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\begin{document}

\title[Joint large deviation  result for empirical measures of the CGRG]
{\Large Joint large deviation  result  for empirical measures of the near intermediate coloured random geometric  graphs}

\maketitle
\thispagestyle{empty}
\vspace{-0.5cm}

\renewcommand{\thefootnote}{}
\footnote{\textit{AMS Subject Classification:} 60F10, 05C80}
\footnote{\textit{Keywords: }  Random geometric graph, Erd\H
os-R\'enyi graph, coloured random  geometric graph, typed graph,
joint large deviation principle, empirical pair measure, empirical
measure, degree distribution, entropy, relative entropy, isolated
vertices .}
\renewcommand{\thefootnote}{1}

\begin{quote}{\small }{\bf Abstract.}
 We prove joint large deviation principle  for the \emph{ empirical pair measure} and  \emph{empirical locality measure}   of  the \emph{near intermediate} coloured
random  geometric graph models, see (Canning \& Penman, 2003), on $n$ points picked uniformly in
$[0,1]^d,$  for  $d\in\N.$ From  this result we obtain large
deviation principles for the \emph{number of  edges per vertex}, the \emph{degree distribution and the
proportion of isolated vertices } for the \emph{near intermediate}
random  geometric graph models.
\end{quote}\vspace{0.3cm}

\vspace{0.3cm}
\section{Introduction}
In this article we study the coloured  geometric  random
graph CGRG,  where
 $n$ points or  vertices or  nodes  are picked uniformly at random in $[0, 1]^d,$   colours or spins are  assigned  independently  from  a  finite  alphabet $\skriu$
and   any two points  with colours $a_1,a_2\in\skriu$ distance at most $r_n(a_1,a_2)$ apart are connected. This random  graph  models,  which  has the geometric  random graph (see Penrose,  2003) as  special  case,    has  been suggested  by  see (Canning \& Penman, 2003) as  a  possible  extension  to  the  coloured  random  graph studied  by  (OConnell ,1998), (Biggins and Penman, 2009), (Doku-Amponsah and
Moerters, 2010), ( Doku-Amponsah, 2006), (Bordenave and Caputo,
2013), (Mukherjee, 2013)  and (Doku-Amponsah, 2014[b]).

Until  recently few  or  no  large  deviation  result  about the CGRG have  been  found. Doku-Amponsah
(2014[b]) proved  some joint large  deviation principle  for empirical pair measure  and  the  empirical locality  measure of the CGRG, where $n$
points are uniformly chosen  in $[0,1]^d$, colours or spins are  assigned by drawing without  replacement from  the  pool of $n\varpi_n(a_1)$ colours and  $n\omega_n(a_1,a_2)$ edges, $a_1,a_2\in\skriu$
are randomly inserted  among  the  points for  some  colour  law  $\varpi_n:\skriu\to[0,1]$  and  edge law $\omega_n:\skriu\times\skriu\to[0,\infty).$

This  article presents a  full large deviation  principle (LDP) for
the empirical pair measure and the  empirical  locality measure  of the near intermediate RGG.Refer to
(Doku-Amponsah and Moerters) for  similar result  for  the coloured  random graphs. From  this  large  deviation results  we  obtain LDP for  graph quantities such as  \emph{number  of  edges per  vertex, the  degree distribution  and the proportion of isolated vertices}  of  geometric random  graphs  in  the  intermediate case.
See,(OConnell ,1998), (Biggins and Penman, 2009), (Doku-Amponsah and
Moerters, 2010), ( Doku-Amponsah, 2006), (Bordenave and Caputo,
2013), (Mukherjee, 2013)  and (Doku-Amponsah, 2014[b]) for similar result for the Erd\"{o}-Renyi
graphs.

In  the  course  of  the  proof  of  this  joint LDP  we  obtain a joint LDP  for
\emph{empirical colour measure} and  \emph{empirical pair
measures} for the CGRG by   the  exponential change  of  measure techniques, see  (Doku-Amponsah \&
 Moerters, 2010) or (Doku-Amponsah, 2006). Our  prove  of  the  main result  uses the  method  of  mixture  idea also used  by  (Doku-Amponsah \&
 Moerters, 2010) or (Doku-Amponsah, 2006).  To  be  specific  about  this  idea,  we  mix the  joint  LDP of  \emph{empirical colour measure} and for \emph{empirical pair
measures},  and joint LDP for  \emph{empirical  pair  measure  and the  empirical  locality  measure} given  the \emph{empirical colour measure} and for \emph{empirical pair
measures},  (see Doku-Amponsah, 2015[b])  to  obtain  the  full joint LDP for  \emph{empirical  pair  measure  and the  empirical  locality  measure} of  CGRG  models.

In  the remainder of the paper we state and prove  our  LDP results.
 In Section~\ref{mainresults}  we   state our LDPs,
 Theorem~\ref{Rgg}, Corollary~\ref{RRiv}and
 Theorem~\ref{main2}.  In Section~\ref{proofmain2} we combine (Doku-Amponsah,Theorem~2,1, 2014[b])
 and (Doku-Amponsah,Theorem~2.1, 2014[c]) to obtain the
Theorem~\ref{main2}, using the setup and result of (Biggins, 2004)
to `mix' the LDPs. The paper concludes with the proofs of our  main
results Theorem~\ref{ERdd} and Corollary~\ref{RRiv} which are given
in Section~\ref{proofcorollaries}.

\section{Statement of the results}\label{mainresults}

\subsection{The joint LDP for empirical pair  measure and  empirical locality  measure  of CGRG.}  In this subsection  we shall look at  a
more general model of random  geometric graphs, the coloured RGGs in
which the connectivity radius depends on  the type or colour or
symbol or  spin of the  nodes. The empirical pair measure and the
empirical locality  measure are our main object of study.

Given   a probability measure $\nu$ on $\skriw$ and a function
$r_n\colon\skriw\times\skriw\rightarrow (0,1]$ we may define the
{\em randomly coloured random geometric graph} or simply
\emph{coloured random geometric graph}~$X$ with $n$ vertices as
follows: Pick vertices  $W_1,...,W_n$  at  random  independently
according to the uniform distribution on $[0,\,1]^2.$ Assign to each
vertex $W_j$ colour $X(W_j)$ independently according to the {\em
colour law} $\mu.$ Given the colours, we join any two vertices
$W_i,W_j$,$(i\not=j)$ by an edge independently of everything else,
if $$\|W_i-W_j\|\le r_n\big[X(W_i),X(W_j)\big].$$ In this  article
we  shall  refer to $r_n(a,b),$   for $a,b\in\skriw$ as a connection
radius,  and always consider
$$X=((X(W_i),X(W_j))\,:\,i,j=1,2,3,...,n),E)$$ under the joint law of
graph and colour. We interpret $X$ as  coloured RGG with vertices
$Y_1,...,Y_n$ chosen at  random uniformly   and independently from
the vertices space $[0,1]^2.$ For  the  purposes of  this  study we
restrict ourselves to  the near intermediate cases .i.e. the
connection radius $r_n$ satisfies the condition $n r_n^d(a,b) \to
C(a,b)$ for all $a,b\in \skriw$, where $C\colon\skriw^2\rightarrow
[0,\infty)$ is a symmetric function, which is not identically equal
to zero.

For any finite or countable set $\skriw$ we denote by
$\skrip(\skriw)$ the space of probability measures, and by
$\tilde\skrip(\skriw)$ the space of finite measures on $\skriw$,
both endowed with the weak topology.  By  convention we write
$\N=\{0,1,2,...\}.$

We associate with any coloured graph $X$ a probability measure, the
\emph{empirical colour measure}~$\skril^1\in\skrip(\skriw)$,~by
$$\skril_X^{1}(a):=\frac{1}{n}\sum_{j=1}^{n}\delta_{X(W_j)}(a),\quad\mbox{ for $a_1\in\skriw$, }$$
and a symmetric finite measure, the \emph{empirical pair measure}
$\skril_X^{2}\in\tilde\skrip_*(\skriw^2),$ by
$$\skril_X^{2}(a,b):=\frac{1}{n}\sum_{(i,j)\in E}[\delta_{(X(W_i),X(W_j))}+
\delta_{((X(W_j),X(W_i))}](a,b),\quad\mbox{ for $(a, b)\in\skriw^2$.
}$$ The total mass $\|\skril_X^2\|$ of the empirical pair measure is
$2|E|/n$.  Finally we define a further probability measure, the
\emph{empirical neighbourhood measure}
$\skrim_X\in\skrip(\skriw\times\N)$, by
$$\skrim_X(a,\ell):=\frac{1}{n}\sum_{j=1}^{n}\delta_{(X(W_j),L(W_j))}(a,\ell),\quad
\mbox{ for $(a,\ell)\in\skriw\times\N$, }$$ where
$L(v)=(l^{v}(b),\,b\in\skriw)$ and $l^{v}(b)$ is the number of
vertices  of colour $b$  connected to vertex $v$.

For any $\mu\in\skrip(\skriw\times\N^{\skriw})$we  denote by $\mu_1$
the $\skriw-$ marginal of $\mu$ and for  every
$(b,a)\in\skriw\times\skriw,$ let $\mu_2$  be the law  of  the pair
$(a,l(b))$ under  the  measure $\mu.$ Define the  measure (finite),
$\langle\mu(\cdot,\ell),\,l(\cdot)\rangle\in\tilde\skrip(\skriw\times\skriw)$
by
$$\skrih_2(\mu)(b,a):=
\sum_{l(b)\in\N}\mu_2(a,l(b))l(b), \quad\mbox{ for $a,b\in\skriw$}$$
and  write $\skrih_1(\mu)=\mu_1.$ We define the function  $\skrih
\colon \skrip(\skriw\times\N^{\skriw}) \to \skrip(\skriw) \times
\tilde\skrip(\skriw \times \skriw)$ by
$\skrih(\mu)=(\skrih_1(\mu),\skrih_2(\mu))$ and note that
$\skrih(\skrim_X)=(\skril_X^1, \skril_X^2).$ Observe  that
$\skrih_1$ is a continuous function  but $\skrih_2$ is
\emph{discontinuous} in the weak topology. In particular, in  the
summation  $\displaystyle \sum_{l(b)\in\N}\mu_2(a,l(b))l(b)$ the
function $l(b)$ may be unbounded and  so  the  functional
$\displaystyle \mu\to\skrih_2(\mu)$ would not be continuous in the
weak topology. We call a pair of measures
$(\varpi,\mu)\in\tilde{\skrip}(\skriw\times\skriw)\times\skrip(\skriw\times\N^{\skriw})$
\emph{sub-consistent} if
\begin{equation}\label{consistent}
\skrih_2(\mu)(b,a) \le \varpi(b,a), \quad\mbox{ for all
$a,b\in\skriw,$}
\end{equation}
and \emph{consistent} if equality holds in \eqref{consistent}. For a
measure $\varpi\in\tilde\skrip_*(\skriw^2)$ and a measure
$\omega\in\skrip(\skriw)$, define
$${{\mathfrak H}_C^{d}}(\varpi\, \| \, \omega ):=
H\big(\varpi\,\|\,\rho(d)C\omega\otimes\omega\big)+\rho(d)\|
C\omega\otimes\omega \| -\|\varpi\|\, ,$$ where the measure
$C\omega\otimes\omega\in\tilde\skrip(\skriw\times\skriw)$ is defined
by $C\omega\otimes\omega(a,b)=C(a,b)\omega(a)\omega(b)$ for
$a,b\in\skriw$. It is not hard to see that $\mathfrak
H_C^{d}(\varpi\,\|\,\omega)\ge 0$ and equality holds if and only if
$\varpi=\rho(d) C\omega\otimes\omega$. For every
$(\varpi,\mu)\in\tilde\skrip_*(\skriw\times\skriw) \times
\skrip(\skriw\times\N)$ define a probability measure
$Q=Q[\varpi,\mu]$ on $\skriw\times\N$ by
$$Q(a\,,\,\ell):=\mu_{1}(a)\prod_{b\in\skriw} e^{-\frac{\varpi(a,b)}{\mu_1(a)}} \,
\frac{1}{\ell(b)!}\,
\Big(\frac{\varpi(a,b)}{\mu_1(a)}\Big)^{\ell(b)}, \quad\mbox{for
$a\in\skriw$, $\ell\in\N$} .$$
We  assume  $d\ge 2$ is finite and
  write
  $$\rho(d)=\sfrac{\pi^{d/2}}{\Gamma\big(\sfrac{(d+2)}{2}\big)},$$
  where  $Gamma$  is  the  gamma  function.

We now  state the principal theorem
in  this section the LDP for the empirical pair measure and the
empirical neighbourhood measure.

\begin{theorem}\label{randomg.LDM}\label{main2}
Suppose that $X$ is a coloured RGG  graph  with colour law $\mu$ and
connection radii $ r_n\colon\skriw\times\skriw\rightarrow[0,1]$
satisfying $n r_n^d(a,b) \to C(a,b)$ for some symmetric function
$C\colon\skriw\times\skriw\rightarrow [0,\infty)$ not identical to
zero. Then, as $n\rightarrow\infty,$ the pair
$(\skril_X^2,\,\skrim_X)$ satisfies an LDP in
$\tilde{\skrip}_*(\skriw\times\skriw)\times\skrip(\skriw\times\N)$
with good rate function
$$\begin{aligned}
J(\varpi,\mu)=\left\{
\begin{array}{ll}H(\mu\,\|\,Q)+H(\mu_1\,\|\,\nu)+
\sfrac{1}{2}\, {\mathfrak H}_C^{d}(\varpi\,\|\,\mu_1 ) & \mbox {if $(\varpi,\mu)$ consistent  and  $\mu_1=\varpi_2,$   }\\
\infty & \mbox{otherwise.}
\end{array}\right.
\end{aligned}$$
\end{theorem}

\begin{remark}

Note that on  typical coloured RGG graph
 we  have, $\omega=\mu_1,$
$\varpi= \rho(d) C \,\mu \otimes \mu$ and
$$\mu(a,\ell)=\nu(a)\prod_{b\in\skriw}e^{-\rho(d) C(a,b)\nu(b)}\, \frac{(\rho(d)C(a,b)\nu(b))^{\ell(b)}}{\ell(b)!},\qquad\mbox{for
all $(a,\ell)\in\skriw\times\N$} .$$ This is the law of a pair
$(a,\ell)$ where $a$ is distributed according to $\mu$ and, given
the value of $a$, the random variables $\ell(b)$ are independently
Poisson distributed with parameter $\rho(d) C(a,b)\nu(b)$.  Hence,
as  $n\to\infty,$  the  empirical neigbourhood  measure
$\skrim_X(a,\ell)$ converges to  $\mu(a,\ell)$  in probability.
\end{remark}

\begin{cor}\label{ERdd}\label{Rgg}
Suppose $D$ is the degree distribution of  the random graph
$\skrig(n,r_n),$  where the connectivity radius $r_n\in (0,1]$
satisfies $n r_n^d \to c \in (0,\infty)$. Then ,as $n\to\infty$, $D$
satisfies an LDP in the space $\skrip(\N \cup \{0\})$ with good rate
function
\begin{equation}\label{randomg.ratedeg}
\begin{aligned}
\eta_1(\delta)= \left\{ \begin{array}{ll}  \frac 12 \, \langle
\delta \rangle\, \log \Big( \sfrac{\langle \delta\rangle}{\rho(d)c}
\Big)
 - \frac 12 \, \langle \delta \rangle  +  \sfrac {\rho(d)c}{2} +H (d\,\|\,q_{\langle \delta\rangle}),
 & \mbox { if $\langle \delta\rangle< \infty,$ }\\[2mm]
\infty & \mbox { if $\langle \delta\rangle= \infty,$ }
\end{array}\right.
\end{aligned}
\end{equation}
where $q_{x}$ is a Poisson distribution with parameter~$x$ and
$\langle \delta\rangle:= \sum_{m=0}^{\infty}m\delta(m)$.
\end{cor}

Next  we give  a  similar result as in O'Connell~\cite{OC98}, the
LDP for the  proportion of isolated vertices of the  RGG.

 \begin{cor}\label{ERiv}\label{RRiv}
Suppose $D$ is the degree distribution of  the random graph
$\skrig(n,r_n),$ where the connectivity radius $r_n\in (0,1]$
satisfies $n r_n^d \to c \in (0,\infty)$. Then, as $n\to\infty$, the
proportion of isolated vertices, $D(0)$ satisfies an LDP in $[0,1]$
with good rate function $$\xi_1(y)=y \log y + \rho(d) c y(1-y/2) -
(1-y) \big[ \log\big (\sfrac {\rho(d)c}{a}\big) -
 \sfrac{(a- \rho(d) c (1-y))^2}{2\rho(d)c(1-y)} \big] \, ,$$
 where $a=a(y)$ is the unique positive solution of $1-e^{-a}=\frac {\rho(d)c}{a}\, (1-y)$.
 \end{cor}

 From  Lemma~\ref{RRiv} we  deduce  that on a typical random  geometric graphs  the
 number  of isolated  vertices will  grow  like  $n
 e^{-\rho(d)c}.$ Thus,  as  $n\to\infty,$  the  number  of isolated
 vertices in the R.G graphs  converges  to $n
 e^{-\rho(d)c}$  in probability.
 In our  last  theorem  in  this  subsection  we  give  the
 LDP for  the  proportion  of  edges to the  number
  of  vertices  of  the  R.G.
  \subsection{The joint LDP  for  the empirical  colour  measure  and  empirical  pair  measure of CGRG}\label{secsub1}
  \begin{theorem}\label{randomg.jointL2L1}\label{main}
Suppose that $X$ is a coloured RGG  graph  with colour law $\nu$ and
connection radii $ r_n\colon\skriy^2\rightarrow[0,1]$ satisfying $n
r_n^d(a,b) \to C(a,b)$ for some symmetric function
$C\colon\skriy^2\rightarrow [0,\infty)$ not identical to zero. Then,
as  $n\rightarrow\infty,$ the pair $(\skril_X^1,\skril_X^2)$
satisfies an LDP  in
$\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2)$ with good rate
function
\begin{equation}\label{randomg.rateL2L1}
I(\varpi,\varpi)=H(\varpi\,\|\,\nu)+\sfrac{1}{2}{\mathfrak{H}_C^{d}}(\varpi\,\|\,\varpi),\,
\end{equation}
where
$${{\mathfrak H}_C^{d}}(\varpi\, \| \, \varpi ):=
H\big(\varpi\,\|\,\Delta(d)C\varpi\otimes\varpi\big)+\Delta(d)\|
C\varpi\otimes\varpi \| -\|\varpi\|\, ,$$ and the measure
$C\varpi\otimes\varpi\in\tilde\skriw(\skriy\times\skriy)$ is defined
by $C\varpi\otimes\varpi(a,b)=C(a,b)\varpi(a)\varpi(b)$ for
$a,b\in\skriy.$

\end{theorem}

Further,  we  state  a  Corollary  of  Theorem~\ref{main} below.

\begin{cor}\label{randomg.L1E}\label{Edgesg}
Suppose that $X$ is a coloured RGG  graph  with colour law $\nu$ and
connection radii $ r_n\colon\skriy^2\rightarrow[0,1]$ satisfying $n
r_n^d(a,b) \to C(a,b)$ for some symmetric function
$C\colon\skriy^2\rightarrow [0,\infty)$ not identical to zero. Then,
as $n\rightarrow\infty,$ the number of edges per vertex $|E|/n$  of
$X$ satisfies an LDP in $[0,\infty)$ with good rate function
$$\zeta(x)= x \log x - x+ \inf_{y>0} \big\{ \psi(y) - x \log(y) + y \big\}, $$
where $\psi(y) = \inf H(\varpi \, \| \, \nu)$ over all probability
vectors $\varpi$ with  $\sfrac 12 \Delta(d)\varpi^{T} C \varpi =y$.
\end{cor}

\begin{remark}By taking $C(a,b)=c$ one will obtain $\psi(y)=0$ for
$y=\sfrac{\Delta(d) }{2}c$, and $\psi(y)=\infty$ otherwise,  which
establishes  that $|E|/n$ obeys an
LDP in $[0,\infty)$ with  good rate function
$$\zeta(x)= x \log x - x+ \inf_{y>0} \big\{\psi(y) - x \log(\sfrac 12 y) + \sfrac 12 y \big\}, $$
where  $\Delta(d)c =y$.

\end{remark}

\section{Proof of Theorem~\ref{main} }\label{proofmain}
\subsection{Change-of-Measure}

For  any two  points  $U_1$  and  $U_2$  uniformly  and
independently chosen  from  the  space $[0,\,1]^d$  write
$$F(t):=\P\Big\{\|U_1-U_2\|\le t\Big\}.$$

Further, given a function $\tilde{f}\colon\skriy\rightarrow\R$  and
a symmetric function $\tilde{g}\colon \skriy^2\rightarrow\R$, we
define the constant $U_{\tilde{f}}$ by
$$U_{\tilde{f}}=\log\sum_{a\in\skriy}e^{\tilde{f}(a)}\nu(a),$$
and the function $\tilde{h}_n\colon\skriy^2\rightarrow\R$ by
\begin{equation}\label{hdef}
\tilde{h}_n(a,b)=\log\Big[\big(1-F(r_n(a,b))+ F(r_n(a,b))
e^{\tilde{g}(a,b)}\big)^{-n}\Big],
\end{equation}
for $a,b\in\skriy.$ We use $\tilde{f}$ and $\tilde{g}$ to define
(for sufficiently large $n$) a new coloured random graph  as
follows:
\begin{itemize}
\item To the $n$ labelled vertices in $V$ we assign colours from
$\skriy$ independently and identically according to the colour law
$\tilde{\nu}$ defined by
$$ \tilde{\nu}(a)=e^{\tilde{f}(a)-B_{\tilde{f}}}\nu(a).$$
\item Given  any two vertices $u,v\in V,$ with $u$ carrying colour $a$ and
$v$ carrying colour $b$, we connect vertex  $u$ to vertex $v$ with
probability
$$F(\tilde{r}_{n}(a,b))=\frac{F(r_n(a,b))e^{\tilde{g}(a,b)}}{1-F(r_n(a,b))+F(r_n(a,b))e^{\tilde{g}(a,b)}}.$$
\end{itemize}
We denote the transformed law by $\tilde{\prob}.$  We observe that
$\tilde{\nu}$ is a probability measure and that  $\tilde{\prob}$ is
absolutely continuous with respect to $\prob$ as, for any coloured
graph $X=((X(v)\colon v\in V),E)$,
\begin{align}
\frac{d\tilde\prob}{d\prob}(X) & = \prod_{u\in
V}\sfrac{\tilde{\nu}(X(Y_u))}{\nu(X(Y_u))}\prod_{(u,v)\in
E}\sfrac{F(\tilde{r}_{n}(X(Y_u),X(Y_v)))}
{F(r_n(X(Y_u),X(Y_v)))}\prod_{(u,v)\not\in E}\sfrac{1-F(\tilde{r}_n(X(Y_u),X(Y_v)))}{1-F(r_n(X(Y_u),X(Y_v)))}\nonumber\\
& = \prod_{u\in
V}\sfrac{\tilde{\nu}(X(Y_u))}{\nu(X(Y_u))}\prod_{(u,v)\in
E}\sfrac{F(\tilde{r}_n(X(Y_u),X(Y_v)))}{F(r_n(X(Y_u),X(Y_v)))}
\times\sfrac{n-nF(r_n(X(Y_u),X(Y_v)))}{n-nF(\tilde{r}_n(X(Y_u),X(Y_v)))}\prod_{(u,v)\in\skrie}\sfrac{n-nF(\tilde{r}_n(X(Y_u),X(Y_v)))}
{n-nF(r_n(X(Y_u),X(Y_v)))}\nonumber\\
& = \prod_{u\in V}e^{\tilde{f}(X(Y_u))-U_{\tilde{f}}}\prod_{(u,v)\in
E}e^{\tilde{g}(X(Y_u),X(Y_v))}\prod_{(u,v)\in\skrie}{e^{\frac 1n\, \tilde{h}_n(X(Y_u),X(Y_v))}}\nonumber\\
& = \exp\big( n\langle \skril_X^1,
\tilde{f}-U_{\tilde{f}}\rangle+n\langle \sfrac{1}{2}\skril_X^2,
\tilde{g}\rangle+n\langle\sfrac{1}{2}\skril_X^1\otimes \skril_X^1,
\tilde{h}_n\rangle-\langle \sfrac{1}{2}L_{\Delta}^{1},
\tilde{h}_n\rangle \big) , \label{randomg.Itransform}
\end{align}
where  $$L_{\Delta}^{1}=\sfrac{1}{n}\sum_{u\in
V}\delta_{(X(Y_u),X(Y_u))}.$$ We write $\langle
g,\varpi\rangle:=\sum_{a,b\in\skriy} g(a,b)\varpi(a,b)$ for
$\varpi\in\tilde{\skriw}(\skriy^2)$, and $\langle
f,\varpi\rangle:=\sum_{a\in\skriy} f(a)\varpi(a)$ for
$\varpi\in\skriw(\skriy)$, and  note that
$$F(r_n(a,b))=\Delta(d)r_n^d(a,b),\,\mbox{ for all $a,b\in\skriy^2$}.$$ i.e. the  volume of a
$d$-dimensional (hyper)sphere with radius $r(a,b)$ satisfying
$nr_n^d (a,b)\to C(a,b).$

The following lemmas will be useful in the  proofs  of  main
Lemmas.

\begin{lemma}[Euler's lemma]\label{randomg.sequence1}
If $nr_n^d (a,b)\to C(a,b)$ for every $a,b\in\skriy$, then
\begin{equation}\label{randomg.sequence}
\lim_{n\rightarrow\infty}\big[1+\alpha
F(r_n(a,b))\big]^{n}=e^{\alpha\Delta(d) C(a,b)},\mbox { for all
$a,b\in\skriy$ and $\alpha\in\R$. }
\end{equation}
\end{lemma}

\begin{Proof}
Observe that, for any $\eps>0$ and for large $n$  we have
$$\Big[1+\sfrac{\alpha \Delta(d)C(a,b)-\eps}{n}\Big]^{n}\le
\Big[1+ \alpha F(r_n(a,b))\Big]^{n}\le \Big[1+\sfrac{\alpha\Delta(d)
C(a,b)+\eps}{n}\Big]^{n},$$ by the pointwise convergence. Hence by
the sandwich theorem and Euler's formula we get
(\ref{randomg.sequence}).
\end{Proof}

We  write
$$P^{(n)}(\varpi) := \prob\big\{\skril_X^1=\varpi\big\}.$$

\begin{lemma}\label{randomg.uniexpotightness}\label{randomg.tightness}  The  family of
measures $({P}^n \colon n\in\N)$  is  exponentially tight on
$\skriw(\skriy)$
\end{lemma}

\begin{Proof}
We use  coupling argument,  see the proof  of \cite[Lemma~5.1]{DM10}
to show that , for every $\theta>0$,  there exists  $N\in\N$  such
that
$$\limsup_{n\to\infty}\frac{1}{n}\P\big\{|E|>nN\big\}\le-\theta.$$
To  begin,  let $c>\max_{a,b\in\skriy}C(a,b)>0.$  Using  similar
coupling  arguments  as  in see the proof  of
\cite[Lemma~5.1]{DM10},  we  can define, for  all sufficiently large
$n,$ a  new  coloured  geometric random  graph  $\tilde{X}$  with
vertices  $Y_1,Y_2, ...,Y_n$ chosen  uniformly  from  the  vertices
space $[0, 1]^d,$  colour  law $\mu$ and  connectivity radius
$\Big(\frac{c}{n}\Big)^{1/d},$  such that  any edge  present  in
$X$ is also present  in  $\tilde{X}.$  Let  $|\tilde{E}|$ be  the
number of edges  of  $\tilde{X}.$  Using  the  binomial formula  and
Euler's formula,  we  have  that

\begin{equation}
\begin{aligned}\nonumber
\prob\Big\{ |\tilde{E}|\ge n l\Big\}\le
e^{-nl}\me\big[e^{|\tilde{E}|}\big]&
 =e^{-nl}\sum_{k=0}^{\frac{n(n-1)}{2}}e^{k} \left(\heap{n(n-1)/2}{
 k}\right)\Big(\frac{\Delta(d)c}{n}\Big)^{k}\Big(1-\frac{\Delta(d)c}{n}\Big)^{n(n-1)/2-k}\\
&= e^{-nl}\Big(
1-\frac{c\Delta(d)}{n}+e\frac{c\Delta(d)}{n}\Big)^{n(n-1)/2} \le
e^{-nl}e^{nc\Delta(d)(e-1+o(1))}.
\end{aligned}
\end{equation}
Now given $\theta>0$ choose $N\in\N$  such that $N>
\theta+\Delta(d)c(e-1)$ and observe that, for sufficiently large
$n,$
\begin{equation}\nonumber
\prob\big\{|E|\ge n N\big\}\le \prob\big\{ |\tilde{E}|\ge
nN\big\}\le e^{-n\theta},
\end{equation}
which implies the statement.
\end{Proof}

\subsection{Proof of the upper bound in Theorem~\ref{main}}\\

We denote by $\skric_1$ the space of functions on $\skriy$ and  by
$\skric_2$ the space of symmetric functions on $\skriy^2$, and
define
$$\hat{I}(\varpi,\varpi)=\sup_{\heap{f\in\skric_{1}}{g\in\skric_{2}}}
\Big\{\sum_{a\in\skriy}\big(f(a)-U_{f}\big)\varpi(a)+\sfrac{1}{2}
\sum_{a,b\in\skriy}g(a,b)\varpi(a,b)+ \sfrac{\Delta(d)}{2}
\sum_{a,b\in\skriy}(1-e^{g(a,b)})C(a,b)\varpi(a)\varpi(b)\Big\}$$
for~$(\varpi,\varpi)\in\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2).$

\begin{lemma}\label{randomg.L2L1uppbound}
For each closed set
$G\subset\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2),$  we  have
$$\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\big\{(\skril_X^1,\skril_X^2)\in
F\big\}\le-\inf_{(\varpi,\varpi)\in F}\hat{I}(\varpi,\varpi).$$
\end{lemma}

\begin{Proof}
First let $\tilde{f}\in\skric_1$ and $\tilde{g}\in\skric_2$ be
arbitrary. Define $\tilde{\beta}\colon\skriy^2\rightarrow\R$ by
$$\tilde{\beta}(a,b)=\Delta(d)(1-e^{\tilde{g}(a,b)})C(a,b).$$
Observe that, by Lemma~\ref{randomg.sequence1},
$\tilde{\beta}(a,b)=\lim_{n\rightarrow\infty}\tilde{h}_n(a,b)$ for
all $a,b\in\skriy$, recalling the definition of $\tilde{h}_n$ from
\eqref{hdef}. Hence, by (\ref{randomg.Itransform}), for sufficiently
large~$n$,
\begin{equation}\nonumber
e^{\max_{a\in\skriy} | \tilde\beta(a,a)| } \ge\int
e^{\langle\frac{1}{2}L_{\Delta}^{1},\,
\tilde{h}_n\rangle}d\tilde{\prob} =\me\Big\{e^{n\langle \skril_X^1,
\tilde{f}-U_{\tilde{f}}\rangle+n\langle\frac{1}{2}\skril_X^2,
\tilde{g}\rangle+n\langle\frac{1}{2}\skril_X^1\otimes \skril_X^1,
\tilde{h}_n\rangle}\Big\},
\end{equation}
where $L_{\Delta}^{1}=\sfrac{1}{n}\sum_{u\in
V}\delta_{(X(Y_u),X(Y_u))}$ and therefore,
\begin{equation}\label{randomg.estP}
\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\me\Big\{e^{n\langle
\skril_X^1,\tilde{f}-U_{\tilde{f}}\rangle+n\langle
\frac{1}{2}\skril_X^2,
\tilde{g}\rangle+n\langle\frac{1}{2}\,\skril_X^1\otimes \skril_X^1 ,
\tilde{h}_n\rangle}\Big\}\le 0.
\end{equation}

Given $\eps>0$ let
$\hat{I}_{\eps}(\varpi,\varpi)=\min\{\hat{I}(\varpi,\varpi),{\eps}^{-1}\}-\eps.$
Suppose  that $(\varpi,\varpi)\in G$ and observe that
$\hat{I}(\varpi,\varpi)>\hat{I}_{\eps}(\varpi,\varpi).$ We now fix
$\tilde{f}\in\skric_1$ and $\tilde{g}\in\skric_2$ such that
$$\langle\tilde{f}-U_{\tilde{f}},\varpi\rangle+ \sfrac 12\,\langle
\tilde{g},\varpi\rangle+ \sfrac 12 \, \langle
\tilde{\beta},\varpi\otimes\varpi\rangle\ge\hat{I}_{\eps}(\varpi,\varpi)
.$$ As $\skriy$ is finite, there exist open neighbourhoods
$B_{\varpi}^{2}$ and $B_{\varpi}^{1}$ of $\varpi,\varpi$ such that
\begin{equation}\nonumber
\inf_{\heap{\tilde{\varpi}\in B_{\varpi}^{1}}{\tilde{\varpi}\in
B_{\varpi}^{2}}}\big\{\langle \tilde{f}-U_{\tilde{f}} ,
\tome\rangle+ \sfrac12 \, \langle
\tilde{g},\tilde\varpi\rangle+\sfrac 12 \, \langle
\tilde{\beta},\tome\otimes\tome\rangle\big\}\ge
\hat{I}_{\eps}(\varpi,\varpi)-\eps.
\end{equation}
Using Chebyshev's inequality and (\ref{randomg.estP}) we have that
\begin{equation}\begin{aligned}
\limsup_{n\rightarrow\infty}\sfrac{1}{n}&
\log\prob\big\{(\skril_X^1, \skril_X^2)\in
B_{\varpi}^{1}\times B_{\varpi}^{2}\big\}\\
&\le\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log
\me\Big\{e^{n\langle
\skril_X^1,\tilde{f}-U_{\tilde{f}}\rangle+n\langle
\frac{1}{2}\skril_X^2,\tilde{g}\rangle+n\langle\frac{1}{2}\skril_X^1\otimes
\skril_X^1,\tilde{h}_n\rangle}\Big\}-\hat{I}_{\eps}(\varpi,\varpi)+\eps\\
&\le -\hat{I}_{\eps}(\varpi,\varpi)+\eps.\label{randomg.LDPballs}
\end{aligned}\end{equation}
Now we use Lemma~\ref{randomg.tightness} with $\theta=\eps^{-1},$ to
choose $N(\eps)\in\N$ such that
\begin{equation}\label{unboundedset}
\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\Big\{|E|> n
N(\eps) \Big\}\le-\eps^{-1}.
\end{equation}
For this $N(\eps),$ define the set $K_{N(\eps)}$ by
$$K_{N(\eps)}=\Big\{(\varpi,\varpi)\in\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2):\|\varpi\|\le
2N(\eps)\Big\},$$ and recall that $\|\skril_X^2\|=2{|E|}/{n}.$ The
set $K_{N(\eps)}\cap F$ is compact and therefore may be covered by
finitely many sets $B_{\varpi_ {r}}^{1}\times B_{\varpi_
{r}}^{2},r=1,\ldots ,m$ with $ (\varpi_{r},\varpi_{r})\in F$ for
$r=1,\ldots ,m.$ Consequently,
\begin{equation}\nonumber
\prob\big\{(\skril_X^1, \skril_X^2)\in
F\big\}\le\sum_{r=1}^{m}\prob\big\{(\skril_X^1, \skril_X^2)\in
B_{\varpi_{r}}^{1}\times
B_{\varpi_{r}}^{2}\big\}+\prob\big\{(\skril_X^1, \skril_X^2)\not\in
K_{N(\eps)}\big\}.
\end{equation}
We may now use (\ref{randomg.LDPballs}) and \eqref{unboundedset} to
obtain, for all sufficiently small $\eps>0$,
\begin{equation}\nonumber
\begin{aligned}
\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\big\{(\skril_X^1,
\skril_X^2)\in F\big\}& \le \max_{r=1}^{m} \Big(
\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\big\{(\skril_X^1,
\skril_X^2)\in
B_{\varpi_{r}}^{1}\times B_{\varpi_{r}}^{2}\big\} \Big) \vee (-\eps)^{-1} \\
& \le \Big( -\inf_{(\varpi,\varpi)\in
G}\hat{I}_{\eps}(\varpi,\varpi)+\eps \Big)\vee (-\eps)^{-1}.
\end{aligned}
\end{equation}
Taking $\eps\downarrow 0$ we get the desired statement.
\end{Proof}
Next, we express the rate function in term of relative entropies,
see for example \cite[(2.15)]{DZ98}, and consequently show that it
is a good rate function. Recall the definition of the function $I$
from Theorem~\ref{main}.

\begin{lemma}\label{randomg.Vrate}\ \\ \vspace{-.6cm}
\begin{itemize}
\item[(i)] $\hat{I}(\varpi,\varpi)=I(\varpi,\varpi),$ for any
$(\varpi,\varpi)\in\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2)$,
\item[(ii)] I is a good rate function and
\item[(iii)] ${\mathfrak{H}_C^d}(\varpi\,\|\, \varpi)\ge 0$ with equality if and only if
$\varpi=\Delta(d)C\varpi\otimes\varpi.$
\end{itemize}

\end{lemma}
\begin{Proof}
(i) Suppose that $\varpi\not\ll\Delta(d) C\varpi\otimes\varpi.$
Then, there exists
 $a_0,b_0 \in\skriy$ with $C\varpi\otimes\varpi(a_0,b_0)=0$ and
 $\varpi(a_0,b_0)>0.$ Define $\hat{g}\colon \skriy^2\rightarrow \R$  by
$$ \hat{g}(a,b)=\log\big[K(\one_{(a_0,b_0)}(a,b)+\one_{(b_0,a_0)}(a,b))+1\big], \mbox{ for
$a,b\in\skriy$ and  $K>0.$ }$$ For this choice of $\hat{g}$ and
$f=0$ we have
\begin{align*}\nonumber
&\sum_{a\in\skriy}\big(f(a)-U_{f}\big)\varpi(a)+\sum_{a,b\in\skriy}
\sfrac 12
\hat{g}(a,b)\varpi(a,b)+\sum_{a,b\in\skriy}\sfrac {\Delta(d)}{2}(1-e^{\hat{g}(a,b)})C(a,b)\varpi(a)\varpi(b)\\
&\ge \sfrac {\Delta(d)}{2} \log(K+1)\varpi(a_{0},b_{0}) \to \infty,
\qquad \mbox{ for $K\uparrow\infty.$ }
\end{align*}
Now suppose that $\varpi\ll C\varpi\otimes\varpi.$ We have
$$\begin{aligned}
\hat{I}(\varpi,\varpi) & = \sup_{f\in\skric_{1}}
\Big\{\sum_{a\in\skriy}\Big(f(a)-\log\sum_{a\in\skriy}e^{{f}(a)}\nu(a)\Big)
\, \varpi(a) \Big\} \\
&\qquad  +  \sfrac {\Delta(d)}{2}\sum_{a,b\in\skriy}
C(a,b)\varpi(a)\varpi(b) +  \sfrac{1}{2} \sup_{g\in\skric_{2}}
\Big\{ \sum_{a,b\in\skriy}g(a,b)\varpi(a,b)-
\Delta(d)\sum_{a,b\in\skriy} e^{g(a,b)}
C(a,b)\varpi(a)\varpi(b)\Big\}.
\end{aligned}$$
By the variational characterization of relative entropy, the first
term equals $H(\varpi \, \| \, \nu)$. By the substitution
$h=\Delta(d)e^{g} \, \frac{C \varpi \otimes \varpi}{\varpi}$ the
last term equals
$$\begin{aligned}
\sup_{\heap{h\in\skric_{2}}{h \ge 0}} & \sum_{a,b\in\skriy} \Big[
\log \Big( h(a,b) \frac{\varpi(a,b)}{\Delta(d)C(a,b)\varpi(a)
\varpi(b)}\Big)
-h(a,b) \Big]  \, \varpi(a,b) \\
& = \sup_{\heap{h\in\skric_{2}}{h \ge 0}} \sum _{a,b\in\skriy} \big(
\log h(a,b) - h(a,b) \big) \, \varpi(a,b) +
\sum _{a,b\in\skriy} \log \Big( \frac{\varpi(a,b)}{\Delta(d)C(a,b) \varpi(a) \varpi(b)} \Big)\, \varpi(a,b) \\
& = - \| \varpi\|  + H(\varpi \, \| \,\Delta(d) C \varpi \otimes
\varpi ),
\end{aligned}$$
where we have used $\sup_{x>0} \log x - x = -1$ in the last step.
This yields that $\hat{I}(\varpi,\varpi)={I}(\varpi,\varpi)$.

(ii) Recall from \eqref{randomg.rateL2L1} and the definition of
${{\mathfrak H}_C}$ that $I(\varpi, \varpi)=H(\varpi \,\|\,\nu)+
\sfrac12\,
H\big(\varpi\,\|\,\Delta(d)C\varpi\otimes\varpi\big)+\sfrac{\Delta(d)}{2}\,
\| C\varpi\otimes\varpi \| - \sfrac12\, \|\varpi\|$. All summands
are continuous in $\varpi, \varpi$ and thus $I$ is a rate function.
Moreover, for all $\alpha<\infty$, the level sets $\{I \le \alpha\}$
are contained in the bounded set
$\{(\varpi,\varpi)\in\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2)\colon
\,{\mathfrak{H}_C^d}(\varpi\,\|\,\varpi)\le\alpha\}$ and are
therefore compact. Consequently, $I$ is a good rate function.

(iii)  Consider the nonnegative function $\xi(x)=x\log x-x+1$, for
$x> 0$, $\xi(0)=1$, which has its only root in $x=1$. Note that
\begin{align}\label{randomg.calculusnot}
{\mathfrak{H}_C^d}(\varpi\,\|\,\varpi)=
\left\{\begin{array}{ll}\int\xi\circ g\,\,d(\Delta(d)C\varpi\otimes
\varpi) &\mbox{ if
$g:=\sfrac{d\varpi}{(d\Delta(d)C\varpi\otimes\varpi)}\ge 0$  exists, }\\
\infty & \mbox{ otherwise.}
\end{array}\right.
\end{align}
Hence ${\mathfrak{H}_C^d}(\varpi\,\|\,\varpi)\ge 0$, and, if
$\varpi=\Delta(d)C\varpi\otimes\varpi,$ then
$\xi(\sfrac{d\varpi}{d(\Delta(d)C\varpi\otimes\varpi}))=\xi(1)=0$
and so
${\mathfrak{H}_C^d}(\Delta(d)C\varpi\otimes\varpi\,\|\,\varpi)=0$.
Conversely, if ${\mathfrak{H}_C^d}(\varpi\,\|\,\varpi)=0$, then
$\varpi(a,b)>0$ implies $C\varpi\otimes\varpi(a,b)>0$, which then
implies $\xi\circ g(a,b)=0$ and further $g(a,b)=1$. Hence
$\varpi=\Delta(d)C\varpi\otimes \varpi,$ which completes the proof
of (iii).
\end{Proof}

\subsection{Proof of the lower bound in Theorem~\ref{main} }\\
We  obtain  the  lower  bound  of Theorem~\ref{main}  from  the
upper bound as  follows:
\begin{lemma}\label{randomg.lowbound1}
For every open set
$O\subset\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2),$  we  have
$$\liminf_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\Big\{(\skril_X^1,\skril_X^2)\in
O\Big\}\ge -\inf_{(\varpi,\varpi)\in O}I(\varpi,\varpi).$$
\end{lemma}

\begin{Proof}
Suppose $(\varpi,\varpi)\in O,$ with  $\varpi\ll
\Delta(d)C\varpi\otimes\varpi$. Define
$\tilde{f}_{\varpi}\colon\skriy\rightarrow \R$  by
\begin{equation}\label{randomg.S1}
\begin{aligned}\nonumber
\tilde{f}_{\varpi}(a)=\left\{\begin{array}{ll}\log\sfrac{\varpi(a)}{\nu(a)},
&\mbox{if $\varpi(a)> 0$,  }\\
0, & \mbox{otherwise.}
\end{array}\right.
\end{aligned}
\end{equation}
and $\tilde{g}_{\varpi}\colon\skriy^2\rightarrow \R$ by
\begin{equation}\label{randomg.S2}
\begin{aligned}\nonumber
\tilde{g}_{\varpi}(a,b)=\left\{\begin{array}{ll}\log\sfrac{\varpi(a,b)}{\Delta(d)C(a,b)\varpi(a)\varpi(b)},&\mbox{if $\varpi(a,b)>0$, }\\
0, & \mbox{otherwise.}
\end{array}\right.
\end{aligned}
\end{equation}
In addition, we let
$\tilde{\beta}_{\varpi}(a,b)=\Delta(d)C(a,b)(1-e^{{\tilde
g}_{\varpi}(a,b)})$ and note that
$\tilde{\beta}_{\varpi}(a,b)=\lim_{n\rightarrow\infty}\tilde{h}_{\varpi,
n}(a,b),$ for all $a,b\in\skriy$  where
\begin{equation}\nonumber
\tilde{h}_{\varpi,
n}(a,b)=\log\Big[\big(1-F(r_n(a,b))+F(r_n(a,b))e^{\tilde{g}_{\varpi}(a,b)}\big)^{-n}\Big].
\end{equation}
Choose $B_{\varpi}^{1},B_{\varpi}^{2}$  open neighbourhoods of
$\varpi,\varpi,$ such that  $B_{\varpi}^{1}\times
B_{\varpi}^{2}\subset O$ and for all
$(\tilde{\varpi},\tilde{\varpi})\in B_{\varpi}^{1}\times
B_{\varpi}^{2}$
$$\langle \tilde{f}_{\varpi},\varpi\rangle+\sfrac 12\, \langle\tilde{g}_{\varpi},\varpi\rangle+
\sfrac 12\,
\langle\tilde{\beta}_{\varpi},\varpi\otimes\varpi\rangle-\eps\le
\langle \tilde{f}_{\varpi},\tome\rangle+ \sfrac 12\, \langle
\tilde{g}_{\varpi},\tilde\varpi\rangle+ \sfrac 12\,
\langle\tilde{\beta}_{\varpi},\tome\otimes\tome\rangle.$$ We now use
$\tilde{\prob},$ the probability measure obtained by transforming
$\prob$ using the functions $\tilde{f}_{\varpi}$,
$\tilde{g}_{\varpi}$. Note that the colour law in the transformed
measure is now $\varpi$, and the connectivity
radii~$\tilde{r}_n(a,b)$ satisfy  $$n \, \tilde{r}_n^d(a,b) \to
{\varpi(a,b)}/(\varpi(a)\varpi(b)) =: \tilde C(a,b), \mbox{ as }
n\to\infty.$$ Using (\ref{randomg.Itransform}), we obtain
\begin{align*}
\prob\Big\{(\skril_X^1,\skril_X^2)& \in
O\Big\}\ge\tilde{\me}\Big\{\sfrac{d\prob}{d\tilde{\prob}}(X)\one_{\{(\skril_X^1,\skril_X^2)\in
B_{\varpi}^{1}\times B_{\varpi}^{2}\}}\Big\}\\
&=\tilde{\me}\Big\{\prod_{u\in
V}e^{-\tilde{f}_{\varpi}(X(Y_u))}\prod_{(u,v)\in
E}e^{-\tilde{g}_\varpi(X(Y_u),X(Y_v))}\prod_{(u,v)\in
\skrie}e^{-\sfrac 1n \, \tilde{h}_{\varpi, n}(X(Y_u),X(Y_v))}
\one_{\{(\skril_X^1,\skril_X^2)\in B_{\varpi}^{1}\times B_{\varpi}^{2}\}}\Big\}\\
&=\tilde{\me}\Big\{ e^{-n\langle \skril_X^1,
\tilde{f}_{\varpi}\rangle-n\frac{1}{2}\, \langle \skril_X^2,
\tilde{g}_{\varpi}\rangle-n\frac{1}{2}\,\langle \skril_X^1\otimes
\skril_X^1, {\tilde{g}}_{\varpi}\rangle+\frac{1}{2}\,\langle L_{\Delta}^{1},\tilde{h}_{\varpi, n}\rangle}\times\one_{\{(\skril_X^1,\skril_X^2)\in B_{\varpi}^{1}\times B_{\varpi}^{2}\}}\Big\}\\
&\ge \exp\big(-n\langle \tilde{f}_{\varpi},\varpi\rangle-n\sfrac 12
\langle\tilde{g}_{\varpi},\varpi\rangle-n \sfrac 12
\langle\tilde{\beta}_\varpi,\varpi\otimes\varpi\rangle
+m-n\eps\big)\times\tilde{\prob}\Big\{(\skril_X^1,\skril_X^2)\in
B_{\varpi}^{1}\times B_{\varpi}^{2}\Big\},
\end{align*}
where  $m:=0 \wedge \min_{a\in\skriy}\tilde\beta(a,a).$ Therefore,
by (\ref{randomg.sequence}), we have
\begin{align}\label{randomg.lowbound2}
&\liminf_{n\rightarrow\infty}\sfrac{1}{n}\log\prob\Big\{(\skril_X^1,\skril_X^2)\in
O\Big\}\nonumber\\
&\ge-\langle \tilde{f}_{\varpi},\varpi\rangle-\sfrac12 \,
\langle\tilde{g}_{\varpi},\varpi\rangle-\sfrac 12\,
\langle{\tilde{{\beta}}_{\varpi}},
\varpi\otimes\varpi\rangle-\eps+\liminf_{n\rightarrow\infty}\sfrac{1}{n}\log\tilde{\prob}
\Big\{(\skril_X^1,\skril_X^2)\in B_{\varpi}^{1}\times
B_{\varpi}^{2}\Big\}.\nonumber
\end{align}

The result follows once we prove that
\begin{equation}\label{randomg.translowbound}
\liminf_{n\rightarrow\infty}\sfrac{1}{n}\log\tilde{\prob}\Big\{(\skril_X^1,\skril_X^2)\in
B_{\varpi}^{1}\times B_{\varpi}^{2}\Big\}= 0.
\end{equation}
We  use the upper bound (but now with the law $\prob$ replaced by
$\tilde{\prob}$) to prove (\ref{randomg.translowbound}). Then we
obtain
\begin{align*}
\limsup_{n\rightarrow\infty}\sfrac{1}{n}\log\tilde\prob\big\{(\skril_X^1,\skril_X^2)
\in (B_{\varpi}^{1}\times B_{\varpi}^{2})^{c} \big\}&\le
-\inf_{(\tome,\tilde{\varpi})\in \tilde{F}}
\tilde{I}(\tome,\tilde\varpi),
\end{align*}
where $\tilde{F}=(B_{\varpi}^{1}\times B_{\varpi}^{2})^{c}$ and
$\tilde{I}(\tilde\varpi, \tilde\varpi):= H(\tilde\varpi \, \| \,
\varpi) + \sfrac 12 {\mathfrak{H}_{\tilde C}^d}(\tilde\varpi\,\|\,
\tilde\varpi)$. It therefore suffices to show that the infimum is
positive. Suppose for contradiction that there exists a sequence
$(\tome_n,\tilde{\varpi}_n)\in\tilde{F}$ with
$\tilde{I}(\tome_n,\tilde\varpi_n)\downarrow 0.$ Then, because
$\tilde{I}$ is a good rate function and its level sets are compact,
and by  lower semi-continuity of the mapping
$({\tome},\tilde{\varpi})\mapsto\tilde{I}({\tome},\tilde{\varpi})$,
we can construct a limit point $(\tome,\tilde\varpi)\in\tilde{F}$
with $\tilde{I}(\tome,\tilde{\varpi})=0$ . By
Lemma~\ref{randomg.Vrate} this implies $H(\tome\,\|\,\varpi)=0$ and
${\mathfrak{H}_C}(\tilde{\varpi}\,\|\,\tome)=0$, hence
$\tome=\varpi,$ and
$\tilde{\varpi}=\tilde{C}\tome\otimes\tome=\varpi$ contradicting
$(\tome,\tilde{\varpi})\in\tilde{F}$.
\end{Proof}

\section{Proof of Theorem~\ref{main2}}\label{proofmain2}
For any $n\in\N$ we define
$$\begin{aligned}
\skrip_n(\skriw) & := \big\{ \omega\in \skrip(\skriw) \, : \, n\omega(a) \in \N \mbox{ for all } a\in\skriw\big\},\\
\tilde \skrip_n(\skriw \times \skriw) & := \big\{ \varpi\in
\tilde\skrip_*(\skriw\times\skriw) \, : \, \sfrac
n{1+\one\{a=b\}}\,\varpi(a,b) \in \N  \mbox{ for all } a,b\in\skriw
\big\}\, .
\end{aligned}$$

We denote by
$\Theta_n:=\skrip_n(\skriw)\times\tilde{\skrip}_n(\skriw\times\skriw)$
and
$\Theta:=\skrip(\skriw)\times\tilde{\skrip}_*(\skriw\times\skriw)$.
With
$$\begin{aligned}
P_{(\omega_n, \varpi_n)}^{(n)}(\mu_n) & := \prob\big\{\skrim_X=\mu_n \, \big| \, \skrih(\skrim_X)=(\omega_n,\varpi_n)\big\}\, ,\\
P^{(n)}(\omega_n,\varpi_n) & :=
\prob\big\{(\skril_X^1,\skril_X^2)=(\omega_n,\varpi_n)\big\}
\end{aligned}$$

the joint distribution of $\skril_X^1, \skril_X^2$ and $\skrim_X$ is
the mixture of $P_{(\omega_n, \varpi_n)}^{(n)}$ with
$P^{(n)}(\omega_n,\varpi_n)$ defined as
\begin{equation}\label{randomg.mixture}
d\tilde{P}^n(\omega_n, \varpi_n, \mu_n):= dP_{(\omega_n,
\varpi_n)}^{(n)}(\mu_n)\, dP^{(n)}(\omega_n, \varpi_n).\,
\end{equation}

(Biggins, Theorem 5(b), 2004) gives criteria for the validity of
large deviation principles for the mixtures and for the goodness of
the rate function if individual large deviation principles are
known. The following three lemmas ensure validity of these
conditions.

\begin{lemma}[Doku-Amponsah, 2014b] \label{randomg.uniexpotightness} The  family of
measures $({P}^n \colon n\in\N)$  is  exponentially tight on
$\Theta$
\end{lemma}

\begin{lemma}[Doku-Amponsah \& Moerters, 2010] \label{randomg.uniexpotightness} The  family of
measures $(\tilde{P}^n \colon n\in\N)$  is  exponentially tight on
$\Theta\times\skrip(\skriw\times\N).$
\end{lemma}

Define the function
$$\tilde{J}\colon{\Theta}\times\skrip(\skriw\times\N)\rightarrow[0,\infty],
\qquad\tilde{J}((\omega,\varpi),\,\mu)=\tilde{J}_{(\omega,\varpi)}(\mu),$$
where

\begin{align}\label{randomg.rateLDprob}
\tilde{J}_{((\varpi,\omega)}(\mu)=\left\{
\begin{array}{ll}H(\mu\,\|\,Q_{poi}) & \mbox
  {if  $(\omega,\mu)$  is  consistent and $\mu_1=\omega_2$ }\\
\infty & \mbox{otherwise.}
\end{array}\right.
\end{align}

\begin{lemma}[Doku-Amponsah \& Moerters, 2010]\label{randomg.convexgoodrate}
$\tilde{J}$ is lower semi-continuous.
\end{lemma}

By (Biggins, Theorem~5(b), 2004) the two previous lemmas and the
large deviation principles we have established in (Doku-Amponsah,
Theorem~2.1, 2014b) and (Doku-Amponsah, Theorem~2.1, 2014c) ensure
that under $(\tilde{P}^n)$ the random variables $(\omega_n,
\varpi_n, \mu_n)$ satisfy a large deviation principle on
$\skrip(\skriw) \times \tilde\skrip_*(\skriw\times\skriw)
\times\skrip(\skriw\times\N)$ with good rate function
$$\hat{J}(\omega, \varpi, \mu)= \left\{ \begin{array}{ll}
H(\omega\,\|\,\nu) + \sfrac12\,{\mathfrak H}_C^d(\varpi\,\|\,\omega)
+
H(\mu\,\| Q_{poi})\, , & \mbox{ if $(\varpi,\mu)$  is  consistent and $\mu_1=\varpi_2,$} \\
 \infty\, , & \mbox{ otherwise.} \\ \end{array}\right. $$
By projection onto the last two components we obtain the large
deviation principle as stated in Theorem~\ref{randomg.LDM} from the
contraction principle, see e.g. (Dembo et al.,1998, Theorem~4.2.1).

\section{Proof of Corollary~\ref{ERdd},~Corollary~\ref{RRiv},~and~Corollary~\ref{randomg.L1E}}\label{proofcorollaries}

We derive the theorems from Theorem~\ref{main2} by applying the
contraction principle, see e.g.~(Dembo \& Zeitouni,Theorem~4.2.1,
1998). In
 fact Theorem~\ref{main2} and the contraction principle imply a large
deviation principle for $D$. It just remains to simplify the rate
functions.

\subsection{Proof of Theorem~\ref{ERdd}.} Note  that,  in the case of an uncoloured
RGG graphs, the function $C$ degenerates to a constant~$c$,
$L^2=|E|/n\in[0,\infty)$ and $M=D\in\skrip(\N\cup\{0\})$.
Theorem~\ref{main2}  and the contraction principle imply a large
deviation principle for $D$ with good rate function
$$\begin{aligned} \eta_1(\delta) & =\inf\big\{J(x,\delta) \colon x\ge 0 \big\}
=\inf\big\{ H(\delta\,\|\,q_{x})+\sfrac1 2 x\log x -\sfrac12 x\log
\rho(d)c+ \sfrac{1}{2}\,\rho(d)c-\sfrac12 x \colon \langle
\delta\rangle \le x\,\big\},
\end{aligned}$$ which is to be understood as infinity if $\langle
d\rangle$ is infinite. We denote by $\eta^{x}(\delta)$ the
expression inside the infimum. For any $\eps>0$, we have
$$\begin{aligned}
\eta^{\langle \delta\rangle+\eps}(\delta)-\eta^{\langle
\delta\rangle}(\delta) & =\sfrac{\eps}{2} +\sfrac{\langle
\delta\rangle-\eps}{2}\log\sfrac{\langle \delta\rangle}{\langle
\delta\rangle+\eps}+\sfrac{\eps}{2}\log\sfrac{\langle
\delta\rangle}{\rho(d)c} \ge \sfrac{\eps}{2} +\sfrac{\langle
\delta\rangle-\eps}{2}\, \big(\sfrac{-\eps}{\langle
\delta\rangle}\big)+\sfrac{\eps}{2}\log\sfrac{\langle
\delta\rangle}{\rho(d)c}
>0,
\end{aligned}$$
so that the minimum is attained at $x=\langle\delta\rangle$.

 \subsection{Proof of Corollary~\ref{RRiv}.} Corollary~\ref{ERiv} follows from Theorem~\ref{ERdd} and the
contraction principle applied to the continuous linear map
$G\colon\skrip(\N\cup\{0\})\rightarrow[0,\,1]$ defined by
$G(\delta)=\delta(0).$ Thus, Theorem~\ref{ERdd} implies the large
deviation
 principle for $G(D)=W$ with the good rate function
$\xi_1(y)=\inf\{\eta_1(\delta) \colon \delta(0)=y, \langle
\delta\rangle < \infty\}.$ We  recall  the  definition  of
$\eta^{x}$ and  observe that $\xi_2(y)$ can be expressed as
$$\xi_1(y)=\inf_{b\ge
0}\inf_{\heap{d\in\skrip(\N\cup\{0\})}{\delta(0)=y,\,
\rho(d)c\langle \delta\rangle=b^2}} \Big\{\sfrac{1}{2}c+y\log
y+\sfrac{b^2}{2\rho(d)c} +\sum_{k=1}^{\infty}\delta(k)\log
\sfrac{\delta(k)}{q_{b}(k)}-b(1-y)\Big\}.$$ Now, using Jensen's
inequality, we have that
\begin{equation}\label{randomg.Jensen}\sum_{k=1}^{\infty}\delta(k)\log
\sfrac{\delta(k)}{q_{b}(k)}\ge
(1-y)\log\sfrac{(1-y)}{(1-e^{-b})},\end{equation}
 with equality if $\delta(k)=\sfrac{(1-y)}{(1-e^{-b})}q_{b}(k),$ for all
$k\in\N.$  Therefore, we have the inequality
$$\inf\big\{\eta(\delta) \colon \delta(0)=y,
\langle \delta\rangle <\infty\big\} \ge \inf\big\{
\sfrac{1}{2}c+y\log y+\sfrac{b^2}{2\rho(d)c}
+(1-y)\log\sfrac{(1-y)}{(1-e^{-b})} -b(1-y) \colon b\ge 0 \big\}.$$
Let $y\in[0,\,1].$  Then, the equation $a(1-e^{-a})=\rho(d)c(1-y)$
has a unique positive solution. Elementary calculus shows that the
global minimum of
 $b\mapsto\sfrac{1}{2}\rho(d)c+y\log y+\sfrac{b^2}{2\rho(d)c} +(1-y)\log\sfrac{(1-y)}{(1-e^{-b})}
-b(1-y)$ on $(0,\infty)$ is attained at the value $b=a$,
 where $a$ is the positive solution of our equation.
We obtain the form of $\xi$  in Corollary~\ref{ERiv} by observing
that
$$\sfrac{a(y)^2+(\rho(d)c)^2 -2\rho(d)ca(y)\big(1-y\big)}{2\rho(d)c}=\sfrac{\rho(d)cy}{2}\big(2-y\big)+\sfrac{1}{2\rho(d)c}\big(a(y)-\rho(d)c(1-y)\big)^2.$$

\subsection{Proof of Corolary~\ref{randomg.L1E}.}
We define the continuous linear map
$W\colon\skriw(\skriy)\times\tilde{\skriw}_*(\skriy^2)
\rightarrow[0,\infty)$  by
$W(\varpi,{\varpi})=\sfrac{1}{2}\|{\varpi}\|,$ and  infer from
Theorem~\ref{randomg.jointL2L1} and the contraction principle that
$W(\skril_X^1,\skril_X^2)=|E|/n$ satisfies a large deviation
principle in $[0,\infty)$ with the good rate function
$$\zeta(x)=\inf\big\{I(\varpi,{\varpi}) \colon W(\varpi,{\varpi})=x\big\}.$$
To obtain the form of the rate in the corollary, the infimum  is
reformulated as unconstrained optimization problem (by normalising
$\varpi$)
\begin{equation}\label{randomg.Jen}
\inf_{\heap{\varpi\in\skriw_*(\skriy^2)}{\varpi\in\skriw(\skriy)}}
\Big\{H(\varpi\,\|\,\nu)+xH(\varpi\,\|\,\Delta(d)C\varpi\otimes\varpi)+x\log2x+
\sfrac{\Delta(d)}{2}\, \| C \varpi \otimes \varpi\| -x\Big\}.
\end{equation}
By Jensen's inequality
$H(\varpi\,\|\,\Delta(d)C\varpi\otimes\varpi)\ge-\log\|\Delta(d)C\varpi\otimes\varpi\|,$
with equality if $\varpi=\sfrac{C\varpi\otimes\varpi}{\|
C\varpi\otimes\varpi\|},$ and hence, by symmetry of $C$ we have
\begin{align*}
\min_{\varpi\in\skriw_*(\skriy^2)}& \Big\{
H(\varpi\,\|\,\nu)+xH(\varpi\,\|\,\Delta(d)C\varpi\otimes\varpi)
+x\log2x+\sfrac{\Delta(d)}{2}\, \| C \varpi\otimes\varpi\|-x\Big\}\\
 &= H(\varpi\,\|\,\nu) -x\log \| \Delta(d)C \varpi\otimes\varpi\|
 +x\log 2x + \sfrac{\Delta(d)}{2}\, \| C \varpi\otimes\varpi\|-x.
\end{align*}

The form given in Corollary~\ref{randomg.L1E} follows by defining
$$y=\Delta(d)\sum_{a,b\in\skriy} C(a,b)\varpi(a)\varpi(b).$$

\bigskip


text:
\begin{document}

\newtheorem{ob}{Observation}[section]
    \newtheorem{lem}{Lemma}[section]
    \newtheorem{cor}{Corollary}[section]
    \newtheorem{theo}{Theorem}[section]
    \newtheorem{defi}{Definition}[section]
    \newtheorem{exam}{Example}[section]
    \newtheorem{rem}{Remark}[section]
	\newtheorem{prop}{Proposition}[section]
\numberwithin{equation}{section}
\title{
\Large
\bf
Convex Neighbourhoods and Complete Finsler Spaces
\footnote{2000 AMS
Mathematics
Subject Classification:\emph{ 53C22, 53C50}
\newline
Key words and phrases:\emph{Finsler spaces,
geodesic connectivity }
\newline Research
supported by  the Italian MURST  60\% and  GNSAGA}
}
 \author{
\small
O. M. Amici
and
  B. C. Casciaro\\
\small  Dipartimento di Matematica\\
\small  Universit{\`a} di Bari\\
\small Campus Universitario\\
\small Via Orabona 4, 70125 Bari, Italy
\\
\small amici@dm.uniba.it
\\
\small casciaro@dm.uniba.it
}
\maketitle
\begin{abstract}
In this paper, it is shown that a large set of
connections on a suitable sub--bundle of the tangent bundle
of a Finsler Manifold can be used to
study all the properties of convex neighbourhoods with respect to the Finsler Metric, which are needed to see that
any Complete Finsler Space is Geodesically Connected.
\end{abstract}
\newpage
\section{Introduction}
Let $M$ be a $C^\infty$--differentiable $n$--dimensional manifold endowed
with a Finslerian metric function $F:TM\to{\mathbb R}$, being $TM$ the tangent bundle
of $M$.

The following properties of the convex neighbourhoods of $M$
with respect to $F$ are
well known (see, \cite{BCS}, \cite{Ma},
see also
\cite{H}).

In order to quote them, let us denote by
\begin{eqnarray*}
B_\rho(0_x)=\{X\in T_x(M)/F(x,X)<\rho\}
\end{eqnarray*}
the
open indicatrix having $\rho$ as its radius and the zero vector
$0_x\in T_x(M)$ as its center, for any $\rho>0$ and $x\in M$.
\par
Moreover, we denote by $Exp_x$ the
exponential mapping of the Finsler metric and $Exp_x$ is defined on an open neighbourhood
of $0_x\in T_xM$ into $T_x(M)$, with $x\in M$. We recall that the mapping $Exp_x$ is defined by setting $Exp_x(X)=c_X(1)$, being $c_X$ the geodesic of $F$ defined by the initial conditions $c_X(0)=x$ and $\dot c_X(0)=X$, with $X$ belonging to a suitable open neighbourhood of $0_x$ in $T_xM$.
\par
With these notations, we have:
\begin{prop}
For each $x\in M$, there exist
two positive real numbers
$\varepsilon=\varepsilon(x)$ and $\eta=\eta(x)$ with $0<\varepsilon<\eta$
such that:

i). $Exp_x:B_\varepsilon(0_x)\to Exp_x(B_\varepsilon(0_x))=
B_\varepsilon(x)$ is a
diffeomorphism defined on $B_\varepsilon(0_x)$, whose degree of differentiability is
$C^\infty$ on $B_\varepsilon(0_x)-\{0_x\}$ and $C^1$ on $B_\varepsilon(0_x)$.
Moreover,
$B_\varepsilon(x)$ is an open neighbourhood of $x$.

ii). For each $y,z\in B_\varepsilon(x)$, there exists a unique geodesic
$c:[0,1]\to M$ lying entirely in
$B_\eta(y)=Exp_y(B_\eta(0_y))$ having length lesser than $\eta$ and such
that $c(0)=y$ and $c(1)=z$.

iii). For $y\in B_\varepsilon(x)$, the mapping $Exp_y:B_\eta(0_y)\to
B_\eta(y)$ is a diffeomorphism defined on $B_\eta(0_y)$, whose degree of
differentiability is $C^\infty$ on $B_\eta(0_y)-\{0_y\}$ and $C^1$ on
$B_\eta(0_y)$ and $B_\eta(y)$ is an open neighbourhood of
$y$ containing $B_\varepsilon(x)$.
\end{prop}

As in the Riemannian case (see, e.g., \cite{Kl}), we consider
the canonical identification
$T_XT_xM=T_xM$, for any $X\in T_xM$ and $x\in M$ and we fix an element $x$ of
$M$. Then, $T'_xM=T_xM-\{0_x\}$ is endowed by the Riemannian Metric
$g_p(X,Y)=g_{(x,p)}(X,Y)$, for each $p\in T'_xM$ and $X,Y\in T_pT_xM=T_xM$,
being $g$ the metric tensor which is induced by $F$.
Moreover, let $\widetilde TM\subseteq TM$ be the open neighbourhood of the zero
section where the exponential map of the Finsler Metric is defined.
Then, we have:
\begin{prop}
Let $x\in M$ and
$X\in \widetilde U_x=\widetilde TM\cap T_xM$. Moreover, let
$\widetilde b:[0,s_1]\to \widetilde U_x$ be
any differentiable curve such that $\widetilde b(0)=0_x$ and
$\widetilde b(s_1)=X$.
Finally let us put $b(s)=Exp_x\widetilde b(s)$, for any
$s\in[0,s_1]$ and $c(t)=Exp_x(tX)$, for any $t\in[0,1]$. Then:

i). $L(c)\leq L(b)$, where $L$ denotes the Finslerian length of
curves.

ii). If $\widetilde b(s)=\tau(s)X$, for any $s\in[0,s_1]$, being
$\tau:[0,s_1]\to[0,1]$ a strictly increasing differentiable mapping, then $L(c)=L(b)$.

iii). If $L(c)=L(b)$ and the total differential
$(DExp_x)_{t\widetilde b(s)}$ has maximal rank, for any $s\in[0,s_1]$ and $t\in[0,1]$, then
there exists a differentiable map $\tau:[0,s_1]\to[0,1]$ such that
$\widetilde b(s)=\tau(s)X$, for each $s\in[0,s_1]$ and $\tau$ is strictly increasing.
\end{prop}

\begin{prop}
Let $\widetilde \varepsilon=\varepsilon/3$, and $\varepsilon$ as in
Proposition 1.1. Then, every $y,z\in B_{\widetilde\varepsilon}(x)$
can be joined by a geodesic $c$,
lying entirely in
$B_{\widetilde\varepsilon}(x)$ and
having Finslerian length equal to the
Finslerian distance between $y$ and $z$. Moreover, any further geodesic joining $y$ and $z$ (if it exists) has points outside of $\overline B_\eta(y)$.
\end{prop}

Finally, we recall the following definition: we say that $F$ is a {\it
Complete Finsler Metric}, if and only if the
distance which is induced by $F$ on $M$ is complete. Then it results:
\begin{prop}
Let $F$ be a Finslerian metric on $M$. Then the
following properties are equivalent:

i). $F$ is a complete Finslerian metric.

ii). There exists a point $p\in M$ such that any geodesic
starting from $p$ can be extended to the whole $\mathbb R$.

iii). Assertion ii) holds, for any $p\in M$.

Moreover, it results:

iv). If Assertion i) holds, any two points of $M$ can be joined by
a geodesic of $F$.
\end{prop}

In 1967, B. T. Hassan (see \cite{H})  proved Proposition 1.1
by using the
{\it Cartan
Connection} relative to $F$. Proofs of more or less complete
versions of Proposition 1.1
are given by several authors by means of connections different from
the Cartan one. In 1993, D. Bao and S. S. Chern proved the same
proposition by using a new connection, called the {\it Chern Connection}
(see \cite{BC}), which coincides with the {\it Rund
Connection} (see \cite{A}).
\par
Finally, in \cite{BCS} the previous
proposition was proved by means of the {\it Cartan Form}.
The proofs of all the
previous propositions can be found in this book.
\par
Generally, all these proofs are different from the corresponding proofs of the
Riemannian case, more difficult than them and this fact explains the number of the
proofs of Proposition 1.1.
\par
Some questions arise for the previous observations. For example: Why
there exist so much proofs for Proposition 1.1? Why they are so different
from the proof needed in the Riemannian case? Does the used
connection play any role?
\par
In \cite{AC} and \cite{AC1}, by using the bundle $T'M=TM-\sigma(M)$, where
$\sigma:M\to TM$ is the zero section, we determined all the connections by means of which
the Euler--Lagrange assumes the simplest form. These connections were called {\it
Finslerian Connections} and they are a large class of connections.
\par
In this paper we prove:
\begin{prop}
Any Finslerian Connection can be used to
prove all the previous propositions.
\end{prop}
We also show that all the proofs of the previous propositions can follow
in the closest way the corresponding proofs
used in the Riemannian case and this fact can be useful for the proof of
further results. We choose to follow \cite{Kl}.
\par
In our opinion, the problems previously listed raise because classical methods used in this context mix up the Finsler Geometry with the geometry of the tangent bundle of the manifold under consideration.
\par
By using the previous considerations and the methods introduced here, in \cite{AC2} we determine all the properties corresponding to the ones considered here for a large class of Lagrangian Functions and in \cite{AC3} we show that the results of \cite{AC2} give a new sufficient condition for the geodesic connectedness of the generalized Bolza problem.

\section{Definitions and Proofs}

Let $M$ be a $C^\infty$--differentiable $n$--dimensional manifold.

The notations, which are more frequently used in the following, are:
\par

$\alpha$). $TM=\cup_{x\in M}T_xM$ is the tangent bundle of $M$ and
$\pi:TM\to M$ its natural projection. Moreover, for each $x\in M$, either
$X_x$ or
equivalently $(x,X)$ will denote the same element of $T_xM$ according to
this element is considered as a tangent vector at $x\in M$ or as a point of
$TM$;
hence $X_x=(x,X)$.
\par
$\beta$). ${\cal I}(M)=\bigoplus_{(r,s)\in{\mathbb N}^2}{\cal I}^r_s(M)$
is the algebra of
tensor fields on $M$,
where ${\cal I}^1_0(M)={\cal X}(M)$ and ${\cal I}^0_0(M)={\cal F}(M)$ are
the Lie algebra of vector fields and the ring of $C^\infty$--differentiable
real valued functions, respectively.
\par
$\gamma$). Let $\sigma:M\to TM$ be the zero section and we set
$T'M=TM-\sigma(M)$. $T'M$ is the open sub--bundle of $TM$ of non--zero tangent
vectors of $M$ and we shall denote by
$\pi':T'M\to M$ its canonical projection.
\par

$\delta$). ${\cal I}_{\pi'}=\bigoplus{\cal I}^r_{s\pi'}$is the ${\cal F}(T'M)$--module
(${\cal F}(M)$--module) of differentiable tensor fields along $\pi'$, with
${\cal I}^1_{0\pi'}={\cal X}_{\pi'}$ and ${\cal I}^0_{0\pi'}={\cal F}(T'M)$.
\par

$\theta$). If $(U,\varphi)$ is a local chart of $M$, we denote by
$(TU,T\varphi)$ the local chart canonically induced on $TM$, with
$TU=\pi^{-1}U$. We set
$\varphi=(x^1,\dots,x^n)$, $T\varphi=(x^1,\dots,x^n,X^1,\dots,X^n)$,
$e_i=\frac{\partial}{\partial x^i}$,
$\varepsilon_{\hat i}=\frac{\partial}{\partial X^{\hat i}},\ e^j=dx^j,\ \varepsilon^{\hat
j}=dX^{\hat j}$, for any $i,j,\hat i,\hat j\in\{1,\dots,n\}$.
\par

We recall that:

A {\it non--linear connection} $\nabla$ ({\it with three
indices}) can be regarded as an ${\mathbb R}$--bilinear mapping
$\nabla:{\cal X}_{\pi'}\times {\cal I}(M)\to {\cal I}_{\pi'}$ such that:
\par

i). $\nabla_{fX}(kY)=fX(k)Y+fk\nabla_XY,\ \forall X\in{\cal X}_{\pi'},\quad
\forall Y\in{\cal X}(M),\ \forall f,k\in{\cal F}(M)\ ;$

ii). $\nabla$ commutes with all the contractions;

iii). for each $X\in{\cal X}_{\pi'}$, with $X$ positively
homogeneous of degree $\rho$, $\nabla_XY$ is positively homogeneous of degree
$\rho$, for any $Y\in{\cal X}(M)$.

Moreover, $\nabla$ defines a horizontal lift $h:{\cal X}_{\pi'}\to{\cal X}(T'M)$
which can be extended in a trivial way to the whole tensor algebra ${\cal I}_{\pi'}$. This
extension is called {\it Matsumoto lift} (see, e.g., \cite{H}).
We also recall that the identity $id:T'M\to T'M$ can be considered as a
vector field along $\pi'$ ($id$ is the so--called {\it fundamental vector
field along $\pi'$}). The vector field $id^h$ will be said
to be {\it associated to $\nabla$} and one of its properties is:
\begin{prop}
$id^h$ can be extended to a spray defined on the
whole $M$, which will be denoted again with $id^h$. This extension is
$C^1$--differentiable on $TM$ and $C^\infty$--differentiable on $T'M$.
Moreover, a
curve $\gamma:[0,1]\to M$ is a non constant geodesic of $id^h$ if and only if
$\gamma$ is a path of $\nabla$.
\end{prop}

The spray $id^h$ defines an exponential map $Exp:\widetilde TM\to M$,
being $\widetilde TM$ an open neighbourhood of $\sigma(M)$. We have
$Exp(x,X)=c_{(x,X)}(1)$, being $c_{(x,X)}:[0,\varepsilon]\to M$, with $\varepsilon>1$, the
geodesic of $id^h$ having $(x,X)$ as initial condition, for all
$(x,X)\in\widetilde TM$.
The exponential map verifies the following properties:

(1). $Exp$ is $C^1$--differentiable on $\widetilde TM$ and
$C^\infty$--differentiable on $\widetilde TM-\sigma(M)$ (see \cite{L}, pg. 72).
\par

(2). $Exp$ has maximal rank in the zero vector $0_x\in T_xM$, for
any $x\in M$ (see \cite{GKM}, 2.8, Satz. (a), pg. 61).
\par

(3). The map $(\pi,Exp):\widetilde TM\to M\times M$ has maximal rank in
$0_x\in T_xM$, for any $x\in M$ (see \cite{GKM}, 2.8, Satz. (c), pg. 61).
\par

(4). The total differential of $Exp$, when it is restricted to $\widetilde TM\cap
T_xM$ and it is calculated in $0_x$, coincides with the identity map, for any
$x\in M$ (see \cite{L}, Th. 8, pg. 72).
\par

Moreover, as in the Riemannian case (see, e.g., \cite{Kl}) from
the previous properties, it follows:
\begin{prop}
There exists an open
neighbourhood $\widetilde W$ of $\sigma(M)$, such that
$Exp(\widetilde W)=W$ is open and:
\par
i). For each $y,z\in W$ there exists a unique geodesic
$c:[0,1]\to M$ such that $c(0)=y,\ c(1)=z$, $\ \dot c(0)\in \widetilde W$ and
$c(t)\in W$, for any $t\in[0,1]$.
\par

ii). For each $y\in W$, the map
$Exp_y:\widetilde W\cap T_yM\to Exp(\widetilde W\cap T_yM)=W(y)$
is a diffeomorphism of class $C^1$, of class $C^\infty$ on
$(\widetilde W\cap T_yM)-\{0_y\}$ and if $y\in W(x)$,
with $x\in M$, then $W(x)\subseteq W(y)$.
\par

iii). The mapping $(\pi,Exp)_{\vert\widetilde W}:\widetilde W\to W\times W$
is a
diffeomorphism of class $C^1$ and of class $C^\infty$ on
$\widetilde W-\sigma(M)$.
\end{prop}

Taking into account the previous proposition, the set $W(0_x)=\widetilde W\cap
T_xM$ is an open neighbourhood of $0_x$ in $T_xM$ and
$W(x)=Exp_x(W(0_x))$ is an open neighbourhood of $x$, for any $x\in M$.

Now, let $F:TM\to{\mathbb R}$
be a Finsler metric and $g$ its metric tensor. Then $g$ is
a family of Riemannian metrics on $M$ depending on $(x,X)\in T'M$ and such
that $g_{(x,X)}(X_x,X_x)=F^2(x,X)$, for any $x\in M$ and $(x,X)=X_x\in TM$.
Locally, we set $g=g_{ij}e^i\otimes e^j$, with
\begin{eqnarray*}
g_{ij}=\frac{1}{2}\frac{\partial^2 F^2}{\partial X^i\partial X^j}\ .
\end{eqnarray*}
Let $V=V(TM)$ be the vertical sub bundle of $TTM$; i.e., $V$ is the subset
of $TTM$ containing all the vectors tangent to the fibres of the tangent
bundle of $M$. We fix a further sub bundle $H$ of $TTM$ such that
$T_{(x,X)}TM=H_{(x,X)}\oplus V_{(x,X)}$, for any $(x,X)\in TM$. Two
canonical projections $P:TTM\to H$ and $Q:TTM\to V$ are associated to the
previous splitting into direct sum of $TTM$. These projections are
determined
by two tensor fields of type (1,1) denoted, by an abuse of notation, again by
$P$ and $Q$. Locally, $P$ and $Q$ are defined by:
\begin{eqnarray*}
P=\delta^i_je_i\otimes e^j-P^{\hat i}_j\varepsilon_{\hat i}\otimes e^j\
\hbox{and}\quad
Q=\delta^{\hat i}_{\hat j}\varepsilon_{\hat i}\otimes \varepsilon^{\hat j}+
P^{\hat i}_j\varepsilon_{\hat i}\otimes e^j\
.
\end{eqnarray*}
We recall that the distribution $H$ is said to be a
{\it non linear--connection with two
indices} if the functions $P^{\hat i}_j$ are positively homogeneous of degree
one (see, e.g.,\cite{Ma}).
Moreover, any connection on $M$ determines a couple of tensor fields
of the previous kind.
\par
Now, we turn to the general case and we set
\begin{eqnarray}
G=g^v=g_{ij}e^i\otimes e^j
\ ,
\end{eqnarray}
being
$v$ the extension to the tensor fields along $\pi'$ of the usual
vertical lift of tensors (see \cite{YI}).
\par
Then, from \cite{AC1}, it follows:
\begin{prop}
There exists a connection
$\nabla$ on $T'M$ such that
\par
i). $C^2_1(P\otimes T)=0$, where $T$ is the torsion tensor field
of $\nabla$ and $C^2_1$ is the contraction of the first contra--variant index
with the second covariant index of $P\otimes T$.
\par
ii). Let us denote by $\pi_*:TTM\to TM$ the total differential of
the canonical projection $\pi:TM\to M$ (see \cite{KN}).
Then, for each vector field
$X\in{\cal X}(T'(M))$, having $\pi_*(X_{(x,Z)})=Z_x$, for any $Z_x=(x,Z)\in
T'(M)$ it results:
\begin{eqnarray*}
(\nabla_XG)(X,Y)=(\nabla_YG)(X,X)=0\ ,\quad\forall Y\in{\cal X}(T'M)
\ .
\end{eqnarray*}

iii). The mapping $\nabla':{\cal X}_{\pi'}\times{\cal I}(M)\to{\cal I}_{\pi'}$
defined by
\begin{eqnarray*}
\nabla'_XY=\pi_*(\nabla_{X^c}Y^c)\ ,\quad\forall X,Y\in{\cal X}(M)
\end{eqnarray*}
being $c$ the complete lift (see \cite{YI}), is a non--linear connection with
three indices.
\end{prop}
A connection $\nabla$ verifying i), ii) and iii) of the previous
proposition is called {\it Finslerian Connection}. Moreover, the non--linear
connection $\nabla'$, defined by iii) of the same proposition, is called
{\it Berwald Connection deduced from $\nabla$}.

From \cite{AC1}, we also get:
\begin{prop}
Let us denote $\nabla$ by a Finslerian Connection
and
let $\nabla'$ be the Berwald Connection deduced from $\nabla$. Let
$Y=(\gamma,\dot\gamma):[a,b]\to T'M$ be a curve.
Then, $\gamma$ is a path of $\nabla'$ if and only if there exists a vector field
$Z\in{\cal X}(T'M)$ such that:

\begin{eqnarray*}
(\nabla_{\dot Y}{\dot Y})(t)=
Q_{(\gamma(t),\dot\gamma(t))}(Z_{(\gamma(t),\dot\gamma(t))})\ ,
\quad\forall
t\in[a,b]
\ .
\end{eqnarray*}
\end{prop}

In \cite{AC1}, it was also proved that the set of all the Finslerian Connections
can be
obtained in the following way.

Let $\widetilde\nabla$ be any connection on $T'M$ and $(U,\varphi)$
be a chart of
$M$. With respect to the chart $(T'U,T'\varphi)$, induced by $(U,\varphi)$
on $T'M$, we set:
\begin{eqnarray*}
& &
\nabla_{e_j}e_k
=
\widetilde\Gamma^{1i}_{jk}e_i+
\widetilde\Gamma^{5\hat i}_{jk}\varepsilon_{\hat i}
\ ,\
\nabla_{\varepsilon_{\hat j}}e_k
=\widetilde\Gamma^{2i}_{\hat jk}e_i+\widetilde\Gamma^{6\hat i}_{\hat jk}\varepsilon_{\hat i}
\nonumber
\\
& &
\nabla_{e_j}\varepsilon_{\hat k}
=\widetilde\Gamma^{3i}_{j\hat k}e_i+
\widetilde\Gamma^{7\hat i}_{j\hat k}\varepsilon_{\hat i}\ ,\
\nabla_{\varepsilon_{\hat j}}\varepsilon_{\hat k}
=\widetilde\Gamma^{4i}_{\hat j\hat k}e_i+
\widetilde\Gamma^{8\hat i}_{\hat j\hat k}\varepsilon_{\hat i}
\ .
\end{eqnarray*}
The $8n^3$ functions
$
(\widetilde\Gamma^{1i}_{jk},
\widetilde\Gamma^{2i}_{\hat jk},
\widetilde\Gamma^{3i}_{j\hat k},
\widetilde\Gamma^{4i}_{\hat j\hat k},
\widetilde\Gamma^{5\hat i}_{jk},
\widetilde\Gamma^{6\hat i}_{\hat jk},
\widetilde\Gamma^{7\hat i}_{j\hat k},
\widetilde\Gamma^{8\hat i}_{\hat j\hat k}
)$
are the local components of $\widetilde\nabla$.
By using the local components of $G$ and $P$ and the previous $8n^3$
functions, we obtain the following new $8n^3$ functions defined on $T'U$:
\begin{eqnarray*}
& &
2\Gamma^{1i}_{jk}=g^{im}(\partial_jg_{mk}+\partial_kg_{mj}-\partial_mg_{jk}+P^{\hat
h}_m\partial_{\hat h}g_{jk}-P^{\hat h}_j\partial_{\hat h}g_{mk}-P^{\hat h}_k\partial_{\hat
h}g_{mj})\ ,
\nonumber
\\
& &
\Gamma^{2i}_{\hat jk}=0\ ,\quad
\Gamma^{3i}_{j\hat k}=0\ ,\quad
\Gamma^{4i}_{\hat j\hat k}=0\ ,
\nonumber
\\
& &
\Gamma^{5\hat i}_{jk}=P^{\hat i}_t\widetilde\Gamma^{1t}_{jk}-P^{\hat i}_t\Gamma^{1t}_{jk}+
\widetilde\Gamma^{5\hat i}_{jk},
\nonumber
\\
& &
\Gamma^{6\hat i}_{\hat jk}=P^{\hat i}_t\widetilde\Gamma^{2t}_{\hat jk}+
\widetilde\Gamma^{6\hat i}_{\hat jk}\ ,\ \
\Gamma^{7\hat i}_{j\hat k}=
P^{\hat i}_t\widetilde\Gamma^{3t}_{j\hat k}+
\widetilde\Gamma^{7\hat i}_{j\hat k}\ ,\ \
\Gamma^{8\hat i}_{\hat j\hat k}=P^{\hat i}_t\widetilde\Gamma^{4t}_{\hat j\hat k}+
\widetilde\Gamma^{8\hat i}_{\hat j\hat k}
\ ,
\end{eqnarray*}
where we used the notations $\partial_if=\frac{\partial f}{\partial x^i}$
and $\partial_{\hat
i}f=\frac{\partial f}{\partial X^{\hat i}}$,
for any $f\in{\cal F}(T'M)$ and any $i,\hat
i\in\{1,\dots,n\}$.

The above functions are the local components of a connection $\nabla$ of
$T'M$. Since $\nabla$ verifies i), ii) and iii) of Proposition 2.3, it is a
Finslerian Connection. The connection $\nabla$ can be used in order to
determine all the Finslerian Connections. In fact, let $\nabla^1$ be a further
connection and let us denote by $N\in{\cal I}^1_2(T'M)$ the tensor field defined by
setting:
\begin{eqnarray*}
N(X,Y)=\nabla_XY-\nabla^1_XY\ ,\quad\forall X,Y\in{\cal X}(T'M)
\ .
\end{eqnarray*}
There exists a unique tensor field, $\widetilde N$, of type $(1,2)$ along $\pi'$,
such that:
\begin{eqnarray*}
\widetilde N(X,Y)=\pi_*N(X^c,Y^c)\ ,\quad\forall X,Y\in{\cal X}(M)
\end{eqnarray*}
$\widetilde N$ is called the {\it tensor field associated to $N$}. Now, we
denote by $d'$ the so called "derivation along the fibres",
which is defined by setting $d'f=(\partial_{\hat i}f)e^{\hat i}$, for any
$f\in{\cal F}(T'M)$ and is extended to the whole tensor algebra ${\cal I}_{\pi'}$ in
the well--known way.
Then, we have:
\begin{prop}
Under the previous assumptions, the connection
$\nabla^1$ is a Finslerian Connection, if and only if the following
conditions hold for $N$:

i). $\widetilde N_{(x,Z)}(Z_x,Z_x,)=0\ ,\quad\forall (x,Z)=Z_x\in T'M$.

ii). $C^1_1(Z_x\otimes C^1_1((d'F\otimes\widetilde N)_{(x,Z)}))=0,
\quad\forall(x,Z)=Z_x\in T'M$.

iii). $PN(X,QY)=0\ ,\quad\forall X,Y\in{\cal X}(T'M)$.

iv). $PN$ is symmetric with respect to the two lower indices and
it is positive homogeneous of degree 0.
\end{prop}
In the literature on Finsler spaces, one can find a lot of tensor fields having the
properties i)--iv) (see, e.g., \cite{R} and \cite{Ma}) and by following
the known
examples many other tensors of the same kind can be constructed.
From $\widetilde N$ one can obtain many tensors like $N$ by
following methods, which are
analogous to the ones previously used for the construction of the connection
$\nabla^1$. Hence, we omit
them for the sake of brevity.

Now, let us consider a point $x\in M$. Then the ({\it closed}) {\it indicatrix having $0_x$
as its center and $\rho>0$ as its radius} is the set
$\overline{B}_\rho(0_x)=\{(x,X)\in T_xM/F(x,X)\leq\rho\}$ and it is the closure
of the open indicatrix $B_\rho(0_x)$, defined in the Introduction.
$\overline{B}_\rho(0_x)$ and $B_\rho(0_x)$ are both convex, that is for any
$X_x,Y_x$ elements of $\overline{B}_\rho(0_x)$ ($B_\rho(0_x)$) the vector
$tX_x+(1-t)Y_x$ belongs to $\overline{B}_\rho(0_x)$ ($B_\rho(0_x)$), for any
$t\in[0,1]$.

Finally, it is easy to see that:
\begin{prop}
Let $x\in M$, the following assertions are
true:

i). The family $\{B_\rho(0_x)\}_{\rho>0}$ is a basic system of open
neighbourhoods of $0_x$ in $T_xM$.

ii). The sets $\widetilde A=\cup_{z\in A}B_\rho(0_z)$, with $A$ open
neighbourhood of $x$ in $M$ and $\rho>0$ form a basic system of open
neighbourhoods of $0_x$ in $TM$.
\end{prop}

Now, we are in position to prove Proposition 1.1:
\begin{proof}
Let $\nabla$ be a Finslerian Connection and let $x\in M$. Since the Berwald
Connection $\nabla'$ induced by $\nabla$ induces a spray,
by Proposition 2.2, there exists an open neighbourhood $\widetilde W$
of the zero section $\sigma(M)$ in $TM$ and an open subset $W$ of $M$
such that
$(\pi,Exp):\widetilde W\to W\times W$ is a diffeomorphism of class $C^1$, where
$Exp$ is the exponential map of the spray defined by $\nabla'$.
Moreover, ii) of Proposition 2.6 ensures us that there exists a positive real
number $\eta$ and an open neighbourhood $A$ of $x$ in $M$, such that
$\widetilde A=\cup_{z\in A}B_\eta(0_z)\subseteq\widetilde W$. Then, by using ii) of
Proposition 2.3 and i) of Proposition 2.6, there also exists a real number
$\varepsilon$, with $0<\varepsilon\leq\eta$ such that $B_\varepsilon(x)\times
B_\varepsilon(x)\subseteq(\pi,Exp)(\widetilde A)\subseteq W\times W$, where
$B_\varepsilon(x)=Exp_x(B_\varepsilon(0_x))$ is an open neighbourhood of $x$. Finally, i)
follows, since $Exp_x:B_\varepsilon(0_x)\to B_\varepsilon(x)$ is a $C^1$--diffeomorphism.

To prove ii), we consider $y,z\in B_\varepsilon(x)$ with $y\not=z$. It results:
\begin{eqnarray*}
0_y\not=(\pi,Exp)^{-1}(y,z)=X_y^z\in B_\eta(0_y)\subseteq\widetilde W
\ .
\end{eqnarray*}
Then, i) of Proposition 1.1, implies that the curve $c(t)=Exp_y(tX^z_y)$,
with $t\in[0,1]$, is the unique geodesic lying entirely in $W$ and joining
$y$ and $z$. The curve $c$ is also the unique geodesic that joins $y$ and
$z$ lying entirely in $B_\eta(y)$. In fact, we shall see that $c$ has
length lesser than $\eta$.
Let $\beta=(c,\dot c):[0,1]\to T'M$ the complete lift of $c$
to $TM$,
for any $t\in[0,1]$. Then, it results
$\dot \beta=\dot c^ie_i+\ddot c^{\hat i}\varepsilon_{\hat i}$. Hence it follows:
\begin{eqnarray*}
g_{(c(t),\dot c(t))}(\dot c(t),\dot c(t))=
G_{\beta(t)}(\dot\beta(t),\dot\beta(t))\ ,\quad\forall t\in[0,1]
\ .
\end{eqnarray*}
Moreover, being $c$ a path of $\nabla'$, we have:
\begin{eqnarray*}
(\nabla_{\dot\beta}{\dot\beta})(t)=Q_{\beta(t)}(Z_{\beta(t)})\ ,
\quad\forall
t\in[0,1]
\end{eqnarray*}
where $Z$ is a suitable vector field defined on an open neighbourhood of
$\beta([0,1])$ in $T'M$.
From the previous identities and ii) of the previous proposition  we obtain:

\begin{eqnarray*}
& &
\frac{d}{dt}g_{(c(t)\dot c(t)}(\dot c(t),\dot c(t))=
(\nabla_{\dot Y} G)_{\beta(t))}(\dot\beta(t),\dot\beta(t))+
\nonumber
\\
& &
2G_{\beta(t)}(\dot\beta(t),Q_{\beta(t)}(Z_{(\beta(t))}))=0\ ,
\ \quad \forall t\in[0,1]
\ .
\end{eqnarray*}
Hence, we have:
\begin{eqnarray}
F^2(c(t),\dot c(t))=g_{(c(t),\dot c(t))}(\dot c(t),\dot
c(t))=F^2(y,X^z_y)<\eta^2
\ ,\quad\forall t\in[0,1]
\ ;
\end{eqnarray}
that is $c$ has length lesser than $\eta$. Moreover,
for $y=z\in B_\varepsilon(x)$
the unique geodesic joining $y$ and $z$ in $W$ is the constant curve.
Now, we observe that, the mapping $Exp_y:B_\eta(0_y)\to B_\eta(y)$ is a
$C^1$--diffeomorphism, because of ii) of Proposition 2.2, by using the
inclusion $B_\eta(0_y)\subseteq\widetilde W\cap T_yM$.
Finally, we have iii), since
the inclusion
$(\pi,Exp)^{-1}(\{y\}\times B_\eta(x))\subseteq B_\eta(0_y)$ implies
$B_\varepsilon(x)\subseteq B_\eta(y)$.
\end{proof}

Now, we recall that the standard identification $T_XT_xM=T_xM$, induces a
Riemannian Metric on $T'_xM$, for any $X\in T_xM$ and any $x\in M$. By
means of this identification, we prove the corresponding of the Gauss Lemma.
\begin{prop}
Let $x\in M$ and $X\in B_\varepsilon(0_x)$, with $\varepsilon$ as
in Proposition 1.1. Then:

i). It results $F(x,X)=F(Exp_x(X),(DExp_x)_XX)$, where we denoted
by
$(DExp_x)_X:T_XT_xM=T_xM\to T_{Exp_x(X)}M$ the total differential of
$Exp_x$.

ii). For any $Y\in T_xM$, such that $g_{(x,X)}(X,Y)=0$, it
results:
\begin{eqnarray*}
g_{(Exp_x(X),(DExp_x)_XX)}((DExp_x)_XX,(DExp_x)_XY)=0.
\end{eqnarray*}
\end{prop}
\begin{proof}
If $X=0_x$, then the assertion is trivially true. Hence, we suppose
$X\not=0_x$.
\par
In order to prove i) let us denote by $c_X:[0,1]\to M$ the geodesic of
the Berwald connection $\nabla'$, then we have:
\begin{eqnarray*}
(DExp_x)_{tX}X=\dot c_X(t)=\beta(t)\in T'M\ ,\quad\forall t\in[01]\ ;
\end{eqnarray*}
and
\begin{eqnarray*}
\frac{d}{dt}g_{(c_X(t),\dot c_X(t))}(\dot c_X(t),\dot c_X(t))=0
\end{eqnarray*}
Hence, from Equation (2.2), it follows:
\begin{eqnarray*}
g_{(c_x(t),\dot c_X(t))}(\dot c_X(t),\dot c_X(t))=F^2(x,X)\ ,\quad\forall
t\in[0,1]
\ .
\end{eqnarray*}
Consequently

\begin{eqnarray}
F(c_X(t),\dot c_X(t))&=&F(Exp_x(tX),(DExp_x)_{tX}X)
=
F(x,X),\
\nonumber
\\
& &
\qquad\forall t\in[0,1]\ ;
\end{eqnarray}
and the assertion follows for $t=1$.

Now, we prove ii). Let $Y\in T'_xM$, such that
$g_{(x,X)}(X,Y)=0$. Consider a curve, denoted again by
$X:(-a,a)\to B_\varepsilon(0_x)$, with
$a>0$ and $\varepsilon$ as in Proposition 1.1,
having $X(0)=X$, $\dot X(0)=Y$ and
$F(x,X(s))=F(x,X)=r<\varepsilon$, for each $s\in(-a,a)$. Being $tX(s)\in
B_\varepsilon(0_x)$, we can define the mapping:
\begin{eqnarray*}
\lambda(s,t)=Exp_x(tX(s))=c_{X(s)}(t)\ ,\quad (s,t)\in(-a,a)\times[0,1]
\ .
\end{eqnarray*}
Then:
\begin{eqnarray*}
(\frac{\partial\lambda}{\partial s})_{(s,t)}=(DExp_x)_{tX(s)}t\dot X(s)
\end{eqnarray*}
and
\begin{eqnarray*}
(\frac{\partial\lambda}{\partial t})_{(s,t)}
=(DExp_x)_{tX(s)}X(s)=\dot c_{X(s)}(t)
\ ,
\end{eqnarray*}
for any $(s,t)\in(-a,a)\times[0,1]$.

Hence, it results:
\begin{eqnarray}
F(\lambda(s,t),(\frac{\partial\lambda}{\partial t})_{(s,t)})=
F(x,X(s))=r\ ,\quad\forall
(s,t)\in(-a,a)\times[0,1]\ .
\end{eqnarray}
Now, we set:
\begin{eqnarray*}
\beta(s,t)=(\lambda(s,t),(\frac{\partial\lambda}{\partial t})_{(s,t)})\in T'M,
\quad\forall(s,t)\in(0,a)\times[0,1]
\ .
\end{eqnarray*}
Then, we have:
\begin{eqnarray*}
& &
\bigl(\frac{\partial}{\partial t}
g_{(\lambda,\frac{\partial\lambda}{\partial t})}
(\frac{\partial\lambda}{\partial s},
\frac{\partial\lambda}{\partial t})\bigr)_{\vert(s,t)}=
G_\beta
\bigl(\nabla_{\frac{\partial\beta}{\partial s}}
\frac{\partial\beta}{\partial t},
\frac{\partial\beta}{\partial t})_{\vert(s,t)}=
\nonumber
\\
& &
\frac{1}{2}\bigl(\frac{\partial}{\partial s}G_\beta
(\frac{\partial\beta}{\partial t},
\frac{\partial\beta}{\partial t})\bigr)_{\vert(s,t)}-
\frac{1}{2}\bigl(\nabla_{\frac{\partial\beta}{\partial s}}
G)(\frac{\partial\beta}{\partial t},
\frac{\partial\beta}{\partial t})\bigr)_{\vert(s,t)}
\ .
\end{eqnarray*}
Hence, because of ii) of Proposition 2.3, recalling the definition of $\beta$,
we have:
\begin{eqnarray*}
\frac{\partial}{\partial s}
F^2(\lambda(s,t),(\frac{\partial\lambda}{\partial t})_{(s,t)})=0\ ,\quad
\forall(s,t)\in(-a,a)\times[0,1]\ .
\end{eqnarray*}
Consequently:
\begin{eqnarray}
& &
g_{(Exp_x(tX(s)),(DExp_x)_{tX(s)}X(s))}(
(DExp_x)_{tX(s)}t\dot X(s),
(DExp_x)_{tX(s)}X(s))=0
\nonumber
\\
& &
\forall(s,t)\in(-a,a)\times[0,1]\ .
\end{eqnarray}
Finally, we obtain ii) by putting $t=1$ and $s=0$ in the previous identity.
\end{proof}
\par
For the sequel, we need the following lemma, which is proved in \cite{H}.
\begin{lem}
Let $x\in M$ and $Z\in T'_xM$. Then for each $Y\in T_xM$
such that $g_{(x,Z)}(Z,Y)=0$ it results $F(x,Y+Z)\geq F(x,Z)$ and equality
holds if and only if $Y=0_x$.
\end{lem}
Then, we can prove the Proposition 1.2.
\begin{proof}

First, we prove i).

We can suppose $\widetilde b(s)\not=0_x$, for any $s\in(0,s_1]$. Then, the
function $r(s)=F(x,\widetilde b(s))$ is not zero, for each $s\in(0,s_1]$.
Hence, we can put:
\begin{eqnarray*}
X(s)=\frac{1}{r(s)}\widetilde b(s)\ ,\quad\forall s\in(0,s_1]
\ .
\end{eqnarray*}
With these notations, it is easy to see that:
\begin{eqnarray}
\widetilde b(s)=r(s)X(s)\ ,
\end{eqnarray}

\begin{eqnarray}
g_{(x,X(s))}(X(s),X(s))=1
\end{eqnarray}

\begin{eqnarray}
\dot b(s)=(DExp_x)_{\widetilde b(s)}\dot r(s)X(s)+
(DExp_x)_{\widetilde b(s)}r(s)\dot X(s))\ ,
\end{eqnarray}

\begin{eqnarray}
g_{(x,X(s)}(X(s),\dot X(s))=0\ ,
\end{eqnarray}
for any $s\in(0,s_1]$.

Then, from ii) of Proposition 2.7 and from (2.4), it follows:
\begin{eqnarray}
g_{(b(s),(DExp_x)_{\widetilde b(s)}\widetilde b(s))}
\bigl((DExp_x)_{\widetilde b(s)}\dot
r(s)X(s),(DExp_x)_{\widetilde b(s)}r(s)\dot X(s)\bigr)=0
\end{eqnarray}
for any $s\in(0,s_1]$.
Moreover, i) of Proposition 2.7 and equations (2.4) and (2.5) imply:
\begin{eqnarray}
F^2(b(s),(DExp_x)_{\widetilde b(s)}\dot r(s)X(s))=\dot r^2(s),
\quad\forall s\in(0,s_1]
\ .
\end{eqnarray}
Then, from (2.7), (2.9), (2.10) and from the Lemma 2.1, we obtain:
\begin{eqnarray}
F^2(b(s),\dot b(s))\geq\dot r^2(s)\ ,
\quad\forall s\in(0,s_1]
\ ,
\end{eqnarray}
hence:
\begin{eqnarray*}
L(b)=\int_\rho^{\sigma_1}\vert\dot r(s)\vert ds\geq\vert r(s_1)-r(\rho)\vert\ ,
\quad \forall \rho\in(0,s_1]
\end{eqnarray*}
and i) is true.

Assertion ii) trivially holds.

Under the assumptions of iii), by using the previous notations, we have
$L(c_X)=L(b)=r(s_1)$. Moreover, being i) true, it results
$L(b_s)\geq L(c_{\widetilde b(s)})=r(s)$, with $b_s=b_{\vert[0,s]}$, for any
$s\in(0,s_1]$. Hence, it follows:
\begin{eqnarray*}
\int_\rho^{s_1}(F(b(s),\dot b(s))-\vert\dot r(s)\vert)ds\leq0\ ,
\quad\forall\rho\in(0,s_1]\ .
\end{eqnarray*}
Since (2.12) holds, the previous inequality implies:
\begin{eqnarray*}
\int_\rho^{s_1}(F(b(s),\dot b(s))-\vert\dot r(s)\vert)ds=0\ ,
\quad\forall\rho\in(0,s_1]\
\end{eqnarray*}
and, for continuity reasons, we have
\begin{eqnarray*}
F(b(s),\dot b(s))=\vert\dot r(s)\vert\ ,\quad\forall s\in[0,s_1]\ .
\end{eqnarray*}
Consequently, by using (2.8), (2.10), (2.11) and the previous Lemma we have:
\begin{eqnarray*}
(DExp_x)_{\widetilde b(s)}r(s)\dot X(s)=0\ ,\quad \forall s\in[0,s_1]\ .
\end{eqnarray*}
Then, $\dot X(s)=0$, for any $s\in[0,s_1]$, because $Exp_x$ has maximal rank
along the curve $\widetilde b$ and $r(s)\not=0$ for any $s\in(0,s_1]$.
Hence, iii) is true.
\end{proof}

From the above proposition, it trivially follows:
\begin{prop}
Let $x\in M$. Moreover, let us consider
$\varepsilon=\varepsilon(x)$ and
$\eta=\eta(x)$ as in Proposition 1.1. Then, a geodesic of
length lesser than $\eta$ starting
from an arbitrary point $y\in B_\varepsilon(x)$
is a curve of minimal length between its end points.
\end{prop}
Let $x\in M$ and $B_\rho(0_x)\subseteq\widetilde W\cap T_xM$. Moreover,
let $u=(u_i)_{1\leq i\leq n}$ be a basis of $T_xM$. Then, $u$ defines a
mapping, which by an abuse of notations, we denote again by $u:{\mathbb R}^n\to
T_xM$ obtained by putting $u(\xi)=\xi^iu_i$, for any $\xi=(\xi^i)_{1\leq
i\leq n}\in{\mathbb R}^n$. Furthermore, the mapping
$\psi=u^{-1}\circ Exp_x^{-1}:B_\rho(x)\to u^{-1}(B_\rho(0_x))$ is a
$C^\infty$--diffeomorphism on $B_\rho(x)-\{x\}$ and $C^1$--differentiable on
$x$. Consequently, $(B_\rho(x),\psi)$ is a chart of the
$C^1$--differentiable manifold structure canonically induced on $M$ by the
considered
structure of $C^\infty$--differentiable manifold on $M$.

Then, we can prove:
\begin{prop}
Let $x\in M$ and $\varepsilon=\varepsilon(x)$ as in Proposition
1.1. There exists $\varepsilon_0=\varepsilon_0(x)\in(0,\eta)$
such that for any non constant
geodesic $c:[0,a]\to M$, with $a>0$, satisfying the conditions
$c(t_0)\in\partial B_\varepsilon((x)=Exp_x(\partial B_\varepsilon(0_x))$ and
$\dot
c(t_0)\in T\partial B_\varepsilon(x)$, for some $t_0\in(0,a)$ and
$\varepsilon\in(0,\varepsilon_0)$
and for each $\mu\in(0,1)$ there exists $\rho=\rho(\mu,\varepsilon)$ with the
following property:
\begin{eqnarray*}
d(x,c(t))\geq d(x,c(t_0))+\mu(t-t_0)^2\ ,\quad \forall
t\in[t_0-\rho,t_0+\rho]\ ,
\end{eqnarray*}
where $d$ is the Finslerian distance function.
\end{prop}
\begin{proof}
We fix a basis of $T_xM$ and consider the $C^1$--differentiable chart\\
$(B_\eta(x),\psi)$, with $\psi=u^{-1}\circ Exp_x^{-1}$. Moreover, we can
suppose $x\not=c(t)\in B_\eta(x)$, for any $t\in(0,a)$. Then all the
derivatives of the function $f(t)=d^2(x,c(t))=F^2(x,\gamma(t))$, being
$\gamma(t)=\psi(c(t))$, for any $t\in(0,a)$, are defined. Hence by
using the homogeneity conditions, we get:
\begin{eqnarray*}
& &
f(t_0)=\varepsilon^2\ ;\quad \dot f(t_0)=
g_{(x,\gamma(t_0))}(\gamma(t_0),\dot\gamma(t_0))\
;
\nonumber
\\
& &
\ddot f(t_0)=
g_{(x,\gamma(t_0))}(\dot\gamma(t_0),\dot\gamma(t_0))+
g_{(x,\gamma(t_0))}(\gamma(t_0),\ddot\gamma(t_0))\ .
\end{eqnarray*}
Therefore, the proof follows as in \cite{Kl} (cf. \cite{H}, too).
\end{proof}

The proof of Proposition 1.3 follows from the previous proposition in a
trivial way.

We can also omit the proof of the following Lemma, because it needs only the
compactness of boundary of convex neighbourhoods, which holds, because an
indicatrix is always compact and the exponential map is a diffeomorphism
near to any point. Hence, its proof follows as in \cite{Kl}.
\begin{lem}
Let $p\in M$ and $\rho>0$ be a positive real number such
that  $Exp_p$ is defined on the ball $B_\rho(0_p)\subseteq T_pM$. Then,
every $q\in M$, with $d(p,q)<\rho$, can be joined to $p$ by a minimizing
geodesic.
\end{lem}

Finally, the proof of Proposition 1.4 is the same as in the Riemannian case
(see, e.g., \cite{Kl}).

config: arxiv


TITLE: Left half open topology
QUESTION [0 upvotes]: Given the topology $\tau$ on X= $\mathbb{R}$, which contains all sets $(a,b]$ with a,b$\in\mathbb{R}$, $a<b$, show that the intervals $(a,b), (a, \infty)$ and $(-\infty,b]$ are also in $\tau$.
Can I say that $(-\infty, b]\cap (a,\infty)=(a,b]$ and thus they are in $\tau$?  Or would one rather construct an infinite union such as $\bigcup_{i \in \mathbb{N}} A_i$ where $A_i=(a, b+i]$ which is open in $\tau$.
How would one formally show that $[a,b)$ is not open in $\tau$ because it cannot be expressed as a union or intersection of open intervals?
I greatly appreciate help with these rather 'easy' topology questions since it's the first week of the topic and I am not yet familiar with it.

REPLY [2 votes]: $$(a,b)=\bigcup_{x \in (a,b)} (a,x]$$
$$(a,+\infty) = \bigcup_{x > a} (a,x]$$
$$(-\infty, b] = \bigcup_{x < b} (x,b]$$
shows that all those types of intervals can be written as unions of basic open sets, and thus are open too.


TITLE: Condition for two elements of An to be conjugated
QUESTION [0 upvotes]: I want to find how the conjugation classes partition the A5, I know I have 24 5-cycles, but since 24 does not divide 60, then those 24 5-cycles are certainly not in the same class of conjugation, a teacher told me that two elements In Sn are conjugated, if and only if it has the same cyclic structure, I imagined that this definition would be equivalent to the number of disjoint cycles in which the permutation can be decomposed, but that would put all 5-cycles in the same class.
Could anyone give a more precise definition and say why these 24 5-cycles divide into two classes of 12 elements each?

REPLY [0 votes]: Here is a more general answer which might be helpful to you. This concerns any index 2 subgroup of a finite group G.
Let $H$ be a index two subgroup of $G$. Since H is necessarily normal it is a union of conjugacy classes. Let $H$ and $Hs$ be two right cosets of $H$ in $G$. Let $x \in H$. Denote $C_{G}(x)$ and $C_{H}(x)$ as the conjugacy class of $x$ in $G$ and $H$ respectively.
Claim: $C_{G}(x)=C_{H}(x) \cup C_{H}(sxs^{-1})$
proof: $C_{G}(x)=\{\ gxg^{-1}|g \in G \}\ $
But since $G=H \cup Hs$ we can write $C_{G}(x)=\{\ hxh^{-1}|h \in H \}\ \cup \{\ hsxs^{-1}h^{-1}|h \in H \}\ $
which clearly implies the claim.
Now, if $sxs^{-1}$ is H-conjugate to $x$, then there exists $h \in H $ s.t $sxs^{-1}=hxh^{-1}$, which implies that  $h^{-1}s$ is in the centraliser of $x$. So,  $sxs^{-1}$ is H-conjugate to $x$ iff $Z_{G}(x) \cap Hs \neq \phi$, where $Z_{G}(x)$ denotes the centraliser of $x$ in $G$. So, the conjugacy class $C_{G}(x)$ will either remain intact inside H if $Z_{G}(x) \cap Hs \neq \phi$, or else if $Z_{G}(x) \cap Hs =\phi$, then it will split into two conjugacy classes having one representative as $x$ and the other representative as $sxs^{-1}$, where $s$ is an arbitrarily chosen non-trivial right coset representative of $H$ in $G$.
Now How can you use this discussion in your question. Clearly $A_{n}$ is index-2 subgroup of $S_{n}$. Suppose $\sigma \in S_{n}$ is a permutation which also belongs to $A_{n}$. You want to find out what happens to its conjugacy class inside $A_{n}$.
There is a great theorem which goes as follows:
If $\sigma \in S_{n}$. Then the cycle type  of $\sigma$ consists of distinct odd integers in its cycle type iff $\sigma$ doesn't commute with any odd permutation.
This  above theorem along with the discussions above shows that if $\sigma \in A_{n}$ , then conjugacy class of $\sigma$ breaks into two classes of equal size iff the cycle type of $\sigma$ consists of distinct odd integers.
In your case $(1 2 3 4 5) \in A_{n}$ has cycle type (5) which obviously has cycle  type comprising of distinct odd integers. So, the conjugacy class breaks into two equal classes in $A_{n}$. You can find the other representative easily by choosing any odd permutation s, say (1 2). Then $(1 2)(1 2 3 4 5)(1 2)^{-1}=(2 1 3 4 5).$
But for $A_{5}$, we don't have to go through the theorem that I have stated. You can just see the above discussion and realise that $(1 2 3 4 5)$ doesn't commute with any odd permutation, hence giving your result.


TITLE: Proof convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$
QUESTION [2 upvotes]: To prove the convergence of $\sum\limits_{n = 1}^\infty\frac {n!} {n^n}$ I used that $\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$ has to be $<1$:
$$\lim\limits_{n\to\infty}|\frac {a_{n+1}} {a_n}|$$
$$=\lim\limits_{n\to\infty}|\frac {\frac {(n+1)!} {(n+1)^{n+1}}} {\frac {n!} {n^n}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)!\cdot n^n} {n!\cdot (n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {(n+1)\cdot n^n} {(n+1)^{n+1}}|$$
$$=\lim\limits_{n\to\infty}|\frac {n^n} {(n+1)^n}|$$
$$=\lim\limits_{n\to\infty}|{(\frac {n} {n+1})}^n|=\frac 1 e$$
Okay... problem solved after thankful advice

REPLY [1 votes]: We can also use the Comparison Test: $$0\le \frac{n!}{n^n}=\frac{1\cdot 2\cdot 3\cdot\ldots\cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}\le \frac{1\cdot 2\cdot n\cdot\ldots \cdot n}{n\cdot n\cdot n\cdot\ldots\cdot n}=\frac{2}{n^2},$$ and $\displaystyle \sum_{n=1}^{+\infty}\frac{2}{n^2}$is convergent, so the given series is convergent.


TITLE: convolution of Poisson kernel with itself
QUESTION [1 upvotes]: I am trying to prove that $p(l,x)*p(u,x)=p(l+u,x)$ where * denotes the convolution of two functions and $p(l,x)=\dfrac{l}{\pi(l^2+x^2)}$. I am having trouble in integrating the left hand side.
$$\int_{-\infty}^\infty \frac{l}{l^2+(x-y)^2}.\frac{u}{u^2+y^2}dy,$$
this is the definition of convolution and is the left hand side of the above equation. Putting the command
Integrate[(l/(l^2+(x-y)^2))*(u/(u^2+y^2)),{y,-infinity,infinity}]
into Mathetica gives me $p(l-u,x)$, instead of $p(l+u,x)$. please help!

REPLY [1 votes]: It's interesting, but this equality can be proved with the help of characteristic functions from the theory of probability.
Let's take 2-sided normal distribution,its density is expressed with a function
$$p(x)=\frac{a}{2}e^{-a\left|x\right|} $$
Now we found characteristic function of distribution (or, as in analysis, Fourier transformation).
$$g(t)=\int_{-\infty}^{+\infty}e^{itx}p(x)dx =\frac{a^2}{a^2+t^2} $$
Then we use the formula of inverse transformation
$$\frac{a}{2}e^{-a\left|t\right|}= \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{-itx}\frac{a^2}{a^2+t^2}dt $$
$$e^{-a\left|t\right|} =\frac{1}{\pi}\int_{-\infty}^{+\infty}e^{itx}\frac{a}{a^2+t^2}dt$$
That means, that $g(t)=e^{-a\left|t\right|}$ is characteristic function of Cauchy distribution. Because of the property of characteristic functions
$$g_{a*b}=g_{a}g_{b} $$
$$g_{p_{l}(x)*p_{u}(x)}(t)=e^{-l\left|t\right|}e^{-u\left|t\right|}=e^{-(l+u)\left|t\right|}, $$
which means that $$p_{l}(x)*p_{u}(x)=p_{u+l}(x) $$


TITLE: Stationary Phase and Propagation Speed (Reference)
QUESTION [1 upvotes]: I'm trying to understand how one can make precise statements about propagation speed for various (linear and nonlinear) PDEs (in particular, ones with infinite propagation speed) and what, if anything, these precise statements can be used to prove (which isn't trivial by other methods).  An example set of questions I'd be interested in learning more about is the following:
Suppose we have a nice initial data $\varphi \in \mathcal{S}(\mathbb{R}^n)$ which is frequency supported in $|\xi| \leq N$.  Let $u$ be the resulting (strong) solution to a linear or nonlinear Schrodinger equation with initial data $\varphi$.  In what sense does $u$ propagate with speed at most $N$ in physical space (as predicted by the formula ''speed of propagation equals minus the gradient of the dispersion relation'' together with the initial frequency localization of $u$)?  If, for example, $u$ at time zero (i.e., $\varphi$) has most of its $L^2$ norm in the set $|x| \leq M$, does most of its $L^2$ norm remain in the set $|x| \leq M + TN$ when $|t| \leq T$?  What about $H^k$ norms (for integral $k \geq 1$, say)?  How is this behavior different in the linear and nonlinear cases?  What if we only assume that $\varphi$ has ''most'' of its frequency support in $|\xi| \leq N$ (as quantified by some weighted $L^2$ norm in frequency space)?
Tao mentions in Nonlinear Dispersive Equations that one can use the method of stationary phase to give precise meaning to statements about propagation speed, but I don't believe he gives any references.  So, I'd be interested in any references which cover questions like the ones indicated above.

REPLY [2 votes]: Here is one response for the case of the linear Schrödinger equation.  In this case, you can write a formula for solutions, namely
$$
u(t,x) = (4\pi i t)^{-d/2} \int_{\mathbb{R}^d} e^{i|x-y|^2/4t} \phi(y)\,dy
$$
is the solution to $i\partial_t u + \Delta u =0$ with initial data $\phi$.
From this you can read off the "Fraunhofer" approximation
$$
u(t,x) \sim u_F(t,x):= (4\pi i t)^{-d/2} e^{i|x|^2/4t} \hat \phi(\tfrac{x}{2t}),
$$
where $\hat\phi$ is the Fourier transform.  You could alternatively derive this asymptotic by solving the PDE with the Fourier transform and then taking the stationary phase approximation in the resulting integral expression.
Comparing with the true formula (and doing a change of variables and using Plancherel), one has
$$
\| u(t) - u_F(t) \|_{L^2} = \|(1-e^{i|y|^2/4t})\phi\|_{L^2} \lesssim t^{-1}\|y^2\phi\|_{L^2} \lesssim t^{-1},
$$
since we assume $\phi$ is Schwartz.
Now, if $\hat\phi$ is supported where $|\xi|\leq N$, then $u_F(t)$ is supported where $|x|\leq 2Nt$.  So, using the approximation above, one finds that
$$
\|u(t)\|_{L^2(|x|>2Nt)} \lesssim t^{-1},
$$
which gives an approximate finite speed of propagation statement (at least for large $t$).


TITLE: Prove that $\mathbb{C}^*$ is isomorphic to a subgroup of $GL_2(\mathbb{R})$
QUESTION [2 upvotes]: Prove that $\mathbb{C}^*$ is isomorphic to the subgroup of $GL_2(\mathbb{R})$ consisting of matrices
  of the form
  $$\begin{bmatrix}
a & b\\
-b &  a
\end{bmatrix}$$

I showed that $\mathbb{C}^*$ and $GL_2(\mathbb{R})$ are homomorphic by proving that $\Phi(z$ o  $w)$ = $\Phi (z)$ . $\Phi (w)$ where $z,w$ are elements of  $\mathbb{C}^*$.
One-to one? I have shown that for any two complex $z$ and $w$ when $$\Phi (z) =\Phi (w) => z=w$$
Onto?
Let
$$\begin{bmatrix}
m & n\\
p &  q
\end{bmatrix} \in GL_2(\mathbb{R})$$
and let $$\Phi (a+bi) = \begin{bmatrix} m & n \\ p & q\end {bmatrix}.$$
Then $$\begin{bmatrix}
a & b\\
-b &  a
\end{bmatrix} = \begin{bmatrix} m & n \\ p & q\end {bmatrix}$$
BUT then what can be said about that to prove that the relation is onto? i am a bit stuck here.

REPLY [4 votes]: Your map $\Phi$ isn't to $GL_2(\Bbb R)$ itself, but rather the subgroup  $G = \left\{\pmatrix{a & b \\ -b & a}: a, b \in \Bbb R\right\}$. The instructions aren't incredibly explicit, but it is very important here that $\Phi$ has codomain $G$, rather than all of $GL_2(\Bbb R)$.
When you pick a matrix $M =\pmatrix{m&n\\p&q}$, you're not necessarily picking something in $G$, the codomain of $\Phi$. It may very well be the case that there's no $z \in \Bbb C^*$ such that $\Phi(z) = M$; this happens if and only if $q = m$ and $p = -n$. So if you pick the $M$ above as your generic matrix and try to find some $z \in \Bbb C^*$ so that $\Phi(z) = M$, you had better not be able to show that $\Phi$ is onto $GL_2(\Bbb R)$ -- it's not!
Instead, to show that $\Phi$ is surjective ("onto"), pick something in the codomain of $\Phi$, something in $G$.
So pick a generic matrix $M = \pmatrix{m&n\\-n&m} \in G$, and find the corresponding $z \in \Bbb C^*$ with $\Phi(z) = M$.

In actuality, defining $\Phi$ as above but making its codomain all of $GL_2(\Bbb R)$ instead of just $G$ isn't a huge problem. It's a general fact that for any group homomorphism $\varphi: H \to H'$, if $\varphi$ is injective, then $H$ is isomorphic to the image $\operatorname{im}(\varphi) = \{\varphi(h) : h \in H\}$ (this is a consequence of the first isomorphism theorem).
Since you've shown that $\Phi$ is an injective homomorphism from $\Bbb C^*$, then $\Bbb C^*$ is necessarily isomorphic to $\operatorname{im}(\Phi)$, which is exactly the $G$ defined in the problem statement. It's just that, unless the codomain of $\Phi$ is exactly $\operatorname{im}(\Phi)$, you won't be able to show that $\Phi$ is surjective.


TITLE: An interesting mathematics task.
QUESTION [0 upvotes]: Find the number of different ways of arrangement of all natural numbers from 1 to 9 inclusive, one in table cells measuring 3 by 3 such that the sum of the numbers in each row and each column are equal. The table can not be turned and reflected.

REPLY [1 votes]: Observations
(1) If we add up by rows, we get a sum of $1+2+\cdots+9=45$. So each row sum (and so also each column sum) must be $15$.
(2) There are $8$ possible sums to $15$ using the distinct numbers given.  They are:
$1+5+9$
$1+6+8$
$2+4+9$
$2+5+8$
$2+6+7$
$3+4+8$
$3+5+7$
$4+5+6$
(3)  In a legal array, each number appears in two sums (one for its row and one for its column).  Because each of the numbers $1, 3, 7, 9$ each appear exactly twice in the above list; the rows and columns of the table must use the sums
$1+5+9$; $1+6+8$; $2+4+9$; $2+6+7$; $3+4+8$; and $3+5+7$.
(4) Once $1$ and its row and column are filled in, the rest of the table is determined.  There are nine choices for $1$; four choices for $5$; and two choices for $6$.  The rest of the positions are determined.  For example
$$\begin{matrix}5&?&?\\1&?&6\\?&?&? \end{matrix}$$
leads to
$$\begin{matrix}5&3&7\\1&8&6\\9&4&2 \end{matrix}$$
Hence there are $9\cdot 4\cdot 2=72$ possible placements, although this counts rotations and reflections.  So this leads to $\frac{72}{8}=9$ possibilities if rotations and reflections are not counted separately.
It is interesting to note that given an arrangement; one may permute rows (6 ways); permute columns (6 ways); and do matrix transposition or not do matrix transposition (2 ways).  So given one arrangement all other arrangements are obtained by row permutations, column permutations, and matrix transposition.


TITLE: differentiable manifold definition
QUESTION [1 upvotes]: For a definition of differentiable manifold, we require transition maps to be diffeomorphisms. Do we also require that for every chart $(U,V,\phi)$ map $\phi: U \to V$ is diffeomorphism, or just a homeomorphism?

REPLY [3 votes]: Given a topological manifold you only require that the charts are homeomorphisms and that the transition functions are diffeomorphisms. Saying that the charts are differentiable does not make sense in the topological setting. But if you have a differentiable atlas you can define that any map from or to the manifold is differentiable if its concatenations with the charts are differentiable. So after you have defined such a differentiable structure the charts are automatically diffeomorphisms.
As for $dφ_p$: After you have defined such a differentiable structure on the manifold, the map $φ$ is just another map between two manifolds where one happens to be $V ⊂ ℝ^n$. Then the notion of tangent space is defined on both of them and for $φ: U → V$ when $p\in U, φ(p) = 0 \in V$ we have that $dφ_p$ is the induced map $T_pU → T_0V \simeq ℝ^n$. Depending on how you construct the tangent spaces also the understanding of $dφ$ will vary. If, like in your MIT text book, you define $T_pU$ to be the space of curves in $U$ through $p$ that “point in the same direction”, all you have to do, to calculate $dφ_p$ is take a representative curve $γ$ in $U$ and push it to $V$ like $φ∘γ$ which gives you a curve in $V$ that represents some element of the tangent space $T_0V$.
Of course in this special case for $φ$ being a chart you can also give a basis of $T_pU$ in terms of $φ$ just by taking the coordinate curves $γ_i : t ↦ (0, …, 0, t, 0, …, 0) \in V$ and then defining $X_i := [φ^{-1}∘γ_i]_{T_pM}$ where $[\cdot]$ stands for taking all curves in the same direction.


TITLE: Closed geodesic in a ball in hyperbolic surface
QUESTION [1 upvotes]: Let S be a hyperbolic surface. Let $\phi : B_{S}(x,R) \rightarrow B_{H}(\phi(x),R)$ is an isometry from a ball in S to a ball in hyperbolic plane.
I claim that there is no locally minimizing closed geodesic in $B_{S}(x,R)$.
This is my idea:
Suppose that there is a locally minimizing closed geodesic in $B_{S}(x,R)$, namely $\gamma$. Then $\phi\circ\gamma$ is a locally minimizing closed geodesic in $B_{H}(\phi(x),R)$. I know that is a contradiction because there is no closed geodesic in $H$ but I can not write down a detail proof for it. Hoping someone can help me! Thank in advance !

REPLY [1 votes]: You need to assume that $R$ is less than the injectivity radius of $S$. If $R$ is greater than the injectivity radius, then $B_S(x,R)$ is not simply connected and there is no isometry from $B_S(x,R)$ to a ball in the hyperbolic plane.
If $R$ is less than the injectivity radius of $S$, then your argument holds. It would work to appeal to the classification of geodesics in one of the common models of the hyperbolic plane, e.g., geodesics of the Poincare disk model of $H$ are exactly the Euclidean circles which are perpendicular to the unit circle.


TITLE: Prove that $x^2 + 6x + 9$ is even iff $x$ is odd
QUESTION [0 upvotes]: Prove. $x^2 + 6x + 9$ is even iff $x$ is odd.
I had first proved this using a contrapositive statement. However, I was told I was not making a distinction between an if and only if statement, and a conditional statement.
I then wrote the problem again and tried it how I was told. I wrote both out both parts of the iff statement, and I started off with assuming $x$ was odd, or $2k+1$. Then I plugged $2k +1$ into $x^2 + 6x + 9$ to get $4k^2 + 16k + 16$. However, after this I am lost.

REPLY [1 votes]: Simply work in $\mathbf Z/2\mathbf Z$ and use lil' Fermat:
$$x^2+6x+9\equiv x+1\bmod 2\equiv \begin{cases}0&\text{if } x\equiv 1\bmod2\\ 1,&\text{if } x\equiv 0\bmod2.\end{cases}
$$


TITLE: How do we know photons have spin 1?
QUESTION [53 upvotes]: Electrons have spin 1/2, and as they are charged, they also have an associated magnetic moment, which can be measured by an electron beam splitting up in an inhomogeneous magnetic field or through the interaction of the electrons's magnetic moment with an external magnetic field in spectroscopic measurements.
On the other hand, a photon is neutral - how can one measure its spin if there's no magnetic moment? How do we know it has spin 1?

REPLY [6 votes]: In Richard Beth's Mechanical Detection and Measurement of the Angular Momentum of Light, bright light from a mercury arc lamp was circularly polarized and passed through a half-wave plate (which reverses the sense of circular polarization) attached to a torsion pendulum.  A bit of clever experimental design sent the light through the half-wave plate twice so that each photon would exchange angular momentum $4\hbar$ with the fiber.  Beth confirmed this prediction, and also showed it was consistent with angular momentum stored in the fields of classical electromagnetism.