that a function $g$ is called a first integral of the system (1.2)
if $\mathcal{X}g=0$. If $g$ is a rational function then $g$ is called a
rational first integral. If the differential system (1.1) has a first
integral, then the system possesses infinitely many invariant algebraic sets.
A proof of this fact can be found on page 102 of [7]. We quote the following
result from [8].

######  Proposition 2.3.

Let $S$ be a regular algebraic hypersurface of degree $d$ in
$\mathbb{R}^{n+1}$. The polynomial vector field $\mathcal{X}$ on $S$ of degree
$m>0$ admits ${n+m\choose n+1}-{n+m-d\choose n+1}+n$ invariant algebraic
hypersurfaces irreducible in $\mathbb{C}[x_{1},x_{2},\cdots,x_{n+1}]$ if and
only if $\mathcal{X}$ has a rational first integral.

##  3\. Degree one and degree two vector fields on $S^{p}\times S^{q}$

In this section, we show that $S^{p}\times S^{q}$ is homeomorphic to the zero
set of a polynomial. Then we characterize degree one and degree two polynomial
vector fields on it. For the remainder of the paper, we fix $n=p+q+1.$

###  3.1. $S^{p}\times S^{q}$ as a hypersurface and its meridians and
parallels

We consider the following hypersurface in $\mathbb{R}^{p+q+1}$ defined by the
polynomial identity

(3.1)
$(x_{1}^{2}+x_{2}^{2}+\cdots+x_{p+1}^{2}-a^{2})^{2}+x_{p+2}^{2}+x_{p+3}^{2}+\cdots+x_{n}^{2}=1,$

where $a>1$ . We denote this hypersurface by $S_{p,q}$. We show that $S_{p,q}$
is homeomorphic to the product $S^{p}\times S^{q}$ of two spheres. Then we
determine when the vector field $\mathcal{X}$ in (1.2) is defined on
$S_{p,q}.$

Let $0\leq c\leq 1$ and $x_{p+2}^{2}+x_{p+3}^{2}+\cdots+x_{n}^{2}=c$. Then

$x_{1}^{2}+x_{2}^{2}+\cdots+x_{p+1}^{2}=\pm\sqrt{1-c}+a^{2}.$

Let

${\scriptstyle
U_{1}~{}~{}~{}:=~{}~{}~{}\left\\{(x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n})\in\mathbb{R}^{n}\big{|}\sum_{i=1}^{p+1}x_{i}^{2}=a^{2}+\sqrt{1-c},\quad\sum_{j=p+2}^{n}x_{j}^{2}=c,\quad
0\leq c\leq 1\right\\}}$

and

${\scriptstyle
U_{2}~{}~{}~{}:=~{}~{}~{}\left\\{(x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n})\in\mathbb{R}^{n}\big{|}\sum_{i=1}^{p+1}x_{i}^{2}=a^{2}-\sqrt{1-c},\quad\sum_{j=p+2}^{n}x_{j}^{2}=c,\quad
0\leq c\leq 1\right\\}.}$

Observe that $U_{1}$ and $U_{2}$ are homeomorphic to $S^{p}\times D^{q}$. They
are identified along their boundary $S^{p}\times S^{q-1}$ via the identity
map, where

$\displaystyle S^{p}\times
S^{q-1}=\left\\{(x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n})\in\mathbb{R}^{n}\big{|}\sum_{i=1}^{p+1}x_{i}^{2}=a^{2},\quad\sum_{j=p+2}^{n}x_{j}^{2}=1\right\\}.$

Thus

$S_{p,q}=~{}~{}~{}U_{1}\bigcup_{S^{p}\times
S^{q-1}}U_{2}~{}~{}~{}=~{}~{}~{}S^{p}\times D^{q}\bigcup_{S^{p}\times
S^{q-1}}S^{p}\times D^{q}\cong~{}~{}~{}S^{p}\times S^{q}.$

Note that the product $S^{p}\times S^{q-1}$ in the above equation is a subset
of $\mathbb{R}^{n}$, whereas the representation of $S^{p}\times S^{q-1}$ as in
(3.1) says $S^{p}\times S^{q-1}$ is a subset of $\mathbb{R}^{n-1}$. These two
representations of $S^{p}\times S^{q-1}$ are different in two different
ambient spaces.

Writing out the condition for $\mathcal{X}$ to be invariant on the
hypersurface $S_{p,q}$, we get the following.

(3.2)
$4\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)\left(\sum_{i=1}^{p+1}x_{i}R_{i}\right)+2\left(\sum_{j=p+2}^{n}x_{j}R_{j}\right)=K\left((\sum_{i=1}^{p+1}x_{i}^{2}-a^{2})^{2}+\sum_{j=p+2}^{n}x_{j}^{2}-1\right),$

for some $K\in\mathbb{R}[x_{1},\ldots,x_{n}]$. One defines the degree vector
of the vector field $\mathcal{X}$ to be
$\overline{m}:=(m_{1},m_{2},\ldots,m_{n})$ where $m_{i}=\deg(R_{i})$ for
$i=1,2,\ldots,n$.

######  Definition 3.1.

  1. (1)

The intersection of $S_{p,q}$ and the hyperplane
$\\{\sum_{i=1}^{p+1}a_{i}x_{i}=0\\}$ where $a_{i}\in\mathbb{R}$, for
$i=1,\ldots,p+1$ is called a ‘meridian’ on $S_{p,q}$.

  2. (2)

The intersection of $S_{p,q}$ and the hyperplane
$\\{\sum_{j=p+2}^{n}a_{j}x_{j}-c=0\\}$ for some $c\in(0,1)$ where
$a_{j}\in\mathbb{R}$, for $j=p+2,\ldots,n$ is called a ‘parallel’ on
$S_{p,q}$.

We say that a meridian on $S_{p,q}$ is invariant by the flow of the polynomial
vector field $\mathcal{X}$ on $S_{p,q}$ if
$\mathcal{X}(\sum_{i}^{p+1}a_{i}x_{i})=K(\sum_{i}^{p+1}a_{i}x_{i})$ where
$a_{i}\in\mathbb{R}$ for $i=1,\ldots,p+1$ and for some
$K\in\mathbb{R}[x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n}]$.

Similarly, one can define an invariant parallel on $S_{p,q}$. We note that
these definitions generalize the definitions of the invariant meridians and
the invariant parallels on $S^{1}\times S^{1}$ of [10].

###  3.2. Degree one vector fields on $S^{p}\times S^{q}$

Let

(3.3) $R_{i}=\sum_{j=1}^{n}c_{ij}x_{j}+c_{i0},$

where $c_{ij}\in\mathbb{R}$ for all $j$ and $i=1,\ldots,n$.

######  Proposition 3.2.

Let $\mathcal{X}=\sum_{i}R_{i}\frac{\partial}{\partial x_{i}}$ be a degree one
vector field defined on $S_{p,q}$ where $R_{i}$s are given by (3.3). Then the
matrix $(c_{ij})_{p+1\times p+1}$ is skew symmetric and $c_{i0}=0$ for all
$i\in\\{1,\ldots,n\\}$.

###### Proof.

For this case, we see that $K$ in (3.2) is a constant since the degree of the
expression on the left is four and the expression in brackets on the right
already has degree four.

In fact, $K$ vanishes since there is a constant term on the right of (3.2),
but all terms on the left have degree at least one. So for linear vector
fields, (3.2) becomes

(3.4)
$4\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)\left(\sum_{i=1}^{p+1}x_{i}R_{i}\right)+2\left(\sum_{j=p+2}^{n}x_{j}R_{j}\right)=0.$

Observe that in the expression for $R_{i}$, $i=1,\ldots,p+1$, there can be no
$x_{k}$s for $k>p+1$. Since if there was, then (3.4) will have terms of the
form $4x_{i}^{2}x_{j}x_{k}$, $k>p+1$ which cannot be canceled out. Hence this
will violate (3.4). Similarly, $R_{j}$ cannot have any terms with $x_{i}$s,
where $1\leq i\leq p+1$ (for $j=p+2,\ldots,n$). By similar reasoning we get
that $c_{i0}=d_{j0}=0$ for all $i$ and $j$.

Substituting $R_{i}$ from (3.3) in (3.4) and collecting coefficients of
$x_{i_{0}}$ for a fixed $i_{0}$, we have

(3.5)
$4(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2})\\{R_{i_{0}}+\sum_{s=1}^{p+1}c_{si_{0}}x_{s}\\}=0.$

Since $R_{i_{0}}=\sum_{j=1}^{p+1}c_{i_{0}j}x_{j}$, then (3.5) becomes

$4\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)\left\\{\sum_{j=1}^{p+1}c_{i_{0}j}x_{j}+\sum_{s=1}^{p+1}c_{si_{0}}x_{s}\right\\}=0.$

Re-indexing and gathering coefficients of $x_{j}$, we get

$(c_{i_{0}j}+c_{ji_{0}})=0,$

since there is a dense open subset of $S_{p,q}$ with
$\sum_{i=1}^{p+1}x_{i}^{2}\neq a^{2}$. From this, we see that the real matrix
determined by $\\{c_{ij}\\}$ is skew-symmetric. ∎

######  Remark 3.3.

We see that every first degree vector field defined on $S_{p,q}$ has a first
integral by (3.4).

###  3.3. Degree two vector fields on $S^{p}\times S^{q}$

Assume that $\mathcal{X}=\sum R_{i}\frac{\partial}{\partial x_{i}}$ is a
degree two vector field. Then we have

(3.6) $R_{i}=\sum_{j\leq
k}\beta_{ijk}x_{j}x_{k}+\sum_{j}\beta_{ij}x_{j}+\beta_{i0},$

where $\beta_{ijk},\beta_{ij},\beta_{i0}$ belong to $\mathbb{R}$ for
$i,j,k\in\\{1,2,\ldots,n\\}$ with $j\leq k$. Since this vector field
$\mathcal{X}$ satisfies (3.2), the polynomial $K$ has to be linear, say
$K=\alpha_{1}x_{1}+\alpha_{2}x_{2}+\cdots+\alpha_{n}x_{n}+\alpha_{0}$ for some
$\alpha_{0},\ldots,\alpha_{n}\in\mathbf{R}$. We get $\alpha_{0}=0$, as there
is no constant term on the left of (3.2).

Since $R_{i}$s in (3.6) satisfy (3.2), equating the degree five terms and
cancelling $\sum_{i=1}^{p+1}x_{i}^{2}$ we have

(3.7) $\displaystyle 4\left(x_{1}\sum_{j\leq
k}\beta_{1jk}x_{j}x_{k}+x_{2}\sum_{j\leq
k}\beta_{2jk}x_{j}x_{k}+\cdots+x_{p+1}\sum_{j\leq
k}\beta_{p+1jk}x_{j}x_{k}\right)=$
$\displaystyle(\alpha_{1}x_{1}+\alpha_{2}x_{2}+\cdots+\alpha_{n}x_{n})(x_{1}^{2}+x_{2}^{2}+\cdots+x_{p+1}^{2}).$

Observe that in (3.7) there are no terms of the form $x_{i}x_{j}x_{k}$ for
$i,j,k$ pairwise distinct on the right hand side, therefore such terms must
sum to zero on the left hand side. This gives

$4(\beta_{ijk}+\beta_{jik}+\beta_{kij})=0,$

when $i,j,k\in\\{1,\ldots,p+1\\}.$ If one of $i,j,k$ is in
$\\{p+2,\ldots,n\\}$, say $k$, then

$4(\beta_{ijk}+\beta_{jik})=0.$

For the remaining terms, we have,

$4\beta_{iii}=\alpha_{i}~{}~{}~{}~{}\mbox{and}~{}~{}~{}~{}4(\beta_{ijj}+\beta_{jij})=\alpha_{i}$

for $i,j\in\\{1,\ldots,p+1\\},$ and assuming without loss of generality that
$i<j$. We also have $4(\beta_{iik})=\alpha_{k},$ for $i\in\\{1,\ldots,p+1\\}$
and $k\in\\{p+2,\ldots,n\\}.$

Observe that there cannot be terms of the form $x_{i}x_{j}x_{k}$ with any two
of $i,j,k$ belonging to $\\{p+2,\ldots,n\\}$.

Notice that there are no degree four terms on the right hand side, this
implies

$4(x_{1}^{2}+\cdots+x_{p+1}^{2})(x_{1}\sum_{j}\beta_{1j}x_{j}+\cdots+x_{p+1}\sum_{j}\beta_{p+1}x_{j})=0.$

This gives

(3.8) $\beta_{ij}+\beta_{ji}=0$

for $i,j\in\\{1,\ldots,p+1\\}.$

Next, equating degree three terms gives

$\displaystyle
4(x_{1}^{2}+\cdots+x_{p+1}^{2})(x_{1}\beta_{10}+\cdots+x_{p+1}\beta_{p+10})-$
$\displaystyle 4a^{2}\left(x_{1}\sum_{j\leq
k}\beta_{1jk}x_{j}x_{k}+x_{2}\sum_{j\leq
k}\beta_{2jk}x_{j}x_{k}+\cdots+x_{p+1}\sum_{j\leq
k}\beta_{p+1jk}x_{j}x_{k}\right)+$ $\displaystyle 2x_{p+2}\sum_{j\leq
k}\beta_{p+2jk}x_{j}x_{k}+\cdots+2x_{n}\sum_{j\leq k}\beta_{njk}x_{j}x_{k}=$
$\displaystyle(\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n})\left(-2a^{2}(x_{1}^{2}+\cdots+x_{p+1}^{2})+x_{p+2}^{2}+\cdots+x_{n}^{2}\right).$

We can rewrite this as, using (3.7),

$\displaystyle
4(x_{1}^{2}+\cdots+x_{p+1}^{2})(x_{1}\beta_{10}+\cdots+x_{p+1}\beta_{p+10})+$
$\displaystyle 2x_{p+2}\sum_{j\leq
k}\beta_{p+2jk}x_{j}x_{k}+\cdots+2x_{n}\sum_{j\leq k}\beta_{njk}x_{j}x_{k}=$
$\displaystyle(\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n})\left(-a^{2}(x_{1}^{2}+\cdots+x_{p+1}^{2})+x_{p+2}^{2}+\cdots+x_{n}^{2}\right).$

In this equation, we see that

$2\beta_{kii}=-a^{2}\alpha_{k},$

for $k\in\\{p+2,\ldots,n\\}$ and $i\in\\{1,\ldots,p+1\\}$. Now we have

(3.9) $4\beta_{i0}=-a^{2}\alpha_{i}$

for $i\in\\{1,\ldots,p+1\\}$. If $i,j,k$ are distinct and all belong to
$\\{p+2,\ldots,n\\}$, then

$\beta_{ijk}+\beta_{jik}+\beta_{kij}=0.$

If one of $i,j,k$, say $j$, belongs to $\\{1,\ldots,p+1\\}$, then

$\beta_{ijk}+\beta_{kji}=0.$

In the remaining cases, we have

(3.10)
$2(\beta_{ijj}+\beta_{jij})=\alpha_{i},~{}~{}~{}~{}\mbox{and}~{}~{}~{}~{}2\beta_{iii}=\alpha_{i}.$

Now equating degree two terms,

$-4a^{2}\left(x_{1}\sum_{j}\beta_{ij}+\cdots+x_{p+1}\sum_{j}\beta_{p+1j}x_{j}\right)+\\\
2x_{p+2}\sum_{j}\beta_{p+2j}x_{j}+\cdots+2x_{n}\sum_{j}\beta_{nj}x_{j}=0.$

Then using (3.8), the above equation becomes

$2x_{p+2}\sum_{j}\beta_{p+2j}x_{j}+\cdots+2x_{n}\sum_{j}\beta_{nj}x_{j}=0.$

This implies that $\beta_{kl}+\beta_{lk}=0$ when $k,l\in\\{p+2,\ldots,n\\}.$

Finally, equating degree one terms, we have

$-4a^{2}(x_{1}\beta_{10}+\cdots+x_{p+1}\beta_{p+10})+2x_{p+2}\beta_{p+20}+\cdots+2x_{n}\beta_{n0}=\\\
(a^{4}-1)(\alpha_{1}x_{1}+\cdots+\alpha_{n}x_{n}).$

This gives

$\beta_{i0}=\frac{(1-a^{4})}{4a^{2}}\alpha_{i},~{}~{}\mbox{and}~{}~{}\beta_{j0}=\frac{(a^{4}-1)}{2}\alpha_{j},$

when $i\in\\{q,\ldots,p+1\\}$ and $j\in\\{p+2,\ldots,n\\}.$ We see that this
is a contradiction to (3.9), hence

(3.11) $\alpha_{i}=0$

for $i\in\\{1,\ldots,p+1\\}.$

Summarizing these computations, we obtain the following characterization.

######  Proposition 3.4.

Let $\mathcal{X}=\sum R_{i}\frac{\partial}{\partial x_{i}}$ be a degree two
vector field on $S_{p,q}$ where $R_{i}$s are given by (3.6). Let
$i,j,k\in\\{1,\ldots,p+1\\}$ and $l,m,n\in\\{p+2,\ldots,n\\},$ all of them
pairwise distinct. Then the co factor in (3.2) is given by

$K=(\alpha_{p+2}x_{p+2}+\cdots+\alpha_{n}x_{n}),$

this means $\alpha_{1}=\alpha_{2}=\cdots=\alpha_{p+1}=0.$

Also, the coefficients of degree two terms of $R_{i}$s satisfy

$\displaystyle\beta_{ijk}+\beta_{jik}+\beta_{kij}=0,\quad\mbox{and}\quad\beta_{lmn}+\beta_{mln}+\beta_{nlm}=0,$
$\displaystyle\beta_{ijl}+\beta_{jil}=0,\quad\mbox{and}\quad\beta_{ljm}+\beta_{mjl}=0,$
$\displaystyle 4(\beta_{ijj}+\beta_{jij})=\alpha_{i},\quad\mbox{and}\quad
2(\beta_{lmm}+\beta_{mlm})=\alpha_{l},$ $\displaystyle
4\beta_{iii}=\alpha_{i},\quad
2\beta_{lii}=-a^{2}\alpha_{l},\quad\mbox{and}\quad 2\beta_{lll}=\alpha_{l}.$

For the coefficients of degree one terms of $R_{i}$s, we have

$\beta_{ij}+\beta_{ji}=0,\quad\mbox{and}\quad\beta_{lm}+\beta_{ml}=0.$

Moreover, the constant terms of $R_{i}$s satisfy

$\beta_{i0}=0,\quad\mbox{and}\quad\beta_{l0}=\frac{(a^{4}-1)}{2}\alpha_{l}.$

##  4\. Invariant hyperplanes on $S^{p}\times S^{q}$

In this section, we are interested in the number of invariant algebraic sets
determined by the intersections of $S_{p,q}$ and hypersurfaces determined by
polynomials of degree one. We compute an upper bound for the number of
invariant meridians and parallels on $S_{p,q}$. We prove a result to compute
the number of invariant meridians of a degree one vector field on $S_{p,q}$.
We show that a degree 2 vector field on $S_{p,q}$ can have at most $p$ many
invariant meridians. Then we discuss some vector fields with the number of
invariant meridians close to an upper bound. Let
$\mathcal{X}=(R_{1},\ldots,R_{n})$ be a polynomial vector field and
$\deg(R_{i})=m_{i}$. Without loss of generality, we may assume that $m_{1}\geq
m_{2}\geq\cdots\geq m_{p+1}$ and that $m_{p+2}\geq m_{p+3}\geq\cdots\geq
m_{n}$.

######  Proposition 4.1.

Let $S^{n+1}$ be the standard unit $(n+1)$-sphere in $\mathbb{R}^{n+2}$ and
$H$ a hyperplane passing through the origin. Then $S^{n+1}\cap H$ is
homeomorphic to $S^{n}$.

######  Proposition 4.2.

On $S_{p,q}$, with $p,q\geq 2$, the meridians and parallels are path
connected.

###### Proof.

Let $\sum_{i=1}^{p+1}a_{i}x_{i}=0$ be the hyperplane $H$, which determines a
meridian and

$\sum_{j=p+2}^{n}x_{j}^{2}=\alpha\in[0,1].$

Then (3.1) gives the following pair of spheres

$\sum_{i=1}^{p+1}x_{i}^{2}=a^{2}\pm\sqrt{1-\alpha}$

unless $\alpha=1$. By Proposition 4.1, the intersection of any of these
spheres with the hyperplane $H$ is a $(p-1)$ dimensional sphere for each
$\alpha\in[0,1)$. Thus, by a similar argument as in the proof that $S_{p,q}$
is homeomorphic to $S^{p}\times S^{q}$, one can show that $S_{p,q}\cap H$ is
homeomorphic to $S^{p-1}\times S^{q}$. Therefore it is path connected since
$p,q>1$.

For the case of parallels, let $\sum_{j=p+2}^{n}b_{j}x_{j}=c$ be the
hyperplane $H_{2}$ and

$\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}=\beta\in[-1,1].$

Then (3.1) can be written as

$\sum_{j=p+2}^{n}x_{j}^{2}=1-\beta^{2}\in[0,1].$

By Proposition 4.1, the intersection of this sphere and $H_{2}-c$ is a $q-1$
dimensional sphere unless $\beta=\pm 1$. If $\beta=\pm 1$, then this
intersection is a point. Therefore, the intersection $S_{p,q}\cap H_{2}$ is
homeomorphic to $S^{p}\times S^{q-1}$ if $\beta\neq\pm 1$ since smooth
homotopies preserve transversality, see Section 6 in [4, Chapter 1]. This is
path connected since $p,q>1$.∎

######  Theorem 4.3.

Suppose that the polynomial vector field $\mathcal{X}$ on $\mathbb{R}^{n}$ has
finitely many invariant algebraic hypersurfaces. If $p,q\geq 2$, then

  1. (1)

the number of invariant meridians of $\mathcal{X}$ in $S_{p,q}$ is at most

${p\choose 2}(m_{1}-1)+\sum_{i=2}^{p+1}m_{i}+1,$

  2. (2)

the number of invariant parallels of $\mathcal{X}$ in $S_{p,q}$ is at most

${q\choose 2}(m_{p+2}-1)+\sum_{j=1}^{q-1}m_{p+j+1}.$

###### Proof.

For (1). An invariant meridian of $\mathcal{X}$ is given by the intersection
of a hyperplane of the form $g:=\sum_{i=1}^{p+1}a_{i}x_{i}=0$ with $S_{p,q}$.
By Proposition 2.2, if this hyperplane is invariant for the vector field
$\mathcal{X}$, then $\sum_{i=1}^{p+1}a_{i}x_{i}$ is a factor of the extactic
polynomial

$\mathcal{E}_{\\{x_{1},x_{2},\ldots,x_{p+1}\\}}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}&\cdots&x_{p+1}\\\
R_{1}&R_{2}&\cdots&R_{p+1}\\\
\mathcal{X}(R_{1})&\mathcal{X}(R_{2})&\cdots&\mathcal{X}(R_{p+1})\\\
\vdots&\vdots&\ddots&\vdots\\\
\mathcal{X}^{p-1}(R_{1})&\mathcal{X}^{p-1}(R_{2})&\cdots&\mathcal{X}^{p-1}(R_{p+1})\end{pmatrix}.$

Since we have chosen degrees of $R_{1},\ldots,R_{p+1}$ in decreasing order, we
see that the term

(4.1)
$\mathcal{X}^{p-1}(R_{p+1})\cdot\mathcal{X}^{p-2}(R_{p})\cdots\mathcal{X}(R_{3})\cdot
R_{2}\cdot x_{1}$

has the least degree in the polynomial
$\mathcal{E}_{\\{x_{1},x_{2},\cdots,x_{p+1}\\}}(\mathcal{X})$. Now the degree
of $\mathcal{X}^{i-1}(R_{i+1})$ is $(i-1)(m_{1}-1)+m_{i+1}$ for $1\leq i\leq
p$. Therefore,

$\deg(\mathcal{X}^{p-1}(R_{p+1})\cdot\mathcal{X}^{p-2}(R_{p})\cdots\mathcal{X}(R_{3})\cdot
R_{2}\cdot x_{1})={p\choose 2}(m_{1}-1)+\sum_{i=2}^{p+1}m_{i}+1.$

The number of invariant meridians cannot exceed the degree of the polynomial
in (4.1) since meridians are determined by linear homogeneous polynomials in
$\\{x_{1},\ldots,x_{p+1}\\}$. This proves (1).

For (2). In this case, the vector space $W$ for $\mathcal{E}_{W}(\mathcal{X})$
is generated by $\\{x_{p+2},\ldots,x_{n},1\\}$. Then the result follows using
the same argument as in the first case. ∎

###  4.1. Invariant Meridians for degree one vector fields on $S^{p}\times
S^{q}$

In this subsection, we give a tight upper bound on the number of invariant
meridians for degree one vector fields on $S_{p,q}$. For these vector fields,
$\mathcal{X}=\sum_{i}R_{i}\frac{\partial}{\partial x_{i}}$ where
$\deg(R_{i})\leq 1$ and at least one of them has degree one. Let $\mathcal{X}$
be invariant on $\sum_{i=1}^{p+1}a_{i}x_{i}=0$. So,

(4.2) $\sum_{i=1}^{p+1}a_{i}R_{i}=K^{\prime}(\sum_{i=1}^{p+1}a_{i}x_{i}),$

for some $K^{\prime}\in\mathbb{R}[x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n}].$
We get that $K^{\prime}$ of (4.2) is a constant since the left hand side has
degree one and the term in the brackets on the right has degree one.

Further substituting for $R_{i}$ from (3.3) in (4.2), we get

$\sum_{i=1}^{p+1}a_{i}\sum_{j=1}^{p+1}c_{ij}x_{j}=K^{\prime}\left(\sum_{s=1}^{p+1}a_{s}x_{s}\right).$

This can be written as

$\sum_{i=1}^{p+1}\sum_{j=1}^{p+1}a_{i}c_{ij}x_{j}=K^{\prime}\left(\sum_{s=1}^{p+1}a_{s}x_{s}\right).$

Reindexing and equating coefficients of $x_{j}$, we have

(4.3) $\sum_{i=1}^{p+1}a_{i}c_{ij}=K^{\prime}a_{j}.$

Thus, $(a_{1},\ldots,a_{p+1})$ is an eigenvector for the real matrix
$A:=(c_{ij})_{1\leq i,j\leq p+1}$ with the eigenvalue $K^{\prime}$. Since the
matrix $(c_{ij})$ is skew symmetric by Proposition 3.2, 0 is the only possible
real eigenvalue (see Exercise 7.(g) in Section 6.6 of [3]). As a consequence,
we get the following which helps to compute the number of invariant meridians
of a degree one vector field on $S_{p,q}$.

######  Theorem 4.4.

A degree one vector field defined on $S_{p,q}$ can have at most as many
invariant meridians as there are linearly independent real eigenvectors of the
matrix $A=(c_{ij})_{1\leq i,j\leq p+1}$ formed by the coefficients of the
polynomials in the vector field $\mathcal{X}$.

######  Remark 4.5.

The real skew symmetric matrix $A=(c_{ij})$ is normal, that is, it commutes
with its adjoint (which, in this case, is the transpose). Hence by the
Spectral Theorem (Theorem 2.5.4 in page 101 of [5]) $A$ is diagonalizable. So
if all eigenvalues of $A$ are zero, $A$ is the zero matrix. Hence $A$ cannot
have all eigenvalues zero, since $\mathcal{X}$ is a non-zero vector field.
Thus, we see that a degree one vector field on $S_{p,q}$ can have at most
$(p-1)$ invariant meridians. In fact, we can readily construct a vector field
with $(p-1)$ invariant meridians starting with a $(p+1)\times(p+1)$ skew-
symmetric matrix with $(p-1)$ eigenvalues zero.

######  Example 4.6.

Let $p=2$ in $S_{p,q}$. Then $\mathcal{X}=(-x_{2},x_{1}-x_{3},x_{2})$ is a
vector field on $S_{p,q}$. So

$A=\begin{pmatrix}0&-1&0\\\ 1&0&-1\\\ 0&1&0\end{pmatrix}.$

Note that the eigenvalues of $A$ are $\pm\sqrt{-2},0$. In this case, the
unique eigenvector is $(1,0,1)$, and hence the corresponding invariant
meridian is given by $\\{x_{1}+x_{3}=0\\}$. Further, we see that this matches
with the bound in Remark 4.5 which is 1 in this case.

###  4.2. General Case

Now we look at the case of vector fields of degree greater than or equal to
four.

######  Theorem 4.7.

Assume $m\geq 4$. We have the following.

  1. (1)

There exists a vector field of degree $m$ with $3m-10$ invariant meridians
counted with multiplicity when $p=2$.

  2. (2)

There exists a vector field of degree $m$ with $6m-21$ invariant meridians
when $p=3$.

###### Proof.

First we construct $R_{j}$s, when $j\in\\{p+2,\ldots,n\\},$ for the invariant
vector field $\mathcal{X}$. Let $R_{j}$s be given by

$\begin{split}&R_{p+2}=x_{1}(x_{p+2}^{2}-1)H,\quad
R_{p+3}=x_{1}x_{p+2}x_{p+3}H,\quad R_{p+4}=x_{1}x_{p+2}x_{p+4}H,\\\
&R_{p+5}=x_{1}x_{p+2}x_{p+5}H,\quad
R_{p+6}=x_{1}x_{p+2}x_{p+6}H,\quad\ldots,\textsf{and},\quad
R_{n}=x_{1}x_{p+2}x_{n}H\end{split}$

where $H\in\mathbb{R}[x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n}]$.

For (1): Let

$D:=(\sum_{i=1}^{3}a_{i}x_{i})^{m-3},$

and

$R_{1}=\frac{1}{2}x_{4}(x_{1}^{2}-a^{2})D,\quad
R_{2}=\frac{1}{2}x_{4}x_{1}x_{2}D\quad\textsf{and}\quad
R_{3}=\frac{1}{2}x_{4}x_{1}x_{3}D.$

Observe that the vector field $\mathcal{X}$ determined by the above choices
for $R_{1},R_{2},R_{3}$ and $R_{4},\ldots,R_{n}$ is invariant on $S_{p,q}$ if
we let $H=D$. To be precise, for these $R_{i}$s , $\mathcal{X}$ satisfies
equation (3.2) with $K=2x_{1}x_{4}$. We shall prove that the vector field
$\mathcal{X}$, defined by these $R_{i}$s have $\sum_{i=1}^{3}a_{i}x_{i}=0$ as
an invariant meridian with multiplicity $3m-10$ for $\mathcal{X}$. The
polynomial $\sum_{i=1}^{3}a_{i}x_{i}$ belongs to the vector space $W:=\textsf{
span}\\{x_{1},x_{2},x_{3}\\}$. We note that if $\sum_{i=1}^{3}a_{i}x_{i}=0$
gives an invariant meridian for this vector field $\mathcal{X}$, then
$\sum_{i=1}^{3}a_{i}x_{i}$ is a factor of the corresponding extactic
polynomial

$\mathcal{E}_{W}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}&x_{3}\\\
\mathcal{X}(x_{1})&\mathcal{X}(x_{2})&\mathcal{X}(x_{3})\\\
\mathcal{X}^{2}(x_{1})&\mathcal{X}^{2}(x_{2})&\mathcal{X}^{2}(x_{3})\end{pmatrix}.$

In our case, this is the following

(4.4) $\displaystyle\mathsf{det}\begin{pmatrix}x_{1}&x_{2}&x_{3}\\\
\frac{1}{2}x_{4}(x_{1}^{2}-a^{2})D&\frac{1}{2}x_{4}x_{1}x_{2}D&\frac{1}{2}x_{4}x_{1}x_{3}D\\\
g_{1}D^{2}+h_{1}E&g_{2}D^{2}+h_{2}E&g_{3}D^{2}+h_{3}E\end{pmatrix}$

where

$E:=\mathcal{X}(D)=(m-3)(\sum_{i=1}^{3}a_{i}x_{i})^{m-4}(\sum_{i=1}^{3}a_{i}x_{\sigma(i)})\cdot
D$ $=(m-3)(\sum_{i=1}^{3}a_{i}x_{i})^{2m-7}(\sum_{i=1}^{3}a_{i}x_{\sigma(i)})$

for some permutation $\sigma$ on $\\{1,2,3\\}$, and
$g_{i},h_{i}\in\mathbb{R}[x_{1},\ldots,x_{3},x_{4},\ldots,x_{n}]$ for
$i\in\\{1,2,3\\}$. The third row consists of the terms like

$\mathcal{X}(\frac{1}{2}x_{4}(x_{1}^{2}-a^{2})D)=\frac{1}{2}(x_{4}^{2}-1)(x_{1}^{2}-a^{2})D^{2}+\frac{1}{2}x_{1}x_{4}^{2}(x_{1}^{2}-a^{2})D^{2}+\frac{1}{2}x_{4}(x_{1}^{2}-a^{2})E$

which we have written as $g_{1}D^{2}+h_{1}E$. Since
$D^{2}=(\sum_{i=1}^{3}a_{i}x_{i})^{2m-6}$, in the third row, one sees that
$(\sum_{i=1}^{p+1}a_{i}x_{i})^{2m-7}$ is a common factor of each term in the
third row of the matrix in (4.4). Also in the second row, $D$ is a common
factor, hence

$\mathcal{E}_{W}(\mathcal{X})=D\cdot E\cdot
h^{\prime}(x_{1},x_{2},x_{3},x_{4})=(\sum_{i=1}^{3}a_{i}x_{i})^{3m-10}h(x_{1},x_{2},x_{3},x_{4})$

for some polynomials $h^{\prime}$ and $h$ in
$\mathbb{R}[x_{1},x_{2},x_{3},x_{4},\ldots,x_{n}]$. Thus (1) is proved, since
$(\sum_{i=1}^{3}a_{i}x_{i})$ divides $\mathcal{E}_{W}(\mathcal{X})$ with
multiplicity $3m-10$.

For (2): Let $R:=(\sum_{i=1}^{4}a_{i}x_{i})^{m-3}$ and

$\displaystyle
R_{1}=\frac{1}{2}x_{5}(x_{1}^{2}-a^{2})R,~{}R_{2}=\frac{1}{2}x_{5}x_{1}x_{2}R,~{}R_{3}=\frac{1}{2}x_{5}x_{1}x_{3}R,~{}\textsf{and},~{}~{}R_{4}=\frac{1}{2}x_{5}x_{1}x_{4}R.$

Observe that $(\sum_{i=1}^{4}a_{i}x_{i})$ is a polynomial in
$W=\textsf{span}\\{x_{1},x_{2},x_{3},x_{4}\\}$ and the corresponding extactic
polynomial is

$\mathcal{E}_{W}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}&x_{3}&x_{4}\\\
\mathcal{X}(x_{1})&\mathcal{X}(x_{2})&\mathcal{X}(x_{3})&\mathcal{X}(x_{4})\\\
\mathcal{X}^{2}(x_{1})&\mathcal{X}^{2}(x_{2})&\mathcal{X}^{2}(x_{3})&\mathcal{X}^{2}(x_{4})\\\
\mathcal{X}^{3}(x_{1})&\mathcal{X}^{3}(x_{2})&\mathcal{X}^{3}(x_{3})&\mathcal{X}^{3}(x_{4})\end{pmatrix}.$

In order to compute $\mathcal{E}_{W}(\mathcal{X})$ in this case, let us first
determine

$T:=\mathcal{X}(R)=(m-3)(\sum_{i=1}^{4}a_{i}x_{i})^{2m-7}(\sum_{i=1}^{4}a_{i}x_{\sigma(i)}),$

and

$U:=\mathcal{X}(T)=(m-3)(2m-7)(\sum_{i=1}^{4}a_{i}x_{i})^{3m-11}(\sum_{i=1}^{4}a_{i}x_{\sigma(i)})^{2}\\\
+2(m-3)(\sum_{i=1}^{4}a_{i}x_{i})^{3m-10}(\sum_{i=1}^{4}a_{i}x_{\sigma^{\prime}(i)})$

where $\sigma$ and $\sigma^{\prime}$ are some permutations on the set
$\\{1,2,3,4\\}$. Then $\mathcal{E}_{W}(\mathcal{X})$ becomes

(4.5)
$\displaystyle\mathcal{E}_{W}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}&x_{3}&x_{4}\\\
\frac{1}{2}x_{5}(x_{1}^{2}-a^{2})R&\frac{1}{2}x_{5}x_{1}x_{2}R&\frac{1}{2}x_{5}x_{1}x_{3}R&\frac{1}{2}x_{5}x_{1}x_{4}R\\\
f_{31}R^{2}+g_{31}T&f_{32}R^{2}+g_{32}T&f_{33}R^{2}+g_{33}T&f_{34}R^{2}+g_{34}T\\\
f_{41}R^{3}+g_{41}RT+h_{41}U&f_{42}R^{3}+g_{42}RT+h_{42}U&f_{43}R^{3}+g_{43}RT+h_{43}U&f_{44}R^{3}+g_{44}RT+h_{44}U\end{pmatrix}$

where the
$f_{ij},g_{lk},h_{st}\in\mathbb{R}[x_{1},\ldots,x_{4},x_{5},\ldots,x_{n}]$ for
$i,l,s\in\\{3,4\\}$ and $j,k,t\in\\{1,2,3,4\\}$.

Notice that $R$ is a factor of each term of the second row,
$(\sum_{i=1}^{4}a_{i}x_{i})^{2m-7}$ is a factor of each term of the third row,
and $(\sum_{i=1}^{4}a_{i}x_{i})^{3m-11}$ is a factor of each term of the
fourth row of the matrix in (4.5). Therefore, the extactic polynomial can be
written as

$\mathcal{E}_{W}(\mathcal{X})=(\sum_{i=1}^{4}a_{i}x_{i})^{6m-21}\cdot
h(x_{1},x_{2},x_{3},x_{4},x_{5})$

for some
$h\in\mathbb{R}[x_{1},\ldots,x_{4},x_{5}]\subset\mathbb{R}[x_{1},\ldots,x_{4},x_{5},\ldots,x_{n}].$
This proves the claim (2), since $(\sum_{i=1}^{4}a_{i}x_{i})$ divides
$\mathcal{E}_{W}(\mathcal{X})$ with multiplicity $6m-21$. ∎

######  Remark 4.8.

For $p=2$ and $p=3$, an upper bound given by Theorem 4.3 is $3m$ and $6m-2$
respectively.

######  Example 4.9.

Suppose that the vector field $\mathcal{X}$ is given by

(4.6)
$R_{1}=\frac{1}{2}x_{p+2}(x_{1}^{2}-a^{2})G,~{}~{}~{}~{}R_{2}=\frac{1}{2}x_{p+2}x_{1}x_{2}G,\ldots,R_{p+1}=\frac{1}{2}x_{p+2}x_{1}x_{p+1}G,\\\
R_{p+2}=x_{1}(x_{p+2}^{2}-1)G,~{}~{}~{}~{}R_{p+3}=x_{1}x_{p+2}x_{p+3}G,~{}~{}~{}~{}R_{p+4}=x_{1}x_{p+2}x_{p+4}G,\\\
R_{p+5}=x_{1}x_{p+2}x_{p+5}G,~{}~{}~{}~{}R_{p+6}=x_{1}x_{p+2}x_{p+6}G,~{}~{}~{}~{}\ldots,~{}~{}~{}~{}R_{n}=x_{1}x_{p+2}x_{n}G,$

where $G\in\mathbb{R}[x_{1},\ldots,x_{p+1},x_{p+2},\ldots,x_{n}]$ is a
polynomial of degree $(\deg\mathcal{X}-3)$. Consider the hyperplane given by
$x_{i}=c$ where $c$ is a constant and $-1<c<1$. We want to look at the
invariant algebraic sets formed by the intersection of these hyperplanes with
$S_{p,q}$. The number of connected components of $\\{x_{i}-c\\}\cap S_{p,q}$
is one, since $\\{x_{i}-c\\}\cap S_{p,q}$ is homeomorphic to $S^{p-1}\times
S^{q}$ for $p\geq 2$, $q\geq 2$. In this case, the extactic polynomial is

$\mathcal{E}_{\\{1,x_{i}\\}}(\mathcal{X})=\mathsf{det}\begin{pmatrix}1&x_{i}\\\
0&R_{i}\end{pmatrix}=R_{i}.$

We see that the maximum possible number of these invariant hyperplanes is
$m_{i}(=\deg R_{i})$. In particular, letting
$G=\prod_{j=1}^{m_{i}-3}(x_{i}-c_{j})$ in the vector field given by (4.6), and
if we regard $x_{i}-0=x_{i}$ also as one of the invariant hyperplanes
($x_{1}-a=0$ and $x_{1}+a=0$ if $i=1$), we see that we have $m_{i}-1$
invariant algebraic sets of the form under consideration for this choice of
the vector field.

Similarly, one can do the computation for $x_{j}=c$, when
$j\in\\{p+2,\ldots,n\\},$ and an almost tight bound can be obtained.

##  5\. Invariant Algebraic sets on $S^{1}\times S^{q}$

In this section, we give a tight upper bound for the number of invariant
hyperplanes of certain types for the vector fields on $S^{1}\times S^{q}$. In
section 3, we showed that $S_{1,q}\cong S^{1}\times S^{q}$.

If the meridians on $S_{1,q}$ are invariant algebraic sets, then
$a_{1}x_{1}+a_{2}x_{2}$ divides the extactic polynomial

(5.1)
$\mathcal{E}_{\\{x_{1},x_{2}\\}}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}\\\
\mathcal{X}(x_{1})&\mathcal{X}(x_{2})\end{pmatrix}=\mathsf{det}\begin{pmatrix}x_{1}&x_{2}\\\
R_{1}&R_{2}\end{pmatrix}=x_{1}R_{2}-x_{2}R_{1}.$

If the parallels on $S_{1,q}$ are invariant algebraic sets, then
$\sum_{j=3}^{n}b_{j}x_{j}-c$ divides the extactic polynomial

$\mathcal{E}_{\\{1,x_{3},\dots,x_{n}\\}}=\mathsf{det}\begin{pmatrix}1&x_{3}&\cdots&x_{n}\\\
0&\mathcal{X}(x_{3})&\cdots&\mathcal{X}(x_{n})\\\
\vdots&\vdots&\ddots&\vdots\\\
0&\mathcal{X}^{q-1}(x_{3})&\cdots&\mathcal{X}^{q-1}(x_{n})\end{pmatrix}.$

######  Proposition 5.1.

The numbers of connected components of the intersections
$\\{a_{1}x_{1}+a_{2}x_{2}=0\\}\cap S_{1,q}$ and
$\\{\sum_{j=3}^{n}b_{j}x_{j}=c\\}\cap S_{1,q}$ are two and one respectively.

###### Proof.

For the case $\\{a_{1}x_{1}+a_{2}x_{2}=0\\}\cap S_{1,q}$, put
$x_{2}=-a_{1}x_{1}/a_{2}$, if $a_{2}\neq 0$, otherwise
$x_{1}=-a_{2}x_{2}/a_{1}$. Then (3.1) may have the following form.

$(x_{1}^{2}(1+\frac{a_{1}^{2}}{a_{2}^{2}})-a^{2})^{2}+\sum_{j=3}^{n}x_{j}^{2}=1.$

That is

$x_{1}^{2}(1+\frac{a_{1}^{2}}{a_{2}^{2}})=a^{2}\pm\sqrt{1-\sum_{j=3}^{n}x_{j}^{2}}.$

For a fixed value of $x_{1}$, we have that $\sum_{j=3}^{n}x_{j}^{2}$ is a
constant, which means we get a copy of $S^{q-1}$. Fixing the sign of $x_{1}$,
we see that when the expression $\sqrt{1-\sum_{j=3}^{n}x_{j}^{2}}$ is zero,
the copies of $S^{q-1}$ corresponding to the plus and minus signs inside the
radical on the right hand side coincide. This implies that they belong to the
same component. The two different possible signs for $x_{1}$ give two distinct
components since $x_{1}$ cannot be equal to zero since $a>1$. The argument for
the case of $(\sum_{j=3}^{n}b_{j}x_{j}=c)\cap S_{1,q}$ is similar to the proof
of the second part of Proposition 4.2. ∎

######  Theorem 5.2.

Let $\mathcal{X}$ be a polynomial vector field of degree $m$ on $S_{1,q}$ such
that there are only finitely many invariant algebraic sets. Then we have the
following.

  1. (1)

There can be at most $2(m-1)$ invariant meridians.

  2. (2)

There exists a polynomial vector field on $S_{1,q}$ with exactly $2(m-1)$
invariant meridians if $(m-1)\geq q$.

###### Proof.

For (1): Let $x_{1}=r\cos\theta$, and $x_{2}=r\sin\theta$ where $r\geq 0$ and
$\theta\in[0,2\pi]$. With these choices of coordinates, the polynomial vector
field becomes

$\begin{split}\mathcal{X}&=\frac{1}{r}(R_{1}(r\cos\theta,r\sin\theta,x_{3},\cdots,x_{n})r\cos\theta\\\
&+R_{2}(r\cos\theta,r\sin\theta,x_{3},\cdots,x_{n})r\sin\theta)\frac{\partial}{\partial
r}\\\
&-\frac{1}{r^{2}}(R_{1}(r\cos\theta,r\sin\theta,x_{3},\cdots,x_{n})r\sin\theta\\\
&-R_{2}(r\cos\theta,r\sin\theta,x_{3},\cdots,x_{n})r\cos\theta)\frac{\partial}{\partial\theta}+\sum_{i=3}^{n}R_{i}\frac{\partial}{\partial
x_{i}}.\end{split}$

This implies that

(5.2)
$x_{1}R_{2}-x_{2}R_{1}=r^{2}\dot{\theta}=(x_{1}^{2}+x_{2}^{2})\dot{\theta}.$

Observe that the maximum degree of the left hand side of (5.2) is $m+1$ and
$x_{1}^{2}+x_{2}^{2}$ is irreducible over $\mathbb{R}$. By Proposition 5.1, we
know that the intersection $\\{a_{1}x_{1}+a_{2}x_{2}\\}\cap S_{1,q}$ has two
connected components. Hence there can be at most $2(m-1)$ invariant meridians
on $S_{1,q}$. This proves the claim (1).

For (2): We consider the vector field $\mathcal{X}$ on $S_{1,q}$ given by

$R_{1}=x_{1}x_{3}\cdots x_{n}-x_{2}G,\quad R_{2}=x_{2}x_{3}\cdots
x_{n}+x_{1}G,$

and

$R_{j}=\frac{2}{q}\left(-a^{2}(x_{1}^{2}+x_{2}^{2})+\sum_{s=3}^{n}x_{s}^{2}-1\right)x_{3}\cdots
x_{j-1}x_{j+1}\cdots x_{n}$

for $j=3,\ldots,n$. Taking

$G=\prod_{i=1}^{m-1}(a_{i}x_{1}+b_{i}x_{2}),$

one can see that $G$ is a factor of $\mathcal{E}_{x_{1},x_{2}}(\mathcal{X})$
of (5.1). We know that the intersection $\\{a_{1}x_{1}+a_{2}x_{2}\\}\cap
S_{1,q}$ has two connected components. Therefore, the number of invariant
meridians for this vector field is $2(m_{1}-1)$. ∎

######  Remark 5.3.

An upper bound for the number of invariant parallels for the vector field
$\mathcal{X}$ on $S_{1,q}$ can be given by a similar calculation as in Theorem
4.3 (2). This upper bound is ${q\choose
2}(m_{p+2}-1)+\sum_{j=1}^{q-1}m_{p+j+1}$.

######  Example 5.4.

Consider the vector field $\mathcal{X}=(R_{1},\ldots,R_{n})$ on $S_{1,q}$
given by

$R_{1}=\frac{x_{1}}{4}(\sum_{s=3}^{n}x_{s}),~{}R_{2}=\frac{x_{2}}{4}(\sum_{s=3}^{n}x_{s}),~{}R_{i}=\frac{x_{i}}{4}(\sum_{s=3}^{n}x_{s})-\frac{a^{2}}{2}(x_{1}^{2}+x_{2}^{2})+\frac{(a^{4}-1)}{2},$

for $i=3,\ldots,n$. One can check that this vector field satisfies (3.1) with
cofactor $K=(x_{3}+x_{4}+\cdots+x_{n}).$ Notice that all meridians are
invariant for this vector field, that is

$\mathcal{X}(a_{1}x_{1}+a_{2}x_{2})=K(a_{1}x_{1}+a_{2}x_{2})$

with $K=\frac{1}{4}(\sum_{s=3}^{n}x_{s}).$ Hence by Proposition 2.3,
$\mathcal{X}$ has a rational first integral.

##  6\. Invariant Algebraic sets on $S^{p}\times S^{1}$

In this section, we study invariant meridians and parallels of the vector
fields defined on $S^{p}\times S^{1}$ for $p\geq 2$.

If a meridian in $S_{p,1}$ is an invariant algebraic set, then
$\sum_{i=1}^{p+1}a_{i}x_{i}$ divides the extactic polynomial

(6.1)
$\mathcal{E}_{\\{x_{1},\dots,x_{p+1}\\}}(\mathcal{X})=\mathsf{det}\begin{pmatrix}x_{1}&\cdots&x_{p+1}\\\
\mathcal{X}(x_{1})&\cdots&\mathcal{X}(x_{p+1})\\\ \vdots&\ddots&\vdots\\\
\mathcal{X}^{p}(x_{1})&\cdots&\mathcal{X}^{p}(x_{p+1})\end{pmatrix}.$

If a parallel in $S_{p,1}$ is an invariant algebraic set, then $x_{p+2}-c$
divides the extactic polynomial

(6.2)
$\mathcal{E}_{\\{1,x_{p+2}\\}}(\mathcal{X})=\mathsf{det}\begin{pmatrix}1&x_{p+2}\\\
\mathcal{X}(1)&\mathcal{X}(x_{p+2})\end{pmatrix}=\mathsf{det}\begin{pmatrix}1&x_{p+2}\\\
0&R_{p+2}\end{pmatrix}=R_{p+2}.$

######  Proposition 6.1.

The numbers of connected components of the intersections
$\\{\sum_{i=1}^{p+1}a_{i}x_{i}=0\\}\cap S_{p,1}$ and $\\{x_{p+2}=c\\}\cap
S_{p,1}$ are one and two respectively for $|c|<1$.

###### Proof.

The argument for the case of $\\{\sum_{i=1}^{p+1}a_{i}x_{i}=0\\}\cap S_{p,1}$
is similar to the proof of the first part of Proposition 4.2.

For the case of $\\{x_{p+2}=c\\}\cap S_{p,1}$, (3.1) becomes

$(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2})^{2}+c^{2}=1~{}~{}\mbox{or},~{}~{}\sum_{i=1}^{p+1}x_{i}^{2}=a^{2}\pm\sqrt{1-c^{2}}.$

Since $c$ is a fixed constant, this gives two different concentric spheres.
Therefore $\\{x_{p+2}=c\\}\cap S_{p,1}$ has two connected components, unless
$|c|\geq 1$. ∎

We note that by the arguments in the proof of Theorem 4.3 (1) we know that the
maximum number of invariant meridians is

${p\choose 2}(m_{1}-1)+\sum_{i=2}^{p+1}m_{i}+1.$

######  Proposition 6.2.

Let $\mathcal{X}$ be a vector field on $S_{p,1}$ such that $\mathcal{X}$ has
only finitely many invariant algebraic sets. Then the maximum number of
invariant parallels for $\mathcal{X}$ is $(m-1)$ where $m=\deg\mathcal{X}$.
Moreover, there is a vector field on $S_{p,1}$ with exactly $(m-1)$ invariant
parallels in $S_{p,1}$.

###### Proof.

The vector field $\mathcal{X}$ is invariant on $S_{p,1}$. So, by definition,
we have

(6.3) $\displaystyle
4\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)\left(\sum_{i=1}^{p+1}x_{i}R_{i}\right)+2x_{p+2}R_{p+2}~{}=~{}K(x_{1},\ldots,x_{p+1},x_{p+2})\left(\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)^{2}+x_{p+2}^{2}-1\right).$

If $x_{p+2}-a_{i}=0$ gives a parallel, then $x_{p+2}-a_{i}$ is a factor of
$R_{p+2}$ by Proposition 2.2 and (6.2). Then $R_{p+2}$ must be of the form

$R_{p+2}=\prod_{i=1}^{\ell}(x_{p+2}-a_{i})\cdot h(x_{1},\ldots,x_{p+1}),$

where $0\leq\ell\leq\deg R_{p+2}\leq m$ and $h$ has no factor of the form
$x_{p+2}-a$.

Suppose that $\ell=m$. Then $h$ is a constant, since $\deg R_{p+2}\leq m.$ Now
putting $x_{1}=x_{2}=\cdots=x_{p+1}=0$ in (6.3), we have the following.

$2x_{p+2}\prod_{i=1}^{\ell}(x_{p+2}-a_{i})\cdot
h=K(0,\ldots,0,x_{p+2})(x_{p+2}^{2}+a^{4}-1),$

where $a>1$. We see that the left hand side is non-zero. So,
$K(0,\ldots,0,x_{p+2})$ is also non-zero. The polynomial
$(x_{p+2}^{2}+a^{4}-1)$ is irreducible over $\mathbb{R},$ since $a>1$. But all
factors in the left hand side have degree one. So, we arrive at a
contradiction. Thus, we have that $\ell\leq(m-1)$ and the proof is complete
for the first part.

Now, consider the vector field $\mathcal{X}$ given by

(6.4)
$\begin{split}&R_{i}=x_{p+2}\frac{\left(\sum_{i=1}^{p+1}x_{i}^{2}-a^{2}\right)}{4}H\\\
&R_{p+2}=\frac{\left(\sum_{i=1}^{p+1}x_{i}\right)(x_{p+2}^{2}-1)}{2}H,\end{split}$

where $R_{i}\in\mathbb{R}[x_{1},x_{2},\ldots,x_{p+1},x_{p+2}]$ for
$i=1,\ldots,p+1$. Observe that $\mathcal{X}$ is a vector field on $S_{p,1}$.
Let $H:=\prod_{i=1}^{m-3}(x_{p+2}-c_{i})$ for $c_{i}\in\mathbb{R}$ with
$|c_{i}|<1$. Then from (6.2), we see that $(x_{p+2}^{2}-1)H$ divides the
extactic polynomial $\mathcal{E}_{\\{1,x_{p+2}\\}}(\mathcal{X})$. Therefore,
the vector field given by (6.4) has $(m-1)$ invariant parallels in $S_{p,1}$.
∎

We remark that there exists $2(m-3)+2=2(m-2)$ connected components in this
case, since each parallel has two connected components if $-1<a_{i}<1$ by
Proposition 6.1.

We note that when $p=q$, then $S_{p,1}$ and $S_{1,q}$ are homeomorphic.
However, their equations say that they are different algebraic subsets of
$\mathbb{R}^{p+2}$.

######  Example 6.3.

Consider the polynomial vector field determined by

$R_{i}=\frac{1}{4}x_{i}x_{p+2}\quad\mbox{and}\quad
R_{p+2}=\frac{x_{p+2}^{2}}{2}-\sum_{i}^{p+1}\frac{a^{2}}{2}x_{i}^{2}+\frac{(a^{4}-1)}{2}$

for $i=1,\ldots,p+1$. One can check that this vector field satisfies (6.3),
with the cofactor $x_{p+2}$. Hence $\mathcal{X}=(R_{1},\ldots,R_{p+1})$ is a
vector field on $S_{p,1}.$ Notice that all meridians on $S_{p,1}$ are
invariant for this vector field, that is

$\mathcal{X}(a_{1}x_{1}+\cdots+a_{p+1}x_{p+1})=K(a_{1}x_{1}+\cdots+a_{p+1}x_{p+1}),$

with $K=\frac{x_{p+2}}{4}.$ Hence by Proposition 2.3, $\mathcal{X}$ has a
rational first integral.

Acknowledgments. The authors thank the anonymous referee for several helpful
comments and suggestions. The first author was supported by a Senior Research
Fellowship of the University Grants Commission of India for the duration of
this work. The second author thanks ‘ICSR office IIT Madras’ for SEED research
grant.

## References

  * [1] Joan C. Artés, Branko Grünbaum, and Jaume Llibre.  On the number of invariant straight lines for polynomial differential systems.  Pacific J. Math., 184(2):207–230, 1998.
  * [2] Yudy Bolaños and Jaume Llibre.  On the number of $N$-dimensional invariant spheres in polynomial vector fields of $\mathbb{C}^{N+1}$.  J. Appl. Anal. Comput., 1(2):173–182, 2011.
  * [3] Stephen H. Friedberg, Arnold J. Insel, and Lawrence E. Spence.  Linear algebra.  Prentice Hall, Inc., Upper Saddle River, NJ, third edition, 1997.
  * [4] Victor Guillemin and Alan Pollack.  Differential topology.  AMS Chelsea Publishing, Providence, RI, 2010.  Reprint of the 1974 original.
  * [5] Roger A. Horn and Charles R. Johnson.  Matrix analysis.  Cambridge University Press, Cambridge, 1990.  Corrected reprint of the 1985 original.
  * [6] Yu. S. Il’yashenko.  Centennial history of Hilbert’s 16th problem [mr1898209].  In Fundamental mathematics today (Russian), pages 135–213. Nezavis. Mosk. Univ., Moscow, 2003.
  * [7] J. P. Jouanolou.  Équations de Pfaff algébriques, volume 708 of Lecture Notes in Mathematics.  Springer, Berlin, 1979.
  * [8] Jaume Llibre and Yudy Bolaños.  Rational first integrals for polynomial vector fields on algebraic hypersurfaces of $\mathbb{R}^{n+1}$.  Internat. J. Bifur. Chaos Appl. Sci. Engrg., 22(11):1250270, 11, 2012.
  * [9] Jaume Llibre and João C. Medrado.  On the invariant hyperplanes for $d$-dimensional polynomial vector fields.  J. Phys. A, 40(29):8385–8391, 2007.
  * [10] Jaume Llibre and João C. Medrado.  Limit cycles, invariant meridians and parallels for polynomial vector fields on the torus.  Bull. Sci. Math., 135(1):1–9, 2011.
  * [11] Jaume Llibre, Rafael Ramírez, and Natalia Sadovskaia.  On the 16th Hilbert problem for algebraic limit cycles.  J. Differential Equations, 248(6):1401–1409, 2010.
  * [12] Jaume Llibre, Rafael Ramírez, and Natalia Sadovskaia.  On the 16th Hilbert problem for limit cycles on nonsingular algebraic curves.  J. Differential Equations, 250(2):983–999, 2011.
  * [13] Jaume Llibre and Salomón Rebollo-Perdomo.  Invariant parallels, invariant meridians and limit cycles of polynomial vector fields on some 2-dimensional algebraic tori in $\mathbb{R}^{3}$.  J. Dynam. Differential Equations, 25(3):777–793, 2013.
  * [14] Jaume Llibre and Xiang Zhang.  Darboux theory of integrability in $\mathbb{C}^{n}$ taking into account the multiplicity.  J. Differential Equations, 246(2):541–551, 2009.
  * [15] Jacek Sokulski.  On the number of invariant lines for polynomial vector fields.  Nonlinearity, 9(2):479–485, 1996.
  * [16] Xiang Zhang.  The 16th Hilbert problem on algebraic limit cycles.  J. Differential Equations, 251(7):1778–1789, 2011.