text:
\begin{document}
\title{UNIQUENESS OF SOLUTIONS TO BOLTZMANN EQUATIONS}
\author{Rafael Galeano Andrades$^{\lowercase{1}}$\,\ and \,\ Mario Almanza Caro$^{\lowercase{2}}$}
\address{Universidad de Cartagena, Facultad de Ciencias Exactas y Naturales}
\address{Departamento de Matemáticas.}
\email{$^1$rgaleanoa@unicartagena.edu.co}
\address{Universidad de Sucre, Facultad de Educaci\'on y Ciencias}
\address{Departamento de Matem\'aticas}
\email{$^2$mario.almanza@unisucre.edu.co}
\subjclass[2010]{35Q20-35B38}
\keywords{Generalized solutions, Boltzmann equation, uniqueness, renormalized solutions, Banach Fixed Point Theorem.}

\maketitle

\begin{abstract}
By means of Banach Fixed Point Theorem, the uniqueness of the Boltzmann equation generalized solutions in Sobolev spaces in $L^{1}\left(\mathbb{R}^{+}\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)$, can be proved,as well as the uniqueness of renormalized solutions of the Boltzmann equation.
\end{abstract}

\section{Introduction}

The problem is posed as follows: Let $f\left(t,x,v\right):\,\mathbb{R}^{+}\times\mathbb{R}^{n}\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{+}$ measurable such that it satisfies:

\begin{equation}
\left\{ \begin{array}{l}
\frac{\partial f}{\partial t}+v\cdot\nabla_{x}f=Q\left(f,f\right)\\
f\left(0,x,v\right)=f_{0}\left(x,v\right),
\end{array}\right.\label{eq:1}
\end{equation}

where

\begin{equation}
Q\left(f,f\right)(v)=\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[f\left(u'\right)f\left(v'\right)-f\left(u\right)f\left(v\right)\right]dwdu,
\end{equation}

as a result of the quadratic operator

\begin{equation}
Q\left(f,g\right)\left(v\right)=\frac{1}{2}\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[f\left(v'\right)g\left(u'\right)+f\left(u'\right)f\left(v'\right)\right.\left.-f\left(u\right)g\left(v\right)-f\left(v\right)g\left(u\right)\right]dwdu
\end{equation}

in the case where $f=g$. It will bemoreover assumed that  $Q\left(f,g\right)\left(v\right)=0$
si $f\neq g$, $u'+v'=u+v$ and $\left|u'\right|^{2}+\left|v'\right|^{2}=\left|u\right|^{2}+\left|v\right|^{2}$. $B$ satisfies the following properties:

\begin{itemize}
	\item[] $B_{0}$) $B$ just depends $\left|v-u\right|$ y $\left|v-u,w\right|$.
	\item[] $B_{1}$) $B\in L_{loc}^{\infty}\left(\mathbb{R}^{n},S^{2}\right)$.
	\item[] $B_{2}$) $B\left(v,w\right)\leq\frac{b_{1}\left|\left(v,w\right)\right|}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)$.
	\item[] $B_{3}$) $\int_{S^{2}}B\left(v,w\right)dw\geq b\left|v\right|\left(1+\left|v\right|^{\mu}\right)^{-1}$.
\end{itemize}

A classical solution $f$ of (\ref{eq:1}) is a function $f\in C^{1}\left(\left[0,T\right]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)$
that satisfies (\ref{eq:1}).

Let $f^{\#}\left(t,x,v\right)=f\left(t,x+vt,v\right)$, as a result

\begin{equation}
 \left\{ \begin{array}{lc}
             \frac{df}{dt}\left(t,x,v\right) =\frac{\partial f^{\#}}{\partial t}\left(t,x,v\right)+v\cdot\nabla_{x}f\left(t,x+vt,v\right)=Q^{\#}\left(f,f\right)(v), \\
             f^{\#}\left(0,x,v\right) =f\left(0,x,v\right)=f_{0}\left(x,v\right).
             \end{array}
   \right.
\end{equation}

If, it is supposed ${\displaystyle f^{\#}\in L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$
y ${\displaystyle Q^{\#}\left(f,f\right)(v)\in L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$,

then

\begin{equation}\label{eq:2}
f^{\#}\left(t,x,v\right)=f_{0}\left(x,v\right)+\int_{0}^{t}Q^{\#}\left(t,f\right)(v)d\tau.
\end{equation}

The functions satisfying (\ref{eq:2}), are called generalized solutions of  (\ref{eq:1}). A solution ${\displaystyle f\in L^{1}\left(\mathbb{R}^{+}\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$ is called a renormalized solution of the Boltzmann equation if:

\[
\frac{Q\left(f,f\right)}{1+f}\in L^{1}\left(\mathbb{R}^{+}\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right),
\]

and if for any continuous Lipchitz function $\beta:\,\mathbb{R}^{+}\rightarrow\mathbb{R}$
satisfying $\left|\beta'(t)\right|\leq\frac{c}{1+t}$ for all $t\geq0$, $c>0$, it follows:

\begin{equation}
\frac{\partial\beta}{\partial t}+v\cdot\nabla_{x}\beta(f)=\beta'(f)Q\left(f,f\right)\label{eq:3}
\end{equation}

in the sense of the distributions. In the particular case where $\beta(f)=f$, it is reduced to the problem studied by Diperna-Lions. In the particular case where $\beta(t)=\ln(1+t)$, $f$ is a Boltzmann's classical solution.

The first existence theorem on time-dependent Boltzmann equation solutions was released in 1932 when Carleman proved the existence of global solutions for the Cauchy problem in the especial homogeneous case (the solution does not depend on $x$). In 1963, Grad built the first local result close to the Maxwellian and in 1974, Ukai built global solutions close to the Maxwellian, extending the result of Grad. The Grad procedure was used to build solutions in $L^{\infty}$ for initial value and boundary problems (for example see Ukai (1976), Nishida (1976), Shizuta (1977), Ukay and Asano (1983), Asano (1985)). In the $L^{1}$ scheme, Diperna and Lions (1989) constructed $L^{1}$-global solutions with minimal hypothesis about the initial data. These solutions are called renormalized solutions. In 2009, a result given by Gressmann, Strain and Arkiv, prove the existence of Boltzmann equation classic solutions. The Boltzmann equation with the term force, adds the term  $F\cdot\nabla_{v}f$ to the left side of the equation. In this sense, in 2011, Galeano and others proved the existence in $L^{1}$ for small data and in 2007, Galeano proved the existence of solutions close to the vacuum.

\section{Desarrollo}

Let ${\displaystyle f_{1}^{\#}\in L^{1}\left(\left[0,T\right]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$ y ${\displaystyle f_{2}^{\#}\in L^{1}\left(\left[0,T\right]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$ 	be Boltzmann equation generalized solution and let  ${\displaystyle g=f_{1}^{\#}-f_{2}^{\#}\in L^{1}\left(\left[0,T\right]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$ be   solutions to the generalized Boltzmann equation, that is to say:

\begin{align*}
f_{1}^{\#}\left(t,x,v\right) & =f_{0}\left(x,v\right)+\int_{0}^{t}Q^{\#}\left(f_{1},f_{1}\right)(v)d\tau,\\
f_{2}^{\#}\left(t,x,v\right) & =f_{0}\left(x,v\right)+\int_{0}^{t}Q^{\#}\left(f_{2},f_{2}\right)(v)d\tau.
\end{align*}

then:

\begin{eqnarray*}
f_{1}^{\#}\left(t,x,v\right)-f_{2}^{\#}\left(t,x,v\right)&=&f_{0}\left(x,v\right)+\int_{0}^{t}Q^{\#}\left(f_{1},f_{1}\right)(v)d\tau-f_{0}\left(x,v\right)\\
&&\hspace{1cm}-\int_{0}^{t}Q^{\#}\left(f_{2},f_{2}\right)(v)d\tau.
\end{eqnarray*}

That is to say:

\begin{align*}
\mathcal{G} \left(t,x,v\right)&=\int_{0}^{t}Q^{\#}\left(f_{2}+g,f_{2}+g\right)(v)d\tau-\int_{0}^{t}Q^{\#}\left(f_{2},f_{2}\right)(v)d\tau,\\
\mathcal{G}(0,x,v)&=0.
\end{align*}

Let

\[
F\left(g\right)\left(t,x,v\right):=\int_{0}^{t}Q^{\#}\left(f_{2}+g,f_{2}+g\right)(v)d\tau-\int_{0}^{t}Q^{\#}\left(f_{2},f_{2}\right)(v)d\tau
\]

be defined in  $L^{1}\left(\left[0,T\right]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)$. It is noticed that $\mathcal{G}=0$ is a fixed point of $F$.It will be proved by means of Banach Fixed Theorem with some additional hypothesis, that zero is the only fixed point of $F$, in order to conclude that $\mathcal{G}\equiv0$ and therefore $f_{1}=f_{2}$.

\begin{thm}
If it is supposed $0\leq L<1$, such that

\begin{eqnarray*}
\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|f_{2}(u)\right|\left|B\left(u,\left|u-v\right|\right)\right|dwdu&\leq&\frac{L\left|v\right|}{b_{1}\left(1+\left|v\right|^{\mu}\right)},\\
\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|g_{i}(u)\right|\left|B\left(w,\left|u-v\right|\right)\right|dwdu&\leq&\frac{L\left|v\right|}{b_{1}\left(1+\left|v\right|^{\mu}\right)},
\end{eqnarray*}

where $i=1,2$. Then $F$ is a continuous lipchitz.
\end{thm}

\begin{proof} \hspace{0.01cm}\\

$\displaystyle\left|F\left(g_{1}\right)\left(t,x,v\right)-F\left(g_{2}\right)\left(t,x,v\right)\right|$
\smallskip
\begin{eqnarray*}
&\leq&\int_{0}^{t}\left|Q^{\#}\left(f_{2}+g_{1},f_{2}+g_{1}\right)(v)-Q^{\#}\left(f_{2},f_{2}\right)\right.(v)\\
& &\hspace{0cm}\left.-Q^{\#}\left(f_{2}+g_{2},f_{2}+g_{2}\right)(v)+Q^{\#}\left(f_{2},f_{2}\right)(v)\right|d\tau\\
&=&\int_{0}^{t}\left|Q^{\#}\left(f_{2}+g_{1},f_{2}+g_{1}\right)(v)\right.\left.-Q^{\#}\left(f_{2}+g_{2},f_{2}+g_{2}\right)(v)\right|d\tau.
\end{eqnarray*}

Then
\medskip

$\left|Q^{\#}\left(f_{2}+g_{1},f_{2}+g_{1}\right)(v)-Q^{\#}\left(f_{2}+g_{2},f_{2}+g_{2}\right)(v)\right|$

\begin{eqnarray*}
&=&\left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[\left(f_{2}+g_{1}\right)\left(t,x+tv,u'\right)\left(f_{2}+g_{1}\right)\left(t,x+tv,v'\right)\right.\right.\\
&&\hspace{1.5cm}-\left(f_{2}+g_{1}\right)\left(t,x+tv,u\right)\left(f_{2}+g_{1}\right)\left(t,x+tv,v\right)\\
		& &\hspace{1.5cm}-\left.\left.\left(f_{2}+g_{2}\right)\left(t,x+tv,u'\right)\left(f_{2}+g_{2}\right)\left(t,x+tv,v'\right)\right.\right.\\
		&&\hspace{1.5cm}\left.+\left(f_{2}+g_{2}\right)\left(t,x+tv,u\right)\left(f_{2}+g_{2}\right)\left(t,x+tv,v\right)\right]dudw\Biggl|\\
&=&\left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[\left(f_{2}\left(t,x+tv,u'\right)+g_{1}\left(t,x+tv,u'\right)\right)\left(f_{2}\left(t,x+tv,v'\right)+g_{1}\left(t,x+tv,v'\right)\right)\right.\right.\\
		&& -\left(f_{2}\left(t,x+tv,u\right)+g_{1}\left(t,x+tv,u\right)\right)\left(f_{2}\left(t,x+tv,v\right)+g_{1}\left(t,x+tv,v\right)\right)\\
		&&-\left(f_{2}\left(t,x+tv,u'\right)+g_{2}\left(t,x+tv,u'\right)\right)\left(f_{2}\left(t,x+tv,v'\right)+g_{2}\left(t,x+tv,v'\right)\right)\\
		&& +\left.\left.\left(f_{2}\left(t,x+tv,u\right)+g_{2}\left(t,x+tv,u\right)\right)\left(f_{2}\left(t,x+tv,v\right)+g_{2}\left(t,x+tv,v\right)\right)\right]dwdu\right.\Biggl|\\
&=& \left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[\left(f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)+g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)\right.\right.\\
&&\hspace{4cm}\left(f_{2}^{\#}\left(t,x+\left(v-v'\right),v'\right)+g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)\\
		&& -\left.\left.\left(f_{2}^{\#}\left(t,x+\left(v-u\right),u\right)+g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)\left(f_{2}^{\#}\left(t,x,v\right)+g_{1}^{\#}\left(t,x,v\right)\right)\right.\right.\\
		&& -\left(f_{2}^{\#}\left(t,x+\left(v-u'\right),u'\right)+g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)\\
		&&\hspace{4cm}\left(f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)+g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)\\
		&&
		+\left(f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)\left(f_{2}^{\#}\left(t,x,v\right)+g_{2}^{\#}\left(t,x,v\right)\right)dwdu\Biggl|\\
& = & \left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\\
&&\hspace{1.5cm}+\left.\left.f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\\
&&\hspace{1.5cm}+g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}+g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}-f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)f_{2}^{\#}\left(t,x,v\right)-f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)\\
&&\hspace{1.5cm}-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)f_{2}^{\#}\left(t,x,v\right)-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)\\
&&\hspace{1.5cm}-f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}-f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}+f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)f_{2}^{\#}\left(t,x,v\right)+f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{2}^{\#}\left(t,x,v\right)\\
&&\hspace{1.5cm}+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)f_{2}^{\#}\left(t,x,v\right)+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{2}^{\#}\left(t,x,v\right)dwdu\Biggl|\\
&=&\left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\right.\\
&&\hspace{5cm}\left.\left.-g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)\right]dwdu\\
\end{eqnarray*}

\begin{eqnarray*}
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right.\right.\\
&&\hspace{5cm}\left.\left.-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)\right]dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\left(g_{2}^{\#}\left(t,x,v\right)-g_{1}^{\#}\left(t,x,v\right)\right)\right]dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[f_{2}^{\#}\left(t,x,v\right)\left(g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)\right]dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right) \left[g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.dwdu\\
&&\hspace{1.5cm}-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{1.5cm}\left.+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{2}^{\#}\left(t,x,v\right)\right]dwdu\Biggl|\\
&=&\left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\\
&&\hspace{3cm}\left.\left.-g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)\right.dwdu\\
&&+\left.\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right.\right.\\
&&\hspace{3cm}\left.\left.-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)\right.dwdu\\
&&+\left.\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\left(g_{2}^{\#}\left(t,x,v\right)-g_{1}^{\#}\left(t,x,v\right)\right)\right.dwdu\\
&&+\left.\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x,v\right)\left(g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)\right.dwdu\\
&&+\left.\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\\
&&\hspace{2cm}-g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{2cm}+g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\\
&&\hspace{2cm}-g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\\
&&\hspace{2cm}+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{2}^{\#}\left(t,x,v\right)\\
&&\hspace{2cm}-g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)+g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)\\
&&\hspace{2cm}\left.\left.+g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)g_{1}^{\#}\left(t,x,v\right)\right]dwdu\right.\Biggl|\\
&=&\left|\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\right.\\
&&\hspace{3cm}\left.-g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right.\\
&&\hspace{3cm}\left.-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)dwdu\\
\end{eqnarray*}

\begin{eqnarray*}
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\left(g_{2}^{\#}\left(t,x,v\right)-g_{1}^{\#}\left(t,x,v\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)f_{2}^{\#}\left(t,x,v\right)\left(g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\right.\\
&&\hspace{3cm}\left.-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right.\\
&&\hspace{3cm}\left.-g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)\left(g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right.\\
&&\hspace{3cm}\left.-g_{2}^{\#}\left(t,x+t\left(v-u'\right),u'\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)\left(g_{2}^{\#}\left(t,x,v\right)-g_{1}^{\#}\left(t,x,v\right)\right)dwdu\\
&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)g_{1}^{\#}\left(t,x,v\right)\left(g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)-g_{1}^{\#}\left(t,x+t\left(v-u\right),u\right)\right)dwdu\\
\end{eqnarray*}

So
\medskip

$\left\Vert F\left(g_{1}\right)-F\left(g_{2}\right)\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}$

\begin{eqnarray*}
	&\leq&\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|f_{2}^{\#}\left(t,x,t\left(v-u\right),u\right)dwdv\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|f_{2}^{\#}\left(t,x,v\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|B\left(w,\left|u-v\right|\right)\right|g_{1}^{\#}\left(t,x,v\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}.\\
\end{eqnarray*}

\begin{eqnarray*}
	&\leq&\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|f_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|f_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|f_{2}^{\#}\left(t,x,v\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|g_{1}^{\#}\left(t,x+t\left(v-u'\right),u'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|g_{2}^{\#}\left(t,x+t\left(v-v'\right),v'\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|g_{2}^{\#}\left(t,x+t\left(v-u\right),u\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\\
	&&+\frac{b_{1}}{\left|v\right|}\left(1+\left|v\right|^{\mu}\right)\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}\left|\left(v-u,w\right)\right|g_{1}^{\#}\left(t,x,v\right)dwdu\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}.\\
\end{eqnarray*}

Then,
\[\left\Vert F\left(g_{1}\right)-F\left(g_{2}\right)\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}\leq L\left\Vert g_{1}^{\#}-g_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}
\]

Therefore, by Banach's Fixed Point Theorem, operator  $F$ has a single Banach fixed point, that is, $\mathcal{G}\equiv0$, then $f_{1}=f_{2}$.
\end{proof}

\section{Renormalized solutions.}

\begin{thm}If it is supposed $\beta_{1}\neq\beta_{2}$, $\beta_{1}\geq\beta_{2}+\sqrt{c}$
y ${\displaystyle \int_{0}^{T}Q^{\#}\left(g,g\right)dt<1}$, then there is only one solution to the Boltzmann renormalized equation.
\end{thm}

\begin{proof} Let $f_{1}$ and $f_{2}$ be two renormalized solutions of Boltzmann's equation, and

	$$
	\frac{\partial\beta\left(f_{1}\right)}{\partial t}+v\cdot\nabla_{x}\beta\left(f_{1}\right)=\beta'\left(f_{1}\right)Q\left(f_{1},f_{1}\right),
	$$

	$$
	\frac{\partial\beta\left(f_{2}\right)}{\partial t}+v\cdot\nabla_{x}\beta\left(f_{2}\right)=\beta'\left(f_{2}\right)Q\left(f_{2},f_{2}\right),
	$$

	\[
	\frac{\partial}{\partial t}\left[\beta\left(f_{1}\right)-\beta\left(f_{2}\right)\right]+v\cdot\nabla_{x}\left[\beta\left(f_{1}\right)-\beta\left(f_{2}\right)\right]=\beta'\left(f_{1}\right)Q\left(f_{1},f_{1}\right)-\beta'\left(f_{2}\right)Q\left(f_{2},f_{2}\right).
	\]

Let $g=f_{1}-f_{2}$, so

\begin{eqnarray*}
\frac{\partial}{\partial t}\left[\beta\left(f_{2}+g\right)-\beta\left(f_{2}\right)\right]+v\cdot\nabla_{x}\left[\beta\left(f_{2}+g\right)-\beta\left(f_{2}\right)\right]&=&\beta'\left(f_{2}+g\right)Q\left(f_{2}+g,f_{2}+g\right)\\
& &\hspace{0.5cm}-\beta'\left(f_{2}\right)Q\left(f_{2},f_{2}\right).
\end{eqnarray*}

Let

\[
\left[\beta^{\#}\left(f_{2}+g\right)-\beta^{\#}\left(f_{2}\right)\right]\left(t,x,v\right)=\beta\left(f_{2}+g\right)\left(t,x+vt,v\right)-\beta\left(f_{2}\right)\left(t,x+vt,v\right),
\]
so,
\[
\frac{d}{dt}\left[\beta^{\#}\left(f_{2}+g\right)-\beta^{\#}\left(f_{2}\right)\right]=\beta'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right).
\]

Then
\begin{eqnarray*}&&\beta^{\#}\left(f_{2}+g\right)\left(t,x,v\right)-\beta^{\#}\left(f_{2}\right)\left(t,x,v\right)\\
&& \hspace{1.7cm}=\beta^{\#}\left(f_{2}+g\right)\left(0,x,v\right)-\beta^{\#}\left(f_{2}\right)\left(0,x,v\right)\\
& &\hspace{1.7cm}+\int_{0}^{T}\left[\beta'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)\right]dt.
\end{eqnarray*}

If it is defined
\begin{eqnarray*}&&F\left[\left(\beta^{\#}\left(f_{2}+g\right)-\beta^{\#}\left(f_{2}\right)\right)\left(t,x+vt,v\right)\right]\\
&& \hspace{1.5cm}=\beta^{\#}\left(f_{2}+g\right)\left(0,x,v\right)-\beta^{\#}\left(f_{2}\right)\left(0,x,v\right)\\
& & \hspace{1.5cm}+\int_{0}^{T}\left[\beta'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)\right]dt,
\end{eqnarray*}

we notice that if $g=0$, then it is a critical point of $F$ . Let's verify that it is the only one, to conclude that  $f_{1}=f_{2}$. For this we will use in Banach's Fixed Point Theorem, in effect:

$F\left[\beta_{1}^{\#}\left(f_{2}+g\right)-\beta_{1}^{\#}\left(f_{2}\right)-\beta_{2}^{\#}\left(f_{2}+g\right)+\beta_{2}^{\#}\left(f_{2}\right)\right]$

\begin{eqnarray*}
&=&F\left[\beta_{1}^{\#}\left(f_{2}+g\right)-\beta_{2}^{\#}\left(f_{2}+g\right)+\beta_{2}^{\#}\left(f_{2}\right)-\beta_{1}^{\#}\left(f_{2}\right)\right]\\
&=&\int_{0}^{T}\left[\beta_{1}'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta_{1}'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)\right]dt\\
& &-\int_{0}^{T}\left[\beta_{2}'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta_{2}'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)\right]dt\\
&=&\int_{0}^{T}\left[\beta_{1}'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)-\beta_{2}'^{\#}\left(f_{2}+g\right)Q^{\#}\left(f_{2}+g,f_{2}+g\right)\right]dt\\
  & &+\int_{0}^{T}\left[\beta_{2}'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)-\beta_{1}'^{\#}\left(f_{2}\right)Q^{\#}\left(f_{2},f_{2}\right)\right]dt\\
&=&\int_{0}^{T}\left[\beta_{1}'^{\#}\left(f_{2}+g\right)-\beta_{2}'^{\#}\left(f_{2}+g\right)\right]Q^{\#}\left(f_{2}+g,f_{2}+g\right)dt\\
&&-\int_{0}^{T}\left[\beta_{2}'^{\#}\left(f_{2}\right)-\beta_{1}'^{\#}\left(f_{2}\right)\right]Q^{\#}\left(f_{2},f_{2}\right)dt\\
&=& \int_{0}^{T}\left[\beta_{1}{}^{\#}\left(f_{2}+g\right)-\beta_{2}{}^{\#}\left(f_{2}+g\right)\right]'Q^{\#}\left(f_{2}+g,f_{2}+g\right)dt\\
&&+\int_{0}^{T}\left[\beta_{2}\#\left(f_{2}\right)-\beta_{1}\#\left(f_{2}\right)\right]'Q^{\#}\left(f_{2},f_{2}\right)dt\\
&\leq&\int_{0}^{T}\frac{c}{1+\beta_{1}^{\#}\left(f_{2}+g\right)-\beta_{2}^{\#}\left(f_{2}+g\right)}Q^{\#}\left(f_{2}+g,f_{2}+g\right)dt\\
&&-\int_{0}^{T}\frac{c}{1+\beta_{2}^{\#}\left(f_{2}\right)-\beta_{1}^{\#}\left(f_{2}\right)}Q^{\#}\left(f_{2}+g,f_{2}+g\right)dt\\
\end{eqnarray*}

\begin{eqnarray*}
&=&\int_{0}^{T}\frac{c}{1+\left(\beta_{1}^{\#}-\beta_{2}^{\#}\right)}\left[Q^{\#}\left(f_{2}+g,f_{2}+g\right)-Q^{\#}\left(f_{2},f_{2}\right)\right]dt\\
&\leq&\int_{0}^{T}\frac{c}{\left(\beta_{1}^{\#}-\beta_{2}^{\#}\right)}\left[Q^{\#}\left(f_{2}+g,f_{2}+g\right)-Q^{\#}\left(f_{2},f_{2}\right)\right]dt\\
&\leq&\int_{0}^{T}\left(\beta_{1}^{\#}-\beta_{2}^{\#}\right)\left[Q^{\#}\left(f_{2}+g,f_{2}+g\right)-Q^{\#}\left(f_{2},f_{2}\right)\right]dt\,\,\,\,\textrm{(si}\,\,\,\beta_{1}\geq\beta_{2}+\sqrt{c}\textrm{).}
\end{eqnarray*}

So that

$$\left\Vert F\left[\beta_{1}^{\#}\left(f_{2}+g\right)-\beta_{1}^{\#}\left(f_{2}\right)-\beta_{2}^{\#}\left(f_{2}+g\right)+\beta_{2}^{\#}\left(f_{2}\right)\right]\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)},$$

is equal or lower than

$$\int_{0}^{T}\left[Q^{\#}\left(f_{2}+g,f_{2}+g\right)-Q^{\#}\left(f_{2}.f_{2}\right)\right]dt\left\Vert \beta_{1}^{\#}-\beta_{2}^{\#}\right\Vert _{L^{1}\left([0,T]\times\mathbb{R}^{n}\times\mathbb{R}^{n}\right)}.$$

Then

$\displaystyle\int_{0}^{T}\left[Q^{\#}\left(f_{2}+g,f_{2}+g\right)-Q^{\#}\left(f_{2}.f_{2}\right)\right]$

\begin{eqnarray*}
& = & \int_{0}^{T}\left[\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left.(f_{2}^{\#}\left(v'\right)f_{2}^{\#}\left(u'\right)+f_{2}^{\#}\left(v'\right)g^{\#}\left(u'\right)\right.\right.\\
		&&+\left.\left.g^{\#}\left(v'\right)f_{2}\left(u'\right)+g^{\#}\left(v'\right)g^{\#}\left(u'\right)+f_{2}^{\#}\left(v\right)f_{2}^{\#}\left(u\right)\right.\right.\\
		&&+\left.f_{2}^{\#}\left(v\right)g^{\#}\left(u\right)-g^{\#}\left(v\right)f_{2}^{\#}\left(u\right)-g^{\#}\left(v\right)g^{\#}\left(u\right)-f_{2}^{\#}\left(v'\right)f_{2}^{\#}\left(u'\right)+f_{2}^{\#}\left(v\right)f_{2}^{\#}\left(u\right)\right]dudwdt\\
&=&\int_{0}^{T}\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[f_{2}^{\#}\left(v'\right)g^{\#}\left(u'\right)+g^{\#}\left(v'\right)f_{2}\left(u'\right)-f_{2}^{\#}\left(v\right)g^{\#}\left(u\right)-g^{\#}\left(v\right)f_{2}^{\#}\left(u\right)\right]dudwdt\\
		&&+\int_{0}^{T}\int_{\mathbb{R}^{n}}\int_{\left|w\right|=1}B\left(w,\left|u-v\right|\right)\left[g^{\#}\left(v'\right)g^{\#}\left(u'\right)-g^{\#}\left(v\right)g^{\#}\left(u\right)\right]dudwdt\\
&=&2\int_{0}^{T}Q^{\#}\left(f_{2},g\right)dt+\int_{0}^{T}Q^{\#}\left(g,g\right)dt\\
&=&\int_{0}^{T}Q^{\#}\left(g,g\right)dt\\
&<&1.
\end{eqnarray*}

So by Banach's Fixed Point Theorem, exists only a single fixed point of  $F$, that is $\beta^{\#}\left(f_{2}+g\right)-\beta^{\#}\left(f_{2}\right)=0$,
	we only are interested in the fixed points of  $\beta^{\#}$, therefore
	$f_{2}+g-f_{2}=0$ implies $g=0$, then $f_{1}=f_{2}$.
    \end{proof}

config: arxiv


TITLE: A relationship between the proof of a renormalizability and gauge fixing conditions?
QUESTION [6 upvotes]: I already know that QCD is renormalizable in several gauges, including the $\xi$ gauge and the background field gauge. That is, the divergence of the quantum effective action is limited by symmetry, and as a result the divergence can be removed by adjusting the coefficients of the original action. However, I feel it odd that the proof of the renormalizability of QCD or some gauge theories as I know it depends on certain gauge fixing conditions. This is because, according to the philosophy of the FP method, a gauge fixing is basically a way to remove the huge non-physical redundancy of gauge degrees of freedom, which in that sense is not an essential component of the theory, and thus these proofs only mean "that the theory is renormalizable in that specific gauge”. Therefore, I would like to ask two questions:

When one show the renormalizability of QCD for example , a particular gauge is chosen. Why is such a "specific gauge dependent" proof widely used? In other word, why is such a gauge dependent proof essential?

Is there a proof of renormalizability that does not depend on gauge fixing conditions? In particular, is the theory renormalizable no matter how strange gauge fixing conditions are imposed? (Intuitively, I don't think that's true.）

REPLY [4 votes]: While OP is right that morally going from one gauge-fixing condition to another is a flow along gauge-orbits, it can in practice be difficult to check that the theory is independent of the gauge-fixing condition: Gauge-fixing in the path-integral may be only partial, and in a quantum average sense, there could be Gribov ambiguities, etc.
Moreover, various objects, such as propagators and vertex-correlators, which we use in the renormalization, explicitly depend on the gauge-fixing condition.
At the formal perturbative level, the Batalin-Vilkovisky formalism has been employed to show renormalization and independence of gauge-fixing, see e.g. Ref. 1.
References:

S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Chapters 15 + 17.


TITLE: How to compute $\int_0^\infty \frac{\log(2+x^2)}{4+x^2}\,\mathrm dx$
QUESTION [1 upvotes]: Evaluate the integral
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
--
I started by stating that the integral from 0 to infinity should be the same as half the integral from $-\infty$ to $\infty$, that is:
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx = \frac{1}{2}\int_{-\infty}^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
and then by stating that this must be equal to
$$\pi i \cdot \sum(\text{residues in upper plane})$$
noting that there is a singularity at $z=2i$ in the upper half plane, and an "issue" (I don't know if it's strictly a singularity) with the log function at $z=\sqrt{2}i$.
The residue at $z=2i$ is easily dealt with, but when I try to deal with the log issue, I can't make any headway. I decided to make my branch cut between $z=\sqrt{2}i$ and $z=-\sqrt{2}i$, and form a contour that goes around this cut and the point, but it doesn't seem to be working out for me.
Suggestions would be appreciated!

REPLY [3 votes]: Using Feynman's trick works:
For $a\geq0$, let $$G(a)=\int_0^\infty \frac{\ln(2+ ax^2)}{4+x^2}\,\mathrm dx.$$
Then the Leibniz rule implies that for $a>0$, $$G'(a)=\int_0^\infty \frac{x^2}{\left(x^2+4\right) \left(a x^2+2\right)}\,\mathrm dx.$$ The last integrand is a rational function and thus has a closed form anti-derivative that can be computed algortihmically. Namely,
$$G'(a)=\left. \frac{\frac{\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{a} x}{\sqrt{2}}\right)}{\sqrt{a}}+2 \tan
   ^{-1}\left(\frac{2}{x}\right)}{2-4 a}\right|_{x=0}^{x=\infty}=\dots=\frac{\pi}{2\sqrt2(1+\sqrt2\sqrt a)\sqrt a}.$$
We know that $G(a)=G(0)+\int_0^a G'(b)\,\mathrm db$. Indeed, $$\int_0^a G'(b)\,\mathrm db = \frac{\pi}{2\sqrt2} \int_0^a \frac1{(1+\sqrt{2}\sqrt b)\sqrt b}\,\mathrm db=\frac\pi2\ln(1+\sqrt2\sqrt a).$$
Also, $$G(0)=\ln(2)\int_0^\infty \frac{1}{4+x^2}\,\mathrm dx=\frac{\ln(2)\pi}4.$$
Hence,
$$\bbox[15px,border:1px groove navy]{
\int_0^\infty \frac{\ln(2+ x^2)}{4+x^2}\,\mathrm dx
=G(1)=\frac{\pi}2\left(\frac{\ln(2)}2+\ln(1+\sqrt2)\right).
}$$

REPLY [2 votes]: $$I=\int_{0}^{\infty} \frac{\log(2+x^2)}{4+x^2}dx=\int_{0}^{\infty} \frac{\log(a^2+x^2)}{4+x^2}dx=I(a)$$
$$\implies \frac{dI}{da}=2a \int_{0}^{\infty} \frac{dx}{(a^2+x^2)(4+x^2)]}=
\frac{2a}{4-a^2}\int_{0}^{\infty} \left( \frac{1}{a^2+x^2}-\frac{1}{4+x^2}\right) dx$$
$$\frac{dI}{da}=\frac{2a}{4-a^2}[\frac{\pi}{2a}-\frac{\pi}{4}]=\frac{\pi}{2(a+2)} \implies I(a)=\frac{\pi}{2} \ln (a+2)+C~~~~(*)$$
$$I(0)=2\int_{0}^{\infty}  \frac{\ln x}{x^2+4}dx=J~~~~(1)$$
Let $x=1/t \implies dx=-dt/t^2$ in $J$, the
$$J=2\int_{0}^{\infty} \frac{-\ln t}{t^2+4}dt=2\int_{0}^{\infty} \frac{-\ln x}{x^2+4}dx~~~~(2)$$
Adding (1) and (2), we get $J=0=I(0)$, from (*) we get $C=-\frac{\pi}{2}\ln 2$
Thus, $$I(a)=\frac{\pi}{2} \ln (a+2)-\frac{\pi}{2} \ln 2= \frac{\pi}{2}\ln\frac{a+2}{2}=\frac{\pi}{2}\ln (1+\frac{1}{\sqrt{2}}),~ as ~a=\sqrt{2}.$$


TITLE: Show that $e^{|xy|}$ is not differentiable
QUESTION [2 upvotes]: I'm a little surprised and am not sure how to show that $e^{|xy|}$ is not differentiable. I am especially surprised because I know that $|xy|$ is differentiable at $(0, 0)$. Any help would be greatly appreciated. Edit: I forgot to add that I am in particular interested in differentiability at $(0, 0)$.

REPLY [4 votes]: In fact this function is differentiable at $(0,0)$. Its derivative at $(0,0)$ is zero. To see this we have to show that $\frac {e^{|xy|}-1} {\sqrt {x^{2}+y^{2}}} \to 0$ as $\|(x,y)\| \to 0$. For this note that that there exists $\delta >0$ such that $0<e^{t} -1\leq 2t$ for $0<t<\delta$. Hence $\frac {e^{|xy|}-1} {\sqrt {x^{2}+y^{2}}} \leq \frac {2|xy|} {\sqrt {x^{2}+y^{2}}}$. Now use the inequality $\sqrt {x^{2}+y^{2}} \geq \sqrt {2|xy|}$ to see that $\frac {e^{|xy|}-1} {\sqrt {x^{2}+y^{2}}} \leq \sqrt 2 \sqrt {|xy|} \to 0$ as $(x,y) \to (0,0)$.


TITLE: On continuity of the Gateaux derivative of p-Laplacian operator
QUESTION [2 upvotes]: Let $\Omega\subset \mathbb{R}^n, N\geq3$, be an open set. For $p\in(1,+\infty)$, define a functional $J:W_0^{1,p}(\Omega)\rightarrow\mathbb{R}$ by
$J(u)=\int_\Omega |\nabla u|^p\,dx.$
Then $J$ is differentiable in $W_0^{1,p}(\Omega)$ and
$J'(u)v=p\int_\Omega |\nabla u|^{p-2}\nabla u\cdot\nabla v\,dx.$
My question is about how to prove the Gateaux derivative $J_G'(u)$ is continuous. Indeed, for any sequence $u_k\rightarrow u$, I can prove that there exists a subsequence $u_{k_j}$ satisfying $J_G'(u_{k_j})\rightarrow J_G'(u)$ in [$W^{1,p}(\Omega)$]', but I can't prove it for the original sequence.
Thanks for any help!

REPLY [2 votes]: Use the following convergence principle:
A sequence is converging to some $x$ if and only if every subsequence contains a subsequence that converges to $x$.


TITLE: What would the water pressure be at this point in a static incompressible fluid?
QUESTION [1 upvotes]: I have a bowl full of water. I invert a glass and place it upside down in the water, leaving a small pocket of air.
My question: what would the water pressure at positions P1 and P2 be? I know P1 = P2 because there's no hydrostatic effect as the water is level, but what is this value?

*This may seem rather trivial but I can't decide between two lines of reasoning. From where my confusion arises:
The atmospheric pressure acts to push the water level in the bowl down and up into the glass. The small pocket of air in the glass has the opposite effect, acting to push the water level in the glass down and increase the level in the bowl. Since the air pressures are equal, the water is at the same level in the bowl as in the glass.
1) Now, since the fluid is in equilibrium, P0 must equal the pressure at P1 for no flow to occur. Likewise, then P2 must equal P0 in the glass. Then P1 = P2 = P0
2) Or the water pressure at P1 and P2 completely zero? And this would be because the pressure from the air in the glass counters the pressure from the air outside acting on the bowl?
Thank you in advance, sorry for the length of this.

REPLY [1 votes]: What you have is a water barometer.  The bowl starts out full of water.  When it is inverted and immersed, the water level in the bowl falls if the bowl has enough height which would be about 10 meters.  If the bowl has less height, it will stay full of water.  It does not fall to the level of the water outside, but will be at a height above the outside water level.  This is because the gas pocket in the bowl formed by the falling water level will be solely water vapor.  There is no air pocket in this problem.  Its pressure will be the vapor pressure of water at whatever temperature the system is, which is not given.
Assuming the temperature is not at or above the boiling point, the vapor pressure will be below atmospheric and the liquid-vapor interface in the bowl will be a degree of vacuum.  We know that the head provided by the column of water in the bowl will be such that the pressure of the water in the bowl is 1 atm at the same level as the water outside the bowl.
In the figure, the pressure will be the same all the way across.


TITLE: find the derivative $f'(0)$
QUESTION [1 upvotes]: Given $$f(x)=\frac{\sqrt{2-e^{2x}}\sqrt[4]{2-e^{4x}}\cdot\ldots\cdot\sqrt[50]{2-e^{50x}}}{(2-e^x)\sqrt[3]{2-e^{3x}}\cdot\ldots\cdot\sqrt[99]{2-e^{99x}}},$$
  find $f'(0)$.

this method was used
$$ \Big(\ln|f|\Big)' = \frac{f'}{f}\quad\Rightarrow\quad f'=f\cdot \Big(\ln|f|\Big)' $$
At the end I came across a harmonic series in the power of the number 2: from 1/2 to 1/50 in the numerator and from 1 to 1/99 in the denominator[. no ideas how to convert further

REPLY [1 votes]: You shouldn’t separate the numerator and denominator when taking $\log$.
We have
$$\ln f =\left(\frac {\ln{(2-e^{2x}})}2 +\frac {\ln{(2-e^{4x}})}4\dots+\frac{\ln(2-e^{50x})}{50}\right)-\left(\ln(2-e^x)+\frac{\ln{(2-e^{3x}})}3+\dots+\frac{\ln(2-e^{99x})}{99}\right)$$
Differentiating, we have $$\frac{f’}f = g(x)=-\left(\frac{e^{2x}}{2-e^{2x}}+\frac{e^{4x}}{2-e^{4x}}+\dots+\frac{e^{50x}}{2-e^{50x}}\right) + \left(\frac {e^x}{2-e^x} +\frac {e^{3x}}{2-e^{3x}}+\dots+\frac{e^{99x}}{2-e^{99x}}\right)$$
And since $f’(0)=f(0)g(0)$ we have $$f’(0)=f(0)\left(50\cdot\frac 1{2-1} - 25\cdot\frac 1{2-1}\right)=25f(0)$$
$$\implies f’(0)=25\left({(2-1)^\frac12(2-1)^\frac14\dots(2-1)^\frac1{50}\over(2-1)(2-1)^\frac13\dots(2-1)^\frac1{99}}\right)=25$$
$$\boxed{f’(0)=25}$$


TITLE: Probability of matching a hand of 13 random cards with cards dealt from a shuffled deck up to position 35?
QUESTION [0 upvotes]: You are given a random assortment of 13 cards from one deck. Another deck is shuffled and cards dealt and called out from the top, the aim is to match all 13 of your cards before reaching 35 cards down into the deck. What is the probability of achieving this?
I realise this is closely related to something like the Birthday Paradox, but I can't seem to use to probability calculator on this site to work out the solution: https://betterexplained.com/articles/understanding-the-birthday-paradox/
The reason I'm asking is because this is a regular thing that happens like a bingo in a pub quiz I frequent and nobody ever wins the prize, which seems unlikely in so many months of playing with a pub full of people all with their own individual random assortment of 13 different cards. After the position of 35 is reached the rest of the deck is read out until someone obviously eventually matches up their cards first, but only win what's in the money pot for the night, rather than big money prize. Me and my friends have worked out various ways to fix the game, but it may be the probability is entirely not in anyone's favour, and the number 35 has been chosen very precisely odds-wise.
Hope someone can provide a solution.

REPLY [1 votes]: If I understand well then your question can be translated into: "if $13$ cards are labeled and then after shuffling $35$ cards are drawn what is the probability that all labeled cards are among them?"
The answer is: $$\frac{\binom{35}{13}}{\binom{52}{13}}=\frac{35!39!}{22!52!}=\frac{23}{40}\times\cdots\times\frac{39}{52}\approx0.002325$$So a very small chance.


TITLE: How the tackle this limited integral?
QUESTION [1 upvotes]: I started with use of a new variable for the things under square root.
I would like to calculate the integral
$$\iint_\Omega\sqrt{\sqrt{x}+\sqrt{y}}~\mathrm{d}x~\mathrm{d}y$$
over the respected area
$$\Omega=\{(x,y)\in\mathbb R^2;x\geq 0,y\geq 0,\sqrt x+\sqrt y\leq 1\}$$

REPLY [1 votes]: HINT:
Transform variables from $(x,y)$ to $(u,v)$ where $x=u^2$ and $y=v^2$.  The Jacobian of transformation is $4uv$ so that
$$\int_0^1\int_0^{\left(1-\sqrt{y}\right)^2}\sqrt{\sqrt{x}+\sqrt{y}}\,dx\,dy=4\int_0^1\,v\,\left(\int_0^{1-v}\,u\,\sqrt{u+v}\,du\right)\,dv$$


TITLE: Why is $A(\cos x)=x/2$ in the unit circle definition of cosine?
QUESTION [3 upvotes]: In the Spivak's calculus, there is the following definition:

If $0 \leq x \leq \pi$, then $\cos x$ is the unique number in $[-1,1]$
  such that
$$A(\cos x)=\frac x 2 \text{ and } \sin x=\sqrt{1-\cos^2 x}$$
  $$\text{where } A(x)=\frac{x \sqrt{1-x^2}}{2}+\int_x^1 \sqrt{1-t^2}dt$$

This seems hardly a definition, but a theorem. Where does $x/2$ come from? It is stated that it is the area but why $x/2$ and not just $x$?

REPLY [3 votes]: Areas contain a one-half factor for the same reason as in a triangle: a circular wedge is narrower at the tip, so area = arc*r/2 (as for the full circle, area is $\frac12 (2\pi r^2)$ but circumference is $2 \pi r$ without the one-half ($2\pi$ is a unit here rather than $\pi$ because it's the full circle).
So, essentially, you are saying that the area is half the arc (in dimensionless units - unit circle), and the area is then expressed with an integral.
Another way of looking at it would be that in the first line, you computed the angle in the polar coordinates,
$$A=\int_0^{x}\int_0^1 r\,dr\,d\phi=\frac {x}{2}$$
and in the second line, you computed exactly the same thing in cartesian coordinates, which has a side-effect of defining $\cos$ and $\sin$ through the transformation between polar and cartesian coordinates.


TITLE: Proving a bounded monotone function has finite one-sided limits
QUESTION [4 upvotes]: all! I got stuck on this question today, although it seemed straight forward when I started. Here is the proposed problem:
Let a and b be extended real numbers with a $\lt$ b. Prove that if f is a bounded, monotone function on the interval (a, b), then $\lim_{x\to a^+}f(x)$ and $\lim_{x\to b^-}f(x)$ both exist and are finite.
I started by using the fact that f has a sup and inf, meaning that the limits must exist and be finite because x $\epsilon$ (a,b). I am just not sure how to prove this rigorously. Thank you!

REPLY [8 votes]: The idea is that monotonicity implies the inf and sup must be the one-sided limits at the end points.
Suppose $f$ is monotonically increasing. Pick $\epsilon > 0$. Let $\sup_{x \in (a,b)}f(x) = M$. Since $M$ is a least upper bound, there exists some $c \in (a,b)$ such that $M - f(c) < \epsilon$. Let $\delta = b - c$. Then if $b - x < \delta$ for $x \in (a,b)$, we have $x > c$, hence $f(x) \geq f(c)$ by monotonicity. Therefore $M - f(x) \leq M - f(c) < \epsilon$.

REPLY [2 votes]: Without loss of generality, suppose that $f$ is increasing.  Let $L = \sup\{f(x) : x \in (a,b)\} < \infty$.  We wish to show that $\lim_{x \to a^-}f(x) = L$.  That is, for any $\epsilon > 0$, there is a $\delta$ such that $a-\delta < x < a \implies |f(x) - L| < \epsilon$.
To that end, let $\delta$ be such that $f(a - \delta) > L - \epsilon$, noting that such a $\delta$ must exist by the definition of a supremum.  Now, to fill in the blanks, show that this $\delta$ satisfies the above requirement.


TITLE: On the unicity of splitting field
QUESTION [10 upvotes]: I have this definition for the splitting field of a polynomial: let $f$ be a polynomial with coefficients in the field $F$. A field $E$ containing $F$ is called a splitting field for $f$ if it satisfies:
i) $f$ splits in $E[X]$, i.e $f(X)=a\displaystyle\prod_i (X-\alpha_i)$, with $a,\alpha_i\in E$
ii) $E$ is generated over $F$ by the roots of $f$, i.e. $E=F[\alpha_1,\ldots,\alpha_n]$
Further in my textbook i can find a proof that any two splitting fields for $f$ are $F$-isomorphic.
I can't understand the meaning of this result. By definition, condition ii) seems to state that a splitting field is minimal among all fields containing $F$ where $f$ completely splits, hence i thought that a splitting field was unique. Why then we need a theorem stating that all splitting fields are isomorphic?

REPLY [9 votes]: Summary of the comments: The splitting field is uniquely determined by the polynomial, if you consider it inside a fixed extension field $E$ of $F$ where the polynomial happens to split. For example, we could use an algebraic closure of $F$ as $E$.
The point of this isomorphism result is to relieve us of the obligation to work inside any previously known field. It allows us to construct the zeros and the extension field any which way we see fit. Secure in the knowledge that the end result will be unique (up to isomorphism).


TITLE: Why $|a_m - a_n| > \frac{a}{3}$?
QUESTION [0 upvotes]: I'm reading Courant's: Differential and Integral Calculus. Here:

I guess I understand the overall idea of the proof: If $|a_m-a_n|$ can be made arbitrarily small when $m,n$ are arbitrarily large and we have two points of accumulation $\eta, \xi$ with a fixed positive distance, we can have an even smaller $|a_m-a_n|$ which implies that the length of this interval will be smaller than the distance from $\eta$ to $\xi$ and this is a contradiction.
I've been able to follow the proof until he argues that $|a_m - a_n| > \frac{a}{3}$. I don't see how he concluded that.

REPLY [1 votes]: Let $|\xi-\eta| = a$
$\{a_i\}$ is cauchy so there is an $M$ where $m,n > M$ implies $|a_m - a_n| < \frac a3$.
$\xi$ is an accumulation point so there is an $N$ where $m > N$ implies $|\xi - a_m| < \frac a3$
$\eta$ is an accumulation point so there is an $K$ where $n>K$ implies $|\eta - a_n| < \frac a3$.
So if $m,n > \max (M,N,K)$ then
$a = |\xi - \eta| \le |\xi - a_m| + |a_m - a_n| + |a_n - \eta| < \frac a3 + \frac a3 + \frac a3= a$
That is a contradiction.
However the author took an slightly different route:
$ |\xi - \eta| \le |\xi - a_m| + |a_m - a_n| + |a_n - \eta|\implies$
$|a_m - a_n| \ge|\xi - \eta| -  |x_i - a_m| - |a_n - \eta|$
and
$|\xi - \eta| -  |x_i - a_m| - |a_n - \eta| > a - \frac a3 - \frac a3= \frac a3$
So $|a_m - a_n|> \frac a3$


TITLE: Evaluating a few complex integrals on the unit circle
QUESTION [0 upvotes]: I'm stuck on a few of these, but I have most of the details worked out:
(i)  $\int_{|r|=1}(z^2-4)^{-1}\,dt=\int_{0}^{2\pi}ie^{i\theta}(e^{2i\theta}-2)^{-1}\,d\theta$
(ii) $\int_{|r|=1}(z^2-2z)^{-1}\,dt=\int_{0}^{2\pi}ie^{i\theta}(e^{2i\theta}-2e^{i\theta})^{-1}\,d\theta$
In both integrals I'm using $z=e^{i\theta}$, and $0 < \theta \leq 2\pi$.
I couldn't think of any integral tricks to make these work out, so I tried putting them into Mathematica.  The first yielded something with ArcTanh.  I became suspicious because my instructor had us explicity skip Hyperbolic trig functions.  The second one I'm not at all sure how to simplify.

REPLY [0 votes]: The first one do not even bother to compute, as its value is zero. And it is zero because
$$
f(z)=\frac{1}{z^2-4}
$$
is analytic in the unit disk, and hence its integral on the unit circle is equal to zero.
In the case of the second one, $f(z)=\dfrac{1}{z^2-2z}$ has a single pole at $z=0$. The only singularity inside the unit circle. Then
$$
\int_{|z|=1}\frac{dz}{z^2-2z}=2\pi i\,\mathrm{Res}\,(z^2-2z, z=0)=2\pi i\lim_{z\to 0} zf(z)=-\pi i.
$$


TITLE: Difference between $\epsilon-n_0$ and $\epsilon-\delta$ limit definition.
QUESTION [6 upvotes]: My Real Analysis course uses the $\epsilon - n_0$ definition of the limit, but I have noticed that the $\epsilon - \delta$ approach seems to be more common. Could someone please explain both formally and informally the difference between the definitions? Is there an advantage to using one over the other? An example of a simple proof using both definitions would be great!
(Sorry if this is a silly question. Looking at the definitions, I don't think that $n_0 = \delta$. but I could be wrong. If $n_0 = \delta$ then at least there is a simple answer, though I'll feel pretty embarrassed.)

REPLY [1 votes]: The $n_0$ is (as noted) for a limit of a sequence, or sometimes for a limit as the variable goes to infinity, as in
$$
\lim_{x\to\infty}\frac{1}{x} = 0
$$
phrased as: for every $\epsilon > 0$ there is $n_0 \in \mathbb R$ such that for all $x > n_0$ we have
$$
\left|\frac{1}{x} - 0\right| < \epsilon .
$$


TITLE: Free subgroup of linear groups
QUESTION [7 upvotes]: Suppose $a,b \in GL(n,\mathbb{C})$, and $\langle a,b\rangle$ is a free group of rank $2$.
Is there a way to choose a $c$ to guarantee that $\langle a,b,c\rangle$ is a free group of rank $3$?

REPLY [6 votes]: Yes.  Note first that it suffices to answer the question in $\textit{SL}(2,\mathbb{C})$.  In particular, if $A,B\in\textit{GL}(2,\mathbb{C})$, let $\hat{A},\hat{B}$ be scalar multiples of $A,B$ that lie in $\textit{SL}(2,\mathbb{C})$, and suppose that there exists a $C\in\textit{SL}(2,\mathbb{C})$ so that $\langle\hat{A},\hat{B},C\rangle$ is a free group of rank $3$.  Then for any word $w(x_1,x_2,x_3)$ in the free group $\langle x_1,x_2,x_3\rangle$, if $w(A,B,C) = I$, then $w(\hat{A},\hat{B},C)$ would have to be a scalar matrix, and hence would lie in the center of $\langle\hat{A},\hat{B},C\rangle$, which is impossible.  We conclude that $w(A,B,C) \ne I$ for all $w$, and hence $\langle A,B,C\rangle$ is free as well.
So suppose $A,B\in\textit{SL}(2,\mathbb{C})$ and $\langle A,B\rangle$ is free of rank $2$.  Consider the group $\textit{SL}(2,R)$, where $R$ is the ring $\mathbb{C}[t_1,t_2,t_3,t_4]/(t_1t_4-t_2t_3-1)$, and let
$$
T \;=\; \begin{bmatrix}t_1 & t_2 \\ t_3 & t_4\end{bmatrix}.
$$
Note that $T$ is invertible over $R$, with
$$
T^{-1} \;=\; \begin{bmatrix}t_4 & -t_2 \\ -t_3 & t_1\end{bmatrix},
$$
and hence $T \in \textit{SL}(2,R)$.  Then $\langle A,B,T\rangle$ must be free, since any element of $\langle A,B\rangle$ can be substituted for $T$, and the free group $\langle A,B\rangle$ has no non-trivial laws.
Now, if $w = w(x_1,x_2,x_3)$ is any non-trivial word in the free group $\langle x_1,x_2,x_3\rangle$, the equation
$$
w(A,B,T) \;=\; I
$$
is equivalent to a nontrivial system of polynomial equations in $t_1,t_2,t_3,t_4$, which define a proper subvariety $V_w$ of $\textit{SL}(2,\mathbb{C})$.  But $\textit{SL}(2,\mathbb{C})$ cannot be expressed as a countable union of proper subvarieties, and hence there exists a matrix $C\in\textit{SL}(2,\mathbb{C})$ that does not lie in any $V_w$.  Then $\langle A,B,C\rangle$ is free.
Practically speaking, the way to choose $C$ is to choose a generic matrix, i.e. a matrix whose entries do not satisfy any polynomial equations over the field generated by the entries of $A$ and $B$.


TITLE: Discrete calculus: is my proof about the difference of two consecutive powers correct?
QUESTION [4 upvotes]: Theorem statement: $\displaystyle \Delta^dF_n(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^{d-1}F_{n-1-k}(x)(-1)^k$

Proof:
$$\begin{align} F_n(x) &= x^n =\Delta^0F(x) \\[2ex] \Delta^1F_n(x) &= x^n -(x-1)^n \\[2ex] &= x^n - \sum_{k=0}^n\binom nk x^{n-k}(-1)^k \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}x^{n-k}(-1)^{k} \\[2ex] &=x^{n-1} -(n-1)x^{n-2} + \ldots \pm (n-1)x^2 \mp 1 \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}\Delta^0F_{n-1-k}(x)(-1)^k\end{align}$$

$$\begin{align}\Delta^2F_n(x) &=\Delta^1F_n(x) -\Delta^1F_n(x-1) \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}x^{n-k}(-1)^k - \sum_{k=0}^{n-1}\binom{n-1}{k}(x-1)^{n-k}(-1)^k \\[2ex] & = x^{n-1}-(x-1)^{n-1}- (n-1)x^{n-2} - (n-1)(x-1)^{n-2} +\ldots \pm(n-1)x^2 -(n-1)(x-1)^2 \\[2ex] & = \sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x)(-1)^k\end{align}$$

$$\begin{align} \Delta^3F_n(x) &= \Delta^2F_n(x) - \Delta^2F_n(x-1) \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x)(-1)^k -\sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x-1)(-1)^k \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k} (\Delta^1F_{n-1-k}(x) - \Delta^1F_{n-1-k}(x-1))(-1)^k \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^2F_{n-1-k}(x)(-1)^k \end{align}$$
$$\therefore \Delta^dF_n(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^{d-1}F_{n-1-k}(x)(-1)^k$$

Is this correct? If not, where did I go wrong?
If it is correct, what can I do with this? Where can I apply the derived identity?
I originally set out to prove that $\Delta^n F_n(x) = c \ \forall x$, but I'm not sure if I'm any closer to this now.

REPLY [3 votes]: There is a mistake early on:
\begin{align*}
x^n - (x - 1)^n &= x^n - \left(x^n - nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} - \cdots \right) \\
&= nx^{n-1} - \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \\
&= \sum_{i = 0}^{n-1} \binom{n}{i}(-1)^{n-i+1}x^i \\
\end{align*}
Whereas you said this is $x^{n-1} -(n-1)x^{n-2} + \ldots \pm (n-1)x^2 \mp 1 = (x - 1)^{n-1}$.
The easiest way to describe $\Delta^k$ of some polynomial is to write the polynomial in the basis $\{x^{\underline i} : i \ge 0\}$ where $x^{\underline i} = x(x-1)(x-2)\cdots(x-i+1)$. This is a nice basis for applying $\Delta$ since $\Delta(x^{\underline i}) = ix^{\underline{i - 1}}$.
The connection coefficients are the Stirling numbers of the second kind:
$$x^n = \sum_{k = 0}^n \begin{Bmatrix} n \\ k \end{Bmatrix} x^{\underline k}.$$
However, if all you care about is the identity $\Delta^n(x^n) = n!$ then it suffices that the coefficient of $x^{\underline n}$ is $1$ and the other coefficients are just some integers.
Now you can show (by induction, formally) that $\Delta^{k}(x^{\underline i}) = i(i - 1) \dots (i - k + 1)x^{\underline{i - k}} = i^{\underline k} x^{\underline{i - k}}$ and that this is $0$ if $k > i$.
It therefore follows that
$$\Delta^{n}(x^n) = \sum_{k = 0}^n \begin{Bmatrix} n \\ k \end{Bmatrix} \Delta^{n}(x^{\underline k}) = \begin{Bmatrix} n \\ n \end{Bmatrix} \Delta^{n}(x^{\underline n}) = n^{\underline n} x^{\underline 0} = n!.$$
Alternatively, you can prove this without the change of basis:

$\Delta^0(x^0) = 0!, \Delta^1(x^1) = x - (x - 1) = 1!$
Assume $\Delta^{i}(x^i) = i!$ for $i < n$
Then $\Delta^{i + 1}(x^i) = 0$ for all $i < n$ and in particular, $\Delta^{n - 1}(x^i) = i$ except $i = n - 1$.
Now apply $\Delta^{n-1}$ to the identity at the very top:

\begin{align*}
\Delta^{n-1}(\Delta^1(x^n)) &= \Delta^{n-1}\left(nx^{n-1} - \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \right) \\
&= \Delta^{n-1}(nx^{n-1}) \\
&= n\Delta^{n-1}(x^{n-1}) \\
&= n(n-1)!
\end{align*}
by induction.


TITLE: Hairy Points in Infinite Graphs (and Peano Continua)
QUESTION [3 upvotes]: I may have to throw this over to overflow, but I figured I would try here, first.  This is a question in continuum theory, but it reduces to a combinatorial/graph-theoretical question.  I am highly confident that this conjecture is both true and already known, so a reference would also be fine.  It might be in Menger's Kurventheorie, but I can't get a pdf nor read German.
We define the simple $n$-od to be any union of $n$ (non-degenerate) paths $A_1, \dots, A_n$ in some space $X$ that are pairwise-disjoint except at a single, common point $p$ (in other words, an $n$-pointed asterisk).  Shown below is a simple $3$-od in the plane.

We define the order of a point $p \in X$ for $X$ a topological space to be the infimum of cardinals $\alpha$ (though here we just care about finite vs. infinite) such that for every neighborhood (equivalently, open neighborhood) $V$ of $p$ there is a neighborhood (equiv. open neighborhood) $U \subset V$ of $p$ such that $\partial(U)$ has at most $\alpha$ points.  For example, the $3$-od as drawn above has order $1$ at the end points, order $2$ at points along edges and order $3$ at the vertex point $p$.
That the topology of the simple $n$-ods agrees with the picture given above is largely contained in this theorem: The order of the vertex point of any simple $n$-od is $n$.  The topology of a simple $n$-od always agrees with the 'standard one' in the plane, so there are no complications here.
A Peano Continuum is a compact, connected, locally connected metric space.  There is a theorem by Karl Menger called the $n$-Beisatz Theorem:
If $X$ is a Peano Continuum and $p \in X$ such that its order is at least $n$, then $X$ contains a simple $n$-od centered at $p$.
Thus any point of infinite order in a Peano Continuum is the vertex of simple $n$-ods for all $n \in \mathbb{N}$.

The space on the left in the figure above is called the Hairy Point (i.e. a 'harmonic fan' the lengths of whose edges converges to $0$), while the space on the right is called the Null Comb.  It's known that every Peano Continuum which isn't a (topological, finite) graph contains a Hairy Point or Null Comb.  But I am also pretty sure the following, which is what I want to prove, is true:
If $p \in X$ has infinite order then $X$ contains a copy of the Hairy Point centered at $p$.
So assuming $p$ has infinite order, then for every $n$ there is a simple $n$-od $T_n$ contained in $X$ centered at $p$.  The way that I have been trying to prove the theorem so far is to construct a sequence $S_k$ of simple $k$-ods centered at $p$ so that $S_{k+1} = S_k$ outside of a small neighborhood of $p$.  Then I can take a limit and prove the theorem in a routine manner.  But constructing $S_{k+1}$ given $S_k$ hasn't worked out so well.  The complication that arises is as follows.
We call the 'main' $n$ arcs in the simple $n$-od to be its chords.  If $A, B$ are two arcs with end point $p$, we say that $A$ is dense along B at $p$ if $p$ is an accumulation point of $A \cap B$, as shown below.

(Easy Case): So if $T_k$ is a simple $k$-od centered at $p$ and there is some $T_n$ with $n > k$ such that it has a chord $A$ which is not dense in any of the chords $A_1, \dots, A_k$ of $T_k$, then after taking a sub-arc of $A$ we can assume that $A$ doesn't intersect any of the $A_j$ at all and thus we obtain a simple $(k+1)$-od $A \cup A_1 \cup \cdots \cup A_k$.
(Hard Case): Thus we may as well assume that for every $n$ with $n > k$ we have that every chord of $T_n$ is dense at $p$ along some chord of $T_k$ (note that it may be dense along several chords of $T_k$, e.g. by spiraling).  Again we can assume that each chord of $T_n$ only intersects those $A_j$ that it's dense along.  By setting $n = (k+1) \cdot 2^k + 1$, then noting that $2^k$ is the number of subsets of chords of $T_k$ by the pigeonhole principle there is some subset $A_{m_1}, \dots, A_{m_n}$ of chords of $T_k$ such that there are at least $k+1$ chords , and thus $m_n + 1$ chords (shown in red below), that are dense along all of $A_{m_1}, \dots, A_{m_n}$.  By substituting the number of ordered subsets of $\lbrace 1, 2, \dots, k \rbrace$ we may assume that they first hit the chords $A_{m_1}, \dots, A_{m_n}$ in the same order (though there may be various manner of backtracking).  Let $R$ be as shown.

Finally, at this point the problem really is just combinatorial.  I have been trying various methods of removing sub-arcs of those relevant chords to obtain an $(m_n + 1)$-od but failing.  Note that the picture is slightly simplified, since for each $A_{m_j}$ we have no control over the order in which the chords of $R$ intersect it.  So I am wondering if the finite picture just isn't sufficient and it's some infinite combinatorial situation.
Anyone have any ideas?  Ultimately, I am trying to generalize the following procedure to arbitrarily many blue chords.

REPLY [0 votes]: If $X$ is a Peano continuum and ord$_X(p) \geq \aleph_0$ then $X$ contains a copy of the hairy point.
Proof: By the $n$-Beisatz Theorem there are simple $n$-ods $T_n \subset X$ centered at $p$.  Fix a $k$; we claim that there is a simple $(k+1)$-od $S \subset X$ centered at $p$ such that $S = T_n$ outside the $\frac{1}{k+1}$-neighborhood of $p$.
(Case 1): First assume there is an $n$ so that some chord $B$ of $T_n$ is not dense at $p$ along any chord in $T_k$, whose chords we denote by $A_1, A_2, \dots, A_k$.  Then by taking a (non-degenerate) sub-arc $B'$ of $B$ if necessary, we may assume that $B'$ is contained in the $\frac{1}{k+1}$-neighborhood of $p$ and $B' \cap A_j = \lbrace p \rbrace$ for all $j$.  Thus $B' \cup T_k = S$ is a simple $(k+1)$-od centered at $p$, $T_k \subset S$ and $S \setminus T_k$ is an arc contained in the $\frac{1}{k+1}$-neighborhood of $p$, as desired.
(Case 2): Now assume that for every $T_n$ with $n > k$ we have that every chord of $T_n$ is dense along at least one of $A_1, \dots, A_k$ at $p$ (such a chord may be dense along more than one $A_j$ at $p$, for example by spiraling).  As before, for each $T_n$ we may replace it by a simple $n$-od $T_n'$ so that every arc $B$ of $T_n'$ is contained in the $\frac{1}{k+1}$-neighborhood of $p$ and only intersects those $A_j$ along which it is dense at $p$.
Noting that $2^k$ is the number of subsets of $\lbrace 1, 2, \dots, k \rbrace$, by the pigeonhole principle if $n = (k+1) \cdot 2^k$ then for at least one subset $\lbrace m_1, \dots, m_\ell \rbrace$ of $\lbrace 1, 2, \dots, k \rbrace$ the simple $n$-od $T_n'$ contains some $k + 1$ chords $C_1, \dots, C_{k+1}$ each dense along all $A_{m_j}$ at $p$.  In particular, there are at least $\ell + 1$ of them $C_1, \dots, C_{\ell + 1}$.  Without loss of generality assume $m_j = j$ for $1 \leq j \leq \ell$, and let $R = C_1 \cup \cdots \cup C_{\ell + 1}$ be the simple $(l+1)$-od centered at $p$.  Recall that its chords only intersect the chords of $T_k$ along which it is dense at $p$, i.e. $A_1, \dots, A_\ell$.

Assume each $A_j$ is ordered so that its initial point is $p$.  Fix an $A_j$.  Among the first $\ell$ chords $C_1, \dots, C_\ell$ of $R$, let $i$ be the index such that $A_j \cap C_i$ is the \textit{last} point of $(C_1 \cup \cdots \cup C_\ell) \cap A_j$ with respect to the order on $A_j$, and denote it by $c(A_j)$.  This index exists since the $C_i$'s are closed and disjoint.  Consider the set $\lbrace c(A_1), \dots, c(A_\ell) \rbrace$.
Assume there is an index $h$, for $1 \leq h \leq \ell$, such that $h \notin \lbrace c(A_1), \dots, c(A_\ell) \rbrace$.  Then some collection of distinct chords $A_{j_1}, \dots, A_{j_h}$ of $T_k$ are last intersected by the same $C_i$ amongst $C_1, \dots, C_\ell$.  Assume it is $i$; let $f_i: I \rightarrow C_i$ be a homeomorphism with $f_i(0) = p$, and consider the sub-arcs $f_i([0,t])$.  If $C_i$ \textit{last} intersects $A_{j_1}, \dots, A_{j_h}$ in that order, by closedness of $C_i$ there is a maximal $t_0 \in (0,1)$ such that $f_i(t_0) \in A_{j_1}$.

By continuity of $f_i$, for $s > 0$ there are only finitely many times the set $f_i([s,1])$ changes which $A_{j_r}$ it last intersects (lest those times have an accumulation point, impossible since the $A_j$'s are mutually separated outside any neighborhood of $p$).  If all of $A_{j_1}, \dots, A_{j_h}$ are still last intersected by $f_i([0,t_0])$ (as opposed to some $C_j$ with $j \neq i$), then since $C_h$ is dense along each $A_{j_m}$ there is an $x \in C_h \cap A_{j_m}$ and a $y \in C_i \cap A_{j_m}$ such that $y < x$ with respect to the ordering on $A_{j_m}$.  Letting $s = f_i^{-1}(x) > 0$ we have, for any $j_m$, that after a finite number of excisions of terminal sub-arcs from $C_i$ we obtain $c(A_{j_m}) \neq i$.  Thus after finitely many such excisions we may assume that $c(A_{j_m}) = i$ for precisely one $m$.
Replacing $C_i$ as above we obtain a new simple $(\ell+1)$-od $R' \subset R$ so that $i$ appears in $\lbrace c_{R'}(A_1), \dots, c_{R'}(A_\ell) \rbrace$ precisely once.  Note that this replacement does not decrease the number of distinct elements of $\lbrace c(A_1), \dots, c(A_\ell) \rbrace$.  If still $h \notin \lbrace c(A_1), \dots, c(A_\ell) \rbrace$ we may repeat this same procedure on some other $C_j$'s from $R'$, obtaining an $R''$ and proceeding inductively.  Since for any $j$ there are only finitely many times $C_j$ changes which $A_i$ it last intersects, after finitely many iterations some $A_i$ will be last intersected by $C_h$.  Thus after a finite number of steps we obtain a simple $(\ell+1)$-od $W$ satisfying $\lbrace c(A_1), \dots, c(A_\ell) \rbrace = \lbrace 1, 2, \dots, \ell \rbrace$.  After taking a sub-arc, assume $C_{\ell+1}$ doesn't intersect any $A_{j_m}$ before one of the others does.

Order the first $\ell$ chords of $W$, which we also call $C_1, \dots, C_\ell$, so that they're in bijection with the chords $A_1, \dots, A_\ell$ such that $C_j$ is the chord of $W$ that last intersects $A_j$.  For each pair $(A_j, C_j)$ let $S_j$ be the arc from $p$ to the other end point of $A_j$ that is the concatenation of $C_j$ up to the last point of $A_j \cap C_j$, followed by the remaining terminal sub-arc of $A_j$ as shown above.  Let $S$ be the union of $C_{\ell+1}$, $S_j$ for $1 \leq j \leq \ell$ and $A_i$ for $i > \ell$.
We show that $S$ is a simple $(k+1)$-od centered at $p$, which will prove the claim.  By construction it is the union of $k+1$ arcs each having $p$ as an end point; we must show this is the only point where any pair of intersect.  If $i > \ell$ then $A_j$ doesn't intersect any other (alleged) chord of $S$ except at $p$ by construction of $T_n'$ and certainly no other $A_j$ does, since $T_k$ is a simple $k$-od.  Since $C_{\ell+1}$ doesn't intersect any $A_i$ for $i > \ell$ except at $p$ by construction of $T_n'$, and doesn't intersect any $S_j$ by construction of $W$, it only remains to show that $S_i \cap S_j = \lbrace p \rbrace$ if $i \neq j$.
Writing $S_h = D_h \oplus B_h$, where $D_h$ is the portion of the chord $C_h$ and $B_h$ is the portion of the chord of $A_h$ comprising $S_h$, for $x \in S_i \cap S_j$, if $x \in D_i \cap D_j$ then $x = p$ since $W$ is a simple $(\ell+1)$-od, and since $B_i \cap B_j = \varnothing$ we may assume $x \in D_i \cap B_j$, without loss of generality.  But then $C_i \cap A_j$ has a point of intersection after $C_j \cap A_j$, impossible by construction of $W$.  Thus $S_i \cap S_j = \lbrace p \rbrace$ for $i \neq j$, proving the claim.
Returning to the main problem, by taking sub-arcs of the chords of $T_n'$ if necessary, assume $T_n'$ doesn't contain any end points (besides $p$) of the chords of $T_k$ (since they are bounded away from $p$).  Thus we may assume each $S_j$ contains a non-trivial sub-arc of $A_j$ $(*)$.  Similarly, we may assume that any $A_i$ is only intersected by a $C_j$ at a time (with respect to the order on $A_i$) before it first intersects the complement of the $\frac{1}{k+1}$-neighborhood $B$ of $p$ ($\dagger$) (the left figure is permitted; the right is not).

Thus there is a sequence of simple $n$-ods $T_n$ centered at $p$ such that, by ($\dagger$), $T_{n+1} = T_n$ outside the $\frac{1}{n+1}$-neighborhood of $p$ - we showed more, namely that by $(*)$ we may assume all but one chord of $T_{n+1}$ is a concatenation of a non-trivial arc and a non-trivial terminal sub-arc $B_i^{(n)}$ of a unique chord of $T_n$ (in Case 1 we simply subdivide each chord of $T_n$ into two non-trivial arcs).  By taking sub-arcs if necessary, we may assume that $T_{n+2} \cap B_i = \varnothing$ for all $i$ such that $1 \leq i \leq n (\dagger \dagger)$.
We claim $\lim_n T_n = G$ is a hairy point centered at $p$.  As $T_{n+1} = T_n$ outside the $\frac{1}{n+1}$-neighborhood of $p$, this limit exists since it's eventually constant outside every neighborhood of $p$, and $p \in T_n$ for all $n$.  Thus $G$ is a continuum containing $p$.  Setting notation, let $H$ be the hairy point with 'chords' $H_1, H_2, \dots$ where $H_n$ is the segment of length $\frac{1}{n}$ emanating from $p$.  If $h_j$ is the linear homeomorphism of $I$ with $H_j$ with $h_j(0) = (0,0)$ then by the dyadics $2^{-n}$ in $H_j$ we mean the images under $h_j$ of the sequence $2^{-n}$ in $I$.
Let $f_0: T_2 \rightarrow H$ map $T_2$ homeomorphically to $H_1 \cup H_2$ with $f_0(p) = (0,0)$ and $f_0(A_j) = H_j$ for its chords $A_1, A_2$.  By assumption, two chords $C_1, C_2$ of $T_3$ terminate in non-trivial sub-arcs $B_1^{(2)}, B_2^{(2)}$ of the chords $A_1, A_2$ of $T_2$ respectively.

Define $f_1: T_3 \rightarrow H$ so that $f_1(C_j) = H_j$ for $j = 1, 2, 3$, while $f_1(B_j) = [2^{-1}, 1]$ in $H_j$ as per the notation defined above for $j = 1, 2$.  We may inductively define $f_n: T_{n+2} \rightarrow H$ so that it sends its chord $C_{n+2}$ homeomorphically onto $H_{n+2}$ with $f_n(p) = (0,0)$ and sends the corresponding $B_j^{(n+1)}$ of $T_{n+1}$ to $[2^{-n},1]$ in $H_j$.  By $(\dagger \dagger)$ we have that $B_j^{(n)} \subset B_j^{(n+1)}$, since $T_{n+2}$ doesn't intersect $B_j^{(n)}$ for any $j$ with $1 \leq j \leq n$.  Thus we may assume $f_n \equiv f_k$ on $B_j^{(k)}$ for all $j$ and for all $n \geq k + 2$.
If $x \in G = \lim_n T_n$ with $x \neq p$ then $x$ is contained in the (closed) complement $E$ of some $\frac{1}{n}$-neighborhood of $p$.  Since $T_k \cap E$ is a sequence of closed sets that is eventually constant, $G \cap E = T_N \cap E$ for some $N$, which thus contains $x$.  Thus $x$ is contained in $B_j^{(N+2)}$ for some $j$, on which the sequence $f_n$ is eventually constant, so $\lim_n f_n(x)$ is well-defined.  If $x = p$ then since $f_n (p) \equiv 0$ we have $\lim_n f(p) = 0$, so $\lim_n f_n = f: G \rightarrow H$ is a well-defined function on $G$.
We wish to show that $f$ is a homeomorphism.  If $x \in G$ with $x \neq p$ then as before $x \in B_j^{(N+2)}$ for some $j$ and for some $N$ sufficiently large; since the $f_n$'s are eventually constant, each a homeomorphism on $B_j^{(N+2)}$, we have that $f = \lim_n f_n$ is continuous at $x$.  If $x_k \rightarrow p$ with $x_k \in G$ then eventually $x_k$ is contained in the $\frac{1}{n+2}$-neighborhood of $p$ and therefore mapped to $[0, 2^{-n}]$ in some $H_{j(k)}$ by definition of $f_n$.  Thus $f$ is continuous at $p$, i.e. $f$ is continuous.
By construction, $f(G)$ contains the end point of $H_j$ for every $j$; since it also contains $p$ and is the continuous image of a connected set, $f(G)$ contains $H_j$ for all $j$.  Since $H = H_1 \cup H_2 \cup \cdots$, we have that $f$ is surjective.  If $x, y \in G$ are two distinct points other than $p$, then for $N$ sufficiently large we have $x \in B_{j(x)}^{(N+2)}$ and $y \in B_{j(y)}^{(N+2)}$.  Since $f(B_{j(x)}^{(N+2)}) \cap f(B_{j(y)}^{(N+2)}) = \varnothing$ if $j(x) \neq j(y)$ and $f$ is a homeomorphism on any given $B_{j(x)}^{(N+2)}$ we have that $f(x) \neq f(y)$.  Similarly, since $(0,0) \notin f(B_{j(x)}^{(N+2)})$ if $x = p$ and $ y \neq p$ we have $f(x) \neq f(y)$.
Thus $f$ is a continuous bijection of the compact $G$ and the Hausdorff $H$, and is therefore a homeomorphism.


text:
\section{Introduction}

This article is a study in the combinatorics of higher categories, in view of applications in higher-dimensional rewriting theory based on polygraphs \cite{burroni1993higher,lafont2007algebra,guiraud2016polygraphs}, also known as computads \cite{street1976limits}. It is technically most indebted to two different, and separate traditions of research:
\begin{enumerate}
	\item combinatorial presentations of strict $\omega$\nbd categories, such as pasting schemes \cite{johnson1989combinatorics,power1991pasting}, parity complexes \cite{street1991parity}, and most of all Steiner's directed complexes \cite{steiner1993algebra} and augmented directed complexes \cite{steiner2004omega};
	\item poset topology, in particular the theory of shellable and CW posets \cite{bjorner1980shellable,bjorner1983lexico,bjorner1984posets}, and the theory of constructible complexes \cite{hachimori2000constructible}.
\end{enumerate}
The two have much in common, yet as far as we know no significant connection has been made before.

The starting point of our work is the observation, made already by Burroni \cite{burroni1993higher}, that a polygraph is a ``formal CW complex'' inside the category $\omegacat$ of strict $\omega$\nbd categories, with certain models of directed $n$\nbd cells standing in for the topological $n$\nbd balls. This analogy has been effectively developed in a series of works \cite{metayer2003resolutions,metayer2008cofibrant}, culminating in the definition, by Lafont, M\'etayer, and Worytkiewicz, of a model structure on $\omegacat$ \cite{lafont2010folk} where polygraphs have the same role, as cofibrant objects, that CW complexes have in the classical Quillen-Serre model structure on topological spaces \cite[Section II.3]{quillen1967homotopical}.

Under some version of Grothendieck's homotopy hypothesis --- that the homotopy theory of topological spaces can be formulated in terms of higher groupoids or, \emph{a fortiori}, higher categories --- we would expect a direct relation between polygraphs and CW complexes, with polygraphs the strictly more general notion, admitting directed as well as undirected cells. However, this comparison is made difficult by the unrestricted combinatorics of $\omega$\nbd categorical (globular) pasting, which even for ``good'' classes of polygraphs that form presheaf categories may allow models of $n$\nbd cells whose natural geometric realisation is non-contractible; see the discussion in \cite[Appendix A]{henry2019non}.

Independently, and at the same time, S.\ Henry and the author came to the \emph{regularity} constraint as a solution \cite[Section 3]{hadzihasanovic2017algebra}. A regular polygraph is one whose $n$\nbd cell shapes have $k$\nbd dimensional boundaries shaped as $k$\nbd dimensional balls. Intuitively, any $\omega$\nbd categorical pasting diagram can be turned into a ball-shaped diagram by inserting enough units or degenerate cells: in this sense, regular polygraphs together with a notion of weak units should be sufficient as a foundation for a theory of higher categories. In \cite{henry2018regular}, Henry proved a version of the homotopy hypothesis for regular polygraphs together with an algebraic notion of ``regular composition'' and a non-algebraic notion of weak units.

In contrast with the algebraic approach of Henry's work, where classes of polygraphs are proved to form presheaf categories before presenting a specific site of definition, we take a combinatorial approach, starting from an explicit description of cell shapes. In higher-dimensional rewriting based on strict $\omega$\nbd categories, the $\omega$\nbd categorical algebra structure is used fruitfully to describe the pasting of generators, but has also unintended effects:
\begin{enumerate}
	\item there is no internal way to distinguish generators from other cells, and one is forced to use complex syntactic arguments to perform basic operations, such as ``surgery'' of cells or reversing the direction of a generator;
	\item one may unwittingly introduce ``Eckmann-Hilton''-type degeneracies.
\end{enumerate}
The second point is related to the well-known fact that strict $\omega$\nbd categories are strictly less general than weak higher categories \cite[Chapter 4]{simpson2009homotopy}, which also limits the contexts in which higher-dimensional theories presented as polygraphs can be intepreted.

We suggest that one should separate the combinatorics of pasting of generators of a polygraph from the algebra of composition in its intended models. In this approach, the combinatorics of pasting are handled by the shape category, so that elements of a presheaf on the shape category correspond to generators of a polygraph, and manipulating generators is easy. Such presheaves are then used as underlying structures of a model of higher categories (which also ``has composition''), in which one can interpret higher-dimensional theories even if they do not internally ``have composition''.

The idea of basing a model of higher categories on combinatorial presentations of pasting diagrams (richer than the ``special'' shapes such as globes, cubes, oriented simplices) was present in Kapranov and Voevodsky \cite{kapranov1991infty}, whose main results are notoriously incorrect, and seems to not have been pursued afterwards. We outline such a model in Appendix \ref{sec:represent}, and develop the idea further in a subsequent paper \cite{hadzihasanovic2019rds}.

\subsubsection*{Constructibility and lax Gray products}

Another motivation of this work is the computation of lax Gray products of polygraphs; in \cite[Chapter 2]{hadzihasanovic2017algebra}, we gave evidence of their relevance to categorical universal algebra and rewriting.

The lax Gray product of $\omega$-categories \cite{al2002multiple}, an oriented, non-symmetric counterpart to the cartesian product of spaces, is notoriously difficult to express in the algebra of globular composition. Steiner's definition \cite{steiner2004omega}, while elegant, requires a long detour via augmented directed complexes, and the proof of its validity has been completed only recently, with some effort, by Ara and Maltsiniotis \cite{ara2016joint}. The \emph{join} is another monoidal structure which can be derived from lax Gray products, and similar considerations apply.

The problem may be characterised as follows. An $\omega$\nbd categorical $n$\nbd cell $x$ has, for $k < n$, an input and output $k$\nbd boundary, $\bord{k}{+}x$ and $\bord{k}{-}x$, which in good cases is $k$\nbd dimensional; a globular pasting of two cells is, topologically, a gluing along an entire $k$\nbd boundary (output for one cell, input for the other). If $x$ is an $n$\nbd cell and $y$ an $m$\nbd cell, their lax Gray product $x \tensor y$ is $(n+m)$\nbd dimensional, and its input $(n+m-1)$\nbd boundary splits as
\begin{equation} \label{eq:laxgray_decomp}
	(\bord{n-1}{-}x \tensor y) \cup (x \tensor \bord{m-1}{\pm}y),
\end{equation}
the second sign depending on the parity of $n$. But this decomposition is \emph{not} a globular pasting: the two parts only share a portion of their $(n+m-2)$\nbd boundary. The situation only becomes more complicated for lower-dimensional boundaries.

With that in mind, we turned to the combinatorics of face posets of regular CW complexes, or CW posets \cite{bjorner1984posets}. Certain conditions that have been studied for CW posets, such as shellability \cite{bjorner1980shellable}, turn out to have clear oriented analogues, suggesting a notion of composition alternative to the globular model.

In particular, the oriented analogue of \emph{constructibility} \cite{hachimori2000constructible} individuates a class of pasting diagrams which are iterated composites of $n$\nbd cells along an $(n-1)$\nbd dimensional diagram in their boundary; we call this kind of composition a \emph{merger}. The decomposition (\ref{eq:laxgray_decomp}) is, in fact, a merger; thus the oriented version of constructibility is very naturally compatible with lax Gray products and joins. In \cite{hadzihasanovic2018weak}, we gave a reformulation of bicategory theory where the algebra of composition is based on mergers instead of globular composition, and showed that it is equivalent to the standard one.

We introduce \emph{constructible directed complexes} as a combinatorial presentation of pasting diagrams obtained by iterated mergers, which we call \emph{constructible molecules}. The terminology is a twist on Steiner's theory of directed complexes \cite{steiner1993algebra}, whose ``molecules'' are pasting diagrams obtained by iterated globular compositions. It is not immediately obvious that constructible molecules are molecules, and that constructible directed complexes are directed complexes; we prove it in Section \ref{sec:omegacat}. Moreover, constructible molecules are regular: the geometric realisation of a constructible $n$\nbd molecule is a regular CW $n$\nbd ball, as shown in Section \ref{sec:geometric}.

We define \emph{constructible polygraphs} as a generalisation of constructible directed complexes: they are presheaves on a category of constructible atoms (``indecomposable'' constructible molecules) and inclusions. The monoidal structures of lax Gray products and joins canonically extend, and so does geometric realisation: the geometric realisation of a constructible polygraph is a CW complex with one cell for each of its elements.

Moreover, constructible polygraphs can be realised as $\omega$\nbd categories, but we do not have a complete understanding of this construction yet: in particular, while we conjecture that the $\omega$\nbd categorical realisation of a constructible polygraph always admits the structure of a polygraph with one generator for each of its elements, we were only able to prove it in restricted cases.

\subsubsection*{Structure of the article}
We start in Section \ref{sec:basic} by recalling some notions of order theory and poset topology. Then, we define an orientation on graded posets, in the form of a labelling of their Hasse diagrams, and progressively introduce the components of the definition of constructible directed complex: oriented thinness (Definition \ref{dfn:thinness}) and constructible molecules (Definition \ref{dfn:globes}). We define the category $\globpos$ of constructible directed complexes and inclusions, and prove some of its properties, most importantly that constructible molecules do not have non-trivial automorphisms (Lemma \ref{lem:noautomorph}). We discuss some special classes of constructible molecules, including a sub-class of the opetopes, and then define the category $\cpol$ of constructible polygraphs.

In Section \ref{sec:properties}, we expand on the combinatorics of constructible directed complexes with the use of two main tools: substitutions (Construction \ref{cons:substitution}) and mergers (Construction \ref{cons:mergers}), where constructible molecules with two maximal elements are sequentially replaced with \emph{atoms}, constructible molecules with a greatest element. We prove that any constructible molecule can be reduced by mergers to an atom (Lemma \ref{lem:atomicmerge}), and we use this to prove that constructible molecules satisfy the globularity condition: their input and output boundary have the same boundaries (Theorem \ref{thm:globularity}). We then consider some consequences of globularity: $n$\nbd dimensional atoms are classified by pairs of constructible $(n-1)$\nbd molecules with isomorphic boundaries (Proposition \ref{prop:atomicnglobe}), and any constructible molecule can be reduced to an atom by a sequence of mergers (Lemma \ref{lem:standardmerge}). We also give a direct characterisation of the $k$\nbd dimensional boundaries of a constructible molecule (Proposition \ref{prop:globelike}).

In Section \ref{sec:monoidal}, we focus on monoidal structures and constructions. We define the suspension (Construction \ref{cons:suspension}), the lax Gray product (Construction \ref{cons:laxgray}), the join (Construction \ref{cons:join}), and the duals (Construction \ref{cons:jdual}) of oriented graded posets, and prove that constructible directed complexes are closed under each of them. We also show that some common classes of higher-categorical cell shapes --- the globes, the cubes, and the oriented simplices --- are constructible molecules. We then study the extensions of lax Gray products and joins to constructible polygraphs, and prove that the first are part of a monoidal biclosed structure, and the second of a locally biclosed structure (Proposition \ref{prop:locallybiclosed}) on $\cpol$. The latter result relies on a \emph{slice} construction for constructible polygraphs (Construction \ref{cons:slices}).

In Section \ref{sec:omegacat}, we first recall the definitions of $\omega$\nbd category and of polygraph. Then, we rephrase Steiner's definition of directed complexes in our notation (Definition \ref{dfn:directedcomplex}), and prove that constructible directed complexes are directed complexes (Theorem \ref{thm:globmolec}). We use this to define a realisation functor for constructible polygraphs in the category $\omegacat$ of $\omega$\nbd categories, and give some partial results on the relation between the monoidal structures of $\cpol$ and those of $\omegacat$.

In Section \ref{sec:polygraphs}, we study the relation between constructible polygraphs and polygraphs as freely generated $\omega$\nbd categories. The realisation of a constructible polygraph in $\omegacat$ admits the structure of a polygraph if constructible directed complexes are ``freely generating'' in the sense of Definition \ref{dfn:freegen} (Theorem \ref{thm:rpolintopol}), which we conjecture to be true, but were not able to prove in the general case. Nevertheless, we give some criteria for free generation which are weaker than loop-freeness as considered by Steiner (Proposition \ref{prop:max_free} and Corollary \ref{cor:flow-conn-split}), and may serve as intermediate steps towards a general proof.

In Section \ref{sec:geometric}, we define the geometric realisation of an oriented graded poset. We prove that the underlying poset of a constructible molecule is recursively dividable in the sense of Hachimori (Proposition \ref{prop:recdivid}), and derive that the geometric realisation of a constructible $n$\nbd molecule is an $n$\nbd ball (Theorem \ref{thm:regularball}). We extend the geometric realisation to constructible polygraphs, and show that the geometric realisation of a constructible polygraph is a CW complex with one cell for each of its elements (Theorem \ref{thm:georegpoly}).

In Appendix \ref{sec:appendix}, we give a more detailed proof of the fact that lax Gray products of constructible molecules are constructible molecules. In Appendix \ref{sec:represent}, we present the first steps in an approach to weak higher categories based on constructible polygraphs, via the key notion of universality of a cell at a constructible submolecule of its boundary (Definition \ref{dfn:universality}), up to the definition of representable constructible polygraph (Definition \ref{dfn:represent}).

\subsubsection*{Outlook and open problems}

Our first concern is to develop the ideas of this work into a good framework for higher-dimensional rewriting, including a suitable model of higher categories to serve as semantics. The main limitation that we see is that operations or rewrites with a ``nullary'' input cannot be directly modelled in a constructible or regular polygraph; this should be overcome with an adequate theory of units or degeneracies simulating nullary inputs. We do that in \cite{hadzihasanovic2019rds}.

We have stated three conjectures of quite different nature. Conjecture \ref{conj:subglobe} is a problem in the same vein as many ``hierarchy problems'' in combinatorial topology, investigating the relation between different properties of posets or simplicial complexes (being shellable, partitionable, Cohen-Macaulay...); it may be rephrased as (oriented) constructibility not being the same as \emph{extendable} constructibility. We believe that a counterexample should exist, but were not able to find one.

On the other hand, Conjecture \ref{conj:monoidality}, that the lax Gray product and join of constructible polygraphs coincide with those of $\omega$\nbd categories \cite{ara2016joint}, is almost obviously true, and the fact that there is no obvious proof says more about how difficult it is to work with these structures on $\omegacat$ than anything else. Similarly, we are almost certain that our positive opetopes (Definition \ref{dfn:pope}) coincide with those defined by Zawadowski \cite{zawadowski2017positive}, and the only effort required is setting up a framework in which to compare the two.

The most important open problem is Conjecture \ref{conj:freegen}. Because of Example \ref{exm:nonsplit}, the strategy used essentially in all previous proofs of free generation does not apply to general constructible directed complexes. Proving the conjecture would be a useful sanity check, reassuring us that whatever we do is a continuation of the general theory of polygraphs. A refutation, on the other hand, would not rule out constructible polygraphs as a notion of ``directed space'', but it would make them somewhat incomparable with strict $\omega$\nbd categories, and indicate that the combinatorics of higher-categorical pasting are even subtler than we think.

Finally, given that constructible polygraphs rely on quite simple data structures, it would be interesting to pursue a formal implementation in the vein of \cite{curien2019syntactic} for opetopes and opetopic sets. A potential obstacle is that our definition of constructible molecules is combinatorial, but it is not enumerative: so far, we have no inductive procedure for enumerating isomorphism classes of constructible molecules without repetition.

\subsubsection*{Errata and notes on an earlier version}
With respect to the first preprint version of this article, we have made extensive changes to the terminology: what are now \emph{constructible molecules} and \emph{constructible directed complexes} used to be called \emph{globes} and \emph{globular posets}, while \emph{constructible polygraphs} used to be \emph{regular polygraphs}.

The fact that constructible directed complexes are directed complexes in Steiner's sense is not obvious, and a good part of this article is devoted to its proof; but once it is established, it seems unnecessary to introduce an entirely separate set of terms. We thus suggest adopting Steiner's ``molecule'' for the most general combinatorial presentation of an $\omega$\nbd categorical composite, and prefixing it with various adjectives. It seems likely that there should be a hierarchy of classes of molecules as rich as that of their unoriented counterparts in poset topology.

We use ``constructible'' here instead of ``regular'' to avoid confusion with S.\ Henry's notion of regular polygraph, which we find is the more general and appropriate use of the term: constructibility implies regularity, but the converse is not true (see Remark \ref{rmk:nonconstructible}). We discuss ``regular directed complexes'' in the follow-up \cite{hadzihasanovic2019rds}.

The first version also contained some mistakes: Lemma 2.23 was wrong, and so were Theorem 4.41 and the lemmas leading to it, whose proof made crucially use of it. The corresponding results in Section \ref{sec:polygraphs} here have now additional hypotheses, which do not hold for all constructible directed complexes. In particular, it is not true, as claimed previously, that all molecules in a constructible directed complex are split. The validity of Theorem 4.43 from the first version, that the $\omega$\nbd categorical realisations of constructible polygraphs are all polygraphs, is now conditional on Conjecture \ref{conj:freegen}, which will need a different proof strategy.

\subsubsection*{Acknowledgements}
This work was supported by a JSPS Postdoctoral Research Fellowship and by JSPS KAKENHI Grant Number 17F17810. Many thanks to Joachim Kock and Jamie Vicary for their feedback on the parts which overlap with my thesis. For the revised version, I would like to thank Dimitri Ara, Yves Guiraud, Simon Henry, and Georges Maltsiniotis for helpful feedback and suggestions.


TITLE: Determine the smallest value of $k$ corresponding to the level of significance $α = 0.01$
QUESTION [0 upvotes]: You have a coin and $p$ is the probability of heads.
  You throw the coin until you obtain heads for the first time.
  You want to test $H_0:p=1/2$ against $H_1:p<1/2$. You reject $H_0$ in favor of $H_1$ if $T ≥ k$, where $T$ is the number of the throw which yielded heads for the first time and $k$ an integer.
  Determine the smallest value of $k$ corresponding to the level of significance $α = 0.01$.

Answer $k=8$
I can't find the correct solution for this question, how it's possible to have $H_1:p<1/2$ and then they say that we reject $H_0$ in favor of $H_1$ if $T ≥ k$, shouldn't they be in the same direction?
I have to compute $P(T>k|H_0)=P(T>k|p=1/2)$ right? How can I compute it?
Can someone help me? Thanks :)

REPLY [1 votes]: You are in the right direction. You need to solve the inequality $P(T\geq k|p=\frac{1}{2})\leq 0.01$.
(I hope this makes sense for you, as this is just the definition of level of the test. Some authors also call it size. But there is a distinction between them. For e.g. look here. To be more precise, if you are given the size for the  test of a simple hypothesis, we would have used equality sign above. )
$T$ being the total no. of throws required for obtaining the first head, its probability distribution can be computed as mentioned in the comment above. I would apply rather an intuitive method to do the same.
\begin{equation}
\begin{aligned}
P(T\geq k|p=\frac{1}{2})&=1-P(T<k)=1-\left[P(T=1)+P(T=2)+\dots_+P(T=k-1)\right]  \\
&=1-\left[\frac{1}{2}+\left(\frac{1}{2}\right)^2\dots+\left(\frac{1}{2}\right)^{k-1}\right]\\&\mbox{(When we say that $T=i, i>1$, we mean that the first $(i-1)$ results in tails.  }\\
&\hspace{2mm} \mbox{Head is obtained only on the $i^{th}$ toss. Since the trials are independent, this  }\\
&\hspace{2mm}\mbox{probability is just $\left(\frac{1}{2}\right)^{i-1}.\frac{1}{2}=\left(\frac{1}{2}\right)^{i}$})\\
&=\left(\frac{1}{2}\right)^{k-1}\leq 0.01
\end{aligned}
\end{equation}
The least value of $k$ for which this holds is $8$.


text:
\section{Conclusion and Future Work}\label{section: future}
In the context of centralized distributed computing environment, we present a novel distributed Newton-type method, named DINO, which enjoys several advantageous properties. DINO is guaranteed to converge under minimal assumption, its analysis is simple and intuitive, it is practically parameter free, and it can be applied to arbitrary non-convex functions and data distributions. Numerical simulations highlight some of these properties.

The following is left for future work.
First, characterizing the relationship between the hyper-parameters $\theta$ and $\phi$ of DINO.
As was discussed, they have intuitive effects on the algorithm and they are easy to tune.
However, there is a non-trivial trade off between them that will be explored in future work.
Second, analysing the connection between DINO and over-parameterized problems.
Finally, extending the theory of DINO to alternative forms of line search, such as having each worker perform local line search and then aggregating this information in a way that preserves particular properties.


TITLE: Smallest positive integer solution
QUESTION [0 upvotes]: I am new to Abstract Algebra, and in the 5th edn. of the book by Hillman,  Alexanderson have come across the below question #25 in section 1.2.
For each of the following integers d, find the smallest positive integer n such that $d|(10^{n} - 1).$
(a) d = 7,
(b) d = 11,
(c) d = 13,
(d) d = 37,
(e) d = 77,
(f) d = 91,
(g) d = 407.
I am unable to make sense in the introductory chapter of this question.
If could provide some link to solve, then would be highly welcome.

REPLY [0 votes]: Hint. By Little Fermat's theorem, if $gcd(d,10) = 1$, then  $d$ divides $10^{\varphi(d)}-1$ where $\varphi$ is the Euler's totient function. The smallest  positive integer $n$ such that $d$ divides $10^{n}-1$ is called the multiplicative order of $10$  modulo $d$ and it is a divisor of $\varphi(d)$.
(a) For $d=7$, $\varphi(7)=7-1=6$ and $10^3-1=999$ is not divisible by $7$. So the order of $10$ modulo $7$ is $6$.
(b) For $d=11$, $\varphi(11)=11-1=10$. Note that $10^2-1=99$ is divisible by $11$, then the order of $10$ modulo $11$ is $2$.
(c) For $d=13$, $\varphi(13)=12$. We have that $10^2-1=99$, $10^3-1=999$, $10^4-1=9999$ are not divisible by $13$,  but
$$10^6-1\equiv (-3)^6-1=(-27)^2-1\equiv (-1)^2-1=0\pmod{13},$$
hence the order of $10$ modulo $13$ is $6$.
Are you able to go on?