text:
\begin{definition}[Definition:Exclusive Or/Truth Table/Matrix Form]
$\begin{array}{c|cc}
\oplus & \T & \F \\
\hline
\T & \F & \T \\
\F & \T & \F \\
\end{array}$
\end{definition}

config: wiki


text:
\begin{document}

\begin{abstract}
We study the analogue of the classical infinitesimal center problem in the plane, but
for zero cycles. We define the displacement function in this context and prove that it is identically zero if and only if the deformation has a composition
factor. That is,  we prove that here the composition conjecture is true, in contrast with the tangential center problem on zero cycles. Finally, we give
examples of applications of our results.
\end{abstract}

\maketitle

\section{Motivation and the main result}

The aim of this paper is to solve the zero dimensional version of the infinitesimal center problem. Zero dimensional means that the problem
is not in terms of families of planar vector fields and one dimensional closed curves (one cycles), but in terms of zero cycles, as we will
explain in the sequel. As the problem can be stated in purely algebraic terms, we first introduce the problem and then, we detail the
motivation and some applications.

\subsection{Infinitesimal center and Hilbert 16th problems}

Given a polynomial $f\in\C[z]$, of degree $m\geq 1$, a zero cycle $C$ of $f$ is a multivalued function
\begin{equation*}
C(t)=\sum_{i=1}^m n_i z_i(t),
\end{equation*}
where the $z_i(t)$ are roots of $f(z_i(t))=t$, and $\sum_{i=1}^m n_i=0$. Consider a small polynomial deformation $f(z)+\epsilon g(z)=t$ of $f = t$ of any degree,
and the deformation induced zero cycle,
\[
C_\epsilon(t)=\sum_{i=1}^m n_i z_i(t,\epsilon),
\]
where $z_i(t,\epsilon)$, $i=1,\ldots,m$, are the multivalued functions determined by $(f+\epsilon g)(z_i(t,\epsilon))=t$, $z_i(t,0)=z_i(t)$.
We shall always assume that the variable $t$ is restricted to non-critical values of the polynomial $f(z)$ so that the deformation is well-defined.
\begin{defi} \label{def:displ}
The \emph{displacement function $\Delta$ of the deformation $f(z)+\epsilon g(z)=t$ along the zero cycle $C$} is defined by
\begin{equation*}
\Delta(t,\epsilon)=\int_{C_\epsilon(t)} f,
\end{equation*}
where by definition
\[
\int_{C_\epsilon(t)} f=\sum_{i=1}^m n_i f(z_i(\epsilon,t)).
\]
Note that, since $\sum_{i=1}^m n_i=0$,
\[
\Delta(t,\epsilon)=\int_{C_\epsilon(t)} f=\int_{C_\epsilon(t)} \left(f+\epsilon g-\epsilon g\right)=-\epsilon\int_{C_\epsilon(t)} g.
\]

\end{defi}

We formulate two problems:

\begin{problem} \label{infcen}\emph{The infinitesimal center problem for zero cycles:}\newline
Caracterize those polynomials $f$, with their cycles $C$ and deformations $g$, such that the displacement function $\Delta$ of the deformation along $C$ is identically
zero.
\end{problem}

\begin{problem}\label{inf16} \emph{The infinitesimal Hilbert 16\textsuperscript{th} problem for zero cycles:}\newline
Bound the number of zeros of the displacement function $\Delta$ of the deformation $f+\epsilon g$ along cycles $C$ of $f$ as a function of the degree of
$f(z)+\epsilon g(z)$ as a polynomial in $z$.
\end{problem}

The two problems are toy examples of the corresponding two problems for small polynomial deformations of Hamiltonian systems in $\C^2$ along one
cycles, which can be traced back to Poincar\'e and Arnold \cite{A} respectively. We are considering only the polynomial case, but the problems
and some of the results can be extended to a more general setting.

\subsection{Solution of the infinitesimal center problem}

The main result we present here is the solution of the infinitesimal center problem for zero cycles. In order to state our result and to relate it with
previous results we need to introduce some notations.

\medskip

Writing the displacement function in power series of $\epsilon$, we obtain
\begin{equation*}
\Delta(t,\epsilon)=\sum_{i=1}^\infty\epsilon^iM_i(t).
\end{equation*}
We call $M_i(t)$ the $i$-th Melikov function.
It is easy to check
that the first Melnikov function is
\[
M_1(t)=-\int_{C(t)}g=\sum_{i=1}^m n_i g(z_i(t)).
\]
That is, a \emph{zero dimensional abelian integral} for the polynomial $g$ along a zero cycle $C(t)=\sum_{i=1}^m n_i z_i(t)$ of $f$, in the terminology of
Gavrilov and Movasati~\cite{GM}.

\medskip

Our problem is analogous to the problem studied by Gavrilov and Movasati in \cite{GM}, but we study the displacement function $\Delta$ whereas they studied
the first Melnikov function $M_1$. They formulated problems corresponding to Problems \ref{infcen} and \ref{inf16} in terms of the first Melnikov function
$M_1$ instead of the displacement function $\Delta$. We call their first order problems the \emph{tangential problems} and reserve the adjective
\emph{infinitesimal} for the study of the displacement function $\Delta$ of a deformation of the identically vanishing function. Example~\ref{exam:1} in
the final section illustrates the two problems.

\medskip

Gavrilov and Movasati~\cite{GM} also provided a solution for the tangential Hilbert 16th problem and a solution of a special case of the tangential center
problem. The complete solution of the tangential center problem is given in \cite{ABM1} (under a generic condition) and in \cite{GP}. See also \cite{ABM2}.

\medskip

A key tool for the solution of the tangential center problem is the composition
condition. Next, we will define it in our context.

Assume that there exist $h,\tilde f,\tilde g\in\C[z]$ such that $f(z)=\tilde f(h(z))$, $g(z)=\tilde g(h(z)$.
Then, for each cycle $C_\epsilon(t)=\underset{i=1}{\stackrel{m}{\sum}} n_i z_i(t,\epsilon)$ of $f+\epsilon g$, we
define the \emph{projected cycle} $h(C_\epsilon(t))$ of $C_\epsilon$ by $h$ as the cycle of the perturbation $\tilde f+\epsilon \tilde g = t$ defined by
\[
h(C_\epsilon(t)) =  \underset{h(z_i(t,\epsilon) )}{\sum} \left( \underset{h(z_j(t,\epsilon) )=h(z_i(t,\epsilon) )}{\sum} n_j \right) h(z_i(t,\epsilon) ).
\]
We say that this projected cycle is \emph{trivial} if
\[
\underset{h(z_j(t,\epsilon))=h(z_i(t,\epsilon))}{\sum} n_i =0,\quad i=1,\ldots,m.
\]

\begin{defi}
We say that $f,g\in\C[z]$ and a cycle $C$ of $f$ satisfy the {\em composition condition} if there exist polynomials
$\tilde f,\tilde g,h \in \mathbb{C}[z]$ such that
\end{defi}
\begin{equation*}
f(z)=\tilde f(h(z)),\quad g(z )=\tilde g(h(z )),
\end{equation*}
\\and for every $\epsilon$, the perturbed cycle $C_\epsilon$ projected by $h$ is trivial. That is,
\begin{equation}\label{eq:cc2}
\sum_{h(z_i(t,\epsilon) )=h(z_j(t,\epsilon) )} n_i = 0, \quad i=1,\ldots,m.
\end{equation}

\medskip

Now, we can state our main result.

\begin{bigtheo}\label{theo:main}
A deformation, $f+\epsilon g = t$, has an infinitesimal center for a cycle $C$ of $f$, i.e., $\Delta(t,\epsilon)\equiv 0$ for all $t$ and all $\epsilon$ small enough, if and only if $f$, $g$, $C$ satisfy the composition condition.
\end{bigtheo}

The sufficiency of the composition condition is easy. Indeed,
by \eqref{eq:cc2} we get that
\[
\Delta(t,\epsilon) = \sum_{j=1}^m n_j f(z_j(t,\epsilon)) = \sum_{j=1}^m n_j \tilde f(h(z_j(t,\epsilon)))
=\int_{h\left(C_\epsilon\right)} \tilde f\equiv 0.
\]
The converse is more difficult and will be proved in the next section.

\begin{rema}
We note that the composition conjecture is {\em not true} for the tangential center problem, as showed by Pakovich's example~\cite{P}.
(See Examples~\ref{exam:1} and \ref{exam:2}.) It is true, however, for the tangential center problem on simple cycles (Theorem 1.7~\cite{CM}.)
\end{rema}

\begin{rema}
In Theorem~\ref{theo:main} we assume that $f,g$ are polynomials, but the problem can also be considered for analytic functions $f,g$. In particular, the
function $f$ could have an infinite number of isolated fibers $z_i(t)$, so the cycle sould be considered with finite support.
\end{rema}

\subsection{Motivation}

The motivation for this paper comes from the study of polynomial vector fields in the plane. Orbits of polynomial vector fields are relatively simple and
most interesting are periodic orbits, which can belong to continuous families of periodic orbits, which we call by abuse a \emph{center}, or can be
isolated, called \emph{limit cycles}.

Two widely open classical problems, the \emph{center problem} (Poincar\'e) and the \emph{Hilbert 16-th problem}, are related to these two situations.

The center problem asks for a geometric caracterization of polynomial vector fields in the plane having a center. The 16-th Hilbert problem asks for a
bound, as a function of the degree, for the number of limit cycles.

Each of these problems has its \emph{infinitesimal version}. One starts with a polynomial system $dF=0$ having a
family of cycles $C$ called a center.
Consider its deformation
\begin{equation}\label{deformation}
dF+\epsilon \omega=0
\end{equation}
of degree $n$, where $\omega$ is a polynomial one-form.

\smallskip
\emph{Infinitesimal center problem on $1$-cycles:}
Find all polynomial deformations \eqref{deformation} of $dF$ preserving the center i.e. for which the family of periodic solutions is tranformed to a
nearby family of periodic solutions.
\smallskip

\emph{Infinitesimal 16-th Hilbert problem} (Arnold~\cite{A}): Give an upper bound for the number of limit cycles born in \eqref{deformation} as a function of the degree $n$.
\smallskip

In order to deal with these problems, one considers a transversal $T$ parametrized by the values of $F$, the (not necessarily closed) trajectory  $C_\epsilon(z)$ of \eqref{deformation} with end points on $T$ deforming $C(z)$ and one defines the \emph{displacement function $\Delta$  of the deformation
\eqref{deformation} by
\begin{equation}\label{eq:displacement}
\Delta(z,\epsilon)=\int_{C_\epsilon}dF=-\epsilon\int_{C_\epsilon}\omega.
\end{equation}}

For the center problem one searches for a characterization when $\Delta$ is identically equal to zero. For the infinitesimal 16-th
Hilbert problem, one searches for a bound for the number of isolated zeros of $\Delta$. Our Definition \ref{def:displ} of the displacement function of  a deformation along zero cycles is directly inspired by \eqref{eq:displacement}.

The previous problems also have their tangential versions, obtained by expanding in series of
functions the function $\Delta$ with respect to powers of $\epsilon$ and asking the analogous questions for the first term $M_1(t)$ of $\Delta(t,\epsilon)$.

A special case is when the planar system can be reduced to a family of Abel equations,
\[
x'=A(t)x^2+\epsilon B(t)x^3,
\]
where $A$, $B$ are trigonometric polynomials. The tangential center problem for special types of cycles first appeared when studying this
family~(see \cite{BFY99,BFY00,BFY00-2}) where, for simplicity, Briskin, Françoise and Yomdin considered $A,B$ polynomials rather than trigonometric polynomials. In that context, the
\emph{composition conjecture} was formulated~\cite{BFY00}, conjecturing that a certain sufficent condition (the composition
condition) for vanishing of the abelian integrals was also necessary. The composition condition defines all the irreducible components of the center
variety in many families of Abel equations~\cite{CGM2}, in some planar systems (see \cite{ZR,ZY}), and accounts for most of the irreducible components when studying the
tangential centers of the Abel equation at infinity~\cite{BRY}.  However, not all centers satisfy the compostion conjecture in the trigonometric Abel equation~\cite{Al},
or even the polynomial Abel equation~\cite{GG}.

The tangential center problem on zero cycles also has appeared when studying hyperelliptic planar systems~\cite{CM}.
In the last section, we show that the infinitesimal center problem also appers in these contexts, although
 in a more general version than the problem solved in this paper.

\section{Monodromy group of perturbations and proof of the main result}

In this section we prove Theorem~\ref{theo:main}. The key of the proof will be to define
a convenient monodromy for the deformation. To that end, we start recalling the monodromy
of a polynomial and then extending it to the deformation of a polynomial.

Given a nonconstant polynomial $f \in \C[z]$, recall that $z_0\in\C$ is a \emph{critical point}
of $f$ if $f'(z_0)=0$, and its associated \emph{critical value} is $t=f(z_0)$. If $t$ is not a
critical value, we say that $t$ is \emph{regular}.

We will denote by $\Sigma'$ the set of all critical values of $f$, which is a finite set. Let $m>1$ be the degree of $f$. Then, for $t \in \C \setminus
\Sigma'$ the set $f^{-1}(t)$ consists of $m$ different points $z_i(t)$, $i=1,\ldots,m$. By the implicit function theorem, one can push locally
each solution $z_i(t)$ to nearby values of $t$ thus defining multi-valued analytic functions $z_i(t)$, $t \in \C \setminus \Sigma'$.

For a polynomial $f$ of degree $m$ with $\Sigma'$ the set of critical values, each loop based at $t_0 \in \mathbb{C} \backslash \Sigma'$ defines a
permutation of the roots $z_1(t_0),\ldots, z_m(t_0)$ of $f(z)=t_0$. Thus, we have a mapping from the fundamental group $\pi_1(\mathbb{C} \backslash
\Sigma', t_0)$ to the automorphism group $Aut(f^{-1}(t_0))$, whose image forms a group $G_f$,  called the \emph{monodromy group} of $f$.

The monodromy group $G_f$ acts transitively on the fibre $f^{-1}(t_0)$ (see~\cite{CM}).
Moreover, $G_f \subseteq S_m$ is the Galois group of the Galois extension of $\mathbb{C}(t)$ by the $m$ pre-images $z_1(t),\ldots, z_m(t)$ of $t
\in \mathbb{C} \backslash \Sigma'$ by $f$ (see, for instance, Theorem~8.12 of \cite{F}). That is,
\[
G_f = \Aut_{\mathbb{C}(t)}\mathbb{C}(z_1,\ldots, z_m).
\]
The monodromy group $G_f$ induces an action on cycles. In fact, reordering the preimages $z_i(t)$ appearing in a cycle $C$ after the action of
the monodromy, we can consider the monodromy as acting on the coefficients $n_i$ of a zero cycle of $f$ by permuting them:
\[
\sigma(C(t)) = \sum_{i=1}^m n_i \sigma(z_i(t)) = \sum_{i=1}^m n_i z_{\sigma(i)}(t) = \sum_{i=1}^m n_{\sigma^{-1}(i)} z_i(t).
\]

\subsection{Monodromy group of a polynomial perturbation}

Now we define the monodromy group $G$ for a family of polynomials $f(z)+\epsilon g(z)$, considered as polynomials in $z$. Let us denote
\[
\begin{array}{rccl}
F: &  \mathbb{CP}\times\C & \to & \mathbb{CP} \\ & (z, \epsilon) & \mapsto & F(z,\epsilon) =f(z)+\epsilon g(z), z\in \mathbb{C},\text{ and }F(\infty,\epsilon)=\infty
\end{array}
\]
and
\[
\begin{array}{rccl}
H: &  \mathbb{CP}\times\C & \to & \mathbb{CP} \times \C \\ & (z, \epsilon) & \mapsto & H(z,\epsilon) =(F(z,\epsilon),\epsilon).
\end{array}
\]

Let $D(t,\epsilon)$ be the discriminant of $f(z)+\epsilon g(z)-t$ as a polynomial in $z$, and $c(\epsilon)$ the coefficient of the monomial of the
highest degree. Define
\[
\Sigma := \{ (t,\epsilon) \in \C^2: c(\epsilon)D(t,\epsilon)=0\}\cup \{\infty\}\times \C.
\]

The fiber of every $(t,\epsilon) \in \mathbb{CP} \times \C \setminus \Sigma$, consists of $n$ different points $(z_i(t,\epsilon),
\epsilon)$, where $n$ is the degree of $F$ as a polynomial in $z$. By Ehresmann's fibration lemma~\cite{E}, the map
 $H: \mathbb{CP}\times\C \setminus
H^{-1}(\Sigma) \to \C^2 \setminus \Sigma$ defines a fibration whose fibers are zero dimensional, given by the union of $n$ distinct points $z_i(\epsilon,t)$.

The fact that $H$ defines a fibration guarantees that any closed path $\gamma$ in the basis $\mathbb{CP}\times\C \setminus \Sigma$ at a point
$(t_0,\epsilon_0)$ lifts to a path  joining two points, $z_i(t,\epsilon)$ and $z_j(t,\epsilon)$, of the fiber. Moreover, closed paths homotopic to
$\gamma$ lift to homotopic paths between the points of the fiber $z_i(t,\epsilon)$ and $z_j(t,\epsilon)$. As $\mathbb{CP}\times\C
\setminus \Sigma$ is path-connected, its fundamental groups with different base poins are conjugated and one can consider the basepoint free homotopy
group. We obtain a well-defined application from $\pi_1(\mathbb{CP}\times\C \setminus \Sigma, (t,\epsilon))$ to $Sym
(H^{-1}(t,\epsilon)) \simeq S_{n}$. We call the subgroup $G$ of $Sym(H^{-1}(t,\epsilon))$ thus obtained as the image of $\pi_1
(\mathbb{CP}\times\C \setminus \Sigma, (t,\epsilon))$ the \emph{monodromy group} $G$ of $f+\epsilon g$.

\medskip

Consider also the projection $p_2\colon \mathbb{CP}\times\C \to \C$, $p_2(t,\epsilon)=\epsilon$. Define
\[
\Sigma_0=\{(t,\epsilon)\in\Sigma\colon dp_{2}|_{\Sigma=0}.\}
\]
Note that since $\Sigma$ is an algebraic variety, the projection of the points where the variety is orthogonal to the projection is just a finite set.

\begin{figure*}[ht]
\begin{center}
\begin{picture}(150,220)
\put(0,0){\includegraphics[width=150pt]{monodromy.pdf}}
\put(145,17){$\epsilon$}
\put(145,40){$\epsilon$}
\put(145,118){$\epsilon$}
\put(10,190){$z$}
\put(10,110){$t$}
\put(100,190){$H^{-1}(\Sigma)$}
\put(120,85){$\Sigma$}
\put(60,72){$\Sigma_0$}
\put(60,16){$p_2(\Sigma_0)$}
\put(155,160){$\mathbb{CP}\times \C$}
\put(170,125){$\Bigg\downarrow$}
\put(180,125){$H$}
\put(155,85){$\mathbb{CP}\times \C$}
\put(170,55){$\Bigg\downarrow$}
\put(180,55){$p_2$}
\put(170,20){$\C$}
\end{picture}
\end{center}
\caption{Diagram of the fibrations}
\end{figure*}

\begin{prop}
Given a closed path $\gamma$ in the basis $\mathbb{CP}\times\C \setminus \Sigma$ and $\epsilon_0\in\C\backslash
p_2(\Sigma_0)$, the path $\gamma$ is homotopic within $\mathbb{CP}\times\C \setminus \Sigma$  to a closed path in
$(\C\setminus \Sigma_{\epsilon_0})\times\{\epsilon_0\}$, where $\Sigma_{\epsilon_0}$ is the set of critical values of $f(z)+\epsilon_0\, g(z)$.
\end{prop}

\begin{proof}
The proof relies on Thom's first isotopy lemma (see e.g.~\cite[p.180]{Ads}), which we apply to
the mapping
\[
p_2 \colon W\to \C\backslash p_2(\Sigma_0),
\]

where $W=(\mathbb{CP}\times \C)\backslash \Sigma_0$, is stratified as
\[
\Sigma\backslash\Sigma_0<W\backslash\Sigma.
\]
The map $H$ restricted to $W$ is proper and its restriction to $\Sigma$ is a submersion. Therefore,
\[
p_2 \colon W\to \C\backslash p_2(\Sigma_0)
\]
is a topologicaly locally trivial fibration compatible with the stratification.
We consider a connection associated with this stratification.

Now, consider a closed path in $W\backslash\Sigma$. Then, we can deform the path by the conection without crossing $\Sigma$ until we obtain a closed path
in $\mathbb{CP}\times\{\epsilon_0\}$, with $\epsilon_0\not\in \Sigma_0$.
\end{proof}

Note that the monodromy induced by the original closed path and the deformed path are conjugated.

For fixed $\epsilon\in\mathbb{C}$, denote by $G_\epsilon$ the monodromy group of $f(z)+\epsilon g(z)$.
The next result shows the relation between $G$ and $G_\epsilon$.

\begin{coro}\label{coro:G}
For every $\epsilon\in\mathbb{C}$ such that $c(\epsilon)\neq 0$, $G_\epsilon$ is a subgroup of $G$, up to conjugation, and for every $\epsilon \not \in
p_2(\Sigma_0)$, $G_\epsilon$ is $G$, up to conjugation. Moreover, $G$ is the Galois group of the
Galois extension of $\mathbb{C}(t,\epsilon)$ by the $n$ preimages $z_1(t,\epsilon),\ldots,z_n(t,\epsilon)$.
\end{coro}

\begin{proof}
Let us consider a base point $(t_0,\epsilon_0) \in \C^2 \setminus \Sigma$ and let $G$ be the monodromy group induced by closed paths with that base point.
The monodromy group $G_{\epsilon_0}$ is generated by the closed paths based at $t_0$, which are in the hyperplane $(\C \setminus \Sigma_{\epsilon_0})
\times \{ \epsilon_0\} \subset \C^2 \setminus \Sigma$. Hence, they generate the same element in $G$ and obviously $G_{\epsilon_0} \subset G$.

On the other hand, by the previous proposition, closed paths in $\mathbb{CP}\times\C \setminus \Sigma$ based at $(t_0,\epsilon_0)$ with
$\epsilon_0\not\in p_2(\Sigma_0)$ can be continuously deformed into closed paths in the hyperplane $\{\epsilon = \epsilon_0\}$. Therefore, permutations
corresponding to closed paths in $G$ also belong to $G_{\epsilon_0}$, so $G\subset G_{\epsilon_0}$.

Let $\bar G$ denote the Galois group of the
Galois extension of $\mathbb{C}(t,\epsilon)$ by the $n$ preimages $z_1(t,\epsilon),\ldots,z_n(t,\epsilon)$, that is
\[
\bar G = \Aut_{\C(t,\epsilon)} (\C(z_1(t,\epsilon),\ldots,z_n(t,\epsilon))).
\]
Since the elements of $G$ are automorphisms of the fibers, then $G$ is a subgroup of $\bar G$.
On the other hand, we have the projection $\bar G\to G_\epsilon$, as any automorphism
can be restricted to a hyperplane. Take $\epsilon,\epsilon'\not\in p_2(\Sigma_0)$, then
the following commutative diagram holds
\[
\begin{matrix}
G_\epsilon & & \longleftrightarrow & & G_{\epsilon'}\\
&\nwarrow &   & \nearrow &\\
&  & \bar G &  &
\end{matrix}
\]
If the image of an element of $\bar G$ is the identity element of $G_\epsilon$,
then the diagram above proves that it is the identity element of $G_{\epsilon'}$
for every $\epsilon'\not\in p_2(\Sigma_0)$ and, in consequence, the element of $\bar G$ is
the identity element. Therefore, the morphism $\bar G\to G_\epsilon =G$ is
injective, and $\bar G$ and $G$ are conjugated.

\end{proof}

\subsection{Proof of the main result}
In order to prove Theorem~\ref{theo:main}, we distinguish different cases, given by the following proposition.
We will follow \cite{CM} (see also \cite{CL} or \cite{ABM1} for more details).

A permutation group $G$ acting on $X$ is said to be {\em imprimitive} if there exists a proper subset $B\subset X$, $\card(B)>1$, such that for any
$\sigma \in G$, $\sigma(B)\cap B=\varnothing$ or $\sigma(B)=B$. If $G$ is not imprimitive, then, it is called {\it primitive}.

\begin{prop}\label{prop:BS}
Let $f,g\in\mathbb{C}[z]$ and let $G$ be the monodromy group of $f+\epsilon g$. Then, one of the following
cases holds:
\begin{enumerate}[(i)]
 \item\label{c1} $G$ is two transitive.
 \item $G$ is isomorphic to the monodromy group of $z^p$ with $p$ prime.
 \item $G$ is isomorphic to the monodromy group of $T_p(z)$ with $p$ prime.
 \item\label{c4} $G$ is imprimitive.
\end{enumerate}
\end{prop}

\begin{proof}
Let $c_k(\epsilon)$, $k=0,\ldots,n:=\max(\deg f,\deg g)$, be the coefficients of the polynomial $f(z)+\epsilon g(z)$ as a polynomial in $z$. That is,
\[
f(z)+\epsilon g(z)=c_n(\epsilon) z^n+c_{n-1}(\epsilon)z^{n-1}+\ldots+c_0(\epsilon),
\]
where $c_k$ are affine in $\epsilon$.

Choose $\epsilon_0$ such that $c_n(\epsilon_0)\neq 0$. If $t$ is large, then $z_k(t,\epsilon_0)$ can be expanded as
\[
z_k(t,\epsilon_0)=\omega^k c(\epsilon_0)^{1/n} t^{1/n} + O(t^{1/n-1}),
\]
where $\omega$ is a primitive nth-root of unity. Hence, if we consider a cycle with $t$
large enough, it produces a cyclic permutation of the $z_k$. Therefore,
there exists a cycle $(1,\ldots,n)\in G$.

Now, applying Burnside-Schur theorem~(see \cite{DM}), either $G$ is imprimitive or, two transitive, or permutationally isomorphic to the affine
group $\Aff(p)$, where $p$ is a prime. If $G$ is two-transitive, then we are in case \eqref{c1}. Assume $G$ is permutationally isomorphic to the
affine group $\Aff(p)$, where $p$ is a prime.

Arguing as in \cite{CM}, we obtain that for any fixed $\epsilon$, the monodromy group of $f+\epsilon g$ is either isomorphic to the monodromy group
of $z^p$ or to the monodromy group of $T_p$, for $p$ prime. By Corollary~\ref{coro:G}, the same holds for $G$ for any $\epsilon\not\in p_2(\Sigma_0)$.

\end{proof}

The case $G$ two-transitive is generic. In particular, it contains the case when all critical points are of Morse type with different critical values.
We first solve the infinitesimal center problem in this case.

\begin{prop}\label{2trans}
Assume that for some polynomials $f,g$, the group  $G$ is two-transitive. Let $C$ be a zero cycle of $f$. Then, $f+\epsilon g$ has an infinitesimal
center for the cycle $C$ if and only if $f, g, C$ satisfy the composition condition.
\end{prop}

\begin{proof}
Fix $\epsilon_0\not\in \Sigma_0$. For any fiber $z_i(t,\epsilon_0)$, denote $G_i$ the stabilizer of the fiber. Acting on the zero cycle by $G_i$,
since $G$ is two-transitive, we get
\[
0=\sum_{\sigma\in G_i} \sum_{j} n_{\sigma(j)} f(z_j)= |G_i| n_i f(z_i) +|G_i|\frac{\sum_{j\neq i} n_j}{n-1}\sum_{j\neq i} f(z_j),
\]
where $n$ is the degree of $f+\epsilon g$ as a polynomial in $z$.
Since $C$ is a cycle then $n_i+\sum_{j\neq i} n_j=0$. Assume that $n_i\neq 0$, dividing by $n_i|G_i|$ in the previous equation, we obtain
\[
0=f(z_i)-\frac{1}{n-1}\sum_{j\neq i} f(z_j).
\]
Assume that $n_1,n_2\neq 0$ (reordering the roots if necessary). Then,
\[
\begin{split}
0&=\left(f(z_1)-\frac{1}{n-1}\sum_{j\neq 1} f(z_j)\right)-\left(f(z_2)-\frac{1}{n-1}\sum_{j\neq 2} f(z_j)\right)
\\&=\frac{n}{n-1}f(z_1)-\frac{n}{n-1}f(z_2).
\end{split}
\]
Arguing as in Proposition~\ref{prop:BS}, by Lüroth's theorem, $L=\C(h(z_1))$, so we obtain $f+\epsilon g=F(h(z),\epsilon)$ for a certain
$F\in \mathbb{C}[z,\epsilon]$, affine in $\epsilon$, and the composition condition holds.
\end{proof}

In the proof of Propostion~\ref{2trans}, we have not used essentially that $f$ and $g$ are polynomials, so the proposition could be generalized
to the class of rational functions.

\medskip

\begin{proof}[Proof of Theorem~\ref{theo:main}]
As we mention is the introduction, the composition condition is
sufficient for $f+\epsilon g$ to have an infinitesimal center for the cycle $C$.

Now, assume that the deformation $f+\epsilon g$ has an infinitesimal center for the cycle $C$.
By Proposition~\ref{prop:BS}, the monodromy group $ G$ of $f+\epsilon g$
is either two-transitive, equivalent to
a monomial $z^p$ or a Chebyshev polynomial $T_p$ with $p$ prime,
or  is imprimitive. If $G$ is two-transitive, we conclude by Proposition~\ref{2trans}. We now deal with each of the remaining cases.

Assume that $G$ is isomorphic to the monodromy group of $z^p$ or $T_p$, with $p$ prime. In the first case,
$f+\epsilon g$ has a unique critical value for every $\epsilon$ (as $G_\epsilon$ is a subgroup of $G$), and
therefore, $f(z)+\epsilon g(z)=K(\epsilon)(z-a(\epsilon))^p$ for some functions $K,a$. But equaling
the coefficients, $a$ must be constant, and $K(\epsilon)$ affine, so $f(z)+\epsilon g(z)=K(\epsilon) (z-a)^p$. In the second case,
arguing analogously, the two critical values must remain constant and therefore, $f$ and $g$
are multiple of the same Chebyshev polynomial.
Note that in all cases, the cycle projected by the composition factor is trivial. To resume, we have obtained that if  the group $G$ is primitive,
then the cycle can by projected to a trivial one by $f$.

Assume that $G$ is imprimitive.
Then there exists a subset of fibers $B$ such that for every $\sigma\in G$, we have either $\sigma(B)\cap B=\varnothing$ or $\sigma(B)=B$. Assume $B$ contains the fiber $z_1$. Denote $G_1$ and $G_B$ the stabilizers of $z_1$ and $B$, respectively. We have
the groups inclusions
\[
G_1\subset G_B\subset G.
\]
Since  $G$ is the Galois group of $f+\epsilon g$, by the fundamental theorem of Galois theory, we have the inclussions of fields
\[
\C(\epsilon)(t)\subset L:=\C(\epsilon)(z_1,\ldots,z_n)^{G_B}\subset \C(\epsilon)(z_1).
\]
Now, applying Lüroth's theorem~(see e.g. \cite{W}), we obtain $f+\epsilon g=F \circ h$ for certain $F, h \in \mathbb{C}(z,\epsilon)$, where $F,h$ have degree $>1$ in $z$. As $F\circ h$ is a polynomial, then for every $\epsilon$ the preimage of $\infty$ by $h$ is a point, $\alpha(\epsilon)$, and the preimage of $\alpha(\epsilon)$ by $F$ is again $\infty$. Then, composing with a convenient Möebius function (see, for instance, Lemma~3.5 of \cite{CM}), we can assume that $\alpha(\epsilon)\equiv \infty$, so $F,h \in \mathbb{C}[z,\epsilon]$. Since the composition $F(h(z,\epsilon),\epsilon)$ has degree one in $\epsilon$, then either $F$ or $h$ does not deppend on $\epsilon$. Moreover, as $\deg F>1$, $F$ must be affine in $\epsilon$, that is, $F(z,\epsilon)=F_0(z)+\epsilon F_1(z)$, for certain $F_0,F_1\in\mathbb{C}[z]$.

That is,  $f(z)+\epsilon g(z)=F(h(z),\epsilon)$, were $F,h$ are polynomials  and $F$ is affine in $\epsilon$,
and consider the projected cycle $h(C_\epsilon)$. We consider the projected problem, that is,
\[
0\equiv \int_{C_\epsilon(t)} f=\int_{h(C_\epsilon(t))} F(\cdot,0).
\]
where the degree of $F(\cdot,0)$ is strictly lower than the degree of $f$.

\medskip

If the projection $h(C_\epsilon)$ is trivial we conclude.

\medskip

If not, we continue projecting until the monodromy group of the projected problem is primitive or until we get a trivial projection. In the first case,
as proved above, the cycle can be proyected to a trivial one, so we assume we are in the second case. Assume that $h,h_1,h_2,\ldots,h_k$ are the succesive
projections, that $F,F_1,F_2,\ldots,F_k$ are the succesive composition factors, and that $h_k(h_{k-1}(\ldots h_1(h(C_\epsilon))))\equiv 0$.
Then,
\[
f(z)+\epsilon g(z)=F_k((h_k\circ h_{k-1}\circ \ldots \circ h_1\circ h)(z),\epsilon),
\]
and the cycle projected by $h_k\circ h_{k-1}\circ \ldots \circ h_1\circ h$ is trivial, so it satisfies the composition condition.

\medskip

\begin{comment}
\medskip

To conclude, we need to prove that $h$ does not deppend on $\epsilon$. Firstly,
note that if $g$ is affine, then $G_\epsilon=S_n$ for any $\epsilon$ small enough, since
after the perturbation, all the singular points are of Morse type associated to
different singular values. In particular, $G$ is two-transitive.

Assume then that $g$ is of degree at least two. Let $k(z)$ be an involution ($g(z)=g(k(z))$). Then,
\[
\begin{split}
f(z)-f(k(z))&=f(z)+\epsilon g(z)-(f(k(z))+\epsilon g(k(z)))
\\&=
F(h(z,\epsilon),\epsilon)-F(h(k(z),\epsilon),\epsilon)
\\&=
\left(h(z,\epsilon)-h(k(z),\epsilon)\right)\tilde F(z,\epsilon),
\end{split}
\]
for certain rational function $\tilde F(z,\epsilon)$. That is, $h(z,\epsilon)-h(k(z),\epsilon)$ does not deppend on $\epsilon${It remains to justify why is this}. In particular,
the coefficients of $h$ as a polynomial in $z$ must be independent of $\epsilon$ except for the independent term, that is,
\[
h(z,\epsilon)=\tilde h(z)+h_0(\epsilon),
\]
where $\tilde h$ is a polynomial. If we denote $\tilde F(z,\epsilon):=F(z+h_0(\epsilon),\epsilon)$, then
\[
f+\epsilon g=\tilde F\circ \tilde h.
\]
\end{comment}
\end{proof}

\begin{rema}
Recall that the composition condition is not necessary for $g$ to be a solution
of the {\em tangential} center problem. This is due to the fact that monodromy in the tangential
center problem is just the monodromy group of $f$, and the problem is
linear in $g$, so it is only required that the summands in $g$ satisfy the
composition condition, while in the infinitesimal
center problem, we are considering a group that contains both the monodromy
groups of $f$ and $g$, and therefore, composition factors considered must be common
composition factors of $f$ and $g$. (See Example \ref{exam:1}.)
\end{rema}

\section{Examples and applications}

In this section, we show some examples and applications to illustrate
the problem and its relation with dynamical systems. The first two
examples show that the problem is not trivial, even in simple cases.
The third example introduces the infinitesimal problem
for planar vector fields and shows how, for certain planar fields,
the one dimensional problem reduces to a zero dimensional problem.
The last two examples shows the same but for the Abel equation.
Moreover, in the last example we shall apply the results of
this paper to obtain a new proof of a recent result.

\begin{exam}\label{exam:1}{\bf Tangential problem.}
Consider $f(z)=z^6$, and a cycle
\[
C(t)=\sum_{j=0}^5 n_j z_j(t),\quad z_j(t)=t^\frac{1}{6} e^{j\frac{2\pi i}{6}},
\]
where $n_0=2$, $n_1=1$, $n_2=-1$, $n_3=-2$, $n_4=-1$, $n_5=1$.

There are two possible decompositions of $f$, with factors $h_1(z)=z^3$ and $h_2(z)=z^2$. Consider a perturbation with $g(x)=z^3+z^2=h_1(z)+h_2(z)$.

The projection of $C(t)$ by $h_1$ consists of identifying the roots $z_j$ such that $h_1(z_j)$ has the same value
and assign the sum of the weights. That is,
\[
h_1(C(t))=(n_0+n_2+n_4) t^{1/2} + (n_1+n_3+n_5) (-t^{1/2})=0.
\]
Analogously, $h_2(C(t))$ is a trivial cycle.

Recall that
\[
\Delta_\epsilon(t)=-\epsilon \int_{C(t)} g+O(\epsilon^2).
\]
It is known (see e.g.~\cite{ABM1}) that for this election of $f,g$ and $C$, the return map
at first order is identically zero, as
\[
\int_{C(t)} g=\int_{C(t)} h_1+h_2=\int_{C(t)} h_1 + \int_{C(t)} h_2=\int_{h_1(C(t))} z + \int_{h_2(C(t))} z=0.
\]

On the other hand, when calculating
\[\Delta(t)=-\epsilon\int_{C_\epsilon}g=-\epsilon\int_{C_{\epsilon}}h_1+h_2,\]
but $C_\epsilon$ is a cycle of $f+\epsilon g$, which has no non-trivial composition factors, so the above argument does not work.

Indeed, as we will prove, the second Melnikov function is not zero, so
the solution for the tangential problem is not a solution of the infinitesimal problem.
Indeed, if we derive in $f(z_j(t,\epsilon))+\epsilon g(z_j(t,\epsilon))=t$ with respect to $\epsilon$, we obtain
\[
\frac{\partial z_j}{\partial \epsilon}(t,\epsilon)=\frac{-g(z_j(t,\epsilon)) g'(z_j(t,\epsilon))}{f'(z_j(t,\epsilon))}.
\]
Then, $\Delta_\epsilon(t)=\epsilon^2 M_2(t) /2+O(\epsilon^3)$, where $M_2$ is obtained differentiating $\int_{C_\epsilon(t)} g(z)$
with respect to $\epsilon$, obtaining
\[
M_2(t)=\int_{C(t)} \frac{-g(z) g'(z)}{f'(z)}.
\]
But, replacing the roots by its value,
\[
M_2(t)=-\sum_{i=0}^6 n_j \frac{g(z_j(t)) g'(z_j(t))}{f'(z_j(t))}=-\frac{17}{12 t}.
\]
Therefore, it is not an infinitesimal center.

\end{exam}

\begin{exam}\label{exam:2}{\bf A simple monomial perturbation.}
Consider $f(z)=z^4$, a perturbation of the form $g(x)=a z^4+b z^2 + c$, and a generic cycle
\[C(t)=\sum_{j=0}^3 n_j z_j(t),\quad z_j(1)=i^j.
\]

By direct computation, it is easy to obtain that the roots of $f(z)+\epsilon g(z)=t$ are
\[
z_j(t,\epsilon)=(-1)^{\lfloor j/2\rfloor}\sqrt{\frac{-\epsilon b+(-1)^j \sqrt{\epsilon^2b^2-4(\epsilon c-t)(1+\epsilon a)} }{2(1+\epsilon a)}}.
\]
Then,
\[
\Delta_\epsilon(t)=\sum_{j=0}^3 n_j f(z_j(t,\epsilon))=\frac{\epsilon b (-n_0+n_1-n_2+n_3)\sqrt{\epsilon^2b^2-4(\epsilon c-t)(1+\epsilon a)} }{2(1+
\epsilon a)^2}.
\]
That is, if $b=0$ or $n_1+n_3=n_0+n_2$, then $\Delta_\epsilon(t)\equiv 0$ for every $t,\epsilon\in\mathbb{C}$. In the first case, $g(z)=a f(z)+c$,
and in the second case, if we denote $h(z)=z^2$, then $f(z)=h^2(z)$, $g(z)=a h^2(z)+b h(z)+c$, and, since $n_0+n_1+n_2+n_3=0$, then $n_2+n_0=0$ and
$n_1+n_3=0$. In particular, the projection of $C_\epsilon$ by $h$ is identically zero.
\end{exam}

\begin{exam}{\bf Infinitesimal center problem and hyper-elliptic planar systems.}
Let us consider a perturbed center problem in the plane, that is, consider
$F\in\mathbb{C}[x,y]$, a polynomial one-form $\omega$, and the deformation of
the foliation $F(x,y)=t$,
\[
dF+\epsilon \omega=0.
\]
Take a regular value $t_0$ of $F$ and $\gamma(t_0)\subset F^{-1}(t_0)$ a closed path. Let $T$ be
a transversal section to the leaves of $F$ parametrized by $t$. Consider the one cycle $\gamma_\epsilon(t)$
obtained by deformation
of $\gamma(t_0)$ with respect to $t$ and $\epsilon$.
Note that $\gamma_\epsilon(t)$ is not necessarily closed, as we are considering it
up to the first intersection with $T$.
Let $\Delta_\epsilon$ be the associated displacement map
\[\Delta_\epsilon(t)=\int_{\gamma_\epsilon(t)} dF=-\epsilon \int_{\gamma_\epsilon(t)} \omega.\]
Then, the deformation preserves the center (foliation with closed paths) defined by $\gamma(t_0)$ if and
only if $\Delta(t,\epsilon)\equiv 0$, for every $t,\epsilon\in\mathbb{C}$.

\medskip

Now, assume that we are in the hyper-elliptic case in the real plane, that is $F(x,y)=y^2+f(x)$, $f\in\mathbb{R}[x]$. Consider a transversal section {$T$} in the axis $y=0$  and $x_0,t_0=f(x_0)$ such
that $f'(x_0)\neq 0$ and the curve $\gamma(t_0)$ is closed. As for $\epsilon=0$ the foliation defined
by $F(x,y)=t$ consists of closed curves, there exists a first time $t_1(t_0)>0$ such that $\gamma(t_1)$ belongs
to the axis $y=0$. We may extend this function to $t,\epsilon$ by the implicit function theorem. Moreover,
as $\gamma(-t_1)$ also belong to the axis $y=0$, we also define a function $t_2(t,\epsilon)$ as the first negative time such that $\gamma(t_2)$ intersects the axis $y=0$. Then, the displacement map can be written as the zero dimensional
integral
\[
\begin{split}
\Delta_\epsilon(t)&=\int_{\gamma_\epsilon(t)} dF=\int_{(x(t_1,\epsilon),0)}^{(x(t_2,\epsilon),0)}dF
=F(x(t_2,\epsilon),0)-F(x(t_1,\epsilon),0)\\&=f(x(t_2,\epsilon))-f(x(t_1,\epsilon)).
\end{split}
\]
The tangential version of this problem has been solved in~\cite{PM} for a vanishing cycle with a
Morse point and in~\cite{GP} for a vanishing cycle in general. Note that in this problem the
deformation is not polynomial, so we can not directly apply our results.

\end{exam}

\begin{exam}{\bf Moment problem.}
Around 2000, in a series of papers~\cite{BFY98,BFY99,BFY00,BFY00-2}, Briskin, Françoise and Yomdim proposed the problem
of determing the trigonometric polynomials $a,b$ such that the
following family of Abel differential equations has a center at the
origin for every $\epsilon\in\mathbb{R}$
\begin{equation}\label{eq:Abel}
x'=a(t) x^2+\epsilon b(t) x^3,
\end{equation}
where an equation of the family is said to have a center at the origin if
every bounded solution is closed.

A necessary condition, called \emph{composition condition} \cite{ALl}, is the existence of functions, $\tilde A,\tilde B, h$,
with $h(0)=h(2\pi)$, such that
\[A(t):=\int_0^t a(s)\,ds=\tilde A(h(t)),\quad B(t):=\int_0^t b(s)\, ds=\tilde B(h(t)).\]

Let us denote $x(t,x_0,\epsilon)$ the solution of \eqref{eq:Abel} determined by
the initial condition $x(0,x_0,\epsilon)=0$. Assume that for $\epsilon=0$, \eqref{eq:Abel}
has a center at the origin. That is $\int_0^{2\pi} a(t)\,dt=0$.
Differentiating $x^{-1}(t,x_0,\epsilon)$
with respect to $t$, evaluating at $t=2\pi$ and denoting $t=1/x(0)$, we obtain
that \eqref{eq:Abel} is a center for every $\epsilon$
if and only if
\begin{equation}\label{eq:AbelInf}
\int_0^{2\pi} \frac{b(t)}{t-\int_0^t a(s)+\epsilon b(s)x(s)\,ds}\,dt\equiv 0,\quad \text{ for every }\epsilon,t\in\mathbb{R}.
\end{equation}
By the change of variables $t\mapsto z=\exp(it)$, and complexifying the variables,
\[
\oint_{|z|=1} \frac{b(z)}{t-\int_0^z a(w)+\epsilon b(w)x(w)\,dw}\,dz\equiv 0,\quad \text{ for every }\epsilon,t\in\mathbb{C}.
\]
By Proposition~3.1 of \cite{ABC}, this is equivalent to
\[
\sum n_i b(z_i(t,\epsilon))\equiv 0,\quad \text{ for every }\epsilon\in\mathbb{R},t\in\mathbb{C},
\]
where $z_i$ are the preimages of $t$ by $\int_0^z a(w)+\epsilon b(w)x(w)\,dw$,
and $n_i$ are related to the branches of $A^{-1}$. Obviously,
the functions involved are not polynomials, not even rational, but we expect some ideas developped here
could be used for that problem.
\end{exam}

\begin{exam}{\bf Polynomial moment problem.}
The same problem can be considered when $a,b$ are polynomials
and a closed solution is a solution $x(t)$ of \eqref{eq:Abel}
such that $x(0)=x(2\pi)$. This polynomial version of the moment problem has been recently
solved by Pakovich~\cite{PSCC}, proving that if \eqref{eq:Abel} has a center for
every $\epsilon\in\mathbb{R}$, then $a,b$ satisfy the composition condition.

Pakovich's solution of the polynomial moment problem is based strongly on the solution of the tangential version of the problem~\cite{PM},
which ask for centers {\it at first order in $\epsilon$}. More precisely, taking $\epsilon=0$ in \eqref{eq:AbelInf},
a necessary condition for \eqref{eq:Abel} to have a center for every $\epsilon$
is that
\begin{equation}\label{eq:tc}
\int_0^{1} \frac{b(t)}{t-\int_0^t a(s)\,ds}\,dt\equiv 0,\quad \text{ for every }t\in\mathbb{R}.
\end{equation}
When \eqref{eq:tc} holds, we say that \eqref{eq:Abel} has a tangential center.

\medskip

A stronger condition is to consider $a(t)=a_0(t)+\epsilon b(t)$, and assume that \eqref{eq:tc} holds for
every $\epsilon\in\mathbb{R}$. It was proposed by Cima, Gasull and Mañosas~\cite{CGM}. They call it \emph{highly persistent center}. They proved that if \eqref{eq:Abel} has a highly persistent
center, then \eqref{eq:Abel} has a composition center.

\medskip

Now, we show an alternative proof of this result, using Theorem~\ref{theo:main}.
Denote $f(z)=\int_0^z a(w)\, dw$, $g(z)=\int_0^z b(w)\, dw$, and $t=1/x(0)$.
By Proposition~8.2 of \cite{ABM1},
\[
\int_0^{1} \frac{b(z)}{t-f(z)}\,dz=\int_{C(t)} g(z),
\]
where $C$ is the cycle
\[
C(t)=n_1\sum_{i=1}^{n_0} z_{a_i}(t)-n_0\sum_{i=1}^{n_1} z_{b_i}(t),
\]
where $z_{a_i}(t)$ are all the solutions of $f(z_{a_i}(t))=t$ with $z_{a_i}(t)$ close to $0$ for $t$ close to $f(0)$,
and analogously for $z_{b_i}(t)$ and $1$.

Now, if $a(z)=a_0(z)+\epsilon b(z)$, then $f(z)=f_0(z)+\epsilon g(z)$, the cycle $C$ becomes $C_\epsilon$.
So, if \eqref{eq:Abel} has a highly persistent center, then
\[
\int_{C_\epsilon(t)} g(z)\equiv 0,\quad \text{for every }t,\epsilon\in\mathbb{C}.
\]
By Theorem~\ref{theo:main}, there exists $F\in\mathbb{C}[\epsilon,z]$, affine in $\epsilon$, and $h\in \mathbb{C}[z]$, such that $f(z)+\epsilon g(z)=F(h(z),\epsilon)$ and $h(C_\epsilon(t))\equiv 0$. This implies
that $h(0)=h(1)$, and that there exists $\tilde f,\tilde g$ such that $f=\tilde f\circ h$, $g=\tilde g\circ h$, so the composition condition holds.

\end{exam}

config: arxiv


TITLE: Is every complete set verifying the property $\mathcal P$ uncountable?
QUESTION [0 upvotes]: The question.
I was wondering whether or not the following assumption is true.

Let $E$ be a complete metric set containing a point verifying the property $\mathcal P$. Then $E$ is uncountable.
A point $x\in E$ verifies the property $\mathcal P(x)$ if for every neighbourhood $V$ of $x$, there exists $y\in V\setminus \{x\}$ and $y$ verifies $\mathcal P(y)$.

In other words, $\mathcal P(x)$ is true if, and only if, $x$ is not isolated, and for every neighbourhood $V$ of $x$, $V\setminus \{x\}$ contain a non isolated point verifying the same property.
What I did.
I tried to prove the veracity of the result since I wasn't able to find a counterexample.
Let $x\in E$ be a point such that $\mathcal P(x)$ is true.
Then we can construct recursively a sequence $(x_n)$:
$$\begin{cases} x_0\ne x \\ \mathcal P(x_0)\text{ is true.}\end{cases}$$
Then assuming $x_0,\ldots,x_n$ are constructed, let's define $r_n$ such that
$$\forall i\in\{1,\ldots,n-1\},\quad x_i\notin B(x_n,r_n).$$
We have a set $X=\{x_i\}$ of points of $E$. Since when can always double the size of $X$ because every $x_i$ verifies $\mathcal P(x_i)$, I would tend to think that we can construct a set $Y$ in bijection with $2^X$, which would have the cardinality $2^{\aleph_0}$, so $Y$ would be uncountable. So $E$ would be.

REPLY [3 votes]: Let $A = \{ x \in E : \mathcal{P}(x)\}$. By assumption, $A \neq \varnothing$. By definition of $\mathcal{P}$, every $x\in A$ is a limit point of $A$. So $A$ is infinite. Let $P = \overline{A}$ (note: since $A = E' \cap E''$, and derived sets are closed in $T_1$-spaces, in fact $P = A$). Then $P$ is a perfect subset of $E$, hence uncountable.


TITLE: P(Y >84 | Y > 72) =?
QUESTION [1 upvotes]: Scores on an examination are assumed to be normally distributed with mean 78 and variance 36.
If it is known that a student’s score exceeds 72, what is the probability that his or her score exceeds 84?
P(Y >84 | Y > 72) = [(Y > 84) and P (Y > 72)] / P(Y > 72) , after this step, I don't know what I need to do.
Would I have to covert this y into a z?
Any help would be appreciated. Thanks!

REPLY [0 votes]: As you wrote:
$$
P(Y >84 | Y > 72) = \frac{P[Y > 84\  \& \ Y > 72]}{P(Y > 72)}
= \frac{P(Y > 84)}{P(Y > 72)}.
$$
Now let us note $Y = 78+  6G$, so that $G$ has a standard gaussian distribution.
$$P(Y >84 | Y > 72) = \frac{P(G > 1)}{P(G > -1)}.$$Eventually, use you tables:
$$
P(Y >84 | Y > 72)\simeq \frac {1-0.8413}{0.8413} \simeq 0.1886$$


TITLE: Natural params and sufficient statistics in multivariate normal distribution
QUESTION [0 upvotes]: Let's represent multivariate normal distribution as an exponential family:
\begin{align}
f_X( x | \Theta ) = h(x)g(\Theta)\exp( \eta(\Theta) \cdot T(x) )
\end{align}
Where natural parameters:
\begin{align}
\eta(\Theta) = \left[
\begin{matrix}
\Sigma^{-1} \mu \\
-\frac12 \Sigma^{-1}
\end{matrix}
\right]
\end{align}
Sufficient statistic:
\begin{align}
T(x) = \left[
\begin{matrix}
x \\
xx^T
\end{matrix}
\right]
\end{align}
Does $\eta(\Theta) \cdot T(x)$ denote a dot product?
If so then we can't get a scalar value. We need some more complex function then dot product - for instance trace: $\operatorname*{Tr}(\Sigma^{-1} xx^T) = \operatorname*{Tr}(x^T\Sigma^{-1} x)= x^T\Sigma^{-1} x$
Update
What are the proper identical transformations to get:
\begin{align}
\eta(\Theta) \cdot T(x) = \\
\left[
\begin{matrix}
\Sigma^{-1} \mu \\
-\frac12 \Sigma^{-1}
\end{matrix}
\right] \cdot \left[
\begin{matrix}
x \\
xx^T
\end{matrix}
\right] = \\
??? = \\
-\frac12 x^T \Sigma^{-1} x + \mu^T \Sigma^{-1} x
\end{align}

REPLY [0 votes]: However, the inner product has an alternative form $x^T\Sigma x=\sum_{i,j} x_i x_j \sigma_{ij}$ where $\Sigma=\{\sigma_{ij}\}$.


TITLE: A Question on Sets of Full Outer Measure
QUESTION [2 upvotes]: I came across this problem whilst studying for a comprehensive exam in real analysis; for reference, see Exercise 1.24(A) in Folland's Real Analysis; it's a modification of that.
Consider the unit interval $I:=\left[0,1\right]$, and let $\mathcal{M}$ be the $\sigma$-algebra of all Lebesgue measurable subsets of $I$.  Denote by $m^*$ the Lebesgue outer measure on $\mathcal{M}$.  Suppose that $E$ is a subset of $I$ such that $m^{*}(E)=1$.  Prove that, if $A,B\in\mathcal{M}$ and $A\cap E=B\cap E$, then $m(A)=m(B)$.
This result is trivial if $E$ is Lebesgue measurable; the sticky wicket occurs if $E$ fails to be so.  I and my colleagues have attempted to solve this using the Carathéodory characterization of Lebesgue measurability, and failed as yet.  We are aware that it need not be the case that $E$ be measurable.  Any hints would be appreciated.

REPLY [0 votes]: Here is one solution written by Xiudi Tang in his course notes:
It suffices to show that $m^*(A \cap E) = m^*(A)$ for any $A \in \mathcal{M}$.
By the measurablity of $A$, we have that $$m^*(I)=m^*(A)+m^*(I-A)$$ $$m^*(E)=m^*(A \cap E)+m^*(E-A).$$
Since $m^*(E)=m^*(I)$ and $E \subset I$, we have that $$m^*(I-A) \geq m^*(E-A),$$ $$m^*(A) \geq m^*(A \cap E)$$ and $$m^*(A)+m^*(I-A)=m^*(A \cap E)+m^*(E-A).$$
Therefore $m^*(A \cap E) = m^*(A)$.


TITLE: The First Homology Group is the Abelianization of the Fundamental Group.
QUESTION [27 upvotes]: I am trying to understand the proof of the following fact from Hatcher's Algebraic Topology, section 2.A.

Theorem. Let $X$ be a path connected space.
  Then the abelianization of $\pi_1(X, x_0)$ is isomorphic to $H_1(X)$.

I am having trouble understanding the last step of the proof.
Step 1. First we define a map $h:\pi_1(X, x_0)\to H_1(X)$ which sends the homotopy class $[f]$ of a loop $f$ based at $x_0$ to the homology class of the cycle $f$. One checks that this is a well-defined group homomorphism.
Further $h$ is surjective.
Step 2. Now since $H_1(X)$ is abelian, the map $h$ factors through the abelianization  $\pi_1(X, x_0)^{ab}$ of $\pi_1(X, x_0)$.
Step 3. The main part is to show that the induced map $\pi_1(X, x_0)^{ab}\to H_1(X)$ is injective.
To do this we need to show that if a loop $f$ based at $x_0$ in $X$ is a boundary, then $f$ is in the commutator subgroup.
Since $f$ is a boundary, we have singular $2$-simplices $\sigma_i$ such that $f=\partial(\sum_i n_i\sigma_i)$.
We have $\partial \sigma_i = \tau_{i0} - \tau_{i1} + \tau_{i2}$, where $\tau_{ij}$'s are singular one simplices obtained by restricting $\sigma_i$ on the edges of the standard $2$-simplex.
Hatcher shows that one may assume that each $\tau_{ij}$ is a loop based at $x_0$.
Step 4. This step is where I am having problem. Hatcher writes "Using the additive notation in the abelian group $\pi_1(X, x_0)^{ab}$, we have the formula $[f]=\sum_{i, j}(-1)^jn_i[\tau_{ij}]$ because of the cancelling pairs of $\tau_{ij}$'s. We can rewrite the summation $\sum_{i, j} (-1)^j n_i[\tau_{ij}]$ as $\sum_i n_i [\partial \sigma_i]$ where $[\partial \sigma_i]=[\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]$. Since $\sigma_i$ gives nullhomotopy of the composed loop $\tau_{i0}-\tau_{i1}+\tau_{i2}$, we conclude that $[f]=0$ in $\pi_1(X, x_0)^{ab}$."
I do not seem to understand anything in the paragraph quoted above. Can somebody elaborate on that? The additive notation is especially confusing. So if possible please use the multiplicative notation. Thank you.

REPLY [14 votes]: I follow the proof by Hatcher that the OP outlines.
The first important point is the following, once you have proved that $f = \Sigma_{i,j}(-1)^jn_i\tau_{i,j}$ for your 1-cycle $f$, you need to remember that the group of 1-chains (which contains the 1-cycles as a subgroup) is the free $\mathbb{Z}$-module (ie the free abelian group) on the set of continuous maps from $\Delta_1$ to $X$. Hence, in particular, every element $f$ of this group as a unique expression of the form $n_1f_1+\ldots +n_pf_p$ with $n_i\in\mathbb{Z}$ and $f_i$ a continuous map from $\Delta_1$ to $X$. So, from the equality $f = \Sigma_{i,j}(-1)^jn_i\tau_{i,j}$ you conclude, as noted by Hatcher (third paragraph from the end), that $f$ is one of the $\tau_{i,j}$'s and the remaining $\tau_{i,j}$'s form canceling pairs. This allows you to state this equality between homotopy classes $[\Sigma_{i,j}(-1)^jn_i\tau_{i,j}] = \Sigma_{i,j}(-1)^jn_i[\tau_{i,j}]$. But  $\Sigma_{i,j}(-1)^jn_i[\tau_{i,j}] = \Sigma_in_i[\partial\sigma_i]$.
The second important point consists in noting the fact that, $\sigma_i$ being a singular 2-simplex with boundary given by $\tau_{i0} -\tau_{i1} +\tau_{i2}$, you can continuously deform this boundary, through $\sigma_i$, into the constant loop at $x_0$. Hence, one has the following equalities between homotopy classes $[\partial \sigma_i] = [\tau_{i0}]-[\tau_{i1}]+[\tau_{i2}]=0$.
Now, third point, you need to realize this last equality means $[\tau_{i2}] = -([\tau_{i0}]-[\tau_{i1}])$, or in multiplicative notation as you wish $[\tau_{i2}]=([\tau_{i0}][\tau_{i1}]^{-1})^{-1}$. Thus $[\partial\sigma_i]=[\tau_{i0}] -[\tau_{i1}] +[\tau_{i2}]$, being an element followed by its inverse, belongs to the commutator subgroup and so $\Sigma_in_i[\partial\sigma_i]=[f]$ does, meaning that $[f]$ is trivial in $\pi(X,x_0)_{ab}$.


TITLE: Semiclassical approximation in Quantum Field Theory
QUESTION [2 upvotes]: I've recently stumbled upon a semiclassical approximation to quantum field theory that I've never heard of and have a hard time understanding. Consider the Hamiltonian,
\begin{equation}
H = \frac{c}{ M ^2  } \bar{\psi} \gamma _\mu \psi \bar{\chi} \gamma ^\mu \chi
\end{equation}
Now the claim is that we can approximation how the $ \chi $ particle will behave in a medium of $ \psi $ particles with the following Hamiltonian,
\begin{equation}
H = \frac{ c }{ M ^2 }  \left\{ \int \,d^3p f ( E _p  ) \left\langle \psi ( p , s ) \right| \bar{\psi} \gamma ^\mu \psi \left| \psi  ( p, s ) \right\rangle  \right\} \bar{\chi} \gamma _\mu \chi
\end{equation}
where $ f ( E _p  ) $ describes the distribution of $ \psi $ field in the medium. This should apparently be an obvious step but I am having trouble wrapping my head around it. How can I justify this step. This is quite different then integrating out the $ \psi $ fields.
Note: Though the context likely isn't very important I stumbled upon this step in a derivation of the MSW effect for neutrino oscillations in matter that can be found here

REPLY [3 votes]: I see this as a partial trace on the $\psi$ space, that is :
Let $H$ be the hamiltonian applying on the $(\psi, \chi)$ space, and $H_{eff}$ the reduced hamiltonian applying on the $\chi$ space. Take some density matrix $\rho$ applying on the $\psi$ space.
Then : $H_{eff} = Tr_\psi(\rho H)$
Here one is taking $\rho = \int d^3p f(E_p) |\psi(p,s)\rangle \langle \psi(p,s)|$


text:
\begin{definition}[Definition:Order Embedding/Definition 3]
$\phi$ is an '''[[Definition:Order Embedding|order embedding]] of $S$ into $T$''' {{iff}} both of the following conditions hold:
:$(1): \quad \phi$ is an [[Definition:Injection|injection]]
:$(2): \quad \forall x, y \in S: x \prec_1 y \iff \map \phi x \prec_2 \map \phi y$
\end{definition}


TITLE: Find all monic polynomials $f(x)$ having integer coefficients such that $f(0)=2020$ and for any irrational $x$, $f(x)$ is also irrational.
QUESTION [7 upvotes]: I have tried to approach the sum with some help from the rational root theorem but apparently it seems to me that for a polynomial with integer coefficients, putting an irrational $x$ would always produce an irrational value of $f(x)$ quite obviously but I cannot prove it rigorously. Can anybody please help me with the solution.

REPLY [0 votes]: Let $p(x)$ be a monic polynomial with integer coefficients such that $p(0) = 2020$. Let $a \in \Bbb{Z}$ be an arbitrary integer. By hypothesis, $y$ irrational implies $p(y)$ irrational, so $p(x) - a$ splits over $\Bbb{Q}$ (and therefore over $\Bbb{Z}$, since it has integer coefficients) regardless of the value of $a$. This means $p(x)$ is a surjective function from $\Bbb{Z}$ onto $\Bbb{Z}$, since $p(x) - a$ splits over $\Bbb{Z}$ only if $a \in p(\Bbb{Z})$.
Now suppose $p(x)$ has degree $n$. The set $$S_k := \{ y \in \Bbb{Z}: |y| \leq k \text{ and } y \in p(\Bbb{Z}) \}$$ grows at the rate $|S_k| = \mathcal{O}(\sqrt[n]{k})$ for $k \geq 1$, and if $p$ is surjective, $|S_k| = 2k+1$. These orders of growth are only compatible if $n=1$, so $p$ is linear, and the only monic linear polynomial with $p(0) = 2020$ is $p(x) = x+2020$.


TITLE: array equalizer by multiplication
QUESTION [0 upvotes]: given set of three numbers and an array of numbers. we can multiply the values in array any number of times by the numbers in set of three numbers. how to detect at any point of time all values in array will be equal or not.
ex. three numbers- [a ,b , c]
array = [1 ,2 ,3 ,4, 5, 6,7]
now we can multiply every value in array any number of times by a , b, c. at any time will the values in the array be equalized or not? what is the condition for that. The three numbers can be any numbers and array can be any array.

REPLY [0 votes]: In order to solve this you need to look at the prime factorisation. So the array of three numbers should contain the unique prime factors of the numbers of the other array. However if some prime(s) only occurs as a common factor raised to some power(s), than that factor can be used as well in the first array. For example 12 and 144, which is $2^2\times3$ and $(2^2\times3)^2$.
In your example there are more prime factors that three, so there will be no solution.