text:
\begin{document}
\begin{abstract}
{
In this paper,
it is shown that, for any equidimensional $C^\infty$ map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$, the map-germ
$F: (\mathbb{R}^n, 0) \to
\mathbb{R}^n\times\mathbb{R}^{\ell}$ defined by
$F(x)=\left(f(x), \mu_1(x){|Jf|^2(x)}, \cdots,
\mu_\ell(x){|Jf|^2(x)}\right)$ is always a frontal;
where $\mu_i$ is a $C^\infty$ function-germ and
$|Jf|$ is the Jacobian-determinant of $f$.
Moreover, it is also shown that when the multiplicity of $f$ is less than or
equal to $3$,
any frontal constructed from $f$ must be $\mathcal{A}$-equivalent to
a frontal $F$ of the above form.
}
\end{abstract}
\subjclass[2010]{57R45, 58K05
}
\keywords{Jacobian-squared,
Frontal, Opening, Ramification module, Equidimensional map-germ.
}

\maketitle
\section{Introduction}\label{section 1}
\noindent
Throughout this paper, let $n, \ell$ be positive integers;
and all map-germs, vector fields and differential forms
are of class $C^\infty$ unless otherwise stated.
\par
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$
 be a map-germ  and let $|Jf|$ denote the Jacobian-determinant of $f$.
The square of $|Jf|$ is called
the \textit{Jacobian-squared} function-germ of $f$.
In the theory of singularities of mappings,
it is well-known that the function-germ $|Jf|$ plays an essential role
to investigate the behavior of $f$
(for instance, see \cite{arnoldguseinzadevarchenko, wall}).
However, to the best of author's knowledge, so far
there have been no literatures
to emphasize the importance of the Jacobian-squared $|Jf|^2$.
In this paper, it is explained that $|Jf|^2$ is very significant
to construct non-trivial frontals from a given equidimensional
map-germ $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$.
\par
Let $F: (\mathbb{R}^n,0)\to (\mathbb{R}^{n+\ell}, 0)$
be a map-germ and let $T\mathbb{R}^{n+\ell}$ be the
tangent bundle of $\mathbb{R}^{n+\ell}$.
A map-germ $\Phi: (\mathbb{R}^n,0)\to T\mathbb{R}^{n+\ell}$
is called a \textit{vector field along $F$} if
the equality $\pi\circ \Phi(x)=F(x)$ holds, where
$\pi: T\mathbb{R}^{n+\ell}\to \mathbb{R}^{n+\ell}$ is the
canonical projection.
Namely, $\Phi$ has the form $\Phi(x)=(F(x), \phi(x))$ where
$\phi(x)\in T_{F(x)}\mathbb{R}^{n+\ell}$.

A map-germ $F: (\mathbb{R}^n,0)\to \mathbb{R}^{n+\ell}$
is called a \textit{frontal} if there exist vector fields
$\Phi_1, \ldots, \Phi_\ell: (\mathbb{R}^n,0)\to T\mathbb{R}^{n+\ell}$
along $F$ such that the following three conditions are satisfied:
\begin{enumerate}
\item[(1)] $\phi_i(x)\cdot tF(\xi)(x)=0$ for any $i$ $(1\le i\le \ell)$
and any $\xi\in \theta(n)$,
where $\Phi_i(x)=\left(F(x), \phi_i(x)\right)$ and the dot in the center
stands for the scalar product of two vectors in
$T_{F(x)}\mathbb{R}^{n+\ell}$.
\item[(2)] $\phi_i(0)\ne 0$ for any $i$ $(1\le i\le \ell)$.
\item[(3)] $\phi_1(0), \ldots, \phi_\ell(0)$ are linearly independent.
\end{enumerate}
\noindent
Here, $tF$ and $\theta(n)$ are the notations
defined by J.~Mather in \cite{mather3} (For details, see \S \ref{section 2}).
\par

Though frontals have been already well-investigated
(for instance see \cite{ishikawa}),

it seems to be desired to obtain how to construct
non-trivial frontals easily and systematically
from a given equidimensional map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ which is not necessarily finite
(for the definition of finite map-germ, see Subsection \ref{subsection 2.1}).

\begin{theorem}\label{theorem 1}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be a map-germ.
For any $i$ $(1\le i\le \ell)$, let
$\mu_i: (\mathbb{R}^n,0)\to \mathbb{R}$
be a function-germ.

Then, the map-germ $F: (\mathbb{R}^n,0)\to \mathbb{R}^{n+\ell}$
defined by
\[
F=\left(f, \mu_1|Jf|^2, \ldots,
\mu_\ell|Jf|^2
\right)
\]
is always a frontal.

\end{theorem}
\begin{definition}[\cite{ishikawaaustralia, ishikawa}]
\label{ramification module}
{\rm
For a given equidimensional map-germ
$f=(f_1, \ldots, f_n): (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$,
the \textit{ramification module} of $f$ (denoted by $\mathcal{R}_f$)
is defined as the $f^*\left(\mathcal{E}_n\right)$-module consisting of
all functions $\varphi$ such that $d\varphi$ is an element of
the $\mathcal{E}_n$-module generated by $df_i$ $(i=1, \ldots, n)$,
where $dh$ for a function-germ $h: (\mathbb{R}^n,0)\to \mathbb{R}$
stands for the exterior differential of $h$.
}
\end{definition}
\noindent
Given an equidimensional map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$,
the \textit{pull-back} $f^*:\mathcal{E}_n\to \mathcal{E}_n$ is defined
by $f^*(\eta)=\eta\circ f$.    The pull-back $f^*$ gives an
$\mathcal{E}_n$-module structure via $f^*$.
In this paper, an $\mathcal{E}_n$-module via $f^*$ is called and
denoted by an $f^*\left(\mathcal{E}_n\right)$-module.     Hence,
$f^*\left(\mathcal{E}_n\right)$ itself is
an $f^*\left(\mathcal{E}_n\right)$-module.
It is clear that $f^*\left(\mathcal{E}_n\right)$ itself is contained in $\mathcal{R}_f$.
By using the notion of the ramification module $\mathcal{R}_f$,
Theorem \ref{theorem 1} is equivalent to assert that
\[
\langle |Jf|^2\rangle_{\mathcal{E}_n}+f^*\left(\mathcal{E}_n\right)
\;\subset\; \mathcal{R}_f,
\]
where $\langle |Jf|^2\rangle_{\mathcal{E}_n}$ is the $\mathcal{E}_n$-module

generated by
$|Jf|^2$.
\begin{definition}[\cite{ishikawaaustralia, ishikawa}]
\label{opening}
{\rm
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be an equidimensional
map-germ and
let $\psi_1, \ldots, \psi_\ell$ be elements of $\mathcal{R}_f$.
Then, the map-germ
\[
(f, \psi_1, \ldots, \psi_\ell) : (\mathbb{R}^n,0)\to
\mathbb{R}^n\times \mathbb{R}^\ell
\]
is called an \textit{opening} of $f$.
}
\end{definition}
\noindent
By definition, for any equidimensional map-germ,
its opening is always a frontal.

Theorem \ref{theorem 1} guarantees that
even for non-finite equidimensional map-germs,
it is constructed automatically infinitely many non-trivial openings.
This is one advantage of Theorem \ref{theorem 1}.
In Subsections \ref{fold singularity}-\ref{opening of $4_k$},
by direct elementary calculations,
it is easily shown that the normal forms of known frontals are actually
$\mathcal{A}$-equivalent to the frontals constructed by
Theorem \ref{theorem 1}.
In other words,
in Subsections \ref{fold singularity}-\ref{opening of $4_k$},
frontals constructed by using Jacobian-squared function-germs
are already very near (with respect to $\mathcal{A}$-equivalence)
to the normal forms of celebrated frontals, and
therefore
criteria for these noticeable frontals are not needed.
Since the normal forms of these frontals seem to be not easy to memorize,
this may be another advantage of Theorem \ref{theorem 1}.

\begin{proposition}[\cite{ishikawa}]\label{proposition frontal}
For any frontal germ
$F: (\mathbb{R}^n,0)\to (\mathbb{R}^{n+\ell}, 0)$,
there exist germs of diffeomorphism
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
$H: (\mathbb{R}^{n+\ell},0)\to (\mathbb{R}^{n+\ell},0)$,
an equidimensional map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
elements $\psi_1, \ldots, \psi_\ell$ of $\mathcal{R}_f$
such that the following equality holds:
\[
H\circ F\circ h = (f, \psi_1, \ldots, \psi_\ell).
\]
\end{proposition}
Based on Proposition \ref{proposition frontal}, it is natural to ask
the converse of Theorem \ref{theorem 1}.
However, it turns out that there exist counterexamples
against the converse of
Theorem \ref{theorem 1} (see Subsection \ref{$A_k$-front singularity}).
Nevertheless, Subsections
\ref{fold singularity}-\ref{opening of $4_k$}
suggest that if $f$ satisfies
$\dim_{\mathbb{R}}Q(f)\le 3$, then asking the converse of
Theorem \ref{theorem 1} is reasonable (for the definition of $Q(f)$, see
Section \ref{section 2}).     This question can be answered affirmatively
as follows.
\begin{theorem}\label{theorem 2}
Let $F: (\mathbb{R}^n,0)\to (\mathbb{R}^{n+\ell}, 0)$ be a frontal germ.
Suppose that there exist germs of diffeomorphism
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
$H: (\mathbb{R}^{n+\ell},0)\to (\mathbb{R}^{n+\ell},0)$,
an equidimensional map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ with
$\dim_{\mathbb{R}}Q(f)\le 3$
and
elements $\psi_1, \ldots, \psi_\ell$ of $\mathcal{R}_f$
such that the following equality holds:
\[
H\circ F\circ h = (f, \psi_1, \ldots, \psi_\ell).
\]
Then, the following holds:
\[
\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_n}+f^*\left(\mathcal{E}_n\right)
=\mathcal{R}_f.
\]
\end{theorem}
\begin{corollary}
\label{corollary 1}
Let $F: (\mathbb{R}^n,0)\to (\mathbb{R}^{n+\ell}, 0)$ be a frontal germ.
Suppose that there exist germs of diffeomorphism
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
$H: (\mathbb{R}^{n+\ell},0)\to (\mathbb{R}^{n+\ell},0)$,
an equidimensional map-germ
$f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ with
$\dim_{\mathbb{R}}Q(f)\le 3$
and
elements $\psi_1, \ldots, \psi_\ell$ of $\mathcal{R}_f$
such that the following equality holds:
\[
H\circ F\circ h = (f, \psi_1, \ldots, \psi_\ell).
\]
Then,  there exist a germ of diffeomorphism
$\widetilde{H}:
(\mathbb{R}^{n+\ell},0)\to (\mathbb{R}^{n+\ell},0)$
and function-germs $\mu_i: (\mathbb{R}^n, 0)\to \mathbb{R}$
$(1\le i\le \ell)$ such that
\[
\widetilde{H}\circ H\circ F\circ h =
 (f, \mu_1|Jf|^2, \ldots, \mu_\ell|Jf|^2).
\]

\end{corollary}

From Theorem \ref{theorem 2},
it is natural to ask the following:

\begin{question}\label{question 1}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be an equidimensional
map-germ.

Then, does there exist a finitely generated $\mathcal{E}_n$-module
$A$ such that the following holds ?
\[
A+f^*\left(\mathcal{E}_n\right)=\mathcal{R}_f.
\]
\end{question}
\noindent
Notice that by Ishikawa
(\cite{ishikawainventiones, ishikawaaustralia},
see also \cite{ishikawa}),
it is known if \lq\lq $f$ is finite and of corank one\rq\rq\;
or \lq\lq it is $\mathcal{A}$-equivalent to
a finite analytic map-germ\rq\rq, then
there exists a finitely generated
$f^*\left(\mathcal{E}_n\right)$-module $B$
satisfying the equality:
\[
B+f^*\left(\mathcal{E}_n\right)=\mathcal{R}_f.
\]
Notice also that in the case of Mather's $\mathcal{A}_e$ tangent space
for a map-germ $g: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$,
the corresponding $\mathcal{E}_n$-module is
nothing but $tg(\theta(n))$ (for the definition of Mather's
$\mathcal{A}_e$ tangent space, see Section \ref{section 2}).
Thus, Question \ref{question 1} asks whether or not
the ramification module $\mathcal{R}_f$ has a similar structure
as $T\mathcal{A}_e(g)$.

\par
\bigskip

This paper is organized as  follows.
In Section \ref{section 2},
preliminaries are given.
Theorem \ref{theorem 1} (resp., Theorem \ref{theorem 2})
is proved in Section \ref{section 3} (resp., Section \ref{section 4}).
Finally, in Section \ref{section 5}, examples concerning Theorem
\ref{theorem 1} and Theorem \ref{theorem 2} are given.

\section{Preliminaries}\label{section 2}

\subsection{Theory of singularities of mappings}\label{subsection 2.1}
\qquad

In this subsection, it is partially
reviewed several well-known notions/terminologies
in the theory of singularities of mappings which are mainly developed by
J.~Mather in \cite{mather3, mather4}.
\cite{wall} is an excellent survey article on notions/terminologies reviewed
in this subsection, which is recommended
to readers.
\par
Two map-germ $f, g: (\mathbb{R}^n,0)\to \mathbb{R}^p$
are said to be $\mathcal{A}$-equivalent if there exist germs of
diffeomorphism
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
$H: (\mathbb{R}^p,f(0))\to (\mathbb{R}^p,g(0))$ such that
$H\circ f\circ h=g$.
\par
Given a map-germ $f: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$,
the $\mathcal{A}$-equivalence class of $f$ is denoted by
$\mathcal{A}(f)$.

Let $\mathcal{E}_n$ be the $\mathbb{R}$-algebra consisting of
function-germs $(\mathbb{R}^n,0)\to \mathbb{R}$ and
let $m_n$ be the unique maximal ideal of $\mathcal{E}_n$.
Let $f: (\mathbb{R}^n, 0)\to (\mathbb{R}^{p}, 0)$ be a
map-germ.
The set consisting of all vector fields along $f$
is denoted by $\theta(f)$.
Notice that $\theta(f)$ is a finitely generated $\mathcal{E}_n$-module
and it is a $\mathcal{E}_p$-module via $f$.
For the identity map-germ
$id_n: (\mathbb{R}^n, 0)\to (\mathbb{R}^{p}, 0)$,
$\theta(id_n)$ is denoted by $\theta(n)$.
The mapping $tf: \theta(n)\to \theta(f)$
(resp., $\omega f: \theta(p)\to \theta(f)$) is defined by
$tf(\xi)=df\circ \xi$ (resp., $\omega f(\eta)=\eta\circ f$).
The set $\mathcal{A}(f)$ may be regarded
as an orbit of $f$ by the direct product
of the following two groups:
\begin{eqnarray*}
\{\mbox{a germ of diffeomorphism }
h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)\},  \\
\{\mbox{a germ of diffeomorphism }
H: (\mathbb{R}^p,0)\to (\mathbb{R}^p,0)\}.
\end{eqnarray*}
Thus, the tangent space of $\mathcal{A}(f)$ at $f$ is naturally defined
as follows:
\[
T\mathcal{A}(f)=tf(m_n\theta(n))+\omega f(m_p\theta(p)).
\]
It is meaningful if $\{\mbox{diffeomorphism }
h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)\}$
(resp., $\{\mbox{diffeomorphism } H: (\mathbb{R}^p,0)\to
(\mathbb{R}^p,0)\}$) is replaced with
$\{\mbox{diffeomorphism } h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,h(0))\}$
(resp., $\{\mbox{diffeomorphism }
H: (\mathbb{R}^p,0)\to (\mathbb{R}^p,H(0))\}$).
In this case, $tf(\theta(n))+\omega f(\theta(p))$
is denoted by $T\mathcal{A}_e(f)$.
\[
T\mathcal{A}_e(f)=tf(\theta(n))+\omega f(\theta(p)).
\]
\par
Two map-germ $f, g: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$
are said to be $\mathcal{R}$-equivalent if there exists a germ of
diffeomorphism
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ such that
$f\circ h=g$.
Given a map-germ $f: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$,
similarly as in the case of $\mathcal{A}$-equivalence class of $f$ ,
$T\mathcal{R}(f)$ and  $T\mathcal{R}_e(f)$ can be naturally defined
as follows:
\begin{eqnarray*}
T\mathcal{R}(f) & = & tf(m_n\theta(n))  \\
T\mathcal{R}_e(f) & = & tf(\theta(n)).
\end{eqnarray*}
Therefore, the condition (1) of the definition of frontal
may be regarded as the condition
that $\phi\in \theta(F)$ is perpendicular to $T\mathcal{R}_e(F)$
in $\theta(F)$.
\par
\smallskip
Given a map-germ $f: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$,
the $\mathbb{R}$-algebra $\mathcal{E}_n/f^*m_p\mathcal{E}_n$ is denoted
by $Q(f)$.     The $\mathbb{R}$-algebra $Q(f)$ is called
the \textit{local algebra} of the map-germ $f$.
Since the local algebra $Q(f)$ is an $\mathbb{R}$-algebra, it is a vector space.
A map-germ $f: (\mathbb{R}^n,0)\to (\mathbb{R}^p,0)$ is said to be
\textit{finite} if the vector space $Q(f)$ is of finite dimension.
The dimension of $Q(f)$ as $\mathbb{R}$ vector space  is called
the \textit{multiplicity} of $f$.

\subsection{Openings}\label{subsection 2.2}

\qquad
In this subsection, it is partially
reviewed several well-known notions/terminologies
in the theory of openings which are mainly developed by
G.~Ishikawa.
\cite{ishikawaaustralia, ishikawa}
are excellent survey articles on notions/terminologies reviewed
in this subsection, which are recommended
to readers.
There is one remark.   In \cite{ishikawa},
openings are defined for a map-germ
$f: (\mathbb{R}^n, 0)\to (\mathbb{R}^m,0)$ $(n\le m)$; and
in \cite{ishikawaaustralia} they are defined even for
any multigerm $f: (\mathbb{R}^n, S)\to (\mathbb{R}^m, 0)$ $(n\le m)$.
However, in this paper,
it is needed only openings of an equidimensional
mono-germ.    Thus, for the sake of clearness, in this subsection,
we concentrate on reviewing notions/terminologies
of opening only for an equidimensional mono-germ.
\par
\begin{definition}\label{versal opening}
{\rm
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be
an equidimensional map-germ.
An opening $F=(f, \psi_1, \ldots, \psi_\ell)$ of $f$ is called a
\textit{versal opening} (resp., \textit{mini-versal opening})
of $f$ if $1, \psi_1, \ldots, \psi_\ell$ form a system
(resp., minimal system) of generators of $\mathcal{R}_f$
as an $f^*\left(\mathcal{E}_n\right)$-module.
}
\end{definition}
\begin{lemma}[\cite{ishikawaaustralia}]
\label{lemma 2.1}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be
an equidimensional map-germ.
Then, the following hold.
\begin{enumerate}
\item[(1)]\quad
$f^*\left(\mathcal{E}_n\right)
\subset \mathcal{R}_f\subset \mathcal{E}_n$.
\item[(2)]\quad
$\mathcal{R}_f$ is a $f^*\left(\mathcal{E}_n\right)$-module.
\item[(3)]\quad
$\mathcal{R}_f$ is a $C^\infty$ subring of $\mathcal{E}_n$.
\item[(4)]\quad
For a germ of diffeomorphism on the target space
$H: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$,
$\mathcal{R}_{H\circ f}=\mathcal{R}_f$ holds.
For a germ of diffeomorphism on the source space
$h: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$,
$\mathcal{R}_{f\circ h}=h^*\left(\mathcal{R}_f\right)$ holds.
\end{enumerate}
\end{lemma}
\begin{proposition}[\cite{ishikawainventiones}]
\label{corank one case}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be
a finite map-germ of corank one.
Then, the following hold.
\begin{enumerate}
\item[(1)]\quad
$\mathcal{R}_f$ is a finitely generated $f^*\left(\mathcal{E}_n\right)$-module.
Therefore, there is a versal opening of $f$.
\item[(2)]\quad
$1, \psi_1, \ldots, \psi_\ell\in \mathcal{R}_f$ generate $\mathcal{R}_f$
as $f(\mathcal{E}_n)$-module if and only if\;
$[1], [\psi_1], \ldots, [\psi_\ell]$ generate the vector space
$\mathcal{R}_f/\left(f^*m_n\right)\mathcal{R}_f$,
where $[a]$ stands for $a+\left(f^*m_n\right)\mathcal{R}_f$.
\end{enumerate}
\end{proposition}
\begin{proposition}
\label{analytic case}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be
a map-germ $\mathcal{A}$-equivalent to a finite
analytic map-germ.
Then, the following hold.
\begin{enumerate}
\item[(1)]\quad
$\mathcal{R}_f$ is a finitely generated $f^*\left(\mathcal{E}_n\right)$-module.
Therefore, there is a versal opening of $f$.
\item[(2)]\quad
$1, \psi_1, \ldots, \psi_\ell\in \mathcal{R}_f$ generate $\mathcal{R}_f$
as $f(\mathcal{E}_n)$-module if and only if\;
$[1], [\psi_1], \ldots, [\psi_\ell]$ generate the vector space
$\mathcal{R}_f/\left(f^*m_n\right)\mathcal{R}_f$.

\end{enumerate}
\end{proposition}
\noindent
For the proof of Proposition \ref{analytic case}, see
\cite{ishikawakodai, ishikawaaustralia}.

\begin{proposition}[\cite{ishikawaaustralia}]
\label{injectivity}
Let $f: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ be a finite
equidimensional map-germ.
Then, every versal opening
$F: (\mathbb{R}^n,0)\to (\mathbb{R}^{n+\ell},0)$ is injective.
\end{proposition}
\noindent
Here, a map-germ is said to be \textit{injective} if it has
an injective representative.

\section{Proof of Theorem \ref{theorem 1}}\label{section 3}

For any $i$ $(1\le i\le \ell)$, let the fundamental vector
$(0, \ldots, 0, 1, 0, \ldots, 0)$ be denoted by ${\bf e}_i$.
The vector ${\bf e}_i$ may be identified with the constant
vector field $\frac{\partial}{\partial X_{n+i}}$
Let $\widetilde{Jf}$ be the cofactor matrix of
the Jacobian matrix $Jf$.

By using $\widetilde{Jf}$,
for any $i$ $(1\le i\le \ell)$,
define the map-germ $\Phi_i: (\mathbb{R}^n,0)\to T\mathbb{R}^{n+\ell}$ as

$\Phi_i=(F, \phi_i)$,
where
\[
\phi_i=\left(
{ }^T\!\left(\left (|Jf|d\mu_i+2\mu_id|Jf|\right)\widetilde{Jf}\right),
-{\bf e}_i
\right).
\]
Here, $d\varphi$ for a function-germ
$h: (\mathbb{R}^n,0)\to \mathbb{R}$ (resp., ${ }^T{\bf v}$ for a
vector field ${\bf v}$) stands for  the gradient vector field of $h$
(resp., the transposed vector field of
${\bf v}$).
Then, it follows that $\Phi_j$ is of class $C^\infty$.
Thus, $\Phi_j$ is a $C^\infty$ vector field along $F$.
It is clear that $\Phi_1, \ldots, \Phi_\ell$ satisfy the condition
(2) and (3) of the definition of frontal.
Therefore, it is sufficient to show that $\Phi_i$ satisfies the condition
(1) for any $i$ $(1\le i\le \ell)$.
Let $\xi$ be an element of $\theta(n)$.
Then,
\begin{eqnarray*}
{ } & { } & \phi_i(x)\cdot tF(\xi)(x) \\
{ } & = &  \left(|Jf|(x)d\mu_i(x)+2\mu_i(x)d|Jf|(x)\right)
\widetilde{Jf}(x)Jf(x)\xi(x) \\
{ } & { } & \qquad\qquad
-
|Jf|(x){ } \left(|Jf|(x)d\mu_i(x)+2\mu_i(x)d|Jf|(x)\right)\xi(x) \\
{ } & = &
|Jf|(x)\left(\left(|Jf|(x)d\mu_i(x)+2\mu_i(x)d|Jf|(x)\right)\xi(x)\right.\\
{ } & { } & \qquad\qquad
-\left.\left(|Jf|(x)d\mu_i(x)+2\mu_i(x)d|Jf|(x)\right)\xi(x)\right) \\
{ } & = & 0.
\end{eqnarray*}
Hence, $F$ is a frontal.
\hfill $\Box$

\section{Proof of Theorem \ref{theorem 2}}
\label{section 4}

\subsection{The case $n=1$}\label{subsection 4.1}
\quad

Set $\delta=\dim_{\mathbb{R}}Q(f)$.
In the case of $n=1$, there exists a germ of diffeomorphism
$h: (\mathbb{R}, 0)\to (\mathbb{R},0)$ such that
$g(x)=f\circ h(x)=\frac{1}{\delta}x^\delta$.
By \cite{ishikawatransaction},
the ramification module of $g$ can be described as follows.
\[
\mathcal{R}_g =
\left\{
\begin{array}{ll}
\left\langle 1 \right\rangle_{g^*\left(\mathcal{E}_1\right)}
=\left\langle 1 \right\rangle_{\mathcal{E}_1}\quad
& (\delta=1), \\
\left\langle 1, x^3 \right\rangle_{g^*\left(\mathcal{E}_1\right)}
\quad
& (\delta=2), \\
\left\langle 1, x^4, x^5 \right\rangle_{g^*\left(\mathcal{E}_1\right)}
\quad
& (\delta=3). \\
\end{array}
\right.
\]
On the other hand, since $|Jg|^2(x)=x^{2(\delta-1)}$,
the following holds.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_1} =
\left\{
\begin{array}{ll}
\left\langle 1 \right\rangle_{\mathcal{E}_1}\quad
& (\delta=1), \\
\left\langle x^2, x^3 \right\rangle_{\mathcal{E}_1}
\quad
& (\delta=2), \\
\left\langle x^4, x^5 \right\rangle_{\mathcal{E}_1}
\quad
& (\delta=3). \\
\end{array}
\right.
\]
Thus, we have the following inclusion.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_1}+
g^*\left(\mathcal{E}_1\right)
\supset \mathcal{R}_g.
\]
Combining this inclusion with Theorem \ref{theorem 1},
the following equality holds.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_1}+
g^*\left(\mathcal{E}_1\right)
= \mathcal{R}_g.
\]
Since $h$ is a germ of diffeomorphism, we have the following by the chain rule.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_1}
=
\left\langle |J(f\circ h)|^2 \right\rangle_{\mathcal{E}_1}
=
h^*\left(\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_1}\right).
\]

It follows
\[
h^*\left(\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_1}+
f^*\left(\mathcal{E}_1\right)\right)
= h^*\left(\mathcal{R}_f\right).
\]
Hence, by pulling both sides back by $\left(h^{-1}\right)^*$,
the desired equality can be obtained as follows.
\[
\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_1}+
f^*\left(\mathcal{E}_1\right)
= \mathcal{R}_f.
\]
\hfill
$\Box$

\subsection{The case $n\ge 2$}\label{subsection 4.2}
\quad

Again in this subsection, $\dim_{\mathbb{R}}Q(f)$ is denoted by $\delta$.
The assumption $\delta\le 3$ implies that $f$ is of corank one.
Since $\frac{1}{2}x^2+t x$ (resp., $\frac{1}{3}x^3+t x$) is
an $\mathcal{R}$-versal unfolding of
$\frac{1}{2}x^2$ (resp., $\frac{1}{3}x^3$), it is deduced that
there exist germs of diffeomorphism
$h, H: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
a function-germ $\alpha: (\mathbb{R}^{n-1}, 0)\to \mathbb{R}$
such that
\[
H\circ f\circ h (x, \lambda_1, \ldots, \lambda_{n-1})
=
\left\{
\begin{array}{ll}
\left(x, \lambda_1, \ldots, \lambda_{n-1}\right)\quad
& (\delta=1), \\
\left(\frac{1}{2}x^2+\alpha(\lambda_1, \ldots, \lambda_{n-1})x,
\lambda_1, \ldots, \lambda_{n-1}\right)\quad
& (\delta=2), \\
\left(\frac{1}{3}x^3+\alpha(\lambda_1, \ldots, \lambda_{n-1})x,
\lambda_1, \ldots, \lambda_{n-1}\right)\quad
& (\delta=3),
\end{array}
\right.
\]
where $(x, \lambda_1, \ldots, \lambda_{n-1})$ is an element of
$\mathbb{R}\times \mathbb{R}^{n-1}=\mathbb{R}^n$.
Set $g=H\circ f\circ h$ and
$\lambda=\left(\lambda_1, \ldots, \lambda_{n-1}\right)$.
By \cite{ishikawatransaction},
the ramification module of $g$ can be described as follows.
\[
\mathcal{R}_g =
\left\{
\begin{array}{ll}
\left\langle 1 \right\rangle_{g^*\left(\mathcal{E}_n\right)}
=\left\langle 1 \right\rangle_{\mathcal{E}_n}\quad
& (\delta=1), \\
\left\langle 1, \frac{1}{3}x^3
+\frac{1}{2}\alpha(\lambda)x^2
\right\rangle_{g^*\left(\mathcal{E}_n\right)}
\quad
& (\delta=2), \\
\left\langle 1, \frac{1}{4}x^4+\frac{1}{2}\alpha(\lambda)x^2,
\frac{1}{5}x^5+\frac{1}{3}\alpha(\lambda)x^3 \right\rangle_{g^*\left(\mathcal{E}_n\right)}
\quad
& (\delta=3). \\
\end{array}
\right.
\]
On the other hand, we have
\[
\left\langle |Jg|^2\right\rangle_{\mathcal{E}_n} =
\left\{
\begin{array}{ll}
\left\langle 1 \right\rangle_{\mathcal{E}_n}\quad
& (\delta=1), \\
\left\langle (x+\alpha(\lambda))^2 \right\rangle_{\mathcal{E}_n}
\quad
& (\delta=2), \\
\left\langle (x^2+\alpha(\lambda))^2 \right\rangle_{\mathcal{E}_n}
\quad
& (\delta=3). \\
\end{array}
\right.
\]
Since
\begin{eqnarray*}
\left(x+\alpha(\lambda)\right)^2
& = &
2\left(\frac{1}{2}x^2+\alpha(\lambda)x\right)+
\left(\alpha(\lambda)\right)^2, \\
x\left(x+\alpha(\lambda)\right)^2
& = &
3\left(\frac{1}{3}x^3+\frac{1}{2}\alpha(\lambda)x^2\right)+
\alpha(\lambda)\left(\frac{1}{2}x^2+\alpha(\lambda)x\right), \\
\left(x^2+\alpha(\lambda)\right)^2
& = &
4\left(\frac{1}{4}x^4+\frac{1}{2}\alpha(\lambda)x^2\right)+
\left(\alpha(\lambda)\right)^2\quad \mbox{and} \\
x\left(x^2+\alpha(\lambda)\right)^2
& = &
5\left(\frac{1}{5}x^5+\frac{1}{3}\alpha(\lambda)x^3\right)+
\alpha(\lambda)\left(\frac{1}{3}x^3+\alpha(\lambda)x\right),
\end{eqnarray*}
the following holds.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_n}+
g^*\left(\mathcal{E}_n\right)
\supset \mathcal{R}_g.
\]
Combining this inclusion with Theorem \ref{theorem 1},
we have the following equality.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_n}+
g^*\left(\mathcal{E}_n\right)
= \mathcal{R}_g.
\]
Since $h, H$ are germs of diffeomorphism, it follows
the following.
\[
\left\langle |Jg|^2 \right\rangle_{\mathcal{E}_n}
=
\left\langle |J(H\circ f\circ h)|^2 \right\rangle_{\mathcal{E}_n}
=
h^*\left(\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_n}\right).
\]
Combining this property with two facts
$g^*\left(\mathcal{E}_n\right)=
h^*\left((H\circ f)^*\left(\mathcal{E}_n\right)\right)
=h^*\left(f^*\left(\mathcal{E}_n\right)\right)
$
and
$\mathcal{R}_g=\mathcal{R}_{H\circ f\circ h}=
h^*\left(\mathcal{R}_{H\circ f}\right)
=h^*\left(\mathcal{R}_f\right)
$,
we have
\[
h^*\left(\langle |Jf|^2 \rangle_{\mathcal{E}_n}+
f^*\left(\mathcal{E}_n\right)\right)
= h^*\left(\mathcal{R}_f\right).
\]
Therefore, also in this case,
we have the following desired equality.
\[
\left\langle |Jf|^2 \right\rangle_{\mathcal{E}_n}+
f^*\left(\mathcal{E}_n\right)
= \mathcal{R}_f.
\]
\hfill
$\Box$

\section{Examples of Theorems \ref{theorem 1} and
\ref{theorem 2}}\label{section 5}

All map-germs $h_i: (\mathbb{R}^n,0)\to (\mathbb{R}^n,0)$ and
$H_j: (\mathbb{R}^{n+\ell},0)\to (\mathbb{R}^{n+\ell},0)$
appearing in this section are
germs of diffeomorphism.

\subsection{Fold Singularity} \label{fold singularity}

Let $f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$ be
the map-germ defined by
$f(x,y)=\left(\frac{1}{2}x^2+xy,\; y\right)$.
The map-germ $f$ is $\mathcal{A}$-equivalent to the map-germ
named \textit{Type 2} in the list of \cite{rieger}.
It is clear that $|Jf|(x,y)=x+y$ and $\dim_{\mathbb{R}}Q(f)=2$.
Set $\mu_1(x,y)=1$.

Then, by Theorem \ref{theorem 1},
the map-germ $F: (\mathbb{R}^2,0)\to (\mathbb{R}^3,0)$
defined by
\begin{eqnarray*}
F(x,y)  & =  & \left(f(x,y), \mu_1(x,y)|Jf|^2(x,y)\right) \\
{ } & = &  \left(\frac{1}{2}x^2+xy,\; y,\; (x+y)^2\right) =
\left(\frac{1}{2}\left(x+y\right)^2-\frac{1}{2}y^2,\; y,\; (x+y)^2\right)
\end{eqnarray*}
is a frontal.
Set $H_1(X,Y,Z)=(X-\frac{1}{2}Y^2,\; Y,\; Z)$.
Then,
\[
H_1\circ F(x,y)  =
\left(\frac{1}{2}(x+y)^2, \; y, \;
(x+y)^2\right).
\]
Secondly, set $h_1(x,y)=\left(x-y, y\right)$ and
$H_2(X,Y,Z)=(2X,\; Y,\; Z-2X)$.   Then,
\[
H_2\circ H_1\circ F\circ h_1(x,y)
=
\left(x^2, \; y, \; 0\right).
\]
In Differential Geometry,
the map-germ $H_2\circ H_1\circ F\circ h_1$ is called
the normal form of {\it fold singularity} (\cite{fujimorietal}).

\subsection{Cuspidal Edge}\label{cuspidal edge}

Not only in this subsection, but also
until Subsection \ref{open folded umbrella},
we start from the same map-germ
$f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$ as in
Subsection \ref{fold singularity}.
Namely, in this subsection, $f$ is the map-germ defined by
$
f(x,y)=\left(\frac{1}{2}x^2+xy,\; y\right)
$.

Set $\mu_1(x,y)=x$.

Then, by Theorem \ref{theorem 1},
the map-germ $F: (\mathbb{R}^2,0)\to (\mathbb{R}^3,0)$
defined by
\begin{eqnarray*}
F(x,y) & = & \left(f(x,y), \mu_1(x,y)|Jf|^2(x,y)\right)\\
{ } & = & \left(\frac{1}{2}x^2+xy,\; y,\; x(x+y)^2\right) \\
{ } & = &  \left(\frac{1}{2}\left(x+y\right)^2-\frac{1}{2}y^2,
\; y,\; (x+y)^3-y(x+y)^2\right)
\end{eqnarray*}
is a frontal.
Set $H_1(X,Y,Z)=(X-\frac{1}{2}Y^2,\; Y,\; Z)$.
Then,
\begin{eqnarray*}
H_1\circ F(x,y)
& = &  \left(\frac{1}{2}(x+y)^2, \; y, \;
(x+y)^3-y(x+y)^2\right).
\end{eqnarray*}
Secondly, set $h_1(x,y)=\left(x-y, y\right)$ and
$H_2(X,Y,Z)=(2X,\; Y,\; Z-2XY)$.   Then,
\[
H_2\circ H_1\circ F\circ h_1(x,y)
=
\left(x^2, \; y, \; x^3\right),
\]
well-known as the normal form of
\textit{cuspidal edge} (for instance, see
\cite{krsuy}).

\subsection{Folded Umbrella(Cuspidal Crosscap)} \label{folded umbrella}

As explained in the last subsection, in this subsection again,
$f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$ is
the map-germ defined by $f(x,y)=\left(\frac{1}{2}x^2+xy,\; y\right)$.

Set $\mu_1(x,y)=x^2$.

Then, by Theorem \ref{theorem 1},
the map-germ $F: (\mathbb{R}^2,0)\to (\mathbb{R}^3,0)$
defined by
\begin{eqnarray*}
F(x,y) & = & \left(f(x,y), \mu_1(x,y)|Jf|^2(x,y)\right)\\
{ } & = & \left(\frac{1}{2}x^2+xy,\; y,\; x^2(x+y)^2\right) \\
{ } & = &  \left(\frac{1}{2}x^2+xy,\; y,\; x^4+2x^3y+x^2y^2\right)
\end{eqnarray*}
is a frontal.
Set $H_1(X,Y,Z)=(X,\; Y,\; Z-X^2)$.
Then,
\begin{eqnarray*}
H_1\circ F(x,y)
& = &  \left(\frac{1}{2}x^2+xy, \; y, \;
\frac{3}{4}x^4+x^3y\right).
\end{eqnarray*}
Secondly, set $h_1(x,y)=\left(x, \frac{1}{2}y\right)$.   Then,
\[
H_1\circ F\circ h_1(x,y)
=
\left(\frac{1}{2}x^2+\frac{1}{2}xy, \; \frac{1}{2}y, \;
\frac{3}{4}x^4+\frac{1}{2}x^3y\right).
\]
Thirdly, set
$H_2(X, Y, Z)=\left(2X, 2Y, \frac{4}{3}Z\right)$.   Then,
we have
\[
H_2\circ H_1\circ F\circ h_1(x,y)
=
\left(x^2+xy, \; y, \;
x^4+\frac{2}{3}x^3y\right),
\]
called the normal form of
{\it folded umbrella} (for instance, see \cite{ishikawa}).
A folded umbrella is also called a
{\it cuspidal crosscap}.
As the normal form of cuspidal crosscap,
Some references adopt
$(x,y)\mapsto \left(x^2,\; y, \; x^3y\right)$
(for instance, see \cite{fujimorisajiumeharayamada}).
Set $h_2(x,y)=\left(\frac{1}{2}(x-y), y\right)$

and

\begin{eqnarray*}
{ } & { } &
H_3(X,Y,Z)\\
{ } & = & \left(4\left(X+\frac{1}{4}Y^2\right), \; Y, \;
-6\left(Z-\left(X+\frac{1}{4}Y^2\right)^2
-\frac{1}{2}\left(X+\frac{1}{4}Y^2\right)Y^2
- \frac{1}{16}Y^4\right)\right).
\end{eqnarray*}
Then,
it is easily confirmed
\[
H_3\circ H_2\circ H_1\circ F\circ h_1\circ h_2(x,y)
 =
\left(x^2,\; y,\; x^3y\right).
\]

\subsection{Open Folded Umbrella}\label{open folded umbrella}

As explained, we start from
the same $f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$
as in Subsection \ref{fold singularity}.
Thus, $f$ is defined by $f(x,y)=\left(\frac{1}{2}x^2+xy,\; y\right)$
in this subsection.

Suppose that $\ell \ge 3$ and
set $\mu_1(x,y)=x^2, \mu_2(x,y)=x^2|Jf|(x,y)$ and
$\mu_i(x,y)=0$ $(3\le i\le \ell)$.

Then,
\begin{eqnarray*}
F(x,y) & = & \left(f(x,y), \mu_1|Jf|^2(x,y), \ldots,
\mu_\ell(x,y)|Jf|^2(x,y)\right) \\
{ }  & = &
\left(\frac{1}{2}x^2+xy,\; y,\; x^2(x+y)^2, \; x^2(x+y)^3, 0, \ldots, 0\right) \\
{ } & = &  \left(\frac{1}{2}x^2+xy,\; y,\; x^4+2x^3y+x^2y^2, \right. \\
{ } & { } & \qquad \left. x^5+3x^4y+3x^3y^2+x^2y^3,
0, \ldots, 0\right).
\end{eqnarray*}
Set
\[
H_1(X,Y,U_1, U_2, U_3, \ldots, U_\ell)
=(X,\; Y,\; U_1-X^2,\; U_2-YU_1,\; U_3, \ldots, U_\ell).
\]
Then,
\begin{eqnarray*}
H_1\circ F(x,y)
& = &  \left(\frac{1}{2}x^2+xy, \; y, \;
\frac{3}{4}x^4+x^3y, \; x^5+2x^4y+x^3y^2, 0, \ldots, 0\right).
\end{eqnarray*}
Set $h_1(x,y)=\left(x, \frac{1}{2}y\right)$.   Then,
\begin{eqnarray*}
H_1\circ F\circ h_1(x,y)
& = &
\left(\frac{1}{2}x^2+\frac{1}{2}xy, \; \frac{1}{2}y, \;
\frac{3}{4}x^4+\frac{1}{2}x^3y, \;
x^5+x^4y+\frac{1}{4}x^3y^2, 0, \ldots, 0\right).
\end{eqnarray*}
Nextly, set
\[
H_2(X, Y, U_1, U_2, U_3, \ldots, U_\ell)=
\left(X,\; Y,\; U_1,\; U_2-YU_1,\; U_3, \ldots, U_\ell\right).
\]   Then,
we have
\begin{eqnarray*}
{ } & { } & H_2\circ H_1\circ F\circ h_1(x,y) \\
{ } & = &
\left(\frac{1}{2}x^2+\frac{1}{2}xy, \; \frac{1}{2}y, \;
\frac{3}{4}x^4+\frac{1}{2}x^3y, \; x^5+\frac{5}{8}x^4y,
0, \ldots, 0\right).
\end{eqnarray*}
Finally, set
\[
H_3(X, Y, U_1, U_2, \ldots, U_\ell)=
\left(2X,\; 2Y,\; \frac{4}{3}U_1,\; U_2, \ldots, U_\ell\right).
\]
Then,
\begin{eqnarray*}
{ } & { } & H_3\circ H_2\circ H_1\circ F\circ h_1(x,y) \\
{ } & = &
\left(x^2+xy, \; y, \;
x^4+\frac{2}{3}x^3y,\; x^5+\frac{5}{8}x^4y,\;
0, \ldots, 0
\right),
\end{eqnarray*}
called the normal form of {\it open folded umbrella} (\cite{arnold2, ishikawa}).

\subsection{Swallowtail}\label{swallowtail}

Let $f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$ be
the map-germ defined by $f(x,y)=\left(\frac{1}{3}x^3+xy,\; y\right)$.
The map-germ $f$ is $\mathcal{A}$-equivalent to the map-germ
named \textit{Type 3} in the list of \cite{rieger}.
It is clear that $|Jf|(x,y)=x^2+y$ and $\dim_{\mathbb{R}}Q(f)=3$.
Set $\mu_1(x,y)=1$.
We consider the map-germ $F: (\mathbb{R}^2,0)\to (\mathbb{R}^3,0)$
defined by
\begin{eqnarray*}
F(x,y) & = & \left(f(x,y), \mu_1(x,y)|Jf|^2(x,y)\right)\\
{ } & = & \left(\frac{1}{3}x^3+xy,\; y,\; (x^2+y)^2\right) \\
{ } & = &  \left(\frac{1}{3}x^3+xy,\; y,\; x^4+2x^2y+y^2\right).
\end{eqnarray*}
By Theorem \ref{theorem 1}, $F$ is a frontal.
Set $H_1(X,\; Y,\; Z)=(X,\; Y,\; Z-Y^2)$,
$H_2(X,\;Y,\; Z)=(-12X,\; 6Y,\; 3Z)$.
Then,
\[
H_2\circ H_1\circ {F}(x, y)=
\left(-4x^3-12xy,6y, 3x^4+6x^2y \right).
\]
Next, set $h_1(x,\; y)=\left(x,\; \frac{1}{6}y\right)$.     Then,
\[
H_2\circ H_1\circ \widetilde{F}\circ h_1(x,\; y)=
\left(-4x^3-2xy,\; y, \; 3x^4+x^2y\right),
\]
well-known as the normal form of
{\it swallowtail} (for instance see \cite{brucegiblin}, 129 page).

\subsection{Open Swallowtail}\label{open swallowtail}

As in Subsection \ref{swallowtail},
let $f: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$
be the map-germ defined by $f(x,y)=\left(\frac{1}{3}x^3+xy,\; y\right)$.
Suppose that $\ell\ge 3$ and
set $\mu_1(x,y)=1, \mu_2(x,y)=x, \mu_i(x,y)=0$ $(3\le i\le \ell)$.
Let $F: (\mathbb{R}^2, 0)\to (\mathbb{R}^{2+\ell},0)$ be the map-germ
defined by
\begin{eqnarray*}
F(x,y) & = & \left(f(x,y), \mu_1|Jf|^2(x,y), \ldots, \mu_\ell|Jf|^2(x,y)\right) \\
{ } & = & \left(\frac{1}{3}x^3+xy,\; y,\; (x^2+y)^2,\; x(x^2+y)^2,\;
0, \ldots, 0\right) \\
{ } & = &  \left(\frac{1}{3}x^3+xy,\; y,\; x^4+2x^2y+y^2,
x^5+2x^3y+xy^2, 0, \ldots, 0\right).
\end{eqnarray*}
Set
\[
H_1(X,Y,U_1, U_2, U_3, \ldots, U_\ell)
=(X,\; Y,\; U_1-Y^2,\; U_2-XY,\; U_3, \ldots, U_\ell).
\]
Then,
\[
H_1\circ F(x,y)
 =   \left(\frac{1}{3}x^3+xy, \; y, \;
x^4+2x^2y,\;x^5+\frac{5}{3}x^3y,\; 0, \ldots, 0\right).
\]
Set $h_1(x,y)=\left(x,\; \frac{1}{3}y\right)$.   Then,
\[
H_1\circ F\circ h_1(x,y)
 =   \left(\frac{1}{3}x^3+\frac{1}{3}xy, \; \frac{1}{3}y, \;
x^4+\frac{2}{3}x^2y\; x^5+\frac{5}{9}x^3y,\; 0, \ldots, 0\right).
\]
Set $H_2(X,Y,U_1, \ldots, U_\ell)
=(3X\; ,3Y,\; U_1, \ldots, U_\ell)
$.   Then,
\[
H_2\circ H_1\circ G\circ h_1(x,y)
 =   \left(x^3+xy, \; y, \;
x^4+\frac{2}{3}x^2y, x^5+\frac{5}{9}x^3y,\; 0,
\ldots, 0\right),
\]
known as the normal form of
{\it open swallowtail} (\cite{ishikawa}).

\subsection{Opening of $4_k$ in the list of \cite{rieger}}
\label{opening of $4_k$}

Let $k$ be an integer greater than $1$ and let
$f_{k, \pm}: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$ be
the map-germ defined by
$f_{k, \pm}(x,y)=\left(\frac{1}{3}x^3\pm xy^k,\; y\right)$.
The map-germ $f_{k, \pm}$ is $\mathcal{A}$-equivalent to the map-germ
named \textit{Type $4_k$} in the list of \cite{rieger}.
It is clear that $|Jf|(x,y)=x^2\pm y^k$ and $\dim_{\mathbb{R}}Q(f)=3$.
Set $\mu_1(x,y)=1$.
We consider the map-germ
$F_{k, \pm}: (\mathbb{R}^2,0)\to (\mathbb{R}^3,0)$
defined by
\begin{eqnarray*}
F_{k, \pm}(x,y) & = & \left(f_{k, \pm}(x,y), \mu_1(x,y)|Jf|^2(x,y)\right)\\
{ } & = & \left(\frac{1}{3}x^3\pm xy^k,\; y,\; (x^2\pm y^k)^2\right) \\
{ } & = &  \left(\frac{1}{3}x^3\pm xy^k,\; y,\; x^4\pm 2x^2y^k+y^{2k}\right).
\end{eqnarray*}
Theorem \ref{theorem 1} guarantees that
$F_{k, \pm}$ is a frontal.
Set $H_1(X,\; Y,\; Z)=(X,\; Y,\; Z-Y^{2k})$ and
$H_2(X,\;Y,\; Z)=(6X,\; 6Y,\; 3Z)$.
Then,
\[
H_2\circ H_1\circ {F_{k, \pm}}(x, y)=
\left(2x^3\pm 6xy^k,\; 6y,\; 3x^4\pm 6x^2y^k\right).
\]
Next, set $h_1(x,\; y)=\left(x,\; \frac{1}{\sqrt[k]{6}}y\right)$
and $H_3(X,\; Y,\; Z)=(X,\; \frac{\sqrt[k]{6}}{6}Y,\; Z)$.
Then,
\[
H_3\circ H_2\circ H_1\circ {F_{k,\pm}}\circ h_1(x,\; y)=
\left(2x^3\pm xy^k,\; y,\; 3x^4\pm x^2y^k\right).
\]
The form of
$H_3\circ H_2\circ H_1\circ {F_{2, -}}\circ h_1$ is
exactly the same as the map-germ called a
\textit{double swallowtail} given in \cite{sajiumeharayamada}.

\subsection{A non-analytic example}
\label{non-analytic example}

Let $\psi: (\mathbb{R},0)\to (\mathbb{R},0)$ be a flat function-germ
(i.e. $j^\infty\psi(0)=0$) and let
$f_{\infty}: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$
be the map-germ defined by
$f_{\infty}(x,y)=\left(\frac{1}{3}x^3\pm x\psi(y),\; y\right)$.
It is clear that $\dim_{\mathbb{R}}Q(f_\infty)=3$.
Thus, by Theorem \ref{theorem 2},
the following equality holds even for the $f_\infty$.
\[
\left\langle |Jf_\infty|^2 \right\rangle_{\mathcal{E}_2}+
f^*_{\infty}\left(\mathcal{E}_2\right)=\mathcal{R}_{f_\infty}.
\]
Therefore, for any opening
$F: (\mathbb{R}^2, 0)\to (\mathbb{R}^{2+\ell},0)$ of $f_\infty$
there exist germs of diffeomorphisms
$h: (\mathbb{R}^2, 0)\to (\mathbb{R}^2,0)$,
$H: (\mathbb{R}^{2+\ell}, 0)\to (\mathbb{R}^{2+\ell},0)$
and function-germs
$\mu_1, \ldots, \mu_\ell: (\mathbb{R}^2,0)\to \mathbb{R}$
such that the following holds.
\[
H\circ F\circ h=
(f_\infty,\; \mu_1|Jf_\infty|^2,\; \ldots,\; \mu_\ell|Jf_\infty|^2).
\]

\subsection{$A_k$-Front Singularity}\label{$A_k$-front singularity}

Let $k$ be an integer greater than $1$ and let
$f_k: (\mathbb{R}^k,0)\to (\mathbb{R}^k,0)$ be the map-germ
defined as follows:
\[
f_k(x_1, \ldots, x_k)
 =
\left(\frac{1}{k+1}x_1^{k+1}+\frac{1}{k-1}x_1^{k-1}x_2
+\cdots + x_1x_k, x_2, \ldots, x_k\right).
\]
The map-germ $f_k$ is a generalization of $f$
given in Subsection \ref{swallowtail}
because $f_2$ is exactly the same as the $f$.
The map-germ $f_k$ is well-known as the normal form of
corank-one isolated stable singularity
(for instance, see \cite{arnoldguseinzadevarchenko, mather4}).
For the $f_k$,
$\dim_{\mathbb{R}}Q(f_k)=k+1$ and
$|Jf_k|^2(x_1,0,\ldots, 0)=x_1^k$.
\par
From now on, suppose that $k$ is greater than $2$.

Let $\mu: (\mathbb{R}^n, 0)\to \mathbb{R}$ be a function-germ.
Since $2k-(k+1)=k-1>1$, for the normal vector field
$\nu$ of the frontal
$(f_k, \mu|Jf_k|^2)$, $\nu(x_1, 0, \ldots, 0)$ must be singular at the origin.
\par
On the other hand, since $f_k$ is a polynomial map-germ of corank one,
$\mathcal{R}_{f_k}$ can be described explicitely
by \cite{ishikawatransaction}.
In particular, there exists an opening $\widetilde{F}_k$ of $f_k$ whose
normal vector field $\nu$ is non-singular with respect to
$x_1$ (A concrete construction of $\nu$ can be found
in \cite{sajiumeharayamadacambridge}).
Therefore, if $k\ge 3$, it is impossible to expect
the converse of Theorem \ref{theorem 1}
for the frontal $\widetilde{F}_k$.

\section*{Acknowledgement}

The author would like to thank Goo Ishikawa
for his helpful suggestions.
This work was

supported
by JSPS KAKENHI Grant Number 17K05245.


TITLE: Are the original Schwarzschild solution and Hilbert's solution exactly the same?
QUESTION [3 upvotes]: Schwarzschild original metric solution can be found here in equation (14): http://old.phys.huji.ac.il/~barak_kol/Courses/Black-holes/reading-papers/SchwarzschildTranslated.pdf
Hilbert's metric solution is the one we are all familiar with, and the one that wikipedia shows under the name "Schwarzschild solution": https://en.wikipedia.org/wiki/Schwarzschild_metric
It is clear that in the metrics (Im not talking about their derivation), they only differ in what they refer as R and r respectively. In Schwarzschild's original metric, $R$ is just an "auxiliary quantity" which follows $R^3=r^3 + α^3$, with $r$ been the distance marker and $\alpha$ being the well-known $α=2GM$. One can easily see that in the Hilbert's metric, he substitues $R$ for $r$. For both, $r$ takes values from zero to infinity. But $R$ and $r$ do not follow a linear relationship!
This has been noted in https://arxiv.org/abs/physics/0310104, and it not only affects the metric but there are also differences between both derivations of the metrics, I quote in page 5: "What is not legitimate, although first done by Hilbert and subsequently handed down to the posterity, is to assume without justification that the range of the “new” $r$ is still $0 < r < ∞$, as it was for the “old” $r$, because this is tantamount to setting sqrt(G(0)) = 0, an arbitrary choice [5], equivalent to setting $ρ = 0$ in Schwarzschild’s result, reported in equation (5)."
Even Schwarzschild himself stated that "Actually Mr. Einstein’s approximation for the orbit goes into the exact solution when one substitutes for $r$ the quantity $R$", suggesting that both $r$-metric and $R$-metric do not result in the exact same orbits for the Mercury problem which Einstein was addresing. You may want to see that 1915 Einstein's aproximation uses $g_{tt}= 1-(α/r)$.
Moreover, I have found other papers claiming that both solutions are different and they do not resemble the same geodesics: https://www.researchgate.net/publication/331936281_Schwarzschild's_family
Thanks for the edit.

REPLY [3 votes]: The solutions are the same: they describe the same spacetime, but using slightly different coordinates. But the early workers, especially Schwarzschild, were not sure what to make of the region inside the horizon, or whether there was such a region. Droste's work showed more insight on this point, I think, and as I understand it his contribution was somewhat more pioneering than Hilbert's (it was a little earlier and quite thorough).


TITLE: Measuring the entropy of a graph representing a transition probability matrix of a first order markov chain
QUESTION [1 upvotes]: There's a research project i'm currently working on which requires me to analyze various aspects of "worlds" represented by transition probability matrices, where the nodes represent objects in the world that our "learner" travels through.
I'm looking for a way to measure the degree to which the structure of the connections in a given a graph provide the learner with any predictive power, for example, in a graph where all the vertices have the same value, the learner has absolutely no ability to predict the next node/nodes given the current one.
The graphs are all weighted and directed, with the sum of outgoing connections from each node normalized to one.
I'm having a bit of trouble coming up with a good way to measure this and would really appreciate some advice,
Thanks,
Ron

REPLY [0 votes]: Here is an example (not very nice though):
$P=\begin{pmatrix}1/6&1/2&1/3\\1/2&0&1/2\\0&1/3&2/3\end{pmatrix}$
Eigenvector for eigenvalue 1: $p=(1, 5/3, 7/2)$
Normalization: $p_\text{normalized}=(36/577, 60/577, 126/577)$
So the entropy (after simplifying a little bit by putting together terms) is:
$h(p,P)=-\frac{6}{577}\cdot\log(1/6)-\frac{54}{577}\cdot\log(1/3)-\frac{78}{577}\cdot\log(1/2)-\frac{84}{577}\cdot\log(2/3)$
This gives an approximate real value of:
$h(p,P)\approx 0.274\ldots$
One final thing: If your matrix $P$ has zero entries, the product $P_{i,j}\log P_{i,j}$ will be zero for the corresponding entry, even so the logarithm would be negative infinity (multiplying by zero wins here).


TITLE: Discrete Uniform random variable calculate mean and var
QUESTION [1 upvotes]: A lottery player decides to use a random variable generator to help him decide how many tickets to buy. He generates a discrete uniform random variable N taking values 1 through 4 with equal probabilities, and then purchases N lottery tickets. If each of the tickets he buys has (independently of the other tickets) the probability p of winning.
How do you calculate the mean and variance of the number of winning tickets that the player buys?
Thanks!

REPLY [1 votes]: Hint: if $Z$ is the number of winning tickets, find $E[Z]$ and $E[Z^2]$ by conditioning on the number of tickets he buys.


TITLE: Force as a Function of Position
QUESTION [0 upvotes]: If given a velocity as a function of position, is force as a function of position just it's derivative times the mass? I'm given the following and I am not sure my above logic is correct:
The speed of a particle with mass $m$ varies with distance $x$ as
$$v\left(x\right)=ax^{-n}$$
You should assume $v\left(x=0\right)=0$ at time $t=0$.
Find Force as a Function of Position
So would $F(x)=m\frac{dv\left(x\right)}{dx}$?

REPLY [0 votes]: please correct your logic
Force  is defined as "rate of change of momentum" and its direction is the direction of  $d\mathbf{p}/dt$ which is a vector as $\mathbf{p}$ is a vector. Momentum ($\mathbf{p}$) is mass times the velocity and your velocity is a function of $x$.
If momentum is a function of position, as in your case,  its rate of change will involve the rate of change of position with time i.e. one can write it as:
$$
F = m \frac{dv}{dt} = m \frac{dv}{dx} \frac{dx}{dt}= m v \frac{dv}{dx}
$$
In your case as  as $v= a. x^{-n}$  ;
$$
dv/dx= a. d/dx(x^{-n})
      =a(-n).x^{-n+1}
      =-a.n.x^{(1-n)}
$$
therefore:
$$
F= m.a.{-n} v. x^{(1-n)} =-m.a.n.a.x^-n.x^{1-n} =- m.a^2. n.(x^ {(1-2n)})
$$
so you get a force which is a function of $x$.
so your conclusion that force will be mass times $dx/dt$ is not correct.


TITLE: $L$-Zariski closure of subgroup $SL_n(F)$ as subset of $M_n(F)$ also a subgroup of $SL_n(F)$
QUESTION [4 upvotes]: Let $F$ be a field, and $SL_n(F)$ be the group of $n \times n$ matrices with determinant $1$. Let $\Gamma \subset SL_n(F)$ be a subgroup. We can consider $\Gamma$ to be a subset of $M_n(F) \cong F^{n^2}$ to define its $F$-Zariski closure. How do I prove that it is also a subgroup of $SL_n(F)$?

REPLY [3 votes]: Note that $SL_n(F)$ is a topological group in the Zariski topology. I'm going to call the subgroup $G$ because I'm typing on a cell phone, and let the closure be $H$. If $g,h\in H$, let $U$ be a neighborhood of $gh$. Since multiplication is continuous, there exist neighborhoods $A$ of $g$ and $B$ of $h$ such that $AB\subseteq U$. Since both $A$ and $B$ contain elements of $G$, so must $U$. Therefore $gh\in H$.  Note also that since inversion is a homeomorphism, if $U$ is a neighborhood of $g$ then $U^{-1}$ is a neighborhood of $g^{-1}$, and since the former neighborhood contains an element of $G$, so does the latter. This $g^{-1}\in H$, so $H$ is a subgroup of $SL_n(F)$.


TITLE: Find the automorphisms for the Galois Group of the minimial polynomial $x^4+1$.
QUESTION [0 upvotes]: Determine the splitting field $L$ for this polynomial over $\mathbb{Q}$.
The splitting field of $x^4+1$ must contain the solutions to $$x^4+1=0,$$ that is, $x^4=-1$. So $x^2=\pm i$, and $x=\pm\sqrt{\pm i}$. So, we have  $$L=\mathbb{Q}\left(\exp\left(\frac{\pi i}{4}\right), \exp\left(\frac{-\pi i}{4}\right) \right).$$ But $\exp\left(\frac{-\pi i}{4}\right)=\exp\left(\frac{7\pi i}{4}\right)$, and so $$L=\mathbb{Q}\left(\exp\left(\frac{\pi i}{4}\right)\right),$$ b. Since we found a spltting field $L$ over $\mathbb{Q}$, we can state that the following extension $L:\mathbb{Q}$ is normal, and it has degree $4$
Determine $\Gamma(L:\mathbb{Q})$.
Let $K=\mathbb{Q}(\zeta_0)$, where $\zeta_0=\exp\left(\frac{\pi i}{4}\right) \in \mathbb{C}$. Now $\zeta_0^4=1$ and $\mathbb{Q}(\zeta_0)$ consists of all elements $$p(x)=c_0+c_1\zeta_0+c_2\zeta_0^2+c_3\zeta_0^3,$$ where $c_i \in \mathbb{Q} \text{ for } i=0, \dots, 3$. So all the automorphisms of the field extension are
\begin{equation*}
\begin{aligned}
\lambda_1 & : & p(x) & \mapsto & c_0+c_1\zeta_0+c_2\zeta_0^2+c_3\zeta_0^3 \\
\lambda_2 & : & p(x) & \mapsto & c_0+c_2\zeta_0+c_3\zeta_0^2+c_1\zeta_0^3 \\
\lambda_3 & : & p(x) & \mapsto & c_0+c_3\zeta_0+c_1\zeta_0^2+c_2\zeta_0^3 \\
\end{aligned}
\end{equation*}
I know I need to four automorphism but I am having trouble determining it.

REPLY [3 votes]: You’re dealing with the four primitive eighth roots of unity. Call any one of them $\zeta$, and let this be fixed. The others now are $\zeta^3$, $\zeta^5$, and $\zeta^7=\zeta^{-1}$. Your four automorphisms are the identity, which takes $\zeta$ to $\zeta$, and the three others, which I suppose you can call $\sigma_3$, $\sigma_5$, and $\sigma_7$, with $\sigma_i$ taking $\zeta$ to $\zeta^i$. You can check, for instance, that $\sigma_3\circ\sigma_5=\sigma_7$ in this way: $$\sigma_3(\sigma_5(\zeta))=\sigma_3(\zeta^5)=[\sigma_3(\zeta)]^5=(\zeta^3)^5=\zeta^{15}=\zeta^7=\sigma_7(\zeta)\,.$$ I leave the rest to you.


TITLE: Simplifying surd of degree 1/4 to 1/2
QUESTION [1 upvotes]: What will be the simplification of

$$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 }  } $$

Thanks in advance!

REPLY [2 votes]: $$\sqrt { 6 } \cdot \sqrt { 2-\sqrt { 3 }  } =\sqrt { 3 } \cdot \sqrt { 2 } \sqrt { 2-\sqrt { 3 }  } =\sqrt { 3 } \sqrt { 4-2\sqrt { 3 }  } =\sqrt { 3 } \sqrt { { \left( \sqrt { 3 } -1 \right)  }^{ 2 } } \\=\sqrt { 3 } \left( \sqrt { 3 } -1 \right) = 3-\sqrt { 3 }  $$


TITLE: Are natural logs, natural in any sense?
QUESTION [2 upvotes]: are natural logarithms of a number natural in any sense? Why are they called natural when the base is e and common when the base is 10?

REPLY [1 votes]: Part of this involves calculus reasons, namely that the derivative of $\ln(x)$ is $1/x$, a claim other logarithms cannot make; so $\ln$ is a "unique" logarithm. Moreover, it's the inverse of $e^x$, which is also special among exponential functions, again for calculus reasons (it is its own derivative).


TITLE: Probability Sum Insurance
QUESTION [1 upvotes]: An insurance company runs three offices, A, B and C. The company’s employees are distributed as follows: 30% work in office A, 20% in office B and 50% in office C. In office A, 10% are managers, in office B, 20% are managers and in office C 5% are managers:
(i) What is the total proportion of managers in the company?
Can someone clarify if I got the answer right?
Calculations for part i:
Office A: (0.30) x (0.10) = 0.03
Office B: (0.20) x (0.20) = 0.04
Office C: (0.20) x (0.05) = 0.01
                    = 0.08

This is what my teacher had?
Is this right? Because Office C should be (0.50) x (0.05) = 0.025

REPLY [0 votes]: Office A and B are fine. For Office C, it should be $0.5\cdot 0.05=0.025$. Thus, the proportion of managers is $0.03+0.04+0.025=0.095$.


TITLE: Proof that all basis holds same number of elements
QUESTION [1 upvotes]: I know that the site has evidence of this statement, but I want to check my own proof.
$\textbf{Lemma 1}$
Uniqueness of the zero vector.
$\textbf{Proof}$: $\overline{\theta_1} = \overline{\theta_1} + \overline{\theta_2} = \overline{\theta_2} + \overline{\theta_1} = \overline{\theta_2}$
$\textbf{Lemma 2}$
For any vector there is a single inverse vector.
$\textbf{Proof}$: $-\overline{x_1} = -\overline{x_1} + \overline{\theta} = - \overline{x_1} + (\overline{x} -\overline{x_2}) = - \overline{x_1} + \overline{x} -\overline{x_2} = \overline{\theta} -\overline{x_2} = -\overline{x_2}$
$\textbf{Lemma 3}$
The product of any vector by the neutral element of the additive group of the field is the zero vector.
$\textbf{Proof}$: $0 \cdot \overline{x} = 0 \cdot \overline{x} + \overline{\theta} = 0 \cdot \overline{x} + (\overline{x} - \overline{x}) = \overline{x}(0 + 1) - \overline{x} = \overline{x} - \overline{x} = \overline{\theta}$
$\textbf{Lemma 4}$:
The product of the zero vector on any field element is equal to the zero vector.
$\textbf{Proof}$: $\alpha\cdot\overline{\theta} = \alpha\cdot(0\cdot\overline{x}) = (\alpha \cdot 0)\cdot \overline{x} = 0\cdot \overline{x}$
$\textbf{Lemma 5}$:
If $\lambda \neq 0$ and $\overline{x} \neq \overline{\theta}$ the $\lambda\cdot\overline{x}\neq\overline{\theta}$.
$\textbf{Proof}$: Suppose it is not a true and we can have $\lambda\cdot\overline{x} = \overline{\theta}$, where $\lambda\neq 0$ and $\overline{x}\neq\overline{\theta}$. $\overline{x} = 1\cdot\overline{x} = (\frac{1}{\lambda}\cdot\lambda)\overline{x} = \frac{1}{\lambda} \cdot (\lambda\cdot \overline{x}) = \frac{1}{\lambda}\cdot \overline{\theta} = \overline{\theta}$ -- contradiction.
$\textbf{Lemma 6}$:
In any set of linearly independent vectors, there is no zero vector.
$\textbf{Proof}$: $\sum_{i=1}^{n}\lambda_i\overline{x_i} = \overline{\theta}$ and one of the $\exists \overline{x_j}\in \{\overline{x_n}\}, \overline{x_j} = \overline{\theta}$ $\Rightarrow$ $\lambda_j$ can be $\neq 0$ $\Rightarrow$ $\{\overline{x_n}\}$ -- linear dependent.
$\textbf{Lemma 7}$
The representation of the zero vector in any basis is unique.
$\textbf{Proof}$: Assume that it is not true: $\sum_{i=1}^{n}\lambda_i\overline{x_i} = \overline{\theta}$, where one of the $\lambda_j \neq 0$ $\Rightarrow$ $\lambda_j\overline{x_j} = 0$, but by Lemma 5 and Lemma 6 it is impossible.
$\textbf{Theorem}$: All basis holds same number of elements.
$\textbf{Proof}$: If we have two basis $\{\overline{e_n}\}$ and $\{\overline{e'_m}\}$, where $m > n$. So we can represent the vectors of the second basis by vectors of the first basis: $\overline{e'_k} = \sum_{i=1}^{n}\lambda_{ik}\overline{e_i}$, where $k=1,\cdots,m$ and for any vector $\overline{x}$ in this space will be true that $\overline{x} = \sum_{k=1}^{m}\lambda_k\overline{e'_k} = \sum_{k=1}^{m}\lambda_k\sum_{i=1}^{n}\alpha_{ik}\overline{e_i} =\sum_{i=1}^{n}(\sum_{k=1}^{m}\lambda_k \alpha_{ik})\overline{e_i} $. The zero vector $\overline{\theta}$ has the unique representation in the basis $\{\overline{e_i}\}$, $\overline{\theta} = \sum_{i=1}^{n}0_i\overline{e_i}$. Therefore, the condition $\sum_{k=1}^{m}\lambda_k\overline{e'_k} = 0$ is equivalent to system of linear equations for $\lambda_k$: $\sum_{k=1}^{m}\lambda_k\alpha_{ik} = 0$, where $i=1,\cdots,n$. Since the number of $\{\lambda_k\}=m$  is greater than the number of equations $=n$, this system has a non-zero solution. This contradicts to the Lemma 7.

REPLY [1 votes]: You should reformulate Lemma 2, since you only proof the uniqueness, not the existence.
In the proof of Lemma 2 it's pretty circuitous that you use these minus signs. Just name the two inverse vectors $x_1$ and $x_2$.
In the proof of Lemma 5 you should formulate the assumption and the contradiction and in the proof of Lemma 6 you're mixing up formulas and sentences.
The proof of Lemma 7 is not correct. You can only follow $$\lambda_j\cdot \overline{x_j}=0$$ from comparing the two representations, if you use that the representation of each basis vector is unique, which uses the assertion you want to proof.
Your proof of the theorem is correct, if Lemma 6 and 7 hold to be true, since there is a solution, hence for the under determined system there must be infinitely many.


text:
\begin{document}

\title[Balanced non-transitive dice]{Possible Probability and irreducibility of balanced non-transitive dice}
\date{April 2020}

\begin{abstract}
We construct irreducible balanced non-transitive sets of $n$-sided dice for any positive integer $n$, which was raised in \cite[Question 5.2]{SS17}. One main tool of the construction is to study so-called fair sets of dice. Furthermore, we also study the distribution of the probabilities of balanced non-transitive sets of dice. For a lower bound, we show that the probability could be arbitrarily close to $\frac{1}{2}$ and for a upper bound, we construct a balanced non-transitive set of dice whose probability is $\frac{1}{2} + \frac{13-\sqrt{153}}{24} \approx \frac{1}{2} + \frac{1}{9.12}.$
\end{abstract}
\maketitle

\section{Introduction}
A non-transitive set of dice, first introduced by Gardner \cite{G70} and further studied in \cite{S94}, is triples of labeled dice, $A$, $B$, and $C$, with the property that all three probabilities that $A$ rolls higher than $B$, that $B$ rolls higher than $C$, and that $C$ rolls higher than $A$ are greater than $\frac{1}{2}$. We write this as $P(A>B)>\frac{1}{2}, P(B>C)>\frac{1}{2}$, and $P(C>A) > \frac{1}{2}$. It is called non-transitive because if we define the relation $X \sim Y$ as $P(X>Y) > \frac{1}{2}$ for $X, Y \in \{A, B, C\}$, then the relation $\sim$ is not transitive. For example, the following set of $6$-sided dice is non-transitive since $P(A>B)=P(B>C)=P(C>A)=\frac{19}{36}$:
\[
\begin{tabular}{ccc}
\cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{18} &                        \\ \hline
\multicolumn{1}{|c|}{13} & \multicolumn{1}{c|}{10} & \multicolumn{1}{c|}{7} \\ \hline
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{5}  &                        \\ \cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{4}  &                        \\ \cline{2-2}
                         & A                       &
\end{tabular}
\ \ \
\begin{tabular}{ccc}
\cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{17} &                        \\ \hline
\multicolumn{1}{|c|}{14} & \multicolumn{1}{c|}{12} & \multicolumn{1}{c|}{9} \\ \hline
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{3}  &                        \\ \cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{2}  &                        \\ \cline{2-2}
                         & B                       &
\end{tabular}
\ \ \
\begin{tabular}{ccc}
\cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{16} &                        \\ \hline
\multicolumn{1}{|c|}{15} & \multicolumn{1}{c|}{11} & \multicolumn{1}{c|}{8} \\ \hline
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{6}  &                        \\ \cline{2-2}
\multicolumn{1}{c|}{}    & \multicolumn{1}{c|}{1}  &                        \\ \cline{2-2}
                         & C                       &
\end{tabular}
\]

In \cite{SS17}, Schaefer and Schweig constructed balanced non-transitive sets of $n$-sided dice for any positive integer $n \geq 3$. Main idea of the construction in \cite{SS17} is to combine several balanced non-transitive sets of dice. Therefore, the sets of dice that are constructed in \cite{SS17} are reducible. (See Definition \ref{def:irreducible} for the meaning of being irreducible.) and authors in \cite{SS17} questions whether there exists an irreducible balanced non-transitive sets of $n$-sided dice for all $n$.
One main purpose of the paper is to construct irreducible non-transitive sets of $n$-sided dice for any positive integer $n$. Our main idea of the construction is to use so-called fair sets of $2$-sided dice. Here, being fair means that probabilities $P(A>B), P(B>C)$, and $P(C>A)$ are all $\frac{1}{2}$. Although we used $2$-sided dice to construct new sets of dice, understanding fair sets of $n$-sided dice for any positive integer $n$ seems to be an important step to understand all irreducible balanced non-transitive dice. We also study fair sets of $n$-sided dice for any positive integer $n$.

The second purpose of the paper is to study possible probabilities of balanced non-transitive sets of $n$-sided dice, i.e., possible value of $P(A>B)$. As far as we know, previous known constructions of balanced non-transitive sets of dice have probability $\frac{1}{2} < P(A>B) < \frac{3}{5}=\frac{1}{2} + \frac{1}{10}$. We first show that there exists a balanced non-transitive set of dice such that $P(A>B)-\frac{1}{2}>0$ becomes arbitrary small, i.e., $\frac{1}{2}$ is a sharp lower bound for balanced non-transitive set of dice. We expect that $\frac{1}{2}+\frac{1}{10}$ is not a sharp upper bound and make a conjecture that $\frac{1}{2} < P(A>B) < \frac{1}{2} + \frac{1}{9}$. To support the conjecture, we first explicitly calculate and prove that both the probability $P(A>B)$ in our paper and the one in \cite{SS17} are less than $\frac{1}{2}+\frac{1}{9}$. We also provide another new construction of a balanced non-transitive set of dice such that $P(A>B) \approx \frac{1}{2} + \frac{1}{9.12}$, which is less than $\frac{1}{2}+\frac{1}{9}$ and is, as far as we know, the maximum among known constructions of balanced non-transitive sets of dice.

Let us now describe the content of our paper. After we introduce notation and preliminaries in Section \ref{notation}, we study fair sets of dice in Section \ref{fair sets}. In Section \ref{irreducible}, we provide construction of irreducible balanced sets of $n$-sided dice for any positive integer $n$ using fair sets of $2$-sided dice. In Section \ref{probability}, we calculate and study possible probability $P(A>B)$ for balanced non-transitive sets of dice.

\section{Notation and preliminaries}\label{notation}

We briefly recall definitions and notations in \cite{SS17} that we are going to use.

\begin{definition}
Fix an integer $n>0$. A set of $n$-sided dice is a collection of three pairwise-disjoint sets $A, B,$ and $C$ with $|A|=|B|=|C|=n$ and $A \cup B \cup C = \{1, \cdots, 3n \}$. We think of die $A, B$, and $C$ as being labeled with the elements of $A, B$, and $C$, respectively and we assume that each dice is fair (i.e. the probability of rolling any one of its numbers is $\frac{1}{n}$.)
\end{definition}

\begin{definition}\label{def} A set of dice is called
\begin{itemize}
    \item  {\it balanced} if $P(A>B)=P(B>C)=P(C>A)$.
    \item  {\it non-transitive} if each of $P(A>B), P(B>C)$, and $P(C>A)$ exceeds $\frac{1}{2}$.
    \item  {\it fair} if $P(A>B)=P(B>C)=P(C>A)=\frac{1}{2}$
\end{itemize}
\end{definition}

\begin{definition}\label{word}
If $D=(A, B, C)$ is a set of $n$-sided dice, define a word $\sigma(D)$ by the following rule: the i$^{th}$ letter of $\sigma(D)$ corresponds to the die on which the number $i$ labels a side.
\end{definition}

For a word $\sigma$, a number of alphabets in the word, which we call the length of $\sigma$, is denoted by $|\sigma|$. A number of $A$ (resp. $B$, $C$) in $\sigma$ is denoted by $|A|_{\sigma}$ (resp. $|B|_{\sigma}$, $|C|_{\sigma}$). For example, if $\sigma$ corresponds to a set of $n$-sided dice, $|A|_{\sigma}=|B|_{\sigma}=|C|_{\sigma} = n$.

Let $D=(A, B, C)$ be a set of $n$-sided dice and $\sigma$ be its corresponding word (Definition \ref{word}). Then we denote by $P_D(A>B)$ or $P_{\sigma}(A>B)$ the probability that the number rolled on A is greater than the number rolled on B when we roll $A$ and $B$. Similarly, we define $P_D(B>C), P_{\sigma}(B>C), P_D(C>A)$, and $P_{\sigma}(C>A)$.

\begin{definition}\label{defs w.r.t. N}
For a set of $n$-sided dice $D$ (and its corresponding word $\sigma$ with length $3n$), we let $N_D(A>B):=n^2P_D(A>B), N_D(B>C):=n^2P_D(B>C)$, and $N_D(C>A):=n^2P_D(C>A)$. Similarly we also define $N_{\sigma}(A>B)$, $N_{\sigma}(B>C)$, and $N_{\sigma}(C>A)$.
\end{definition}

It is by definition that
$D=(A,B,C)$ is balanced if $N_D(A>B)=N_D(B>C)=N_D(C>A)$. And $D=(A,B,C)$ is non-transitive if $N_D(A>B), N_D(B>C), N_D(C>A) > \frac{n^2}{2}.$

\begin{remark}
$N_D(A>B)$ in Definition \ref{defs w.r.t. N} represents the number of consequences that a die $A$ wins a die $B$ when we consider all possible outcomes when we roll $A$ and $B$. Note that $N_D(A>B)$ is the same as the notation $\displaystyle\sum\limits_{s_i=A} q_{\sigma(D)}^+(s_i)$ in \cite{SS17}.
\end{remark}

\begin{example}
Let $D$ be the following set of $3$-sided dice:
\begin{eqnarray*}
A &=& 9 \ 5 \ 1 \\
B &=& 8 \ 4 \ 3 \\
C &=& 7 \ 6 \ 2.
\end{eqnarray*}
Then $\sigma(D)=ACBBACCBA$ and $P(A>B)=P(B>C)=P(C>A)= \frac{5}{9}$. Therefore this set of dice is balanced and non-transitive.
\end{example}

\begin{definition}
The concatenation of two words of $\sigma$ and $\tau$ is simply the word $\sigma$ followed by $\tau$, denoted by $\sigma \tau$.
\end{definition}
We recall a recursive relation of $N(A>B), N(B>C)$, and $N(C>A)$. The following is in the proof of Lemma 2.4 in \cite{SS17}.
\begin{lemma}[(1) in \cite{SS17}]
Let $\sigma$ and $\tau$ be two words that correspond to two sets of dice, respectively. Let $|\sigma|=3m$ and $|\tau|=3n$ (i.e., corresponding dice are $m$-sided and $n$-sided, respectively). Then we have

\begin{equation}\label{recursion formula}
    N_{\sigma\tau}(A>B) = N_{\sigma}(A>B) + N_{\tau}(A>B) + mn.
\end{equation}
\end{lemma}

\section{Fair sets of dice}\label{fair sets}

In the following two lemmas ``probabilities" means $P(A>B)$, $P(B>C)$ and $P(C>A)$ and $x$, $y$ or $z$ is one of $A$, $B$ or $C$.
\begin{lemma}\label{symm exchange}
Assume that $\sigma(D)$ has $xy$ and $yx$, i.e., $\sigma(D)= \cdots xy \cdots yx \cdots$. Then exchanging the orders of $xy$ and $yx$ at the same time, i.e., $\sigma(\tilde{D})= \cdots yx \cdots xy \cdots$ does not change the ``probabilities".
\end{lemma}
\begin{lemma}\label{xyz exchange}
Assume that $\sigma(D)$ is a word having three different letters, say $x$, $y$ and $z$. Then $xyz \, \sigma(D)$ and $\sigma(D)\, xyz$ have the same ``probabilities" if and only if $D$ has the same numbers of $x$, $y$ and $z$.
\end{lemma}

\begin{definition}\label{similar}
Two words are called {\it similar}, $\sigma(D_1) \sim \sigma(D_2)$, if  $\sigma(D_1)$ can be re-written to $\sigma(D_2)$ by exchanging the orders in Lemmas \ref{symm exchange} and \ref{xyz exchange}.
\end{definition}
Note that if $\sigma(D_1) \sim \sigma(D_2)$, then two words have the same length and  $P_{D_1}(A>B)=P_{D_2}(A>B), P_{D_1}(B>C)=P_{D_2}(B>C)$, and $P_{D_1}(C>A)=P_{D_2}(C>A)$.

\begin{example}
Lemma \ref{symm exchange} and Lemma \ref{xyz exchange} indicate that $\cdots xy \cdots yx \cdots \sim \cdots yx \cdots xy \cdots$ and $xyz \, \sigma(D) \sim \sigma(D)\, xyz$ if and only if  $|x|_{\sigma(D)}=|y|_{\sigma(D)}=|z|_{\sigma(D)}$.
\end{example}

\begin{conjecture}[Fair dice]\label{fair dice}
If $D$ is fair, then the length of $D$ is a multiple of 6 and it is similar to a product of $xyzzyx$.
\end{conjecture}
Note that $\tau:=ABCCBA$ is a fair set of $2$-sided dice, which will be used in Section \ref{irreducible}.

\begin{example}
Let $\sigma(D)$ be $AABBCCCCBBAA$. It is fair and
\[
AABBCCCCBBAA \sim ABABCCCCBABA \sim ABACBCCBCABA
\]
\[\sim ABCABCCBACBA \sim ABCCBAABCCBA =(ABCCBA)^2.
\]
\end{example}

To support Conjecture \ref{fair dice}, let us consider only {\it two dice} case, i.e., a fair word with only two letters $A$ and $B$. In this case a set of dice and a corresponding word are called fair if $P(A>B)=P(B>A)=\frac{1}{2}.$
\begin{theorem}
Assume that a given fair word $\sigma_f$ has $4m$ letters with $|A|=|B|=2m$. Then $\sigma_f\sim (ABBA)^m$.
\end{theorem}
\begin{proof}
We may  assume that $\sigma_f$ starts with $A$ (if not, we exchange the notations of $A$ and $B$). Then let us claim that the factor $ABBA$ can be extracted to the front without changing probabilities, i.e., $\sigma_f\sim ABBA\, \omega_f$ where $\omega_f$ is a fair word having $4(m-1)$ letters with $|A|=|B|=2(m-1)$. Mathematical induction with this fact leads us to the result that $\sigma_f\sim (ABBA)^m$.

Observe first that $\sigma_f\sim AB\,\omega_0$ where $\omega_0$ is a word such that $AB\,\omega_0$ is fair. If $\sigma_f=AB\,\omega_0$, then we are done. Suppose not, i.e., $\sigma_f=A\cdots A B \,\omega_1$. We would like to see that $\omega_1$ has a sequel $BA$. If $\omega_1$ would not have $BA$, then the given fair dice should be expressed by $\sigma_f=A\cdots A B A\cdots A B\cdots B$, which is not fair at all. Due to having $BA$ in $\omega_1$, exchanging $AB$ and $BA$ together will hold probabilities, i.e., $\sigma_f\sim A\cdots BA\,\omega_2$, where $\omega_2$ is the word obtained from $\omega_1$ by replacing $BA$ by $AB$. A similar argument works to reveal that $\omega_2$ has $BA$ and therefore $\sigma_f\sim A\cdots BAA\,\omega_3$. Keep doing this procedure until $\sigma_f\sim AB\,\omega_0$.

Next see that $\sigma_f\sim ABBA \,\omega_f$ as claimed. If $\omega_0$ starts with $BA$, then we are done. If not, there are three cases, that is, $\omega_0$ is one of $AB\,\omega_4$, $A\cdots AB\,\omega_5$, $B\cdots BA\,\omega_6$. Since a similar argument can be applied, let us consider the first case when $\omega_0=AB\,\omega_4$. If $\omega_4$ would not have $BA$, then it should be expressed by $A\cdots A B\cdots B$. This means that $\sigma_f\sim ABABA\cdots A B\cdots B$, which is impossible to be fair. So $\omega_4$ has a sequel $BA$. Exchanging $BA$ (in $\omega_4$) to $AB$ (in front of $\omega_4$) shows that $\sigma_f\sim ABBA\,\omega_f$. As mentioned, in the other two cases, a similar argument reveals that $\sigma_f\sim ABBA\,\omega_f$ as claimed.
\end{proof}

\section{Irreducible set of dice}\label{irreducible}
In this section, we construct irreducible balanced non-transitive sets of $n$-sided dice for any positive integer $n$. Note that this answers Question 5.2 in \cite{SS17}.

\begin{definition}\label{def:irreducible}
A balanced non-transitive word (and a corresponding set of dice) is called irreducible if there do not exist balanced non-transitive words $\sigma_1$ and $\sigma_2$ (both nonempty) such that $\sigma = \sigma_1\sigma_2$.
\end{definition}
\begin{lemma}\label{baby construction}
Let $\sigma$ be a balanced non-transitive word and $\tau=ABCCBA$. Then $\tau\sigma$ is balanced and non-transitive.
\end{lemma}
\begin{proof}
$\tau$ is balanced but not non-transitive since $P(A>B)=P(B>C)=P(C>A)= \frac{1}{2}$. Lemma 2.4 in \cite{SS17} implies that $\tau\sigma$ is balanced. It remains to prove that it is non-transitive. Let $|\sigma|=3n$. For $x=A, B,$ or $C$, the equation \eqref{recursion formula} implies
\[
N_{\tau\sigma}(A>B) = 2 + N_{\sigma}(A>B) + 2n > 2 + \frac{n^2}{2} + 2n = \frac{(n+2)^2}{2}.
\]
Therefore $\tau\sigma$ is also non-transitive (Definition \ref{defs w.r.t. N}).
\end{proof}

\begin{lemma}\label{main construction}
Let $\sigma$ be a balanced non-transitive word that corresponds to a set of either $3$-sided or $4$-sided dice and $\tau=ABCCBA$. Then $(\tau)^k\sigma$ is an irreducible balanced non-transitive word.
\end{lemma}

\begin{proof}
Applying Lemma \ref{baby construction} $k$ times, we conclude that $(\tau)^k\sigma$ is balanced and non-transitive. It remains to prove that it is irreducible. Suppose that $(\tau)^k\sigma$ is reducible. Then we can write $(\tau)^k\sigma$ as $\pi\pi'$ where $\pi$ is an irreducible balanced non-transitive word and $\pi'$ is a balanced non-transitive word. Since $(\tau)^n$ is fair (so not non-transitive) and $(\tau)^n ABC$ is not balanced for any non-negative integer $n$, $\pi$ should be of the form $(\tau)^k \pi''$ ($\pi''$ is not empty). We write $(\tau)^k\sigma$ as $(\tau)^k\pi'' \pi'''$. Since $|\pi'''| < 12$, $\pi'''$ is irreducible. This implies that $\sigma=\pi''\pi'''$ corresponds to a balanced non-transitive set of $4$-sided dice and when it partially has a balanced non-transitive word $\pi'''$ of length less than $12$, which is impossible.
\end{proof}

We are now ready to construct an irreducible balanced non-transitive set of $n$-sided dice.
\begin{theorem}\label{main construction2}
For any $n \geq 3$, there exists an irreducible balanced non-transitive set of $n$-sided dice.
\end{theorem}
\begin{proof}
We first consider the following two sets of dice:
$$
\begin{matrix}
A_3=9 & 5 & 1\\
B_3=8 & 4 & 3\\
C_3=7 & 6 & 2\\
\end{matrix}
\qquad\textrm{and}\qquad
\begin{matrix}
A_4= 10 & 7 & 5 & 4\\
B_4= 12 & 9 & 3 & 2\\
C_4= 11 & 8 & 6 & 1\\
\end{matrix}.
$$
Both $D_3=(A_3, B_3, C_3)$ and $D_4=(A_4, B_4, C_4)$ are irreducible, balanced and non-transitive. As before we consider $\tau=ABCCBA$.
We now construct an irreducible balanced non-transitive set of $n$-sided dice for any $n \geq 3$.
We already constructed such set of dice when $n=3$ and $n=4$.
When $n \geq 5$ is odd, we write $n = 3+ 2k, k \geq 1$. Then $(\tau)^kD_3$ is an irreducible balanced non-transitive set of $n$-sided dice due to Lemma \ref{main construction}. When $n \geq 5$ is even, we write $n = 4 + 2k, k \geq 1$. Then $(\tau)^kD_4$ is an irreducible balanced non-transitive set of $n$-sided dice due to Lemma \ref{main construction}.
\end{proof}

\section{Possible probability}\label{probability}
Let $(A,B,C)$ be a balanced non-transitive set of $n$-sided dice with $P(A>B)=P(B>C)=P(C>A)> \frac{1}{2}$. In this section our interest is on a possible probability $P(A>B)$.

Let us state what is in our mind.
\begin{conjecture}\label{conjecture prob}
If $(A, B, C)$ is a balanced non-transitive set of $n$-sided dice such that $P(A>B) > \frac{1}{2}$, then
$$ \left( \frac{1}{2} < \,\right) \,\, P(A>B) < \frac{1}{2} + \frac{1}{9}.$$
\end{conjecture}

\subsection{Probabilities of our construction in Theorem \ref{main construction2}}
To support Conjecture \ref{conjecture prob}, we first calculate all possible probabilities of the set of dice which are constructed in Section \ref{irreducible}.

\begin{lemma}\label{recursion formula2}
Let $\sigma$ (resp. $\tau$) be a balanced non-transitive word of length $3m$ (resp. $3n$). Then
\[
P_{\sigma\tau}(A>B) = \frac{1}{2} + \frac{\left(N_{\sigma}(A>B) - \frac{m^2}{2}\right) + \left(N_{\tau}(A>B) - \frac{n^2}{2}\right)}{(m+n)^2}.
\]
Similarly, we have
\[
P_{\sigma\tau}(B>C) = \frac{1}{2} + \frac{\left(N_{\sigma}(B>C) - \frac{m^2}{2}\right) + \left(N_{\tau}(B>C) - \frac{n^2}{2}\right)}{(m+n)^2}
\]
and
\[
P_{\sigma\tau}(C>A) = \frac{1}{2} + \frac{\left(N_{\sigma}(C>A) - \frac{m^2}{2}\right) + \left(N_{\tau}(C>A) - \frac{n^2}{2}\right)}{(m+n)^2}.
\]
\end{lemma}

\begin{proof}
It is an easy consequence of the equation \eqref{recursion formula} since $N_{\sigma\tau}(A>B) = (m+n)^2 P_{\sigma\tau}(A>B)$.
\end{proof}
\begin{remark}
Note that $N_{\sigma}(A>B)$ represents the number of consequences that a die $A$ wins a die $B$. Therefore, $\left(N_{\sigma}(A>B) - \frac{m^2}{2}\right)$ in Lemma \ref{recursion formula2} represents how far it is away from being fair for a word $\sigma$. For example, if $N_{\sigma}(A>B) - \frac{m^2}{2}=0$, then $P_{\sigma}(A>B)= \frac{1}{2}$.
\end{remark}

\begin{lemma}\label{recursion formula3}
Let $\sigma$ and $\tau$ be as above. Assume that $\frac{1}{2} <P_{\sigma}(A>B)<a$ and $\frac{1}{2}<P_{\tau}(A>B)<b$. Then $\frac{1}{2} < P_{\sigma\tau}(A>B) < max \{ a, b \}$.
\end{lemma}
\begin{proof}
\[
(m+n)^2 P_{\sigma\tau}(A>B) = m^2 P_{\sigma}(A>B) + n^2 P_{\tau}(A>B)+mn <
\]
\[ max \{a, b\} (m^2+n^2 + \frac{1}{ max\{a, b\}}mn) <  max \{a, b\} (m+n)^2
\]
since $\frac{1}{2} < max \{a, b\}$.
\end{proof}

We are now ready to calculate the probability $P(A>B)$. Let $\sigma$ be a word that is constructed in Theorem \ref{main construction2}. Then $\sigma$ is either $(\tau)^k\tau_1$ or $(\tau)^k\tau_2$ for some non-negative integer $k$, where
\[
D(\tau_1)=
\left\{ \begin{matrix}
9 & 5 & 1\\
8 & 4 & 3\\
7 & 6 & 2\\
\end{matrix} \right., \ \
D(\tau_2)=
\left\{ \begin{matrix}
10 & 7 & 5 & 4\\
12 & 9 & 3 & 2\\
11 & 8 & 6 & 1
\end{matrix} \right., \ \
D(\tau)=
\left\{ \begin{matrix}
6 & 1\\
5 & 2\\
4 & 3
\end{matrix} \right.
\]
Note that $(\tau)^k$ is a fair word (i.e., $P_{\tau^k}(A>B)=\frac{1}{2}$).
For $i=1, 2$, Lemma \ref{recursion formula} implies that
\[
(2k+n_i)^2P_{\sigma}(A>B) =  (2k)^2P_{\tau^k}(A>B) + (n_i)^2 P_{\tau_i}(A>B) + 2kn_i=
\]
\[
\frac{(2k)^2}{2} + \big[ \frac{n_i^2+2}{2}\big] + 2kn_i = \big[ \frac{(n_i+2k)^2+2}{2}\big]
\]
which is closest integer greater than $\frac{(n_i+2k)^2}{2}$. Here, $3n_i$ is the length of $\tau_i$ and, therefore, $n_i = i+2$.

Therefore, Conjecture \ref{conjecture prob} is true in this case.
\begin{remark}
Similarly, we can also calculate possible probabilities of the set of dice that is constructed in \cite{SS17}.
Let $\sigma$ be a word that is constructed in Theorem 2.1 of \cite{SS17}. Then $\sigma$ is a product of several $\tau_1, \tau_2$, and $\tau_3$, where
\[
D(\tau_1)=
\left\{ \begin{matrix}
9 & 5 & 1\\
8 & 4 & 3\\
7 & 6 & 2\\
\end{matrix} \right., \ \
D(\tau_2)=
\left\{ \begin{matrix}
12 & 10 & 3 & 1\\
9 & 8 & 7 & 2\\
11 & 6 & 5 & 4
\end{matrix} \right., \ \
D(\tau_3)=
\left\{ \begin{matrix}
15 & 11 & 7 & 4 & 3\\
14 & 10 & 9 & 5 & 2\\
13 & 12 & 8 & 6 & 1
\end{matrix} \right.
\]
$P_{\tau_i}(A>B) = \frac{[ \frac{n_i^2 +2}{2} ]}{n_i^2}$ for $i=1, 2, 3$, where $n_i = (i+2)^2$. Therefore, Lemma \ref{recursion formula3} implies that $
P_{\sigma}(A>B) \leq \frac{5}{9} = \frac{1}{2} + \frac{1}{18}$. Therefore, Conjecture \ref{conjecture prob} is true in this case.
\end{remark}

\subsection{Best bound for probability}

We first show that $\frac12$ is the greatest lower bound of $P(A>B)$, which means that the inequality $P(A>B)>\frac12$ is optimal.
\begin{theorem}
There exists a balanced non-transitive set of dice such that $P(A>B)-\frac12>0$ becomes arbitrary small.
\end{theorem}
\begin{proof}
Let $n=2m+1\,\,\,(m\in\mathbb{N})$. Then we construct a word $\sigma$ (or a corresponding dice) as follows. First consider the word $(ABCCBA)^m (BAC)$ which is not balanced, since the number of the events for $C$ to beat $A$ or $B$ is one more than the ones of the events for $A$ or $B$ to do $C$ (therefore $C$ always beats both $A$ and $B$ at least in probability). To change this word to a balanced one, let us replace $BC$ by $CB$ on the first factor $ABCCBA$. In all, the word $\sigma$ obtained by
\begin{equation*}
 \sigma:=(ACBCBA)(ABCCBA)^{m-1}(BAC)
\end{equation*}
is balanced. A direct computation of $P(A>B)$ on $\sigma$, or more precisely,
\begin{equation*}
    N_{\sigma(D)}(A>B)=0+2+\{(2+4)+\cdots+(2(m-1)+2m)\}+(2m+1)=2m^2+2m+1,
\end{equation*}
shows that
 \begin{equation*}
     P(A>B)=\frac{2m^2+2m+1}{(2m+1)^2}=\frac12+\frac{0.5}{n^2},
 \end{equation*}
which goes to $\frac12$ as $n\to\infty$. This indicates that  $\frac12$ is the greatest lower bound of $P(A>B)$ and the inequality $P(A>B)>\frac12$ could not get better.
\end{proof}
\medskip

For the upper bound of $P(A>B)$, recall that Conjecture \ref{conjecture prob} states that
\begin{equation*}
  P(A>B) < \frac{1}{2} + \frac{1}{9}.
\end{equation*}
That is, the probability $P(A>B)$ cannot exceed (nor be equal to) $\frac{11}{18}$.
To support this inequality, we provide three new constructions of sets of dice of which probability is close to $\frac12 + \frac19$.
For simplicity, assume that $n=6p$ (later we discuss when $n=6p+2$ or $n=6p+4$ which is similar to the presenting case).
Let us start with the most unmixed fair $n$-sided dice $\sigma_f$,
\begin{equation}\label{unmixed fair}
   \sigma_f:= A\cdots A \, B\cdots B \, C\cdots C \, C\cdots C \, B\cdots B \, A\cdots A,
\end{equation}
where ``$\cdots$" means that all the letters are the same as the boundary letters and the numbers of letters in ``$\cdots$" are the same, i.e., $|A\cdots A|=|B\cdots B|=|C\cdots C|=3p$.

We apply the following algorithm to construct most probability $P(A>B)$:
\noindent
\begin{algorithm} \quad
\begin{itemize}
    \item[(Step1)]  If there is no triple $AB, BC, CA$ in a word, then replace $xy$ and $yx$ in a word for $x, y \in \{A, B, C\}$ so that the resulting word contains $AB, BC, CA$.
    \item[(Step2)] If you find $AB, BC, CA$, then replace $AB, BC, CA$ by $BA, CB, AC$ respectively, but all together.
   \end{itemize}
\end{algorithm}

\noindent
\begin{remark} \quad
\begin{itemize}
    \item Replacement in (Step1) does not change the probability due to Lemma \ref{symm exchange}.
    \item Replacement in (Step2) increases the number of events for $A$ (resp. $B$ or $C$) to beat $B$ (resp. $C$ or $A$) by 1. In particular, doing this replacement keeps the words balanced and non-transitive. Therefore we may expect that keeping replacing in this way would lead most probability of $P(A>B)$.
\end{itemize}
\end{remark}

Since there is no $CA$ in $\sigma_f$, we first apply (Step1) (exchange of $BC$ and $CB$) several times to $\sigma_f$ in order to obtain a word $\sigma_1$ which is still fair but having sequels $CA$ by

\begin{eqnarray}\label{sigma1}
  \sigma_1:= &&  A\cdots A \, A\cdots A \, A\cdots A \, B\cdots B \, B\cdots B \, C\cdots C \\ \nonumber
  && \,\, C\cdots C \, C\cdots C \, {\bf {\color{red} B\cdots B}} \, C\cdots C \, C\cdots C \, B\cdots B \\  \nonumber
  && \,\,\, B\cdots B \, B\cdots B \, {\bf {\color{red} C\cdots C}} \, A\cdots A \, A\cdots A \, A\cdots A \nonumber
\end{eqnarray}
(i.e., the last one third of the first $B\cdots B$ in \eqref{unmixed fair} moves to the middle and the last one third of the second $C\cdots C$ in \eqref{unmixed fair} does to the front of the second $A\cdots A$ (the moved ones are in red and bold above). Here, $|A \cdots A|=|B \cdots B|=|C \cdots C|=p$ in \eqref{sigma1}. Therefore the numbers of replacing $BC$ and $CB$ by $CB$ and $BC$ respectively are the same as $3p^2$.)

We now apply (Step2) $3p^2$ times to obtain a balanced non-transitive word $\sigma_2$ by

\begin{eqnarray*}
  \sigma_2:= && {\bf {\color{red} B\cdots B}} \, A\cdots A \, A\cdots A \, A\cdots A \,  C\cdots C \, C\cdots C \\
  && \,\, C\cdots C \, {\bf {\color{red} B\cdots B}} \, B\cdots B \, C\cdots C \,  C\cdots C  \, B\cdots B \\
  && \,\,\, B\cdots B \, B\cdots B  \, A\cdots A \, A\cdots A \, A\cdots A \, {\bf {\color{red} C\cdots C}}
\end{eqnarray*}
(i.e., in \eqref{sigma1} the first $B\cdots B$ moves in front, the second $B\cdots B$ does to almost middle and the last $C\cdots C$ does to the end. The moved ones are in red and bold as before.) Here the probability $P(A>B)$ of $\sigma_2$ increases to  $1/2+3p^2/(6p)^2=7/12$.

Since there is no $AB$ and $CA$ in $\sigma_2$, we apply (Step1) $6p$ times as follows;

we move $B$ and $C$ in $\sigma_2$ to the ones in $\sigma_3$ (all of which are in bold below) in order to produce the products $AB$ and $CA$.
\begin{eqnarray*}
  \sigma_2:= && B\cdots B \, A\cdots A \, A\cdots A \, A\cdots A \,  C\cdots C \, C\cdots C \\
  && \,\, C\cdots C \, {\bf {\color{red} B}}\cdots B \, B\cdots B \, C\cdots C \,  C\cdots {\bf {\color{red} C}}  \, B\cdots B \\
  && \,\,\, B\cdots B \, B\cdots B  \, A\cdots A \, A\cdots A \, A\cdots A \, C\cdots C
\end{eqnarray*}
\begin{eqnarray*}
  \sigma_3:= && B\cdots B \, A\cdots A \, A\cdots A \, A\cdots A \, {\bf {\color{red} B}}\, C\cdots C \, C\cdots C \\
  && \,\, C\cdots C \, {\dot B}\cdots {\color{blue}{{\ddot B} \, {\ddot B} \cdots {\ddot B} \, {\ddot C}\cdots {\ddot C} \,  {\ddot C}}}\cdots {\dot C}  \, B\cdots B \\
  && \,\,\, B\cdots B \, B\cdots B \, {\bf {\color{red} C}} \, A\cdots A \, A\cdots A \, A\cdots A \, C\cdots C\\
  && (\textrm{${\dot B}$ and ${\dot C}$ have been moved to ${\bf {\color{red} B}}$ and ${\bf {\color{red} C}}$, respectively.})
\end{eqnarray*}
This means that we need to exchange $BC$ to $CB$ in the middle part (which is blue and has two dots above the character on $\sigma_3$) $6p$ times ($3p$ to move $B$ and another $3p$ to do $C$). Then to increase $P(A>B)$, one more exchange from $BC$ to $CB$ in the middle part is necessary.

In addition to this observation, we would like to measure how many times we can apply (Step2) to $\sigma_3$ in terms of $p$. (This is because the denominator of $P(A>B)$ is $(6p)^2$, so we would better express in $p$ in order to examine asymptotic behavior of $P(A>B)$.)

 Put by $m$ the number of further replacement such that $m$-many $B$'s move between $B\cdots B$ and $A\cdots A$ in front. For this we need extra $3pm$-many exchanges of $BC$ and $CB$ in the middle part. Since the remaining of each $B$'s or $C$'s in the middle is $2p-m$, the possible exchange of $BC$ to $CB$ is $(2p-m)^2$, which should be at least $9pm(=3pm+3pm+3pm)$ (the last $3pm$ is due to move $m$-many $B$'s in the further replacement). As a summary, we have that
\begin{eqnarray*}
 && (2p-m)(2p-m)\ge 9pm \quad (m\leq 2p) \\
 && \quad \Longrightarrow \quad  m\leq \frac{13-\sqrt{154}}{2} \, p.
\end{eqnarray*}
Therefore the largest probability on $P(A>B)$ in the argument above is
\begin{equation*}
    \frac12+\frac{3p^2+\frac{13-\sqrt{154}}{2} \, p \cdot 3p}{(6p)^2}=\frac12+\frac{15-\sqrt{154}}{24} \approx \frac12+\frac{1}{9.25} < \frac12+\frac19.
\end{equation*}
Note that the maximum $m$ (in $\mathbb N$) means that, since all $BC$ are exhausted in the last word, the probability $P(A>B)$ is not able to be larger in this construction.
\medskip

The cases when $n=6p+2$ and $n=6p+4$ can be investigated by a similar procedure. By applying (Step1) (or exchanging $BC$ by $CB$) and (Step2) (or replacing $AB$, $BC$ and $CB$ by $BA$, $CB$ and $BC$), the probability $P(A>B)$ increases by
\begin{equation*}
    \frac{p(3p+1)}{[2(3p+1)]^2} \quad (\textrm{when } n=6p+2) \,\,\textrm{ or }\,\, \frac{p(3p+2)}{[2(3p+2)]^2} \quad (\textrm{when } n=6p+4)
\end{equation*}
(which is similar to the increase $p(3p)/(6p)^2$ when $n=6p$).
In these cases, after doing this procedure, new words are expressed by
\begin{eqnarray*}
 n=6p+2: \,\,\sigma_2= && B\cdots B A\cdots A A\cdots A A\cdots A A {\bf {\color{red}B}} C\cdots C C\cdots C  \\
 && \,\, C\cdots C C B\cdots B B\cdots B C\cdots C C\cdots C C \cdots B\\
 && \,\,\,\, B\cdots B B\cdots B B A\cdots A A\cdots A A\cdots A A C\cdots C \\
 n=6p+4: \,\,\sigma'_2= && B\cdots B A\cdots A A\cdots A A\cdots A AA {\bf {\color{red}BB}} C\cdots C C\cdots C  \\
 && \,\, C\cdots C CC B\cdots B B\cdots B C\cdots C C\cdots C CC \cdots B\\
 && \,\,\,\, B\cdots B B\cdots B BB A\cdots A A\cdots A A\cdots A AA C\cdots C
\end{eqnarray*}
(Here each $\cdots$ has exactly $p-2$ letters. For example $|A\cdots A|=p-2+2=p$.)

Next, to make $P(A>B)$ most, we do (Step2) further as in the previous case. Shortly speaking, when $n=6p+2$, we have that
\begin{eqnarray*}
 && (2p-m)(2p+1-m)-(3p+1) \ge 3m(3p+1) \quad (m\leq 2p) \\
 && \quad \Longleftrightarrow \quad m^2-(13p+4)m+(4p^2-p-1) \ge 0\\
 && \quad \Longrightarrow \quad  m\leq \frac{13p+4-\sqrt{153p^2+108p+20}}{2}.
\end{eqnarray*}
Hence the increased probability on $P(A>B)$ in the argument (i.e., except $1/2$) is
\begin{equation*}
    \frac{\left(p+\frac{13p+4-\sqrt{153p^2+108p+20}}{2}\right)(3p+1)}{(6p+2)^2} \xrightarrow{p\to\infty}\frac{1}{12}+\frac{13-\sqrt{153}}{24} <\frac{1}{9.12} <\frac19.
\end{equation*}
Note that, since the numerator $13p+4-\sqrt{153p^2+108p+20}$ is increasing in $p$, the last inequality holds for all $p$.

Similarly when $n=6p+4$, we have that
\begin{eqnarray*}
 && (2p-m)(2p+2-m)-2(3p+2) \ge 3m(3p+2) \quad (m\leq 2p) \\
 && \quad \Longleftrightarrow \quad m^2-(13p+8)m+(4p^2-2p-4) \ge 0\\
 && \quad \Longrightarrow \quad  m\leq \frac{13p+8-\sqrt{153p^2+216p+80}}{2}.
\end{eqnarray*}
Hence the increased probability on $P(A>B)$ for $n=6p+4$ (again except $1/2$) is
\begin{equation*}
    \frac{\left(p+\frac{13p+8-\sqrt{153p^2+216p+80}}{2}\right)(3p+2)}{(6p+4)^2} \xrightarrow{p\to\infty}\frac{1}{12}+\frac{13-\sqrt{153}}{24} < \frac{1}{9.12} < \frac19.
\end{equation*}
Note again that the numerator $13p+8-\sqrt{153p^2+216p+80}$ increases in $p$.
\medskip

Let us summarize the argument above. Start with the most unmixed fair word $\sigma_f$ (which is \eqref{unmixed fair}). By replacing $AB$, $BC$, $CA$ by $BA$, $CB$, $AC$, i.e., (Step2) as much as possible, the probability $P(A>B) \,(=P(B>C)=P(C>A))$ increases most. Then the computation above tells us that
$$
 P(A>B) < \frac{1}{2} + \frac{1}{9},
$$
which upholds Conjecture \ref{conjecture prob}.

\noindent
\begin{remark} \quad
\begin{itemize}
    \item Let us tell why we think the last constructed word in this argument provides the most probability on $P(A>B)$. First, observe that we have considered the only case when $n$ was even (when $n=6p$, $6p+2$ or $6p+4$). We, however, believe that this would be enough due to the following observation. A direct computation shows that there is only one possible probability $P(A>B)$ when $n=3$, or equivalently, any three balanced, non-transitive dice with 3 sides can be expressed by a balanced, non-transitive word which is similar to $\sigma_3=CBABACACB$. In this case $P_{\sigma_3}(A>B)=5/9$. Then add $CBA$ on the end of $\sigma_3$ and then exchange $CA$ (in $\sigma_3$) to $AC$. Then the constructed word $\sigma_4$ is $\sigma_4=CBABA{\color{red}AC}CB{\color{blue}CBA}$, which becomes balanced and non-transitive. For this, $P_{\sigma_4}(A>B)=9/16$, which is greater than $P_{\sigma_3}(A>B)=5/9$. A similar computation can be done for more general $n$. Therefore, with this trick (however we do not know if we can do this trick at all times), the probability of $P(A>B)$ would get greater for even $n$'s than odd ones.
    \item Second, proving that the above probability is the most probability is related to construction of all balanced non-transitive sets of dice and it also seems to be related to property of fair sets of dice, i.e., Conjecture \ref{fair dice} and we leave this for future work.
\end{itemize}
\end{remark}

\begin{remark}
In \cite{S18}, Schaefer further generalize the results in \cite{SS17} to sets of $n$-sided $m$ dice for any positive integer $m$. It will be very interesting to generalize our results to sets of $m$ dice for $m \geq 4$ and we also leave this for future work.
\end{remark}

\medskip

\subsection*{Acknowledgement}
The first-named author was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Korea government  $($NRF-2019R1F1A1061300$)$. The second-named author was partially supported by Chonnam National University (Grant number: 2018-0978), and by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP) (No.2017R1C1B2010081).

config: arxiv


TITLE: Why does $f(x,y)= \frac{xy^2}{x^2+y^4}$ have different limits when approaching $(0,0)$ along straight lines vs. along the curve $(1/t^2,1/t)$?
QUESTION [3 upvotes]: I have a stupid question about continuity in higher dimensions.
There are maps, for example, $f(x,y)=\frac{xy^2}{x^2+y^4}$, when $(x,y)\neq (0,0)$ and $f(x,y)=(0,0)$ when $(x,y)=(0,0)$, when we approach $(0,0)$ along every straight line, the limit of the function is $0$, but when along a curve, for example $(\frac{1}{t^2},\frac{1}{t})$, the limit of $f$ is not $0$.
But, it feels like all the straight lines can cover a neighbourhood of $(0,0)$, so every point on a curve is also on a different straight line. Why is it that when the same points are arranged in a different way the limit changes?

REPLY [1 votes]: This has something to do with the uniform continuity around 0. Given a fixed $\epsilon > 0$, if there exist an universal constant $\delta_u$ such that
$$|(x,y)|<\delta_u \implies |f(x,y)|<\epsilon$$
then your intuition is right: along any curve go to 0, $f$ will have limit 0.
However, for this example the origin is not uniformly continuous. Along lines  $y=ax$, $a \in\mathbb{R}$ we can define a class of fucntions $f_a(r) = \frac{a^2 x}{1+a^4 x^2}$. Suppose the universal constant $\delta_u$ exist, then for every $a$ we must have
$$ |x|<\delta_u \implies |f_a(x)|<\epsilon $$
However, let $a=\frac{1}{\sqrt{\delta_u}}$ we have $f_a(\delta_u) = 1/2$. Therefore, above argument cannot be correct, and the origin is not uniform continuous with respect to $\{f_a(x)\}$.


TITLE: "The Conjugate of a matrix"
QUESTION [1 upvotes]: I am having some trouble understanding a definition/question in my linear algebra text book.
The question states " If $A$ is a square matrix, a matrix of the form $P^{-1}AP$ where $P$ is invertible is called a  'Conjugate ' of $A$.
I am having trouble understanding that, is there any examples? It then asks is $P^{-1}AP=A$ and I'm not sure what direction to take. I tried $P(P^{-1}AP=A)$ , giving $AP=PA$ and then $(AP=PA)P^{-1}$ giving $A=PAP^{-1}$. But I'm not sure if this is a correct proof/ not sure what I am showing. It also asks if some conjugate of $A$ is invertible show that $A$ is invertible, not sure where to begin on that either.
Thanks so much for any help

REPLY [0 votes]: 1) $A$ is conjugate to itself, simply choose $P = I$ the identity matrix.
2) Suppose some conjugate of $A$ is invertible $\implies \det(P^{-1}AP) \neq 0 \implies \det(A) \neq 0$ which shows that $A$ is also invertible.


TITLE: Elementary Proof for the number of quadratic residues for $n = p_1^{i}p_2^{j}p_3^k...p_\alpha^i$
QUESTION [2 upvotes]: I would just like to get some help in proving that the number of QRs for $n = p_1^{i}p_2^{j}p_3^k...p_\alpha^i$ is $\frac{\phi(n)}{2^\alpha} $.
Intuitively, I understand that each prime $p^i$ 'brings' $\frac{\phi(p^i)}{2} $ QRs to $n$ and obviously multiplying each of these together gives $\frac{\phi(n)}{2^\alpha} $.
I know this question is similar to:
On the number of quadratic residues $\pmod{pq}$ where$p$ and $q$ are odd primes.
But could someone help me with how I would make my intuition a more formal generalised proof?
Many thanks!

REPLY [1 votes]: Here is a heuristic justification that will give you grounds for more formal justification.
First of all, it should make sense that there are $\frac{p-1}{2}$ quadratic residues $\pmod{p}$, because the congruence $x^{2} \equiv a \pmod{p}$ should have two solutions: $x$ and $-x$, and there are $p-1$ non-zero residues $\pmod{p}$, so we're just pairing each $x$ with a $-x$ to leave $\frac{p-1}{2}$ solutions. Now consider the congruence $x^{2} \equiv a \pmod{n}$, where $n = pq$ and $p$ and $q$ are both odd primes. Remember that $x^{2} \equiv a \pmod{n} \iff x^{2} = a + nk$ for some $k \in \mathbb{Z}$. But remember that since $p \vert n$ and $q \vert n$, so we could get the same $a$ just by dividing $x^2$ by $p$ or $q$.
Well, first of all, it should make sense that there are $\frac{p-1}{2}$ quadratic residues $\pmod{p}$, because the congruence $x^{2} \equiv a \pmod{p}$ should have two solutions: $x$ and $-x$, and there are $p-1$ non-zero elements $\pmod{p}$, so we're just taking half of these. Now consider the congruence $x^{2} \equiv a \pmod{n}$, where $n = pq$ and $p$ and $q$ are both odd primes. Remember that $x^{2} \equiv a \pmod{n} \iff x^{2} = a + nk$ for some $k \in \mathbb{Z}$. But remember that since $p \vert n$ and $q \vert n$, so we could get some of the same $a$ just by dividing $x^2$ by $p$ or $q$. Also note that the set of quadratic residues $\pmod{p}$ is closed under multiplication (because $x^2y^2 = (xy)^2$). So if there are $\frac{p-1}{2}$ residues that we already know will be quadratic $\pmod{n}$ because we know $p \vert n$, it should make sense that we would be able to make more quadratic residues by multiplying each of the $\frac{p-1}{2}$ quadratic residues by the $\frac{q-1}{2}$ quadratic residues as a result of this closure. Because $n$ has only two prime divisors, we know that we can't do this again. So we know that there should then be exactly $\frac{p-1}{2} \cdot \frac{q-1}{2} = \frac{(p-1)(q-1)}{2^2}$ quadratic residues $\pmod{n}$. It should make sense why we can make more or less quadratic residues by this multiplication process which is dependent on the number of prime divisors of $n$.


TITLE: Radical Field Extension Example
QUESTION [3 upvotes]: I want to find an example of fields $\mathbb{Q} \lt E \lt K$ such that the extension $K:\mathbb{Q}$ is radical but the extension $E:\mathbb{Q}$ is not radical and provide justification.

REPLY [5 votes]: Let’s see. Try adjoining a  (primitive) seventh root of $1$, call it $\zeta$, and call this field $K$. Now look at the cubic extension $E=\Bbb Q(\zeta+\zeta^{-1})\supset\Bbb Q$. (The minimal polynomial for this number is $X^3+X^2-2X-1$.) Since $K$ is abelian over $\Bbb Q$, so is $E$. And $E$ is normal over $\Bbb Q$, too.
Now if $E'$ is any cubic extension of $\Bbb Q$ that is radical, it should be of form $E'=\Bbb Q(\sqrt[3]n\,)$, but these cubic extensions of $\Bbb Q$ are not normal, since you’ll need the cube roots of unity to get all cube roots of $n$. So our $E$ above is not radical.
EDIT: You have asked for another explanation, not dependent on Galois Theory. I should say that I assume that what you meant by “radical extension” was a field of the form $\Bbb Q(\sqrt[m]n\,)$, for an integer $m$ and a rational number $n$, where for convenience’s sake we assume that $n$ contains no $m$-th powers in its expansion as a product of prime powers.
There are two cases: first, the cyclotomic case where $n=\pm1$, and you’re adjoining the $m$-th roots of unity if $n=1$ and the $2m$-th roots of unity if $n=-1$. The second case covers all other values of $n$. In the first case, the degree $[\Bbb Q(\zeta_n):\Bbb Q]$ is equal to $\phi(n)$, the Euler phi-function, which is even for all values of $n$ except $1$ and $2$. This shows that the cubic extension above is not $\Bbb Q(\zeta_k)$ for any (primitive) $k$-th root of unity $\zeta_k$.
How do I show that it’s not $\Bbb Q(\sqrt[m]n\,)$ for any $n\ne\pm1$? Those radical extensions, if not real quadratic (so with degree $2$) all have embeddings into $\Bbb C$ that are not into $\Bbb R$, since the minimal polynomial for the generator has nonreal roots. In the technical language, those fields are not “totally real”. But the cubic extension I constructed for you is totally real, because the roots of $X^3+X^2-2X-1$ are all real.


TITLE: Strong Induction for Fibonacci number related identity $f_{n-m} = f_{m}f_{n+1} + f_{m-1} f_n$
QUESTION [1 upvotes]: Let $f_n$ be the $n^{th}$ Fibonacci number. Let $m$ be a fixed strictly positive integer.
Prove by strong induction that for all $n\ge 0$,
$$f_{n+m} = f_{m}f_{n+1} + f_{m-1} f_n$$
edit: $f_{n+m} = f_{m}f_{n+1} + f_{m-1} f_n$ instead of $f_{n-m} = f_{m}f_{n+1} + f_{m-1} f_n$
(so sorry lol)

REPLY [2 votes]: Hint:
\begin{align*}
f_{n+1-m} & =f_{n-m}+f_{(n-1)-m}\\
 & =f_{m}f_{n+1}+f_{m-1}f_{n}+f_{m}f_{n}+f_{m-1}f_{n-1}\\
 & =f_{m}(f_{n+1}+f_{n})+f_{m-1}(f_{n}+f_{n-1})\\
 & =\cdots
\end{align*}


TITLE: Sum of geometric progression $z^{-m}$?
QUESTION [1 upvotes]: We know that the sum of the geometric progression of $z^{-n}$ where $n$ starts at zero and goes to infinity is $1/(1-z)$, but is there a way to show the sum when $n$ starts at $m$? I.e. what is
$$\sum_{n=m}^\infty \frac{1}{z^n}$$

REPLY [0 votes]: $\frac{1}{1-z}$ is the sum of $z^k$ (when $|z|<1$), NOT $z^{-k}$ (in this case the sum would be $\frac{z}{z-1}$ and $|z|$ would be greater than $1$).
However if $|z|<1$ we have that
$$
\sum_{k=m}^{+\infty}z^{k}=
\sum_{k=0}^{+\infty}z^{k}-\sum_{k=0}^{m-1}z^{k}=\frac{1}{1-z}-\frac{1-z^m}{1-z}
=\frac{z^m}{1-z}
$$
If you mean $z^{-k}$ changes only a few: as already said $|z|>1$ and you'd have just to turn $z$ into $\frac{1}{z}$ in the above formula:
$$
\sum_{k=m}^{+\infty}z^{-k}=\frac{\frac1{z^m}}{1-\frac1{z}}=\frac{1}{z^m-z^{m-1}}
$$


TITLE: limit of factorial divided by sum of factorials
QUESTION [0 upvotes]: I'd like to determine the limit as $n\rightarrow \infty$ following function:$$f\left(n\right)=\frac{n!}{\sum_{k=0}^{n}k!}.$$The following calculations suggest the answer is 1:
$$
\begin{align*}
f\left(10\right)&\approx0.898682 \\\\
f\left(30\right)&\approx0.966627 \\\\
f\left(70\right)&\approx0.985711.
\end{align*}
$$
But this isn't very satisfying. Any input on how I might proceed in finding this limit would be appreciated.

REPLY [1 votes]: If you write
$$\sum_{k=0}^n k!=n!\Bigg[\frac 11 +\frac 1 {n}+ \frac 1 {n(n-1)}+ \frac 1 {n(n-1)(n-2)}+\frac 1 {n(n-1)(n-2)(n-3)}+\cdots \Bigg]$$ Expanding as series for large values of $n$
$$\sum_{k=0}^n k!=n!\Bigg[1+\sum_{k=1}^\infty \frac {B_k}{n^k}\Bigg]$$where appear Bell numbers. So, using long division
$$f_n=\frac{n!}{\sum_{k=0}^{n}k!}=1-\frac 1n-\frac 1{n^3}-\frac 2{n^4}+O\left(\frac{1}
   {n^5}\right)$$
Using the above truncated expansion for $n=10$ as you did
$$f_{10}=\frac{2247}{2500}=0.898800$$ to be compared to the exact value
$$f_{10}=\frac{1814400}{2018957}=0.898682$$corresponding to a relative error of $0.013$%.