text
stringlengths 0
6.23M
| __index_level_0__
int64 0
419k
|
|---|---|
\begin{document}
\title{On Gaussian multiplicative chaos}
\author{Alexander Shamov\thanks{E-mail address: \protect\href{mailto:trefoils@gmail.com}{trefoils@gmail.com}}}
\date{July 16, 2014\\
revised Mar 08, 2016}
\maketitle
\noindent
\begin{lyxgreyedout}
\global\long\def\op#1{\operatorname{#1}}
\global\long\def\midmid{\mathrel{}\middle|\mathrel{}}
\global\long\def\NN{\mathbb{N}}
\global\long\def\ZZ{\mathbb{Z}}
\global\long\def\QQ{\mathbb{Q}}
\global\long\def\RR{\mathbb{R}}
\global\long\def\CC{\mathbb{C}}
\global\long\def\HH{\mathbb{H}}
\global\long\def\KK{\mathbb{K}}
\global\long\def\FF{\mathbb{F}}
\global\long\def\P{\op{\mathsf{P}}}
\global\long\def\Q{\op{\mathsf{Q}}}
\global\long\def\E{\op{\mathsf{E}}}
\global\long\def\OMEGA{\op{\mathsf{\Omega}}}
\global\long\def\I{\op{\mathsf{1}}}
\global\long\def\Var{\op{\mathsf{Var}}}
\global\long\def\H{\op{\mathsf{H}}}
\global\long\def\Law{\op{\mathsf{Law}}}
\global\long\def\Cov{\op{\mathsf{Cov}}}
\global\long\def\Pc#1#2{\P\left\{ #1\midmid#2\right\} }
\global\long\def\Pcsq#1#2{\P\left[#1\midmid#2\right]}
\global\long\def\Ec#1#2{\E\left[#1\midmid#2\right]}
\global\long\def\Ecm#1#2#3{\E_{#1}\left[#2\midmid#3\right]}
\global\long\def\Varc#1#2{\Var\left[#1\midmid#2\right]}
\global\long\def\Hc#1#2{\H\left[#1\midmid#2\right]}
\global\long\def\Lawc#1#2{\Law\left[#1\midmid#2\right]}
\global\long\def\Lawcm#1#2#3{\Law_{#1}\left[#2\midmid#3\right]}
\global\long\def\Covc#1#2{\Cov\left[#1\midmid#2\right]}
\global\long\def\Wick#1{{:\!#1\!:}}
\global\long\def\norm#1{\left\Vert #1\right\Vert }
\global\long\def\const{\op{const}}
\global\long\def\diam{\op{diam}}
\global\long\def\diag{\op{diag}}
\global\long\def\det{\op{det}}
\global\long\def\per{\op{per}}
\global\long\def\pf{\op{pf}}
\global\long\def\sgn{\op{sgn}}
\global\long\def\Leb{\op{Leb}}
\global\long\def\tr{\op{tr}}
\global\long\def\span{\op{span}}
\global\long\def\supp{\op{supp}}
\global\long\def\Spec{\op{Spec}}
\global\long\def\rank{\op{rank}}
\global\long\def\im{\op{im}}
\global\long\def\coker{\op{coker}}
\global\long\def\coim{\op{coim}}
\global\long\def\colim{\op{colim}}
\global\long\def\cone{\op{cone}}
\global\long\def\cyl{\op{cyl}}
\global\long\def\Hom{\op{Hom}}
\global\long\def\Ext{\op{Ext}}
\global\long\def\Tor{\op{Tor}}
\global\long\def\RKHS{\op{RKHS}}
\global\long\def\pr{\op{pr}}
\global\long\def\id{\op{id}}
\global\long\def\Sets{\op{\mathbf{Sets}}}
\global\long\def\FinSets{\op{\mathbf{FinSets}}}
\global\long\def\Groups{\op{\mathbf{Groups}}}
\global\long\def\AbGroups{\op{\mathbf{AbGroups}}}
\global\long\def\Mod{\op{\mathbf{Mod}}}
\global\long\def\Modl#1{_{#1}\mathbf{Mod}}
\global\long\def\Modr#1{\Mod_{#1}}
\global\long\def\Modb#1#2{_{#1}\mathbf{Mod}_{#2}}
\global\long\def\Top{\op{\mathbf{Top}}}
\end{lyxgreyedout}
\begin{comment}
Available macros: \textbackslash{}op \textbackslash{}midmid \textbackslash{}NN
\textbackslash{}ZZ \textbackslash{}QQ \textbackslash{}RR \textbackslash{}CC
\textbackslash{}HH \textbackslash{}KK \textbackslash{}FF \textbackslash{}P
\textbackslash{}Q \textbackslash{}E \textbackslash{}OMEGA \textbackslash{}I
\textbackslash{}Var \textbackslash{}H \textbackslash{}Law \textbackslash{}Cov
\textbackslash{}Pc \textbackslash{}Pcsq \textbackslash{}Ec \textbackslash{}Ecm
\textbackslash{}Varc \textbackslash{}Hc \textbackslash{}Lawc \textbackslash{}Lawcm
\textbackslash{}Covc \textbackslash{}Wick \textbackslash{}norm \textbackslash{}const
\textbackslash{}diam \textbackslash{}sgn \textbackslash{}diag \textbackslash{}det
\textbackslash{}per \textbackslash{}pf \textbackslash{}Leb \textbackslash{}tr
\textbackslash{}span \textbackslash{}supp \textbackslash{}Spec \textbackslash{}rank
\textbackslash{}im \textbackslash{}coker \textbackslash{}coim \textbackslash{}colim
\textbackslash{}cone \textbackslash{}cyl \textbackslash{}Hom \textbackslash{}Ext
\textbackslash{}Tor \textbackslash{}RKHS \textbackslash{}pr \textbackslash{}id
\textbackslash{}Sets \textbackslash{}FinSets \textbackslash{}Groups
\textbackslash{}AbGroups \textbackslash{}Mod \textbackslash{}Modl
\textbackslash{}Modr \textbackslash{}Modb \textbackslash{}Top
\end{comment}
\begin{abstract}
We propose a new definition of the Gaussian multiplicative chaos and
an approach based on the relation of subcritical Gaussian multiplicative
chaos to randomized shifts of a Gaussian measure. Using this relation
we prove general results on uniqueness and convergence for subcritical
Gaussian multiplicative chaos that hold for Gaussian fields with arbitrary
covariance kernels.
Keywords: Gaussian multiplicative chaos; Random measures; Gaussian
measures
MSC 2010 subject classification: 60G15, 60G57, 60B10
\end{abstract}
\tableofcontents{}
\section{Introduction}
\subsection{The object of interest}
Let $\left(\mathcal{T},\mu\right)$ be a finite measure space, and
let $X=\left(X\left(\omega,t\right)\right)_{\omega\in\OMEGA,t\in\mathcal{T}}$
be a Gaussian field parametrized by $t\in\mathcal{T}$ and defined
on a probability space $\left(\OMEGA,\P\right)$. With this data one
can associate the following random measure:
\begin{equation}
M\left(dt\right):=\exp\left[X\left(t\right)-\frac{1}{2}\E\left|X\left(t\right)\right|^{2}\right]\mu\left(dt\right).\label{eq:FormalGMC}
\end{equation}
The Gaussian multiplicative chaos (GMC) is the natural generalization
of such a random measure to the setting when the field $\left(X\left(t\right)\right)$
is defined in a distributional sense rather than pointwise, i.e. via
a family of formal ``integrals'' against test functions from an
appropriate class. Obviously, for such generalized Gaussian fields
\eqref{eq:FormalGMC} does not make sense literally, since $X\left(t\right)$
need not be well-defined as a random variable for any particular $t$.
Accordingly, in nontrivial cases $M$ is almost surely $\mu$-singular,
so the density $M\left(dt\right)/\mu\left(dt\right)$ is not well-defined
either.
The commonly used ways of interpreting \eqref{eq:FormalGMC} rigorously
and constructing such random measures proceed by approximating the
field $X$ by Gaussian fields $X_{n}$ that are, unlike $X$, defined
pointwise. One defines a GMC $M$ as a limit, in an appropriate topology,
of random measures
\begin{equation}
M_{n}\left(dt\right):=\exp\left[X_{n}\left(t\right)-\frac{1}{2}\E\left|X_{n}\left(t\right)\right|^{2}\right]\mu\left(dt\right).\label{eq:ApproxGMC}
\end{equation}
This approach leads naturally to the following problems, both of which
will be addressed in this paper.
\begin{problem}
Find conditions on the approximation $X_{n}\to X$ that are sufficient
for convergence of $M_{n}$. \label{prob:Convergence}
\end{problem}
\begin{problem}
Prove that the limit is independent of the approximation procedure.
\label{prob:Uniqueness}
\end{problem}
As far as we know, in previous works these problems have only been
partially solved under unnecessarily restrictive assumptions. Below
we provide an overview of the commonly used approximation procedures.
\subsubsection{Martingale approximation \cite{KahaneSur}}
The martingale approximation is employed in Kahane's original work
on GMC in \cite{KahaneSur}. In his construction the increments $X_{n}-X_{n-1}$
are independent (and $X_{0}:=0$), which implies that $\left(M_{n}\right)$
is a positive measure-valued martingale. The martingale property guarantees
that $M_{n}$ converges to a random measure $M$ in the sense that
$M_{n}\left[A\right]\to M\left[A\right]$ almost surely for any fixed
measurable set $A$. Moreover, $\E M=\mu$ iff the martingale $\left(M_{n}\left[\mathcal{T}\right]\right)$
is uniformly integrable, in which case the limit $M$ is taken as
the interpretation of \eqref{eq:FormalGMC}.
One intuitively expects the martingale approximation to yield the
``right'' and completely general notion of subcritical GMC. However,
Kahane's work falls short of establishing the basic setup in sufficient
generality.
The construction in \cite{KahaneSur} takes as its input a function
$K:\mathcal{T}\times\mathcal{T}\to\RR_{+}\cup\left\{ \infty\right\} $
--- thought of as the covariance kernel of the Gaussian field $X$.
$K$ is assumed to be decomposable into a sum
\begin{equation}
K\left(t,s\right):=\sum_{n}p_{n}\left(t,s\right)\label{eq:SigmaPositive}
\end{equation}
of kernels $p_{n}:\mathcal{T}\times\mathcal{T}\to\RR_{+}$ that are
both positive definite and \emph{positive in the pointwise sense}
(i.e. $p_{n}\left(t,s\right)\ge0,\forall t,s$). We may assume that
$K$ is finite $\mu\otimes\mu$-almost everywhere, but it may explode
on the diagonal. Unlike the sum $K$, the kernels $p_{n}$ are only
allowed to take finite values, so that indeed there are independent
Gaussian fields $X_{n}-X_{n-1}$ with covariance
\[
\E\left(X_{n}\left(t\right)-X_{n-1}\left(t\right)\right)\left(X_{n}\left(s\right)-X_{n-1}\left(s\right)\right):=p_{n}\left(t,s\right).
\]
The kernels $K$ that admit a representation \eqref{eq:SigmaPositive}
(with $p_{n}$ continuous with respect to a given compact metrizable
topology on $\mathcal{T}$) are said to be of $\sigma$-positive type.
Under this $\sigma$-positivity assumption it is proved in \cite{KahaneSur}
that the \emph{law} of the limiting random measure $M$ is independent
of the decomposition \eqref{eq:SigmaPositive}.
One problem with this approach is that due to the pointwise positivity
assumption $p_{n}\left(t,s\right)\ge0$, $\sigma$-positivity is both
unnecessarily restrictive and hard to check in practice. Another problem
is that while $M$ is naturally defined on the same probability space
with the underlying Gaussian randomness, Kahane only proves uniqueness
in law rather than uniqueness of $M$ as a function of the Gaussian
field. That is, his result does not rule out the possibility that
different decompositions of $X$ yield different random measures with
the same law.
\subsubsection{Mollifying operators \cite{RVRevisited,DuShKPZ,RVReview}}
This approximation technique restricts the generality to an important
special case where $\mathcal{T}$ is a domain in $\RR^{d}$, $\mu$
is the Lebesgue measure, and the covariance kernel $K$ of the field
$X$ has the special form
\begin{equation}
K\left(t,s\right):=\gamma^{2}\log^{+}\norm{t-s}^{-1}+g\left(t,s\right),\label{eq:LogKernel}
\end{equation}
where $\log^{+}:=\max\left(\log,0\right)$, the function $g:\mathcal{T}\times\mathcal{T}\to\RR$
is bounded and continuous, and $\gamma^{2}<2d$. In this case $X$
is obviously well-defined as a random distribution, i.e. it can be
integrated against smooth test functions. The fields $X_{n}$ are
constructed by convolution:
\begin{equation}
X_{n}\left(t\right):=\intop_{\mathcal{T}}X\left(t^{\prime}\right)\psi_{1/n}\left(t-t^{\prime}\right)dt^{\prime},\label{eq:Convolution}
\end{equation}
\[
\psi_{1/n}\left(x\right):=n^{d}\psi\left(nx\right),
\]
where $\psi:\RR^{d}\to\RR_{+}$, $\intop\psi\left(t\right)dt=1$,
subject to appropriate smoothness conditions. This approximation method
was used in \cite{RVRevisited} for stationary fields on $\mathcal{T}=\RR^{d}$,
and according to \cite{RVReview}, the same techniques apply to the
non-stationary setting. Unlike in Kahane's approach, for convolution
approximations the convergence of $M_{n}$ is a nontrivial fact. In
\cite{RVRevisited} it is proved that $M_{n}$ converges in law to
some $M$, and that $\Law M$ is independent of the choice of the
mollifier $\psi$. Naturally, one expects the stronger result that
$M_{n}$ converges almost surely or at least in probability rather
than just in law. Similarly, the random measure $M$ should be unique
not just in law but as a function of the Gaussian field $X$.
In \cite{DuShKPZ} a related construction with circle averages was
used in the special case where $X$ is the Gaussian free field in
dimension 2. In this very special setting the authors prove almost
sure convergence of $M_{n}$.
\subsection{The new definition}
In this paper we introduce a new definition of GMC that is based on
our view of $M$ as a function of the field $X$ rather than a standalone
random measure.
Recall that a Cameron-Martin vector (or admissible shift) of the Gaussian
field $X$ is a deterministic function $\xi$ on $\mathcal{T}$, such
that the distribution of $X+\xi$ is absolutely continuous with respect
to that of $X$\footnote{The Cameron-Martin space is dual to the Hilbert space of measurable
linear functionals of the field, i.e. ``test functions''. For the
reason explained in Appendix in our setting the space of test functions
contains $L^{2}\left(\mathcal{T},\mu^{\prime}\right)$ for some equivalent
measure $\mu^{\prime}\sim\mu$, so that Cameron-Martin vectors are
representable by $\mu$-equivalence classes of functions.}. We denote by $H$ the space of Cameron-Martin vectors.
Our starting point is the following basic observation: the ``exponential''
behavior \eqref{eq:FormalGMC} of a GMC $M$ can be characterized
by the way $M$ changes when the field $X$ is shifted by Cameron-Martin
vectors. Namely, for all $\xi\in H$ the following should hold almost
surely:
\begin{equation}
M\left(X+\xi,dt\right)=e^{\xi\left(t\right)}M\left(X,dt\right).\label{eq:DefGMC4Field}
\end{equation}
This property is taken as our definition of GMC. A GMC is called \emph{subcritical}
if $\E M$ is $\sigma$-finite, in which case we will often assume
for convenience that $\E M=\mu$. This does not restrict generality,
as will be explained in Remark \ref{rem:EM}. In this paper we only
deal with subcritical GMC theory.
One obvious feature of our definition is that unlike the previous
ones, it is not tied to any particular \emph{construction} of GMC.
On the other hand, for any particular construction that exhibits $M$
as a function of $X$ it is typically easy to check that it satisfies
\eqref{eq:DefGMC4Field}, at least for a dense subspace of Cameron-Martin
shifts, which turns out to be enough. This facilitates the comparison
of different constructions; in particular, the seemingly complicated
problem of independence of the approximation procedure reduces to
the uniqueness problem for our notion of subcritical GMC, which turns
out to be remarkably easy.
In terms of generality, our notion of GMC includes both Kahane's GMC
\cite{KahaneSur} and the subcritical and critical GMC over logarithmic
fields as constructed in \cite{RVRevisited,DuShKPZ,RVReview}, with
the caveat that it retains information about the dependence on the
underlying Gaussian field. On the other hand, it does not include
distributional limits of GMCs that are not measurable with respect
to the Gaussian field even when properly coupled to it, most notably
the atomic supercritical GMC \cite{MRVSupercritical,BJRVKPZ}, nor
does it include the complex GMC \cite{LRVComplex}, which is not a
random positive measure but rather a complex-valued field of almost
surely infinite variation.
\subsection{Approximation}
The main result of the paper is a general approximation theorem for
subcritical GMC. A version of it can be stated as follows.
We assume that $X$ is defined on test functions in $L^{2}\left(\mu\right)$,
so that the Cameron-Martin space $H$ is embedded into $L^{2}\left(\mu\right)$.
Let $A_{n}:H\to L^{2}\left(\mu\right)$ be bounded operators that
converge strongly to the identity embedding $H\to L^{2}\left(\mu\right)$.
We use them to define the approximating fields $X_{n}:=A_{n}X$:
\[
\intop X_{n}\left(t\right)f\left(t\right)\mu\left(dt\right):=\intop X\left(t\right)A_{n}^{\ast}f\left(t\right)\mu\left(dt\right).
\]
Assume that the covariance kernels of the fields $X$ and $X_{n}$
are Hilbert-Schmidt, i.e. there are functions $K_{n},K\in L^{2}\left(\mathcal{T}\times\mathcal{T},\mu\otimes\mu\right)$,
such that for all $f\in L^{2}\left(\mu\right)$ we have
\[
\E\left(\intop X_{n}\left(t\right)f\left(t\right)\mu\left(dt\right)\right)^{2}=\intop K_{n}\left(t,s\right)f\left(t\right)f\left(s\right)\mu\left(dt\right)\mu\left(ds\right),
\]
and similarly for $X_{n}$ and $K_{n}$ replaced by $X$ and $K$.
Assume that $K_{n}\to K$ in measure ($\mu\otimes\mu$).
\begin{thm}
If there exist subcritical GMCs $M_{n}$ over the fields $X_{n}$
with the same expectation $\mu$, and $\left\{ M_{n}\left[\mathcal{T}\right]\right\} $
are uniformly integrable then there exists a subcritical GMC $M$
over $X$ with expectation $\mu$, and $M_{n}\to M$ in the sense
that for every $f\in L^{1}\left(\mu\right)$
\[
\intop f\left(t\right)M_{n}\left(X_{n},dt\right)\overset{L^{1}\left(\OMEGA,\P\right)}{\to}\intop f\left(t\right)M\left(X,dt\right).
\]
\label{thm:ApproximationA}
\end{thm}
Later we restate and prove the approximation theorem in different
notation as Theorem \ref{thm:Approximation}. The assumptions there
are only marginally more general --- most notably, the approximating
fields are only assumed to be jointly Gaussian, not necessarily measurable
with respect to the limiting field. We also show in Theorem \ref{thm:HilbertSchmidtMoments}
and its Corollary \ref{cor:Moments} that the assumption that the
covariances $K_{n}$ and $K$ are Hilbert-Schmidt, at least with respect
to some equivalent measure $\mu^{\prime}\sim\mu$, follows from the
existence of a subcritical GMC over $X_{n}$, and therefore does not
need to be included as a separate clause.
In spite of the generality of Theorem \ref{thm:Approximation}, even
for logarithmic fields \eqref{eq:LogKernel} and convolution approximations
$A_{n}\xi:=\xi\ast\psi_{1/n}$ as in \eqref{eq:Convolution} our result
is stronger than the approximation theorem of Robert and Vargas \cite{RVRevisited}
in that we assert convergence in probability rather than just in distribution
and identify the limit as a function of the Gaussian field. In this
special case the only condition of Theorem \ref{thm:Approximation}
that is nontrivial to check is the uniform integrability of $M_{n}\left[\mathcal{T}\right]$,
and we will see in Section \ref{sub:ApplicationLog} that it follows
from the results of \cite{KahaneSur}, namely the existence of a GMC
for some specific logarithmic field and Kahane's comparison inequality.
Other approaches to the problems of convergence and independence of
the mollifier for logarithmic GMCs were explored in the papers \cite{JSUniqueness,BerElementary},
both of which appeared after our initial preprint. Unlike our result,
\cite{JSUniqueness} also covers the critical case.
\subsection{Randomized shifts}
A central idea that we employ throughout the paper is to study the
random measure $M$ by considering the measure
\[
\Q\left(d\omega,dt\right):=\P\left(d\omega\right)M\left(X\left(\omega\right),dt\right)
\]
on $\mathcal{\OMEGA}\times\mathcal{T}$, where $\left(\OMEGA,\mathcal{F},\P\right)$
is the underlying probability space. Since by Cameron-Martin theorem,
multiplying the Gaussian measure by an exponential of a linear functional
amounts to shifting the measure, and the GMC is a generalization of
exponentials, the following fact should come as no surprise.
\begin{thm}
A random measure $M$ is a subcritical GMC over a field $X$ iff for
every positive function $f=f\left(X,t\right)$, measurable with respect
to the field $X$ and $t\in\mathcal{T}$, we have
\begin{equation}
\E\intop f\left(X,t\right)M\left(X,dt\right)=\E\intop f\left(X+K\left(t,\cdot\right),t\right)\mu\left(dt\right),\label{eq:Peyriere}
\end{equation}
where $K$ is the covariance of the field $X$.
\end{thm}
The left-hand side of \eqref{eq:Peyriere} can be written as $\intop f\left(X\left(\omega\right),t\right)\Q\left(d\omega,dt\right)$,
so its right-hand side characterizes the measure $\Q$ on the $\sigma$-algebra
$\sigma\left(X,t\right)$ in terms of the field $X$ and the expectation
$\mu=\E M$. On the other hand, the random measure $M$ can be recovered
from $\Q$ by disintegrating $\Q\left(d\omega,dt\right)$ with respect
to the variable $\omega$, and in fact only the restriction of $\Q$
to $\sigma\left(X,t\right)$ matters, since $M$ is measurable with
respect to $X$. This leads to an important corollary:
\begin{cor}
The subcritical GMC with a given expectation $\mu$ over a given Gaussian
field $X$ is unique whenever it exists.
\end{cor}
In particular, this means that all approximation-based constructions
of subcritical GMC yield the same limit.
There is another point of view on \eqref{eq:Peyriere} that is in
some ways more natural and more appropriate for our purposes, and
which we adopt for the rest of the text. Note that \eqref{eq:Peyriere}
implies that whenever $X$ and $t\in\mathcal{T}$ are sampled independently,
the latter according to $\mu$, the distribution of $X+K\left(t,\cdot\right)$
is absolutely continuous with respect to that of $X$, with density
equal to the total mass $M\left[\mathcal{T}\right]$. We express this
by calling $K\left(t,\cdot\right)$ a \emph{randomized shift} of $X$.
Intuitively, randomized shifts generalize deterministic Cameron-Martin
shifts in the same way as subcritical GMCs generalize exponentials
of Gaussian random variables. More precisely, there is a bijective
correspondence between the two. The relation between subcritical GMCs
and randomized shifts is certainly not a new idea --- for example,
a special case of it is mentioned in \cite{STTranslation} --- but
we have not seen it stated in its proper generality in the literature.
A clean statement of the bijection requires a notational twist, which
we explain next.
Suppose that instead of a Gaussian field we are given an abstract
Gaussian random vector $X$ ``in'' some abstract real separable
Hilbert space $H$, with no a priori relation to the space $\left(\mathcal{T},\mu\right)$.
Then the additional ``field'' structure that is needed in order
to make sense of the definition of \eqref{eq:DefGMC4Field} is a way
to map vectors $\xi\in H$ into ($\mu$-equivalence classes of) functions
on $\mathcal{T}$ --- in other words, a continuous linear operator
$Y:H\to L^{0}\left(\mathcal{T},\mu\right)$, where $L^{0}$ is the
space of $\mu$-equivalence classes of functions equipped with the
topology of convergence in measure. We write $Y$ applied to $\xi\in H$
as $\left\langle Y,\xi\right\rangle \in L^{0}\left(\mathcal{T},\mu\right)$,
and the value of $\left\langle Y,\xi\right\rangle $ at a point $t\in\mathcal{T}$
is written as $\left\langle Y\left(t\right),\xi\right\rangle $. The
operator $Y$ can be viewed as a ``generalized $H$-valued function
on $\mathcal{T}$'', with $\left\langle Y\left(t\right),\xi\right\rangle $
being the ``scalar product'' of the generalized value ``$Y\left(t\right)$''
with $\xi$. Note that $\left\langle Y\left(t\right),\xi\right\rangle $
is only defined for $\mu$-almost all $t$ for every fixed $\xi$,
not for all $\xi$ simultaneously, so ``$Y\left(t\right)$'' may
fail to be a true vector in $H$. In the same way the ``standard
Gaussian'' $X$ itself is not a true random vector in $H$ (unless
it is finite-dimensional), but rather defined as an operator $X:H\to L^{0}\left(\OMEGA,\P\right)$
that takes any $\xi\in H$ to a Gaussian random variable of variance
$\norm{\xi}^{2}$.
We define a generalized random vector in $H$, defined on a probability
space $\left(\OMEGA,\P\right)$ (or $\left(\mathcal{T},\mu\right)$)
as an operator $H\to L^{0}\left(\OMEGA,\P\right)$ (resp. $H\to L^{0}\left(\mathcal{T},\mu\right)$).
For the sake of concreteness, the ``value'' $X\left(\omega\right)$
(resp. $Y\left(t\right)$) of such a vector may be identified with
its sequence of coordinates with respect to a fixed orthonormal basis
$\left\{ e_{n}\right\} \subset H$, i.e. scalar products $\left\langle X\left(\omega\right),e_{n}\right\rangle $
and $\left\langle Y\left(t\right),e_{n}\right\rangle $.
From the point of view described above the objects $X$ and $Y$ are
treated on equal grounds as ``generalized random vectors''. The
defining property of GMC over such a pair $\left(X,Y\right)$ is rewritten
as
\[
M\left(X+\xi,dt\right)=e^{\left\langle Y\left(t\right),\xi\right\rangle }M\left(X,dt\right).
\]
Finally, the relation between GMCs and randomized shifts can be stated
as follows:
\begin{thm}
There exists a subcritical GMC $M$ over $\left(X,Y\right)$ iff $Y$
is a randomized shift, i.e. the distribution of $X+Y\left(t\right)$
when $t$ is sampled independently according to $\mu$ is absolutely
continuous with respect to that of $X$. If the subcritical GMC $M$
does exist then for every $\left(X,t\right)$-measurable function
$f$, we have
\[
\E\intop f\left(X,t\right)M\left(X,dt\right)=\E\intop f\left(X+Y\left(t\right),t\right)\mu\left(dt\right).
\]
\end{thm}
To summarize, we arrive at two points of view on a Gaussian field.
The conventional one is that a field is defined by formal integrals
against test functions, and the other one is that a field is a pair
$\left(X,Y\right)$, where $X$ is a standard Gaussian in some Hilbert
space $H$ and $Y$ is some generalized random vector in $H$ indexed
by $\left(\mathcal{T},\mu\right)$. Intuitively, the relation between
them is that the ``value'' of the Gaussian field at a point $t\in\mathcal{T}$
should be ``$\left\langle X,Y\left(t\right)\right\rangle $''. That
these two points of view are equivalent follows from a nontrivial
result of functional analysis --- a factorization theorem due to Maurey
and Nikishin \cite{Nikishin,Maurey1,Maurey2}, as stated in the Appendix.
The ``$\left(X,Y\right)$'' notation for Gaussian fields clarifies
not only the ``GMC $\leftrightarrow$ randomized shift'' relation
but also the approximation theorem. The sequence of jointly Gaussian
fields corresponds to a sequence of couples $\left(X,Y_{n}\right)$
with the same $X$ representing the underlying Gaussian randomness,
and the convergence condition is $Y_{n}\to Y$ in the strong operator
topology, i.e. $\left\langle Y_{n},\xi\right\rangle \overset{L^{0}}{\to}\left\langle Y,\xi\right\rangle $
for each $\xi\in H$. In other words, it is really the randomized
shift $Y$ that is being approximated rather than the integrals of
the Gaussian fields against any particular test function. In the notation
of Theorem \ref{thm:ApproximationA} this corresponds to requiring
that $A_{n}\overset{s}{\to}A$ rather than $A_{n}^{\ast}\overset{s}{\to}A^{\ast}$
(where $\overset{s}{\rightarrow}$ denotes convergence in the strong
operator topology).
\subsection{Kernel regularity}
It is important in the formulation of our approximation theorem that
the covariance kernels of the fields are representable by functions
up to $\mu\otimes\mu$-equivalence. In GMC theory it is customary
to assume this from the beginning, and we are not aware of any known
results that justify this assumption.
Our Theorem \ref{thm:HilbertSchmidtMoments} and its Corollary \ref{cor:Moments}
provide such a result in the case of subcritical GMC. The statement
that we prove there is that the existence of a subcritical GMC implies
that the covariance kernel of the field is indeed a function (up to
$\mu\otimes\mu$-equivalence), and moreover, this function has polynomial
moments with respect to some equivalent measure $\mu^{\prime}\otimes\mu^{\prime}$:
\begin{equation}
K\in\bigcap_{p}L^{p}\left(\mu^{\prime}\otimes\mu^{\prime}\right).\label{eq:PolyMoment}
\end{equation}
Note that the existence of the \emph{$1$-exponential} moment
\begin{equation}
\intop e^{K\left(t,s\right)}\mu\left(dt\right)\mu\left(ds\right)<\infty\label{eq:ExpMoment}
\end{equation}
is already sufficient for the existence of a subcritical GMC. Indeed,
formally, the $1$-exponential moment of $K$ equals $\E\left[\left(M\left[\mathcal{T}\right]\right)^{2}\right]$,
and \eqref{eq:ExpMoment} ensures that Kahane's martingale approximations
are bounded in $L^{2}$, and therefore uniformly integrable.
In relation to this it is appropriate to mention Kahane's $\frac{1}{2}$-exponential
moment conjecture that states that a subcritical GMC exists iff for
some $\mu^{\prime}\sim\mu$ we have
\begin{equation}
\intop e^{\frac{1}{2}K\left(t,s\right)}\mu^{\prime}\left(dt\right)\mu^{\prime}\left(ds\right)<\infty.\label{eq:ExpMoment2}
\end{equation}
This conjecture turned out to be false --- namely, as Sato and Tamashiro
demonstrated in \cite{STTranslation}, for any $\varepsilon>0$ even
\[
\intop e^{\left(1-\varepsilon\right)K\left(t,s\right)}\mu\left(dt\right)\mu\left(ds\right)<\infty
\]
is not sufficient for the existence of a GMC. To find a correct replacement
for Kahane's condition appears to be an open problem.
\subsection{Organization of the paper}
\begin{itemize}
\item In Section \ref{sec:AbstractNonsense} we introduce relevant notion
of ``generalized Gaussian fields'' and the definition of GMC.
\item In Section \ref{sec:MainResults} we formulate the main results of
the paper --- the GMC $\leftrightarrow$ randomized shift bijection
(Theorem \ref{thm:GMCShifts}) in Section \ref{sub:RandomizedShifts},
the kernel regularity theorem (Theorem \ref{thm:HilbertSchmidtMoments}
and Corollary \ref{cor:Moments}) in Section \ref{sub:KernelRegularity}
and the approximation theorem (Theorem \ref{thm:Approximation}) in
Section \ref{sub:Approximation}. In the remaining Section \ref{sub:ApplicationLog}
we apply the approximation theorem to the convolution approximations
of logarithmic fields (Theorem \ref{thm:ApproximationLog}).
\item In Section \ref{sec:GMCShifts} we prove the GMC $\leftrightarrow$
randomized shift bijection (Theorem \ref{thm:GMCShifts}).
\item In Section \ref{sec:KernelRegularity} we prove the kernel regularity
theorem (Theorem \ref{thm:HilbertSchmidtMoments} and Corollary \ref{cor:Moments}).
The reader only interested in the approximation theorem can safely
skip this, as long as (s)he is willing to assume that both the approximating
fields and the limiting field have Hilbert-Schmidt covariances.
\item In Section \ref{sec:Approximation} we prove our main approximation
theorem (Theorem \ref{thm:Approximation}).
\item In the Appendix we state the Maurey-Nikishin factorization theorem
and explain the relation between the ``$\left(X,Y\right)$'' and
the test function point of view on Gaussian fields.
\end{itemize}
\subsection{Notation and standard assumptions}
We always denote by $H$ a separable infinite-dimensional real Hilbert
space; vectors in $H$ are denoted by $\xi,\eta,\dots$, generalized
random vectors (Definition \ref{def:SVect}) --- by uppercase $X,Y,Z,\dots$.
Among the latter, $X$ is reserved for a standard Gaussian in $H$
(Example \ref{ex:StdGaussian}), defined on a standard probability
space $\left(\OMEGA,\mathcal{F},\P\right)$ (i.e. one isomorphic to
a Polish space equipped with a Borel probability measure). $Y,Z,\dots$
are defined on a standard measurable space $\mathcal{T}$ equipped
with a finite or $\sigma$-finite positive measure $\mu$. This $\mathcal{T}$
serves as the parameter space for generalized Gaussian fields.
Assuming $Y$ is defined on $\mathcal{T}$, a Gaussian multiplicative
chaos over the field $\left(X,Y\right)$ (Definition \ref{def:GMC})
is denoted by $M\left(X,dt\right)$ or, in cases of ambiguity, $M_{Y}\left(X,dt\right)$.
The notation $\Law$ is used for the distribution of a generalized
random vector, i.e. the joint distribution of linear functionals of
it. Modifiers like $\Law_{\mu},\Law_{\Q}$ are used when the underlying
probability space is equipped with probability measures $\mu,\Q$.
Similarly, $\E_{\mu},\E_{\Q}$ are used for the expectation with respect
to $\mu,\Q$.
\section{The setup \label{sec:AbstractNonsense}}
\subsection{Generalized Gaussian fields}
For any standard probability space $\left(\mathcal{T},\mu\right)$
we denote by $L^{0}\left(\mathcal{T},\mu\right)$ the space of $\mu$-equivalence
classes of functions on $\mathcal{T}$, equipped with the topology
of convergence in measure.
Throughout the text we fix a real separable Hilbert space $H$.
\begin{defn}
A \emph{generalized $H$-valued function} defined on $\left(\mathcal{T},\mu\right)$
is a continuous linear operator $Y:H\to L^{0}\left(\mathcal{T},\mu\right)$.
Generalized $H$-valued functions defined on the distinguished probability
space $\left(\OMEGA,\P\right)$ are also called \emph{generalized
random vectors} (in $H$). \label{def:SVect}\end{defn}
\begin{rem}
The notion of generalized $H$-valued function only depends on the
equivalence class of measures. Indeed, if $\mu^{\prime},\mu^{\prime\prime}$
are two equivalent probability measures then the topologies of convergence
in measure are the same for $\mu^{\prime}$ and $\mu^{\prime\prime}$,
so $L^{0}\left(\mu^{\prime}\right)$ and $L^{0}\left(\mu^{\prime\prime}\right)$
can be identified in a canonical way. This also allows us to define
$L^{0}$ over a (nonzero) $\sigma$-finite measure as $L^{0}$ over
any equivalent probability measure. For the sake of having the right
definitions in the trivial case, for $\mu=0$ we set $L^{0}\left(\mu\right):=\left\{ 0\right\} $.
\label{rem:L0}
\end{rem}
We use the ``scalar product'' notation $\left\langle Y,\xi\right\rangle $
for a generalized $H$-valued function $Y$ and a vector $\xi\in H$
to denote the corresponding (equivalence class of) function. By an
abuse of notation we will write the value $\left\langle Y,\xi\right\rangle \left(t\right)$
as $\left\langle Y\left(t\right),\xi\right\rangle $.
For the sake of concreteness we may treat any generalized $H$-valued
function $Y$ as a ``true'' equivalence class of a function with
values in $\RR^{\infty}$ by taking its coordinates in a fixed orthonormal
basis $\left(e_{n}\right)$ of $H$. Namely, the ``value $Y\left(t\right)$''
is identified with its sequence of coordinates $Y^{\left(n\right)}\left(t\right):=\left\langle Y\left(t\right),e_{n}\right\rangle $,
which is well-defined up to equality almost everywhere. Not all $\RR^{\infty}$-valued
functions can serve as generalized $H$-valued functions. The necessary
and sufficient condition for a sequence $Y^{\left(n\right)}$ to come
this way from a generalized $H$-valued function is that for all $\xi\in\ell^{2}$
the series $\sum_{n}\xi_{n}Y^{\left(n\right)}$ should converge in
$L^{0}$, so that it is possible to define $\left\langle Y,\xi\right\rangle $
almost everywhere as a measurable linear functional.
The choice of $\RR^{\infty}\supset H$ as the space to host the values
of all generalized $H$-valued functions is highly arbitrary and actually
irrelevant for our purposes. It is, however, important for $Y\left(t\right)$
to have \emph{some} Frechet (thus: standard Borel) space to live in.
We identify true $H$-valued functions with a special case of generalized
$H$-valued functions. Accordingly, ``$Y\left(t\right)\in H$ for
almost all $t$'' means that $Y$ is a true $H$-valued function.
\begin{example}
We call a generalized random vector $X$ \emph{standard Gaussian}
in $H$ if all $\left\langle X,\xi\right\rangle ,\xi\in H$ are centered
Gaussian random variables with variance $\E\left\langle X,\xi\right\rangle ^{2}=\norm{\xi}^{2}$.
\label{ex:StdGaussian}
\end{example}
We say that a generalized $H$-valued function $Y$ on a probability
$\left(\mathcal{T},\mu\right)$ has a weak first moment (with respect
to the measure $\mu$) if for any $\xi\in H$ the function $\left\langle Y,\xi\right\rangle $
is in $L^{1}\left(\mu\right)$. In this case, by a standard application
of the closed graph theorem, there exists a vector $\intop Y\left(t\right)\mu\left(dt\right)\in H$
defined in the obvious way:
\[
\left\langle \intop Y\left(t\right)\mu\left(dt\right),\xi\right\rangle :=\intop\left\langle Y\left(t\right),\xi\right\rangle \mu\left(dt\right).
\]
We may replace the symbol $\intop\dots\mu\left(dt\right)$ by $\E_{\mu}$,
and $\intop\dots\P\left(d\omega\right)$ by $\E$.
\begin{defn}
A \emph{generalized Gaussian field} on a standard measure space $\left(\mathcal{T},\mu\right)$
is a couple $\left(X,Y\right)$, where $X$ is the standard Gaussian
random vector in $H$ (defined on $\left(\OMEGA,\P\right)$) and $Y$
is a generalized $H$-valued function defined on $\left(\mathcal{T},\mu\right)$.
\label{def:GaussianField}
\end{defn}
As mentioned in the introduction, this point of view on generalized
Gaussian fields is equivalent to the more conventional one in terms
of integration against $L^{2}$ test functions. The equivalence between
the two is nontrivial and involves a factorization theorem due to
Maurey and Nikishin (Theorem \ref{thm:Factorization}). The translation
in both directions is explained in the Appendix.
\subsection{The definition of a GMC}
By a random measure on a measure space $\left(\mathcal{T},\mu\right)$
we always mean a random \emph{positive finite} measure $M$, such
that $\E M$ is $\mu$-absolutely continuous (notation: $\E M\ll\mu$).
The measure $\E M$ is defined by $\left(\E M\right)\left[A\right]:=\E\left(M\left[A\right]\right)$
for all measurable subsets $A\subset\mathcal{T}$. Note that $\E M$
need not be a $\sigma$-finite measure; however, it is equivalent
to a finite one --- namely, $\E\left[\left(M\left[\mathcal{T}\right]\vee1\right)^{-1}M\right]$
--- in the sense that their classes of null sets coincide. Similarly,
the condition $\E M\ll\mu$ is understood in the sense that for any
measurable $A\subset\mathcal{T}$, such that $\mu\left[A\right]=0$,
we have $\E M\left[A\right]=0$, or equivalently, $M\left[A\right]=0$
a.s.
Note that even though $\E M$ is $\mu$-absolutely continuous, $M$
itself may be almost surely $\mu$-singular.
We refer to the vectors $\xi\in H$ as Cameron-Martin shifts (of the
Gaussian $X$). For the necessary background on them the reader is
referred to \cite[Theorem 14.1]{Janson}.
\begin{defn}
A random measure $M$ on $\left(\mathcal{T},\mu\right)$ is called
a \emph{Gaussian multiplicative chaos} (GMC) over the Gaussian field
$\left(X,Y\right)$ if
\begin{enumerate}
\item $\E M\ll\mu$
\item $M$ is measurable with respect to $X$ (which allows us to write
$M=M\left(X\right)$);
\item For all vectors $\xi\in H$
\begin{equation}
M\left(X+\xi,dt\right)=e^{\left\langle Y\left(t\right),\xi\right\rangle }M\left(X,dt\right)\text{ a.s.}\label{eq:DefGMC}
\end{equation}
\end{enumerate}
The GMC is called \emph{subcritical} if $\E M$ is $\sigma$-finite.
\label{def:GMC}
\end{defn}
Instead of ``GMC over the Gaussian field $\left(X,Y\right)$'' we
may also say ``GMC associated to $Y$'' with the Gaussian $X$ understood
implicitly.
Formula \eqref{eq:DefGMC} requires a couple of comments. First, $X+\xi$
is a shifted Gaussian, so by the Cameron-Martin theorem, its distribution
is equivalent to that of $X$, which makes $M\left(X+\xi\right)$
well-defined. Second, even though $\left\langle Y,\xi\right\rangle $
is only defined almost everywhere with respect to $\mu$, and in the
interesting cases $M$ is almost surely $\mu$-singular, $e^{\left\langle Y,\xi\right\rangle }M$
is still well-defined, precisely because $\E M$ is $\mu$-absolutely
continuous. Indeed, if $\varphi,\tilde{\varphi}$ are two measurable
functions on $\mathcal{T}$ that are equal $\mu$-almost everywhere
to $e^{\left\langle Y,\xi\right\rangle }$ then for $\P$-almost all
$\omega$ and $M\left(X\left(\omega\right),dt\right)$-almost all
$t$ we have $\varphi\left(t\right)=\tilde{\varphi}\left(t\right)$,
thus almost surely $\varphi M=\tilde{\varphi}M$.
\begin{example}
If $Y\left(t\right)\in H$ for almost all $t$ then
\begin{equation}
M\left(X,dt\right):=\exp\left[\left\langle X,Y\left(t\right)\right\rangle -\frac{1}{2}\norm{Y\left(t\right)}^{2}\right]\mu\left(dt\right)\label{eq:TrivialGMC}
\end{equation}
is a subcritical GMC over the Gaussian field $\left(X,Y\right)$ with
expectation $\E M=\mu$. \label{ex:TrivialGMC}\end{example}
\begin{rem}
Obviously, in the definition of GMC only the equivalence class of
$\mu$ matters. However, in the subcritical case we will sometimes
assume that $\E M=\mu$, which is no loss of generality because a
GMC on $\left(\mathcal{T},\mu\right)$ is also a GMC on $\left(\mathcal{T},\E M\right)$.
Furthermore, another common assumption will be that $\E M=\mu$ is
finite, which is also no loss of generality because our theory is
essentially local, i.e. the statements for the whole space $\left(\mathcal{T},\mu\right)$
reduce to those for its subsets of finite measure. \label{rem:EM}
\end{rem}
Finally we would like to remark there exist GMCs that are not subcritical.
So far the only examples known to the author are the critical GMCs
over logarithmic fields and their hierarchical counterparts \cite{HuShiMinimal,DSRVCriticalConvergence,DSRVRenormalization}.
For these critical GMCs we have for any measurable set $A\subset\mathcal{T}$
\[
\E M\left[A\right]=\begin{cases}
0, & \mu\left[A\right]=0\\
\infty, & \mu\left[A\right]>0
\end{cases}
\]
so that it is impossible to normalize them by expectation. In these
cases $\E M$ is a non-$\sigma$-finite measure that we agree to call
$\mu$-absolutely continuous; its density with respect to $\mu$ is
almost everywhere infinite.
\section{Main results \label{sec:MainResults}}
\subsection{Randomized shifts \label{sub:RandomizedShifts}}
\begin{defn}
A generalized $H$-valued function $Y$ defined on $\left(\mathcal{T},\mu\right)$
is called a \emph{randomized shift} if
\[
\Law_{\P\otimes\mu}\left[X+Y\right]\ll\Law_{\P}X.
\]
\end{defn}
Note that being a randomized shift only depends on the equivalence
class of $\Law_{\mu}Y$.
\begin{example}
[Trivial shifts] By the Cameron-Martin theorem every $H$-valued
function $Y:\mathcal{T}\to H$ (not ``generalized''!) is a randomized
shift.
\end{example}
These Cameron-Martin shifts are viewed as ``trivial''. There are
less trivial ones:
\begin{example}
[Gaussian shifts] By the Hajek-Feldman theorem \cite[Theorem 6.3.2]{Bog},
a Gaussian generalized random vector in $H$ (i.e. that for which
all linear functionals are Gaussian) is a randomized shift iff its
covariance is Hilbert-Schmidt. Note also that being ``trivial''
in the above sense is equivalent to the covariance being trace class.
\end{example}
In Theorem \ref{thm:GMCShifts} we describe the relation between subcritical
GMC over Gaussian fields with parameter space $\left(\mathcal{T},\mu\right)$
and randomized shifts defined on $\left(\mathcal{T},\mu\right)$.
Here it is convenient to view $\left(\mathcal{T},\mu\right)$ as an
additional source of randomness, so functions on $\OMEGA\times\mathcal{T}$
are treated as random variables --- in particular, the projection
map $\OMEGA\times\mathcal{T}\to\mathcal{T}$ is treated as ``the''
random point $t$ in $\mathcal{T}$. Accordingly, we use the notation
$\Law_{\mu},\Law_{\P\otimes\mu},\dots$ for the law of a generalized
random vector defined on the probability space $\left(\mathcal{T},\mu\right),\left(\OMEGA\times\mathcal{T},\P\otimes\mu\right),\dots$.
\begin{thm}
There exists a subcritical GMC $M$ over the Gaussian field $\left(X,Y\right)$
with expectation $\mu$ iff $Y$ is a randomized shift, in which case
under the Peyrière measure on $\OMEGA\times\mathcal{T}$
\begin{equation}
\Q\left(d\omega,dt\right):=\P\left(d\omega\right)M\left(X\left(\omega\right),dt\right)\label{eq:DefQ}
\end{equation}
we have
\begin{equation}
\Law_{\Q}\left[X,t\right]=\Law_{\P\otimes\mu}\left[X+Y\left(t\right),t\right].\label{eq:QShift}
\end{equation}
\label{thm:GMCShifts}
\end{thm}
Note that \eqref{eq:QShift} characterizes uniquely the measure $\Q$
on the $\sigma$-algebra generated by $\left(X,t\right)$, so by disintegration
with respect to $X$ it also characterizes $M\left(X\right)$. Therefore,
we have the following important corollary: a subcritical GMC is unique.
\begin{cor}
A subcritical GMC with a given expectation $\mu$, associated to a
given $Y$, is unique, whenever it exists. $M$ can be recovered from
$\mu$ and $Y$ as follows:
\[
M\left[\mathcal{T}\right]=\Law_{\P\otimes\mu}\left[X+Y\left(t\right)\right]/\Law X,
\]
\[
\left(M\left[\mathcal{T}\right]\right)^{-1}M\left(x,dt\right)=\Lawcm{\P\otimes\mu}{t\in dt}{X+Y\left(t\right)=x},
\]
where $\dots/\Law X$ denotes the Radon-Nikodym density viewed as
a function of $X$, and $\Lawc{\cdot}{\cdot}$ denotes the conditional
distribution. \label{cor:Uniqueness}
\end{cor}
Note that unlike in the previous approaches to GMC theory, we have
proven uniqueness of $M$ as a function of $X$ rather than just in
law.
It is instructive to note the role of the subcriticality of $M$ in
Theorem \ref{thm:GMCShifts}. For every GMC, not necessarily a subcritical
one, we can construct the measure $\Q$ as in \eqref{eq:DefQ}. In
general, $\Q$ is only $\sigma$-finite, and subcriticality is equivalent
to the finiteness of $\mu$-almost all fiber measures $\Q_{t}$ in
the disintegration
\[
\Q\left(d\omega,dt\right)=\Q_{t}\left(d\omega\right)\mu\left(dt\right).
\]
Indeed, the density of $\E M$ with respect to $\mu$ at $t\in\mathcal{T}$
is equal to the total mass of the fiber measure $\Q_{t}$. The proof
of Theorem \ref{thm:GMCShifts} proceeds essentially by verifying
that the fiber $\Q_{t}$ behaves under shifts by $\xi\in H$ like
the standard Gaussian measure shifted by $Y\left(t\right)$ in the
sense that it satisfies the corresponding Cameron-Martin formula.
Note that among \emph{finite} measures there is a unique one (up to
scaling) that satisfies the Cameron-Martin formula, namely the Gaussian
itself, so that the behavior of the GMC under shifts completely characterizes
$\Q$. However, the uniqueness argument fails without the subcriticality
assumption because there are many different $\sigma$-finite measures
that satisfy the Cameron-Martin formula. For example, for any positive
sequence $\left(C_{n}\right)$, such that $\sum_{n}e^{-C_{n}^{2}}<\infty$
there is a unique $\sigma$-finite measure $\gamma$ on $\RR^{\infty}$
that satisfies the Cameron-Martin formula for all shifts in $H:=\ell^{2}$
and whose restriction to $\prod_{n}\left[-C_{n},C_{n}\right]$ is
$\gamma^{\op{res}}:=\bigotimes_{n}\left(Z_{n}^{-1}\I\left\{ \left|x_{n}\right|\le C_{n}\right\} e^{-x_{n}^{2}/2}dx_{n}\right)$,
$Z_{n}:=\intop_{-C_{n}}^{C_{n}}e^{-x^{2}/2}dx$. One can construct
such a measure by gluing together the ``compatible'' measures $e^{-\left\langle \cdot,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}\left(S_{\xi}\right)_{\ast}\gamma^{\op{res}}$
(where $S_{\xi}$ is the shift $x\mapsto x+\xi$). The resulting measure
$\gamma$ is singular to the Gaussian iff $\sum_{n}C_{n}^{-1}e^{-C_{n}^{2}/2}=\infty$.
In fact, by choosing different sequences $\left(C_{n}\right)$ one
can produce a continuum of mutually singular $\sigma$-finite measures
that satisfy the Cameron-Martin formula. Due to this difficulty the
uniqueness problem for GMCs without the subcriticality assumption
remains open.
\subsection{Regularity of the kernel \label{sub:KernelRegularity}}
A priori for a Gaussian field $\left(X,Y\right)$ one can define the
covariance kernel, ``$K\left(t,s\right)=\left\langle Y\left(t\right),Y\left(s\right)\right\rangle $'',
as the formal kernel of the bilinear form on the $L^{2}\left(\mu^{\prime}\right)$
test functions (see the Appendix). Namely,
\[
\begin{aligned}\intop K\left(t,s\right)f\left(t\right)f\left(s\right)\mu^{\prime}\left(dt\right)\mu^{\prime}\left(ds\right) & :=\norm{\intop f\left(t\right)Y\left(t\right)\mu^{\prime}\left(dt\right)}^{2}\\
& =\E\left\langle X,\intop f\left(t\right)Y\left(t\right)\mu^{\prime}\left(dt\right)\right\rangle ^{2}.
\end{aligned}
\]
Not all bounded operators in $L^{2}$ are integral operators, so neither
are such $K$'s represented by ``true'' functions on $\mathcal{T}\times\mathcal{T}$.
However, we will prove that in the GMC theory this pathology does
not happen, i.e. whenever a Gaussian field admits a subcritical GMC,
the covariance kernel $K$ is actually representable by a function
on $\mathcal{T}\times\mathcal{T}$. Furthermore, this function has
all moments with respect to some equivalent measure $\mu^{\prime}\otimes\mu^{\prime}$.
The proof proceeds by translating this property to an equivalent statement
about the randomized shift $Y$ and relies on the factorization theorem
(Theorem \ref{thm:Factorization}) for the construction of $\mu^{\prime}$.
For a Hilbert space $H$ we denote by $H^{\otimes n}$ its Hilbert
$n$-th tensor power, also called the space of Hilbert-Schmidt tensors.
By definition, it is the completion of the algebraic tensor power
$H_{\op{alg}}^{\otimes n}$ with respect to the scalar product
\[
\left\langle \xi_{1}\otimes\dots\otimes\xi_{n},\eta_{1}\otimes\dots\otimes\eta_{n}\right\rangle :=\left\langle \xi_{1},\eta_{1}\right\rangle \dots\left\langle \xi_{n},\eta_{n}\right\rangle ,
\]
extended from decomposable tensors to all tensors by multilinearity.
We denote by $\norm{\cdot}_{2}$ the Hilbert space norm corresponding
to this scalar product.
For a standard Gaussian $X$ in $H$ there is a well-known theory
of random variables that are polynomial in $X$, also known as the
Wick calculus (see \cite[Chapter III]{Janson}). Its basic construction
is the Wick product of jointly Gaussian random variables, denoted
by $\Wick{\left\langle X,\xi_{1}\right\rangle \dots\left\langle X,\xi_{n}\right\rangle }$,
and defined by the polarization of the identity
\[
\Wick{\left\langle X,\xi\right\rangle \dots\left\langle X,\xi\right\rangle }=\norm{\xi}^{n}h_{n}\left(\norm{\xi}^{-1}\left\langle X,\xi\right\rangle \right),
\]
where $h_{n}$ is the $n$-th Hermite polynomial
\[
h_{n}\left(x\right):=e^{-\frac{1}{2}\frac{\partial^{2}}{\partial x^{2}}}x^{n}=x^{n}-\frac{1}{2}n\left(n-1\right)x^{n-2}+\dots
\]
The basic fact is that $\left\langle \Wick{X^{\otimes n}},\lambda\right\rangle $,
defined initially for finite-rank tensors (i.e. elements of the algebraic
tensor product) $\lambda\in H_{\op{alg}}^{\otimes n}$ by
\[
\left\langle \Wick{X^{\otimes n}},\xi_{1}\otimes\dots\otimes\xi_{n}\right\rangle :=\Wick{\left\langle X,\xi_{1}\right\rangle \dots\left\langle X,\xi_{n}\right\rangle },
\]
extends by $L^{2}$-continuity to all Hilbert-Schmidt tensors $\lambda\in H^{\otimes n}$.
In our language this is stated as follows:
\[
\Wick{X^{\otimes n}}\text{ is a generalized random vector in }H^{\otimes n}.
\]
It turns out that this implies a corresponding property for a randomized
shift $Y$ --- with the crucial difference that $Y$, unlike $X$,
does not need the Wick renormalization:
\[
Y^{\otimes n}\text{ is a generalized random vector in }H^{\otimes n}.
\]
This is the content of our Theorem \ref{thm:HilbertSchmidtMoments},
and it turns out to be equivalent to the existence of the $n$-th
moment of the covariance kernel $K$ with respect to $\mu^{\prime}\otimes\mu^{\prime}$
for some equivalent measure $\mu^{\prime}\sim\mu$.
\begin{thm}
Let $X$ be a standard Gaussian in $H$, and let $\left(\mathcal{T},\mu\right)$
be a standard probability space. Let $Y$ be a randomized shift, defined
on $\left(\mathcal{T},\mu\right)$. Then for every $n\in\NN$ there
is an equivalent measure $\mu_{n}^{\prime}\sim\mu$ on $\mathcal{T}$,
such that under $\mu_{n}^{\prime}$ all $\left\langle Y,\xi\right\rangle $
have finite absolute $n$-th moment, and the symmetric tensor $\E_{\mu_{n}^{\prime}}Y^{\otimes n}$,
defined by the polarization of $\xi\mapsto\E_{\mu_{n}^{\prime}}\left\langle Y,\xi\right\rangle ^{n}$,
is Hilbert-Schmidt. \label{thm:HilbertSchmidtMoments}\end{thm}
\begin{cor}
In the setting of Theorem \ref{thm:HilbertSchmidtMoments} the quadratic
form
\[
f\mapsto\norm{\intop f\left(t\right)Y\left(t\right)\mu_{n}^{\prime}\left(dt\right)}^{2}
\]
is Hilbert-Schmidt. Thus there exists a unique symmetric function
$K\in L^{2}\left(\mu_{n}^{\prime}\otimes\mu_{n}^{\prime}\right)$,
such that
\[
\norm{\intop f\left(t\right)Y\left(t\right)\mu_{n}^{\prime}\left(dt\right)}^{2}=\intop K\left(t,s\right)f\left(t\right)f\left(s\right)\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)
\]
for all $f\in L^{2}\left(\mu_{n}^{\prime}\right)$. Moreover,
\[
\intop\left(K\left(t,s\right)\right)^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)<\infty.
\]
\label{cor:Moments}\end{cor}
\begin{rem}
Yet another interpretation of this result could be: the existence
of a subcritical GMC, which is the Wick exponential of the Gaussian
field, implies the existence of all Wick powers ``$\Wick{\left(X\left(t\right)\right)^{n}}$''
of that field, due to the formal identity
\begin{equation}
\E\left|\intop\Wick{\left(X\left(t\right)\right)^{n}}\mu_{n}^{\prime}\left(dt\right)\right|^{2}=\intop\left(K\left(t,s\right)\right)^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)<\infty.\label{eq:WickL2}
\end{equation}
\end{rem}
\begin{rem}
For a randomized shift $Y$ there is a single measure $\mu^{\prime}\sim\mu$,
such that
\[
\intop\left(K\left(t,s\right)\right)^{n}\mu^{\prime}\left(dt\right)\mu^{\prime}\left(ds\right)<\infty
\]
for all $n$ simultaneously. Such a measure can be constructed as
follows:
\[
\mu^{\prime}\left(dt\right):=\inf_{n}\left(r_{n}\mu_{n}^{\prime}\left(dt\right)/\mu\left(dt\right)\right)\cdot\mu\left(dt\right),
\]
where $\left(r_{n}\right)$ is a sequence of positive numbers that
decreases fast enough (namely, so that $\mu\left(dt\right)/\mu_{n}^{\prime}\left(dt\right)=O\left(r_{n}\right),n\to\infty$
for $\mu$-almost all $t$). This fact is not used anywhere in the
text.
\end{rem}
\begin{rem}
Instead of the factorization theorem, which only implies that $K^{n}$
is a bounded bilinear form on $L^{2}\left(\mu^{\prime}\right)$ test
functions for some $\mu^{\prime}$, one can use a sharp bound, replacing
$L^{2}$ by the Orlicz space $L\left(\log L\right)^{n/2}$. This follows
by Orlicz space duality from the a priori $\exp\left(-\const\cdot x^{2/n}\right)$
tail decay of the distribution of random variables of the form $\left\langle \lambda,\Wick{X^{\otimes n}}\right\rangle $,
$\lambda\in H^{\otimes n}$ (see e.g. \cite[Theorem 6.7]{Janson}),
which carries over with a change of measure to $\left\langle \lambda,\Wick{\left(X+cY\right)^{\otimes n}}\right\rangle $
and thus also to $\left\langle \lambda,Y^{\otimes n}\right\rangle $.
The resulting bound,
\[
\intop\left(K\left(t,s\right)\right)^{n}f\left(t\right)g\left(s\right)\mu^{\prime}\left(dt\right)\mu^{\prime}\left(ds\right)\le\const\cdot\norm f_{L\left(\log L\right)^{2/n}}\norm g_{L\left(\log L\right)^{n/2}},
\]
yields the following estimate: if $\mathcal{T}=\left[0,1\right],\mu^{\prime}=\op{Lebesgue}$,
$f:=\frac{1}{\varepsilon}\I\left[a,a+\varepsilon\right],g:=\frac{1}{\varepsilon}\I\left[b,b+\varepsilon\right]$
for some $a,b\in\left[0,1-\varepsilon\right]$, then we have
\[
\op{ess}\sup\left(K^{n}\ast\left(\varepsilon^{-1}\I\left[0,\varepsilon\right]\right)^{\otimes2}\right)=O\left(\left|\log\varepsilon\right|^{n}\right),\varepsilon\to0.
\]
By considering the logarithmic kernels one can see that this bound
is sharp. This will not be used in the paper.
\end{rem}
\begin{rem}
A weaker bound, $K\in L^{2}\left(\mu^{\prime}\otimes\mu^{\prime}\right)$
for some $\mu^{\prime}\sim\mu$, can be proved without the subcriticality
assumption using a different approach. This will be presented elsewhere.
\end{rem}
\subsection{Approximation \label{sub:Approximation}}
Let $Y_{n},n\ge1$ be randomized shifts defined on a probability space
$\left(\mathcal{T},\mu\right)$. Let $K_{Y_{n}Y_{n}}\left(t,s\right):=\left\langle Y_{n}\left(t\right),Y_{n}\left(s\right)\right\rangle $
be the corresponding kernel, which, by Corollary \ref{cor:Moments},
is well-defined as a function on $\left(\mathcal{T}\times\mathcal{T},\mu\otimes\mu\right)$.
Let $M_{Y_{n}}$ be the subcritical GMC associated to $Y_{n}$ with
expectation $\mu$.
Our main result on the approximation of subcritical GMC is as follows:
\begin{thm}
Assume that:
\begin{itemize}
\item The family of random variables $\left\{ M_{Y_{n}}\left[\mathcal{T}\right]\right\} $
is uniformly integrable;
\item There exists a generalized $H$-valued function $Y$ defined on $\left(\mathcal{T},\mu\right)$
that is the limit of $Y_{n}$ in the sense that
\begin{equation}
\forall\xi\in H:\left\langle Y_{n},\xi\right\rangle \overset{L^{0}\left(\mu\right)}{\to}\left\langle Y,\xi\right\rangle .\label{eq:ConvergenceOfShifts}
\end{equation}
\end{itemize}
Then $Y$ is a randomized shift. If, furthermore,
\begin{itemize}
\item the kernels $K_{Y_{n}Y_{n}}$ converge to $K_{YY}$ in $L^{0}\left(\mu\otimes\mu\right)$,
\end{itemize}
then the subcritical GMC $M_{Y}$ (associated to $Y$ with expectation
$\mu$) is the limit of $M_{Y_{n}}$ in the sense that
\begin{equation}
\forall f\in L^{1}\left(\mu\right):\intop f\left(t\right)M_{Y_{n}}\left(X,dt\right)\overset{L^{1}}{\to}\intop f\left(t\right)M_{Y}\left(X,dt\right).\label{eq:ConvergenceOfGMC}
\end{equation}
\label{thm:Approximation}
\end{thm}
\subsection{Application to logarithmic kernels \label{sub:ApplicationLog}}
Let $\mathcal{T}\subset\RR^{d}$ be a bounded domain, let $\mu$ be
the Lebesgue measure on $\mathcal{T}$, and let $K$ be a positive
definite Hilbert-Schmidt kernel on $\left(\mathcal{T},\mu\right)\times\left(\mathcal{T},\mu\right)$,
such that for some $\delta>0$
\begin{equation}
K_{YY}\left(t,s\right)\le\left(2d-\delta\right)\log\norm{t-s}^{-1}+O\left(1\right),\forall t,s\in\mathcal{T}.\label{eq:BoundOnK}
\end{equation}
Consider also a bounded function $\psi$ on $\RR^{d}$ with compact
support, such that $\psi\ge0$, $\intop\psi\left(x\right)dx=1$, and
denote $\psi_{\varepsilon}\left(x\right):=\varepsilon^{-d}\psi\left(\varepsilon^{-1}x\right)$.
Take any generalized random vector $Y$ with $\left\langle Y\left(t\right),Y\left(s\right)\right\rangle :=K_{YY}\left(t,s\right)$.
In order to construct one we may start with a Gaussian field on $\left(\mathcal{T},\mu\right)$
defined by its integrals against test functions. Namely, to every
test function $f\in L^{2}\left(\mu\right)$ we associate the Gaussian
variable $\left\langle X,Af\right\rangle $, where $A:L^{2}\left(\mu\right)\to H$
is a bounded linear operator, such that
\[
\left\langle Af,Ag\right\rangle =\intop K_{YY}\left(t,s\right)f\left(t\right)g\left(s\right)\mu\left(dt\right)\mu\left(ds\right),
\]
and we construct $Y$ as the composition $H\overset{A}{\to}L^{2}\left(\mathcal{T},\mu\right)\overset{\id}{\to}L^{0}\left(\mathcal{T},\mu\right)$.
\begin{thm}
Let
\[
Y_{\varepsilon}\left(t\right):=\intop_{\mathcal{T}}Y\left(t^{\prime}\right)\psi_{\varepsilon}\left(t-t^{\prime}\right)dt^{\prime},
\]
\[
K_{Y_{\varepsilon},Y_{\varepsilon}}\left(t,s\right):=\left\langle Y_{\varepsilon}\left(t\right),Y_{\varepsilon}\left(s\right)\right\rangle =\intop_{\mathcal{T}\times\mathcal{T}}K_{YY}\left(t^{\prime},s^{\prime}\right)\psi_{\varepsilon}\left(t-t^{\prime}\right)\psi_{\varepsilon}\left(s-s^{\prime}\right)dt^{\prime}\,ds^{\prime},
\]
\[
M_{Y_{\varepsilon}}\left(dt\right):=\exp\left[\left\langle X,Y_{\varepsilon}\left(t\right)\right\rangle -\frac{1}{2}K_{Y_{\varepsilon}Y_{\varepsilon}}\left(t,t\right)\right]dt.
\]
Then there exists a subcritical GMC $M_{Y}$ over $\left(X,Y\right)$,
and $M_{Y_{\varepsilon}}\to M_{Y}$ in probability (the space of measures
is equipped with the weak topology). This $M_{Y}$ does not depend
on the function $\psi$ used for approximation. \label{thm:ApproximationLog}
\end{thm}
Note that in this case $K_{Y_{\varepsilon}Y_{\varepsilon}}$ is a
continuous kernel, so the Gaussian field $\left(\left\langle X,Y_{\varepsilon}\left(t\right)\right\rangle \right)_{t\in\mathcal{T}}$
has well-defined values at each $t\in\mathcal{T}$. The field $\left\langle X,Y_{\varepsilon}\left(\cdot\right)\right\rangle $
is obtained by convolving our generalized field $\left\langle X,Y\left(\cdot\right)\right\rangle $
with $\psi_{\varepsilon}$, which makes sense, since $\psi_{\varepsilon}$
is allowed as a test function.
We will see that Theorem \ref{thm:ApproximationLog} follows from
Theorem \ref{thm:Approximation} once we have a way of verifying the
uniform integrability assumption of the latter. This is done using
a known result on existence of GMC for specific logarithmic kernels
and Kahane's comparison inequality \cite{KahaneSur}.
\begin{thm}
[See \cite{KahaneSur}] Consider the following kernels on $\mathcal{T}\times\mathcal{T}$:
\[
\tilde{K}_{C,\gamma}\left(t,s\right):=\gamma^{2}\intop_{1}^{C}e^{-u\norm{t-s}}\frac{du}{u}=\gamma^{2}\log\left(C\wedge\norm{t-s}^{-1}\right)+O\left(1\right).
\]
Then for $\gamma<\sqrt{2d}$ the family of GMCs with these kernels
is uniformly integrable. \label{thm:ExistenceClassical}
\end{thm}
\begin{thm}
[See \cite{KahaneSur}] Let $\left(X,Y_{1}\right)$ and $\left(X,Y_{2}\right)$
be Gaussian fields on $\mathcal{T}$ with continuous (or, more generally,
trace class) covariance kernels $K_{1},K_{2}$. Assume that
\[
\forall t,s:K_{1}\left(t,s\right)\le K_{2}\left(t,s\right).
\]
Then for every convex function $f:\RR_{+}\to\RR_{+}$
\[
\E f\left(\intop\exp\left[\left\langle X,Y_{1}\left(t\right)\right\rangle -\frac{1}{2}K_{1}\left(t,t\right)\right]dt\right)\le\E f\left(\intop\exp\left[\left\langle X,Y_{2}\left(t\right)\right\rangle -\frac{1}{2}K_{2}\left(t,t\right)\right]dt\right).
\]
\label{thm:KahaneInequality}\end{thm}
\begin{proof}
[Proof of Theorem \ref{thm:ApproximationLog}] All assumptions of
Theorem \ref{thm:Approximation} except for the uniform integrability
are quite trivial to check.
The convergence assumption $Y_{\varepsilon}\to Y$ amounts to the
following. For every $\xi\in H$ we associate the function $\xi\left(t\right):=\left\langle \xi,Y\left(t\right)\right\rangle $
which belongs to the Cameron-Martin space of the field (equivalently,
the reproducing kernel Hilbert space associated to $K$). The assumption
$Y_{\varepsilon}\to Y$ is equivalent to
\[
\xi\left(\cdot\right)\ast\psi_{\varepsilon}\to\xi\left(\cdot\right)
\]
for every such function, which is trivial, since $\xi\left(\cdot\right)\in L^{2}$.
The condition $K_{\varepsilon}\to K$ in $L^{0}\left(\mu\otimes\mu\right)$
is also trivially satisfied.
To verify uniform integrability we use Kahane's comparison inequality
(Theorem \ref{thm:KahaneInequality}) and the reference family of
kernels $K_{C,\gamma}$ for which uniform integrability is known.
It follows from \eqref{eq:BoundOnK} that there exists a constant
$C_{0}$, such that for every $\varepsilon>0$ there exists $C\left(\varepsilon\right)$,
such that
\[
\forall t,s:K_{\varepsilon}\left(t,s\right)\le\tilde{K}_{C\left(\varepsilon\right),\gamma}\left(t,s\right)+C_{0},
\]
where $\gamma=\sqrt{2d-\delta}$. Now by Theorem \ref{thm:ExistenceClassical}
GMCs with kernels $\tilde{K}_{C\left(\varepsilon\right),\gamma}\left(t,s\right)$,
and thus also $\tilde{K}_{C\left(\varepsilon\right),\gamma}\left(t,s\right)+C_{0}$,
are uniformly integrable. Therefore, by la Vallée Poussin's theorem
and Kahane's inequality (Theorem \ref{thm:KahaneInequality}), GMCs
with kernels $K_{\varepsilon}\left(t,s\right)$ are also uniformly
integrable. Therefore, all the assumptions of Theorem \ref{thm:Approximation}
are verified, and we have convergence $M_{\varepsilon}\overset{L^{0}}{\to}M$.
That $M$ does not depend on the approximation follows from our uniqueness
result (Corollary \ref{cor:Uniqueness}).
\end{proof}
\section{Proof of Theorem \ref{thm:GMCShifts} \label{sec:GMCShifts}}
\begin{proof}
Assume first that there exists a subcritical GMC $M$.
Define a measure $\Q$ on $\OMEGA\times\mathcal{T}$ by \eqref{eq:DefQ}.
We are going to prove \eqref{eq:QShift} by computing the conditional
Laplace transform of $X$ given $t$ under the measure $\Q$.
Let $\xi\in H$ and let $\varphi$ be an arbitrary positive measurable
function on $\mathcal{T}$ (defined $\mu$-almost everywhere). Then
\begin{equation}
\begin{aligned}\E_{\Q}\varphi\left(t\right)\exp\left\langle \xi,X\right\rangle & \underset{\left(1\right)}{=}\E\intop\varphi\left(t\right)\exp\left\langle \xi,X\right\rangle M\left(X,dt\right)\\
& \underset{\left(2\right)}{=}\exp\frac{1}{2}\norm{\xi}^{2}\cdot\E\intop\varphi\left(t\right)M\left(X+\xi,dt\right)\\
& \underset{\left(3\right)}{=}\exp\frac{1}{2}\norm{\xi}^{2}\cdot\E\intop\varphi\left(t\right)\exp\left\langle \xi,Y\left(t\right)\right\rangle M\left(X,dt\right)\\
& =\E\exp\left\langle \xi,X\right\rangle \cdot\E_{\mu}\varphi\left(t\right)\exp\left\langle \xi,Y\left(t\right)\right\rangle \\
& =\E_{\P\otimes\mu}\varphi\left(t\right)\exp\left\langle \xi,X+Y\left(t\right)\right\rangle .
\end{aligned}
\label{eq:BunchOfEqualities}
\end{equation}
``$\underset{\left(1\right)}{=}$'' follows from the definition
of $\Q$, ``$\underset{\left(2\right)}{=}$'' is an application
of the Cameron-Martin theorem \cite[Theorem 14.1]{Janson} to the
shift $\xi$, and ``$\underset{\left(3\right)}{=}$'' is the definition
of GMC. The equality of the left-hand side and the right-hand side
of \eqref{eq:BunchOfEqualities} for all $\xi$ and $\varphi$ implies
\eqref{eq:QShift}, since it amounts to equality of conditional Laplace
transforms of $X$ conditioned on $t$, together with the tautology
$\Law_{\Q}t=\Law_{\P\otimes\mu}t=\mu$.
Note that if $M$ exists then \eqref{eq:QShift} implies that $Y$
is a randomized shift. Indeed, $\Law_{\P\otimes\mu}\left[X+Y\right]\ll\Law X$
with density $M\left[\mathcal{T}\right]$.
Conversely, assume that $Y$ is a randomized shift. Then define a
measure $\Q^{\prime}$ on $\OMEGA\times\mathcal{T}$ equipped with
the $\sigma$-algebra $\sigma\left(X,t\right)$ by
\[
\Law_{\Q^{\prime}}\left[X,t\right]:=\Law_{\P\otimes\mu}\left[X+Y\left(t\right),t\right].
\]
The absolute continuity property in the definition of randomized shift
amounts to saying that the $\OMEGA$-projection of $\Q^{\prime}$
is absolutely continuous with respect to $\P$ on $\sigma\left(X\right)$,
so, in particular, one can define a random measure $M^{\prime}\left(X,dt\right)$
via disintegration:
\[
\P\left(d\omega\right)M^{\prime}\left(X,dt\right):=\Q^{\prime}\left(d\omega,dt\right)
\]
(this all happens on $\left(\OMEGA\times\mathcal{T},\sigma\left(X,t\right)\right)$,
so $M^{\prime}$ is automatically measurable with respect to $X$).
$\E M^{\prime}$ is the $\mathcal{T}$-projection of $\Q^{\prime}$,
so $\E M^{\prime}=\mu$ is $\sigma$-finite.
To check that $M^{\prime}$ is a GMC, introduce a measure class preserving
action $S_{\xi},\xi\in H$ of the additive group of $H$ on $\left(\OMEGA\times\mathcal{T},\sigma\left(X,t\right),\Q^{\prime}\right)$
by
\[
S_{\xi}\left(X,t\right):=\left(X+\xi,t\right).
\]
Since it really acts only on $X$ and does not change $t$, there
is also an action $X\mapsto X+\xi$ on $\left(\OMEGA,\sigma\left(X\right),\P\right)$,
which we also denote by $S_{\xi}$.
$S_{\xi}$ is measure class preserving ($\Q^{\prime}$), since it
preserves the measure class of almost all fibers in the disintegration
of $\Q^{\prime}$ with respect to $t$. Indeed, this amounts to saying
that $\Lawcm{\P\otimes\mu}{X+Y\left(t\right)+\xi}t\ll\Lawcm{\P\otimes\mu}{X+Y\left(t\right)}t$
for $\mu$-almost all $t$, which is obvious from the Cameron-Martin
theorem. Moreover, the same argument gives an expression for the density:
\begin{equation}
\frac{\left(S_{\xi}\right)_{\ast}\Q^{\prime}\left(d\omega,dt\right)}{\Q^{\prime}\left(d\omega,dt\right)}=\exp\left[\left\langle X\left(\omega\right)-Y\left(t\right),\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}\right]\label{eq:Density1}
\end{equation}
($\left(S_{\xi}\right)_{\ast}\Q^{\prime}$ is the pushforward of $\Q^{\prime}$
by the map $S_{\xi}$, i.e. $\left(S_{\xi}\right)_{\ast}\Q^{\prime}\left[A\right]=\Q^{\prime}\left[S_{\xi}^{-1}\left[A\right]\right]$
for measurable sets $A\in\sigma\left(X,t\right)$).
Now we compute the very same density in a different way, by disintegrating
$\Q^{\prime}$ with respect to $X$ instead of $t$. This shows how
$M^{\prime}$ behaves with respect to shifts:
\begin{equation}
\begin{aligned}\frac{\left(S_{\xi}\right)_{\ast}\Q^{\prime}\left(d\omega,dt\right)}{\Q^{\prime}\left(d\omega,dt\right)} & =\frac{\left(S_{\xi}\right)_{\ast}\left[\P\left(d\omega\right)M^{\prime}\left(X\left(\omega\right),dt\right)\right]}{\P\left(d\omega\right)M^{\prime}\left(X\left(\omega\right),dt\right)}\\
& =\frac{\left(S_{\xi}\right)_{\ast}\P\left(d\omega\right)}{\P\left(d\omega\right)}\cdot\frac{M^{\prime}\left(X\left(\omega\right)-\xi,dt\right)}{M^{\prime}\left(X\left(\omega\right),dt\right)}\\
& =\exp\left[\left\langle X\left(\omega\right),\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}\right]\cdot\frac{M^{\prime}\left(X\left(\omega\right)-\xi,dt\right)}{M^{\prime}\left(X\left(\omega\right),dt\right)}.
\end{aligned}
\label{eq:Density2}
\end{equation}
By comparing \eqref{eq:Density1} to \eqref{eq:Density2} we see that
$M^{\prime}$ is a GMC.
\end{proof}
\section{Proof of Theorem \ref{thm:HilbertSchmidtMoments} and Corollary \ref{cor:Moments}
\label{sec:KernelRegularity}}
The following basic observation will be useful in the proof:
\begin{lem}
Let $M_{Y}$ be the subcritical GMC associated to $Y$. Then there
exists a subcritical GMC associated to $cY$ for any $\left|c\right|\le1$,
namely,
\[
M_{cY}\left(X\right):=\Ec{M_{Y}\left(cX+\left(1-c^{2}\right)^{1/2}X^{\prime}\right)}X
\]
where $X^{\prime}$ is an independent copy of $X$. \label{lem:CY}\end{lem}
\begin{proof}
Shifting $X$ by $\xi\in H$ in the left-hand side amounts to shifting
the argument $cX+\left(1-c^{2}\right)^{1/2}X^{\prime}$ in the right-hand
side by $c\xi$, so that the GMC property for $M_{cY}$ follows trivially
from the one for $M_{Y}$.
\end{proof}
\begin{proof}
[Proof of Theorem \ref{thm:HilbertSchmidtMoments}] Fix $n$, and
consider a Hilbert-Schmidt symmetric tensor $\lambda\in H^{\otimes n}$.
Denote the Wick polynomial corresponding to $\lambda$ by
\[
P_{\lambda}\left(X\right):=\Wick{\left\langle \lambda,X^{\otimes n}\right\rangle }.
\]
It is well-known (see, e.g., \cite{Janson}) that $\E\left|P_{\lambda}\left(X\right)\right|^{2}=n!\norm{\lambda}_{2}^{2}$,
so in particular the family of random variables
\[
\left\{ P_{\lambda}\left(X\right),\norm{\lambda}_{2}\le1\right\}
\]
is bounded in probability. Now by Lemma \ref{lem:CY} we know that
for every $\left|c\right|\le1$ the measure $\Law_{\P\otimes\mu}\left[X+cY\right]$
is absolutely continuous with respect to $\Law X$. Therefore for
fixed $c$ the family of random variables
\[
\left\{ P_{\lambda}\left(X+cY\right)\,\middle|\,\norm{\lambda}_{2}\le1\right\}
\]
is bounded in probability. Since $c\mapsto P_{\lambda}\left(X+cY\right)$
is an $n$-th degree polynomial, we can extract its $n$-th degree
coefficient in $c$ (denoted by $\left[c^{n}\right]P_{\lambda}\left(X+cY\right)$)
by taking an appropriate linear combination of its values at different
$c$. For instance, we can use the $n$-th iterated difference with
step $\frac{1}{n}$:
\begin{equation}
\left[c^{n}\right]P_{\lambda}\left(X+cY\right)=\frac{n^{n}}{n!}\sum_{k=0}^{n}{n \choose k}\left(-1\right)^{n-k}P_{\lambda}\left(X+\frac{k}{n}Y\right).\label{eq:nCoeff}
\end{equation}
Next we argue that the random variables $\left\langle \lambda,Y^{\otimes n}\right\rangle $
are well-defined and continuously depend on $\lambda\in H^{\otimes n}$.
To this end note that they are well-defined a priori for $\lambda$
of finite rank (i.e. $\lambda\in H_{\op{alg}}^{\otimes n}$). Also
note that the Wick product and the ordinary product only differ in
lower degree terms, therefore for any finite rank $\lambda$ we have,
\[
\left\langle \lambda,Y^{\otimes n}\right\rangle =\left[c^{n}\right]P_{\lambda}\left(X+cY\right).
\]
Now for fixed $n$, by \eqref{eq:nCoeff}, this family of random variables
is bounded in probability as $\norm{\lambda}_{2}\le1,\rank\lambda<\infty$.
This implies that the $L^{0}$-valued operator $Y^{\otimes n}$, defined
on finite rank tensors, is bounded in the Hilbert-Schmidt norm, so
it can be extended by continuity to an operator defined on all Hilbert-Schmidt
tensors $\lambda$. Therefore, by Theorem \ref{thm:Factorization},
for some measures $\mu_{n}^{\prime}$ equivalent to $\mu$ all linear
functionals of $Y^{\otimes n}$ have second (therefore, first) moments,
and $\norm{\E_{\mu_{n}^{\prime}}Y^{\otimes n}}_{2}<\infty$.
\end{proof}
\begin{proof}
[Proof of Corollary \ref{cor:Moments}] If $Y$ is an $H$-valued
function, we have:
\[
\begin{aligned}\intop\left(K\left(t,s\right)\right)^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right) & =\intop\left\langle Y\left(t\right),Y\left(s\right)\right\rangle ^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)\\
& =\intop\left\langle \left(Y\left(t\right)\right)^{\otimes n},\left(Y\left(s\right)\right)^{\otimes n}\right\rangle \mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)\\
& =\norm{\E_{\mu_{n}^{\prime}}Y^{\otimes n}}_{2}^{2}
\end{aligned}
\]
which is finite for the measure $\mu_{n}^{\prime}$ constructed in
Theorem \ref{thm:HilbertSchmidtMoments}. Below we extend this computation
to the case where $Y$ is a generalized random vector.
Fix an orthonormal basis $\left(e_{i}\right)$ in $H$, and denote
by $Y_{i}$ the $i$-th coordinate of $Y$ in this basis, i.e. $Y_{i}=\left\langle Y,e_{i}\right\rangle $.
Fix also a basis $\left(\varepsilon_{p}\right)$ in $L^{2}\left(\mu_{n}^{\prime}\right)$.
Then the Hilbert-Schmidt norm of $K$ may be rewritten as
\[
\begin{aligned}\sum_{p,q}\left\langle \E_{\mu_{n}^{\prime}}\varepsilon_{p}Y,\E_{\mu_{n}^{\prime}}\varepsilon_{q}Y\right\rangle _{H}^{2} & =\sum_{p,q}\left(\sum_{i}\E_{\mu_{n}^{\prime}}\varepsilon_{p}Y_{i}\cdot\E_{\mu_{n}^{\prime}}\varepsilon_{q}Y_{i}\right)^{2}\\
& =\sum_{p,q}\sum_{i,j}\E_{\mu_{n}^{\prime}}\varepsilon_{p}Y_{i}\cdot\E_{\mu_{n}^{\prime}}\varepsilon_{q}Y_{i}\cdot\E_{\mu_{n}^{\prime}}\varepsilon_{p}Y_{j}\cdot\E_{\mu_{n}^{\prime}}\varepsilon_{q}Y_{j}\\
& =\sum_{i,j}\left(\sum_{p}\E_{\mu_{n}^{\prime}}\varepsilon_{p}Y_{i}\cdot\E_{\mu_{n}^{\prime}}\varepsilon_{p}Y_{j}\right)^{2}\\
& =\sum_{i,j}\left(\E_{\mu_{n}^{\prime}}Y_{i}Y_{j}\right)^{2}=\norm{\E_{\mu_{n}^{\prime}}Y^{\otimes2}}_{2}^{2},
\end{aligned}
\]
which is finite by Theorem \ref{thm:HilbertSchmidtMoments}.
It follows by standard theory that
\[
\norm{\intop f\left(t\right)Y\left(t\right)\mu_{n}^{\prime}\left(dt\right)}^{2}=\intop K\left(t,s\right)f\left(t\right)f\left(s\right)\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)
\]
for some $K\in L^{2}\left(\mu_{n}^{\prime}\otimes\mu_{n}^{\prime}\right)$.
Now that $K$ is actually a function, we may express it as the $L^{2}\left(\mu_{n}^{\prime}\otimes\mu_{n}^{\prime}\right)$-convergent
sum
\[
K\left(t,s\right)=\sum_{i}Y_{i}\left(t\right)Y_{i}\left(s\right).
\]
Without limitation of generality we may assume that $n$ is even.
By Fatou's lemma,
\[
\begin{aligned}\intop\left(K\left(t,s\right)\right)^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right) & \le\liminf_{N\to\infty}\intop\left(\sum_{i\le N}Y_{i}\left(t\right)Y_{i}\left(s\right)\right)^{n}\mu_{n}^{\prime}\left(dt\right)\mu_{n}^{\prime}\left(ds\right)\\
& =\liminf_{N\to\infty}\sum_{i_{1}\le N}\dots\sum_{i_{n}\le N}\left(\E_{\mu_{n}^{\prime}}Y_{i_{1}}\dots Y_{i_{n}}\right)^{2}\\
& =\norm{\E_{\mu_{n}^{\prime}}Y^{\otimes n}}_{2}^{2},
\end{aligned}
\]
which is finite by Theorem \ref{thm:HilbertSchmidtMoments}.
\end{proof}
\section{Proof of Theorem \ref{thm:Approximation} \label{sec:Approximation}}
\subsection{Outline}
The idea of the proof of Theorem \ref{thm:Approximation} can be outlined
as follows.
By a compactness argument (Lemma \ref{lem:Compactness}) we extract
a subsequence $\left(n^{\prime}\right)$, such the couple $\left(X,M_{Y_{n^{\prime}}}\right)$
converges in law to $\left(X,M\right)$ for some random measure $M$,
defined on an extended probability space and not necessarily measurable
with respect to $X$. By a general argument (Lemma \ref{lem:MeasurabilityImpliesConvergence})
in order for $M_{Y_{n}}$ to converge to $M$ in probability rather
than just in law it is enough to show that $M$ is measurable with
respect to $X$.
On the other hand, we show in Lemma \ref{lem:CoupledGMC} that the
coupling of the limiting measure $M$ to the Gaussian $X$ exhibits
the following property: for every $\xi\in H$ the distribution of
$\left(X+\xi,e^{\left\langle \xi,Y\right\rangle }M\right)$ is absolutely
continuous with respect to that of $\left(X,M\right)$ with the usual
Cameron-Martin density $e^{\left\langle \cdot,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}$
--- in particular, for any bounded continuous function $G:\RR_{+}\to\RR_{+}$
\begin{equation}
\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}G\left(M\left[\mathcal{T}\right]\right)=\E G\left(\intop e^{\left\langle \xi,Y\left(s\right)\right\rangle }M\left(ds\right)\right).\label{eq:GtrickXi}
\end{equation}
This property implies, in particular, that $\Ec MX=:M_{Y}$ is a subcritical
GMC associated to $Y$, which is, therefore, a randomized shift.
The heart of the proof consists in establishing the measurability
of $M$ with respect to $X$, or, in other words, that $M=\Ec MX=M_{Y}$.
This is done by replacing the deterministic shift $\xi$ in \eqref{eq:GtrickXi}
by the randomized shift $Y$, in two ways:
\begin{equation}
\begin{aligned}\E M\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right) & \underset{(1)}{=}\E\intop\mu\left(dt\right)G\left(\intop e^{\left\langle Y\left(t\right),Y\left(s\right)\right\rangle }M\left(ds\right)\right)\\
& \underset{(2)}{\le}\E M_{Y}\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right).
\end{aligned}
\label{eq:GtrickY}
\end{equation}
This is the content of Claim \ref{cl:Gtrick}. On the one hand, ``$\underset{(1)}{=}$''
is a property of the law of $M$ alone, and it is proved by passing
to the limit the corresponding property for the approximating GMCs
$M_{Y_{n}}$ (Lemma \ref{lem:ExpK}). On the other hand, ``$\underset{(2)}{\le}$''
is a property of the coupling $\left(X,M\right)$ that follows by
randomizing $\xi$ in \eqref{eq:GtrickXi}.
Finally, we apply \eqref{eq:GtrickY} with $G\left(x\right):=\frac{x}{1+x}$.
It follows that $M\left[\mathcal{T}\right]=M_{Y}\left[\mathcal{T}\right]$
by simple manipulations with Jensen's inequality, using the fact that
$G$ is strictly concave and $x\mapsto xG\left(x\right)$ is strictly
convex.
The rest of the section is structured as follows. In Section \ref{sub:Body}
we state explicitly the Lemmas alluded to above and show how Theorem
\ref{thm:Approximation} follows from them. In Section \ref{sub:ExpK}
we prove the central technical Lemma \ref{lem:ExpK}, which allows
to replace deterministic shifts by randomized shifts in the definition
of GMC and is crucial to the proof of \eqref{eq:GtrickY} above. In
Section \ref{sub:USAC} we introduce the notion of uniform stochastic
absolute continuity that is used in the approximation procedures.
In Sections \ref{sub:CoupledGMC}-\ref{sub:Basic} we prove all the
lemmas announced in Section \ref{sub:Body}.
After the proof, in Section \ref{sub:Remark} we briefly discuss another
use of Lemma \ref{lem:ExpK} and the related nonatomicity problem.
\subsection{Key lemmas and proof of Theorem \ref{thm:Approximation} \label{sub:Body}}
Throughout this section we use the notation and assumptions of Theorem
\ref{thm:Approximation}.
We begin with the basic distributional compactness result, Lemma \ref{lem:Compactness}.
The important assumptions for it are the boundedness of expectations
$\E M_{Y_{n}}=\mu$, which ensures that the sequence $\left\{ M_{Y_{n}}\left[\mathcal{T}\right]\right\} $
is tight, and the uniform integrability of $M_{Y_{n}}\left[\mathcal{T}\right]$
which implies that the limiting random measure also has expectation
$\mu$. Although in our setting the space $\mathcal{T}$ carries no
canonical topology, in the proof of Lemma \ref{lem:Compactness} we
reduce it to the classical weak compactness of the space of measures
on a compact metrizable space by introducing an auxiliary topology.
\begin{lem}
There exists a subsequence $\left(n^{\prime}\right)$, and a random
measure $M$ on $\left(\mathcal{T},\mu\right)$ with $\E M=\mu$,
possibly defined on an extended probability space, such that
\begin{equation}
\forall\xi\in H,\forall f\in L^{1}\left(\mu\right):\left(\left\langle X,\xi\right\rangle ,\intop f\left(t\right)M_{Y_{n^{\prime}}}\left(dt\right)\right)\overset{\Law}{\to}\left(\left\langle X,\xi\right\rangle ,\intop f\left(t\right)M\left(dt\right)\right).\label{eq:JointLimit}
\end{equation}
\label{lem:Compactness}
\end{lem}
We abbreviate \eqref{eq:JointLimit} to ``$\left(X,M_{Y_{n^{\prime}}}\right)\overset{\Law}{\to}\left(X,M\right)$''.
In the sequel we restrict attention to a convergent subsequence without
further mention and reserve the letter $M$ for the limiting measure.
Next we reduce the convergence $M_{Y_{n^{\prime}}}\to M$ in probability
to the measurability of $M$ with respect to $X$.
\begin{lem}
Assume that $M$ is measurable with respect to $X$. Then $M_{Y_{n}}\overset{L^{1}}{\to}M$
in the sense that
\[
\forall f\in L^{1}\left(\mu\right):\intop f\left(t\right)M_{Y_{n}}\left(dt\right)\overset{L^{1}}{\to}\intop f\left(t\right)M\left(dt\right).
\]
\label{lem:MeasurabilityImpliesConvergence}
\end{lem}
The rest of the proof concerns identifying GMC-like properties of
$M$ by passing to the limit the corresponding properties of $M_{Y_{n}}$.
\begin{lem}
For any $\xi\in H$
\[
\Law\left[X+\xi,e^{\left\langle Y,\xi\right\rangle }M\right]\ll\Law\left[X,M\right],
\]
and for any positive random variable $g$, measurable with respect
to $\left(X,M\right)$, we have
\begin{equation}
\E g\left(X+\xi,e^{\left\langle Y,\xi\right\rangle }M\right)=\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}g\left(X,M\right).\label{eq:CoupledGMC}
\end{equation}
Furthermore, $Y$ is a randomized shift, and $\Ec MX=:M_{Y}$ is the
GMC associated to it. \label{lem:CoupledGMC}
\end{lem}
The property \eqref{eq:CoupledGMC} can be viewed as the natural extension
of the definition of GMC to random measures that are not necessarily
measurable with respect to $X$.
The remaining part of the statement of Theorem \ref{thm:Approximation}
is the measurability of $M$ with respect to $X$, or equivalently,
$M=\Ec MX$. The key role in proving that is played by the following
claim:
\begin{claim}
Let $G:\RR_{+}\to\RR_{+}$ be any bounded continuous function. Then
\[
\E M\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right)=\E\intop\mu\left(dt\right)G\left(\intop e^{\left\langle Y\left(t\right),Y\left(s\right)\right\rangle }M\left(ds\right)\right)\le\E M_{Y}\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right).
\]
\label{cl:Gtrick}
\end{claim}
Note that a part of the claim is that $\intop e^{\left\langle Y\left(t\right),Y\left(s\right)\right\rangle }M\left(ds\right)<\infty$
almost surely for $\mu$-almost all $t$. This is also contained in
Lemma \ref{lem:ExpK} below.
Finally we show how Theorem \ref{thm:Approximation} follows from
the above lemmas.
\begin{proof}
[Proof of Theorem \ref{thm:Approximation}] We use Lemma \ref{lem:Compactness}
to extract a subsequence $\left(n^{\prime}\right)$ and a limiting
couple $\left(X,M\right)$. Apply Claim \ref{cl:Gtrick} with the
function $G\left(x\right):=\frac{x}{1+x}$ to obtain the following
chain of inequalities:
\begin{equation}
\begin{aligned}\E M\left[\mathcal{T}\right]G\left(\Ec{M\left[\mathcal{T}\right]}X\right) & =\E\left[\Ec{M\left[\mathcal{T}\right]}XG\left(\Ec{M\left[\mathcal{T}\right]}X\right)\right]\\
& \le\E M\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right)\\
& \underset{(!)}{\le}\E\left[\Ec{M\left[\mathcal{T}\right]}XG\left(M\left[\mathcal{T}\right]\right)\right]\\
& =\E\left[M\left[\mathcal{T}\right]\Ec{G\left(M\left[\mathcal{T}\right]\right)}X\right]\\
& \le\E\left[M\left[\mathcal{T}\right]G\left(\Ec{M\left[\mathcal{T}\right]}X\right)\right].
\end{aligned}
\label{eq:TrickJensen}
\end{equation}
The inequality ``$\underset{(!)}{\le}$'' is the content of Claim
\ref{cl:Gtrick}. The first and the last inequalities are both instances
of Jensen's inequality, applied to the strictly convex function $x\mapsto xG\left(x\right)$
and the strictly concave function $G$ respectively. Equality in Jensen's
inequality implies that $M\left[\mathcal{T}\right]=\Ec{M\left[\mathcal{T}\right]}X=M_{Y}\left[\mathcal{T}\right]$,
so that $M\left[\mathcal{T}\right]$ is measurable with respect to
$X$.
The same argument works with $M$ replaced by $f\cdot M$ for any
nonnegative bounded $f$ (and $\mu$ by $f\cdot\mu$ accordingly),
which amounts to replacing $M\left[\mathcal{T}\right]$ by $\intop f\left(t\right)M\left(dt\right)$.
Thus in fact Claim \ref{cl:Gtrick} implies that $M$ is measurable
with respect to $X$, i.e. $M=M_{Y}$.
Now we use Lemma \ref{lem:MeasurabilityImpliesConvergence} to assert
that $M_{Y}$ is a limit in $L^{1}$ of the chosen subsequence. By
Lemma \ref{lem:Compactness}, any subsequence has a convergent subsequence,
and since the limit is the unique (Corollary \ref{cor:Uniqueness})
subcritical GMC $M_{Y}$, it does not depend on the choice of subsequences.
This means that the original sequence converges.
\end{proof}
\subsection{The ``$\exp K$ lemma'' \label{sub:ExpK}}
A central role in the proof of Theorem \ref{thm:Approximation} is
played by Lemma \ref{lem:ExpK} stated below.
\begin{lem}
Let $Z$ and $W$ be randomized shifts, defined on the probability
space $\left(\mathcal{T},\mu\right)$, and let $M_{Z},M_{W}$ be the
corresponding subcritical GMCs with expectation $\mu$. Let $K_{ZW}\left(t,s\right):=\left\langle Z\left(t\right),W\left(s\right)\right\rangle $.
Then:
\begin{equation}
M_{Z}\left(X+W\left(s\right),dt\right)=\exp K_{ZW}\left(t,s\right)\cdot M_{Z}\left(X,dt\right)\ \left(\P\otimes\mu\text{-a.s.}\right)\label{eq:GMCProperty}
\end{equation}
and
\begin{equation}
\E\left[M_{Z}\left(X\right)\otimes M_{W}\left(X\right)\right]=\exp K_{ZW}\cdot\mu\otimes\mu.\label{eq:SecondMoment}
\end{equation}
\label{lem:ExpK}
\end{lem}
Since both covariances $K_{ZZ}$ and $K_{WW}$ are Hilbert-Schmidt
by means of Corollary \ref{cor:Moments}, $K_{ZW}$ is also Hilbert-Schmidt.
Indeed, $\left[\begin{matrix}K_{ZZ} & K_{ZW}\\
K_{WZ} & K_{WW}
\end{matrix}\right]$ is positive definite and the diagonal blocks are Hilbert-Schmidt,
therefore so are the off-diagonal blocks.
We will use Lemma \ref{lem:ExpK} in the case $Z:=W:=Y_{n}$ or $Z:=W:=Y$,
but for the proof of Lemma \ref{lem:ExpK} it is convenient to separate
$Z$ and $W$.
In the case of ``trivial'' GMC (in the sense of Example \ref{ex:TrivialGMC})
both statements of Lemma \ref{lem:ExpK} are elementary, so in general
they may look formally ``obvious''. However, the real content of
Lemma \ref{lem:ExpK} lies in the implied absolute continuity:
\begin{equation}
M_{Z}\left(X+W\left(s\right)\right)\ll M_{Z}\left(X\right)\text{ (\ensuremath{\P\otimes\mu}-a.s.)},\label{eq:GMCPropertyAC}
\end{equation}
\begin{equation}
\E\left[M_{Z}\otimes M_{W}\right]\ll\E M_{Z}\otimes\E M_{W}.\label{eq:SecondMomentAC}
\end{equation}
In particular, \eqref{eq:SecondMomentAC} is a property specific to
GMCs that is far from being true for general random measures.
Another immediate observation following from Lemma \ref{lem:ExpK}
is that $\E\left[M_{Z}\otimes M_{W}\right]$ is $\sigma$-finite ---
since the right-hand side of \eqref{eq:SecondMoment} obviously is.
This measure may, however, fail to be locally finite in any reasonable
sense --- for example, it may assign infinite mass to every product
set $A\times B\subset\mathcal{T}\times\mathcal{T}$, such that $\mu\left[A\right],\mu\left[B\right]>0$.
In the proof of Lemma \ref{lem:ExpK} we will need the following general
fact:
\begin{lem}
Let $Z_{n}$ and $Z$ be randomized shifts of $X$ defined on the
same probability space $\left(\mathcal{T},\mu\right)$. Assume that
the family of random variables $\left\{ \intop_{\mathcal{T}}M_{Z_{n}}\left(X,dt\right),n\ge1\right\} $
is uniformly integrable. Also assume that $Z_{n}\to Z$ in the following
sense:
\[
\forall\xi\in H:\left\langle Z_{n},\xi\right\rangle \overset{L^{0}\left(\mathcal{T},\mu\right)}{\to}\left\langle Z,\xi\right\rangle .
\]
Then for every $f\left(X\right)\in L^{0}\left(\OMEGA,\sigma\left(X\right),\P\right)$
we have
\[
f\left(X+Z_{n}\right)\overset{L^{0}\left(\left(\OMEGA,\P\right)\otimes\left(\mathcal{T},\mu\right)\right)}{\to}f\left(X+Z\right).
\]
\label{lem:ContinuityRandomized}\end{lem}
\begin{proof}
For $f\left(X\right):=\exp i\left\langle \xi,X\right\rangle $, $\xi\in H$,
this follows from the assumptions. It is well-known that linear combinations
of exponentials $\exp i\left\langle \xi,X\right\rangle $ are dense
in $L^{0}\left(\OMEGA,\sigma\left(X\right)\right)$. Let $f\in L^{0}\left(\OMEGA\right)$
and let $f_{n}\left(X\right)\to f\left(X\right)$ be a sequence of
linear combinations of exponentials approximating $f\left(X\right)$
in $L^{0}\left(\OMEGA\right)$.
We claim that
\begin{equation}
f_{n}\left(X+Z_{m}\right)\overset{L^{0}}{\to}f\left(X+Z_{m}\right),n\to\infty,\text{ uniformly in }m.\label{eq:UniformL0}
\end{equation}
Indeed, for any $\varepsilon$ we have
\[
\P\left\{ \left|f_{n}\left(X\right)-f\left(X\right)\right|>\varepsilon\right\} \to0,n\to\infty.
\]
Due to the uniform integrability of $\left\{ M_{Z_{m}}\left[\mathcal{T}\right]\right\} $,
this implies
\[
\sup_{m}\E M_{Z_{m}}\left[\mathcal{T}\right]\I\left\{ \left|f_{n}\left(X\right)-f\left(X\right)\right|>\varepsilon\right\} \to0,n\to\infty,
\]
so that
\[
\sup_{m}\E_{\P\otimes\nu}\I\left\{ \left|f_{n}\left(X+Z_{m}\right)-f\left(X+Z_{m}\right)\right|>\varepsilon\right\} \to0,n\to\infty,
\]
which implies \eqref{eq:UniformL0}. Similarly, we have
\begin{equation}
f_{n}\left(X+Z\right)\overset{L^{0}}{\to}f\left(X+Z\right).\label{eq:fnf}
\end{equation}
Now by a standard argument \eqref{eq:UniformL0}, \eqref{eq:fnf}
and the fact that
\[
\forall n:f_{n}\left(X+Z_{m}\right)\overset{L^{0}}{\to}f_{n}\left(X+Z\right),m\to\infty
\]
implies the statement of the lemma.
\end{proof}
\begin{proof}
[Proof of Lemma \ref{lem:ExpK}] As a first step, we prove \eqref{eq:GMCProperty}
under the assumption that
\[
\left(1+\delta\right)W\text{ is a randomized shift for some }\delta>0,
\]
which we refer to as ``strict subcriticality''.
We approximate the shift $W$ by its projections $P_{n}W$, where
$\left(P_{n}\right)$ is an increasing sequence of finite-dimensional
orthogonal projection operators in $H$, converging strongly to $1$.
Note that for $P_{n}W$ both statements of Lemma \ref{lem:ExpK} are
satisfied by the definition of GMC $M_{Z}$, since $P_{n}$ are finite-dimensional
and $P_{n}W$ are just $H$-valued functions.
Let $f\in L^{\infty}\left(\mathcal{T},\mu\right)$ be a positive function.
By applying Lemma \ref{lem:ContinuityRandomized} to $W_{n}:=cP_{n}W$,
$c\in\left\{ 1,1+\delta\right\} $ , we get:
\[
\intop f\left(t\right)M_{Z}\left(X+cP_{n}W\left(s\right),dt\right)\overset{L^{0}\left(\P\otimes\mu\right)}{\to}\intop f\left(t\right)M_{Z}\left(X+cW\left(s\right),dt\right).
\]
In the left-hand side, since $P_{n}$ is finite-dimensional,
\begin{equation}
M_{Z}\left(X+cP_{n}W\left(s\right),dt\right)=e^{c\left\langle P_{n}W\left(s\right),Z\left(t\right)\right\rangle }M_{Z}\left(X,dt\right).\label{eq:ExpKFiniteDim}
\end{equation}
By applying \eqref{eq:ExpKFiniteDim} to $c=1+\delta$, we see that,
along a deterministic subsequence, the integrals
\[
\intop e^{\left(1+\delta\right)\left\langle P_{n}W\left(s\right),Z\left(t\right)\right\rangle }M_{Z}\left(X,dt\right)
\]
converge $\P\otimes\mu$-almost surely, and thus stay $\P\otimes\mu$-almost
surely bounded as $n$ increases. Therefore, the subsequence of functions
$t\mapsto e^{\left\langle P_{n}W\left(s\right),Z\left(t\right)\right\rangle }$
is $\P\otimes\mu$-almost surely uniformly integrable against the
random measure $M_{Z}$. Therefore, by the Lebesgue convergence theorem,
along that subsequence
\[
\intop f\left(t\right)e^{\left\langle P_{n}W\left(s\right),Z\left(t\right)\right\rangle }M_{Z}\left(X,dt\right)\to\intop f\left(t\right)e^{\left\langle W\left(s\right),Z\left(t\right)\right\rangle }M_{Z}\left(X,dt\right).
\]
This implies that $M_{Z}\left(X+W\left(s\right),dt\right)=e^{\left\langle W\left(s\right),Z\left(t\right)\right\rangle }M_{Z}\left(X,dt\right)$.
Next we reduce the general case to the strictly subcritical one. For
this we approximate a general randomized shift $W$ by shifts $\left(1-\varepsilon\right)W$
as $\varepsilon\to0$. All $\left(1-\varepsilon\right)W$ are obviously
strictly subcritical, so we already know that
\[
M_{Z}\left(X+\left(1-\varepsilon\right)W\right)=e^{\left(1-\varepsilon\right)K_{ZW}}M_{Z}\left(X\right).
\]
Now on the one hand, by Lemma \ref{lem:ContinuityRandomized}, for
any bounded $f\ge0$
\[
\intop f\left(t\right)M_{Z}\left(X+\left(1-\varepsilon\right)W,dt\right)\overset{L^{0}\left(\P\otimes\mu\right)}{\to}\intop f\left(t\right)M_{Z}\left(X+W,dt\right),\varepsilon\to0.
\]
On the other hand,
\[
\intop f\left(t\right)e^{\left(1-\varepsilon\right)K_{ZW}}M_{Z}\left(X,dt\right)\to\intop f\left(t\right)e^{K_{ZW}}M_{Z}\left(X,dt\right)
\]
almost surely. Indeed, one can split $\intop f\left(t\right)e^{\left(1-\varepsilon\right)K_{ZW}}M_{Z}\left(X,dt\right)$
into the integral over $\left\{ K_{ZW}\ge0\right\} $ and the integral
over $\left\{ K_{ZW}<0\right\} $; on $\left\{ K_{ZW}\ge0\right\} $
the integrand increases as $\varepsilon\downarrow0$, so the monotone
convergence theorem applies, and on $\left\{ K_{ZW}<0\right\} $ it
is dominated by $1$, which is integrable against $M_{Z}$. Thus we
have proved \eqref{eq:GMCProperty} in the general case.
\eqref{eq:SecondMoment} is deduced from \eqref{eq:GMCProperty} by
taking the expectation of both sides multiplied by $\mu\left(ds\right)$.
Indeed, on the one hand, by Theorem \ref{thm:GMCShifts},
\[
\E\left[M_{Z}\left(X+W\left(s\right),dt\right)\mu\left(ds\right)\right]=\E\left[M_{Z}\left(X,dt\right)\otimes M_{W}\left(X,ds\right)\right],
\]
and on the other hand,
\[
\E\left[\exp K_{ZW}\left(t,s\right)\cdot M_{Z}\left(X,dt\right)\mu\left(ds\right)\right]=\exp K_{ZW}\left(t,s\right)\cdot\mu\left(dt\right)\otimes\mu\left(ds\right).
\]
\end{proof}
\subsection{Uniform stochastic absolute continuity \label{sub:USAC}}
In order to deal with the approximations of random measures involved
in the proof of Claim \ref{cl:Gtrick} we need a measure-theoretic
tool (Lemma \ref{lem:StochasticLebesgue}) for proving convergence
of integrals of functions against random measures. This lemma can
be seen as a stochastic analogue of Lebesgue's convergence theorem,
and just like Lebesgue's theorem, it comes with a related notion of
``uniform integrability''.
\begin{defn}
A family of random measures $\left\{ M_{\alpha}\right\} $ on a measurable
space $\mathcal{T}$ is called \emph{uniformly stochastically absolutely
continuous} with respect to a deterministic probability measure $\mu$
on $\mathcal{T}$ if
\begin{itemize}
\item $\forall\alpha:\E M_{\alpha}\ll\mu$
\item For every $c>0$ we have
\[
\sup_{\substack{A\subset\mathcal{T}\\
\mu\left[A\right]\le\varepsilon
}
}\sup_{\alpha}\P\left\{ M_{\alpha}\left[A\right]>c\right\} \to0,\varepsilon\to0.
\]
\end{itemize}
\end{defn}
\begin{example}
If $\E M_{\alpha}=\mu$ for all $\alpha$ then $\left\{ M_{\alpha}\right\} $
is uniformly stochastically absolutely continuous. Indeed, if $\mu\left[A\right]<\varepsilon$
then $M_{\alpha}\left[A\right]$ are uniformly small in $L^{1}$,
therefore uniformly small in probability. \label{ex:USACL1}\end{example}
\begin{lem}
Let $\left(M_{n}\right)$ be a sequence of random measures on $\mathcal{T}$,
such that $\forall\alpha:\E M_{\alpha}\ll\mu$, and let $F_{n},F\in L^{0}\left(\mathcal{T},\mu\right)$.
Assume that:
\begin{itemize}
\item $M_{n}\to M$ in the sense that
\begin{equation}
\forall f\in L^{\infty}\left(\mathcal{T},\mu\right):\intop f\left(t\right)M_{n}\left(dt\right)\overset{L^{0}}{\to}\intop f\left(t\right)M\left(dt\right);\label{eq:ConvergenceOfRandomMeasures}
\end{equation}
\item $F_{n}\overset{L^{0}\left(\mu\right)}{\to}F$
\item For all $n$ we have
\[
\intop\left|F_{n}\left(t\right)\right|M_{n}\left(dt\right)<\infty;
\]
\item The family of random measures $\left\{ \left|F_{n}\left(t\right)\right|M_{n}\left(dt\right)\right\} $
is uniformly stochastically absolutely continuous with respect to
$\mu$.
\end{itemize}
Then
\[
\intop\left|F\left(t\right)\right|M\left(dt\right)<\infty
\]
and
\begin{equation}
\intop F_{n}\left(t\right)M_{n}\left(dt\right)\overset{L^{0}}{\to}\intop F\left(t\right)M\left(dt\right).\label{eq:ConvergenceOfIntegrals}
\end{equation}
The same is true if we replace convergence in $L^{0}$ by convergence
in law both in \eqref{eq:ConvergenceOfRandomMeasures} and in \eqref{eq:ConvergenceOfIntegrals}.
\label{lem:StochasticLebesgue}\end{lem}
\begin{proof}
By passing to a subsequence, we may assume that $F_{n}\to F$ almost
everywhere. Fix $\varepsilon>0$. By Egorov's theorem, there is a
set $A_{\varepsilon}\subset\mathcal{T}$, such that $\mu\left[\mathcal{T}\setminus A_{\varepsilon}\right]\le\varepsilon$
and $F_{n}\to F$ uniformly on $A_{\varepsilon}$.
\begin{multline*}
\left|\intop F_{n}\left(t\right)M_{n}\left(dt\right)-\intop_{A_{\varepsilon}}F\left(t\right)M_{n}\left(dt\right)\right|\\
\le\intop_{\mathcal{T}\setminus A_{\varepsilon}}\left|F_{n}\left(t\right)\right|M_{n}\left(dt\right)+\mathop{\mathrm{ess}\sup}_{A_{\varepsilon}}\left|F_{n}-F\right|\cdot M_{n}\left[\mathcal{T}\right].
\end{multline*}
The first term is small in probability uniformly in $n$ whenever
$\varepsilon$ is small due to the uniform stochastic absolute continuity.
The second term is small in probability for fixed $\varepsilon$ and
large $n$ because the family of random variables $\left\{ M_{n}\left[\mathcal{T}\right]\right\} $
is bounded in probability and $\mathop{\mathrm{ess}\sup}_{A_{\varepsilon}}\left|F_{n}-F\right|\to0,n\to\infty$.
On the other hand, $\intop_{A_{\varepsilon}}F\left(t\right)M_{n}\left(dt\right)$
is close in probability (resp. in law) to $\intop_{A_{\varepsilon}}F\left(t\right)M\left(dt\right)$
by assumption.
\end{proof}
The next statement is based entirely on Lemma \ref{lem:ExpK}.
\begin{lem}
Let $\left\{ Y_{\alpha}\right\} $ and $\left\{ Z_{\beta}\right\} $
be families of randomized shifts on $\left(\mathcal{T},\mu\right)$,
and let $\left\{ M_{Y_{\alpha}}\left[\mathcal{T}\right]\right\} $
be uniformly integrable. Let $T$ be a random point in $\mathcal{T}$
with law $\mu$, independent of $X$. Then the family of random measures
$N_{\alpha\beta}$ on $\mathcal{T}\times\mathcal{T}$ given by
\[
N_{\alpha\beta}:=\exp K_{Y_{\alpha}Z_{\beta}}\left(T,\cdot\right)\cdot\delta_{T}\otimes M_{Z_{\beta}}\left(X\right)
\]
is uniformly stochastically absolutely continuous with respect to
$\mu\otimes\mu$. \label{lem:USACGMC}\end{lem}
\begin{proof}
Fix $\varepsilon>0$. Consider a measurable subset $A\subset\mathcal{T}\times\mathcal{T}$,
and denote by $A_{t},t\in\mathcal{T}$ its $t$-section, i.e.
\[
A_{t}:=\left\{ s\in\mathcal{T}:\left(t,s\right)\in A\right\} \subset\mathcal{T}.
\]
By Lemma \ref{lem:ExpK}
\[
\begin{aligned}\P\left\{ N_{\alpha\beta}\left[A\right]>c\right\} & =\P\left\{ \intop_{A_{T}}\exp K_{Y_{\alpha}Z_{\beta}}\left(T,s\right)M_{Z_{\beta}}\left(X,ds\right)>c\right\} \\
& =\P\left\{ \intop_{A_{T}}M_{Z_{\beta}}\left(X+Y_{\alpha}\left(T\right),ds\right)>c\right\} \\
& =\E\intop M_{Y_{\alpha}}\left(X,dt\right)\I\left\{ \intop_{A_{t}}M_{Z_{\beta}}\left(X,ds\right)>c\right\} \\
& \le\mu\left\{ t:\nu\left[A_{t}\right]>\varepsilon\right\} \\
& \quad+\E\intop M_{Y_{\alpha}}\left(X,dt\right)\I\left\{ \nu\left[A_{t}\right]\le\varepsilon\right\} \I\left\{ \intop_{A_{t}}M_{Z_{\beta}}\left(X,ds\right)>c\right\} .
\end{aligned}
\]
The first term is small whenever $\left(\mu\otimes\nu\right)\left[A\right]$
is small enough. The second term is small whenever $\varepsilon$
is small enough, since for $t$, such that $\nu\left[A_{t}\right]\le\varepsilon$,
we have
\[
\P\left\{ \intop_{A_{t}}M_{Z_{\beta}}\left(X,ds\right)>c\right\} \le c^{-1}\E\intop_{A_{t}}M_{Z_{\beta}}\left(X,ds\right)=c^{-1}\nu\left[A_{t}\right]\le c^{-1}\varepsilon,
\]
and $\left\{ \intop M_{Y_{\alpha}}\left(X,dt\right)\right\} $ is
assumed to be uniformly integrable.
\end{proof}
\subsection{Proof of Lemma \ref{lem:CoupledGMC} \label{sub:CoupledGMC}}
\begin{proof}
It is enough to check \eqref{eq:CoupledGMC} for functions $g\left(X,M\right)$
of the form
\[
g\left(X,M\right):=\exp\left[\left\langle X,\eta\right\rangle -\frac{1}{2}\norm{\eta}^{2}\right]\cdot h\left(M\right)
\]
for all $\eta\in H$ and all bounded measurable $h$ that depend continuously
on finitely many integrals $\intop f_{1}\left(t\right)M\left(dt\right),\dots,\intop f_{m}\left(t\right)M\left(dt\right)$.
First we consider the case $\eta=0$. For the functions of such form,
it follows from the convergence in law $\left(X,M_{Y_{n}}\right)\overset{\Law}{\to}\left(X,M\right)$
that
\[
\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}h\left(M_{Y_{n}}\right)\to\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}h\left(M\right).
\]
On the other hand, by the Cameron-Martin formula and the definition
of GMC,
\[
\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}h\left(M_{Y_{n}}\right)=\E h\left(M_{Y_{n}}\left(X+\xi\right)\right)=\E h\left(e^{\left\langle \xi,Y_{n}\right\rangle }M_{Y_{n}}\right).
\]
The random measures $e^{\left\langle \xi,Y_{n}\right\rangle }M_{Y_{n}}$
are uniformly stochastically absolutely continuous by a degenerate
special case of Lemma \ref{lem:USACGMC} where one of the shifts is
the deterministic shift $\xi$. Thus by Lemma \ref{lem:StochasticLebesgue}
applied to the measures $M_{Y_{n}}\overset{\Law}{\to}M$ we have
\[
\E h\left(e^{\left\langle \xi,Y_{n}\right\rangle }M_{Y_{n}}\right)\to\E h\left(e^{\left\langle \xi,Y\right\rangle }M\right).
\]
Therefore,
\begin{equation}
\E e^{\left\langle X,\xi\right\rangle -\frac{1}{2}\norm{\xi}^{2}}h\left(M\right)=\E h\left(e^{\left\langle Y,\xi\right\rangle }M\right).\label{eq:CoupledGMCEta0}
\end{equation}
Now consider the case where $\eta$ is not necessarily $0$. \eqref{eq:CoupledGMC}
amounts to proving that
\[
\E e^{\left\langle X,\xi+\eta\right\rangle -\frac{1}{2}\norm{\xi}^{2}-\frac{1}{2}\norm{\eta}^{2}}h\left(M\right)=\E e^{\left\langle X+\xi,\eta\right\rangle -\frac{1}{2}\norm{\eta}^{2}}h\left(e^{\left\langle Y,\xi\right\rangle }M\right).
\]
By \eqref{eq:CoupledGMCEta0}, the left-hand side above equals $e^{\left\langle \xi,\eta\right\rangle }\E h\left(e^{\left\langle Y,\xi+\eta\right\rangle }M\right)$.
The right-hand side is the same by \eqref{eq:CoupledGMCEta0} applied
to the function $\tilde{h}\left(M\right):=h\left(e^{\left\langle Y,\xi\right\rangle }M\right)$.
\end{proof}
\subsection{Proof of Claim \ref{cl:Gtrick}}
Claim \ref{cl:Gtrick} consists of two statements which we prove separately:
\begin{equation}
\E M\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right)=\E_{\P\otimes\mu}G\left(\intop\exp\left\langle Y\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right),\label{eq:TrickEq1}
\end{equation}
\begin{equation}
\E M_{Y}\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right)\ge\E_{\P\otimes\mu}G\left(\intop\exp\left\langle Y\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right).\label{eq:TrickEq2}
\end{equation}
\begin{proof}
[Proof of Claim \ref{cl:Gtrick}: \eqref{eq:TrickEq1}]
By uniform integrability of $\left\{ M_{Y_{n}}\left[\mathcal{T}\right]\right\} $
and convergence $M_{Y_{n}}\left[\mathcal{T}\right]\overset{\Law}{\to}M\left[\mathcal{T}\right]$
we have
\[
\E M\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right)=\lim_{n\to\infty}\E M_{Y_{n}}\left[\mathcal{T}\right]G\left(M_{Y_{n}}\left[\mathcal{T}\right]\right).
\]
By Theorem \ref{thm:GMCShifts} and Lemma \ref{lem:ExpK}, both applied
to a single randomized shift $Y_{n}$, we have
\[
\begin{aligned}\E M_{Y_{n}}\left[\mathcal{T}\right]G\left(M_{Y_{n}}\left[\mathcal{T}\right]\right) & =\E_{\P\otimes\mu}G\left(M_{Y_{n}}\left(X+Y_{n}\left(t\right)\right)\left[\mathcal{T}\right]\right)\\
& =\E_{\P\otimes\mu}G\left(\intop\exp K_{Y_{n}Y_{n}}\left(t,s\right)M_{Y_{n}}\left(X,ds\right)\right).
\end{aligned}
\]
Thus to prove the claim it is enough to show that for a random point
$T$ in $\mathcal{T}$ with distribution $\mu$, independent of $X$,
we have
\begin{equation}
\intop\exp K_{Y_{n}Y_{n}}\left(T,s\right)M_{Y_{n}}\left(X,ds\right)\overset{\Law}{\to}\intop\exp K_{YY}\left(T,s\right)M\left(ds\right).\label{eq:ConvergenceExpKM}
\end{equation}
We rewrite both integrals tautologically in a way that involves random
measures and deterministic (i.e. not dependent on $T$) functions:
\[
\intop_{\mathcal{T}}\exp K_{Y_{n}Y_{n}}\left(T,s\right)M_{Y_{n}}\left(X,ds\right)=\intop_{\mathcal{T}\times\mathcal{T}}\exp K_{Y_{n}Y_{n}}\left(t,s\right)\cdot\delta_{T}\left(dt\right)\otimes M_{Y_{n}}\left(X,ds\right),
\]
\[
\intop_{\mathcal{T}}\exp K_{YY}\left(T,s\right)M\left(ds\right)=\intop_{\mathcal{T}\times\mathcal{T}}\exp K_{YY}\left(t,s\right)\cdot\delta_{T}\left(dt\right)\otimes M\left(ds\right).
\]
Now we apply Lemma \ref{lem:USACGMC} to the randomized shifts $\left\{ Y_{\alpha}\right\} :=\left\{ Y_{n}\right\} ,\left\{ Z_{\beta}\right\} :=\left\{ Y_{n}\right\} $
and deduce that $\left\{ \exp K_{Y_{n}Y_{n}}\cdot\delta_{T}\otimes M_{Y_{n}}\right\} $
is uniformly stochastically absolutely continuous. Then apply Lemma
\ref{lem:StochasticLebesgue} to the random measures $\delta_{T}\otimes M_{Y_{n}}$
and functions $\exp K_{Y_{n}Y_{n}}$. Note that $M_{Y_{n}}\overset{\Law}{\to}M$
trivially implies $\delta_{T}\otimes M_{Y_{n}}\overset{\Law}{\to}\delta_{T}\otimes M$,
and that by the assumptions of Theorem \ref{thm:Approximation}, $\exp K_{Y_{n}Y_{n}}\overset{L^{0}\left(\mu\otimes\mu\right)}{\to}\exp K_{YY}$,
so indeed Lemma \ref{lem:StochasticLebesgue} is applicable in this
case, yielding \eqref{eq:ConvergenceExpKM}, and therefore also the
claim.
\end{proof}
\begin{proof}
[Proof of Claim \ref{cl:Gtrick}: \eqref{eq:TrickEq2}] The strategy
is to randomize the $\xi$ in Lemma \ref{lem:CoupledGMC} and thus
approximate the randomized shift $Y$. By applying \eqref{eq:CoupledGMC}
conditionally, we have for every measurable vector-valued function
$\xi:\mathcal{T}\to H$
\[
\E_{\P\otimes\mu}\exp\left[\left\langle \xi\left(t\right),X\right\rangle -\frac{1}{2}\norm{\xi\left(t\right)}^{2}\right]G\left(M\left[\mathcal{T}\right]\right)=\E_{\P\otimes\mu}G\left(\intop\exp\left\langle \xi\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right).
\]
Take any increasing sequence $\left(P_{n}\right)$ of finite-dimensional
projections in $H$ that converge strongly to $1$. Since $P_{n}$
has finite-dimensional range, $P_{n}Y$ is in fact a vector-valued
function. Therefore, we can take $\xi\left(t\right):=P_{n}Y\left(t\right)$
above and obtain
\begin{multline*}
\E_{\P\otimes\mu}\exp\left[\left\langle P_{n}Y\left(t\right),X\right\rangle -\frac{1}{2}\norm{P_{n}Y\left(t\right)}^{2}\right]G\left(M\left[\mathcal{T}\right]\right)\\
=\E_{\P\otimes\mu}G\left(\intop\exp\left\langle P_{n}Y\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right).
\end{multline*}
Since the kernel $\left\langle Y\left(t\right),Y\left(s\right)\right\rangle $
is Hilbert-Schmidt, the function $\left\langle P_{n}Y\left(t\right),Y\left(s\right)\right\rangle $
converges in measure ($\mu\otimes\mu$) to $\left\langle Y\left(t\right),Y\left(s\right)\right\rangle $.
Therefore, by Fatou's lemma
\begin{multline}
\E_{\P\otimes\mu}G\left(\intop\exp\left\langle Y\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right)\\
\le\liminf_{n\to\infty}\E_{\P\otimes\mu}G\left(\intop\exp\left\langle P_{n}Y\left(t\right),Y\left(s\right)\right\rangle M\left(ds\right)\right).\label{eq:EqTrick2LimInf}
\end{multline}
On the other hand, $t$ and $M\left[\mathcal{T}\right]$ are conditionally
independent given $X$, so
\begin{multline*}
\E_{\P\otimes\mu}\exp\left[\left\langle P_{n}Y\left(t\right),X\right\rangle -\frac{1}{2}\norm{P_{n}Y\left(t\right)}^{2}\right]G\left(M\left[\mathcal{T}\right]\right)\\
=\E\left[\intop\exp\left[\left\langle P_{n}Y\left(t\right),X\right\rangle -\frac{1}{2}\norm{P_{n}Y\left(t\right)}^{2}\right]\mu\left(dt\right)\cdot G\left(M\left[\mathcal{T}\right]\right)\right]\\
=\E M_{P_{n}Y}\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right).
\end{multline*}
Since $M_{P_{n}Y}\left[\mathcal{T}\right]=\Ec{M_{Y}\left[\mathcal{T}\right]}{P_{n}X}$
is a uniformly integrable martingale that converges to $M_{Y}\left[\mathcal{T}\right]$,
we have
\[
\liminf_{n\to\infty}\E_{\P\otimes\mu}\exp\left[\left\langle P_{n}Y\left(t\right),X\right\rangle -\frac{1}{2}\norm{P_{n}Y\left(t\right)}^{2}\right]G\left(M\left[\mathcal{T}\right]\right)=\E M_{Y}\left[\mathcal{T}\right]G\left(M\left[\mathcal{T}\right]\right).
\]
Together with \eqref{eq:EqTrick2LimInf}, this proves the claim.
\end{proof}
\subsection{Proof of Lemmas \ref{lem:Compactness} and \ref{lem:MeasurabilityImpliesConvergence}
\label{sub:Basic}}
\begin{proof}
[Proof of Lemma \ref{lem:Compactness}]
By our assumptions $\left(\mathcal{T},\mu\right)$ is a standard measure
space, so we may assume that $\mathcal{T}=\left[0,1\right]$ with
its Borel $\sigma$-algebra. We can also identify $X$ (and other
generalized random vectors in $H$) with a random element in the Polish
space $\RR^{\infty}$, equipped with the isomorphism $H\simeq\ell^{2}\subset\RR^{\infty}$.
The family $\left\{ M_{Y_{n}}\left[\mathcal{T}\right]\right\} $ is
tight, therefore the family $\left\{ \left(X,M_{\alpha}\right)\right\} $
of random elements of $\RR^{\infty}\times\op{Measures}\left(\mathcal{T}\right)$
is tight when the space of measures is equipped with the weak topology.
Thus for some subsequence $\left(n^{\prime}\right)$ there is a distributional
limit $\left(X,M\right)$ of $\left(X,M_{Y_{n^{\prime}}}\right)$,
possibly on an extended probability space. This implies \eqref{eq:JointLimit}
for continuous $f$ and $\xi\in H$ with finitely many nonzero coordinates.
It is easy to see that the family of maps
\[
H\times L^{1}\left(\mu\right)\to L^{2}\left(\P\right)\times L^{1}\left(\P\right),
\]
\[
\left(\xi,f\right)\mapsto\left(\left\langle X,\xi\right\rangle ,\intop f\left(t\right)M_{Y_{n^{\prime}}}\left(dt\right)\right),
\]
is equicontinuous, so \eqref{eq:JointLimit} follows for all $\xi\in H,f\in L^{1}$.
The equality $\E M=\mu$ follows from $\E M_{Y_{n^{\prime}}}=\mu$
together with the uniform integrability of $\left\{ M_{Y_{n}}\left[\mathcal{T}\right]\right\} $.
\end{proof}
\begin{proof}
[Proof of Lemma \ref{lem:MeasurabilityImpliesConvergence}]
Consider the sequence of triples $\left(X,M_{Y_{n}},M\right)$. By
the same reasoning as in Lemma \ref{lem:Compactness}, it has a subsequential
distributional limit. Since $\left(X,M_{Y_{n}}\right)\overset{\Law}{\to}\left(X,M\right)$,
this distributional limit has the form $\left(X,M,M^{\prime}\right)$,
where $\left(X,M\right)$ and $\left(X,M^{\prime}\right)$ have the
same joint distribution. But since $M$ (and therefore also $M^{\prime}$)
is a function of $X$, we have $M=M^{\prime}$. This implies, in particular,
that $\left(M_{Y_{n}},M\right)\overset{\Law}{\to}\left(M,M\right)$,
so for every function $f\in L^{1}\left(\mu\right)$
\[
\intop f\left(t\right)M_{Y_{n}}\left(dt\right)-\intop f\left(t\right)M\left(dt\right)\overset{\Law}{\to}\intop f\left(t\right)M\left(dt\right)-\intop f\left(t\right)M\left(dt\right)=0.
\]
Convergence in law to a constant is equivalent to convergence in probability,
so $\intop f\left(t\right)M_{Y_{n}}\left(dt\right)\overset{L^{0}}{\to}\intop f\left(t\right)M\left(dt\right)$.
By uniform integrability, this upgrades to convergence in $L^{1}$.
\end{proof}
\subsection{A remark on the ``$\exp K$ lemma'' \label{sub:Remark}}
Lemma \ref{lem:ExpK}, which is of central importance in the proof
of Theorem \ref{thm:Approximation}, has other uses as well and may
be of independent interest. In particular, note that the property
\eqref{eq:SecondMomentAC} with $Z=W$ implies immediately that if
$\mu$ has no atoms then almost surely $M_{Z}$ has no atoms. Indeed:
\[
\E\sum_{t\in\left\{ \text{atoms of }M_{Z}\right\} }\left(M_{Z}\left\{ t\right\} \right)^{2}=\E\intop_{\op{diag}\mathcal{T}}M_{Z}\otimes M_{Z}=\intop_{\diag\mathcal{T}}\E\left[M_{Z}\otimes M_{Z}\right]=0,
\]
where $\diag\mathcal{T}:=\left\{ \left(t,t\right)\midmid t\in\mathcal{T}\right\} \subset\mathcal{T}\times\mathcal{T}$.
The measure $\E\left[M_{Z}\otimes M_{Z}\right]$ may ``explode along
the diagonal'', yet it assigns mass $0$ to it.
Recently nonatomicity was proven for some \emph{critical} GMCs ---
namely, those over critical ($\gamma^{2}=2d$) logarithmic fields
\cite{DSRVCriticalConvergence} and the related hierarchical fields,
also known as multiplicative cascades \cite{BKNSWCritical}. In the
general case (i.e. without the subcriticality assumption) nonatomicity
remains an open question.
\section{Acknowledgments}
The author wishes to thank Ofer Zeitouni for many valuable comments
that helped shape the manuscript.
The work is partially supported by Israel Science Foundation grants
111/11 and 147/15.
\appendix
\section{Appendix: the Maurey-Nikishin factorization theorem}
The following factorization theorem follows trivially from the $q:=2$
case of \cite[Théorème 3 b)]{Maurey2}:
\begin{thm}
[Nikishin, Maurey] Let $H$ be a Hilbert space, let $\left(\mathcal{T},\mu\right)$
be a standard probability space. Let $Y:H\to L^{0}\left(\mu\right)$
be a continuous linear operator. Then there exists a probability measure
$\mu^{\prime}$ equivalent to $\mu$, and a bounded operator $Y^{\prime}:H\to L^{2}\left(\mu^{\prime}\right)$,
such that $Y$ factors through the tautological embedding $\id:L^{2}\left(\mu^{\prime}\right)\to L^{0}\left(\mu\right)$
as follows:
\[
\xymatrix{H\ar[rr]^{Y}\ar@{-->}[dr]_{Y^{\prime}} & & L^{0}\left(\mu\right)\\
& L^{2}\left(\mu^{\prime}\right)\ar[ur]_{\id}
}
\]
\label{thm:Factorization}
\end{thm}
The factorization theorem is the ultimate reason behind all occurrences
of ``an equivalent measure $\mu^{\prime}\sim\mu$'' throughout the
text.
Using this theorem we show that the definitions of Gaussian fields
in terms of a couple $\left(X,Y\right)$ and a jointly Gaussian family
of integrals against $L^{2}\left(\mu^{\prime}\right)$ test functions
are equivalent.
Let $X$ and $Y$ be a standard Gaussian in $H$ and a generalized
$H$-valued function defined on $\left(\mathcal{T},\mu\right)$, respectively.
Then by Theorem \ref{thm:Factorization} there exists an equivalent
measure $\mu^{\prime}\sim\mu$, such that
\begin{equation}
\forall\xi\in H:\intop\left|\left\langle Y\left(t\right),\xi\right\rangle \right|^{2}\mu^{\prime}\left(dt\right)<\infty.\label{eq:FactorizationY}
\end{equation}
Now fix some function $f\in L^{2}\left(\mu^{\prime}\right)$. It follows
from \eqref{eq:FactorizationY} that for every $\xi\in H$ the function
$f\cdot\left\langle Y,\xi\right\rangle $ is in $L^{1}$. In other
words, $fY$ has a weak first moment $\E_{\mu^{\prime}}fY$. Thus
one can define for every test function $f\in L^{2}\left(\mu^{\prime}\right)$
a Gaussian random variable $\left\langle X,\E_{\mu^{\prime}}fY\right\rangle $,
which is to be interpreted as the integral of the ``Gaussian field
$\left(\left\langle X,Y\left(t\right)\right\rangle \right)$'' against
the test function $f$.
Conversely, suppose that we have a map that takes any test function
$f\in L^{2}\left(\mu^{\prime}\right)$ to a measurable linear functional
of some Gaussian vector $X$, i.e. a variable of the form $\left\langle X,Af\right\rangle $
for some $Af\in H$, such that the operator $A:L^{2}\left(\mu^{\prime}\right)\to H$
is bounded. Then one can define a generalized random vector $Y$ as
the composition $H\overset{A^{\ast}}{\to}L^{2}\left(\mu^{\prime}\right)\overset{\id}{\to}L^{0}\left(\mu\right)$.
Thus we have constructions that produce a couple $\left(X,Y\right)$
from a field defined by test functions and vice versa. The verification
that they are inverse to each other reduces to the routine unraveling
of the definitions which is left to the reader.
\bibliographystyle{abbrv}
\bibliography{MultiplicativeChaos}
\end{document}
| 68,680
|
Frequently asked questions…
Lithium battery technology is relatively new when compared to lead acid batteries which have been around for 100 years or so. Despite lithium batteries being used in phones and laptops for decades there is a lot of misinformation in the public domain around lithium batteries in particular surrounding their safety. So, we have compiled a comprehensive FAQ database which will hopefully answer your most common questions or concerns.
Battery Management Systems
Our Battery Management System (BMS) is made up of a printed circuit board which monitors and manages the 4x 3.2V lithium cells to ensure they are charged and discharged together and at the same rate.
It also provides protective features to prevent over charge and over discharge of the cells, keeping the battery within a safe operating envelope. Our innovative BMS is also designed to be able to draw up to 200A continuously and 500A for 30 seconds on the load side and charge at up to 150A.
A Battery Management System (BMS) is a key part of a lithium battery. Unlike lead-acid batteries which do not require a BMS, lithium batteries do.
BMS primarily is a circuit board which manages, monitors, controls and protects the lithium cells from being over charged or over discharged, which can lead to permanent damage or in the worst case lead to fire. It also needs to be able to disconnect your battery before the cells go outside of their safe operating limits.
Therefore it is vital that your investment in a lithium battery is not compromised by using an inferior BMS. That is why we have opted to design and engineer our very own BMS right here in Brisbane, Australia – so that we can control its functionality, quality and reliability.
Lithium Technology
Yes you can but it is less of a drop in voltage.
Lithium batteries when fully charged rest on 13.3V-13.4V for a nominal 12V system and they hold their voltage right until the end of cycle and only then drop by approximately 0.5V-1.0V when they reach close to 90% discharged.
Lead based chemistry in lead acid and AGM batteries have a more linear rate of voltage drop and by half way through the cycle they have dropped 1V and by end of cycle they are 2-3V lower. This is why inverters often go into error because of voltage drop at the end of cycle with lead acid batteries.
Lithium batteries hold their voltage better, sit at a higher resting voltage and are a better option to avoid inverter errors for heavier loads.
You can still infer the state of charge with lithium battery voltages but the voltage drop is subtle and less accurate than the greater drop in lead acid batteries.
A remote display is an option listed on the datasheet for most of our batteries batteries. The remote display is a LCD screen control panel (connected via cable to our battery) which gives you access to vital information like SOC (state of charge), hours remaining at current usage rate, voltage, temperature etc.
In most cases yes, but it is best to contact us first to confirm. We will let you know if your charger, whether it be solar, DC from the alternator or AC mains charger is compatible with our lithium batteries.
Our 100Ah lithium batteries can be charged in one hour with the correct chargers which is 2.5 times faster than lead acid batteries. You can charge our lithium batteries at 100A total, which could be a combination of solar, DC and AC chargers.
Our lithium batteries deliver at least 6 times more usable amp hours over their lifetime compared to lead acid batteries. This reduces the cost of replacement, installation and inconvenience every 3 years, saving you money in the long run. You have to look at the total life cycle of the battery and not just the upfront cost.
There are numerous advantages that lithium batteries have over lead acid batteries with respect to recreational vehicles such as caravans, motorhomes and camper trailers. These can be summarised as:
- Weight – lithium batteries are half the weight as lead acid equivalents which is important in an RV. A 100Ah lead acid battery weighs around 30kg, whereas our 100Ah lithium battery is 15kg.
- Depth of discharge (DOD) – lithium batteries can be safely discharged to 80% DOD compared to typically 50% DOD for lead acid batteries. This means for a 100Ah battery you have 80Ah available to use with a lithium battery and only 50Ah with a lead acid.
- Constant voltage – lithium batteries discharge at more or less a constant voltage until they reach around 80% DOD, after which the voltage begins to drop off. Compare this to a lead acid battery where voltage begins to drop as soon as power is drawn. This means that with a lithium battery, low voltage disconnects to protect appliances are not activated before the battery is empty like they are with a lead acid battery.
- Cycles – lithium batteries last a lot longer than lead acid batteries, in that you can cycle them more times before they need to be replaced. Our lithium batteries are rated at 2,000 cycles times at 80% DOD compared to a lead acid battery at 500 cycles.
- Power output – our 100Ah lithium batteries can discharge at up to 250A which is made possible by our BMS. Many other 100Ah lithium batteries used in RV’s limit you to 100A. 100Ah AGM batteries struggle to provide even 100A. Being able to discharge at 250A allows the RV owner to run high power appliances such as microwaves, kettles, coffee machines etc.
No not LBS lithium batteries. Some lithium batteries have been dangerous because they cut corners, use a cheap BMS and/or use a different lithium chemistry to the one that we use.
It’s similar to the fuel in the fuel tank of a car containing stored energy. Petrol sloshing around in a cheap bucket is dangerous but protected by a reputable fuel tank manufacturer, the fuel is perfectly safe to use and store. Well designed lithium battery systems are safer than Lead Acid chemistry which can vent flammable hydrogen gas and swell up when overcharged. Lithium batteries vent no gases.
At Lithium Battery Systems we take safety as the primary design criteria and have delivered a solution based on our “Triple Guard” methodology.
1) We have chosen the safest chemistry of lithium cells being Lithium Ferrous (iron) Phosphate (LFP) and we only use prsimatic cells that are encased in a safe Aluminium casing.
2) We electrically protect the system using our proprietary Battery Management System which disconnects, isolates and balances cells to keep them electrically safe.
3) We then place the whole battery assembly into another Aluminium outer enclosure. Lightweight yet robust to protect the entire package from knocks and accidental puncture.
Lithium Battery Systems lithium batteries are lightweight, safe and powerful.
| 44,565
|
-
-
-
: Bollywood
Namaste Ms. Zinta
When I first discovered Bollywood, Preity Zinta danced across the streets of New York in Kal Ho Naa Ho(2003) and Kabie Alvida Na Kehna(2006) as the quintessential NRI (Non Resident Indian) working girl. Even in her début in one of my all-time …
When Greed Is Good
While Anglo-Americans flock to theaters to watch Tom Cruise conduct impossible missions or Daniel Craig test drive M’s devices, the rest of the world seeks their thrills as Shah Rukh Khan completes perfect crimes as Don. These are not films …
Posted in Bollywood, Diversity, History, Travel Tagged Daniel Craig, Don, Don 2, Europe, Good, Greed, India, Islam, Priyanka Chopra, Shah Rukh Khan, Stereotypes, Tom Cruise Leave a comment
Hope Hota Hai
My blogging idol, John Fea, threw down the gauntlet and demanded a statement on hope. When I stood under a palm tree and watched two strangers exchange wedding vows, I knew what I needed to write. The Scandinavian-American groom arrived …
Posted in Academic Life, Biography, Bollywood, Diversity, History, Motherhood, Travel Tagged Family, Kajol, marriage, Shah Rukh Khan 1 Comment
Hot Off the Virtual Press: What if you could do anything?
Posted in Academic Life, Biography, Bollywood, Diversity, History, Motherhood, Travel Leave
Bertrand’s Bollywood
If Bollywood was good enough for renowned British philosopher Bertrand Russell, surely it’s good enough for you! With thanks to Northwestern University’s Medill School of Journalism alumna and India’s leading film critic, Anupama Chopra:
Taming a Tiger: How Khan & Chopra Chastise Chua
A …
Posted in Academic Life, Biography, Bollywood, Diversity, Motherhood, Travel Tagged 3 Idiots, Aamir Khan, Amy Chua, Engineering, Himalayas, Shah Rukh Khan, Swades, Taare Zameen Par, Vinod Chopra 2 Comments
| 320,292
|
TITLE: Basic integration as introduction to Fourier analysis: Showing orthonormality of the Fourier basis
QUESTION [1 upvotes]: I'm starting with Fourier analysis and in the first chapter the following exercise is stated:
Prove that if $n,m \geq 1$, we have
$\frac{1}{\pi} \int_{-\pi}^{\pi} \cos(nx) \cos(mx) dx$ equals $0$ if $n \neq m$ and equals $1$ if $n=m$.
The latter I've shown using the fact that $\cos^2(x)$=$\frac{1}{2}+\frac{1}{2}\cos(2x)$, that works just fine.
But I'm stuck at the first part, when $n \neq m$. I know it's pretty much basic calculus but I'm stuck all the same.
REPLY [2 votes]: Let's do it by integration by parts (IPP). Let $n,m\geq 1$ with $n\neq m$; further, since we want to show the integral is equal to $0$, I'll omit the factor $\frac{1}{\pi}$. Then (coloring in red the factor I integrate, and blue the one I differentiate),
$$\begin{align}
\int_{-\pi}^\pi \color{red}{\cos nx}\color{blue}{\cos mx} dx &= \color{red}{\frac{1}{n}}[\color{red}{\sin nx}\color{blue}{\cos mx}]_{-\pi}^\pi- \frac{\color{blue}{-m}}{\color{red}{n}}\int_{-\pi}^\pi \color{red}{\sin nx}\color{blue}{\sin mx} dx\\
&= \frac{m}{n}\int_{-\pi}^\pi \sin nx\sin mx dx \tag{1}
\end{align}$$
using the fact that $\sin k\pi = 0$ for every $k\in\mathbb{Z}$ to see that the first term cancels.
Further, again by IPP,
$$\begin{align}
\int_{-\pi}^\pi \color{red}{\sin nx}\color{blue}{\sin mx} dx
&= \frac{1}{n}[-\cos nx \sin mx] + \frac{m}{n}\int_{-\pi}^\pi \cos nx\cos mx dx \\
&= \frac{m}{n}\int_{-\pi}^\pi \cos nx\cos mx dx \tag{2}
\end{align}$$
Combining (1) and (2), we get
$$
\int_{-\pi}^\pi \cos nx\cos mx dx = \frac{m^2}{n^2}\int_{-\pi}^\pi \cos nx\cos mx dx \tag{3}
$$
and since $\frac{m^2}{n^2} \neq 1$ since implies
$$
\int_{-\pi}^\pi \cos nx\cos mx dx = 0\,.\tag{4}
$$
| 6,320
|
TITLE: Definition and Doubt: Universal $\mathrm{C}^*$-Algebras given by generators and relations
QUESTION [0 upvotes]: Let $A:=\mathrm{C}^*(G,R)$ be the (defined) universal $\mathrm{C}^*$-algebra generated by generators $G$ and relations $R$. As far as I can discern, there are two ways of defining such objects --- either using category theory or non-commutative polynomials and ideals.
The big ticket is that given any $\mathrm{C}^*$ algebra $A$ with $|G|$ generators that satisfy $R$ there is a unique surjective $*$-homomorphism $\mathrm{C}^*(G,R)\rightarrow A$. This is the universal property.
I have an informal way of thinking about universal $\mathrm{C}^*$-algebras (that will shortly come into conflict with my second question).
My informal intuition for universal $\mathrm{C}^*$-algebras is that we more or less form a large direct sum of all $\mathrm{C}^*$-algebras that have $|G|$ generators which satisfy $R$ and that is where the universal property comes from (or sums of) projections $\bigoplus_\alpha A_\alpha\rightarrow A_\beta$.
Question 1: How bad is this intuition? Is there something big that it misses (forget if it is formally incorrect)?
That is more or less the question on definition. Now for my doubt. We can consider a particular universal $\mathrm{C}^*$-algebra $B$ which is generated by a magic unitary $u=(u_{ij})_{i,j=1}^4$. That $u$ is a magic unitary is to say that each entry is a projection $u_{ij}=u_{ij}^2=u_{ij}^*$ and that each row and column is a partition of unity:
$$\sum_{k=1}^4u_{ik}=1=\sum_{k}u_{kj}.$$
In the compact quantum group community (say p.8) this universal $\mathrm{C}^*$-algebra is the setting for the algebra of continuous functions on the compact quantum group of permutations. The definition (of compact quantum group) requires a $*$-homomorphism $\Delta :B\rightarrow B\underset{\min}{\otimes} B$ and how this is constructed in this example is to define
$$\Delta(u_{ij})=\sum_{k=1}^4u_{ik}\otimes u_{kj},$$
and the way it is shown that this is a $*$-homomorphism is to show that the matrix:
$$\left[\Delta(u_{ij})\right]_{i,j=1}^4\in M_4(B\otimes B)$$
is a magic unitary, and so $\Delta:B\rightarrow B\otimes B$ is a *-homomorphism. Then the confusion comes in because in my informal intuition from a above $B$ is the largest object with magic unitaries... but here we have $B\otimes B$ which in my understanding is LARGER again than $B$ is also a sub-object of it...
Question 2: Is there a way to resolve this confusion with my informal intuition of what a universal $\mathrm{C}^*$-algebra (perhaps by arguing that the $\mathrm{C}^*$-algebra generated by $\left[\Delta(u_{ij})\right]_{i,j=1}^4$ while living in a large space $B\otimes B$, is in fact smaller than $B$... or is my informal intuition messing up my understanding here?
REPLY [1 votes]: Question 1: Your intuition is not only correct, but it also underlies the proof of the existence of the universal algebra.
Question 2: The universal property is only relevant in the domain. All that is asked about the co-domain is that the relations be satisfied.
| 112,106
|
TITLE: How often was the most frequent coupon chosen?
QUESTION [22 upvotes]: In the coupon collector's problem, let $T_n$ denote the time of completion for a collection of $n$ coupons. At time $T_n$, each coupon $k$ has been collected $C_k^{n}\geqslant 1$ times. Consider how often the most frequently chosen coupon, was chosen, that is, the random variable $$C^*_n=\max_{1\leqslant k\leqslant n}C_k^{n}.
$$
Can one compute $E(C^*_n)$? What is a simple asymptotics of $E(C^*_n)$ when $n$ grows large? Does $C_n^*/E(C^*_n)$ converge in distribution and, if it does, what is its limit?
Note that $C_1^n+C_2^n+\cdots+C_n^n=T_n$. Since $E(T_n)=nH_n$ where $H_n=\sum\limits_{k=1}^n\frac1k$ denotes the $n$th harmonic number, such that $H_n=\log n+O(1)$, one has $E(C_n^*)\geqslant\log n$ and $E(C_n^*)=O(n\log n)$.
In a somewhat more ambitious version of this question, consider the nondecreasing rearrangement $C_{(1)}^n\leqslant C_{(2)}^n\leqslant\cdots\leqslant C_{(n)}^n$ of $(C_1^n,C_2^n,\ldots,C_n^n)$. Thus, $C_{(1)}^n=1$ and $C_{(n)}^n=C_n^*$.
Can one compute (or, get some simple asymptotics of) each $E(C^n_{(k)})$? And what is the "profile" of the random vector $(C_{(1)}^n,C_{(2)}^n,\ldots,C_{(n)}^n)$ when $n$ grows large? To be specific:
Does the random vector $$\left(\frac{C_{(1)}^n}{C_{(n)}^n},\frac{C_{(2)}^n}{C_{(n)}^n},\ldots,\frac{C_{(n)}^n}{C_{(n)}^n}\right)$$ converge in distribution and, if it does, what is its limit?
Edit: Amy N. Myers and Herbert S. Wilf (Some New Aspects of the Coupon Collector's Problem, SIAM Review 48(3), 2006) provide explicit formulas for the distribution and the mean of the number $S_n$ of singletons. In the notations above, $S_n$ is the size of the set of $1\leqslant k\leqslant n$ such that $C^n_k=1$, and also the maximum of the set of $1\leqslant k\leqslant n$ such that $C^n_{(k)}=1$. Myers and Wilf show that, for every $i$, $$P(S_n=i)=i{n\choose i}\int_0^\infty x^{i-1}(e^x-1-x)^{n-i}e^{-nx}dx,$$ and they deduce from this the esthetically pleasing identity $$E(S_n)=H_n.$$
REPLY [1 votes]: For your first question Michael's second heuristic (that the $C_1^*$ is approximately $e \log n$) is correct. This follows from combining the observation that $T_n$ is with high probability very close to $n \ln n$ with a result of Raab and Steger (Case 2 of Theorem 1 here) that gives precise asymptotics for the behavior of the most collected coupon at time $c n \ln n$.
This question was also asked at Math Overflow, and my answer there gives more details (in particular, why you should expect enough concentration around the mean to turn "with high probability" into a statement about the expectation).
| 48,785
|
TITLE: Split numbers and linear independence of multiplicative inverses
QUESTION [1 upvotes]: My understanding is that the product of ijk may equal +1 when working with split-quaternions. What are good examples of systems defined such that the product of two (not three) linearly independent (i.e., not jj or kk) elements of unit magnitude may equal +1? I have been poking around http://math.chapman.edu/~jipsen/structures/doku.php/index.html but I have yet to find/recognize the answer to my question via that resource.
REPLY [1 votes]: You don't actually specify what type of algebraic structure you're considering, so let's figure it out from context:
You ask for linear independence of elements, which implies you have an underlying vector space structure. You then want to take products of those elements, so you have some sort of algebra over tha vector space (possibly non-associative). You reference the element "1" as a unit, so I'll assume your algebra is unital. You want those elements to be of "unit magnitude", so you also have some sort of norm on this algebra.
I suspect the general category you want to be asking about is Clifford algebras. In any case, my example is from the complex numbers, which should be an allowable space no matter what additional requirements come up.
If $z,w$ are two elements from your algebra, and you want $z w = 1$, that means that $z$ and $w$ are units and are inverse of each other. And for any reasonable definition of "magnitude", you should be able to scale your answer so that both of the elements are of unit magnitude.
So to turn your question around, you're asking for an element of your algebra that is linearly independent from its inverse. There are many examples, perhaps the simplest being the primitive cubic roots of unity $z = -1/2 + i\sqrt{3}/2$ and $w = z^2 = -1/2 - i\sqrt{3}/2$. They are linearly independent, of unit magnitude, and inverses.
Edit: Additional requirements have now been added: The system should be a finite dimensional unital normed algebra over the reals (the vector space has to be over a field, so it can't be over the non-negative reals), $z$ and $w$ must have no real part (in other words, ${1, z, w}$ must be linearly independent), and $z$ and $w$ must have $+1$ as a coefficient (with respect to what generating set it is unclear, since they will always have $+1$ as a coefficient in any basis of which they are members).
With these additional requirements, I believe no example exists. Let's take a slightly stronger version of your question and suppose that we're in a composition algebra, which only additionally assumes that the norm comes from a non-degenerate quadratic form (all of your named examples so far are composition algebras). From every such composition algebra, you can define a symmetric bilinear form $\langle v, w \rangle$ that gives rise to the norm, as well as a conjugate $\bar{v} = 2 \langle v, 1 \rangle - v$. In particular, when $v$ "has no real part", that means that $\langle v, 1 \rangle = 0$, and so $\bar{v} = - v$. But we also have from Prop 2.7 on my linked page that when $N(v) = 1$, $v^{-1} = \bar{v}$. Putting these two requirements together gives that $v^{-1} = -v$, and thus, $\{v, v^{-1}\}$ can never be linearly independent. Thus, no such $j$ and $k$ can exist in such an algebra.
So if there are any examples, you would need a finite dimensional unital normed algebra over the reals where the norm doesn't come from a quadratic form, which I think will not look sufficiently close to the "systems" you're considering. (Depending on how you define "norm", there might be none at all.)
| 188,240
|
Explaining Fast Programs In ultius reviews essaysrescue
As you’ll be able to see that essay creating might seem to be a smaller process, but it surely needs to be carried out wit sure consciousness or else you will have to encounter the audio of staying careless in your essay crafting. Assist corporations with a staff of vastly competent writers can give dependable composing options. If you order 10 content material, they may provide ten articles. If an individual author will not be in a position to fulfill the goal, a further writer could make up the shortfall.
The Latest On Convenient ultius review essaysrescue Secrets
Must you get customized writing help from ultius essay writing services evaluations examine them and choose which one you’ll be able to trust your essay paper rank. Ultius, ensure that to online dissertation on-line how one paper writing companies ultius usa does not start when jobs are roadmaps. Evaluation essay writing providers reviews for you. Outline for writing a paper. Why worry in regards to the assessment. Essay on affirmative motion – blog ultius. The revision is the problem.
You have to be misplaced. This sub shouldn’t be for endlessly debating ethical dilemmas with a closed thoughts. It’s for writers and writing opportunities. You possibly can have your opinion as as to whether tutorial writing is skilled writing or not, however it’s, without a doubt, a distinct segment of writing for freelancers, which is what this sub is all about. Please stop ultius reviews picking fights each time educational writing is talked about. Anytime you decide that fight, you’re arguing with a closed mind and you’re arguing with another person who’s likewise close-minded. You are not going to change anyone’s opinion here, so it seems like you are simply trolling.
Probably the most treasured aspects together with the standard is the fast supply of the orders. I’m a scholar in school and as many others students understand the importance of meeting the stated deadlines. Related Post: killer deal Nowadays ultius reviews, when everybody has many companies and is at all times in a rush, it’s laborious to do all the pieces in time. For that reason, online writing services are principally appreciated because of their on-time delivery.
The lowering authorized consuming age essays minimal authorized drinking age. We are going to write a custom essay sample on lower ingesting age to 18 particularly for. Decreasing the consuming age in america. No, you’ll have to wait three more years for that, till your twenty. Well i for one consider that the legal ingesting age should ultius be lowered from twenty. If we are considered an grownup and anticipated to behave like one at age 18 it isn. T right to restrict us to a ingesting age of 21. In the essay, engs believes that the drinking age must be lowered to 18 or 19. She makes the argument that the majority faculty college students devour alcohol while.
Elements Of ultius essaysrescue Across The Uk
For this you’ll be able to refer the testimonials, feedbacks, ultius companies opinions so forth evaluate of ultius a wide range of resources. Simply go surfing to our internet site and you’ll come across our freelance essay composing buyer care to listen ultius review at your instructions. Content material may be employed to develop one way hyperlinks to your web website, and this gets the targeted visitors you require. So do acceptable investigation and purchase an essay or low-cost essays from a reputed organization.
Comparing Convenient ultius essaysrescue Solutions
presents a live chat feature on their web site, however requires you to share private data, similar to your email handle and identify before you will get any answers to your questions. You need to also give a purpose for contacting the brokers before you’re related with one ultius reviews of them. Calling the site obtained us an agent fairly shortly, but they can not share progress studies in your paper. You are also unable to directly contact the writer working on your paper. If you happen to want a great amount of help, this is probably not the company to offer your order.
If we consider prices of the writing from scratch, the lowest worth shall be for a high school degree with the deadline of 20 days. It will value $17.50. The highest value is required for the master’s stage written just in three hours. Related Post: see this page It will value $seventy eight. If a consumer needs to decide on graduate writer it might value 20% more, requested writer would value 15% more.
Sadly, there are many web sites that supply customized writing companies however fail to deliver by way of high quality, service and buyer outcomes. Ultius proudly differentiates itself by offering a reputable and bonafide service that is trusted by clients all over the world. We’re accredited with the Better Business Bureau ( and have an A+ ranking ), have nearly 700 constructive evaluations on impartial evaluation platforms and are a genuine American firm that doesn’t outsource writing overseas. When prospects are dissatisfied and depart poor reviews, we reply to each one among them and provide a resolution. Given our robust track file of service since 2010, we hope you will take into account giving us a chance before giving up your search.
We’ll dive a bit more deeply into every of these qualities in subsequent sections. But for now, consider these the fundamental conceptual building blocks behind the expansion of the dishonest enterprise. There’s ultius review some prevailing query as as to if or not the explosion of on-line courses has led to an increase in tutorial dishonesty. It seems probable but there’s additionally no irrefutable evidence that this is the case.
Thank you for taking time to depart this evaluation. We appreciate the suggestions you’ve provided right here and want to handle a few important deserves. In your review, you famous the positive facets to writing with us including writing in your schedule and using this chance to gas creative power ultius reviews. We are pleased to hear that this has been your expertise up to now. You also famous that deadlines are agency and sometimes work should be completed in a restricted timeframe. We hope you keep in mind that you are not obligated to just accept work with urgent deadlines, however should you do, you’re asked to adhere to the set deadline.
| 90,925
|
There will be a record amount of highway and bridge construction in progress in Illinois this summer and next year, Gov. James Thompson announced Thursday.
The state`s construction plans for the next 14 months include work on most Chicago area expressways, which should discourage motorists who moaned last summer that the number of roads under construction couldn`t possibly increase.
| 206,859
|
If you're looking for a flexible way to fit education into your life, Northeast Online can help. Not only do we offer a variety of complete degree programs and individual classes, we're a great value.
We have reasonable tuition rates and don't charge more for online classes. Compare that to other colleges and you'll see why we're a great choice.
Thank you, Northeast, for giving me a jump start into what has developed...
Patsy B. -- Louisville, NE
A new Learning Management System (LMS) will be implemented for the Summer B session that begins on July 11. If you have taken an online class in the past from Northeast, you will now have a new user name and password in our new system.
In order to login in to our new LMS called "My Classes Online", you will use your new student ID as your user name. Your password will be emailed to you when you are enrolled in the course. Please be sure to change your password after you have logged into the course the first time by clicking "Account" and then "Modify Details" and keying in a new password.
If you need assistance with getting logged into your online course, contact Online Technology Support at helpdesk@northeast.edu or 402-844-7741.
Hours of operation:
M-F – 8am – 5pm
We also monitor email and voice mail until 10 pm during the week and from noon – 5 pm on weekends.
Northeast Community College, 801 East Benjamin Ave, Norfolk,
Nebraska 68701
phone: 402-371-2020 | tollfree: 1-800-348-9033 | email:
help@northeast.edu
Privacy
| 322,271
|
RISK
DROTT – EKTA – STAMPE
(Jamila Drott, Daniel "EKTA" Götesson, Joakim Stampe)
Galleri Thomassen, Gothenburg. August 2012
FIVE BOOKS PHOTOGRAPHED, PRINTED AND BOUND IN FIVE DAYS
On the opening day (August 14, 2012) the gallery space was completely empty. All the material and tools needed to create the show had been delivered and placed on the street outside. The invited visitors helped bringing everything inside the gallery, which became the three artist's studio during five days, open for anyone to come by and watch the progress.
As part of the show photographer Per Englund documented the working process and produced five 24 page booklets in five days inside the gallery.
He also documented the documentation by pointing a surveillance camera towards himself, which broadcasted live on the publisher's website (svenskbokproduktion.se).
On August 19 the public was invited back to take part of the the finished result of the five days work. The handmade booklets were displayed and sold in a slipcase in an edition of 25 numbered copies.
-------------------------------------------------------------------------------------
Photographer – Per Englund
Publisher – Svensk Bokproduktion
Illustrations – 83 color photographs
Pages – 120 (5 booklets, 24 pages each)
Dimensions – 105 x 148 mm
Printing – Samsung CLP-325, laser printer
Binding – Husqvarna sewing machine, black cotton thread
Slipcase – Cardboard with title in black foil. Premade by Norrmalms Kartong
Paper – Scandia smooth, 100 gsm
Book design – Logotype & headlines by Museum Studio
Idea & concept – Per Englund / Svensk Bokproduktion
ISBN – 978–91–980137–2–6
Edition – 25 numbered copies
Printed and bound by Per Englund / Svensk Bokproduktion
Galleri Thomassen, Gothenburg – 2012
the finished books on display
logotype designed by museum studio
per englund's working station
some of the tools being used
bookbinding with sewing machine
glueing of the slipcases
assembling the books
example spreads
artist biographies. book no. 5
photographs from the series
day 1
day 2
day 3
day 4
day 5
| 403,383
|
\begin{document}
\selectlanguage{english}
\title[Topological and frame properties of certain pathological $C^*$-algebras
]
{Topological and frame properties of certain pathological $C^*$-algebras
}
\author{D.V. Fufaev}
\thanks{The work was supported by the Foundation for the
Advancement of Theoretical Physics and Mathematics ``BASIS''}
\address{Moscow Center for Fundamental and Applied Mathematics,
\newline
Dept. of Mech. and Math., Lomonosov Moscow State University}
\email{denis.fufaev@math.msu.ru, fufaevdv@rambler.ru}
\begin{abstract}
We introduce a classification of locally compact Hausdorff topological spaces with respect to the behavior of $\s$-compact subsets, and relying on this classification
we study properties of corresponding $C^*$-algebras
in terms of frame theory and the theory of $\A$-compact operators in Hilbert $C^*$-modules, some pathological examples are constructed.
\end{abstract}
\maketitle
\section*{Introduction}
Originally the concept of frames was introduced for the Hilbert spaces theory, and then by Frank and Larson in
\cite{FrankLarson1999} and \cite{FrankLarson2002} was generalized for the case of Hilbert $C^*$-modules.
During the last decade it has been developing intensively
(see, e.g., \cite{ArBak2017}, \cite{Bak2019}).
It turns out that there is a connection between frame theory and the theory of $\A$-compact operators and uniform structures.
It is well-known that bounded linear operator in Hilbert space is compact (i.e. can be approximated in norm by operators of finite rank) if and only if it maps the unit ball to a totally bounded set with respect to the norm. For the case of Hilbert $C^*$-modules, i.e. if we consider some $C^*$-algebra instead of the scalar field $\mathbb C$ (in this case operators are called $\A$-compact) it is not true: indeed, even in the case of any infinite-dimensional unital $C^*$-alebra $\A$ the identity operator has finite rank (which is equal to one), but the unit ball is not totally bounded. So a question how to describe the $\A$-compactness in geometric terms arose.
First steps were done in papers \cite{KeckicLazovic2018}, \cite{lazovic2018}.
In \cite{Troitsky2020JMAA} by E.V. Troitsky a significant development was obtained, namely, a specific uniform structure was constructed and it was proved that if $\cN$ is a countably generated Hilbert $C^*$-module then an adjointable operator
$F:\M\to\cN$ is $\A$-compact if and only if the image of the unit ball of $\M$ is totally bounded in $\cN$ with respect to this uniform structure.
In \cite{TroitFuf2020} it was proved that $\A$-compactness of an operator implies totally boundedness of the image of the unit ball with respect to this uniform structure for any Hilbert $C^*$-module $\cN$. The inverse statemant was proved for modules $\cN$
which could be represented as an orthogonal direct summand in the standard module over the unitalization algebra $\dot{\A}$ (which is equal to $\A$ itself in unital case) for some cardinality --- that is, in the module of the form $\bigoplus_{\l\in\Lambda}\dot{\A}$.
In particular it holds for modules which could be represented as an orthogonal direct summand in $\bigoplus_{\a\in\Lambda}{\A}$ in case when $C^*$-algebra $\A$ is countably generated as a module over itself (this in fact is equivalent to the fact that $\A$ is $\s$-unital, i.e. it has a countable approximate unit, see. \cite[Proposition 2.3]{Asadi2016}).
But in the case when $\A$ is not $\s$-unital in \cite{Fuf2021faa} a counterexample was constructed, namely, an algebra $\A$, considered as a module over itself, such that the identity operator from $\A$ into itself is not $\A$ -compact, but the unit ball is totally bounded with respect to the introduced uniform structure (and even with respect to some stronger one).
The constructed algebra is commutative, so this result
makes the study of underlying topological space
interesting.
Also it turns out that the existence of a representation of a module $\cN$ as an orthogonal direct summand in $\bigoplus_{\l\in\Lambda}\dot{\A}$ is equivalent to the existence of a standard frame in $\cN$, so there is a connection between stated problem on $\A$-compact operators and the frame theory in Hilbert $C^*$-modules, and we will study constructed $C^*$-algebras from this point of view too.
More precisely, the existence of a standard frame is sufficient to satisfy the $\A$ -compactness criterion. Hence, by proving that the criterion is not satisfied for some module, we, as a consequence, show that there is no standard frame for this module
(which, by the way, does not mean that there is no outer standard frame in sense of \cite{ArBak2017}, see \cite[Remark 3.3]{Fuf2021faa}).
In \cite{HLi2010}, \cite{Asadi2016}, \cite{AmAs2016},\cite{FrAs2020} there is also a searching for modules without frames, more precisely, for algebras $\A$ such that it is possible to say surely is there an $\A$-module without frames or not.
We introduce a scale of classes of locally compact Hausdorff topological spaces that decreases with respect to some property.
$\KI$ --- $\s$-compact spaces (i.e. spaces which could be covered by countable family of compact subsets);
$\KII$ --- non-$\s$-compact spaces which have a dense $\s$-compact subset;
$\KIII$ --- spaces in which no $\s$-compact subset is dense, that is, the complement to any $\s$-compact subset has an inner point, but the point at infinity (in a one-point compactification) may not be inner for the complement; equivalently, there exists a $\s$-compact not precompact subset;
$\KIV$ --- spaces in which the complement to any $\s$-compact subset has an inner point, and (in a one-point compactification) the point at infinity is always inner for the complement; equivalently, every $\s$-compact subset is precompact.
In \S1 some preliminaries on Hilbert $C^*$-modules, frames and topological spaces are given and some properties of $\KI$ and $\s$-unital algebras are established.
In \S2 we obtain some properties of algebras $C_0(K)$ for $K\in\KII$: we prove that such algebras never have standard frames (theorem \ref{ii_nostfr}).
Nevertheless, this does not imply any conclusion about the existence of a non-standard frame (which we will define later): it is proved that in the case of separable space it really does not exist (theorem \ref{ii_nofr}), but also an example of a space when such a frame exists is given (example \ref{ii_exfr}).
Also an example when the $\A$-compactness criterion does not hold is given, since the absence of a standard frame is not enough for this; there is also no non-standard frames for this example (example \ref{ii_exnocr}, theorems \ref{ii_nocr} and \ref{ii_nocr_nofr}).
In \S3 we study the algebras $ C_0(K) $ for the class $ \KIII $.
This case seems to be worse (from the point of view of the behavior of $\s$-compact subsets), but the situation is better than in cases $ \KII $ and $ \KIV $: we can find an example of an algebra for which there exists a standard frame (example \ref{iii_good}), and hence the $ \A $-compactness criterion is satisfied (one can say that in this context algebraic properties ``are not monotonic with respect to topological properties''). On the other hand, there is an example for which the criterion does not hold (and so there is no standard frame), and there is also no non-standard frame (example \ref{iii_bad}, theorem \ref{iii_bad_th}). However, there is also an intermediate example, for which there is no standard frames, but a non-standard one exists (example \ref{iii_bg}).
In \S4 it is established that for the class $ \KIV $ there are no non-standard frames in the algebra $ C_0(K) $. Earlier (in \cite{Fuf2021faa}) it was proved that for this class the Troitsky's criterion never holds, which also means that there are no standard frames.
Author is grateful to E.V.Troitsky, V.M. Manuilov, K.L. Kozlov, A.Ya. Helemskii and A.I.Korchagin for helpful discussions.
\section{Preliminaries and properties of $\KI$}
Let us recall basic notions and facts about
Hilbert $C^*$-modules and operators in them, which one can find in
\cite{Lance},\cite{MTBook},\cite{ManuilovTroit2000JMS}.
\begin{dfn}
\rm
A (right) pre-Hilbert $C^*$-module over a $C^*$-algebra $\A$
is an $\A$-module equipped with an $\A$-\emph{inner product}
$\<.,.\>:\M\times\M\to \A$ being a sesquilinear form on the
underlying linear space such that, for any $x,y\in\M$, $a\in\A$ :
\begin{enumerate}
\item $\<x,x\> \ge 0$;
\item $\<x,x\> = 0$ if and only if $x=0$;
\item $\<y,x\>=\<x,y\>^*$;
\item $\<x,y\cdot a\>=\<x,y\>a$.
\end{enumerate}
A \emph{Hilbert $C^*$-module} is a pre-Hilbert $C^*$-module over $\A$ , which
is complete w.r.t. its norm $\|x\|=\|\<x,x\>\|^{1/2}$.
A pre-Hilbert $C^*$-module $\M$ is called \emph{countably generated}
if there is a countable collection of its elements such that
their $\A$-linear combinations are dense in $\M$.
The \emph{Hilbert sum} of Hilbert
$C^*$-modules in the evident sense will be denoted by $\oplus$ .
\end{dfn}
\begin{dfn}\label{dfn:operators}
\rm
An \emph{operator} is a bounded $\A$-homomorphism.
An operator having an adjoint (in an evident sense) is
\emph{adjointable} (see \cite[Section 2.1]{MTBook}).
We will denote the Banach space of all operators
$F: \M\to \cN$ by ${\mathbf{L}}(\M,\cN)$
and the Banach space of adjointable operators
by ${\mathbf{L}}^*(\M,\cN)$.
\end{dfn}
\begin{dfn}\label{dfn:Acompact}
\rm
Denote by $\theta_{x,y}:\M\to\cN$, where $x\in\cN$
and $y\in\M$, an \emph{elementary} $\A$-\emph{compact} operator,
which is defined by formula $\theta_{x,y}(z):=x\<y,z\>$.
Then the Banach space ${\mathbf{K}}(\M,\cN)$
of $\A$-\emph{compact} operators
is the closure of the subspace generated by all
elementary $\A$-compact operators in ${\mathbf{L}}(\M,\cN)$.
\end{dfn}
To define a uniform structure, that is, the system of pseudometric, we need to remind the notions of the multiplier algebra and multiplier module (see \cite{BakGul2002}, \cite{BakGul2004} for more details).
Recall that $M(\A)$ is a $C^*$-algebra of multipliers of $C^*$-algebra $\A$ (see \cite[Chapter 2]{Lance}, for example), $M(\A)=\A$ if $\A$ is unital.
Also, if $K$ is locally compact Hausdorff topological space and $\A=C_0(K)$
then $M(\A)=C_b(K)$ --- the algebra of all continuous bounded functions on $K$.
For every Hilbert $\A$-module $\cN$ there exists a Hilbert $M(\A)$-module $M(\cN)$ (which is called the multiplier module of the module $\cN$) containing $\cN$ as an ideal submodule associated with $\A$, i.e. $\cN=M(\cN)\A$. Moreover, $\<x,y\>\in\A$ holds for any $x\in\cN$, $y\in M(\cN)$. $M(\cN)=\cN$ if the algebra $\A$ is unital.
Also, since each element of $x\in\cN$ can be represented as $y\cdot a$ for some $y\in\cN$, $a\in\A$ (see \cite[ 1.3.10]{MTBook}), the module $\cN$ and any its submodule can be considered as $M(\A)$-modules.
If we consider $\cN=C_0(K)$ as a module over itself, then, as in the case of the multiplier algebra, $M(C_0(K))=C_b(K)$.
The uniform structures on submodules of $\cN$ are defined as follows (see \cite{Fuf2021faa} and \cite{Troitsky2020JMAA} for details).
\begin{dfn}\label{dfn:admissyst}
\rm
Consider a Hilbert $C^*$-module $\cN$ over $\A$. A countable system $X=\{x_{i}\}$ of elements of the multiplier module $M(\cN)$ is called (outer) \emph{admissible} for a (possibly not closed) submodule $\cN^0\subseteq \cN$
(or outer $\cN^0$-\emph{admissible}), if
\begin{enumerate}
\item[1)]
for each $x\in\cN^0$
the series $\sum_i \<x,x_i\>\<x_i,x\>$ is convergent in $\A$;
\item[2)]
its sum is bounded by $\<x,x\>$, that is, $\sum_i \<x,x_i\>\<x_i,x\> \le \<x,x\>$;
\item[3)]$\|x_i\|\le 1$ for any $i$.
\end{enumerate}
\end{dfn}
\begin{dfn}\rm
Denote by $\F$ a countable collection $\{\f_1,\f_2,\dots\}$
of states on $\A$ (i.e. positive linear functionals of norm 1). For each pair $(X,\F)$
with an outer $\cN^0$-admissible $X$,
consider a non-negative function defined by
the equality
$$
\nu_{X,\F}(x)^2:=\sup_k
\sum_{i=k}^\infty |\f_k\left(\<x,x_i\>\right)|^2,\quad x\in \cN^0.
$$
It can be checked that this is a seminorm on the module $\cN^0$.
Denote the system of all these functions by
$\mathbb{OSN}(\cN,\cN^0)$. Also we will write $(X,\F)\in
\mathbb{OA}(\cN,\cN^0)$
for pairs with outer admissible $X$.
\end{dfn}
\begin{dfn}\label{dfn:pseme}
Consider for $(X,\F)\in \mathbb{OA}(\cN,\cN^0)$
the following function
$d_{X,\F}:\cN^0\times \cN^0\to [0,+\infty)$
$$
d_{X,\F}(x,y)^2:=\nu_{X,\F}(x-y)^2=
\sup_k
\sum_{i=k}^\infty |\f_k\left(\<x-y,x_i\>\right)|^2,\quad x,y\in \cN^0.
$$
We will write $d_{X,\F}\in \mathbb{OPM}(\cN,\cN^0)$.
\end{dfn}
Evidently,
$d_{X,F}$ are \emph{pseudometrics} in sense of \cite[Definition 2.10]{Troitsky2020JMAA} (and \cite[Chapter IX, \S 1]{BourbakiTop2}), so they form a uniform structure.
If $X$ contains only elements of the module $\cN$, the word ``outer'' is not used, and in this case one may write $\mathbb{SN}$, $\mathbb{A}$ and $\mathbb{PM}$ instead of $\mathbb{OSN}$, $\mathbb{OA}$ and $\mathbb{OPM}$.
The definition of \emph{totally bounded} sets
for the uniform structure under consideration
(or for the system $\mathbb{OPM}(\cN,\cN^0)$) takes the following form.
\begin{dfn}\label{dfn:totbaundset}
\rm
A set $Y\subseteq \cN^0 \subseteq \cN(\subseteq M(\cN))$ is \emph{totally bounded}
with respect to this
uniform structure, if for any $(X,\F)$,
where $X \subseteq M(\cN)$ is outer $\cN^0$-admissible,
and any
$\e>0$ there exists a finite collection $y_1,\dots,y_n$
of elements of $Y$ such that the sets
$$
\left\{ y\in Y\,|\, d_{X,\F}(y_i,y)<\e\right\}
$$
form a cover of $Y$. This finite collection is called an
$\e$\emph{-net in $Y$ for} $d_{X,\F}$.
If so, we will say that $Y$ is \emph{externally} $(\cN,\cN^0)$-\emph{totally bounded} (or $(M(\cN),\cN^0)$-totally bounded).
\end{dfn}
In these terms the $\A$-compactness of operators for some class of modules can be describe by following:
\begin{teo}(\cite[Theorem 3.5]{TroitFuf2020})
Suppose, $\M$, $\cN$ and ${\mathcal K}$ are Hilbert $\A$-modules, $\cN\oplus{\mathcal K}\cong\bigoplus\limits_{\l\in\Lambda}\dot{\A}$ for some $\L$,
$F:\M\to\cN$ is an adjointable operator
and $F(B)$ is $(\cN,\cN)$-totally bounded, where $B$ is the unit ball of $\M$.
Then $F$
is $\A$-compact as an operator from $\M$ to $\cN$.
\end{teo}
The inverse statement holds for arbitrary modules (\cite[Theorem 2.4]{TroitFuf2020}). For the case when $\cN$ is a countably generated module a similar result was stated and proved as criterion of $\A$-compactness by E.V. Troitsky (\cite[Theorem 2.13]{Troitsky2020JMAA}).
Evidently $(M(\cN),\cN^0)$-totally bounded set is $(\cN,\cN^0)$-totally bounded, so all results which state that $(\cN,\cN)$-totally boundedness of the image of the unit ball implies $\A$-compactness of corresponding adjointable operator are still valid if we consider $(M(\cN),\cN)$-totally boundedness.
Let us recall a notion of a frame in Hilbert $C^*$-module (see, e.g., \cite{FrankLarson1999}, \cite{FrankLarson2002}).
Among all frames, standard ones are also considered.
\begin{dfn}\label{dfn:fr}
Let $\cN$ be a Hilbert $C^*$-module over an unital $C^*$-algebra $\A$ and $J$ be some set. A family $\{x_j\}_{j\in J}$ of elements of $\cN$ is said to be a standard frame in $\cN$ if there exist positive constants $c_1, c_2$ such that
for any $x\in\cN$
the series
$\sum\limits_j \<x,x_j\>\<x_j,x\>$
converges in norm in $\A$
and the following inequalities hold:
$$c_1\<x,x\>\le \sum\limits_j \<x,x_j\>\<x_j,x\>\le c_2\<x,x\>.$$
A frame is called tight if $c_1=c_2$ and normalized if $c_1=c_2=1$.
If the series converges only in the ultraweak topology (also known as $\s$-weak) to some element of the universal enveloping von Neumann algebra $\A''$, then the frame is said to be non-standard.
Unlike the case of a standard frame, in this case
the number of nonzero elements of the series can be uncountable, the convergence in this case is considered as the convergence of a net consisting of all finite partial sums (see remarks before \cite[1.2.19]{KadRin1} and \cite[5.1.5]{KadRin1}).
We will write just ``frame'' if it is at least non-standard.
\end{dfn}
If the algebra $\A$ is not unital, then $\cN$ can be considered as a module over its unitalization $\dot{\A}$ and frame can be defined in $\cN$ as in an $\dot{\A}$-module.
Further we will assume that frames are defined in this way.
\begin{rk}
For a system $\{x_j\}_{j\in J}$ there is a connection between the so-called reconstruction formula and the property of being a frame: if for any $x\in\cN$ it holds that
$x=\sum\limits_{j\in J}x_j\<x_j,x\>$,
then $\{x_j\}_{j\in J}$ is a normalized frame, and the convergence takes place in the same sense (see \cite[Example 3.1]{FrankLarson2002})
\end{rk}
\begin{rk}\label{rk_bes}
If there exists a positive constant $c$ such that for any $x\in\cN$ and
for any partial sum of the considered series the inequality
$\sum\limits_j \<x,x_j\>\<x_j,x\>\le c\<x,x\>$ holds (that is, the right side of the inequality for the frame), then the system $\{x_j\} _{j\in J}$ is called the Bessel system. Due to \cite[2.4.19]{BratRob} the series with respect to the Bessel system always converges in the ultrastrong topology and, as a consequence, also in the ultraweak one.
\end{rk}
Recall the following characterization of non-standard frames:
\begin{lem}\label{fr_cr}
(\cite[Proposition 3.1]{HLi2010})
A system $\{x_j\}_J$ is a frame in $\cN$ if and only if there exist positive constants $c_1, c_2$ such that for any $x\in\cN$ and any state $\f$ on $\A$
the following inequalities hold:
$$
c_1\f(\<x,x\>)\le
\sum\limits_j \f(\<x,x_j\>\<x_j,x\>)\le
c_2\f(\<x,x\>)
$$
\end{lem}
From this property and the definition of a standard frame it is obvious that any frame in the algebra $C_0(K)$ as a module over itself must separate points of the space $K$.
In our context the main structural result of the frame theory is the following:
from the results of Frank and Larson (\cite[3.5, 4.1 and 5.3]{FrankLarson2002}, see also \cite[Theorem 1.1]{HLi2010}) it follows that the Hilbert $C^*$-module
$\cN$ over the $C^*$-algebra $\A$ can be represented as an orthogonal direct summand in the standard module of some cardinality over the unitalization algebra $\bigoplus\limits_{\l\in\Lambda}\dot{\A}$ if and only if there exists a standard frame in $\cN$.
Kasparov's stabilization theorem (\cite{Kasp}, or \cite[Theorem 1.4.2]{MTBook}) implies that every countably generated module has a standard frame. The construction of a module that does not have a standard frame will show that the stabilization property of the Kasparov type is not satisfied for this module.
Note that frames are well-defined even for the case when the algebra is not unital,
but to use this stabilization property we need to take its unitalization.
\begin{rk}
If an algebra $\A$ has a standard frame as a module over itself, then any $\A$-module $\cN$ which can be represented as an orthogonal direct summand in the standard module $\bigoplus\limits_{\l\in\Lambda}{\A}$ also has a standard frame (and hence can be represented as an orthogonal direct summand in the standard module $\bigoplus\limits_{\l\in\Lambda'}\dot{\A}$). For some special case it was noted in \cite[remark 3.6]{TroitFuf2020}.
\end{rk}
Let us recall now some preliminaries from topology.
For a locally compact Hausdorff topological space $K$ we denote by $\a{K}=K\cup\{t_\infty\}$ its one-point compactification (see \cite[29.1]{Munkr}). We will also often use the fact that the $C^*$-algebra $C_0(K)$ of continuous functions vanishing at infinity (see \cite[436I]{Frem4}) is isomorphic to an ideal in $C^*$ -algebra $C(\a{K})$ consisting of functions vanishing at the point $t_\infty$ (\cite[Lemma 3.44]{Weaver}).
We denote by $\be K$ the Stone-\v{C}ech compactification of $K$; it is known that $C_b(K)\cong C(\be K)\cong M(C_0(K))$ (see \cite[Chapter 1]{Walker}).
\begin{lem}\label{extrest} (\cite[Corollary 1.3]{Fuf2021faa})
Let $K$ be a locally compact Hausdorff space, $A$ be a closed subset of $K$ and
$f\in C_0(A)$,
then $f$ can be extended to a function from $C_0(K)$. Moreover, $f$ is bounded and the extended function can be chosen to be bounded by the same constant.
In particular, for every compact set $K'\subset K$ there is a function $g\in C_0(K)$ such that $g=1$ on $K'$ and $|g|\le1$ on $K$ .
\end{lem}
We need the following useful examples of topological spaces:
$[0,\om_1]$ --- the space of all ordinals $\a$ such that $\a\le\om_1$ with order topology, where $\om_1$ is the first uncountable ordial; this space is uncountable. Also, $[0,\om_1)\in\KIV$.
$[0,\om_0]$ is the space of all ordinals $\a$ such that $\a\le\om_0$ with order topology, where $\om_0$ is the first infinite ordial (this space is homeomorphic to $\{\frac{1}{n}\}_{n\in\N}\cup \{0\}$, or $\N\cup\{\infty\}$). See \cite{Fuf2021faa} or \cite[chapters VI-VII]{KurMos} and \cite[3.1.27]{Engel} for details.
The following well-known for specialists statement is useful for our goal.
\begin{lem}
The $C^*$-algebra $C_0(K)$ is $\s$-unital if and only if $K$ is $\s$-compact.
\end{lem}
\begin{proof}
Recall that $\s$-unitality is equivalent to the existence of a strictly positive element, in our case --- an everywhere positive function.
If the algebra $C_0(K)$ is $\s$-unital, then it contains an everywhere positive function $f$. The sets $\{t\in K:|f(t)|\ge\frac{1}{n}\}$ are compact (as closed subsets of such corresponding compact sets outside of which $|f(t)|<\frac {1}{n}$), and therefore form a countable cover of $K$ by compacts.
Conversely, if there are compact sets $K_n$ such that $K=\bigcup\limits_{n\in\N}K_n$, then there exist functions $f_n\in C_0(K)$ such that $f_n(t)=1 $ on $K_n$ and $|f_n(t)|\le1$ everywhere on $K$. The function $\sum\limits_{n\in\N}\frac{1}{2^n}|f_n(t)|$ is everywhere positive on $K$, since for any point there is a compact set $K_n$ in which it is contained.
\end{proof}
\begin{lem}\label{frp1}
Let $K$ be a locally compact, not $\s$-compact space. Then for any countable family of functions $\{f_n\}\subset C_0(K)$ the set $F=\{t\in K: f_n(t)=0\,\, \forall n\in\N\}$ is non-empty, and even uncountable.
In particular, a countable family of functions cannot separate points of the space $K$.
\end{lem}
\begin{proof}
Indeed, if the set $F$ is countable and equals to $\{z_n\}_{n\in \N}$, then for every $n\in\N$ there exists a function $g_n\in C_0(K)$ such that $ g_n(z_n)=1$, $|g_n(t)|\le1$,
and then the function $g(t)=\sum\limits_{n\in\N}\frac{1}{2^n}\frac{|f_n(t)|}{1+|f_n(t)|}+\sum\limits_{n\in\N}\frac{1}{ 2^n}|g_n(t)|$ is everywhere positive on $K$, which contradicts the fact that $K$ is not $\s$-compact.
The proof is also valid for the cases when $F$ is finite or empty.
\end{proof}
\begin{lem}\label{frp}
Let $K$ be a locally compact, not $\s$-compact space, $\{K_n\}_{n\in\N}$ be a countable collection of compact sets from $K$. Then for any countable family of functions $\{f_n\}\subset C_0(K)$ the set $F=\{t\in K\setminus \bigcup\limits_{n\in\N}K_n: f_n(t)=0 \,\, \forall n\in\N\}$ is non-empty, and even uncountable.
\end{lem}
\begin{proof}
Similarly to the previous lemma, consider the function $g(t)=\sum\limits_{n\in\N}\frac{1}{2^n}\frac{|f_n(t)|}{1+|f_n(t) |}+\sum\limits_{n\in\N}\frac{1}{2^n}|g_n(t)|+\sum\limits_{n\in\N}\frac{1}{2^ n}h_n(t)$, where $h_n\in C_0(K)$ is a non-negative function which equals to 1 on $K_n$ and bounded by 1.
\end{proof}
It is known (see \cite[Example 3.5]{FrankLarson2002}) that every $\s$-unital algebra as a module over itself has a normalized standard frame.
Let us construct an example of a non-$\s$-unital algebra that has a frame with the same properties.
Recall that $\A=c_0-\sum\limits_{\l\in\L}\A_\l$ is a $c_0$-direct sum of the algebras $\A_\l$ (see \cite[\S 1.4]{Averson} or \cite[\S 1]{FrM2020}), the elements $x=(x_\l)_{\l\in\L}$ of this sum are enumerated by indices from $\L$ sets such that $x_\l\in\A_\l$,
there are at most countable set $\{x_{\l_m}\}_{m\in\N}$ of non-zero elements and $\lim\limits_{m\to\infty}||x_{\l_m}||_{\A_{\l_m}}
=0$.
The norm in this algebra is given by the formula $||x||_\A=\sup\limits_{\l\in\L}||x_\l||_{\A_\l}$.
In fact, this algebra is obtained by completing an algebra whose elements are non-zero only for a finite number of indices.
\begin{teo}\label{iii_stfr}
Let $\A=c_0-\sum\limits_{\l\in\L}\A_\l$,
where each $\A_\l$ has a normalized standard frame for which the reconstruction formula is valid (for example, due to \cite[Proposition 2.3]{Asadi2016}, when each $\A_\l$ is $\s$-unital).
Then $\A$ as a $\dot{\A}$-module has a normalized standard frame.
\end{teo}
\begin{proof}
Denote by $\{x_j\}_{j\in J_\l}$ the frame in $\A_\l$.
For each $\l\in\L$, the elements of the frame can be considered as elements of the entire algebra $\A$ by extending them by zero
outside the corresponding index. Let us show that $\{x_j\}_{j\in \bigcup\limits_{\l\in\L}J_\l}$ is a frame in $\A$.
Let $x\in\A$. Fix arbitrary $\e>0$. There exists an element $x^m\in\A$ which is non-zero only in a finite number of indices $\l_1,\dots,\l_m$ such that $||x-x^m||<\e$. For each $\l_k$, $k=1,\dots,m$, there exists a finite set $\{x_j\}_{j\in J_{\l_k}'}$ of elements of the frame such that
$$||x^m_{\l_k}-\sum\limits_{j\in J_{\l_k}'}x_j\<x_j,x^m_{\l_k}\>||<\e,$$ hence
$$||x^m-\sum\limits_{j\in \bigcup\limits_{k=1}^mJ_{\l_k}'}x_j\<x_j,x^m\>||<\e$$ and
$$||x-\sum\limits_{j\in \bigcup\limits_{k=1}^mJ_{\l_k}'}x_j\<x_j,x^m\>||<2\e.$$
So we have
$x=\sum\limits_{j\in J}x_j\<x_j,x\>$,
and the series converges in norm in $\A$. Hence
$\<x,x\>=\sum\limits_{j\in J}\<x,x_j\>\<x_j,x\>$.
\end{proof}
\begin{cor}
Since $\A$ has a frame, it can be represented as an orthogonal direct summmand in the standard module
$\bigoplus\limits_{\l\in\Lambda}\dot{\A}$
of some cardinality, and so, by \cite[Theorem 3.5]{TroitFuf2020}, it satisfies the $\A$-compactness criterion.
Moreover, every $\A$-module which can be represented as an orthogonal direct summmand in the standard module $\bigoplus\limits_{\l\in\Lambda'}{\A}$ also can be represented
in the standard module
$\bigoplus\limits_{\l\in\Lambda}\dot{\A}$
of some cardinality, and hence it satisfies the $\A$-compactness criterion too.
\end{cor}
\begin{rk}\label{iii_dop}
If $\A$ also is a commutative algebra, then $\A\cong c_0-\sum\limits_{\l\in\L}C_0(K_\l)\cong C_0(\bigsqcup\limits_{\l\in \L}K_\l)$, and $C_0(K_\l)$ is $\s$-unital if and only if $K_\l$ is $\s$-compact; if $\L$ is uncountable, then $K=\bigsqcup\limits_{\l\in\L}K_\l$ is not $\s$-compact, because a compact set in $K$ intersects only a finite number of $K_\l$. If $\A_\l$ is a unital algebra, then as a frame in $\A_\l$ we can take just one element, the identity of the algebra (it corresponds to a function which identically equals to one in the commutative case).
\end{rk}
\section{The properties of $\KII$}
Let us introduce several examples of spaces from the class $\KII$.
\begin{ex}\label{rat_seq}
$K$ --- the set of real numbers with rational sequence topology (\cite[\S 65]{Exampl}). Moreover, it is separable.
\end{ex}
\begin{ex}\label{beN}
$K=\be\N\setminus\{t'\}$, where $t'\in\be\N\setminus\N$ --- the Stone-\v{C}ech compactification of natural numbers without an arbitrary
point from the
growth.
It is not $\s$-compact since \cite[9.6]{GilJer}, and obviously it is separable.
More generally, we can take instead of $\N$ any separable non-compact space (or just $\s$-compact, but we can lost separability).
\end{ex}
\begin{ex}
$K=\a{P}\times [0,\om_0]\setminus\{(p_\infty,\om_0)\}$, where $P$ is a locally compact, non-$\s$-compact Hausdorff space, $\a{P}=P\cup\{p_\infty\}$ --- its one-point compactification.
\end{ex}
\begin{teo}\label{ii_nostfr}
Let $K\in\KII$. Then there is no standard frame in the $\dot{C_0}(K)$-module $C_0(K)$.
\end{teo}
\begin{proof}
Assume that there exists a frame $\{x_j\}_{j\in J}$ of elements from $C_0(K)$. It contains an uncountable number of nonzero elements, since countable cannot separate the points of $K$.
Let $\{K_n\}_{n\in\N}$ be a sequence of compact sets in $K$ such that $\overline{\bigcup\limits_{n\in\N}K_n}=K$.
For every $n\in\N$ there exists a function $g_n\in C_0(K)$ such that $g_n(t)=1$ on $K_n$, $|g_n(t)|\le1$ on $K$.
There is a non-empty at most countable set $\{x_j\}_{j\in J_n}$ of elements of the frame such that $x_j(t)\ne0$ identically on $K_n$ because the series $\sum\limits_j \<g_n,x_j\>\<x_j,g_n\>(t)=\sum\limits_j |x_j(t)|^2$ converges uniformly on $K_n$. That is, if $j\in J\setminus J_n$, then $x_j(t)=0$ on $K_n$.
Hence, there is at most countable set $\{x_j\}_{j\in \bigcup\limits_{n\in\N}J_n}$ such that $x_j(t)\ne0$ identically on $\bigcup\limits_{n\in\N}K_n$. Hence, for every $j\in J\setminus \bigcup\limits_{n\in\N}J_n$ we have $x_j(t)=0$ for $t\in\bigcup\limits_{n\in\N} K_n$, but due to the fact that $\overline{\bigcup\limits_{n\in\N}K_n}=K$ it also holds for $t\in K$. Hence, only a countable set of frame elements is not identically zero.
A contradiction.
\end{proof}
\begin{cor}
From this result it follows that $C_0(K)$ cannot be represented as an orthogonal direct summand of a standard module $\bigoplus\limits_{\l\in\Lambda}\dot{C_0}(K)$.
\end{cor}
Let now $K$ be
moreover separable, that is, finite sets can be taken as compacts in the definition of $\KII$.
Spaces from examples \ref{rat_seq} or \ref{beN} can be taken as such spaces.
Let us show that in this case
non-standard frames also don't exist.
\begin{teo}\label{ii_nofr}
Let $K\in\KII$ and $K$ is separable.
Then there is no frames in the $\dot{C_0}(K)$-module $C_0(K)$.
\end{teo}
\begin{proof}
Assume that there exists a frame $\{x_j\}_{j\in J}$ of elements from $C_0(K)$. It must contain an uncountable number of nonzero elements, since countable cannot separate the points of $K$.
Let $\{t_n\}_{n\in\N}$ be a countable dense subset of $K$.
For each $n\in\N$ there exists a function $g_n\in C_0(K)$ such that $g_n(t_n)=1$, $|g_n(t)|\le1$ on $K$.
There is a non-empty at most countable set $\{x_j\}_{j\in J_n}$ of elements of the frame such that $x_j(t_n)\ne0$,
because
taking $x=g_n$ and $\f$ --- evaluation at the point $t_n$, due to lemma \ref{fr_cr} we have that
the series $\sum\limits_j \<g_n,x_j\>\<x_j,g_n\>(t)=\sum\limits_j |x_j(t)|^2$ converges at the point $t_n$. That is, if $j\in J\setminus J_n$, then $x_j(t_n)=0$.
Hence, there is at most countable set $\{x_j\}_{j\in \bigcup\limits_{n\in\N}J_n}$ such that $x_j(t)\ne0$ identically on $\bigcup\limits_{n\in\N}\{t_n\}$. Hence, for every $j\in J\setminus \bigcup\limits_{n\in\N}J_n$ we have that $x_j(t)=0$ for $t\in\bigcup\limits_{n\in\N} \{t_n\}$, but due to the fact that $\overline{\bigcup\limits_{n\in\N}\{t_n\}}=K$ it also holds for $t\in K$. Hence, only a countable set of frame elements is identically zero.
A contradiction.
\end{proof}
Despite the previous two theorems, it is possible to construct an example of a space $K\in\KII$ such that a non-standard frame exists in $C_0(K)$.
\begin{ex}\label{ii_exfr}
Let $P$ be a non-$\s$-compact space such that $C_0(P)$ has a normalized frame $\{u_\be\}_{\be\in B}$.
We know that such a space exists (see remark \ref{iii_dop}).
Take $K=\a{P}\times [0,\om_0]\setminus\{(p_\infty,\om_0)\}$ and define $y_\be\in C_0(K)$, $\be\in B$, by the formula $y_\be(p,n)=u_\be(p)$, where $(p,n)=t\in K$, i.e. $\{y_\be\}_{\be\in B}$ is a ``copying'' of functions $\{u_\be\}_{\be\in B}$ on each ``row'' ${ P}\times\{n\}$, $n\in[0,\om_0]$. Also consider $\{w_n\}_{n\in[0,\om_0)}$ such that $w_n=1$ on $\a{P}\times\{n\}$ and $w_n$ vanishes outside $\a{P}\times\{n\}$
(obviously, $w_n\in C_0(K)$ for any $n\in[0,\om_0)$, since every $\a{P}\times\{n\}$ is a clopen set).
Define $\{x_j\}_{j\in J}=\{y_\be\}_{\be\in B}\cup\{w_n\}_{n\in[0,\om_0)}$ .
\end{ex}
\begin{teo}\label{ii_fr}
In example \ref{ii_exfr} the system $\{x_j\}_{j\in J}$ is a (non-standard) frame in $C_0(K)$.
\end{teo}
\begin{proof}
Take an arbitrary $x\in C_0(K)$. For any partial sum of the series $\sum\limits_j \<x,x_j\>\<x_j,x\>(t)=\sum\limits_j|x(t)|^2|x_j(t)|^2$ we have $\sum\limits_j|x(t)|^2|x_j(t)|^2\le2|x(t)|^2=2\<x,x\>(t)$ (actually, if $p=p_\infty$ or $n=\om_0$, where $t=(p,n)$, then two can be replaced by one), which means that the series converges in the ultrastrong topology.
Hence we obtain the right side of the inequality
of lemma \ref{fr_cr} with $c_2=2$ for any state $\f$ (in particular, the corresponding series converges).
Indeed, if for some state $\f$ we have $\sum\limits_j \f(\<x,x_j\>\<x_j,x\>)>
2\f(\<x,x\>)$,
then there exists some partial sum of the series for which it also holds that
$\sum\limits_{j\in J'} \f(\<x,x_j\>\<x_j,x\>)>
2\f(\<x,x\>)$. And then we have $\f(2\<x,x\>-\sum\limits_{j\in J'} \<x,x_j\>\<x_j,x\>)<0$, which contradicts the inequality
$2\<x,x\>\ge\sum\limits_{j\in J'} \<x,x_j\>\<x_j,x\>$ and positivity of $\f$.
Let us show that the left side holds too.
Let $\f$ be a state on $C_0(K)$, that is, there is a Radon measure $\mu$ on $K$ such that $\f(x)=\int\limits_{K}x( t)d\mu(t)$ (see \cite[436K]{Frem4}).
Then $\mu$ can be represented as a sum of the Radon measures $\mu_1$ which support is $P\times \{\om_0\}$ and $\mu_2$ which support is $\a{P}\times [0,\om_0) $.
Then for any $x\in C_0(K)$ we have
$\f(\<x,x\>)=\int\limits_{P\times \{\om_0\}}|x(t)|^2d\mu_1(t)+\int\limits_{\a{ P}\times [0,\om_0)}|x(t)|^2d\mu_2(t)$.
The representation of a measure as a sum corresponds to the representation of $\f$ as a sum of states $\f_1$
and $\f_2$.
Identify the restrictions of $y_\be(p,n)$ on $P\times \{\om_0\}$ with $u_\be(p)$, then $\{u_\be\}_{\be\in B }$ is a normalized frame in $C_0(P\times \{\om_0\})$.
For
the restriction of $x$ to $P\times \{\om_0\}$ (and hence for $x$ itself, since $\mu_1$ vanishes outside this subset) we have
$\f_1(\<x,x\>)\le
\sum\limits_{\be\in B}\f_1(\<x,u_\be\>\<u_\be,x\>)
=\sum\limits_{j}\f_1(\<x,x_j\>\<x_j,x\>)
$
($\f_1$ can be considered as a state on $C_0(P\times \{\om_0\})$).
$P\times \{\om_0\}$ is a closed set in $K$, so $\a{P}\times [0,\om_0)$ is open and hence it is Borel set. Therefore $\mu_2$ is a Radon measure on $\a{P}\times [0,\om_0)$ (\cite[416R(b)]{Frem4}).
By the monotone convergence theorem (see, for example, \cite[Theorem 2.25]{Weaver} or \cite[Proposition 8.7(b)]{Yeh})
we have
$$
\int\limits_{\a{P}\times [0,\om_0)}|x(t)|^2d\mu_2(t)=
\int\limits_{\a{P}\times [0,\om_0)}\sum\limits_{n=1}^\infty|x(t)|^2|w_n(t)|^2d\mu_2(t)=
\sum\limits_{n=1}^\infty\int\limits_{\a{P}\times [0,\om_0)}|x(t)|^2|w_n(t)|^2d\mu_2(t),
$$
hence
\begin{multline*}
\f_2(\<x,x\>)=\int\limits_{\a{P}\times [0,\om_0)}|x(t)|^2d\mu_2(t)=
\sum\limits_{n=1}^\infty\int\limits_{\a{P}\times [0,\om_0)}|x(t)|^2|w_n(t)|^2d\mu_2(t)=\\
=\sum\limits_{n=1}^\infty\f_2(\<x,w_n\>\<w_n,x\>)\le
\sum\limits_{j}\f_2(\<x,x_j\>\<x_j,x\>).
\end{multline*}
By summing up the obtained inequalities, we have
$
\f(\<x,x\>)\le
\sum\limits_{j}\f(\<x,x_j\>\<x_j,x\>)
$.
Thus, for every state $\f$ on $C_0(K)$ and every $x\in C_0(K)$
the inequalities $\f(\<x,x\>)\le\sum\limits_j \f(\<x,x_j\>\<x_j,x\>)\le2\f(\<x,x\> )$ hold, so $\{x_j\}_{j\in J}$ is a frame in $C_0(K)$ with constants 1 and 2.
\end{proof}
\begin{rk}
It is clear that the proof of the previous theorem is still valid if instead of normalized frame in $C_0(P)$ we take a frame with arbitrary frame constants.
\end{rk}
The existence of a standard frame is a sufficient but not necessary condition for the $\A$-compactness criterion to be satisfied.
Therefore we cannot assert that for any topological space from the
class $\KII$
the criterion is not valid,
but we can construct an example of a space (more precisely, some subclass of spaces) for which the criterion actually fails.
First, let us consider several properties of topological spaces (it is assumed everywhere that $K$ is a locally compact Hausdorff space).
1) $\be K=\a K$ (in other words, every continuous bounded function has a limit at infinity).
2) Any $\s$-compact subset of $K$ is precompact in $K$ (that is, $K\in\KIV$).
2.1) Any continuous function which tends to zero at infinity is constant outside some compact $K'$.
2.2) Any continuous function that has a limit at infinity is constant outside some compact $K'$.
3) Any continuous function on $K$ is constant outside some compact $K'$.
Let's observe the relationships between these properties.
\begin{lem}
The properties 2), 2.1), 2.2) are equivalent.
\end{lem}
\begin{proof}
Indeed, 2.1) obviously follows from 2.2).
Let's prove 2.2) if 2.1) is true.
Let $f$ be a continuous function which has a limit at infinity equals to $f_\infty$. Then the continuous function $f-f_\infty$ tends to zero at infinity, so it vanishes outside some compact set, so the function $f$ is constant and equals to $f_\infty$ outside the same compact set.
Let's prove 2) if 2.1) is true.
Let $\{K_n\}_{n\in\N}$ be a sequence of compact sets in $K$. For every $n\in\N$ there exists a function $g_n\in C_0(K)$ such
that $g_n(t)=1$ on $K_n$, $|g_n(t)|\le1$ on $K$. Define the function $g\in C_0(K)$ by formula $g(t)=\sum\limits_{n\in\N}\frac{1}{2^n}g_n(t)$. It tends to zero at infinity, so it vanishes outside some compact $K'$, and moreover, it is nonzero at $\bigcup\limits_{n\in\N}K_n$, so $\bigcup\limits_ {n\in\N}K_n\subset K'$. Q.E.D.
Implication 2) $\Rightarrow$ 2.1) was proved in \cite[Lemma 1.5]{Fuf2021faa}.
\end{proof}
Obviously, 3) implies 2). The converse, in general, is false; it suffices to consider a disjoint union of sets with property 2), for example, $[0,\om_1)$. The same example shows that 2) does not imply 1).
From 3) it also obviously follows that 1) holds. The converse is not true, as will follow from the example we will construct. Also, this example will not satisfy 2). Also, this example will represent a class of spaces for which the criterion of $\A$-compactness fails.
\begin{ex}\label{ii_exnocr}
Take $K=\a{P}\times [0,\om_0]\setminus\{(p_\infty,\om_0)\}$, where $P\in\KIV$ (that is, $P$ satisfies the property 2)),
$\a{P}=P\cup\{p_\infty\}$ --- its one-point compactification.
\end{ex}
A special case of this construction is the deleted Tychonoff plank (\cite[\S 87]{Exampl}) if $P=[0,\om_1)$, $\a P=[0,\om_1]$.
This space does not satisfy 2) (and, as a consequence, does not satisfy 3)), since it contains a countable dense family of compact sets $\{\a P\times \{n\}\}_{n\in\N }$. Let us show that it satisfies 1)
(this generalizes the properties of the deleted Tychonoff plank, see \cite[8.20]{GilJer}).
\begin{teo}
Let $K$ be the space from the example \ref{ii_exnocr}. Then $\a K=\be K$, that is, every continuous bounded function has a limit at infinity. Moreover, $K$ is pseudo-compact, that is, every continuous function on $K$ is bounded.
\end{teo}
\begin{proof}
Let $f\in C(K)$. Then for each $n\in\N$ the restriction of $f$ to $\a P\times \{n\}$ is continuous, and the restriction of $f$ to $P\times \{n\}$ is a continuous function that has a limit at infinity. Hence, by property 2), for every $n\in\N$ there is a compact set $P_n\subset P$ such that $f$ is constant (and equals to some $p_{n,\infty}$) outside the compact $ P_n \times \{n\}\subset P\times \{n\}$. $\{P_n\}_{n\in\N}$ is a countable family of compact sets in $P$, so there exists a compact set $P'\subset P$ which contains all of them. So outside $P'\times [0,\om_0)$ the function $f$ depends only on the number $n\in\N$ and is equals to $p_{n,\infty}$ on the ``$n$th row'' $P\times\{n\}$.
Consider $(p,\om_0)\in P\times\{\om_0\}$ with $p\notin P'$. Since $f$ is continuous at $(p,\om_0)$, we have $f(p,\om_0)=\lim\limits_{n\to\infty}f(p,n)$, this limit does not depend on $p$ outside $P'$, hence the function $f(p,\om_0)$ is constant outside $P'$, so the function $f$ can be extended by continuity at the point $(p_\infty,\om_0)$ by $f(p_\infty,\om_0)=\lim\limits_{n\to\infty}f(p_\infty,n)=\lim\limits_{p\to p_\infty}\lim\limits_{n \to\infty}f(p,n)$.
Thus, we have proved the property 1).
\end{proof}
\begin{cor}
For $K$ from the example \ref{ii_exnocr} we have
$
C(K)=C_b(K)=
\dot{C_0}(K)=M(C_0(K))\cong C(\a K)
=C(\be K)
$ (except the first equality, this is also true for any space with property 1)).
\end{cor}
To show that Troitsky's theorem does not hold for $C_0(K)$ as a module over itself with such $K$, we need one more intermediate step.
\begin{teo}
A system $\{x_j\}\subset C_b(K)$ is $(C_b(K),C_0(K))$-admissible if and only if it is $(C_b(K),C_b(K)) $-admissible.
\end{teo}
\begin{proof}
An implication $\Leftarrow$ is obvious; let us prove the inverse.
Let $\{x_j\}\subset C_b(K)$ be $(C_b(K),C_0(K))$-admissible, i.e., for every $x\in C_0(K)$ we have
\begin{enumerate}
\item[1)]
the series $\sum_i \<x,x_i\>\<x_i,x\>$ converges in norm (i.e., uniformly);
\item[2)]
its sum is bounded by $\<x,x\>$;
\item[3)]$\|x_i\|\le 1$ for any $i$.
\end{enumerate}
Let us take an arbitrary function $x\in C_b(K)$ and show that these conditions are also satisfied for it (it suffices to show 1) and 2), obviously).
Similar to the previous proof, there exists a compact set $P'\subset P$ such that the function $x$ and all the functions $x_j$ are constant outside $P'\times[0,\om_0]$ on each ``row''. There exists $p'\in P\setminus P'$, denote $P''=P'\cup\{p'\}$.
There exists a function $g\in C_0(P)$ such that $g(p)=1$ on $P''$, $|g(p)|\le1$ on $P$.
Define
$\widetilde{x}\in C_0(K)$ by the formula $\widetilde{x}(t)=x(t)g(p)$, where $t=(p,n)\in K$. $\widetilde{x}$ satisfies conditions 1) and 2) on $K$, and hence on the set $P''\times [0,\om_0]$ on which $\widetilde{x}=x$ .
Outside $P''$ we have
$\sum\limits_i \<x,x_i\>\<x_i,x\>(p,n)=
\sum\limits_i \<x,x_i\>\<x_i,x\>(p'',n)
$. That is, outside $P''$ conditions 1) and 2) are satisfied, since they are satisfied for $p=p'$.
If the uniform convergence on each of two sets holds, then it also holds on their union; if the inequality holds on sets, then it also holds on their union. Hence, conditions 1), 2) are satisfied on the whole $K$ for any $x\in C_b(K)$. Q.E.D.
\end{proof}
\begin{cor}
A set $Y\subset C_0(K)$ is $(C_b(K),C_0(K))$-totally bounded if and only if it is $(C_b(K),C_b(K))$-totally bounded.
\end{cor}
\begin{rk}
Using \cite[4.2]{Walker} one can construct more complex examples of spaces with the described properties, by taking instead of $[0,\om_0]$ an arbitrary infinite compact set and
choosing for $P$ instead of $\om_1$
a sufficiently large ordinal if it necessary --- to use the condition that on each ``row'' the function is eventually constant.
\end{rk}
\begin{teo}\label{ii_nocr}
The unit ball in $C_0(K)$ (and hence the image of the unit ball with respect to the identity operator $Id:C_0(K)\to C_0(K)$) is $(C_b(K),C_0(K))$- totally bounded, but the identity operator is not $C_0(K)$-compact.
\end{teo}
\begin{proof}
The unit ball in $C_0(K)$ is a subset of the unit ball in $C_b(K)$, which is $(C_b(K),C_b(K))$-totally bounded since it is the image of the unit ball with respect to the identity operator $Id:C_b(K)\to C_b(K)$, which is $C_b(K)$-compact because $C_b(K)$ is unital and it is countably generated as a module over itself. Hence, the unit ball in $C_0(K)$ is also $(C_b(K),C_b(K))$-totally bounded, and by the previous corollary it is $(C_b(K),C_0(K))$-totally bounded.
The idenitity operator is not $C_0(K)$-compact since the image of $\A$-compact operator must be countably generated (\cite[Lemma 1.10]{Troitsky2020JMAA}), but $C_0(K)$ is not.
\end{proof}
Let us also prove that for the constructed example there is no non-standard frames, and we must start with the following useful lemma.
\begin{lem}\label{frame_restrict}
Let $K$ be a locally compact Hausdorff space, $A$ its closed subset. If there is a frame $\{x_j\}_{j\in J}$ (standard or not) in $C_0(K)$, then its restriction to $A$ $\{y_j\}_{j\in J }$ is a frame in $C_0(A)$ in the same sense.
\end{lem}
\begin{proof}
Since uniform convergence on a set implies uniform convergence on a subset, the proposition is obvious for standard frames. Let us prove it for non-standard.
Take $x\in C_0(A)$ and let $\f$ be a state on $C_0(A)$,
i.e. a Radon measure $\mu$ on $A$. It can be extended to a measure on whole $K$ by zero outside $A$ --- that is, to the state $\f'$ on $C_0(K)$.
The function $x$ can be extended to the function $x'\in C_0(K)$ due to Lemma \ref{extrest}.
Since $\{x_j\}_{j\in J}$ is a frame, we have
$$
c_1\f'(\<x',x'\>)\le
\sum\limits_j \f'(\<x',x_j\>\<x_j,x'\>)\le
c_2\f'(\<x',x'\>).
$$
Since $\f'$ is a measure which is actually calculated on the restrictions of functions on $A$, for $x$ we have
$$
c_1\f(\<x,x\>)\le
\sum\limits_j \f(\<x,y_j\>\<y_j,x\>)\le
c_2\f(\<x,x\>),
$$
i.e. $\{y_j\}_{j\in J}$ is a non-standard frame in $C_0(K)$.
\end{proof}
\begin{teo}\label{ii_nocr_nofr}
For $K$ from the example \ref{ii_exnocr} there are no non-standard frames in $C_0(K)$.
\end{teo}
\begin{proof}
Suppose that there exists a non-standard frame $\{x_j\}_{j\in J}$ in $C_0(K)$.
Then its restriction $\{y_j\}_{j\in J}$ to $P\times\{\om_0\}$ is a non-standard frame in $C_0(P\times\{\om_0\})$, so $C_0(P)$ also has a frame, which cannot be, as we will see later (theorem \ref{iv_nofr}). A contradiction.
\end{proof}
\section{The properties of $\KIII$}
\begin{ex}\label{iii_good}
Due to \ref{iii_stfr} and \ref{iii_dop} as a ``good'' example when there exists a standard frame it suffices to take $K=\bigsqcup\limits_{\l\in\L}K_\l$ with uncountable $\L$, where all $K_\l$ are $\s$-compact. Indeed, any compact set in $K=\bigsqcup\limits_{\l\in\L}K_\l$ intersects only a finite number of $K_\l$, which means that a $\s$-compact set intersects only a countable number of them. Hence, the complement to any $\s$-compact subset contains some $K_\be$, which is an open set in $K$.
\end{ex}
Let us now introduce an example when the $\A$-compactness criterion is not satisfied, which implies that there is no standard frame; there is also no frames for it at all.
\begin{ex}\label{iii_bad}
Let $K=P_1\sqcup P_2$,
where $P_1\in\KIV$, $P_2\in\KI$, $\KII$ or $\KIII$. A $\s$-compact set in $K$ is a union of $\s$-compact sets from $P_1$ and $P_2$ respectively. The complement to a $\s$-compact set in $P_1$ is open, so the same is true for $K$. However, one can reach a point at infinity with a countable set of compact sets from $P_2$, so $K\in\KIII$.
\end{ex}
\begin{teo}\label{iii_bad_th}
Let $K$ be a space from example \ref{iii_bad}. Then the operator $F:C_0(K)\to C_0(K)$ of multiplication by the identity function on $P_1$ and by zero function on $P_2$ is not $C_0(K)$-compact, but the image of the unit ball with respect to this operator is $(C_b(K),C_0(K))$-totally bounded. Obviously, this operator is adjointable.
Also, $C_0(K)$ has no frames.
\end{teo}
\begin{proof}
The image of this operator is an uncountably generated module $C_0(P_1)$, so the operator cannot be $C_0(K)$-compact.
The image of the unit ball is the unit ball in $C_0(P_1)$, and since the restriction of the Radon measure to a measurable subset is the Radon measure (\cite[416R(b)]{Frem4}), the seminorm $\nu_{X,\F}$ on the image has the following form
$$
\nu_{X,\F}(x)^2=
\sup_k
\sum_{i=k}^\infty |
\int\limits_K\overline{x(t)}\cdot x_i(t)d\mu_k(t)
|^2
=
\sup_k
\sum_{i=k}^\infty |
\int\limits_{K'}\overline{x(t)}\cdot x_i(t)d\mu_k(t)
|^2
=$$
$$=
\sup_k
\sum_{i=k}^\infty |
\int\limits_{P_1\sqcup P_2}\overline{x(t)}\cdot x_i(t)d\mu_k(t)
|^2
=
\sup_k
\sum_{i=k}^\infty |
\int\limits_{P_1}\overline{x(t)}\cdot x_i(t)d\mu_k(t)
|^2,
$$
that is, the seminorm on the image is calculated as a seminorm on $C_0(P_1)$. Obviously, the restriction to $P_1$ of any $(C_b(K),C_0(K))$-admissible system is $(C_b(P_1),C_0(P_1))$-admissible.
Hence, the unit ball in $C_0(P_1)$
is $(C_b(K),C_0(K))$-totally bounded by reasons similar to \cite[Theorem 2.5]{Fuf2021faa} because the unit ball in $C_0(P_1)$ is $(C_b(P_1),C_0(P_1))$-totally bounded
(more specifically, since the elements of the $\e$-net, which are functions on $P_1$, can be extended to whole $K$).
If there exists a frame in $C_0(K)$, its restriction to $P_1$ would also be a frame, but as we will see later (theorem \ref{iv_nofr}), $C_0(P_1)$ has no frames, so there are no frames in $C_0(K)$ too.
\end{proof}
There is also an intermediate example of space: there is no standard frame, but a non-standard one exists.
But first let us prove the following another one useful lemma.
\begin{lem}
Let $P_1, P_2$ be locally compact Hausdorff spaces, and both $C_0(P_1)$ and $C_0(P_2)$ have frames $\{x_j\}_{j\in J}$ (with constants $ d_1,d_2$) and $\{y_i\}_{i\in I}$ (with constants $c_1,c_2$) respectively.
Then in $C_0(K)$, where $K=P_1\sqcup P_2$ there also exists a frame $\{w_g\}_{g\in G}=\{x_j\}_{j\in J}\cup \{y_i\}_{i\in I}$. If each of the original frames is standard, then $\{w_g\}_{g\in G}$ is standard too.
\end{lem}
\begin{proof}
Uniform convergence on a finite number of sets implies uniform convergence on their union. The same is true for the inequalities, so the case when both frames are standard is obvious. Let us prove for non-standard.
Let $\f$ be a state on $C_0(K)$, i.e. a measure on $K$. It can be represented as the sum of measures on $P_1$ and $P_2$ respectively, i.e. $\f=\f_1+\f_2$, where $\f_1, \f_2$ are states on $P_1, P_2$ respectively. It is also possible to represent in such a way the function $w\in C_0(K)$, $w=x+y$ and
$\<w,w\>=|w|^2=\<x,x\>+\<y,y\>$. Hence we get that
$$
\sum\limits_g \f(\<w,w_g\>\<w_g,w\>)=
\sum\limits_j \f_1(\<x,x_j\>\<x_j,x\>)+\sum\limits_i \f_2(\<y,y_i\>\<y_i,y\>),
$$
and hence
$$
\sum\limits_g \f(\<w,w_g\>\<w_g,w\>)\le
d_2\f_1(\<x,x\>)+c_2\f_2(\<y,y\>)=
$$
$$=
d_2\f(\<x,x\>)+c_2\f(\<y,y\>)\le
\max\{d_2, c_2\}(\f(\<x,x\>)+\f(\<y,y\>))=
\max\{d_2, c_2\}\f(\<w,w\>).
$$
Similarly, we have that
$$
\min\{d_1, c_1\}\f(\<w,w\>)
\le
\sum\limits_g \f(\<w,w_g\>\<w_g,w\>)\le
\max\{d_2, c_2\}\f(\<w,w\>),
$$
i.e. $\{w_g\}_{g\in G}$ is a frame in $C_0(K)$. Q.E.D.
\end{proof}
\begin{ex}\label{iii_bg}
Let $K=P_1\sqcup P_2$,
where $P_1$ is the space from example \ref{iii_good}, $P_2$ is the space from example \ref{ii_exfr}. Similar to the previous discussion, $K\in\KIII$.
\end{ex}
\begin{teo}
There is no standard frame in $C_0(K)$, but there exists a non-standard one.
\end{teo}
\begin{proof}
If $C_0(K)$ has a standard frame, its restriction to $P_1$ would also be a standard frame, but since $P_2\in\KII$, $C_0(P_2)$ has no standard frame (theorem \ref {ii_nostfr}), hence $C_0(K)$ also does not have.
Let us show that there exists a non-standard one. We know that in $C_0(P_1)$ there is a normalized standard frame $\{x_j\}_{j\in J}$, and in $C_0(P_2)$ there is a frame $\{y_i\}_{ i\in I}$ with constants $c_1=1$, $c_2=2$.
By the previous lemma
their union $\{w_g\}_{g\in G}=\{x_j\}_{j\in J}\cup\{y_i\}_{i\in I}$ is a frame in $ C_0(K)$.
\end{proof}
\section{Non-existence of non-standard frames in $\KIV$}
\begin{teo}\label{iv_nofr}
Let $K\in\KIV$. Then the $\dot{C_0}(K)$-module $C_0(K)$ has no frame.
\end{teo}
\begin{proof}
Assume that there exists a frame $\{x_j\}_{j\in J}$ in $C_0(K)$.
Take an arbitrary point $t_1\in K$.
There exists a function $g_1\in C_0(K)$ such that $g_1(t_1)=1$, $|g_1(t)|\le1$ on $K$.
There is a non-empty at most countable set $\{x_j\}_{j\in J_1}$ of elements of the frame such that $x_j(t_1)\ne0$ because by
taking $x=g_1$ and $\f={\delta_{t_1}}$ --- evaluation at the point $t_1$, due to lemma \ref{fr_cr} we have that
the series $\sum\limits_j \<g_1,x_j\>\<x_j,g_1\>(t)=\sum\limits_j |x_j(t)|^2$ converges at the point $t_1$. That is, if $j\in J\setminus J_1$, then $x_j(t_1)=0$.
For every $j\in J_1$ there is a compact set $K_{1,j}\subset K$ such that $x_j=0$ outside $K_{1,j}$. $J_1$ is at most a countable set, so there is a compact set $K_1$ such that $\bigcup\limits_{j\in J_1}K_{1,j}\subset K_1$. That is, $x_j=0$ outside $K_1$ for any $j\in J_1$.
Assume that we have already found points $t_1,\dots,t_n$, compact sets $K_1,\dots,K_n$ and index sets
$J_1,\dots,J_n\subset J$ such that $t_i\notin \bigcup\limits_{l=1}^{i-1}K_l$ for $i=2,\dots,n$, $x_j (t_l)\ne0$ only for $j\in J_l$ (as a consequence, different sets $J_l$ do not intersect), $x_j=0$ outside $K_l$ for $j\in J_l$.
Take an arbitrary point $t_{n+1}\in K\setminus\bigcup\limits_{l=1}^{n}K_l$. As in the case when $n=1$, there exists a non-empty at most countable set $\{x_j\}_{j\in J_{n+1}}$ of elements of the frame such that $x_j(t_{n+1})\ne0$ for $j\in J_{n+1}$ (and hence $J_{n+1}$ does not intersect any $J_{l}$, $l=1,\dots,n$, since the functions $x_j$ for $j\in \bigcup\limits_{l=1}^{n}J_l$ vanishes outside $\bigcup\limits_{l=1}^{n}K_l$).
There also exists a compact set $K_{n+1}$ such that $x_j=0$ outside $K_{n+1}$ for any $j\in J_{n+1}$. By induction, we can continue this construction for any $n\in\N$.
The sequence $\{t_n\}_{n\in\N}$ is a $\s$-compact set, so there exists a compact set $K'$ containing this sequence.
Hence the sequence has a limit point $t_0\in K'$. Let us show that $x_j(t_0)=0$ for all $j\in J$, which will contradict the fact that $\{x_j\}_{j\in J}$ is a frame.
First let it be that $j\in J\setminus\bigcup\limits_{l=1}^{\infty}J_{l}$. Then $x_j(t_n)=0$ for all $n\in\N$. Hence,
$x_j(t_0)=0$, because otherwise if $x_j(t_0)=q\ne0$ then in any neighborhood of the point $t_0$ there is a point $t_n$ such that $|x_j(t_0)-x_j(t_n)|=|q|>0$ --- a contradiction with the continuity of $x_j$.
Let now $j\in\bigcup\limits_{l=1}^{\infty}J_{l}$, i.e. $j\in J_k$ for some $k\in\N$, and suppose that $x_j(t_0)\ne0$. Then $t_0\in K_k$ (because $x_j=0$ outside $K_k$). Hence, $x_j(t_{k+l})=0$ for all $l\in\N$ (because $t_{k+l}\notin K_k$) and $t_0$ is still a limit point for the sequence $\{t_n\}_{n=k+1}^\infty$, and then $x_j(t_0)=0$ similarly to the previous case.
Hence $\{x_j\}_{j\in J}$ is not a frame. Q.E.D.
\end{proof}
| 179,884
|
TITLE: Find the probablity that at least 2 balls are placed in corresponding number boxes?
QUESTION [2 upvotes]: There are six balls and six boxes numbered 1 to 6.
So I have to find the probablity that at least 2 balls are placed in corresponding number boxes.
My approach.
Out of 6 balls , select 2 which will be placed in right boxes.
So it will be $\binom62$.
Now remaining 4 balls can be arranged in any ways so it will be .
So total will be $\binom62 \times 4!$.
And the denominator will be $6!$.
So answer comes to be $0.5$ .
But the answer is $1/60$ .
Where am i wrong ?
Because i think that i have done it a right way.
Thanks in advance.
REPLY [2 votes]: You’re counting many arrangements more than once. As an extreme case, consider the arrangement that has every ball in the right box: it gets counted $\binom62=15$ times, once for every pair of balls.
It’s probably easier to count the arrangements that don’t have at least two balls in the right boxes; here’s a start. Those arrangements are of two types: the derangements of the $6$ balls $-$ the arrangements that have no ball in the right box $-$ and the arrangements that have exactly one ball in the right box. The derangements are a bit messy to count, but the link has a good deal of information to help you with that. Once you know how to count derangements, you can count the arrangements that have exactly one ball in the right box: multiply $6$ ways to choose the ball that’s correctly placed by the number of derangements of the other $5$ balls.
| 174,109
|
HOLLYWOOD -- State transportation officials on Wednesday said they were left with no choice but to select Sheridan Street and Interstate 95 as the site of a future park-and-ride lot, even though city commissioners rejected that location a year ago.
``We couldn`t find sites that were acceptable to the city and acceptable to us,`` said Bruce Seiler, a planner with the state Department of Transportation.
Seiler said DOT staff considered seven sites between Pembroke Road and Sheridan Street and concluded that the Sheridan location was the best compromise.
Commissioner Suzanne Gunzburger, a member of the tri-county commuter rail organization, had wanted the parking lot near Hollywood Boulevard, where a rail stop is planned.
``I am very unhappy,`` Gunzburger said. ``I feel five to seven years from now the station`s location will change because of this decision.``
Commissioners last month voted 3-2 against a park-and-ride garage inside Stan Goldman Park near I-95 between Hollywood Boulevard and Johnson Street after residents opposed it.
The planned 1,000-space lot near the southwest corner of the intersection would be accessible only from Sheridan Street when it opens. Seiler said eventually DOT would like to build a flyover bridge that would connect I-95 with the lot. Building that bridge may force DOT to take some land at the nearby Colony Mobile Home Park.
The parking lot would be on 14 acres at the site of a Mack Industries cement operation owned by Hollywood Inc. Seiler said DOT will try to neogotiate a sale with the company and will condemn the property if necessary.
| 184,377
|
We offer meet and greet service on all Manchester airport pickups
AIRPORT TRANSFERS
With appointments to keep and places to be, planning your next trip can be
stressful enough without worrying about the journey to and from the airport, which is where our airport transfer service is the solution.
We offer a meet and greet service on all airport pickups. On your arrival,
our drivers will be waiting in the airport arrival hall with your name on a sign board.
We monitor the status of all flights. This way we can ensure that even if a flight arrives earlier than expected or is delayed your driver will be there waiting for you. GMEC will ensure peace of mind whether you are travelling to a well-deserved holiday or business trip.
You can call us on 07948 200 222 or email us on bookings@gmexec.co.uk .
Also you can fill out our contact form
| 273,081
|
TITLE: Prove the opposite angles of a quadrilateral are supplementary implies it is cyclic.
QUESTION [4 upvotes]: There is a well-known theorem that a cyclic quadrilateral (its vertices all lie on the same circle) has supplementary opposite angles.
I have a feeling the converse is true, but I don't know how to prove it. The converse states:
If a quadrilateral's opposite angles are supplementary then it is cyclic.
Should I approach this proof by contradiction? Or is it possible to prove by construction?
REPLY [5 votes]: A proof by contradiction is a good approach. Suppose you have a quadrilateral $ABCD$ whose opposite angles are supplementary, but it is not cyclic. The vertices $A,B,C$ determine a circle, and the point $D$ does not lie on this circle, since we assume the quadrilateral is not cyclic.
Suppose for instance that $D$ lies outside the circle, and so the circle intersects $ABCD$ at some point $E$ on $CD$ (try drawing a picture to see this if needed.) Now $D$ is supplementary to $B$, and since $E$ is the opposite angle of $B$ in the cyclic quadrilateral $ABCE$, $E$ is supplementary to $B$ by the theorem you already know, and so $D$ and $E$ are congruent. But this contradicts the fact that an exterior angle cannot be congruent to an interior angle, which proves the converse. A similar method works if $D$ lies inside the circle as well. (I abuse notation a bit and refer to a vertex and the angle at that vertex by the same letter.)
| 135,189
|
TITLE: Let $f:[0,1] \to R$ be continuous such that $|f(x)| \le \int_0^xf(t)dt$ for all $x \in [0,1]$.
QUESTION [3 upvotes]: Let $f:[0,1] \to R$ be continuous such that $|f(x)| \le \int_0^xf(t)dt$ for all $x \in [0,1]$. Then,
a. Such $f$ does not exist
b. $f(x)=0$ for all $x \in [0,1]$
c. $f(x)=c$ for all $x \in [0,1]$ and some constant $c$
d. None of the above is true.
What I've tried doing :
$$
f(x) \le |f(x)| \le \int_0^xf(t)dt
$$
and hence
$$
f(x)-\int_0^xf(t)dt \le 0
$$
Let $g(x)=f(x)-\int_0^xf(t)dt$
$$
g'(x)=f'(x)-f(x)
$$
I have no idea what to do now. Any hints?
REPLY [5 votes]: Note that the derivative of $e^{-x}h(x)$ is $e^{-x}(h'(x)-h(x))$ for any differentiable function $h$. (Remember this trick whenever you see an expression of the form $h'(x)-h(x)$.) Here, that gives us
$$\frac{d}{dx}\left(e^{-x}\int_0^xf(t)\,dt\right)=e^{-x}\left(f(x)-\int_0^xf(t)\,dt\right)$$
The right-hand side is $\le 0$ on $[0,1]$, because $f(x)-\int_0^xf(t)\,dt\le|f(x)|-\int_0^xf(t)\,dt$. So the function $e^{-x}\int_0^xf(t)\,dt$ is non-increasing on $[0,1]$. And it takes the value $0$ at $x=0$, so it must be $\le 0$ on the whole of $[0,1]$. $e^{-x}$ is strictly positive, so we have $\int_0^xf(t)\,dt \le 0$ for $x\in[0,1]$.
Hence $0\le|f(x)|\le\int_0^xf(t)\,dt \le 0$ on $[0,1]$, which gives immediately $f(x)=0$ for all $x\in[0,1]$.
| 87,421
|
After not having been to Bratislava in a while, my schedule allowed me an afternoon/evening to make the trip into Slovakia to visit Lucy. The weather was beautiful, and a walk along the majestic Danube always take my mind back in time to my own experiences connected to this river as well as it huge role in European history. Lucy was doing great. Her life both materially and spiritual seems very well in order as she continues to be a pillar of faith in this part of the world.
| 193,499
|
Updated: 25/5/2020 at 10:00 (Local time)
Data: Hong Kong Inter-National Airport
Was it hotter in Tsuen Wan last year? Was it colder? Data reported by the weather station on May 25, 2019. Such day as today a year ago.
You can see all the historical data and other dates in:
| 28,824
|
TITLE: How can I prove that a function has a certain bound on an interval?
QUESTION [0 upvotes]: To preface this, I have had calculus, but not analysis. I am working on a discrete dynamical system which gives rise to the function
$f(x) = (-x^2-x+1)(e^{-x^2})$.
I need to show that $|f(x)|<1$ for all values $x > 0$. I know that this is true from plotting the function, but I have no idea how to begin approaching this formally (probably because I haven't had any training in analysis). How do I approach this proof?
REPLY [2 votes]: Since $e^{x^2}$ is positive, we can multiply through on both sides to see that
$$
|f(x)|\leq 1\iff |1-x-x^2|\leq e^{x^2}.
$$
For $x\geq 0$ (the domain you are interested in), obviously $1-x-x^2\leq 1$. Since $1\leq e^{x^2}$ for all $x$,
$$
1-x-x^2\leq 1\leq e^{x^2},
$$
proving half of the inequality. For the other half, we must show that
$$
x^2+x-1\leq e^{x^2},
$$
which can by handled by using the inequality $e^y\geq y+1$ valid for all $y\geq 0$.
Specifically, for $x\leq 1$ we can take $y=x^2$ to conclude that
$$
x^2+x-1\leq x^2+1\leq e^{x^2},
$$
whereas for $x\geq 1$ we can take $y=x^2-1$ to conclude that
$$
x^2+x-1\leq 2x^2\leq ex^2\leq e\cdot e^{x^2-1}=e^{x^2},
$$
as desired.
| 212,160
|
\begin{document}
\maketitle
\begin{abstract}
For a group $G$ and a natural number $m$, a subset $A$ of $G$ is called $m$-thin if, for each finite subset $F$ of $G$, there exists a finite subset $K$ of $G$ such that $|Fg\cap A|\leqslant m$ for every $g\in G\setminus K$. We show that each $m$-thin subset of a group $G$ of cardinality $\aleph_n$, $n= 0,1,\ldots$ can be partitioned into $\leqslant m^{n+1}$ 1-thin subsets. On the other side, we construct a group $G$ of cardinality $\aleph_\omega$ and point out a 2-thin subset of $G$ which cannot be finitely partitioned into 1-thin subsets.
\end{abstract}
Let $G$ be a group, $\kappa$ and $\mu$ be cardinals, $|G|\geqslant \kappa \geqslant \aleph_0$ and $\mu \leqslant \kappa$, $[G]^{<\kappa} = \{X\subset G: |X| < \kappa\}$.
We say that a subset $A$ of $G$ is {\it $(\kappa,\mu)$-thin } if, for every $F\in [G]^{<\kappa}$, there exists $K\in [G]^{<\kappa}$ such that
$$ |Fg\cap A|\leqslant \mu$$
for each $g\in G\setminus K$.
If $\kappa$ is regular then $A$ (see Lemma 1) is $(\kappa,1)$-thin if and only if, for each $g\in G$, $g\ne e$, $e$ is the identity of $G$, we have
$$|\{a\in A: ga\in A\}|< \kappa.$$
An $(\aleph_0,1)$-thin subset is called {\it thin }. For thin subsets, its modifications and applications see \cite{b1}, \cite{b2}, \cite{b3}, \cite{b4}, \cite{b5}, \cite{b6}, \cite{b7}. For $m\in \mathbb{N}$, the $(\aleph_0, m)$-thin subsets appeared in \cite{b3} under name {\it $m$-thin} in attempt to characterize the ideal in the Boolean algebra of subsets of $G$ generated by the family of thin subsets of $G$. If a subset $A$ of $G$ is a union of $m$ thin subsets then $A$ is $m$-thin. On the other hand, if $G$ is countable and $A$ is $m$-thin then $A$ can be partitioned into $\leqslant m$ thin subsets. Thus, the ideal generated by thin subsets of a countable group $G$ coincides with the family of all $m$-thin, $m\in \mathbb{N}$ subsets of $G$. Does this characterization remain true for all infinite groups? In other words, can every $m$-thin subset of an uncountable group $G$ be partitioned in $m$ (finitely many) thin subsets? In this paper we give answer to these questions.
The paper consists of 5 sections. In the first section we see that the thin subsets can be defined in the much more general context of balleans, the counterparts of the uniform topological spaces. From this point of view, a thin subset is a counterpart of a uniformly discrete subset of a uniform space. As a corollary of some ballean statement (Theorem 1), we get that, for each infinite regular cardinal $\kappa$ and each $m\in \mathbb{N}$, every $(\kappa, m)$-thin subset of a group $G$ of cardinality $\kappa$ can be partitioned into $\leqslant m$ $(\kappa,1)$-thin subsets.
In section 2 we extend this statement showing (Theorem 3) that, for every infinite regular cardinal $\kappa$, $m\in \mathbb{N}$ and $n\in \omega$, each $(\kappa, m)$-thin subset of a group $G$ of cardinality $\kappa^{+n}$ can be partitioned into $\leqslant m^{n+1}$ $(\kappa,1)$-thin subsets. Here, $\kappa^{+0} = \kappa, \kappa^{+(n+1)}= (\kappa^n)^+$. In particular, every $m$-thin subset of a group $G$ of cardinality $\aleph_n$ can be partitioned into $\leqslant m^{n+1}$ thin subsets. Clearly, in this case the ideal generated by thin subsets also coincides with the family of all $m$-thin subsets, $m\in \mathbb{N}$. In Theorem 4 we describe $(\kappa, \mu)$-thin groups that can be partitioned into $\mu$ $(\kappa,1)$-subsets.
In section 3 one can find two auxiliary combinatorial theorems (of independent interest!) on coloring of the square $G\times G$ of a group $G$ which will be used in the next section.
Answering a question from \cite{b3}, G.~Bergman constructed a group $G$ of cardinality $\aleph_2$ and a 2-thin subset $A$ of $G$ which cannot be partitioned into two thin subsets. With kind permission of the author, we reprint in section 4 his letter with this remarkable construction (Example 1). Then we modify the Bergman's construction to show (Example 2) that for each natural number $m\geqslant 2$ there exist a group $G_n$ of cardinality $\aleph_n$, $ n = \frac{m(m+1)}{2} -1$, and a 2-thin subset $A$ of $G$ which cannot be partitioned into $m$-thin subsets. And finally (Example 3), we point out a group $G$ of cardinality $\aleph_\omega$ and a 2-thin subset of $G$ which cannot be finitely partitioned into thin subsets.
We conclude the paper with some observations on interplay between thin subsets and ultrafilters in section 5.
\section{Ballean context}
A {\it ball structure} is a triple $\mathcal{B}=(X,P,B)$, where $X$, $P$ are non-empty sets and, for
any $x\in X$ and $\alpha\in P$, $B(x,\alpha)$ is a subset of $X$
which is called a \emph{ball of radius} $\alpha$ around $x$. It is
supposed that $x\in B(x,\alpha)$ for all $x\in X$ and $\alpha\in P$.
The set $X$ is called the {\it support} of $\mathcal{B}$, $P$ is
called the {\it set of radii}. Given any $x\in X, A\subseteq X,
\alpha\in P$ we put
$$
B^*(x,\alpha)=\{y\in X:x\in B(y,\alpha)\},\
B(A,\alpha)=\bigcup_{a\in A}B(a,\alpha).
$$
Following \cite{b8}, we say that a ball structure $\mathcal{B}=(X,P,B)$ is a {\it ballean} if
\begin{itemize}
\item for any $\alpha,\beta\in P$,
there exist $\alpha',\beta'$ such that, for every $x\in X$,
$$B(x,\alpha)\subseteq B^*(x,\alpha'),\ B^*(x,\beta)\subseteq B(x,\beta');$$
\item for any $\alpha,\beta\in P$,
there exists $\gamma\in P$ such that, for every $x\in X$,
$$B(B(x,\alpha),\beta)\subseteq B(x,\gamma).$$
\end{itemize}
We note that a ballean can also be defined in terms of entourages of diagonal in $X\times X$. In this case it is called a coarse structure \cite{b9}.
A ballean $\B$ is called {\it connected} if, for any $x,y\in X$, there exists $\alpha\in P$ such that $y\in B(x,\alpha)$. All balleans under consideration are supposed to be connected. Replacing each ball $B(x,\alpha)$ to $B(x,\alpha) \cap B^*(x,\alpha)$, we may suppose that $B(x,\alpha) = B^*(x,\alpha)$ for all $x\in X,\alpha\in P$. A subset $Y\subseteq X$ is called {\it bounded} if there exist $x\in X$ and $\alpha\in P$ such that $Y\subseteq B(x,\alpha)$.
We use a preordering $\leqslant$ on the set $P$ defined by the rule: $\alpha \leqslant \beta$ if and only if $B(x,\alpha)\subseteq B(x,\beta)$ for every $x\in X$. A subset $\P' \subseteq P$ is called {\it cofinal} if, for every $\alpha\in P$, there exists $\alpha' \in P'$ such that $\alpha \leqslant \alpha'$. A ballean $\B$ is called {\it ordinal} if there exists a cofinal subset $P'\subseteq P$ well ordered by $\leqslant$.
Let $\B = (X,P,B)$ be a ballean, $\mu$ be a cardinal. We say that a subset $A\subseteq X$ is {\it $\mu$-thin} if, for every $\alpha \in P$, there exists a bounded subset $Y\subseteq X$ such that $|B(x,\alpha)\cap A|\leqslant \mu$ for every $x\in G\setminus Y$. A 1-thin subset is called {\it thin}.
\begin{lemma}
Let $\B = (X,P,B)$ be a ballean, $\mu$ be a cardinal. A subset $A\subseteq X$ is $\mu$-thin if and only if the set
$$\{a\in A: |B(a,\alpha) \cap A| > \mu\}$$
is bounded.
\end{lemma}
\begin{proof}
The "if" part is evident. To verify the "only if", we take an arbitrary $\alpha\in P$ and choose $\beta \in P$ such that $B(B(x,\alpha),\alpha)\subseteq B(x,\beta)$ for each $x\in X$. By the assumption, the set $Y = \{a\in A: |B(a,\alpha) \cap A| > \mu\}$ is bounded. We put $Z = B(Y,\alpha)$ and take an arbitrary $x\in X\setminus Z$. If $|B(x,\alpha)\cap A| > \mu$ and $a \in B(x,\alpha)\cap A$ then $|B(a,\beta)\cap A| > \mu$ because $B(x,\alpha)\subseteq B(a,\beta)$. Hence, $a\in Y$ and $x\in Z$. This contradiction shows that $|B(x,\alpha)\cap A| \leqslant \mu$ and $A$ is $\mu$-thin.
\end{proof}
\begin{theorem}
Let $\B = (X,P,B)$ be a ballean, $\mu$ be a cardinal, $A\subseteq X$. Then the following statements hold
\begin{itemize}
\item[(i)] if $A$ is a union of $\mu$ thin subsets and a union of $\mu$ bounded subsets of $X$ is bounded then $A$ is $\mu$-thin;
\item[(ii)] if $\B$ is ordinal and $A$ is $\mu$-thin, $\mu\in\mathbb{N}$ then $A$ can be partitioned into $\leqslant \mu$ thin subsets.
\end{itemize}
\end{theorem}
\begin{proof}
(i) Let $A=\bigcup_{\lambda \leqslant \mu} A_\lambda$ and each $A_\lambda$ is thin, $\alpha \in P$. For each $\lambda \leqslant \mu$, we pick a bounded subset $Y_\lambda$ such that $|B(x,\alpha)\cap A|\leqslant 1$ for each $x\in X\setminus Y_\lambda$. We put $Y=\bigcup_{\lambda \leqslant \mu} Y_\lambda$. By the assumption, $Y$ is bounded. Clearly, $|B(x,\alpha)\cap A| \leqslant \mu$ for each $x\in X\setminus Y$ so $A$ is $\mu$-thin.
(ii) Apply Lemma 1 and \cite[Theorem 1.2]{b3}.
\end{proof}
\begin{theorem}
Let $G$ be a group, $\kappa$ be an infinite regular cardinal, $|G| = \kappa$. Then the following statements hold
\begin{itemize}
\item[(i)] each $(\kappa, m)$-thin subset $A$ of $G$, $m\in \mathbb{N}$ is a union of $\leqslant m$ $(\kappa,1)$-thin subsets;
\item[(ii)] the ideal generated by the family of $(\kappa,1)$-thin subsets coincides with the family of all $(\kappa, m)$-thin subsets, $m\in \mathbb{N}$.
\end{itemize}
\end{theorem}
\begin{proof}
Clearly, (ii) follows from (i). To prove (i), we consider a ballean $\B(G,\kappa) = (G, [G]^{<\kappa}, B)$ where $B(x,F) = Fx\cup \{x\}$ for all $x\in G, F\in [G]^{<\kappa}$. We enumerate $G=\{g_\alpha: \alpha<\kappa\}$ and put $F_\alpha = \{g_\beta: \beta < \alpha\}$. Since $\kappa$ is regular, $\{F_\alpha: \alpha <\kappa \}$ is cofinal in $[G]^{<\kappa}$ so $\B(G,\kappa)$ is ordinal. To apply Theorem 1(i), it suffices to note that $A$ is $(\kappa, m)$-thin if and only if $A$ is $m$-thin in the ballean $\B(G,\kappa)$.
\end{proof}
In view of \cite[Chapter 1]{b8}, the balleans can be considered as asymptotic counterparts of the uniform spaces. For uniform spaces see \cite[Chapter 8]{b10}. Now we describe the uniform counterparts of thin subsets.
Let $\UU$ be a uniformity on a set $X$. For an entourage $U\in \UU$ and $x\in X$, we put $U(x) =\{y\in X: (x,y)\in U\}$. Let $A$ be a subset of $X$, $\mu$ be a cardinal. We say that $A$ is {\it $(\UU, \mu)$-discrete } if there exists $U\in \UU$ such that $|U(x) \cap A| \leqslant \mu$ for each $x\in X$, a $(\UU,1)$-discrete subset is called {\it $\UU$-discrete}. We show that each $(\UU, \mu)$-discrete subset $A$ of $X$ can be partitioned into $\leqslant \mu$ $\UU$-discrete subsets.
We fix an entourage $U\in \UU$ such that $|U(x)\cap A|\leqslant \mu$ for each $x\in X$ and choose a symmetric entourage $V\in \UU$ such that $V^2 \subseteq U$. Then we consider a graph $\Gamma$ with the set of vertices $A$ and the set of edges $E$ defined by the rule: $(x,y)\in E$ if and only if there exists $z\in X$ such that $x,y\in V(z)$. Since $|U(x)\cap A|\leqslant \mu$ for each $x\in X$ and $V^2 \subseteq U$, each unit ball in $\Gamma$ is of cardinality $\leqslant \mu$. Hence, the chromatic number $\chi(\Gamma)$ does not exceed $\mu$. We take a partition $\mathcal{P}$ of $A$ such that $|\mathcal{P}| = \chi(\Gamma)$ and each $P\in \mathcal{P}$ has no incident vertices. If $x\in P$ then $V(x)\cap P = \{x\}$. It follows that $P$ is $\UU$-discrete.
\section{Partitions}
\begin{lemma}
Let $G$ be a group, $\kappa$ be an infinite cardinal, $\kappa \leqslant |G|$, $m\in\mathbb{N}$. Let $A$ be a $(\kappa,m)$-thin subset of $G$, $S\subseteq G$, $|S|\geqslant \kappa$. Then there exists a subgroup $H$ of $G$ such that $S\subseteq H$, $|H| =|S|$ and $|Hx\cap A|\leqslant m$ for each $x\in G\setminus H$. In particular, $A$ is $(\kappa',m)$-thin for each $\kappa' \geqslant \kappa$, $\kappa' \leqslant |G|$.
\end{lemma}
\begin{proof}
We may suppose that $S$ is a subgroup. Let $H_0 = S$, $[H_0]^{m+1} = \{X\subset H_0: |X| = m+1\}$, $|S| = \kappa'$. For each $X\in [H_0]^{m+1} $, we choose $K_0(X) \in [G]^{<\kappa} $ such that $|Xg\cap A| \leqslant m$ for every $g\in G\setminus K_0(X)$. We put $K_0 = \cup\{K_0(X): X\in [H_0]^{m+1}\}$ and note that $|K_0|\leqslant \kappa'$ and $|H_0x\cap A|\leqslant m$ for each $x\in G\setminus K_0$.
We consider a subgroup $H_1$ of $G$ generated by $H_0\cup K_0$. Clearly, $|H_1| = \kappa'$. For each $X\in [H_1]^{m+1}$, we take $K_1(X) \in [G]^{<\kappa}$ such that $|Xg\cap A| \leqslant m$ for every $g\in G\setminus K_1(X)$. We put $K_1 = \cup\{K_1(X): X\in [H_1]^{m+1}\}$ and note that $|H_1x\cap A|\leqslant m$ for each $x\in G\setminus K_1$.
After $\omega$ steps we get an increasing sequence of $\{H_n: n\in \omega\}$ of subgroups of $G$ and a sequence $\{K_n:n\in \omega\}$ of subsets of $G$ such that $|H_n| = \kappa'$, $|K_n| \leqslant \kappa'$. Since $\cup_{n\in \omega} K_n \subseteq \cup_{n\in \omega} H_n$, for the subgroup $H = \cup_{n\in \omega} H_n$ we get a desired statement.
To show that $A$ is $(\kappa',m)$-thin, we take $S\in [G]^{<\kappa'}$. If $|S|<\kappa$ then there exists $K\in [G]^{<\kappa}$ such that $|Sx\cap A|\leqslant m$ for each $x\in G\setminus K$ because $A$ is $(\kappa,m)$-thin. If $|S|\geqslant \kappa$, we apply the previous statement.
\end{proof}
For a cardinal $\kappa$ and $n\in\omega$, we use the following notations from \cite{b11}: $\kappa^{+0} = \kappa, \kappa^{+(n+1)}= (\kappa^{+n})^+$. In particular, $\aleph_0^{+n} = \aleph_n$ for each $n\in \omega$.
\begin{theorem}
Let $\kappa$ be an infinite regular cardinal, $m\in\mathbb{N}$, $n\in\omega$, $G$ be a group of cardinality $\kappa^{+n}$. Each $(\kappa, m)$-thin subset $A$ of $G$ can be partitioned into $\leqslant m^{n+1}$ $(\kappa,1)$-thin subsets.
\end{theorem}
\begin{proof}
We use an induction by $n$. For $n=0$, this is Theorem 2. Let $A$ be a $(\kappa,m)$-thin subset of $G$ and $|G|=\kappa^{+(n+1)}$. By Lemma 2, $A$ is $(\kappa^{+(n+1)}, m)$-thin. Applying Theorem 2, we can partition $A$ in $\leqslant m$ $(\kappa^{+(n+1)}, 1)$-thin subsets. We suppose that $A$ itself is $(\kappa^{+(n+1)}, 1)$-thin and show that $A$ can be partitioned in $\leqslant m^{n+1}$ $(\kappa,1)$-thin subsets.
Since $A$ is $(\kappa^{+(n+1)}, 1)$-thin, we use Lemma 2 to write $G$ as a union of increasing chain of subgroups $\{H_\alpha: \alpha < \kappa^{+(n+1)}\}$ such that $H_0 =\{e\}$, $e$ is the identity of $G$, $|H_\alpha| = \kappa^{+n}$ for each $\alpha >0$, $H_\beta = \cup_{\alpha < \beta} H_\alpha$ for each limit ordinal $\beta < \kappa^{+(n+1)}$. Clearly, $G\setminus \{e\} = \cup_{\alpha < \kappa^{+(n+1)}} H_{\alpha+1}\setminus H_\alpha$.
For each $\alpha < \kappa^{+(n+1)}$, $\alpha >0$, we put $A_\alpha = A\cap H_\alpha$. Since $A_\alpha$ is $(\kappa,m)$-thin and $|H_\alpha| = \kappa^{+n}$, by the inductive assumption, each $A_\alpha$ can be partitioned in $k_\alpha \leqslant m^{n+1}$ $(\kappa,1)$-thin subsets of $H_\alpha$. Admitting empty sets of the partition, we suppose that $k_\alpha = m^{n+1}$ for each $\alpha < \kappa^{+(n+1)}$ and write
$$A_\alpha = A_\alpha(1) \cup \ldots \cup A_\alpha(m^{n+1}),$$
where each $A_\alpha(i)$ is $(\kappa,1)$-thin.
For all $\alpha < \kappa^{+(n+1)}$ and $i\in \{1,\ldots , m^{n+1} \}$, we put
$$B_\alpha(i) = A_{\alpha+1}(i) \setminus A_\alpha(i), \quad B_i = \bigcup_{\alpha < \kappa^{+(n+1)}}B_\alpha(i).$$
Since $A\setminus \{e\} = \cup \{B_i: i\in \{1,\ldots , m^{n+1} \} \}$, it suffices to verify that each subset $B_i$ is $(\kappa,1)$-thin. It turns out, since $\kappa$ is regular, in view of Lemma 1, it suffices to show that, for each $g\in G$, $g\ne e$,
$$|\{x\in B_i: gx \in B_i\}|< \kappa.$$
We take the minimal $\alpha < \kappa^{+(n+1)}$ such that $g\in H_{\alpha+1} \setminus H_\alpha$. If $x\in B_i\setminus H_{\alpha +1}$ then $gx\notin B_i$ by the choice of $H_{\alpha +1}$. Since $A_{\alpha+1}(i)$ is $(\kappa,1)$-thin, $|\{x\in A_{\alpha+1}(i) \setminus A_\alpha(i) : gx\in A_{\alpha+1}(i) \setminus A_\alpha(i)\}|< \kappa$. If $x,y\in A_{\alpha+1}(i) \setminus A_\alpha(i)$ and $gx,gy\in A_\alpha(i)$ then $x^{-1}y \in A_\alpha(i)$ so, by the choice of $H_{\alpha }$, $x=y$. If $x,y\in A_\alpha(i)$ and $gx, gy \in A_{\alpha+1}(i) \setminus A_\alpha(i)$, replacing $g$ to $g^{-1}$, we get the previous case.
Theorem is proved.
\end{proof}
Given an infinite group $G$ and infinite cardinal $\kappa$, $\kappa \leqslant |G|$, we denote by $\mu(G,\kappa)$ the minimal cardinal $\mu$ such that $G$ can be partitioned in $\mu$ $\kappa$-thin subsets. For a cardinal $\gamma$, $cf\;\gamma$ is a cofinality of $\gamma$, $\gamma^+$ is the cardinal successor of $\gamma$. By \cite{b5},
$$
\mu(G,\kappa) = \left\{
\begin{array}{ll}
\gamma,& \mbox{if $|G|$ is non-limit cardinal and $|G| = \gamma^+$;}\\
|G|,& \mbox{if $|G|$ is a limit cardinal and either}\\
&\mbox{$\kappa < |G|$ or $|G|$ is regular} \\
cf\; |G|, &\mbox{if $|G|$ is singular, $\kappa = |G|$ and $cf\; |G|$}\\
&\mbox{is a limit cardinal} \\
\end{array} \right.
$$
If $|G|$ is singular, $\kappa = |G|$ and $cf\; |G|$ is a non-limit cardinal, $cf|G|= \gamma^+$ then $\mu(G,\kappa)\in \{\gamma, \gamma^+\}$.
Now let $\gamma$ be a cardinal, $\gamma \leqslant \kappa$. Then $G$ is $(\kappa, \gamma)$-thin if and only if $\kappa = \gamma^+$. Applying above formulae for $\mu(G,\kappa)$, we get the following statement.
\begin{theorem}
Let $G$ be a group, $\gamma$ be an infinite cardinal, $\kappa = \gamma^+$, $|G|\geqslant \gamma$. Then $G$ can be partitioned in $\gamma$ $(\kappa,1)$-thin subsets if and only if $|G|=\kappa$.
\end{theorem}
\section{Colorings}
For a group $G$ and $g\in G$, we say that $\{G\}\times \{g\}$ is a {\it horizontal line} in $G\times G$, $\{g\}\times G$ is a {\it vertical line} in $G\times G$, $\{(x,gx): x\in G\}$ is a {\it diagonal} in $G\times G$.
The statement (i) in the following theorem was proved by G.~Bergman, (ii) by the first author.
\begin{theorem}
For a group $G$ with the identity $e$, the following statements hold
\begin{itemize}
\item[(i)] if $|G| \geqslant \aleph_2$ and $\chi: G\times G \to \{1,2,3\} $ then there is $g\in G$, $g\ne e$ such that either some horizontal line $G\times \{g\}$ has infinitely many points of color 1, or some vertical line $\{g\}\times G$ has infinitely many points of color 2, or some diagonal $\{(x,gx): x\in G\}$ has infinitely many points of color 3;
\item[(ii)] if $|G| \leqslant \aleph_1$ then there is a coloring $\chi: G\times G \to \{1,2,3\} $ such that each horizontal line has only finite number of points of color 1, each vertical line has only finite number of points of color 2, each diagonal has only finite number of points of color 3.
\end{itemize}
\end{theorem}
\begin{proof}
(i) We suppose the contrary and fix a corresponding coloring $\chi: G\times G \to \{1,2,3\}$. Let $G_0, G_1$ be subgroups of $G$ such that $G_0\subset G_1$, $|G_0|=\aleph_0$, $|G_1|=\aleph_1$. Since the set $G\times G_1$ has at most $\aleph_1$ points of color 1, there is $g\in G$, $g\ne e$ such that $gG_1 \times G_1$ has no points of color 1. The set $gG_0 \times G_1$ has at most $\aleph_0$ points of color 2, so there is $h\in G_1$ , $h\ne g$, $h\ne e$ such that $gG_0\times hG_0$ has no points of color 2. Hence, the set $\{(gx, hx): x\in G_0\}$ consists of color 3 and is contained in the diagonal $\{(y, g^{-1}hy): y\in G\}$.
(ii) We proceed in three steps.
Step 1. Let $X$ be a countable set, $\{A_n:n\in \omega\}$, $\{B_n:n\in \omega\}$ be partitions of $X$ such that $A_n\cap B_m$ is finite for all $n,m\in\omega$. Then there is a coloring $\chi:X->\to\{1,2\}$ such that each subset $A_n$ has only finite number of elements of color 1 and each subset $B_n$ has only finite number of elements of color 2.
We color $A_0$ in 2, $B_0\setminus A_0$ in 1, $A_1\setminus B_0$ in 2, $B_1\setminus (A_0\cup A_1)$ in 1, $A_2\setminus (B_0\cup B_1)$ in 2, $B_2\setminus(A_0\cup A_1 \cup A_2)$ in 1, and so on.
Step 2. Let $H$ be a countable group, $K$ be a subgroup of $H$. Applying Step 1, we define a coloring $\chi:((H\times H)\setminus (K\times K))\to \{1,2,3\}$ such that
\begin{itemize}
\item $\chi((H\setminus K)\times(H\setminus K))=\{1,2\}$ and each horizontal line in this set has only finite number of points of color 1, and each vertical line has only finite number of points of color 2.
\item $\chi(K\times(H\setminus K))=\{1,3\}$ and each horizontal line in this set has only finite number of points of color 1, and each diagonal has only finite number of points of color 3.
\item $\chi((H\setminus K)\times K)=\{2,3\}$ and each vertical line in this set has only finite number of points of color 2, and each diagonal has only finite number of points of color 3.
\end{itemize}
Step 3. To prove (ii), we may suppose that $|G| = \aleph_1$ so write $G$ as a union of an increasing chain $\{G_\alpha:\alpha < \omega_1\}$, $G_0=\{e\}$, $e$ is the identity of $G$, such that $G_\beta=\bigcup_{\alpha<\beta} G_\alpha$ for each limit ordinal $\beta<\omega_1$. We put $\chi_0(e)=1$ and, for each $\alpha < \omega_1$ use a coloring $\chi_\alpha((G_{\alpha+1}\times G_{\alpha+1})\setminus(G_\alpha\times G_\alpha))$ defined on Step 2. We put $\chi=\bigcup_{\alpha<\omega_1}\chi_\alpha$ and verify that $\chi: G\times G\to \{1,2,3\}$ is thin.
Clearly, $(G\times\{e\})\cap \chi^{-1}(1)\subseteq G_1\times\{e\}$, $(\{e\}\times G)\cap \chi^{-1}(2)\subseteq \{e\}\times G_1$. If $g\in G_{\alpha+1}\setminus G_\alpha$ then
$$
(G\times\{g\})\cap \chi^{-1}(1)\subseteq G_{\alpha+1}\times\{g\},\quad (\{g\}\times G)\cap \chi^{-1}(2)\subseteq \{g\}\times G_{\alpha+1}.
$$
Thus, each horizontal line in $G\times G$ has only finite number of points of color 1, and each vertical line in $G\times G$ has only finite number of points of color 2.
At last, if $(x,y)\in (G_{\alpha+1}\setminus G_\alpha)\times G_\alpha$ or $(x,y)\in G_\alpha\times (G_{\alpha+1}\setminus G_\alpha)$ then $x^{-1}y\in G_{\alpha+1}\setminus G_\alpha$. It follows that each diagonal has only finite number of points of color 3.
\end{proof}
In \cite{b12}, \cite{b13} R.~Davies proved the following theorem (see also \cite[Theorem 1.7]{b11}).
For every $n\in \mathbb{N}$, the following statements are equivalent
\begin{enumerate}
\item $2^{\aleph_0}\leqslant \aleph_n$;
\item There is a sequence $L_0,\ldots L_{n+1}$ of lines in the plane $\mathbb{R}^2$ and a coloring $\chi: \mathbb{R}^2 \to \{0,\ldots, n+1\}$ such that, for each $i\in \{0,\ldots, n+1\}$, every line in $\mathbb{R}^2$ parallel to $L_i$ intersects $\chi^{-1}(i)$ in finitely many points.
\end{enumerate}
We note that the group $\mathbb{R}$ of real numbers is the direct sum of $2^{\aleph_0}$ copies of the group $\mathbb{Q}$ of rational numbers and use a part of this theorem in the following form.
\begin{theorem}
Let $n\in \mathbb{N}$, $H=\oplus_{\aleph_n}\mathbb{Q}$, $a_0, \ldots ,a_n$, $b_0, \ldots ,b_n$ be rational numbers such that, for each $i$, either $a_i \ne 0$ or $b_i \ne 0$. Then, for every coloring $\chi: H\times H\to \{0,\ldots, n\}$, there exist $i\in \{0,\ldots, n\}$, $h\in H$, $h\ne 0$ and infinitely many pairs $a,b\in H$ such that $\chi(a,b)=i$ and $a_ia+b_ib = h$.
\end{theorem}
\begin{proof}
For $i\in \{0,\ldots, n\}$, let $L_i = \{(x,y)\in H\times H : a_ix + b_iy = 0\}$. Apply Davies' theorem to the lines $L_0,\ldots L_{n}$.
\end{proof}
\section{Examples}
With minor changes, the first example is a reprint of the original Bergman's letter to the first author (15 May 2011).
\begin{example}
Let $H,K$ be groups of cardinality $\aleph_2$, $G = H\times K$. We construct a 2-thin subset $A$ of $G$ which cannot be partitioned into two thin subsets.
One can find a thin subset $X\subset K$ of cardinality $\aleph_2$. For instance, do a recursion over $\aleph_2$, selecting at each step an element not in the subgroup generated by those that precede. Since $X$ has cardinality $\aleph_2$, we can index it by pairs of elements of $H$: $X = \{x_{\{a,b\}}: a,b\in H\}$. After choosing such an indexing, we let
$$A = \{x_{\{a,b\}}, ax_{\{a,b\}}, bx_{\{a,b\}}: a,b \in H\}, \quad A\subseteq H\times K = G.$$
I claim first that $A$ is 2-thin. For this it suffices to show that for every 3-element subset $F$ of $G$, only finitely many right translates of $F$ lie in $A$. In proving this, we may, by an initial right translation assume that $e\in F$, $e$ is the identity of $G$.
Assume that $F$ lay in $HK$ but not in $H$. Then every one of its right translates $Fg: g\in G$ has elements lying in more than one left coset of $H$; hence if such a right translate is contained in $A$, its elements do not all have the same second coordinate in $X$. Since $X$ is thin in $K$, we have $\{g\in G: Fg\subset A\}$ is finite.
We are left with the case $F\subset H$. In this case, it is not hard to see that $F$ has exactly 6 right translates contained in $A$; namely, these are obtained by taking the 6 arrangement of the elements of $F$ as an ordered 3-tuple, applying to each the right translate by a member of $H$ that puts it in the form $(e,a,b)$ and then right translating this by $x_{\{a,b\}}$ to get a 3-tuple of members of $A$.
Finally, let us show that $A$ cannot be partitioned into two thin subsets, $A_1$ and $A_2$. Suppose we had such a partition. Then let us color $H\times H$ as follows. Color an element $(a,b)\in H\times H$
\begin{itemize}
\item with color 1 if $x_{\{a,b\}}$ and $ax_{\{a,b\}}$ lie in the same one of $A_1$ and $A_2$, and $bx_{\{a,b\}}$ lies in the other one.
\item with color 2 if $x_{\{a,b\}}$ and $bx_{\{a,b\}}$ lie in the same one of $A_1$ and $A_2$, and $ax_{\{a,b\}}$ lies in the other one.
\item with color 1 if $ax_{\{a,b\}}$ and $bx_{\{a,b\}}$ lie in the same one of $A_1$ and $A_2$, and $x_{\{a,b\}}$ lies in the other one.
\item with any of these three colors if $x_{\{a,b\}}$, $ax_{\{a,b\}}$, and $bx_{\{a,b\}}$ all lie in the same set $A_1$ or $A_2$.
\end{itemize}
Now if some horizontal line $\{a\} x H$ in $H\times H$ had infinitely many points of color 1, then there would be infinitely many $b$ such that $x_{\{a,b\}}$ and $ax_{\{a,b\}}$ lay in the same one of $A_1$ and $A_2$. Hence one of the latter sets, say $A_i$, contains infinitely many 2-element sets $\{ x_{\{a,b\}}, ax_{\{a,b\}}\}$. This gives infinitely many right translates of the pair $\{1, a\}$ in $A_i$, contradicting the assumption of thinness.
If some horizontal line $H\times \{a\}$ or diagonal $\{(h,ah): h\in H\}$ had infinitely many points of color 2, respectively 3, we would get a contradiction in the same way. The case of horizontal line is like that of vertical line, so let us check the diagonal case. Suppose that for some $a$ infinitely many of the pairs $hx_{\{h,ah\}}$ and $ahx_{\{h,ah\}}$ lay in the same of our thin sets. Then at least one of those sets would contain infinitely many of these pairs; but this shows that the set would contain infinitely many right translates of the pair $\{e,a\}$, contradicting thinness.
The above arguments show that our coloring of $H\times H$ contradicts Theorem 5(i); so $A$ cannot, as assumed, be decomposed into two thin sets.
\end{example}
\begin{example}
For each natural number $m\geqslant 2$, we construct an Abelian group $G$ of cardinality $\aleph_n$, $ n = \frac{m(m+1)}{2} -1$, and find a 2-thin subset $A$ of $G$ which cannot be partitioned into $m$-thin subsets.
We put $H=\oplus_{\aleph_n}\mathbb{Q}$, take an arbitrary Abelian group $K$ of cardinality $\aleph_n$ and let $G=K\oplus H$. Then we choose a thin subset $X$ of $K$, $|X| = \aleph_2$ and enumerate $X$ by the pairs of elements of $H$: $X = \{x_{\{a,b\}}: a,b\in H\}$. After choosing such an indexing, we let
$$A = \{x_{\{a,b\}} + ka + k^2b: a,b \in H , k \in \{0,\ldots,m\} \}.$$
To see that $A$ is 2-thin, it suffices to show that, for any two distinct non-zero elements $x,y\in G$, the set
$$A(x,y)=\{a\in A: a+x\in A, a+y\in A\}$$
is finite. We write $x = x_1 + x_2$, $y = y_1 + y_2$, $x_1,x_2\in K$, $y_1,y_2\in H$. If either $x_1 = 0$ or $x_2 = 0$, $A(x,y)$ is finite because $X$ is thin. Let $x_1 = x_2 = 0$. If $x_{\{a,b\}} + ia + i^2b\in A(x,y)$ then
$$
\left\{
\begin{array}{lll}
x_2 + ia + i^2b & = & ja+j^2b \\
y_2 + ia + i^2b & = & ka+k^2b\\
\end{array} \right.
$$
for some distinct $i,k\in \{0,\ldots,m\} $. In this system of relations, $a,b$ are uniquely determined by $i,j,k$. Since we have only finite number of possibilities to choose $i,j,k$, $A(x,y)$ is finite.
Now assume that $A$ is partitioned $A=A_1\cup \ldots \cup A_m$. To show that at least one cell of the partition is not thin, we define a coloring $\chi: H\times H\to \{(k,l): 0\leqslant k < l \leqslant m\}$ by the following rule: for $a,b\in H$, we choose $k,l$ so that $x_{\{a,b\}} + ka + k^2b$, $x_{\{a,b\}} + la + l^2b$ lie in the same cell of the partition $A_1\cup \ldots \cup A_m$. We note that
$$(x_{\{a,b\}} + ka + k^2b) - (x_{\{a,b\}} + la + l^2b) = (k-l)a + (k^2-l^2)b.$$
Since $\frac{m(m+1)}{2} = n+1$, by Theorem 6, there exist $h\in H, h\ne 0, k<l$, and infinitely many monochrome pairs $a,b$ such that
$$(k-l)a + (k^2-l^2)b = h.$$
By the definition of $\chi$, there are a cell $A_i$ of the partition and infinitely many pairs $a,b$ such that
$$x_{\{a,b\}} + ka + k^2b \in A_i, x_{\{a,b\}} + ka + k^2b + h\ in A_i,$$
so $A_i$ is not thin.
\end{example}
\begin{example}
We construct a group $G$ of cardinality $\aleph_\omega$ and point out a 2-thin subset $A$ of $G$ which cannot be partitioned into $m$ thin subsets for each $m\in\mathbb{N}$.
For each $m\geqslant 2$, we take a group $G_n$, $ n = \frac{m(m+1)}{2} -1$ from Example 2, put $N = \{ \frac{m(m+1)}{2} -1: m\geqslant 2 \}$, take a 2-thin subset $A_n$ of $G_n$ which cannot be partitioned into $m$-thin subsets, and denote
$$G = \oplus_{n\in N} G_n,\quad A = \cup_{n\in N}A_n.$$
We take any distinct $x,y\in G\setminus\{0\}$ and, in notation of Example 2, show that $A(x,y)$ is finite, so $A$ is 2-thin. If $x,y\in G_n$ for some $n$ then $A(x,y)$ is 2-thin because $A_2$ is 2-thin. If $x\notin G_n$ for each $n$ then $|A\cap (A+x)|\leqslant 1$ so $A(x,y)$ is also finite. By the choice of $\{A_n: n\in N\}$, $A$ cannot be finitely partitioned into thin subsets.
\end{example}
\section{Ultrafilter context}
Let $G$ be a discrete group, $\beta G$ be the Stone-\v{C}ech compactification of $G$.
We take the elements of $\beta G$ to be ultrafilters on $G$ identifying $G$ with the set of principal ultrafilters, so $G^* = \beta G\setminus G$ is the set of free ultrafilters. The topology of $\beta G$ can be defined by the family $\{\overline{A}: A\subseteq G \}$ as a base for open sets, $\overline{A} = \{p\in \beta G: A\in p\}$.
The multiplication on $G$ can be naturally extended to $\beta G$ (see \cite[Chapter 4]{b14}). By this extension, the product $pq$ of ultrafilters $p$ and $q$ can be defined as follows. Take an arbitrary $P\in p$ and, for each $g\in P$, pick $Q_g\in q$. Then $\cup_{g\in P} gQ_g \in pq$ and each member of $pq$ contains a subset of this form. In particular, if $g\in G$ and $q\in \beta G$ then $gq = \{gQ: Q\in q\}$.
The proofs of all propositions in this section can be easily extracted from corresponding definitions.
\begin{proposition} For each $m\in \mathbb{N}$, a subset $A$ of a group $G$ is $m$-thin if and only if $|Gp\cap \overline{A}|\leqslant m$ for each $p\in G^*$.
\end{proposition}
By \cite{b1}, a subset $A$ of a group $G$ is {\it sparse} if for any infinite subset $X\subset G$ there exists a finite subset $F$ such that $\cap_{g\in F}gA$ is finite. An ultrafilter $p\in G^*$ has a sparse member if and only if $p\notin \overline{G^*G^*}$.
\begin{proposition} A subset $A$ of a group $G$ is sperse if and only if the set $Gp\cap \overline{A}$ is finite for each $p\in G^*$.
\end{proposition}
\begin{proposition} A subset $A$ of a group $G$ can be partitioned in finite number of thin subsets if and only if each ultrafilter $p\in \overline{A}$ has a thin member.
\end{proposition}
Now take a group $G$ of cardinality $\aleph_\omega$ from Example 3 and corresponding 2-thin subset $A$. Since $A$ cannot be finitely partitioned into thin subsets, by Proposition 3, there is $p\in \overline{A}$ with no thin members. Hence, $p$ has a base consisting of 2-thin but not thin subsets.
| 185,200
|
Startups and Securities
Startups and Security Laws
Small businesses need capital to startup, operate, and grow. Seeking capital from outside investors is instrumental when the entrepreneur lacks the ability or desire to fund the venture from personal assets.
Back To: BUSINESS TRANSACTIONS, ANTITRUST, & SECURITIES LAW
Obtaining a loan from an outside institution is one manner of seeking funds; however, the ability to borrow substantial amounts of money is often limited for startups.
Entrepreneurs must turn to alternative methods of financing, such as selling an ownership interest (equity) in their business venture. If a business entity (Issuer) offers or sells securities it must comply with and meet the procedural requirements of the 33 Act.
Specifically, Section 5 of the 33 Act prohibits persons from using any method of interstate commerce to buy, sell, deliver, or offer to buy, sell, or delivery any security, except in accordance with the provisions stated therein.
Violating Section 5 carries potential civil and criminal penalties. Most notably among the securities laws is the requirement to register with the SEC any sale of securities. Registering a securities issuance is a detailed and burdensome information-filing process.
Congress recognizes the importance of balancing investor protection with the need to foster economic activity by promoting the growth and development of startup ventures. Few people deny the significant impact that startup ventures have on the economy.
In light of the above-stated objectives, Congress and the SEC promulgated several exemptions, both statutory and rule-based, to the extremely burdensome registration requirements.
The exemptions allow non-public companies (companies not registered with the SEC as publicly traded) to raise capital through a non-public sale or private placement of equity. The exemptions, however, can be difficult for entrepreneurs to navigate.
Entrepreneurs must understand the requirements to register a security, the applicable registration exemptions, the types of investors the entrepreneur can solicit, the amount of funds available under any exemption.
These considerations play an important role in the entrepreneurs decision on how, if at all, to pursue startup financing.
| 251,017
|
![endif]-->
Separate?
I'm guessing it's doable on the GT500s only?
My salesman set it up for me this way, and I liked it :D
Well I have them already, I know how they look, put a set on my friend's red 05GT :flag:
But I was wondering if the silicone might be a better...
Only two things that annoy me so far about the 05-07s:
1) The hood seems eternally long when driving the thing.
2) The way the corners of the...
[IMG]
Favorite so far
[IMG][IMG]
quieter,...
The stock GT500 vents are open underneath, water is not an issue.
Same goes for 03 Cobras who opened up their factory vents.
:cheers:
No, it really doesn't..:lol:
:flag:
How would protesting against a Message Board help the cause? :canada:
Yup, Shelby doesn't touch the GT500s.
IF you're on the KR list, I expect MSRP to be $15+ over a normal GT500. Easy.
The only performance mod I made so far was the FR 'loud' axle-back.
I've also done the Cobra power outlet plug and 20% tint, and have the painted...
I think I'm still around 88 or so.. :nonono:
| 263,698
|
After hearing the awesome reviews of the newest scary movie, Marvin and Henry devise a way to watch it without parental supervision.
Embed this Video
Marvin is feeling Literally Awesome Mega Excellent!.
From best buttkicker to funniest star, it's time to pick your faves!
Kick back and enjoy some non-stop hilarity!
| 322,131
|
\begin{document}
\maketitle
\begin{abstract}
Space mission planning and spacecraft design are tightly coupled and need to be considered together for optimal performance; however, this integrated optimization problem results in a large-scale Mixed-Integer Nonlinear Programming (MINLP) problem, which is challenging to solve. In response to this challenge, this paper proposes a new solution approach to this MINLP problem by iterative solving a set of coupled subproblems via the augmented Lagrangian coordination approach following the philosophy of Multi-disciplinary Design Optimization (MDO). The proposed approach leverages the unique structure of the problem that enables its decomposition into a set of coupled subproblems of different types: a Mixed-Integer Quadratic Programming (MIQP) subproblem for mission planning and one or more Nonlinear Programming (NLP) subproblem(s) for spacecraft design. Since specialized MIQP or NLP solvers can be applied to each subproblem, the proposed approach can efficiently solve the otherwise intractable integrated MINLP problem. An automatic and effective method to find an initial solution for this iterative approach is also proposed so that the optimization can be performed without the need for a user-defined initial guess. In the demonstration case study, a human lunar exploration mission sequence is optimized with a subsystem-level parametric spacecraft design model. Compared to the state-of-the-art method, the proposed formulation can obtain a better solution with a shorter computational time even without parallelization. For larger problems, the proposed solution approach can also be easily parallelizable and thus is expected to be further advantageous and scalable.
\end{abstract}
\section{Nomenclature}
{\renewcommand\arraystretch{1.0}
\noindent\begin{longtable*}{lll}
$\mathcal{A}$ &\quad=\quad& Set of arcs \\
$\boldsymbol{a}_{vijt}$ &\quad=\quad& Cost coefficient matrix of commodity\\
${a'}_{vijt}$ &\quad=\quad& Cost coefficient matrix of spacecraft\\
$\boldsymbol{d}_{it}$ &\quad=\quad& demand vector\\
$\boldsymbol{e}_v$ &\quad=\quad& Spacecraft design variable vector \\
$\mathcal{F}(-)$ &\quad=\quad& Spacecraft sizing function\\
$f$ &\quad=\quad& Objective function (subproblem) \\
$\boldsymbol{g}$ &\quad=\quad& Inequality constraint \\
$\boldsymbol{h}$ &\quad=\quad& Equality constraint \\
$H_{vij}$ &\quad=\quad& Concurrency matrix \\
$\mathcal{J}$ &\quad=\quad& Objective function \\
$k$ &\quad=\quad& Decomposed subproblem index\\
$L$ &\quad=\quad& Number of subsystems in the dry mass\\
$M$ &\quad=\quad& Number of subproblems in a quasi-separable MDO problem\\
$m$ &\quad=\quad& Mass of spacecraft subsystems \\
$m_d$ &\quad=\quad& Spacecraft dry mass \\
$m_f$ &\quad=\quad& Spacecraft propellant capacity \\
$m_p$ &\quad=\quad& Spacecraft payload capacity \\
$N$ &\quad=\quad& Number of types of spacecraft \\
$\mathcal{N}$ &\quad=\quad& Set of nodes \\
$n$ &\quad=\quad& dimension of variables \\
$Q_{vijt}$ &\quad=\quad& Commodity transformation matrix \\
$q$ &\quad=\quad& Iteration count\\
$\mathcal{T}$ &\quad=\quad& Set of time steps \\
$t_{mis}$ &\quad=\quad& Mission length \\
$\Delta t_{ij}$ &\quad=\quad& Time of Flight (ToF)\\
$u_{vijt}$ &\quad=\quad& Spacecraft flow variable \\
$\mathcal{V}$ &\quad=\quad& Set of spacecraft \\
$W_{ij}$ &\quad=\quad& Launch time window \\
$\boldsymbol{x}_{vijt}$ &\quad=\quad& Commodity flow variable\\
$\boldsymbol{y}$ &\quad=\quad& Shared variables \\
$\boldsymbol{z}$ &\quad=\quad& Local variables \\
$\zeta$ &\quad=\quad& Propellant type \\
$\phi$ &\quad=\quad& Penalty function \\ \\
\emph{Subscipt}&&\\
$i$ &\quad=\quad& Node index (departure) \\
$j$ &\quad=\quad& Node index (arrival) \\
$k$ &\quad=\quad& Subproblem index \\
$l$ &\quad=\quad& Subsystem index \\
$t$ &\quad=\quad& Time index \\
$v$ &\quad=\quad& Vehicle index \\
\end{longtable*}}
\section{Introduction}
\lettrine{A}{s} we pursue sustainable presence in space, a framework to optimize large-scale, long-term space missions efficiently is imperative. A number of studies on space logistics that incorporates the transportation network in large-scale space mission design have been developed, including SpaceNet \cite{Shull2007MS}, the interplanetary logistics model \cite{taylor2007logistics}, and the extensive literature on space logistics optimization frameworks based on the generalized multicommodity network flow \cite{ishimatsu2016gmcnf, ho2014time-expanded, ho2016FlexiblePath}. Utilizing the linear nature of such space logistics or transportation network optimization problems, researchers have developed frameworks that can efficiently optimize the mission design as Mixed-Integer Linear Programming (MILP) problems \cite{chen2018MILP, chen2018regular, chen2019isruMars, takubo2021HRL}.
However, due to the nonlinear nature of spacecraft design, a naive integration of spacecraft design into space mission/campaign planning (a transportation scheduling or resource distribution) would result in a large-scale Mixed-Integer Nonlinear Programming (MINLP) problem, which is oftentimes computationally prohibitive.
Since the concurrent optimization of space mission planning and spacecraft design is highly desired in practice, each community took different approaches to bridge these two domains.
In the space logistics community, spacecraft design has been considered as a high-level nonlinear sizing model and has been integrated into mission planning either by separating the nonlinear part from the mission planning optimization or by piecewise linearization of the spacecraft model. Taylor \cite{taylor2007phd} developed a parametric spacecraft sizing model which determines the spacecraft dry mass from its payload capacity and propellant capacity. Based on this model, Simulated Annealing (SA) or a similar metaheuristic optimization algorithm optimizes the spacecraft design variables, while the linear programming (LP) or MILP solver evaluates the constraints and determines transportation flow variables. In this way, the LP or MILP solver is embedded into SA, and thus it was called the embedded optimization methodology.
Using the same spacecraft sizing model, Chen and Ho \cite{chen2018MILP} employed piecewise linear (PWL) approximation of the nonlinear model to approximate the entire MINLP problem as a MILP problem that can be efficiently solvable. However, this approach is an approximation model, and the resulting solution is not guaranteed to be feasible nor optimal in the original nonlinear problem.
On the other hand, aerospace vehicle design has been tackled by the Multidisciplinary Design Optimization (MDO) community. Despite various optimization and sizing methods that can deal with the high-dimensional nonlinear design of aircraft or spacecraft \cite{sobieszczanski1997MDOsurvey}, few studies integrated the mission-level analysis or optimization. One of the few studies that tackled the integrated mission planning and spacecraft design is Ref. \cite{beauregard2021lunarMDO} by Beauregard et al., which proposed an MDO architecture for a lunar lander design with a lunar mission sequence architecture analysis. This architecture connects the mission planning and spacecraft design problem using a sequential procedure without a feedback structure (i.e., the mission architecture is first chosen and fixed, then the lunar lander MDO is performed); therefore, strictly speaking, the mission and spacecraft are not simultaneously optimized and spacecraft design is neglected when selecting the mission architecture. In addition, the candidates of the mission architectures are given \emph{a priori} and discrete (combinatory). These two factors limit the design space and make this approach not suitable for the integrated space mission design.
This paper proposes an efficient decomposition-based optimization scheme for integrated space mission planning and spacecraft design. The key idea is to decompose the integrated MINLP problem into multiple coupled subproblems of different types: the Mixed-Integer Quadratic Programming (MIQP) subproblem for space mission planning and the Nonlinear Programming (NLP) subproblem(s) for spacecraft design. Since specialized efficient MIQP or NLP optimizers (e.g., Gurobi \cite{gurobi} for MIQP; IPOPT \cite{IPOPT} for NLP) can be utilized to solve each subproblem, the proposed method can solve the otherwise intractable integrated MINLP problem efficiently.
The iterative coordination between each subproblem can be achieved using an MDO approach \cite{martins2013MDOsurvey,sobieszczanski1997MDOsurvey}. Specifically, the Augmented Lagrangian Coordination (ALC) approach \cite{tosserams2007ALC} with the Analytical Target Cascading (ATC) structure \cite{ATC,ATCextended} is chosen for the proposed method. This architecture fits our problem well because (1) it allows us to decompose the original complex problem into the subproblems with different and simpler types (MIQP or NLP), each of which can be efficiently solvable with specialized solvers; (2) it has a robust convergence property; and (3) it allows the complex hierarchical structure for the spacecraft design subproblem(s) and can be easily parallelizable (and thus scalable) if needed. Since the nonlinear optimization solvers generally require a good initial guess, we further develop an automated initial guess generation method based on PWL approximation to the MINLP problem so that no user-defined initial guess is needed for the optimization.
The remainder of this paper proceeds as follows. In Section \ref{Problem Definition}, the problem definition of the integrated space mission planning and spacecraft design as an all-in-one optimization problem formulation is described. Section \ref{OurALC} illustrates the solution procedure for the proposed problem based on the decomposition-based method. Section \ref{CaseStudy} introduces a case study of human lunar exploration missions and compares the computational efficiency of the proposed method and existing method. Finally, Section \ref{conclusion} states the conclusion.
\section{Problem Definition: Integrated Space Mission Planning and Spacecraft Design} \label{Problem Definition}
The goal of this research is to optimize the transportation scheduling (referred to as space mission planning) and vehicle design (referred to as spacecraft design) for a long-term space campaign that can potentially comprise multiple missions. This section introduces the formulation for this integrated space mission planning and spacecraft design problem (referred to as the all-in-one formulation). The idea behind this formulation is to consider space mission planning as a transportation network optimization problem for which the design of vehicles is also part of the decision variables. In the network, the nodes correspond to the orbital or surface locations and the arcs correspond to the trajectories connecting the nodes. The decision variables include both the commodities that flow over the network and the design parameters for the vehicles that carry these commodities. The optimization formulation is listed as follows, and the list of variables and parameters is included in Table~\ref{tab_SLvar}.
\input{Eqns/ProbDef}
\input{Tables/ProbDef_var_list}
Equation~\eqref{SL_obj} indicates the objective function, which can be the lifecycle cost or launch mass, depending on the application context. In this research, we set the coefficients $\boldsymbol{a}_{vijt}$ and ${a}_{vijt}$ so that the objective function corresponds to the sum of initial mass at low-earth orbit (IMLEO).
Equations~\eqref{SL_constr1}-\eqref{SL_constr3} are the constraints for space mission planning.
First, Eq.~\eqref{SL_constr1} is the mass balance constraint that guarantees that the inflow (supply) of the commodity is larger than the sum of the outflow and demand. $Q_{vijt}$ is the transformation matrix, which indicates the transformation of the commodity during the spaceflight; for example, the relationship of impulsive propellant consumption can be illustrated using this constraint.
Next, Eq.~\eqref{SL_constr2} is the concurrency constraint. This indicates that the commodity loaded on each spacecraft is constrained by the dimension of the spacecraft. Specifically in this paper, the payload and propellant flow is limited: the amount of propellant is lower than the propellant capacity of the spacecraft, and the sum of other payloads is lower than the payload capacity.
Finally, Eq.~\eqref{SL_constr3} is the time window constraints. The commodity flow is allowed only if the time $t$ belongs to the launch window vector $W_{ij}$, and for the remaining time steps, the commodity flow is conserved to be zero.
Equation~\eqref{SL_vehicle_sizing} indicates an abstract representation of the spacecraft design constraints, which describes the constraints between the properties of the vehicle. This can take a wide range of complexity, including an explicit or implicit relationship of the subsystems or design parameters of the spacecraft; when the spacecraft requires multiple disciplines or multiple subsystems, an MDO problem can be embedded in this constraint.
Along with Table~\ref{tab_SLvar}, Eqs. \eqref{xdef}, \eqref{udef}, and \eqref{evdef} show the definitions and domains of commodity flow variables, spacecraft flow variables, and spacecraft design variables, respectively.
This integrated mission planning and spacecraft design problem results in a constrained MINLP problem, one of the most challenging optimization problem types to solve. Namely, this problem contains both discrete and continuous variables as well as both linear and nonlinear constraints. Specifically, the discrete variables represent the definition of the commodity flow and the number of spacecraft on the mission planning side of the problem. In addition, the nonlinearity appears in two ways: (1) the spacecraft design relationship in Eq.~\eqref{SL_vehicle_sizing}; (2) the quadratic terms in the mass balance constraint (Eq.~\ref{SL_constr1}) and concurrency constraint (Eq.~\ref{SL_constr2}) for mission planning (Note: both $\boldsymbol{e}_v$ and $u_{vijt}$ are variables). Fortunately, this second nonlinearity can be converted into an equivalent linear relationship through the big-M method, as explained in Ref.~\cite{chen2018MILP}, so that the nonlinearity only exists on the spacecraft design side of the problem.
Therefore, as a result, the problem contains two coupled problems: one for space mission planning which is linear with integer variables, and the other for spacecraft design which is nonlinear with continuous variables.
Our approach leverages this unique structure of the problem and proposes a new approach to solve this problem efficiently.
\section{Proposed Approach: Decomposition-Based Optimization with Augmented Lagrangian Coordination}
\label{OurALC}
Decomposition-based optimization is often used to decompose an MDO problem in terms of disciplines or subsystems. Leveraging the unique feature of the integrated space mission planning and spacecraft design problem, we apply this approach to decompose the large-scale MINLP problem (Fig.~\ref{fig:1a}) into coupled MIQP and NLP subproblems (Fig.~\ref{fig:1b}), each of which is significantly easier to solve with specialized solvers compared to the original MINLP problem. The space mission planning subproblem can be solved using a MIQP solver, and the spacecraft design subproblem can be solved using an NLP solver without any integer variables. The coupled subproblems are solved iteratively using the ALC-based coordination until convergence is reached. To enable the optimization without a user-defined initial guess, an automated and effective initial solution generation approach is also proposed.
\begin{figure}[h]
\centering
\begin{subfigure}[t]{1.8in}
\centering
\includegraphics[scale=0.5]{Figures/MINLP_SL.PNG}
\subcaption{All-in-one formulation}\label{fig:1a}
\end{subfigure}
\begin{subfigure}[t]{4.5in}
\centering
\includegraphics[scale=0.36]{Figures/ALC_proposed2.PNG}
\subcaption{Proposed decomposition-based formulation based on \cite{tosserams2007ALC}}\label{fig:1b}
\end{subfigure}
\caption{Solution strategy for integrated space mission planning and spacecraft design.}\label{fig:1}
\end{figure}
\subsection{Derivation of Decomposed Problems with Augmented Lagrangian Coordination}
\label{OriginalALC}
We first start with deriving the formulations of the decomposed problems with ALC. ALC tackles complex MDO optimization problems that are quasi-separable and thus can be decomposed into a set of coupled subproblems. ALC is attractive because of (1) its ability to break down our MINLP problem into MIQP and NLP problems; (2) its robust convergence property; and (3) its flexibility with the hierarchical structure of the problems. For an extensive discussion on ALC, refer to Ref.~\cite{tosserams2007ALC}.
The formulation for the quasi-separable MDO problem with $M$ subproblems is given as follows:
\begin{equation}
\begin{array}{rl}
\underset{\boldsymbol{y}, \boldsymbol{z}_{0}, \ldots, \boldsymbol{z}_{M-1}} {\text{ min }} \quad
\displaystyle{\sum_{k=0}^{M-1} f_{k}\left(\boldsymbol{y}, \boldsymbol{z}_{k}\right)} \\
\text { subject to }
\quad \boldsymbol{g}_{k}\left(\boldsymbol{y}, \boldsymbol{z}_{k}\right) \leq \boldsymbol{0} & k=0, \ldots, M-1 \\
\quad \boldsymbol{h}_{k}\left(\boldsymbol{y}, \boldsymbol{z}_{k}\right)=\boldsymbol{0} & k=0, \ldots, M-1
\end{array}
\end{equation}
\noindent where $\boldsymbol{y} \in \mathbb{R}^{n^{y}}$ indicates the shared variables, $\boldsymbol{z}_k \in \mathbb{R}^{n_{k}^{z}}$ indicates the local variables for subproblem $k$. The shared variables $\boldsymbol{y}$ can be common variables over multiple subproblems. $f_k:\mathbb{R}^{n_{k}} \mapsto \mathbb{R}$ indicates the local objective function, $\boldsymbol{g}_k$ and $\boldsymbol{h}_k$ indicate the equality and inequality constraints for each subproblem. The dimension of the total design variable $\boldsymbol{s}=\left[\boldsymbol{y}^{T}, \boldsymbol{z}_{0}^{T}, \ldots, \boldsymbol{z}_{M-1}^{T}\right]^{T}, \boldsymbol{s} \in \mathbb{R}^{n}$ is $n = n^{y} + \sum_{k=0}^{M-1} n_{k}^{z}$. The dimension of the local design variable is $n_j = n^{y} + n_{k}^{z}$.
The decomposition-based approach for this problem follows the following steps. First, we introduce the auxiliary variables and consistency constraints so that the local constraints, $\boldsymbol{g}_k$ and $\boldsymbol{h}_k$, are only dependent on the auxiliary variables $\boldsymbol{y}_k$ and independent of the shared variables $\boldsymbol{y}$.
\begin{equation}
\begin{alignedat}{2}
\min_{\boldsymbol{y}, \boldsymbol{y}_{0}, \boldsymbol{z}_{0}, \ldots, \boldsymbol{y}_{M-1}, \boldsymbol{z}_{M-1}} \quad &\sum_{k=0}^{M-1} f_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) & \\
\text { subject to }
\quad &\boldsymbol{g}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) \leq \boldsymbol{0} &\quad k=0, \ldots, M-1 \\
&\boldsymbol{h}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right)=\boldsymbol{0} &\quad k=0, \ldots, M-1 \\
&\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)=\boldsymbol{0} &\quad k=0, \ldots, M-1
\end{alignedat}
\end{equation}
\noindent With the consistency constraints $\boldsymbol{c}_{k}$, which ensures that the auxiliary variables $\boldsymbol{y}_{k}$ are the same as the shared variables $\boldsymbol{y}$, the shared variables are separated from the local variables while representing the same problem as the original one. Next, the relaxation of the consistency constraints is introduced with the local Lagrangian penalty function:
\begin{equation}
\begin{alignedat}{1}
\min_{\boldsymbol{y}, \boldsymbol{y}_{0}, \boldsymbol{z}_{0}, \ldots, \boldsymbol{y}_{M-1}, \boldsymbol{z}_{M-1}} \quad &\sum_{k=0}^{M-1} f_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) +\sum_{k=0}^{M-1} \phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)\right) \\
\text { subject to }
\quad &\boldsymbol{g}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) \leq \boldsymbol{0} \quad k=0, \ldots, M-1 \\
\quad &\boldsymbol{h}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right)=\boldsymbol{0} \quad k=0, \ldots, M-1
\end{alignedat}
\end{equation}
\noindent The augmented Lagrangian penalty function for subproblem $k$, $\phi_k$, is defined as follows.
\begin{equation}
\phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)\right)=\boldsymbol{v}_{k}^{T}\left(\boldsymbol{y}-\boldsymbol{y}_{k}\right)+\left\|\boldsymbol{w}_{k} \circ\left(\boldsymbol{y}-\boldsymbol{y}_{k}\right)\right\|_{2}^{2}
\end{equation}
\noindent where $\boldsymbol{v}$ is the vector of Lagrange multiplier estimates, and $\boldsymbol{w}$ is the vector of penalty weights. Here, $\circ$ represents the element-wise product of matrices or vectors, also known as the Hadamard product. By moving the consistency constraints into the local objective functions, the local subproblems can be completely separated.
The bi-level decomposition-based problem is now formulated by establishing the master problem above the subproblems. The master problem minimizes the penalty function and updates the shared variables $\boldsymbol{y}$. Note that even though the bi-level formulation is employed here, the ALC has the capability to handle multi-level hierarchical formulation as well.
(1) Master Problem
\begin{equation}
\min _ {\boldsymbol{y}} \quad \sum_{k=0}^{M-1} \phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)\right)
\end{equation}
(2) Subproblem $k$
\begin{equation}
\begin{aligned}
\underset{\boldsymbol{y}_k, \boldsymbol{z}_{k}} {\text{ min }} \quad &f_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) +\phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)\right) \\
\text { subject to }
\quad&\boldsymbol{g}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) \leq \boldsymbol{0} \\
&\boldsymbol{h}_{k}\left(\boldsymbol{y}_{k}, \boldsymbol{z}_{k}\right) =\boldsymbol{0}
\end{aligned}
\end{equation}
Adopting the above approach to our problem of the integrated space mission planning and spacecraft design with $N$ vehicle types, Fig.~\ref{fig:1b} represents the decomposition-based optimization architecture. We have one space mission planning subproblem (Subproblem $0$) and multiple spacecraft design subproblems (Subproblems $1,\ldots,N$), where $N$ is the number of spacecraft types; thus, we have $N+1$ subproblems in total (i.e., $M=N+1$). The shared variables among them include the vehicle design parameters $\boldsymbol{y} = [\boldsymbol{y}_1^T, \ldots, \boldsymbol{y}_N^T]^T$, where $\boldsymbol{y}_v = [m_{p_v}, m_{f_v}, m_{d_v}]^T$ for each vehicle $v$ where $m_p$, $m_f$, $m_d$ respectively represent payload capacity, propellant (fuel) capacity, and dry mass of the spacecraft.
First, the space mission planning problem ($P_0$ in Fig. \ref{fig:1b}) is different from the all-in-one formulation outlined in Section \ref{Problem Definition} with respect to the following two points: the nonlinear vehicle sizing constraint (Eq. \eqref{SL_vehicle_sizing}) is not included, and the quadratic penalty function is added to the objective function as Eq. \eqref{SubSL_obj} shows. Due to the quadratic objective function, this subproblem is a MIQP problem.
\begin{equation}
\begin{aligned}
\label{SubSL_obj}
\min_{\boldsymbol{x}_{vijt}, u_{vijt},\boldsymbol{y}_0} \quad &\sum_{t\in \mathcal{T}} \sum_{(v,i,j)\in \mathcal{A}} (\boldsymbol{a}_{vijt}^{T} \boldsymbol{x}_{vijt} + {a'}_{vijt}^{T} m_{d_{v}} u_{vijt}) + \phi_{0}\left(\boldsymbol{c}_{0}\left(\boldsymbol{y}, \boldsymbol{y}_{0}\right)\right) \\
\text{subject to} \quad & \text{Eqs.~\eqref{SL_constr1}--\eqref{SL_constr3} and \eqref{xdef}--\eqref{evdef} }\\
\text{where} \quad &
\boldsymbol{y} = [\boldsymbol{y}_1^T, \ldots , \boldsymbol{y}_N^T]^T \quad \text{and} \quad \boldsymbol{y}_v = [m_{p_v}, m_{f_v}, m_{d_v}]^T
\end{aligned}
\end{equation}
Next, for the spacecraft design subproblems ($P_v$ in Fig. \ref{fig:1b}), the penalty function is minimized and the vehicle sizing constraint ($m_{d_v} = \mathcal{F}(m_{p_v}, m_{f_v})$) is enforced. This subproblem contains various interacting subsystems and a hierarchical structure can be used to provide detailed subsystem-level design if needed. The subproblem for $v$-th type of vehicle can be expressed as Eq. \eqref{SizingSub}. Due to the nonlinear constraint, the subproblem is an NLP problem and can be solved by an NLP solver.
\begin{equation}
\begin{aligned}
\label{SizingSub}
\min_{\boldsymbol{y}_v} \quad &\phi_{v}\left(\boldsymbol{c}_{v}\left(\boldsymbol{y}, \boldsymbol{y}_{v}\right)\right) \\
\text{subject to} \quad &m_{d_v} = \mathcal{F}(m_{p_v}, m_{f_v}) \\
\text{where} \quad &
\boldsymbol{y}_v = [m_{p_v}, m_{f_v}, m_{d_v}]^T
\end{aligned}
\end{equation}
\subsection{Solution Algorithm and Iteration Scheme}
\label{iterscheme}
This subsection introduces the iterative solution algorithm for the decomposition-based algorithm introduced in Section \ref{OriginalALC}. The formulated decomposed optimization problems with ALC can be solved iteratively in two loops: the outer loop updates the augmented Lagrangian penalty parameters ($\boldsymbol{v}$, $\boldsymbol{w}$), while the inner loop solves the master problem and each subproblem to update the variables. The iteration continues until the convergence (i.e., all subproblems are consistent, or $\boldsymbol{c}_k$ is near zero, within a tolerance). The following describes the details of each loop.
For the updates for the outer loop, the solution from the inner loop is used~\cite{tosserams2007ALC}. Specifically, at $q$-th iteration, $\boldsymbol{v}$ is updated as follows:
\begin{equation}
\boldsymbol{v}^{q+1}=\boldsymbol{v}^q+2\boldsymbol{w}^q\circ\boldsymbol{w}^q\circ\boldsymbol{c}^q
\end{equation}
In addition, for $r$-th consistency constraint $c_r$, the corresponding penalty weight $w_r$ is updated as follows:
\begin{equation}
w^{q+1}_r=
\begin{cases}
w^q_r &\text{if} \quad |c^q_r|\leq\gamma_2|c^{q-1}_r|\\
\gamma_1 w^q_r &\text{if} \quad |c^q_r|>\gamma_2|c^{q-1}_r|
\end{cases}
\end{equation}
\noindent where $\gamma_1>1$ and $0<\gamma_2<1$. The initial penalty parameter values can take $\boldsymbol{v}^1 = \bm{0}$ and $\boldsymbol{w}^1 \approx \bm{1}$.
The updates for the inner loop is performed by alternating between solving the master problem and the subproblems with the fixed penalty parameters. While each subproblem can be solved using the specialized numerical optimizer for MIQP or NLP, the master problem can be solved analytically as follows.
\begin{equation}
\boldsymbol{y}= \underset{\boldsymbol{y}}{\operatorname{argmin}} \sum_{k=0}^{N} \phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{y}, \boldsymbol{y}_{k}\right)\right)=\frac{\sum_{k=0}^{N}\left(\boldsymbol{w}_{k} \circ \boldsymbol{w}_{k} \circ \boldsymbol{y}_{k}\right)-\frac{1}{2} \sum_{k=0}^{N} \boldsymbol{v}_{k}}{\sum_{k=0}^{N}\left(\boldsymbol{w}_{k} \circ \boldsymbol{w}_{k}\right)}
\end{equation}
For our problem, we make an additional heuristics-based modification to the master problem to facilitate the convergence. Namely, the aforementioned master problem updates all the shared variables at the same time at every iteration, but this approach does not work effectively in our problem. This is because, the space mission planning, with no knowledge of the constraints behind the spacecraft design, can return an aggressive or infeasible spacecraft design, which can deteriorate the convergence performance. Therefore, we propose to only update the spacecraft payload capacity and the propellant capacity in the master problem, while passing the spacecraft dry mass from the spacecraft design subproblem directly to the next iteration, as shown in Fig. \ref{ALC_2}. Mathematically, we separate the shared variables $\boldsymbol{y}$ into the regular shared variables $\boldsymbol{\alpha} = [m_{p_1}, m_{f_1}, \ldots, m_{p_N}, m_{f_N}]$ and the prioritized shared variables $\boldsymbol{\beta}=[m_{d_1}, \ldots, m_{d_N}]$ (i.e., $\boldsymbol{y} = [\boldsymbol{\alpha, \beta}]$), and only $\boldsymbol{\alpha}$ is updated in the master problem.
\begin{equation}
\label{ModMaster}
\begin{aligned}
\min _ {\boldsymbol{\alpha}} \quad &\sum_{k=0}^{N} \phi_{k}\left(\boldsymbol{c}_{k}\left(\boldsymbol{\alpha}, \boldsymbol{\alpha}_k\right)\right)
\end{aligned}
\end{equation}
Note that, in the space mission planning subproblem, the spacecraft dry mass remains a variable, not a fixed parameter, and is subject to the penalty function. It indicates that the resultant dry mass $\beta_0$ is not used in the entire optimization architecture but only used to facilitate the convergence of the whole optimization problem.
\begin{figure}[hbt!]
\centering
\includegraphics[width=.5\textwidth]{Figures/ALC_proposed.PNG}
\caption{Proposed decomposition-based optimization architecture with prioritized shared variables.}
\label{ALC_2}
\end{figure}
\subsection{Automatic Initial Solution Generation}
\label{guess}
For the above iterative algorithm to perform effectively, a good initial guess of the shared variable is necessary. Thus, there is a need to develop an automatic and effective method that does not require a user-defined initial guess. To this end, we propose to use the PWL approximation of the nonlinear optimization spacecraft design problem, and convert the entire MINLP into a MILP problem, which can be solved using a specialized solver \cite{chen2018MILP}. Although the PWL approximation does not necessarily return an optimal or even feasible solution to the original MINLP problem, the returned shared variables can be used as a good initial guess for the iterative approach.
Another advantage is that the MILP problem can be solved to the global optimum for the approximated nonlinear model \cite{chen2018MILP}. Thus, the MILP-based initial guess is not only automatically generated but also likely to be close to the nonlinear global optimum.
Specifically, in our problem, nonlinearity exists on the spacecraft design side of the problem. Thus, we choose a series of equally-spaced "mesh" points over the feasible ranges of the spacecraft design parameters and use them as breakpoints for the PWL function generation. Note that since the dry mass is an (implicit) function of the payload capacity and propellant capacity, we only used the latter two for breakpoint generation. The breakpoint increment (or the number of breakpoints) is a key hyperparameter; a smaller increment or more breakpoints would lead to a more accurate initial guess, but it will also require a longer computational time.
\section{Case Study: Human Lunar Exploration Campaign}
\label{CaseStudy}
To demonstrate the effectiveness of the proposed approach, we perform a demonstration case study and compare our approach with the state-of-the-art method. We first introduce the case study settings, followed by the results and the computational performance analysis.
\subsection{Case Study Settings}
A human lunar exploration with two missions is considered here for the case study.
The mission network model, parameters, commodity demand and supply used in this case study are presented in Fig. \ref{LunarMission}, Table \ref{MisisonParameter}, and Table \ref{LunarMissionDemand}, respectively. Note that only one type of spacecraft, which is a single-stage lunar lander, is considered for simplicity. It means that the lander sizing constraint is applied to other vehicles such as in-space transfer vehicles. As landers are typically heavier than other spacecraft due to their landing structure, the optimization result might represent a conservative design. In addition, in-situ resource utilization (ISRU) is also considered as an option in the formulation, although it is never chosen by the optimizer in this case study due to the considered short time horizon.
\begin{figure}[hbt!]
\centering
\includegraphics[width=.6\textwidth]{Figures/LunarMission.png}
\caption{Lunar campaign network model \cite{chen2018MILP}.}
\label{LunarMission}
\end{figure}
\input{Tables/MissionParameter}
\input{Tables/LunarMissionDemand}
The subsystem-level spacecraft model used as the spacecraft design constraint in Eq. \eqref{SL_vehicle_sizing} in this study is developed by the least square curve fitting to the data from the lunar lander design database in Ref.~\cite{isaji2018landerdata, isaji2020lander}. The following set of equations shows the spacecraft model used in the case study.
\input{Tables/sizing_result}
Note that, beyond the payload capacity and propellant capacity, there are some additional parameters in these equations: $n_{stg}$ is the number of stages (either 1 or 2), $\rho_f$ is the propellant density in kg/m\textsuperscript{3}, $t_{mis}$ is the surface time of the lunar mission in days, $n_{crew}$ is the number of crew, $c_{misc}$ is the miscellaneous mass fraction. The miscellaneous mass fraction $c_{misc}$ represents how much of the dry mass is categorized as the miscellaneous mass. It can range from 0 to 0.15, meaning 0\% to 15\% of the dry mass is the miscellaneous mass. The higher $c_{misc}$, the heavier and more conservative the vehicle design becomes. All mass properties are defined in kg.
As shown in Eq.~\eqref{sizing}, the model captures the subsystem-level interactions to return the relationship between the payload capacity, propellant capacity, and dry mass of the spacecraft. Particularly, the subsystem interactions are captured \emph{through} the dry mass. For instance, an increase in any subsystem mass will raise the dry mass. Since each subsystem mass is dependent on the dry mass, their mass should increase as well, which will further raise the dry mass. The 'balanced' dry mass with such subsystem circular references can be found by solving Eq. \eqref{sizing} for the dry mass, $m_d$. More details on this model can be found in Appendix A.
In the case study, the computational time for all problems is measured on a platform with Intel Core i7-10700 (8 Core at 2.9 GHz). In the proposed decomposition-based method, Gurobi 9.1 solver \cite{gurobi} is used for the initial MILP problem and MIQP subproblem, and IPOPT \cite{IPOPT} is chosen for the NLP subproblem.
\subsection{Optimization Results by the Proposed Decomposition-Based Formulation}
\label{ALC_CS}
This subsection introduces the optimization results by the proposed decomposition-based formulation. Since the performance of the proposed method is dependent on the breakpoint increment (or the number of breakpoints) for the PWL approximation of the MILP-based initial solution generation, five different increments are tested. The results are shown in Table \ref{ALCresutls}.
Although the optimizer's computational time involves some randomness depending on the individual problems, there are several general trends that can be observed. First, when the increment is too large (too few breakpoints, e.g., 10,000 kg increment with 13 mesh points), the initial solution quality becomes poor, and thus the final solution IMLEO is also poor.
Second, the computational time to solve the initial MILP problem rapidly increases when the increment is too small (too many breakpoints, e.g., 625 kg increment with 1,595 mesh points), resulting in a long total computational time.
In summary, we can observe the expected trend that a smaller increment (more breakpoints) leads to a better initial guess at the cost of computational time.
Thus, the most efficient strategy is to use an increment that can generate a reasonably accurate initial solution and leave the rest to the proposed decomposition-based optimization.
Although this hyperparameter needs to be chosen for the proposed algorithm, it is worth noting that the computational performance is not very sensitive against the choice of its value except for the extreme cases.
Also, note that, theoretically speaking, if we reduce the increment to zero (an infinite number of breakpoints), the solution would match with the global optimum; however, this is impractical as it requires infinite computational time. The proposed decomposition-based formulation can take the reasonable approximate solution by PWL formulation and offer a better computational efficiency to achieve a high-quality solution.
\input{Tables/ALCresults}
\subsection{Benchmarking with State-of-the-Art Method: Embedded Optimization}
\label{embedded_CS}
Although our formulation of the integrated mission planning and subsystem-level spacecraft design has not been directly tackled in the literature, we can extend straightforwardly a state-of-the-art approach for a similar problem as a benchmark to evaluate our newly proposed method.
The identified state-of-the-art approach is the embedded optimization method by Taylor \cite{taylor2007phd}, which was demonstrated to be more efficient than directly solving the original integrated MINLP problem using a global optimizer.
With the embedded optimization method, the spacecraft variables are separated from the whole problem and determined by a metaheuristics algorithm. At every iteration, the metaheuristics algorithm picks the payload and propellant capacity of $N$ vehicles, and the corresponding spacecraft dry mass is then calculated as a function of them, following the spacecraft subproblem procedure.
After obtaining the feasible vehicle design, these values are fed to the space mission planning problem, which is solved by the MILP optimizer. Unlike the all-in-one formulation, the vehicle parameters are fixed within the space mission planning part. Then, the corresponding objective function value is returned to the metaheuristic optimizer for the evaluation for the next iteration.
As a result, the metaheuristics only handles an optimization problem with $2N$ variables (i.e., the payload capacity and propellant capacity for each spacecraft), where the evaluation of the constraints and the determination of the remaining variables are handled by the embedded MILP solver. The problem to be optimized by the metaheuristic solver is expressed as Eq. \eqref{heuristics}.
\begin{equation}
\label{heuristics}
\begin{aligned}
\min_{\boldsymbol{\alpha}} \quad &\text{IMLEO}(\boldsymbol{\alpha}, \mathcal{F}(\boldsymbol{\alpha}))\\
\text{where} \quad &\boldsymbol{\alpha} = [m_{p_1}, m_{f_1}, \ldots, m_{p_N}, m_{f_N}], \quad\boldsymbol{\alpha} \in \mathbb{R}^{2N}
\end{aligned}
\end{equation}
Since the performance of the embedded optimization would depend on the choice of the metaheuristics algorithm, three different metaheuristics algorithms are tested: the extended Ant Colony Optimization (ACO) \cite{AntColony}, the Genetic Algorithm (GA) \cite{GA}, and the Particle Swarm Optimization (PSO) \cite{PSO}. The optimization is terminated when a predefined number of generations are populated; different termination generation numbers are tested for each algorithm to explore the tradeoff between the computational time and accuracy. Furthermore, due to the random nature of the metaheuristic optimizers, the optimization is run three times with the same algorithm and generation number.
Table \ref{EmbeddedHighlight} shows each algorithm's best results with 10, 50, and 100 generations. Note that 'inf' indicates that no feasible solutions can be found. The complete set of results is given in Table \ref{EmbeddedFullTable2} in Appendix. In many cases, especially with low numbers of generations, the optimizers fail to even reach a feasible IMLEO solution. As the number of generations increases, the computational time increases, a feasible solution is more likely to be found, and the solution tends to be better, although such trends might not always hold due to the random nature of the metaheuristic algorithms.
\input{Tables/embedded_highlight}
\subsection{Performance Analysis and Discussions}
As we compare the optimization results by the proposed decomposition-based optimization method in Table \ref{ALCresutls} and the state-of-the-art embedded optimization method in Table \ref{EmbeddedHighlight}, it is clear that the proposed method can achieve a better solution (lower IMLEO) at a less computational time. Even with respect to the best embedded optimization case in Table \ref{EmbeddedHighlight}, which is the PSO case with 100 generations (IMLEO 677,221kg, computational time 793.9s), the proposed decomposition-based optimization can achieve a better solution with a substantially shorter computational time (676,930kg, computational time 16.27s). Note that the computational time by the two methods is measured without any parallelization under a fair setting.
Beyond the numerical solution comparison, one substantial advantage of the proposed method is its deterministic and thus repeatable performance. This is in contrast to the metaheuristics that returns different results every run, varying from near-optimal results to infeasible results. The proposed formulation can consistently achieve better solutions than those that the metaheuristics optimizers would find "by chance."
Another advantage of the proposed formulation is that more complicated spacecraft design problems, such as models with more constraints or even MDO problems, can be integrated in a scalable way. Namely, if more subproblems are considered, they can be parallelized to further reduce the computational time. When complex MDO problems are included as subproblems, a multi-level hierarchical ALC formulation can also be utilized \cite{ATCextended}.
Overall, the case study demonstrates that the higher computational performance of the proposed method compared to the state-of-the-art embedded optimization method. The proposed formulation can consistently obtain a better solution in a shorter computational time. It also has greater room for potential improvement and extension, such as parallelization and MDO subproblem integration.
\section{Conclusion}
\label{conclusion}
This paper tackles the challenging problem of integrated space mission planning and spacecraft design. The all-in-one formulation is presented as an MINLP problem, and an efficient solution approach is developed leveraging the unique structure of the problem and following the philosophy of MDO. Namely, the all-in-one MINLP problem is decomposed into the space mission planning subproblem (MIQP) and the spacecraft design subproblem(s) (NLP) so that they can be solved iteratively using the ALC approach to find the optimal solution for the original MINLP problem. Furthermore, an automatic and effective approach for finding an initial solution for this iterative process is proposed leveraging a piecewise linear approximation of the nonlinear vehicle model, so that no user-defined initial guess is needed.
The case study results demonstrate that the proposed method achieves a better result in less time compared to the state-of-the-art embedded optimization method. The combination of the unique problem structure, the iterative algorithms for shared variables, and the efficient initial solution generation method leads to this computational efficiency even without parallelization. The parallelizable nature of the algorithm is expected to make the proposed method even more advantageous for large-scale problems. Due to the flexibility of the ALC method, the proposed formulation can also integrate more complex vehicle design models, which is left for future work.
\section*{Acknowledgments}
This material is based upon work supported by the National Science Foundation under Grant No. 1942559.
\section*{Appendix A: Spacecraft Design Model}
\label{App_A}
This appendix provides more details on the parametric sizing model for the spacecraft used in the case study.
In the considered model, the subsystems of single-stage landers and their relations to the dry mass are defined as Eq. \eqref{drymass_eq}.
\begin{equation}
\label{drymass_eq}
m_{d}=\sum m_{sub} = m_{str}+m_{prop}+m_{power}+m_{avi}+m_{ECLSS}+m_{misc}
\end{equation}
\noindent where $m_{sub}$ indicates the mass of subsystem. $m_{str}$ indicates the structure and thermal protection system (TPS), which includes all subsystems that are attached to support or connect other components. This is not limited but includes landing legs and truss, TPS for the reentry to the earth, and docking mechanism. $m_{prop}$ is the propulsion system, such as propellant tanks, reaction control system (RCS), and hardware of engines. $m_{power}$ is the power system, which contains batteries, fuel cells, solar panels, or other electrical systems. $m_{avi}$ indicates the avionics, and $m_{ECLSS}$ indicates the environmental and life control system (ECLSS) that supports the crew's lives such as consumables (food, water, air) or related piping and tankage. Finally, we also consider other miscellaneous required components, expressed as $m_{misc}$. Through the dry mass, each subsystem interacts with every other subsystem, and this relation is visualized in Fig.~\ref{N2} as an N\textsuperscript{2} diagram.
\begin{figure}[hbt!]
\centering
\includegraphics[width=.45\textwidth]{Figures/N2diagram.PNG}
\caption{Relationship of domains in a single-stage lunar lander.}
\label{N2}
\end{figure}
For the defined subsystems, mass estimation relationships (MERs) are developed as functions of payload capacity, propellant capacity, propellant type $f$, and some other known parameters. If the propellant type is fixed, the subsystems MERs and dry mass are dependent on the payload capacity and propellant capacity only, and thus serves as the vehicle sizing constraint (Eq. \eqref{SL_vehicle_sizing}, $m_d = \mathcal{F}(m_p, m_f$)). Each subsystem MER is developed by the least square curve fitting to the data from the lunar lander design database in Ref.~\cite{isaji2018landerdata, isaji2020lander}, which includes both existing and elaborated conceptual design.
The form of each subsystem's MER is manually determined to be a sufficiently simple yet accurate form.
The resultant MERs are shown in Eq. \eqref{sizing}.
Table \ref{tab_SizingVar} summarizes the independent variables, the $R^2$ values for curve fitting, number of data points used for curve fitting ($N_{data}$), average errors against the data points, and the maximum errors. Note that only a small number of data points are used for the propulsion system MER since two-stage lander data are excluded as their propulsion systems with staging are too distinct from those of single-stage ones. One may also see that relatively poor correlations are obtained for the power systems and avionics mass as they simply might not be strong functions of the dry mass or vehicle size. However, since they typically account for small portions of the dry mass, the poor correlation does not have a significant effect on the validation process.
The limitation of this sizing model should also be noted. Because the MERs are developed from the existing data points, a solution for vehicles that are significantly heavier than the ones in the database would either be a low-fidelity model or infeasible. In other words, $m_d$ that satisfies Eq. \eqref{drymass_eq} might not exist for certain weight classes. Specifically, the upper bound of the dry mass is approximately 23,000 kg. When $t_{mis}$ is 3 days, $n_{crew}$ is 4, $c_{misc}$ is 0.05, and the propellant is LH2/LOX, the upper bound are found at 500 kg payload and 75,500 kg propellant, or at 10,000 kg payload and 45,500 kg propellant.
\begin{table}[h]
\caption{\label{tab_SizingVar} Summary of subsystem MERs }
\centering
\begin{tabular}{p{0.17\textwidth}p{0.09\textwidth}p{0.2\textwidth}p{0.08\textwidth}p{0.08\textwidth}p{0.1\textwidth}p{0.1\textwidth}}
\hline\hline Subsystem & Notation & Independent Variables & $R^2$ & $N_{data}$ & Avg. Error & Max. Error\\
\hline
Structure + TPS & $m_{str}$ & $m_{d},n_{stg},m_p$ & 0.9254 & 17 & 7.379\% & 24.31\% \\
Propulsion System & $m_{prop}$ & $m_d, m_p, \rho_p$ & 0.9279 & 8 & 7.429\% & 11.16\%\\
Power System & $m_{power}$ & $m_d$ & 0.7182 & 13 & 16.24\% & 36.68\%\\
Avionics & $m_{avi}$ & $m_{power}(m_d), t_{mis}$ & 0.6204 & 22 & 36.42\% & 75.94\%\\
ECLSS & $m_{ECLSS}$ & $m_d, n_{crew}, n_{stg}, t_{mis}$ & 0.9293 & 12 & 11.93\% & 38.09\% \\
Miscellaneous & $m_{misc}$ & $m_d$ & - & - & - & - \\
\hline\hline
\end{tabular}
\end{table}
\section*{Appendix B: Summary of the Embedded Optimization Results}
Table \ref{EmbeddedFullTable2} includes the full results obtained from the embedded optimization.
\input{Tables/embedded_highlight2}
\clearpage
\bibliography{sample}
\end{document}
| 5,468
|
\begin{document}
\title[A Class of Distal Functions$\cdots$ ]{A Class of distal Functions on Semitopological
Semigroups}
\author{A. Jabbari }
\address{ Department of Mathematics, Ferdowsi University of
Mashhad, P. O. Box 1159, Mashhad 91775,
Iran}\email{shahzadeh@math.um.ac.ir}
\author{H.\ R.\ E.\ Vishki}
\address{Department of Mathematics, Ferdowsi University of Mashhad\\
P. O. Box 1159, Mashhad 91775, Iran; \newline Centre of Excellence
in Analysis on Algebraic Structures (CEAAS), Ferdowsi University of
Mashhad, Iran.} \email{vishki@ferdowsi.um.ac.ir}
\subjclass[2000]{ 54H20, 54C35, 43A60}
\keywords{semitopological semigroup, distal function, strongly almost periodic
function, semigroup compactification, m-admissible subalgebra.}
\date{}
\dedicatory{}
\commby{}
\begin{abstract}
The norm closure of the algebra generated
by the set $\{n\mapsto {\lambda}^{n^k}:$ $\lambda\in{\mathbb {T}}$
and $k\in{\mathbb{N}}\}$ of functions on $({\mathbb {Z}}, +)$ was
studied in \cite{S} (and was named as the Weyl algebra). In this
paper, by a fruitful result of Namioka, this algebra is generalized
for a general semitopological semigroup and, among other things, it
is shown that the elements of the involved algebra are distal. In
particular, we examine this algebra for $({\mathbb {Z}}, +)$ and
(more generally) for the discrete (additive) group of any countable
ring. Finally, our results are treated for a bicyclic semigroup.
\end{abstract}
\maketitle
\section{Introduction} Distal functions on topological groups were extensively studied by Knapp ~\cite{K}.
The norm closure of the algebra generated by the set $F=\{n\mapsto
{\lambda}^{n^k}$: $\lambda\in{\mathbb {T}}$ and $k\in{\mathbb{N}}\}$
of functions on $({\mathbb {Z}}, +)$ was called the Weyl algebra by
E. Salehi in \cite{S}. Knapp, ~\cite{K}, showed that all of the
elements of $F$ are distal on $({\mathbb {Z}}, +)$. Also Namioka
~\cite[Theorem 3.6]{N} proved the same result by using a very
fruitful result (\cite[Theorem 3.5]{N}) which played an important
role for the construction of this paper. By the above mentioned
results of Knapp and Namioka, all elements of the Weyl algebra are
distal, however it dose not exhaust all distal functions on
$({\mathbb {Z}}, +)$; \cite[Theorem 2.14]{S}. In this paper, we
generalize the notion of Weyl algebra to an arbitrary
semitopological semigroup and also we show that all elements of the
involved algebra are distal. In particular, our method provides a
convenient way to deduce a result of M. Filali~\cite{Fi} on the
distality of the functions $\chi(q(t))$, where $\chi$ is a character
on the discrete additive group of a (countable) ring $R$ and $q(t)$
is a polynomial with coefficients in $R$.
\section{Preliminaries}
For the background materials and notations we follow Berglund et al. \cite{BJM} as
much as possible. For a semigroup $S$, the right translation $r_{t}$ and the
left translation $l_{s}$ on $S$ are defined by
$r_{t}(s)=st=l_{s}(t)$, ($s, t\in S$). The semigroup $S$, equipped with a topology, is said to be
right topological if all of the right translations are continuous,
semitopological if all of the left and right translations are
continuous. If $S$ is a right topological semigroup then the set
$\Lambda(S)=\{s\in S: l_{s}$ is continuous$\}$ is called the
topological centre of $S$.\\
Throughout this paper, unless otherwise stated, $S$ is a
semitopological semigroup. The space of all bounded continuous
complex valued functions on $S$ is denoted by $C(S)$. For
$f\in{{C}}(S)$ and $s\in S$ the right (respectively, left)
translation of $f$ by $s$ is the function $R_{s}f=f\circ r_{s}$
(respectively,
$L_{s}f=f\circ l_{s}$).\\
A left translation invariant unital $C^\ast$-subalgebra $F$ of $C(S)$
(i.e., $L_sf\in F$ for all $s\in S$ and $f\in F$) is called $m$-admissible if the function
$s\mapsto (T_\mu f)(s)=\mu(L_sf)$ belongs to $F$ for all $f\in F$
and $\mu\in S^F$(=the spectrum of F). If $F$ is $m$-admissible then $S^F$ under the
multiplication $\mu\nu=\mu\circ T_\nu$ ($\mu, \nu\in S^F$),
furnished with the Gelfand topology is a compact Hausdorff right topological semigroup and it makes $S^F$ a
compactification (called the $F$-compactification) of $S$.\\
Some of the usual $m$-admissible subalgebras of $C(S)$, that are
needed in the sequel, are the left multiplicatively continuous,
strongly almost periodic and distal functions on $S$. These are
denoted by $LMC(S)$, $SAP(S)$ and $D(S)$, respectively. Here and
also for other emerging spaces when there is no risk of confusion,
we have suppressed the letter $S$ from the notation.\\
The interested reader may refer to \cite {BJM} for ample information about these $m$-admissible subalgebras and the properties of their corresponding compactifications.
\section{Main results}
The idea of defining our new algebras $W_k$ and
$W$, in the form given
below, came from a nice result of Namioka~\cite[Theorem 3.5]{N}.\\
Let $\Sigma=\{T_{\mu}
:LMC(S)\rightarrow LMC(S); \mu\in S^{{LMC}}\}$. Let $F_0$ be the set
of all constant functions of modulus 1. For every $k\in\mathbb{N}$
assume that we have defined $F_i$ for $i=1,2,\ldots,k-1$ and define
$F_k$ by
\[F_{k}=\{f\in
LMC : |f|=1 {\rm \ and \ for \ every \ } \sigma\in\Sigma,
\sigma(f)=f_{\sigma}f, {\rm \ for \ some \ } f_{\sigma}\in
F_{k-1}\};\] It is clear from definitions that
${{F}}_{k}\subseteq{{F}}_{k+1}$, for all $k\in\mathbb{N}\cup\{0\}$.
Let ${{W}}_{k}$ and $W$ be the norm closure of the algebras
generated by ${{F}}_{k}$ and ${\bigcup}_{k\in\mathbb {N}}^{} F_k$ in
${{C}}(S)$, respectively; then trivially
${{W}}_{k}\subseteq{{W}}_{k+1}\subseteq W$. Hence, $W$ is the
uniform closure of the algebra $\bigcup_{k\in{\mathbb{N}}}W_k$. It
is also readily verified that $W$ is the direct limit of the family
$\{W_i:~i\in{\mathbb{N}}\}$ (ordered by inclusion, and with the
inclusion maps as morphisms). \noindent From now on, we assume that
$k\in\mathbb{N}$ is arbitrary. We leave the following simple
observations without proof.
\begin{proposition}\label{I}
$(i)$ For every $f\in F_k$ and $\sigma\in\Sigma$ the function $f_{\sigma}$ with the
property $\sigma(f)=f_{\sigma}f$ is unique.
$(ii)$ For every $f, g\in
{{F}}_{k}$ and $\sigma\in\Sigma$,
$(fg)_{\sigma}=f_{\sigma}g_{\sigma}$. In particular,
${{F}}_{k}$ is a multiplicative subsemigroup of $LMC$.
$(iii)$ $F_k$ is conjugate closed; in other words, if $f\in F_k$ then $\overline{f}\in F_k$.
$(iv)$ $F_k$ contains the constant functions.
\end{proposition}
\begin{lemma}\label{II} The set $F_k$ is left translation invariant and it is also invariant under $\Sigma$; in other words, $L_S(F_k)\subseteq F_k$ and $\Sigma(F_k)\subseteq F_k$.
\end{lemma}
\begin{proof} A direct verification reveals that $F_k$ is left translation invariant. For the invariance under $\Sigma$ let $f\in {{F}}_{k}$ and $\sigma\in\Sigma$, the equality ${\sigma}(f)=f_{\sigma}f$ for some $f_\sigma\in F_{k-1}$ implies that $|\sigma(f)|=1$ and so for every $\tau\in\Sigma$, \[\tau(\sigma(f))=\tau(f_\sigma f)=\tau(f_\sigma)\tau(f)=(({f_\sigma})_\tau f_\sigma)(f_\tau f)=({(f_\sigma})_\tau f_\tau)(f_\sigma f)=(({f_\sigma})_\tau f_\tau)\sigma(f).\] Since $({f_\sigma})_\tau f_\tau\in F_{k-1}$ we have $\sigma(f)\in F_k$; in other words $\Sigma(F_k)\subseteq F_k$, as required.
\end{proof}
\begin{lemma}\label{III} All
elements of ${{F}}_{k}$ remain fixed under the
idempotents of $\Sigma$.
\end{lemma}
\begin{proof} It is easily seen that the result holds for $k=1$. Assume that $k>1$ and that the result holds for $k-1$. Let $f\in
F_{k}$ and let $\varepsilon\in\Sigma$ be an idempotent, then
$\varepsilon(f)=f_{\varepsilon}f$ for some $f_{\varepsilon}\in
F_{k-1}$. Therefore
\[f_{\varepsilon}f=\varepsilon(f)={\varepsilon}^2
(f)={\varepsilon}({\varepsilon}(f))={\varepsilon}(f_{\varepsilon}f)=\varepsilon(f_\varepsilon)\varepsilon(f)=f_\varepsilon(f_\varepsilon
f)={f_{\varepsilon}}^2f;\] hence ${f_{\varepsilon}}=1$ and
${\varepsilon}(f)=f$, as claimed.
\end{proof}
\begin{lemma}\label{IV}
${{F}}_{k}\subseteq {{D}}$ .
\end{lemma}
\begin{proof} Let $f\in
{{F}}_{k}$. To show that $f\in{{D}}$, using \cite[Lemma 4.6.2]{BJM}, it is
enough to show that $\varepsilon\sigma(f)=\sigma(f)$ for each
$\sigma\in\Sigma$ and each idempotent $\varepsilon$ in
$\Sigma$. By Lemma~\ref{II}, $\sigma(f)\in
{{F}}_{k}$, so that Lemma~\ref{III} implies that
$\varepsilon(\sigma(f))=\sigma(f)$, as required.
\end{proof}
Using parts $(iii)$ and $(iv)$ of Proposition~\ref{I}, $W_k$ and $W$
are unital $C^*-$subalgebras of $C(S)$. The following result shows
that these are indeed $m-$admissible subalgebras of $D$.
\begin{theorem}\label{iV} For every semitopological semigroup $S$, ${{W}}_{k}$ and ${W}$ are
$m$-admissible subalgebras of $D(S)$.
\end{theorem}
\begin{proof} For the $m-$admissibility of $W_k$ it is enough to show that it is left translation invariant and also invariant under $\Sigma$.
Let $\langle{F_k}\rangle$ be the algebra generated by $F_k$.
Lemma~\ref{II} implies that $L_S(\langle{F_k}\rangle)\subseteq
\langle{F_k}\rangle$ and also $\Sigma(\langle{F_k}\rangle)\subseteq
\langle{F_k}\rangle$. For every $f\in{{W}}_{k}$ there exists a
sequence $\{f_n\}\subseteq \langle{F_k}\rangle$ which converges (in
the norm of $C(S)$) to $f$. Let $\sigma\in\Sigma$ and $s\in S$, then
the inequalities $\|L_s(f_n)-L_s(f)\|\leq\|f_{n}-f\|$ and
$\|\sigma(f_n)-\sigma(f)\|\leq\|f_{n}-f\|$ imply that
$L_s(f_{n})\rightarrow L_s(f)$ and $\sigma(f_n)\rightarrow
\sigma(f)$, respectively. Since for each $n\in \mathbb {N}$,
$L_s(f_n)$ and $\sigma(f_n)$ lie in $\langle{{{F}}_{k}}\rangle$, we
have $L_s(f)\in W_k$ and also $\sigma(f)\in{{W}}_{k}$. It follows
that ${{W}}_{k}$ is $m$-admissible. A similar argument may apply for
the $m-$admissibility of $W$. The fact that $W_k$ and $W$ are
contained in $D$ follows trivially from Lemma~\ref{IV}.
\end{proof}
The next result gives $S^{{W}}$ in terms of the subdirect product of
the family $\{S^{{{W}}_{k}}: k\in{\mathbb {N}}\}$. For a full
discussion of the subdirect product of compactifications one may
refer to \cite[Section 3.2]{BJM}.
\begin{proposition} The compactification $S^{{W}}$ is the subdirect product of the family
$\{S^{{{W}}_{k}}: k\in{\mathbb
{N}}\}$; in symbols, $S^{{W}}=\bigvee\{S^{{{W}}_{k}}: k\in{\mathbb {N}}\}$.
\end{proposition}
\begin{proof} The family (of homomorphisms)
$\{\pi_k:S^{{W}}\rightarrow S^{{{W}}_{k}}; k\in{\mathbb {N}}\}$,
where for each $\mu\in S^{{W}}$, $\pi_k(\mu)=\mu |_{{{W}}_{k}}$,
separates the points of $S^{{W}}$, because for given $\mu,\nu\in
S^W$ with $\mu\neq\nu$ (on $W$) one has $\mu\neq\nu$ of
$F=\bigcup_{k\in{\mathbb{N}}}F_k$, hence there exists a natural
number $j$ and an element $f\in F_j$ such that $\mu(f)\neq\nu(f)$.
Therefore $\mu |_{{{W}}_{j}}\neq\nu |_{{{W}}_{j}}$, that is
$\pi_j(\mu)\neq\pi_j(\nu)$. Now the conclusion follows from
\cite[Theorem 3.2.5]{BJM}.
\end{proof}
\begin{proposition}\label{V}
$(i)$ For every abelian semitopological semigroup $S$, $SAP(S)\subseteq
W_k(S)$.\\
\indent $(ii)$ For every abelian semitopological semigroup $S$ with a left
identity, ${{W}}_1(S)={SAP(S)}$.
\end{proposition}
\begin{proof} $(i)$ Since $S$ is abelian ${{SAP}}(S)$ is the
closed linear span of the set of all continuous characters on $S$; see \cite[Corollary 4.3.8]{BJM}.
Let $f$ be any continuous character on $S$ and let $\sigma\in\Sigma$. Then there exists a net $\{s_\alpha\}$ in $S$ such that $\sigma(f)=\lim_\alpha R_{s_\alpha}f$. By passing to a subnet, if necessary, we may assume that $f(s_\alpha)$ converges to some $\lambda_\sigma\in\mathbb{T}$. Therefore for every $s\in S$, $\sigma(f)(s)=\lim_\alpha R_{s_\alpha}f(s)=\lim_\alpha f(ss_\alpha)=f(s)\lim_\alpha f(s_\alpha)=f(s)\lambda_\sigma$. Hence $\sigma(f)=\lambda_\sigma f$ or equivalently $f\in F_1$.
$(ii)$ By part $(i)$ it is enough to show that ${{W}}_1\subset{{SAP}}$. Indeed we are going to show that $F_1\subset{{SAP}}$; for this end, let $f\in{{F}}_1$ and let $s\in S$, then
$R_sf=\lambda_s f$ for some $\lambda_s$ in ${\mathbb{T}}$. Let $e$
be a left identity of $S$, then for each $s$ in $S$,
$f(s)=R_sf(e)=\lambda_s f(e)$. Let $h=\frac{1}{f(e)}f$, then $h$ is
a continuous character on $S$. But $f=f(e)h$, now using the fact
that ${{SAP}}$ is the closed linear span of continuous characters
of $S$ we have $f\in{{SAP}}$, as required.
\end{proof}
As a consequence of the latter result we have
\begin{corollary} For any compact abelian
topological group $G$, $W_k(G)=W(G)={{C}}(G)$.
\end{corollary}
\begin{proof} Since for every compact topological group
$G$, ${{SAP}}(G)={{C}}(G)$, \cite[Theorem 4.3.5]{BJM}, the result follows from
the last proposition.
\end{proof}
\section{Examples}
{\bf Example (a).} Here we restrict our discussion to the
discrete group $(\mathbb{Z}, +)$ and examine $W$ and $W_k$ for this
particular case, which were studied extensively by Salehi in
~\cite{S}. Note that although we would deal with countable discrete
rings in part (b), but the proofs on ${\mathbb{Z}}$ are more
interesting and characterizations of $F_k({\mathbb{Z}})$ are more
explicit. We commence with the next key lemma which characterizes
$F_k$ in $l^\infty(\mathbb{Z})$.
\begin{lemma} The set ${{F}}_{k}({\mathbb {Z}}, +)$ is the
(multiplicative) sub-semigroup of $l^{\infty}({\mathbb {Z}})$
generated by the set $\{n\mapsto {\lambda}^{n^i}:$
$\lambda\in{\mathbb {T}}$,
$i=0, 1,..., k\}$.
\end{lemma}
\begin{proof} For each $k\in{\mathbb{N}}$, let
$A_k$ denote the multiplicative sub-semigroup of
$l^{\infty}({\mathbb {Z}})$ generated by the set $\{n\mapsto
{\lambda}^{n^i}$, $\lambda\in{\mathbb {T}}$ and $i=0, 1,..., k\}$.
For $k=1$ a direct verification reveals that $A_1\subseteq F_1$; for
the reverse inclusion let $f\in F_1$. Then for some
$\lambda\in{\mathbb {T}}$, $R_{1}f={\lambda}f$, hence
$f(1)=R_{1}f(0)={\lambda}f(0)={\lambda}\lambda_1$, in which
$\lambda_1=f(0)$. Also
$f(2)=R_{1}f(1)={\lambda}f(1)={\lambda}^{2}\lambda_1$, by induction
it is easily proved that for each $n\in{\mathbb {N}}$,
$f(n)={\lambda}^{n}\lambda_1$. Let $R_{-1}f={\beta}f$, where
$\beta\in{\mathbb{T}}$, then $f(-1)=R_{-1}f(0)={\beta}f(0)$. But
$f(1)=R_{-1}f(2)={\beta}f(2)={\beta}{\lambda}^{2}\lambda_1$,
therefore ${\lambda}\lambda_1={\beta}{\lambda}^{2}\lambda_1$, hence
$\beta={\lambda}^{-1}$ and for all $n\in{\mathbb {Z}}$,
$f(n)={\lambda}^{n}\lambda_1$. Thus
$f\in A_1$, and so $F_1=A_1$.\\
Let $k\geq 2$ and assume that $A_{k-1}=F_{k-1}$. Let $n\in{\mathbb
{Z}}$ and $\lambda\in{\mathbb {T}}$ and let $f\in A_k$ and assume
(without loss of generality) that $f(n)={\lambda}^{n^{k}}$, then for
given $\sigma={\lim}_{\alpha}m_{\alpha}\in\Sigma$ we have
$\sigma(f)(n)={\lim}_{\alpha}R_{m_{\alpha}}f(n)={\lim}_{\alpha}{\lambda}^{(n+m_{\alpha})^{k}}=
f(n)f_{\sigma}(n)$, in which
$f_{\sigma}(n)={\mu}_{1}^{n^{k-1}}{\mu}_{2}^{n^{k-2}}...{\mu}_{k-1}^{n}{\mu}_{k}$,
where (by going through a sub-net of $\{m_\alpha\}$, if necessary)
${\mu}_{i}={\lim}_{\alpha}{\lambda}^{({}_{i}^{k}){m_{\alpha}}^{i}}$,
for $i=1, 2,..., k$. But $f_{\sigma}\in A_{k-1}=F_{k-1}$, so $f\in
F_k$. Hence $A_k\subseteq F_k$. Now let $f\in F_k$, we have to show
that $f\in A_k$. We have $R_{1}f=f_{1}f$, for some $f_{1}\in
F_{k-1}=A_{k-1}$. Since $f_1\in A_{k-1}$ we may assume that
$f_{1}(n)={\lambda}_{1}^{n^{k-1}}{\lambda}_{2}^{n^{k-2}}...{\lambda}_{k-1}^{n}{\lambda}_{k}$,
where ${\lambda}_{1}, {\lambda}_{2},..., {\lambda}_{k}\in{\mathbb
{T}}$. Then $f(1)=R_{1}f(0)=f_{1}(0)f(0)$ and
$f(2)=R_{1}f(1)=f_{1}(1)f(1)=f_{1}(1)f_{1}(0)f(0)$, and by
induction, \bea
f(n)&=&({\lambda}_{1}^{(n-1)^{k-1}}{\lambda}_{2}^{(n-1)^{k-2}}...\lambda_{k-1}^{n-1}{\lambda}_{k})
({\lambda}_{1}^{(n-2)^{k-1}}{\lambda}_{2}^{(n-2)^{k-2}}...{\lambda}_{k-1}^{n-2}{\lambda}_{k})
...({\lambda}_{1}\lambda_2...{\lambda}_{k})
({\lambda}_{k})f(0)\\
&=&{\lambda}_{1}^{{\sum}_{j=1}^{n-1}j^{k-1}}{\lambda}_{2}^{{\sum}_{j=1}^{n-1}j^{k-2}}
...{\lambda}_{k-1}^{{\sum}_{j=1}^{n-1}j}{\lambda}_{k}^{n}f(0). \eea
So for each $i=0,1,2,\ldots,k-1$ the power of $\lambda_{k-i}$ is a
polynomial in $n$ of degree $i+1$. Hence the power of $\lambda_1$
(which has the maximum degree) is a polynomial of degree $k$. It
follows that $f\in A_k$ and the proof is complete by induction.
\end{proof}
As an immediate consequence of the latter lemma we have the next
theorem which characterizes our algebras for the additive group
$\mathbb Z$.
\begin{theorem}
$(i)$ $W_{k}({\mathbb {Z}}, +)$ coincides with the norm closure of
the algebra generated by the set of functions $\{n\mapsto
{\lambda}^{n^i}:$ $\lambda\in{\mathbb {T}}$, $i=0, 1,..., k\}$.
$(ii)$ $W({\mathbb {Z}}, +)$ coincides with the norm closure of the
algebra generated by the set of functions
$\{n\mapsto{\lambda}^{n^k}: \lambda\in{\mathbb {T}}$, and
$k\in{\mathbb {N}}\}$, that is, $W({\mathbb {Z}}, +)$ coincides with
the Weyl algebra.
\end{theorem}
{\bf Example (b).} Let $R$ be a countable discrete ring.
Let $\chi$ be an arbitrary character on the additive group
$(R_d,+)$, where $R_d$ denotes $R$ with the discrete topology.
Without loss of generality, assume that $R$ is abelian. We are going
to show that for each $s$ in $R$ the function $f(t)=\chi(st^3)$
belongs to $F_3(R_d,+)$. To this end, let $\sigma\in\Sigma$. Thus
there exists a sequence $s_n$ in $R$ such that for each $h\in
l^\infty(R_d)$, $\sigma(h)(t)=\lim_n
R_{s_n}h(t)=\lim_n h(t+s_n)$.\\
Thus $\sigma f(t)=\lim_n
f(t+s_n)=\lim_n\chi(s{(t+s_n)}^3)=f(t)f_\sigma(t)$, in which
$f_\sigma(t)=\lim_n\chi(ss_n^3+3ss_n t^2+3ss_n^2 t)$. Since $R$ is
countable, by the diagonal process, there exists a subsequence, say
$s_n$, of the sequence $s_n$ such that, for all $t$ in $R$, all of
the limits $\lim_n\chi(ss_n^3)$, $\lim_n\chi(ss_n t^2)$ and
$\lim_n\chi(ss_n^2 t)$ exist. (In fact, one may first choose a
subsequence of $s_n$, if necessary, such that $\lim_n\chi(ss_n^3)$
exist. Let $R=\{x_1,x_2,x_3,\ldots\}$. Choose a subsequence of
$s_n$, say $s_{1,n}$ such that both limits $\lim_n\chi(ss_{1,n}
x_1^2)$ and $\lim_n\chi(ss_{1,n}^2 x_1)$ exist. This time, choose a
subsequence of $s_{1,n}$, say $s_{2,n}$, such that both limits
$\lim_n\chi(ss_{2,n} x_2^2)$ and $\lim_n\chi(ss_{2,n}^2 x_2)$ exist.
Continue this process and choose the resulting sequence $s_{n,n}$ on
the diagonal, which is eventually our desired subsequence, say
$s_n$). Hence for each $t\in R$,
\[f_\sigma(t)=\lim_n\chi(ss_n^3)\lim_n\chi^3(ss_n^2
t)\lim_n\chi^3(ss_n t^2).\] By definition, to prove that $f\in F_3$
it is enough to show that $f_\sigma\in F_2$. To see this, let
$\tau\in\Sigma$ be arbitrary. Then there exists a sequence $t_m$ in
$R$ such that for each $h\in l^\infty(R_d)$, $\tau(h)(t)=\lim_m
R_{t_m}h(t)=\lim_m h(t+t_m)$.\\
Thus $\tau f_\sigma(t)=\lim_m
f_\sigma(t+t_m)=f_\sigma(t)f_{\sigma\tau}(t)$, where
\[f_{\sigma\tau}(t)={(f_\sigma)_\tau}(t)=\lim_m\lim_n\chi^3(ss_n^2
t_m)\lim_m\lim_n\chi^3(ss_n t_m^2+2ss_n t_m t).\] $R$ is countable,
hence by going through a subsequence of $t_m$ (by using the diagonal
process) one may assume that for all $t$ in $R$ the limits
$\lim_m\lim_n\chi(ss_n t_m^2)$ and $\lim_m\lim_n\chi(ss_n t_m t)$
exist. Therefore
\[f_{\sigma\tau}(t)=\lim_m\lim_n\chi^3(ss_n^2
t_m)\lim_m\lim_n\chi^3(ss_n t_m^2)\lim_m\lim_n\chi^6(ss_n t_m t).\]
Again by definition, to prove that $f_\sigma\in F_2$ it is enough to
show that $f_{\sigma\tau}\in F_1$. Let $\xi=\lim_l u_l\in\Sigma$,
then it follows from the above equation that
\[\xi(f_{\sigma\tau})(t)=f_{\sigma\tau}(t)\lim_l\lim_m\lim_n\chi^6(ss_n
t_m u_l).\] That is, $\xi(f_{\sigma\tau})=\lambda f_{\sigma\tau}$
where $\lambda=\lim_l\lim_m\lim_n\chi^6(ss_n t_m u_l)\in
F_0={\mathbb{T}}$. Hence $f_{\sigma\tau}\in F_1$ and so $f_\sigma\in
F_2$ and this implies that $f\in F_3$. Our claim is now established.
By using the above method, one may prove that for each
$k\in{\mathbb{N}}$ and $s\in R$ the function
$t\rightarrow\chi(st^k)$ is an element of $F_k$.\\
Briefly speaking, the above argument and Lemma ~\ref{IV} imply that:
\begin{corollary}
If $R$ is a countable discrete ring, then for each character $\chi$
on the discrete additive group of $R$ the function $\chi(q(t))$, in
which $q(t)$ is a polynomial with coefficients in $R$, belongs to
$W(R_d,+)$ and is also a distal function. \end{corollary}
It should be remarked that the distality of the functions $\chi(q(t))$ was
first proved by Filali~\cite{Fi}
without the countability condition on $R$.\\
{\bf Example (c).} $(i)$ If $S$ contains a right zero element, i.e.
there exists $t\in S$ such that $st=t$ for all $s\in S$, then for
$f\in F_1$ there exists $\lambda_t\in{\mathbb{T}}$ such that
$R_tf=\lambda_tf$, hence for all $s\in S$,
$f(t)=f(st)=R_tf(s)=\lambda_tf(s)$, that is
$f=f(t)/\lambda_t\in{\mathbb{T}}$. Therefore $F_1={\mathbb{T}}$ and
so $F_k={\mathbb{T}}$ for all $k\in{\mathbb{N}}$. It follows that
for such a semigroup $S$, $W_k(S)=W(S)=$ the set of constant
functions.
$(ii)$ If $S$ is a left zero semigroup (i.e. $st=s$ for all $s,t\in
S$), then for each function $f\in LMC(S)$ we have $\sigma(f)=f$ for
all $\sigma$ in $\Sigma$, and so if $|f|=1$, then $f\in F_1$. Hence
for all $k\in {\mathbb{N}}$, $W=W_k=W_1=LMC$.\\
Now we examine some of the newly defined algebras on a non-trivial
non-group
semigroup.\\
{\bf Example (d).} Let $S$ be the bicyclic semigroup of
\cite[Example 2.10]{BJM}, i.e. $S$ is a semigroup generated by
elements $1$, $p$ and $q$, where $1$ is the identity and $p$ and $q$
satisfy $pq=1\neq qp$. The relation $pq=1$ implies that any member
of $S$ may be written uniquely in the form $q^m p^n$, where
$m,n\in{\mathbb{Z}}^+$ and $p^0=q^0=1$. \\
We are going to show that
\[F_1(S)=\{q^mp^n\mapsto\mu^{r}\nu^{1-r}:~~r=m-n~and
~\mu,~\nu\in{\mathbb{T}}\}\] To see this, let $f\in F_1$, then for
each $s\in S$, $R_sf=\lambda_sf$ for some
$\lambda_s\in{\mathbb{T}}$. Hence $f(q)=\lambda_qf(1)$,
$f(p)=\lambda_pf(1)$ and $f(1)=f(pq)=R_qf(p)=\lambda_qf(p)$,
therefore $\lambda_p\lambda_q=1$. It is also readily seen that
$\lambda_{q^mp^n}=\lambda_q^{m-n}$. One may use induction to simply
prove that for each $n\in{\mathbb{Z}}^+$,
$f(qp^n)=f(q)(\frac{f(p)}{f(1)})^n$, and then use the latter to show
(again by induction on $m$) that
$f(q^mp^n)=f(p)^{n-m}f(1)^{1-(n-m)}$. But $f(p)f(q)=f(1)^2$, so
$f(q^mp^n)=f(q)^{m-n}f(1)^{1-(m-n)}$. Hence it is enough to take
$\mu=f(q)$ and $\nu=f(1)$. The converse inclusion is easily
verified.\\
To prove the next theorem, the following lemma is needed.
\begin{lemma}\label{l:5}
Let $S$ be the bicyclic semigroup described above. If $f\in F_1(S)$,
then $f(p)f(q)=f(1)^2$.
\end{lemma}
\begin{proof}
Let $f\in F_1(S)$, then there exist constants $\lambda_p$ and
$\lambda_q\in{\mathbb{T}}$ such that $R_pf=\lambda_pf$ and
$R_qf=\lambda_qf$. Hence $f(p)f(q)=\lambda_p\lambda_qf(1)^2$. Thus
it is enough to show that $\lambda_p\lambda_q=1$. But,
$f(1)=f(pq)=R_qf(p)=\lambda_qf(p)=\lambda_q\lambda_pf(1)$, that is
$\lambda_q\lambda_p=1$, and the result follows.
\end{proof}
\begin{theorem}\label{t:5} Let $S$ be the bicyclic semigroup generated by $1, p$ and $q$, where $1$ is the identity and $pq=1\neq qp$.
Then $W_1(S)$ and $W_2(S)$ are the norm closure of the algebras
generated by the sets \[\{q^mp^n\mapsto\mu^{r}\nu^{1-r}:~~r=m-n~and
~\mu,~\nu\in{\mathbb{T}}\}\] and
\[\{q^mp^n\mapsto\lambda^{\frac{r^2-r}{2}}\mu^{\frac{r^2+r}{2}}\nu^{1-r^2},~~r=m-n~
and~\lambda,~\mu,~\nu\in{\mathbb{T}}\}\] respectively.
\end{theorem}
\begin{proof} By what we already discussed, the result is clear for
$W_1(S)$. To complete the proof, it is enough to show that
\[F_2(S)=\{q^mp^n\mapsto\lambda^{\frac{r^2-r}{2}}\mu^{\frac{r^2+r}{2}}\nu^{1-r^2},~~r=m-n~
and~\lambda,~\mu,~\nu\in{\mathbb{T}}\}\] Let $A$ denote the right
hand side of the above equation and let $f\in A$. Then there exist
$\lambda,\mu,\nu\in{\mathbb{T}}$ such that for all
$m,n\in{\mathbb{Z}}^+\cup\{0\}$,
$f(q^mp^n)=\lambda^{\frac{r^2-r}{2}}\mu^{\frac{r^2+r}{2}}\nu^{1-r^2}$
with $r=m-n$. By choosing suitable $m,n\in{\mathbb{Z}}^+\cup\{0\}$
we derive that $\lambda=f(p)$, $\mu=f(q)$ and $\nu=f(1)$. To prove
$f\in F_2(S)$ is to prove that there exist elements $f_p$ and $f_q$
in $F_1(S)$ such that $R_pf=f_pf$ and $R_qf=f_qf$. In fact it is
enough to take
\[f_p(q^mp^n)=[f(q)^{-1}f(1)]^r[f(p)f(1)^{-1}]^{1-r}\] and
\[f_q(q^mp^n)=[f(p)f(q)^2f(1)^{-3}]^r[f(q)f(1)^{-1}]^{1-r}\]
where $r=m-n$. Our discussion preceding the theorem reveals that
both $f_p$ and $f_q$ are elements of $F_1(S)$, therefore $f\in
F_2(S)$.\\
Conversely, let $f\in F_2(S)$. To show $f\in A$ is to show that for
all $m,n\in{\mathbb{Z}}^+\cup\{0\}$,
\[f(q^mp^n)=f(p)^{\frac{r^2-r}{2}}f(q)^{\frac{r^2+r}{2}}f(1)^{1-r^2},~where~r=m-n.~ \ \ \ \ \ (*)\]
Since $f\in F_2$, there exists
$f_q\in F_1(S)$ such that $R_qf=f_qf$. Therefore, (from the above
Lemma) \[f_q(p)f_q(q)=f_q(1)^2.~ \ \ \ \ \ (I)\] On the other hand,
$f_q(1)=f(q)f(1)^{-1}$, and also $f(1)=f(pq)=R_qf(p)=f_q(p)f(p)$,
thus $f_q(p)=f(p)^{-1}f(1)$. Hence it follows from $(I)$ that
\[f_q(q)=f(p)f(q)^2f(1)^{-3}.~ \ \ \ \ \ (II)\] Fix $n\in{\mathbb{N}}$, then
by using induction on $m$ and the fact that
$f(q^{m+1}p^n)=R_qf(q^mp^n)=f_q(q^mp^n)f(q^mp^n)=f_q(q)^{m-n}f_q(1)^{1-(m-n)}f(q^mp^n)$,
we deduce from $(II)$ and $(*)$ that
\[f(q^{m+1}p^n)=f(p)^{\frac{s^2-s}{2}}f(q)^{\frac{s^2+s}{2}}f(1)^{1-s^2}, ~where~ s=(m+1)-n.\]
The theorem is now established by induction.
\end{proof}
\begin{corollary}
Let $S$ be the bicyclic semigroup generated by $1, p$ and $q$, where
$1$ is the identity and $pq=1\neq qp$. If either $p$ or $q$ is an
idempotent, then $W(S)=W_i(S)={\mathbb{C}}$, for all $i$.
\end{corollary}
\begin{proof}
It is enough to show that $F_1(S)={\mathbb{T}}$. To this end, let
$f\in F_1(S)$, and assume that $p^2=p$. Then there exist
$\lambda_p\in{\mathbb{T}}$ such that $R_pf=\lambda_pf$, hence
$f(p)=f(p^2)=R_pf(p)=\lambda_pf(p)$ and so $\lambda_p=1$. That is,
$R_pf=f$ and $f(p)=f(1)$. It follows from Lemma~\ref{l:5} that
$f(q)=f(1)$. Now the first part of the above theorem implies that
for all $m,n\in{\mathbb{Z}}^+\cup\{0\}$,
$f(q^mp^n)=f(q)^{m-n}f(1)^{1-(m-n)}=f(1)$. Hence
$f=f(1)\in{\mathbb{T}}$. The proof for the case where $q$ is an
idempotent is similar.
\end{proof}
{\bf Remarks.} $(i)$ By the results ~\ref{iV} and ~\ref{V}, for
every abelian semitopological semigroup $S$, $W_k$ and also $W$ lie
between $SAP$ and $D$. It would be desirable to study the structure
of the (right topological abelian group) compactifications $S^{W_k}$
and $S^W$. In particular, it would be more desirable if one could
investigate the size of the topological centres of $S^{W_k}$ and
$S^W$. The latter problem is of particular interest among some
authors, (the interested reader is referred to~\cite{J} and
~\cite{DL}). For an elegant characterization of the topological
centre of the largest compactification of a locally compact group
one may refer to ~\cite{LP} and also~\cite{NU}.
$(ii)$ In \cite[Theorem 2.13]{S} and~\cite[Corollary 3.3.3]{J}, (by
using different methods) it is proved that all elements of the Weyl
algebra $W(\mathbb{Z}, +)$ are uniquely ergodic. One may seek the
same result for $W(S)$, where $S$ is an arbitrary semitopological
semigroup.
$(iii)$ It would be quite interesting to find a general formula for
$F_k(S)$ in Theorem~\ref{t:5}.
\section*{acknowledgment}
The first author would like to thank Professor Anthony Lau, at the
University of Alberta, for his encouragement and support through his
NSERC grant A7679. And the authors would like to thank the kind
referee for the helpful suggestions.
| 148,267
|
Product 10/29
This product was added to our catalog on Thursday 22 August, 2013.
Ordered a squad on a whim to try out, received exceptional quality models with very reasonably priced shipping (USA to AUS is normally really...Read More ->
I cannot recommend this guy enough. I just bought a squad for my new Mordian Iron Guard army for Warhammer 40k. The detail is AMAZING!! Defiantly...Read More ->
| 207,505
|
With iconic Californian routes such as the 101 and PCH, there’s another hwy worthy of a road trip: Highway 395 to the Eastern Sierras. I grew up taking HWY 395 from LA to June Mountain (Mammoth’s sister mountain), but as a high schooler focused on friends and snowboarding, I guess I never even bothered to look out the window to see just how majestic everything else other than skiing is in this part of California.
I’ve dreamed of hopping on the #vanlife movement, and found an amazing rental company out of Costa Mesa that has a massive selection of VW vans in just about every color. Vintage Surfari Wagons is run by Southern California natives Bill and Diane Staggs who have a deep love for adventure and the VW bus. We were lucky enough to have the Hale Pua, a 1982 Vanagan Westfalia dream as our home away from home for our road trip up the 395.
Big Pine
First, we went to Big Pine with the intention of hiking up to Second Lake (original name, I know). I had seen pictures on Instagram of this glacial lake that looked like something you would find in Patagonia. I had to see it for myself! It's a 13 mile hike which we weren't really prepared for but so determined to make it. The views were freakin' epic the entire way, especially once we got into the snow covered portion. Since it was December, the days were shorter, which meant we had to turn around by 1PM to make it back before the sun set. By 1PM we still hadn't made it to First Lake, so we had to turn around by 1:30PM... ok 1:45 PM... ok 2:00PM... wait... 2:30PM! I'm telling you we were so determined to make it to Second Lake since we had come so far! And I needed to see if the grams were real. By 2:30 PM, we had made it up to Second Lake! It's one of the most stunning places I've visited in California. The lake was frozen and would make almost a howling noise in the silence. The photos barely do this place justice.
Well, it was nearly 3PM and we just booked it down that mountain! The sun had set but thankfully there was still enough light by the time we got to our car (insert Praise Hand Emoji here).
Mammoth
Next up were the Mammoth Hot Springs, which is what had inspired this road trip to begin with. There are natural hot springs all around Mammoth that you can just drive up to, get out, and soak for however long you want with the best views of the snow covered Eastern Sierras. We made it to three different ones. Wild Willy's, Shepherd, and Rock Hot Tub. I loved Wild Willy's because there is a hot spring shaped like a heart! Shepherd is for 4 people max which was so cute and darling. And Rock Hot Tub is for two people max, but not nearly as warm as it should be to be enjoyable in the middle of winter :). Another favorite stop in the Mammoth Area was Geological Hot Creek. It's out of this world beautiful. The furthest we made it up was Mammoth but you can continue up to the Travertine Hot Springs and Bodie Ghost town on your way to Tahoe.
Alabama Hills
Next was the Alabama Hills right next to Lone Pine, CA. Honestly, I had no idea just about every Hollywood Western was filmed here with many of the local residents acting as extras in these films. I definitely recommend going into the Lone Pine Film History Museum to get a better history of the area. Also definitely recommend stopping at Gus’ Really Good Fresh Jerky in Olancha on your way in.
Perhaps it was the time of year (Mid December), but the Alabama Hills felt untouched. We felt like we were the only ones there! Our camp spot held the perfect backdrop of Mt. Whitney.
When it came to finding good food + coffee there are a number of cute spots along the road. Schat’s Bakery is a must for baked goods, Black Velvet Coffee for your caffeine mix, Good Earth Yogurt for a healthier meal and Bishop Burger Barn or Mahagony’s Smoked Meats for an All-American meal.
I couldn’t believe just how stunning the landscape is on the other side of the Sierras. It was magical watching the sun rise and set every day turning the mountains pink in the morning and a purple blue in the evening. Don’t get me wrong, I love the snowboarding and skiing that you have access to in Mammoth and June mountain, but the terrain surrounding these resorts are every bit adventurous as they are gorgeous.
| 223,459
|
TITLE: $\mathcal V(y^2-x^2(x-\lambda))$ is rational
QUESTION [0 upvotes]: I want to show, that $$C_\lambda = \mathcal V(y^2-x^2(x-\lambda))$$ is rational. Where $\lambda$ is in a field $K$ and $\lambda \ne 0$. I am given that a parametrization of $C_\lambda$ is the inverse of $$\phi: C_\lambda \dashrightarrow \mathbb A^1, (x,y)\mapsto \frac x y.$$
So there must be $p(t),q(t),u(t),v(t)\in K[t]$, such that $$\phi^{-1}: \mathbb A^1 \dashrightarrow \mathbb A^2, t \mapsto \left(\frac{p(t)}{q(t)}, \frac{u(t)}{v(t)}\right).$$
There are $(\pm\lambda, 0), (0,0)\in C_\lambda$ for certain. But $\phi$ is not defined for any of these points. How can I find $\phi^{-1}$? Is there any geometric interpretation like in the case of stereographic projection?
Any help is appreciated.
REPLY [1 votes]: You can actually arrange for the parameterization of $C_\lambda$ to be given by polynomials, i.e. there's a regular map $\psi:\Bbb A^1\to C_\lambda$ which is an isomorphism except at finitely many points. (In particular, you can get an isomorphism away from the singular point $(0,0)$.)
Bezout's theorem and some basic intersection theory provides a geometric explanation of why this works. Any line in the projective plane intersects the projectivization of $C_\lambda$ with multiplicity 3, and the intersection multiplicity at $(0,0)$ is at least two. So the third point of intersection of a line through $(0,0)$ with $C_\lambda$ is uniquely determined by the slope of the line through $(0,0)$, and it remains to check that this gives us a rational parameterization.
Let $t$ be the slope of a line $\ell = V(y=tx)$ through $(0,0)$. Substituting this equation in to the equation of $C_\lambda$ to eliminate $y$, we get $t^2x^2=x^2(x-\lambda)$, or $x^3-(\lambda+t^2)x^2=0$. The sum of the roots of this equation is $\lambda+t^2$, and since $x=0$ is two of those roots, we find that the third root is $x=\lambda+t^2$. So $(\lambda+t^2,\lambda t+t^3)$ is the third point on the curve, meaning we have a parameterization $\Bbb A^1\to C_\lambda$ given by $$t\mapsto (\lambda+t^2,\lambda t+t^3),$$ and this is a rational inverse to the rational map $C_\lambda \dashrightarrow \Bbb A^1$ given by sending $(x,y)\mapsto \frac{y}{x}$.
If you want to see this process on a graph, click here to play around with it!
| 201,645
|
TITLE: How do I know that gauge fields are bosons?
QUESTION [1 upvotes]: QED and the Dirac equation have field operators $\psi$ interact with a gauge field $A^{\mu}$.
We identify $\psi$ as a fermionic field and $A^{\mu}$ as a gauge boson - the photon.
Do we or can we know that one is a fermion and the other is a boson?
Or do we get that information from the commutation relations when we quantise the theory?
REPLY [0 votes]: The possible Lorentz invariant wave equations is a relatively limited set of equations. Most wave equations one can write will not be Lorentz invariant under any transformation law you can think of.
We know all the possible finite dim Lorentz invariant field equations. One of the properties of these equations is that each of them have a distinct number associated with them that relates to their angular momentum. We need some representation theory to exactly explain what I mean by this, but trust me that there is a number. This number represents how rotations effect the equations, and it is in fact the "Equation's spin". When quantised, this is the particle's spin.
So once you see a particle's wave equation, you can always tell the spin of the particle. From this point you can look at the gauge field's differential equation and immediately tell what is the gauge particle's spin. You can also ask the reverse question - what are the possible wave equations of a spin 7/2 particle?
The main point is, that the possible Lorentz invariant field equations is such a limited set, that we can tell from the wave equations themselves everything there is to know about the particles they represent. If the particle violates these rules, it cannot be Lorentz Invariant.
The kinetic term of the gauge fields, i.e. $ F_{\mu \nu}F^{\mu \nu}$, is only Lorentz invariant if it has spin 1. You can write different Lagrangians with different kinetic terms, and have gauge fields with different spins.
(The technical name for the "equation's spin" is the total angular momentum of the representation of the Lorentz group of the field)
| 144,451
|
TITLE: Which of the following properties does a process with independent increments really admit?
QUESTION [2 upvotes]: Let $E$ be a normed $\mathbb R$-vector space and $(X_t)_{t\ge0}$ be an $E$-valued adapted process on a filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$ such that $X_t-X_s$ is independent of $\mathcal F_s$ for all $t\ge s\ge0$.
Let $s\ge0$. I wonder which of the following statements we are able to show:
$\left(\sigma\left(X_{t_n}-X_{t_{n-1}}\right),\ldots,\sigma\left(X_{t_1}-X_{t_0}\right),\mathcal F_s\right)$ is independent for all $n\in\mathbb N$ and $t_n>\cdots>t_0\ge s$;
$(X_t-X_s)_{t\ge s}$ is independent of $\mathcal F_s$.
If $u\ge s$, then $(X_u-X_t)_{t\in[s,\:u]}$ is independent of $\mathcal F_s$.
Assuming that $X$ is càdlàg: If $t>s$, then $\Delta X_t:=X_t-X_{t-}$, where $X_{t-}:=\lim_{r\to t-}X_r$, is independent of $\mathcal F_s$.
Assuming that $X$ is càdlàg: If $B\in\mathcal B(E)$ with $0\not\in\overline B$ and $t\ge s$, how can we show that $$\sum_{\substack{r\in(s,\:t]\\\Delta X_r\:\ne\:0}}1_B(\Delta X_r)$$ is independent of $\mathcal F_s$?
(1.) can easily be shown. Regarding 2.: Let $Y_t:=X_t-X_s$ for $t\ge s$. If $n\in\mathbb N$ and $t_n>\cdots>t_1\ge t_0:=s$, then $$\underbrace{X_{t_n}-X_{t_{n-1}}}_{=\:Y_{t_n}\:-\:Y_{t_{n-1}}},\ldots,\underbrace{X_{t_2}-X_{t_1}}_{=\:Y_{t_2}\:-\:Y_{t_1}},\underbrace{X_{t_1}-X_{t_0}}_{=\:Y_{t_1}}$$ are independent of $\mathcal F_s$ by (1.). Noting that $$\Sigma_n:E^n\to E^n\;,\;\;\;x\mapsto(x_1,x_1+x_2,\ldots,x_1+\cdots+x_n)$$ is Borel measurable, we can conclude that $$\left(Y_{t_1},\ldots,Y_{t_n}\right)=\Sigma_n\left(Y_{t_1},Y_{t_2}-Y_{t_1},\ldots,Y_{t_n}-Y_{t_{n-1}}\right)$$ is independent of $\mathcal F_s$. This should immediately yield (2.).
Regarding 3.: We should be able to argue in a similar way. Let $Z_t:=X_u-X_t$ for $t\in[s,u]$. If $n\in\mathbb N$ and $u=:t_{n+1}\ge t_n>\cdots>t_1\ge s$, then $$\underbrace{X_{t_{n+1}}-X_{t_n}}_{=\:Z_{t_n}},\underbrace{X_{t_n}-X_{t_{n-1}}}_{=\:Z_{t_{n-1}}\:-\:Z_{t_n}},\ldots,\underbrace{X_{t_2}-X_{t_1}}_{=\:Z_{t_1}\:-\:Z_{t_2}}$$ is independent of $\mathcal F_s$ by (1.) and hence $$\left(Z_{t_n},\ldots,Z_{t_1}\right)=\Sigma_n\left(Z_{t_n},Z_{t_{n-1}}-Z_{t_n},\ldots,Z_{t_1}-Z_{t_2}\right)$$ is independent of $\mathcal F_s$, from which it should again immediately that (3.) holds.
Regarding 4.: Let $(t_n)_{n\in\mathbb N}\subseteq(s,t)$ with $t_n\xrightarrow{n\to\infty}t$, then $$\Delta X_t=\lim_{n\to\infty}\left(X_t-X_{t_n}\right).$$ By (3.), $\left(X_t-X_{t_n}\right)_{n\in\mathbb N}$ is independent of $\mathcal F_s$ and since independence is preserved under almost sure limits (even under limits in probability), the claim should follow.
Regarding 5.: With this one I'm quite lost. The problem is the following: By (4.), each term $1_B(\Delta X_r)$ is independent of $\mathcal F_s$ for all $r\in[s,t]$. Now, we know that $X(\omega)$ has only countably many jumps on $[s,t]$ for each fixed $\omega$, but since this countable set of times depends on $\omega$, I cannot simply conclude (as before) by arguing that independence is preserved under limits (of sequences). (I've asked a related, more general question, on this topic on mathoverflow: https://mathoverflow.net/q/413062/91890).)
Is there any issue with my proofs of (2.), (3.) or (4.)? And what can we do to show (5.)?
REPLY [1 votes]: You don't need $\Sigma _n$ to prove 1. (and I dont see how you can use it to prove it), and 2. is already stated in the definition of $(X_t)_t$. Just note that as $(\mathcal{F}_t)_t$ is a filtration then if $Y$ is $\mathcal{F}_s$ measurable and $t>s$ then, as $\mathcal{F}_s\subset \mathcal{F}_t$ you have that $Y$ is also $\mathcal{F}_t$ measurable. Then it follows that $X_t-X_s$ is $\mathcal{F_t}$ measurable, and so for $r\geqslant t$ we have that $X_r-X_t$ is independent of $\sigma (X_t-X_s)\subset \mathcal{F}_t$, then 1. follows easily from here.
Similarly, as $\mathcal{F}_s\subset \mathcal{F}_t$ and $X_r-X_t$ is independent of $\mathcal{F}_t$, it follows that its also independent of $\mathcal{F}_s$, this proves your last question.
| 36,519
|
Top Adelaide Best Digital Marketing Secrets
Global or getaway flights may well should be requested even supplemental beforehand. The chart underneath exhibits The most cost effective flight rates by thirty day period for Adelaide to Hobart, Australia flights.
Noted for concentrating on profits in place of just prospects, this company’s take on “perspective that matters” is winning the appropriate minds. Deal with
Adelaide Best Online Marketing Companies OptionsBecause of this, You will find a big turnover level in community marketing since Many individuals stop trying and do not want to maintain bothering their relations. The wonderful thing about community marketing is you could get it done anytime and anyplace with practically zero constraints!
All of us get pleasure from strolling down the shopping mall, and procuring all around all day extended every so often returning dwelling Whilst using the vacant bag.
You for this reason can realize that the solutions supplied by Sortlist are unique and of of outstanding nature. Consequently never sit on fence any longer and put in place straightaway by distributing your campaign. We've been assured that the best agency for your personal exercise is within our listing.
"Just A quick Be aware to increase our many thanks with the endeavours and capabilities in generating our new modern Web-site. Your crew really grasped this what we are already attempting to get by remodelling our out-of-date dinosaur.
Adelaide Best Online Marketing Things To Know Before You BuyJust before investing any time or cash by using a network marketing company, glimpse the company up by means of the higher Business enterprise Bureau (BBB) or consult people today you realize that have practical experience With all the company.
We don’t utilize a painter to try and do The endeavor of a computer scientist. That’s why our in home group consists of academics and pros at the best in their respective Professions.
Adelaide Best Marketing Companies Fundamentals ExplainedIt can be advised you adhere to This method extremely carefully, particularly when you happen to be starting in your business. Many distributors make the crucial mistake of never ever providing any help for their recruits. This will stagnate your online business as the dimensions of your achievements is tremendously depending on the achievement of Other folks. view Achievement in any revenue centered organization is usually during the follow-up.
Intermediate / Practitioner Find out with the specialists all you need to know about digital marketing.
Adelaide Best Online Marketing Companies OptionsNetwork marketing instruction is presented- Commonly, if you Read Full Article join a network marketing company, that you are offered many of the resources and MLM teaching you might want to get started down The trail to network marketing success.
Akendi produces merchandise & company designs knowledgeable by in-depth men and women analysis. These are obsessed with offering you intentional encounters.
The 2-Minute Rule for Adelaide Best Digital MarketingWe offer high-quality Internet Internet-sites, Online search engine marketing procedure, e-commerce and online marketing therapies for enterprises in the course of practical web-site Australia.
Please email errors@quora.com if you suspect This is often an mistake. Please involve your IP deal with in your e-mail.
Comments on “Not known Details About Adelaide Best Online Marketing Companies”
| 200,307
|
TITLE: Fields for which there exist multivariable polynomials vanishing at single specified point
QUESTION [9 upvotes]: For which fields $k$ does the following hold for all $n \geq 1$? Let $(a_1,\ldots,a_n) \in k^n$. Then there exists a polynomial $f(x_1,\ldots,x_n) \in k[x_1,\ldots,x_n]$ such that $f(a_1,\ldots,a_n) = 0$ but $f(b_1,\ldots,b_n) \neq 0$ for all $(b_1,\ldots,b_n) \neq (a_1,\ldots,a_n)$?
This is clearly impossible for $k$ algebraically closed.
It does hold for $k$ a subfield of $\mathbb{R}$; indeed, in that case we can take
$$f(x_1,\ldots,x_n) = (x_1-a_1)^2 + (x_2-a_2)^2 + \cdots + (x_n-a_n)^2.$$
REPLY [20 votes]: If $k$ is not algebraically closed, such a polynomial always exists (the opposite is also true and is mentioned in the post).
We may assume that $a_i=0$ for all $i$. Take an irreducible polynomial $g(x)$ of degree $d>1$, then for the homogeneous form $G(x,y)=y^dg(x/y)$ we have $G(x,y)=0$ if only if $x=y=0$. This solves the case $n=2$, for $n=3$ consider the polynomial $G(G(x,y),z)$, it takes zero value only when $x=y=z=0$, and so on.
REPLY [2 votes]: I guess finite fields $k = \mathbb{F}_{q}$ satisfy this property, namely we can take $$ f = ((x_{1}-a_{1})^{q-1}-1) \dotsb ((x_{n}-a_{n})^{q-1}-1) - (-1)^{n} $$ for any $n$.
In fact we have a Lagrange interpolation formula for finite fields, see this answer.
| 43,905
|
Recently, I have seen lots of articles and videos circulating over the social media showing professional photographers trying to explain to potential wedding clients on why the budget allocated wasn’t enough, eventhough the potential wedding clients have spend a fortune on dresses, limos, props for ceremony/reception and etc. Photographers even go to the extend of explaining that they have invested in expensive gears, workshops, post-processing skills and etc. I am only speaking on my behalf and from my personal point of view.
With the current influx of newbie photographers in the market trying to get their portfolio, they would offer free or provide a very cheap service to clients. I have nothing against them as I was one of them when I first started. Lots of professional photographers are against those newbies from undervaluing their work. From my point of view, I dont think they are undervaluing their work, the real value comes from experience, creativity, technical expertise and etc. Those new photographers who are just trying to break in the market would not have these real value and they are putting their clients at a risk by not having sufficient redundancy, technical experience and experience in dealing with challenges on or before the wedding day. That is why, I do not think they are under charging their clients, I dont even think client should be paying for their service as they are just using the wedding clients as hands on training and to develop skills in photography in the process. These new photographers are necessary to help those who can’t afford to hire a professional wedding photographer. Those who are just planning a very small wedding event, tight on budget and just needed some basic photos for their memory, as not everyone can afford high quality photos, that is the reason why we have small and cheap compact cars, and luxury sports car. Having options of clients I think is a good thing.
On another note, I dont think professional photographers should explain and convince their clients that they are investing their valuable time, experience, gears, workshop, traning and etc..At the end of the day, its not what expensive gears you have, what training you attended, or how much time you put in. Its the end result regardless on what you have or invested. Some people need a degree or PhD in art to be an artist and some were born talented with just basic training would be sufficient. I think the more photographers trying to explain, the worst it would become. I think and believe that when I walk in to an art gallery in Musee du Louvre, Paris and walking into a local smalltown art gallery, the experience is entirely different. The lightings on the artwork, the large, elegant and artistic open space, exclusive large framing, one of a kind quality and creativity of work, finess in details and not to mention, the famous names who created the masterpiece who took years to develop their style or ” brand “. Now, if i had the money to funish my beautiful house/mansion, I would invest on the best masterpiece to suit my style and elegance. Do I need to wake Da Vinci up to convince me that he used the most expensive brush in the world, the highest grade canvas, the best colours and etc?, I dont really think so. Everyone’s taste varies and if i see his work and love it so much, I would pay the price for it, without a salesman trying to convince me, why I am paying the price for it. Of course, unless the artwork you are producing is not up to standard and you would need a salesman to convince you about what brush they use and etc..Or if the clients don’t think the masterpiece is worth it, its just their taste, nothing agaist personal taste, as I may not like all the famous artwork as well. Its all back to personal choice.
If they love it, they would not need hard selling salesman to convince them. They would just get it. Again, I am just speaking from my personal point of view and experience. Its just in general and may not apply to all.
Thanks for reading :)
| 251,705
|
ADVISORY,rd & Broadway – Broadcast Studio
When:
Tuesday, February 11, 2014 – 9:15 a.m. to 9:30 a.m. ET
Matt Clawson
(949) 474-4300
matt@allencar Foundation Medicine, Inc. [FMI]::
| 51,112
|
Another breakthrough - making sense out of the non-sensical.
Just working along the other day, this time in studying copywriting - and found an interesting approach to things.
For too long, I've isolated myself from many things which simply didn't make sense to me. A lot of social conventions such as drugs, popular music, sports - these I either ignored or tolerated if I had to. (Drugs, much easier to stay away from - well, maybe if you exclude coffee and donuts.)
Recently, it came to me - especially going down some classic works in copywriting and advertising - that there is a definition of "sense" which has been missing.
Of course, it's taken a decade of relative isolation to come to that conclusion - unclouded by all the intense amount of jabbering that takes place in cities (well, excepting that last U.S. election cycle - which was disappointing to everyone for none of the same reasons.)
This can be looked at with both right- and left-brained logic (which are obviously metaphors.)
When you get these both working in sync, then you have something which enables you to survive better. This is the road to happiness - or at least, a general feeling of satisfaction.
And now we also have a real meaning for "common sense." This is based on observing what is going on around you, but that sensing is done on many levels. It brings up Serge Kahili King's 4 levels of Analysis - Objective, Subjective, Holistic, and Symbolic.
Sorting out things around you with those four will tend to sync things for you - as long as they all align. Of course, you don't have to do that, but these give you options of how to process things happening around you.
Common sense would be a set of patterns where your own experience is sorted out completely. You have patterns which are very survival for the area you live in and around. City folks lack sense in the country (I recall high-heels being worn on gravel paths in winter.) And rural residents will often shy away from much travel in cities as they are confronted by so much to process so quickly - their established patterns won't keep up easily.
But you can live anywhere.
This means a global common sense is possible - not for every single thing that happens, but for continuing trends.
I'll leave you with that for now.
Luck (and peace) to us all.
PS. Look for a few new books coming out - one on the underlying story-telling premise of marketing...
| 201,138
|
propertyaround.com was registered 7 years 2 months ago. It has an alexa rank of #447,070 in the world. It is a domain having .com extension. Current estimated worth of $ 14,400.00 and have a daily income of around $ 15.00. As no active threats were reported recently, propertyaround.com is SAFE to browse.
Open Magazyn.pl
Official Site of Jack Link's protein snacks. Find product and nutritional information, Messin' With Sasquatch videos, promotions and more.
| 250,890
|
What are the health benefits of cosmetic dentistry?
Most of us have something we’d like to change about our smile if we could, but maybe we hesitate because it seems like an impractical expense. When it comes to cosmetic dentistry in Seattle, WA, however, some treatments can actually improve your wellness—both your oral health and your emotional well-being.
When teeth are missing, your jawbone shrinks. Dental implants behave like tooth roots, stimulating your jawbone to remain viable. Dental implants also stop your teeth from shifting, which can lead to future wear and further tooth loss.
A dental implant is a small titanium post implanted in your gum. This post bonds with your jawbone over a period of months and becomes the foundation for a natural-looking tooth restoration.
It’s also important to consider your emotional health. If you feel bad about your smile and are not confident enough to laugh out loud, this impacts your self-esteem.
At Phinney Ridge Dental, we provide beautiful restoration teeth in Seattle, WA for dental implants, and we offer cosmetic dentistry options to fit most budgets.
These options include:
- Professional teeth whitening
- Dental bonding
- Tooth-colored fillings
- Porcelain veneers
Get in Touch with Us
If you’re considering dental implants in Seattle, WA, our top cosmetic dentist is here to help. Contact us today to schedule a consultation with our Seattle, WA dentist.
| 311,873
|
\begin{document}
\numberwithin{equation}{section}
\title[Counting the minimal number of inflections of a plane curve]
{Counting the minimal number of inflections of a plane curve}
\author[G. Nenashev]{Gleb Nenashev}
\address{ Chebyshev Laboratory, St. Petersburg State University, 14th Line, 29b, Saint Petersburg, 199178 Russia.}
\thanks{The author is supported by Rokhlin grant,
by the Chebyshev Laboratory (Department of Mathematics and Mechanics, St. Petersburg State University) under RF Government grant 11.G34.31.0026
and by JSC "Gazprom Neft".}
\email{glebnen@mail.ru}
\begin{abstract}
Given a plane curve $\ga: S^1\to \bR^2$, we consider the problem of determining the minimal
number $I(\ga)$ of inflections which curves $\diff(\ga)$ may have, where $\diff$
runs over the group of diffeomorphisms of $\bR^2$. We show that if $\ga$ is an immersed curve with $D(\ga)$
double points and no other singularities,
then $I(\ga)\leq 2D(\ga)$. In fact, we prove the latter result for the so-called plane doodles which are finite collections of closed immersed plane curves whose only singularities are double points.
\end{abstract}
\maketitle
\section{\bf Introduction}
It is obvious that any plane curve $\ga:S^1\to \bR^2$ diffeomorphic to the figure-eight must have at least two inflection points.
Generalizing this observation, B.~Shapiro posed in \cite{Sh} the problem of finding/estimating the minimal number
of inflection points of a given immersed plane curve having only double points
under the action of the group of
diffeomorphisms of the plane.
He obtained a number of results for the class of the so-called tree-like curves characterized by the
property that removal of any double point makes the curve disconnected.
\begin{definition}
A tree-like curve
is a closed immersed
plane curve with follow property:
removal of any double point with its neighborhood makes the curve disconnected
\end{definition}
In particular, using a natural plane tree associated to any tree-like curve, he got lower and upper bounds for the number
of inflections for such curves and also found a criterion
when a tree-like curve can be drawn without inflections.
When we say that a curve $\gamma$ can be drawn with a certain number
of inflection points we mean that there is a plane diffeomorphism
$\diff$
such that $\diff(\gamma)$ has
that many inflections.
Respectively drawing is $\diff(\gamma)$.
In what follows we shall work with the following natural generalization of immersed plane curves with at most double points, comp. e.g. \cite{Me1}.
\begin{definition} A doodle is a union of a finite number of closed immersed plane curves
without triple intersections.
\end{definition}
The main result of this note is as follows.
\begin{theorem}\label{th:main}
Any doodle with $n$ double points can be drawn with at most $2n$ inflection points.
\end{theorem}
We conjecture the following stronger statement.
\begin{conjecture}\label{conj:main}
Any closed plane curve with $n$ double points can be drawn with at most $n+1$ inflection points.
\end{conjecture}
This conjecture is true for tree-like curves.
\begin{theorem}\label{th:tree}
Any tree-like curve with $n$ double points except figure-eight can be drawn with at most $n$ inflection points.
\end{theorem}
The bound from Theorem~\ref{th:tree} is tight. There are examples with $2k$ double points, which can not be drawn with less than $2k$ inflections. We must take the closed curve with alternating $2k$ loops by turn outward and inward.
In complement to Theorems~\ref{th:main} and~\ref{th:tree},
we present in \S3 an infinite family of topologically distinct minimal fragments forcing
an inflection point which implies that the problem of defining the exact minimal
number of inflection points of a given doodle is algorithmically very hard. Therefore there is no chance to obtain an explicit formula for the latter number except for some very special families of plane curves.
Our results seem to support the general principle that invariants of curves and knots of geometric origin
are difficult to calculate even algorithmically. Observe that algebraic invariants of doodles similar
to Vassiliev invariants of knots were introduced by V.~I.~Arnold in \cite{Ar} and
later considered by number of authors. See especially, \cite{Me1}, \cite{Me2}.
\medskip
\noindent
{\bf Acknowledgement.} The author is grateful to the Mathematics Department
of Stockholm University for the financial support of his visit to
Stockholm in November 2013 and to Professor B.~Shapiro for the formulation of the problem.
\bigskip
\section{\bf Proofs}
\begin{proof}[Proof of Theorem~\ref{th:main}] Assume the contrary, i.e. that there exists a doodle with $n$ double points
which can not be drawn with less than $2n+1$ inflections.
Let us consider a counterexample with the minimal number of double points.
Obviously our counterexample is not an embedded circle and it is connected.
Consider this doodle as an (obvious) planar graph $G$
with possible
multiple
edges and loops.
Double points are the vertices of this graph, and the arcs
connecting double points are the edges.
By faces of a doodle we mean the bounded faces of the (complement to the) planar graph.
By the length of a face we mean the number of edges
in its boundary.
\begin{lemma}
A minimal counterexample has the following properties:
\begin{itemize}
\item $a)$ there are no faces of length $1$.
\item $b)$ there are no faces of length $2$.
\item $c)$ there are no edges of multiplicity $\geq 3$.
\end{itemize}
\begin{proof}
{\bf a)}
{\it Assuming that there exists a face of length $1$}; remove temporarily
its boundary and remove the resulting vertex of valency $2$
by gluing two edges into one. (It might happen that there will be
no vertices left.) Then we obtain a graph corresponding to a doodle
with $n-1$ double points.
Thus we can draw a new doodle with at most $2n-2$ inflection points.
Then by returning back the removed face we add no more than $2$ inflection points, see Fig.~\ref{return-pet}.
Contradiction with the minimality assumption.
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.4]{pet.eps}
\caption{Returning the face of length $1$.}
\label{return-pet}
\end{figure}
}
{\bf b)} {\it Assuming that there exists a face of length $2$}, denote the vertices of
this face by $A, B$, and its edges by $l_1, l_2$. Vertices $A$ and $B$ are distinct, since otherwise this common vertex
would have valency $4$, and therefore there exist edges joining this face with other vertices.
But then our doodle has just one double point; it is easy to check that this can
not be a counterexample.
Remove edges $l_1,l_2$ and contract $A$ and $B$ to one vertex called $\widehat{AB}$. We obtain a new
doodle with $n-1$ double points. By the minimality of our counterexample we can draw it with no more than $2n-2$ inflection points.
Ungluing the double vertex $\widehat{AB}$ and smoothing the resulting picture we add exactly two new inflection points, see Fig.~\ref{return-kr}.
Contradiction with the minimality assumption.
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.55]{kr1.eps}
\caption{Returning pairs of double edges bounding the face of length $2$.}
\label{return-kr}
\end{figure}
}
{\bf c)} {\it Assume that there exists a triple edge}.
Consider edges $l_1,l_2,l_3$, forming this triple edge and connecting a pair of vertices called $A,B$.
(Observe that $A$ and $B$ are distinct, since otherwise their valency should be $6$, but the maximal valency is $4$.)
Edges $l_1,l_2,l_3$ divide the plane in two finite domains and one infinite. Let us denote the finite domains by $\sigma_1, \sigma_2$.
Both vertices $A$ and $B$ have exactly one additional edge each.
Either both these edges go inside $\sigma_i$ ($i=1,2$), or none of them goes inside $\sigma_i$.
(Otherwise in the graph induced by all vertices inside $\sigma_i$ one vertex will have valency $3$ and
the remaining will have valency $4$, but the sum of all valencies must be even!).
Thus edges can not go into $\sigma_1$ and $\sigma_2$ simultaneously.
Without loss of generality assume that these edges do not go into $\sigma_1$. But then
either the doodle is disconnected which is impossible, or
$\sigma_1$ is a domain with empty interior which is impossible by b).
\end{proof}
\end{lemma}
Notice that in our doodle there still might be double edges or loops with non-empty interior.
Let us split each loop into three subedges by adding two fake vertices.
Additionally in each double edges we split one of them into two subedges by adding one fake vertex.
Denote by $G'$ the obtained planar graph;
it does not contain multiple edges or loops.
By Fary's theorem it has a drawing $ \zeta $ in which all the edges are straight segments
and $ \zeta $ is equivalent to the original drawing.
Denote by $ \zeta '$ the drawing of the graph $G$ obtained by a smoothening of the angles between the edges at each vertex in the drawing $ \zeta $ (see Fig.~\ref{sgl}).
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.5]{sgl.eps}
\caption{Smoothing a vertex of valency $4$.}
\label{sgl}
\end{figure}
}
\begin{lemma}
In the drawing $ \zeta '$ each edge of the graph $ G $ contains at most one inflection.
\begin{proof}
If we do not split an edge, then obviously it has at most one inflection.
If an edge is split into three subedges, then it is a loop. Call it $ ABC $, where $ B, C $ are the fake vertices.
Vertex $A$ is the original and hence its valency is $4$. Thus it has exactly two other edges.
Either both other edges go inside the triangle $\triangle ABC$ or both go outside this triangle.
If they go outside, then either this loop is a face of length $1$ or our doodle is disconnected.
Hence, both edges go inside $\triangle ABC $ (see Fig.~\ref{ABC}~$ I $) and then the loop $ABC$ has no inflections.
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.4]{ABC.eps}
\caption{}
\label{ABC}
\end{figure}
}
It remains to consider the case when the edge splits into two subedges. Call it $ ABC $ with the fake vertex $ B $.
Then $AC$ is an edge in the graph $G$.
Consider the triangle $\triangle ABC $.
If we go along the edge $ABC$ across $C$ in the doodle we go inside a triangle or along the edge $CA$.
Then the part $BC$ of the edge $ABC$ has no inflection. Hence, there is at most one inflection on the edge $ABC$.
In the remaining case we go outside of the triangle; then the fourth edge of $ C $
also goes outside it (see Fig.~\ref{ABC}~$II$).
Similarly, we need to consider the case when other edges of vertex $A$ go outside $\triangle ABC$.
Then these edges do not go inside $\triangle ABC$, hence, either there is a face of length 2
or the doodle is disconnected. Both case are impossible.
Since we covered all possible cases, the lemma is proved.\end{proof}
\end{lemma}
Lemma~3 implies that the number of inflections does not exceed the number of edges,
hence it is at most $2n$. Theorem~1 is proved. \end{proof}
\bigskip
\begin{proof}[Proof of Theorem~\ref{th:tree}]
We prove that minimal number of inflections is not more than number of double points plus $1$.
The idea of the proof without plus $1$ can be found in the remark~\ref{m1}.
Now assume the contrary, i.e. that there exists a tree-like curve with $n$ double points
which can not be drawn with less than $n+2$ inflections.
Let us consider a counterexample with the minimal number of double points. Obviously, $n>1$.
Consider the tree that corresponds to our curve.
We split our curve into $n+1$ closed parts of curve, these parts corresponds to vertices of the tree
and points of tangency of these parts corresponds to edges of the tree (see fig.~\ref{tree}, more information about appropriate tree see in~\cite{Sh}).
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.4]{tree.eps}
\caption{}
\label{tree}
\end{figure}
}
If the outer face corresponds to vertex of the tree, then we call this vertex {\it bad}, all other vertices are called {\it good}.
Let $v_1\dots v_k$ be the longest path in the tree.
\begin{lemma} For a minimal counterexample the following conditions are impossible.
\begin{itemize}
\item $a)$ The vertex $v_2$ ($v_{k-1}$) has degree $deg(v_2)>2$ and it is adjacent to $deg(v_2) - 1 $ good leaf vertices.
\item $b)$ The vertex $v_2$ ($v_{k-1}$) has degree $deg(v_2)=2$, $v_1,v_2$ are good and $v_1\cup v_2$ is not boundary
of outer face.
\end{itemize}
\begin{proof}
{\bf a)} The vertex $v_2$ is adjacent to $deg(v_2)$ vertices and $deg(v_2)-1$ of them are good leaves. Consequently, there are two good leaves which are attached in a sequence.
Removing these two leaves, we obtain a smaller tree-like curve, hence, it is not a counterexample.
We can draw this curve with number of inflections is less that number of double points plus $1$
and after that we return two deleted leaves with addition no more than two inflections (see fig.~\ref{del-2pet}, left 1-3).
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.2]{del-2pet.eps}
\caption{}
\label{del-2pet}
\end{figure}
}
{\bf b)} The vertex $v_2$ is adjacent only to vertices $v_1$ and $v_3$, furthermore $v_3$ is attached to outer side of $v_2$.
Removing vertices $v_1$ and $v_2$, we obtain a smaller tree-like curve, hence, it is not a counterexample.
We can draw this curve with number of inflections is less that number of double points plus $1$
and after that we return two deleted vertices with addition no more than two inflections (see fig.~\ref{del-2pet}, 4-7).
\end{proof}
\end{lemma}
Now return to the proof of our theorem. Consider the next case, let $deg(v_2)>2$.
The vertex $v_2$ is adjacent to at least $deg(v_2)-1$ leaves. Hence (by $a)$), one of these leaves is bad and $k>3$ (otherwise $v_2$ is adjacent to $deg(v_2)$ leaves).
Then $v_{k-1}$,~$v_{k}$ are good and $v_{k-1}\cup v_{k}$ is not a boundary of outer face, hence, $deg(v_{k-1})>2$ (otherwise we have a contradiction to $b)$).
Similarly, $v_ {k-1}$ is adjacent to the bad vertex too. Then the bad vertex has degree at least two, but it is a leaf in this case. Hence, this case is not possible.
Then $v_2$ has degree $2$ and, analogically, the vertex $v_{k-1}$ has degree $2$. Furthermore, $v_1$, $v_2$ or $v_1\cup v_2$ is the boundary of outer face (otherwise we have a contradiction to $b)$).
Similarly, $v_{k-1}$, $v_{k}$ or $v_{k-1}\cup v_{k}$ is the boundary of outer face. Hence, $k=3$ and $v_2$ is a bad vertex. Then all vertices except $v_2$ are attached to the inner side of $v_2$, but this tree-like curve can be drawn without inflections. This is a contradiction. We consider all possible cases, the theorem is proved.
\end{proof}
\begin{remark}
\label{m1}
To prove the bound without plus $1$ we must prove that tree-like curves with $3$ double points are not counterexamples, because our proof is based on the step from $n$ to $n-2$.
\end{remark}
\bigskip
\section{\bf On minimal fragments forcing an inflection.}
\begin{definition} A {\it fragment} is the union of a finite number of immersed plane curves without triple intersections
(up to diffeomorphisms).
\end{definition}
Obviously, if a doodle $\ga$ has $k$ disjoint
fragments forcing an inflection (see next definition), then any drawing of $\ga$
contains at least $k$ inflections.
\begin{definition} A fragment is called {\it a minimal fragment forcing an inflection} if the following two conditions are satisfied (see Fig.~\ref{min}):
\begin{itemize}
\item any drawing of this fragment necessarily contains an inflection point.
\item removing any double point or any curve or cutting any curve (between two double points) we obtain a fragment which can be drawn without inflection points.
\end{itemize}
\end{definition}
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.5]{min.eps}
\caption{Fragments forcing an inflection point. $a$ -- non-minimal, $b,c$ -- minimal.}
\label{min}
\end{figure}
}
\begin{remark}Obviously, any
minimal fragment forcing an inflection
is
connected.
\end{remark}
In this section we construct an infinite series of minimal fragments forcing an inflection. Additionally,
this construction implies the following result:
\begin{theorem}
\label{exp}
There exists $c>0$ such that the number of fragments
forcing an inflection with at most $n$ double points is at least $e^{cn}$.
\end{theorem}
The above theorem is true even for fragments consisting of curves without self-intersections, but for $n$ at least some $N_0$.
That fact in its turn makes it very hard not only to count the minimum number of inflections of a given doodle but also to find a
criterion when a doodle can be drawn without inflections.
Now let us construct a series of minimal fragments.
\begin{definition}
{\it A key} $b$ for the curve $z$ is a curve shown in Fig.~\ref{key}.
\end{definition}
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.5]{key.eps}
\caption{Key $b$ for a curve $z$.}
\label{key}
\end{figure}
}
\begin{lemma}
\begin{enumerate}
\item If a drawing of a curve $z$ has no inflections,
then its key determines the direction of convexity of the curve $z$.
\item If a curve $z$ is convex in the right direction, then its key can always be drawn without inflections.
\item If a part of key is removed,
then the remaining parts of the key can always be drawn without inflections.
\end{enumerate}
\begin{proof}
Items $1$ and $3$ are obvious. In the right-hand of Fig.~\ref{key} it is shown how to draw a key in item $2$.
\end{proof}
\end{lemma}
Now we present an infinite series of distinct minimal fragments forcing an inflection.
It consists of fragments having the following form:
\begin{itemize}
\item $k\geq 3$ curves bound a domain in which each curve intersects only with its neighbors and goes after crossing inside the domain
(see Fig.~\ref{ex}, left).
\item Each of these curves has either a key of type $II$ or $III$ or a loop close to one of its endpoints (see Fig.~\ref{ex}).
\end{itemize}
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.4]{ex.eps}
\caption{Minimal fragments forcing inflection points.}
\label{ex}
\end{figure}
}
\begin{theorem}
\label{omega}
Fragments in the above series are minimal fragments forcing inflections.
\begin{proof}
Consider a fragment consisting of $k$ curves (excluding keys).
This fragment must contain an inflection point, because otherwise
all curves are convex
inwards
(due to the presence of keys or loops)
and the "vertices" of the $k$-gon have the same convexity, but this is impossible.
It remains to prove that this fragment is minimal.
{$1^\circ$} {\it If we remove something from at least one key or a loop or cut a loop or a key of type~$II$.}
Then we can draw one curve convex
outwards
and
the others convex inwards
(see Fig.~\ref{ex-del}, left).
After that we can draw all loops and other keys without inflections.
{$2^\circ$} {\it If we remove a part of a curve inside a key of type~$III$.}
Then we can draw the part of a curve convex
outwards,
the other part of this curve and other curves
convex inwards,
and
this key of type~$III$
without inflections (see Fig.~\ref{ex-del}, middle and right).
After that we can draw all loops and keys without inflections.
{$3^\circ$} {\it If we remove a part of curve inside a key of type~$II$.}
This case can be proved
by combining the ideas of cases $1^\circ$ and $2^\circ$.
We draw "big" part of the curve convex outwards and other $k-1$ curves inwards (see Fig.~\ref{ex-del}, left)
and later we draw the key of type~$II$ with second part of this curve on the end of "big" part (see Fig.~\ref{ex-del}, right).
After that we can draw all loops and keys without inflections.
{$4^\circ$} {\it If we cut a curve in the boundary of k-gon outside a key of type~$III$.}
This case
is obvious. We can do all $k$ curves with convexity in the correct direction, because
we should not build a "$k$-gon".
{\begin{figure}[htb!]
\centering
\includegraphics[scale=0.3]{ex-del.eps}
\caption{}
\label{ex-del}
\end{figure}
}
We
have
considered all possible cases,
so
the theorem is proved.
\end{proof}
\end{theorem}
\begin{proof}[Proof of Theorem~\ref{exp}]
We will use only loops (similarly, we could use only keys of type $II$ and $III$). Fixing $k>0$, we have 2 possibilities for each loop.
Hence, we have at least $2^k/k$ minimal fragments, because each fragment is considered at most $k$ times.
They have exactly $2k$ double points.
Now it is obvious that there exists desired $c>0$.
\end{proof}
| 92,351
|
TITLE: Weak*-closure of finite rank operators on dual space
QUESTION [14 upvotes]: Given a Banach space $X$, we consider the space $B(X^*)$ of bounded, linear operators on $X^*$ with the weak*-topology from its canonical predual $B(X^*)_*=X^*\hat{\otimes}X$. What is $\overline{F(X^*)}^{wk*}$, the weak*-closure of the finite rank operators on $X^*$? Since this is rather vague, here are some concrete questions:
Q1: Do we always have $K(X^*)\subseteq\overline{F(X^*)}^{wk*}$, i.e., is every compact operator in the weak*-closure of finite-rank operators?
Q2: Is there a characterization when $B(X^*)=\overline{F(X^*)}^{wk*}$, i.e., when the finite-rank operators are weak*-dense?
Q3: Is there a characterization when $B(X^*)=\overline{K(X^*)}^{wk*}$, i.e., when the compact operators are weak*-dense?
Considering $B(X)$ as a subalgebra of $B(X^*)$ in the usual way, we may ask related questions in connection with $F(X)$ and $K(X)$. The principle of local reflexivity implies $\overline{F(X)}^{wk*}=\overline{F(X^*)}^{wk*}$ in $B(X^*)$. However, it is not clear to me if we always have $\overline{K(X)}^{wk*}=\overline{K(X^*)}^{wk*}$ (I guess not). Therefore, we may also ask:
Q4: Do we always have $K(X)\subseteq\overline{F(X^*)}^{wk*}$?
REPLY [8 votes]: Regarding Q2: If and only if $X$ has the approximation property. I'll use Ryan's book "Introduction to tensor products of Banach spaces" as a reference, see Prop 4.6 (but this is all standard stuff).
Theorem: $X$ has the approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \mu_n(x) x_n = 0$ for all $x\in X$, then $u=0$.
As we have $\sum_n \|\mu_n\| \|x_n\|<\infty$, the condition is equivalent to $\sum_n \mu_n(x) \mu(x_n)=0$ for all $x\in X,\mu\in X^*$ (Hahn-Banach) and hence also equivalent to $\sum_n \mu(x_n)\mu_n = 0$ for all $\mu\in X^*$, and so finally also equivalent to $\sum_n \mu(x_n) f(\mu_n) = 0$ for all $\mu\in X^*, f\in X^{**}$.
This in turn is equivalent to $\langle u, F \rangle=0$ for all finite rank operators $F$ on $X^*$, under your dual pairing between $X^*\hat\otimes X$ and $F(X^*)$.
Finally, observe that $F(X^*)$ is weak$^*$ dense in $B(X^*)$ if and only if the only element of $X^*\hat\otimes X$ which annihilates all of $F(X^*)$ is $0$.
There is a related definition of the "compact approximation property". If I recall it correctly, then you can adapt the proof, and get
Thm: $X$ has the compact approximation property if and only if, whenever $u=\sum_{n\geq 1} \mu_n\otimes x_n \in X^*\hat\otimes X$ is such that $ \sum_n \langle\mu_n,T(x_n)\rangle = 0$ for all $T\in K(X)$, then $u=0$.
Thus, if $X$ has the compact approximation property, but not the approximation property (I think there is an example due to Willis) then we can find $u\in X^*\hat\otimes X$ which annihilates all of $F(X^*)$ but is not zero. There is then $T\in K(X)\subseteq K(X^*)$ with $\langle T,u\rangle \not=0$, and as
$$ \overline{F(X^*)}^{wk^*} = \{ T\in B(X^*) : \langle T,u\rangle=0
\text{ for all }u\in X^*\hat\otimes X\text{ with } \langle S,u\rangle=0
\text{ for all } S\in F(X^*) \} $$
we conclude that $K(X)$ is not contained in the weak$^*$-closure of $F(X^*)$. So Q4 is a negative.
| 218,911
|
TITLE: How many sequences of $k$ elements in ascending order from a set $S$?
QUESTION [0 upvotes]: Suppose I have a set $S = \{1,2,\ldots,n\}$. How many sequence $r$ contains $k$ elements from set $S$ in ascending order if:
$r$ contains repetition, i.e. an element in $S$ can appear several time in $r$.
$r$ does not contain repetition.
This is actually from a programming question in topcoder, and I wonder if there exist formulas to count the number of those sequences, rather than using a brute-force algorithm to count.
REPLY [1 votes]: For the second one (the first one is clear by now I guess) - choose $k$ elements from the set $S$ and arrange them in ascending order - that's it. So, it is $n\choose k$.
| 129,441
|
\begin{document} \maketitle
\begin{abstract}
This paper concerns towers of curves over a finite field with many
rational points, following Garcia--Stichtenoth and Elkies. We present
a new method to produce such towers. A key ingredient is the study of
algebraic solutions to Fuchsian differential equations modulo $p$. We
apply our results to towers of modular curves, and find new
asymptotically good towers.
\noindent 2000 {\em Mathematics Subject Classification}. Primary
11G20. Secondary: 14H35, 14G05, 14G50. \end{abstract}
\section{Introduction}\label{introsec}
Let $p$ be a prime and $q=p^a$, for some $a>0$. Consider a projective
smooth curve $X$ of genus $g$, defined over $\FF_q$, and write
$N_q(X)$ for the number of $\FF_q$-rational points of $X$. We write
$N_q(g)$ for the maximum of $N_q(X)$, taken over all curves $X$ of
genus $g$ which are defined over $\FF_q$. The
Drinfel'd--Vl\u{a}du\c{t} bound \cite{DrinfeldVladut} states that
\[
A(q):=\limsup_{g\to\infty}\frac{N_q(g)}{g}\leq \sqrt{q}-1.
\]
Moreover, $A(q)> c\log(q)$, where $c>0$ is a constant
\cite{Serre83}.
Garcia--Stichtenoth \cite{GS} constructed many examples of infinite
towers of curves $\cdots \to X_{m+1}\to X_{m}\to\cdots \to X_0$
defined over a finite field $\FF_q$ such that the limit of
$N_{q^2}(X_m)/g(X_m)$ is $q-1$. Such towers are called {\sl
asymptotically optimal}. If this limit is positive, the tower is
called {\sl asymptotically good}. Asymptotically optimal towers have, for
example, interesting applications to coding theory
(\cite{Goppa,tsfa-vlad}). For these applications it is important to
have explicit equations for the curves $X_m$. Garcia--Stichtenoth
define towers of curves recursively, starting from a correspondence
$(g,h):X_0\rightrightarrows X_{-1}$, by taking suitable (normalized)
Cartesian products. A nice feature of this recursive definition is
that one obtains explicit equations for all curves $X_m$ starting from
an equation for $(g,h)$. One has to choose the correspondence very
carefully for the corresponding tower to have many rational
points. Garcia--Stichtenoth find correspondences that work, but they
do not give a systematic method for finding such correspondences.
Elkies (\cite{Elkies97}) applies this approach to correspondences
$(g,h): X_0(\ell^2)\rightrightarrows X_0(\ell)$, for certain small
values of $\ell$. Here $g$ is the natural projection and $h$ is the
composition of $g$ with the Atkin--Lehner involutions on both
sides. The corresponding tower is $\cdots \to X_0(\ell^{m+1})\to
X_0(\ell^m)\to \cdots $. This gives equations for modular curves
$X_0(\ell^m)$ starting from equation from
$(g,h):X_0(\ell^2)\rightrightarrows X_0(\ell)$. This is a second
important application of the theory.
Elkies also constructs other asymptotically optimal towers of curves,
starting from correspondences between other Shimura varieties, such as
Drinfel'd modular curves. These Shimura varieties are moduli spaces
of curves, surfaces etc, and the correspondences are an analog of the
Hecke correspondences for modular curves. Elkies shows that all
asymptotically optimal towers constructed by Garcia--Stichtenoth et.\
al.\ are of this form (\cite{Elkies97}, \cite{Elkies01}). Elkies
suggests that all asymptotically optimal towers arise in this way
(\cite[Fantasia]{Elkies97}).
In this paper we give a new method for constructing asymptotically
good towers. We extract the essential ingredients
from the approach of Garcia--Stichtenoth and Elkies, and formulate a
general set-up. Our approach is concrete and not just applicable to
towers of modular curves. This allows for a more systematic search for
asymptotically optimal towers.
We start from a correspondence $(g,h): X_0\rightrightarrows X_{-1}$
over a finite field $\FF_q$, together with a Fuchsian differential
equation on $X_{-1}$. We say that the correspondence $(g,h)$ is
adapted to the differential equation if the pull back via $g$
is equivalent to the pull back via $h$ (see Section \ref{desec}
for precise definitions). The correspondence $(g,h)$ gives rise to a
tower of curves ${\mathcal T}_{g,h}=(X_m)_{m\geq 0}$. Under some technical
assumptions we show the following.
\bigskip\noindent {\bf Theorem} \ref{thm:asgood}
{\sl The tower
${\mathcal T}_{g,h}$ is asymptotically good. This means that the limit
of $N_{q}(X_m)/g(X_m)$ is positive.}
\bigskip\noindent One of the assumptions we make is that $g$ and $h$
are tame, i.e.\ the characteristic of the ground field does not divide
the ramification indices. We also give a criterion for the tower
${\mathcal T}_{g,h}$ to be asymptotically optimal (Theorem
\ref{thm:asopt}).
The reason why such towers have many rational points is roughly the
following. We suppose that the differential equation has an algebraic
solution $\Phi$. After extending the field of definition $\FF_q$ of
the correspondence, we may assume that the zeros and poles of $\Phi$
are $\FF_q$-rational. The set of these zeros and poles has a subset
${T}$ with the following property. For every $P\in
{T}$ and every $m$, the inverse image of $P$ in $X_m$
consists of unramified and $\FF_q$-rational points.
It appears that all known examples of tame asymptotically optimal
towers can be reformulated in these terms. The reason is that, by
Elkies' work, the known examples come from certain correspondences
between Shimura curves. (In fact, the tame towers are all towers of
modular curves.) Such moduli spaces come naturally equipped with a
differential equation: the Picard--Fuchs differential equation of a
versal family of the objects it parameterizes. For towers of modular
curves we work this out in Section \ref{modularsec}. We expect that it
is possible to generalize (parts of) our method to wildly ramified
towers.
The idea for using differential equations for studying the growing
behavior of rational points in a tower came from \cite{GS}. In that
paper Gau\ss' hypergeometric differential equation was used to prove a
property for the Deuring polynomial. We show that the arguments of
\cite{GS} vastly simplify and generalize if one makes a more
systematic use of differential equations. Our method is also related
in spirit to older work of Ihara (see \cite{Ihara} for a
survey). However, Ihara's work only applies to towers of Shimura
curves. Moreover, it uses $p$-adic uniformization to count points. We
work purely in characteristic $p$ which is more convenient in practice.
To find new examples of asymptotically good towers, we construct
correspondences via pull back. Given a correspondence
$(g,h):X_0\rightrightarrows X_{-1}$ and an arbitrary map $f:Y_{-1}\to
X_{-1}$, we define a new correspondence $(\tilde{g},
\tilde{h}):Y_0\rightrightarrows Y_{-1}$. This gives a systematic
construction of the towers of modular curves found by Elkies in
\cite[Appendix]{LMS}. This allows to find very many asymptotically
good towers.
The situation for asymptotically optimal towers is more
complicated. We give a criterion for the pull back of an
asymptotically optimal tower to be again asymptotically optimal
(Theorem \ref{thm:pullbackopt}). Since our approach does not use the
interpretation of the curves we consider as Shimura varieties, one
might expect to find counter examples to Elkies' Conjecture. However,
we did not find such an example. The reason is that in Theorem
\ref{thm:pullbackopt} there is one condition which is hard to control.
In a later paper we will come back to the question whether this idea
can be used as evidence for Elkies' Conjecture.
The organization of the paper is as follows. In Section \ref{desec} we
review and extend known results on Fuchsian differential equations on
curves in positive characteristic. In Section \ref{pointsec} we give
the recursive definition of a tower of curves corresponding to a
correspondence and establish basic properties. We estimate how the
genus and the number of rational points grow in the tower, and prove
the criterion for a tower to be asymptotically good. In Section
\ref{pullbacksec} we develop the construction of correspondences via
pull back and construct new examples. Section \ref{modularsec}
reformulates and extends some results of Elkies on towers of modular
curves.
\section{Fuchsian differential equations}\label{desec}
In this section we recall some standard results on Fuchsian
differential equations. For proofs and more details we refer to
\cite[Section 11]{Katz} and \cite{Honda}. Let $k$ be a field of
characteristic $p>0$ and $X/k$ a smooth projective curve. Let $K=k(X)$
be the function field of $X$. Suppose that $M$ is a finite-dimensional
vector space over $K$.
\begin{defn}\label{connectiondef}
A $k$-connection $\nabla$ on $M$ is an additive map
\[
\nabla:M\to \Omega^1_{K/k}\otimes_K M
\]
satisfying the Leibniz rule
\[
\nabla(fm)={\rm d}f\otimes m+f\nabla(m),
\]
for $f\in K$ and $m\in M$.
\end{defn}
Equivalently (\cite[Section 1.0]{Katz}), $\nabla$ corresponds to a
$K$-linear map \[ \nabla:\Der(K/k)\to \End_k(M) \] such that
$\nabla(D)(fm)=D(f)m+f\nabla(D)m$ for $D\in {\Der}(K/k)$, $f\in
K$ and $m\in M$. A {\sl horizontal morphism} from $(M_1, \nabla_1)$
to $(M_2, \nabla_2)$ is a morphism $ \varphi:M_1\to M_2$ of $K$-vector
spaces which is compatible with the connections, i.e.\
$\varphi(\nabla_1(D) m)=\nabla_2(D)(\varphi(m))$. We write $\MC(X)$
for the category of $K$-modules with connection.
\begin{defn}\label{singdef}
Let $P$ be a place of $K/k$ and $t$ a local parameter at $P$. For a
$K$-basis ${\mathbf e}$ of $M$, we write $\nabla({\rm d}/{\rm d}t)
{\mathbf e}=A_{\mathbf e} \cdot {\mathbf e}$ with $A_{\mathbf e}\in
M_n(K)$, where $n=\dim_K M$. We say that $P$ is a {\sl singular point}
of $(M, \nabla)$ if the matrix $A_{\mathbf e}$ has a pole at $P$ for
every basis ${\mathbf e}$.
\end{defn}
\begin{defn}\label{cyclicdef}
We say that $(M, \nabla)$ is {\sl cyclic} if there exists a vector
$m\in M$ and a nonzero derivation $D\in {\Der}(K/k)$ such
that $m, \nabla(D)m, \ldots \nabla^{d-1}(D)m$ span $M$ over $K$, where $d=\dim_K M$.
\end{defn}
It is shown in \cite[11.4]{Katz} that the notion of a cyclic module is
independent of the choice of the derivation $D$. In the rest of this
section, we suppose that $M$ is cyclic, and of $K$-dimension $d=2$.
A $K$-basis $m, \nabla(D)m$ of $M$ is called a {\sl cyclic basis}.
If $P$ is place of $K/k$, we write ${\mathcal O}_P$ (resp.\
${\mathfrak m}_P$) for the local ring (resp.\ the maximal ideal) at
$P$. Let
\[
\Der_P(K/k)=\{D\in {\Der}(K/k)\, |\,
D({\mathfrak m}_P)\subset {\mathfrak m}_P\}.
\]
If $t=t_P$ is a local parameter of $X$ at $P$, then ${\Der}_P(K/k)$ is
a free ${\mathcal O}_P$-module with basis $t\,{\rm d/d}t$.
\begin{defn}\label{regsingdef}
Let $P$ be a singular point of $(M,\nabla)$ and $t$ a local parameter
at $P$. For a $K$-basis ${\mathbf e}$ of $M$, we write $\nabla(t\,{\rm
d}/{\rm d}t) {\mathbf e}=B_{\mathbf e} \cdot {\mathbf e}$ with
$B_{\mathbf e}\in M_2(K)$. We say that $P$ is a {\sl regular
singularity} if there exists a $K$-basis ${\mathbf e}$ of $M$ such
that $B_{\mathbf e}$ is holomorphic at $P$. If all singularities of
$(M,\nabla)$ are regular, we say that $(M, \nabla)$ is a {\sl Fuchsian
module}.
Suppose that $P$ is a regular singularity of $(M,\nabla)$, and let
$B_{\mathbf e}\in M_n({\mathcal O}_P)$ be as above. Write $B_{\mathbf
e}(0)$ for the value of $B_{\mathbf e}$ at $t=0$. The characteristic
polynomial of $B_{\mathbf e}(0)$ is called the {\sl indicial
equation}. Its roots are the {\sl local exponents.}
\end{defn}
If $P$ is not a singularity its local exponents are $0,1=\dim_K
M-1$. The converse need not be true. Singularities with local
exponents $0,1=\dim_K M-1$ are called {\sl apparent
singularities}.
We now associate to $(M, \nabla)$ a $2$nd order differential equation.
Let $P$ be a regular singularity of $(M, \nabla)$ and $t$ a local
parameter at $P$. Let
${\mathbf e}=(e_1, e_2:=\nabla({\rm d}/{\rm d}t)(e_1))$ be a cyclic basis
of $M$. Write
\begin{equation}\label{cyclicbasiseq}
\nabla(\frac{\rm d}{{\rm d}t}){\mathbf e}=A\cdot {\mathbf e}, \qquad\mbox{
with }\qquad A=\begin{pmatrix}0&-a_2\\
1&-a_1\end{pmatrix} .
\end{equation}
It is easy to check that the fact that $P$ is a regular singularity
means that we may choose ${\mathbf e}$ such that $a_i$ has a pole
of order at most $i$ at $P$ for $i=1,2$.
Let $M^\ast=\Hom_{K}(M, K)$ be the $K$-linear dual of $M$. We define a
$k$-connection $\nabla^\ast$ on $M^\ast$ by requiring that
\[
\langle \nabla(D) (m), m^\ast\rangle+\langle m,
\nabla^\ast(D)(m^\ast)\rangle=D(\langle m, m^\ast\rangle),
\]
for $D\in \Der(K/k)$, $m\in M$ and $m^\ast\in M^\ast$. One easily
checks that $\nabla^\ast$ is a connection. In fact, with respect to
the dual basis ${\mathbf e}^\ast$ of $M^\ast$, we have
$\nabla^\ast({\rm d}/{\rm d} t){\mathbf e}^\ast=-A^t{\mathbf
e}^\ast$. Here $A^t$ is the transpose of $A$.
Write $\hat{M}^\ast_P=M^\ast\otimes_K\hat{K}_P$, where $\hat{K}_P$ is the completion of $K$ at $P$. Denote by
$(\hat{M}^\ast_P)^{\nabla^\ast}$ the horizontal sections, i.e.\
\begin{equation}\label{deeq}
f_1e_1^\ast+f_2e_2^\ast\in (\hat{M}^\ast_P)^\nabla\qquad \mbox{
iff }\qquad
\left\{\begin{array}{l}
f_2=f_1',\\
L(f_1):=f_1''+a_1 f_1'+a_2f_1=0.
\end{array}\right.
\end{equation}
This is the {\sl differential equation corresponding to} $(M,
\nabla)$. Giving $(M,\nabla)$ is equivalent to giving the differential
equation $(\ref{deeq})$. We sometimes call $(M, \nabla)$ itself a
differential equation.
One computes that
\[
\nabla(t\frac{{\rm d}}{{\rm d}t})(e_1, te_2)=
\begin{pmatrix}0&-t^2a_2 \\
1& 1-ta_1
\end{pmatrix}(e_1, te_2).
\]
Writing $a_i =c_i t^{-i}+t^{-1}(\cdots)$ with $c_{i}\in k$, we find
that the indicial equation is
\begin{equation}\label{indicialeq}
X^2+(-1+c_1)X+c_2=0.
\end{equation}
The local exponents $\gamma_1, \gamma_2$ are the roots of this
equation. Note that our notion of local exponents agrees with the
classical ones. This is the reason for taking the differential
equation corresponding to the horizontal sections of $\hat{M}_P^\ast$
rather than $\hat{M}_P$.
Our next topic is algebraic solutions of Fuchsian differential
equations in positive characteristic, following Honda
\cite{Honda}. Let $(M,\nabla)\in \MC(X)$ be a cyclic module of
dimension $2$, and let $P\in X$ be a regular singularity with local
parameter $t$. Choose a cyclic basis ${\mathbf e}$ of $M$. Let $L/K$
be an algebraic extension. We say that $u\in L$ is an {\sl algebraic
solution} of $(M,\nabla)$ if it is a solution of the corresponding
differential equation (\ref{deeq}). This is equivalent to the fact
that $ue_1^\ast+u'e_2^\ast\in ({ M}^\ast\otimes_K L)^{\nabla^\ast}$. In what
follows we mainly consider solutions in $K$.
\begin{prop}\label{degsolprop}
Let $(M,\nabla)\in \MC(X)$ be a cyclic module of dimension $2$, and let
$u\in K$ be an algebraic solution (with respect to some choice of a
cyclic basis ${\mathbf e}$).
\begin{itemize}
\item[(i)] Suppose that $P$ is a regular singularity. Write $\gamma_1,
\gamma_2$ for its local exponents. Then
\[
\ord_P(u)\equiv \gamma_i\bmod{p},
\]
for some $i\in\{1,2\}$. In particular, $\gamma_i\in \FF_p$.
\item[(ii)] If $P\in X$ is a zero of $u$ we have
\[
\ord_P(u)\equiv 1\bmod{p}\qquad \mbox{ or }\qquad \ord_P(u)\equiv 0\bmod{p}.
\]
\end{itemize}
\end{prop}
\proof Let $u$ be an algebraic solution. Suppose that $P$ is a regular
singularity. Choose a local parameter $t$ at $P$. Put
$\delta:=\ord_P(u)$. In the complete
local ring $\hat{\O}_P$, we may write $u=t^\delta(\sum u_i t^i)$ with
$u_0$ a unit of $K$. By assumption, $u$ satisfies
\[
u''+a_1 u'+a_2 u=0,
\]
where $a_i $ has a pole of order at most $i$, since $P$ is a regular
singularity. Write
\[
a_1=c_{1}t^{-1}+\cdots,\qquad
a_2=c_{2}t^{-2}+\cdots .
\]
Substituting this in the differential equation and taking the
coefficient of $t^{\delta-2}$, we find that
$\delta^2+\delta(c_1-1)+c_2$. Since the indicial equation
(\ref{indicialeq}) is $X^2+(c_1-1)X+c_2$, (a) follows.
If $P$ is a regular point, we may suppose that $a_i $ does not have
a pole at $P$ for $i=1,2$. Hence (b) immediately follows from the
differential equation. \Endproof
\begin{prop}\label{polysolprop}
Let $(M,\nabla)\in \MC(X)$ be a cyclic module of dimension $2$, and
let $u_1, u_2\in K$ be algebraic solutions (with respect to some
choice of a cyclic basis ${\mathbf e}$). We suppose that
$\ord_P(u_1)\equiv \ord_P(u_2)\bmod{p}$ for some regular singularity
$P$. We write $\Div(u_i)$ for the divisor of $u_i$ on $X$. Then
\[
\Div(u_1)\equiv \Div(u_2)\bmod{p}.
\]
\end{prop}
\proof
This proof follows \cite[Proposition 5.1]{Honda}.
Consider the set $\Sigma_\delta$ of algebraic solutions $u\in K$ of
$u''+a_1u'+a_2u=0$ which are holomorphic at $P$ and such that
$\ord_P(u)\equiv \ord_P(u_1)\bmod{p}$.
Let $v_1\in \Sigma_\delta$ be a solution whose order at $P$ is minimal
and write $\delta:=\ord_P(v_1)$. Suppose there exists $v_2\in
\Sigma_\delta$ such that $\Div(v_1)\not \equiv \Div(v_2)\bmod{p}$. It
is no restriction to suppose that $v_2$ has minimal order at $P$ among
all algebraic solutions with this property. Write
$\ord_P(v_2)=\delta+p\nu$. Then there exists a constant $c\in \bar{k}$
such that the order of $w:=v_2-ct^{p\nu}v_1\in \Sigma_\delta$ at $P$
is strictly less than $\delta+p\nu$. This contradicts the choice of
$v_2$. Therefore every $u\in \Sigma_\delta$ differs by a $p$th power
from $v_1$. This proves the proposition.
\Endproof
\begin{exa}\label{GMexa}
A key example of a Fuchsian differential equation we will be
interested in in this paper, is the one coming from the Gau\ss--Manin
connection on the modular curve $X(2)$. We recall the situation from
\cite{Katz84}. The statements are easy to generalize to other modular
curves (Section \ref{modularsec}). Let $S=\Spec(\ZZ[\lambda,
1/2\lambda(\lambda-1)])$ and write ${\mathcal E}\to S$ for the
elliptic curve over $S$ given by $y^2=x(x-1)(x-\lambda)$.
We denote by $M:=H^1_\dR({\mathcal E}/S)$ the first de Rham
cohomology group, and by $\nabla:M\to
\Omega^1_S\otimes M$ the Gau\ss--Manin
connection. Write
\[
\omega=\frac{{\rm d} x}{y}=\frac{ {\rm d}
x}{[x(x-1)(x-\lambda)]^{1/2}},\qquad
\omega':=\nabla(\frac{\partial}{\partial \lambda})\, \omega= \frac{{\rm
d} x}{2[x(x-1)]^{1/2}(x-\lambda)^{3/2}}.
\]
Then $\omega$ and $\omega'$ form a basis of $M$. One computes that
\[
\nabla(\frac{\partial}{\partial \lambda})(\omega, \omega')=
\begin{pmatrix}
0&\displaystyle{ -\frac{1}{4\lambda(\lambda-1)}}\\
1&\displaystyle{ -\frac{2\lambda-1}{\lambda(\lambda-1)}}
\end{pmatrix}(\omega, \omega').
\]
The corresponding differential equation (\ref{deeq}) is
\begin{equation}\label{Gausseq}
\lambda(\lambda-1)u''+(2\lambda-1)u'+\frac{1}{4}u=0.
\end{equation}
The differential equation (\ref{Gausseq}) is Gau\ss ' hypergeometric
differential equation. It has three singularities $0,1,\infty$ with
local exponents $0,0;0,0;1/2,1/2$. Working out the statement of
Proposition \ref{degsolprop} for the singularity $P=\infty$, we obtain
the following. Let $u\in k(\lambda)$ be an algebraic solution. After
multiplying $u$ with a $p$th power, we may suppose that $u$ is a
polynomial. Then $\deg(u)\equiv -\gamma_i\bmod{p}$, were $\gamma_1,
\gamma_2$ are the local exponents at $\infty$. In our case we find
therefore that $\deg(u)\equiv -1/2\bmod{p}$.
The Deuring polynomial (or Hasse
invariant)
\begin{equation}\label{Hasseeq}
\Phi:=\sum_{i=0}^{(p-1)/2}\binom{(p-1)/2}{i}^2\lambda^i
\end{equation}
is a solution $\bmod{p}$ of this differential equation of degree
$(p-1)/2$. Proposition \ref{polysolprop} implies that every other
algebraic solution in characteristic $p$ is of the form $\psi^p \Phi$, for some $\psi\in K$.
\end{exa}
In general, a module $(M,\nabla)\in \MC(X)$ does not have algebraic
solutions. Honda \cite[appendix]{Honda} shows that $(M,\nabla)$ has
``sufficiently many solutions in a weak sense'' if and only if the
$p$-curvature of $(M,\nabla)$ is nilpotent. The notion of sufficiently
many solutions in a weak sense is stronger than just the existence of
an algebraic solution; we refer to \cite{Honda} for a
definition. However, if $(M,\nabla)$ is cyclic, has dimension two, and
its singularities are $\FF_p$-rational than the two notions are
equivalent \cite[Cor.\ 1 to Prop.\ 2.3]{Honda}. Katz \cite{Katz} shows
that in a ``geometric'' context (like in Example \ref{GMexa}) the
$p$-curvature is always nilpotent, in particular, the corresponding
differential equation has an algebraic solution in some extension
$L/K.$
Suppose that $(M_1,\nabla_1)$ and $(M_2, \nabla_2)$ are elements of
$\MC(X)$. We define a $k$-connection $\nabla$ on $M=M_1\otimes_K M_2$ by
putting $\nabla(D)(m_1\otimes m_2)=(\nabla_1(D)m_1)\otimes
m_2+m_1\otimes (\nabla(D)m_2)$, for $D\in \Der(K/k)$ and $m_i\in M_i$.
\begin{defn}\label{equivdef}
Let $(M_1, \nabla_1)$, $(M_2, \nabla_2)\in \MC(X)$ be modules with
$\dim_K M_1=\dim_K M_2$. We say that $(M_1, \nabla_1)$ is {\sl
equivalent} to $(M_2, \nabla_2)$ is there exists a one-dimensional
module $(M_3, \nabla_3)\in \MC(X)$ such that:
\begin{itemize}
\item $(M_1, \nabla_1)\otimes_K(M_3, \nabla_3)\simeq (M_2, \nabla_2)$,
\item $(M_3, \nabla_3)$ has an algebraic solution $\theta$,
\item the set of singularities of $(M_3, \nabla_3)$ is contained in
the set of singularities of $(M_1, \nabla_1)$.
\end{itemize}
\end{defn}
In terms of local coordinates this definition means the following. Let
$(M_1, \nabla_1)\in \MC(X)$ be a cyclic, Fuchsian module of dimension
two and let $(M_3, \nabla_3)\in \MC(X)$ be a one-dimensional Fuchsian
module. Let $P$ be a regular singularity of both $(M_1, \nabla_1)$ and
$(M_3, \nabla_3)$ with local parameter $t$. Choose a cyclic basis
$(e_1, e_2)$ for $M_1$ as in (\ref{cyclicbasiseq}), i.e.\ we write
\[
\nabla_1({\rm d}/{\rm d}t)(e_1, e_2)=
\begin{pmatrix}0&-a_2 \\1&-a_1 \end{pmatrix}(e_1, e_2).
\]
We identify $M_3$ with $K$, and write $\nabla_3({\rm d}/{\rm
d}t)=B\cdot 1. $ Then with respect to the basis $\xi_1=e_1\otimes 1$
and $\xi_2=Be_1\otimes 1+e_2\otimes 1$ of $M_2=M_1\otimes_K M_3$ we
find
\[
\nabla_2({\rm d}/{\rm d}t)(\xi_1, \xi_2)=
\begin{pmatrix} 0&\displaystyle{-a_2
+B a_1+B'-B^2}\\
1&\displaystyle{-a_1+2B}
\end{pmatrix}(\xi_1, \xi_2).
\]
The corresponding differential equation is
\[
y''+(a_1-2B)y'+(-B'-B a_1+a_2+B^2)y=0.
\]
One checks that if $u$ is an (algebraic) solution of the differential
equation corresponding to $(M_1, \nabla_1)$ then $\theta u$ is an
(algebraic) solution of $(M_2, \nabla_2)$. Here $\theta$ is an algebraic solution of $M_3$ (Definition \ref{equivdef}).
Let $\gamma$ (resp\ $\gamma_1, \gamma_2$) be the local exponents of
$(M_3, \nabla_3)$ (resp.\ $(M_1, \nabla_1)$) at $P$. One computes that
the indicial equation of $(M_2,\nabla_2)$ at $P$ is
$(X-\gamma_1-\gamma) (X-\gamma_2 -\gamma)$.
Suppose we are given $(M,\nabla)\in \MC(X)$ and a cover $f:Y\to X$
defined over $k$, i.e.\ $f$ is a finite separable map between smooth
and absolutely irreducible curves. Let $L=k(Y)$ be the function field
of $Y$.
\begin{defn}\label{pullbackdef}
We define the {\sl pull back} $( M_f, \nabla_f)$ on $Y$ as
follows. Write $\nabla(m)=m\otimes{\rm d}g$, where ${\rm d} g\in
\Omega^1_K$. Then $ M_f=M\otimes_K L$ and $ \nabla_f:M\otimes_K L\to
\Omega^1_L\otimes (M\otimes_K L)$ is defined by $m\otimes 1\mapsto
f^\ast(m\otimes {\rm d}g):=m\otimes {\rm d} (g\circ f)\in M\otimes_K
L\otimes_L\Omega^1_L=M_f\otimes_L\Omega^1_L.$
\end{defn}
In local coordinates this may be described as
follows. Let $Q$ be a point of $Y$ and $P$ its image in $X$. Choose a
local parameter $s$ of $Q$ and let $t=s^e$ be a local parameter of
$P$. Here $e$ is the ramification index of $Q$ in $f$. Write $f'(s)\in
\O_Q$ for the derivative of $f$ at $Q$. Choose an appropriate basis
${\mathbf e}=(e_1, e_2)$ of $M$ at $P$, and write
\[
\nabla({\rm d}/{\rm d}t){\mathbf e}=
\begin{pmatrix}0&-a_2 \\ 1&-a_1 \end{pmatrix}{\mathbf e},
\]
as above. Then
\[
\nabla_f({\rm d}/{\rm d}s)(e_1, f'(s)e_2)=
\begin{pmatrix}0&\displaystyle{-(f'(s))^2 a_2(f(s))}\\
1&\displaystyle{-f'(s)a_1(f(s))+f''(s)/f'(s)}
\end{pmatrix}(e_1, f'(s)e_2).
\]
The corresponding differential equation is
\[
L_f(v)=\left(\frac{\rm d}{f'(s){\rm d}s}\right)^2
v+a_1(f(s))\left(\frac{\rm d}{f'(s){\rm
d}s}\right)v+a_2(f(s)) v=0.
\]
One easily checks that if $(M,\nabla)$ is Fuchsian, then $(M_f,
\nabla_f)$ is Fuchsian also. Moreover, with notation as above, if $P$
is a regular singularity with local exponents $(\gamma_1, \gamma_2)$
then $(e\gamma_1, e\gamma_2)$ are the local exponents at $Q$. Note
that it may happen that $P$ is a singularity but $Q$ is not. (It is
easy to characterize this in terms of the local monodromy, but we do
not need this here.)
\begin{notation}\label{polysolnot}
Let $(M, \nabla)\in \MC(X)$ and write $S$ for its set of
singularities. Suppose that $(M, \nabla)$ has an algebraic solution
$\Phi$. Let $f:Y\to X$ be a cover and $(M_f, \nabla_f)$ the pull back
module. We write $\Phi_f:=\Phi\circ f$ for the corresponding algebraic
solution of $(M_f,
\nabla_f)$.
\end{notation}
\begin{defn}\label{corrdef}
Let $(M, \nabla)\in \MC(X)$. A {\sl
correspondence adapted to} $(M, \nabla)$ is a pair of (separable)
covers $g, h:Y\rightrightarrows X$ between smooth and absolutely
irreducible curves such that the pull back modules $(M_g, \nabla_g)$
and $(M_h, \nabla_h)$ on $Y$ are equivalent. The correspondence is
{\sl trivial} if there exists an automorphism
$\sigma:X\stackrel{\sim}{\to}X$ such that $h=\sigma\circ g$. A
correspondence $(g,h)$ is called {\sl tame} if the covers $g$ and $h$
are tame.
\end{defn}
A correspondence $(g,h):Y\rightrightarrows X$ may equivalently be
described by giving a curve $C\subset X\times X$. Here $C=\{( g(P),
h(P))\, |\, P\in Y \}$ is the {\sl curve of correspondence.} The {\sl
degree of the correspondence } is the cardinality of $\{ (x, y)\in
C\}$, where $x\in X$ is a fixed, sufficiently general point. We will
be particularly interested in correspondences of degree one. In this
case the map $Y\to C$ defined by $P\mapsto (g(P), h(P))$ is
generically an isomorphism.
\begin{prop}\label{feprop}
Let $(M,\nabla)\in \MC(X)$ be a Fuchsian, cyclic module of dimension
two. Let $S$ be its set of singularities. Suppose that
\begin{itemize}
\item $(M,\nabla)$ has an algebraic solution $\Phi$, and
\item there exists a point $P\in S$ whose local exponents $\gamma_1,
\gamma_2$ are equal.
\end{itemize}
Let $(g,h):Y\rightrightarrows X$ be a correspondence adapted to
$(M,\nabla)$. Write ${\mathfrak S}$ for the set of singularities of
the pull back module $(M_g, \nabla_g)$. Then there exists a divisor
$D=\sum_{P_i\in {\mathfrak S}} n_i P_i$ such that
\[
\Div(\Phi_g)\equiv \Div(\Phi_h)+D\bmod{p}.
\]
\end{prop}
\proof Note that ${\mathfrak S}$ is also the set of singularities of
$(M_h, \nabla_h)$. The fact that $(M_g, \nabla_g)$ and $(M_h,
\nabla_h)$ are equivalent means that there exists a one-dimensional
module $(N, \nabla_N)$ such that $(M_g, \nabla_g)\otimes (N,
\nabla_N)\simeq (M_h, \nabla_h)$. It is no restriction to suppose that
we have equality. Recall that there exists an algebraic function
$\theta$ such that $\theta\Phi_g$ is a solution of $(M_h,
\nabla_h)$. Moreover, the poles and zeros of $\theta$ are in
${\mathfrak S }$. This function is an algebraic solution of the module
$(N, \nabla_N)$. Put $D:=\Div(\theta)$.
Let $P$ be as in the statement of the proposition and let $Q$ be a
point of $Y$ with $g(Q)=P$. Write $e_g$ (resp.\ $e_h$) for the
ramification index of $Q$ in $g$ (resp.\ $h$). Let $\gamma$ be the local
exponent of $(N, \nabla_N)$ at $P$. By comparing the local exponents,
we see that the sets $\{e_g\gamma_1+\gamma, e_g\gamma_2+\gamma\}$ and
$\{e_h\gamma_1, e_h\gamma_2\}$ are equal. By assumption, $\gamma_1=
\gamma_2$. Therefore Proposition \ref{degsolprop} implies that
$\ord_Q(\theta\Phi_g)\equiv e_g\gamma_1+\gamma\equiv e_h\gamma_1\equiv
\ord_Q(\Phi_h)\bmod{p}$. The statement now follows from Proposition
\ref{polysolprop}.
\Endproof
\begin{lem}\label{pullbacklem}
Let $(M, \nabla)\in \MC(X)$ and $g,h:Y\rightrightarrows X$ be a
correspondence adapted to $(M, \nabla)$. Then, for every map
$\phi:Z\to Y$, the modules $(M_{g\circ\phi}, \nabla_{g\circ\phi})$ and
$(M_{h\circ\phi}, \nabla_{h\circ\phi})$ are equivalent also.
\end{lem}
\proof Straight forward.
\Endproof
\section{Estimates for the number of points and the genus in a tower} \label{pointsec}
In this section we define a tower of curves from a tame correspondence
adapted to a differential equation $(M, \nabla)$. We also estimate the
genus (Proposition \ref{genusprop}) and number of points (Proposition
\ref{countprop}) in the tower. The results are easiest to understand
in the well-known case of towers of modular curves (Section
\ref{modularsec}). It may be helpful to look at this case before
reading the proofs in the general case.
Let $(M,\nabla)$ be a Fuchsian differential equation of rank $2$ with
set of singularities $S$. We always suppose that $M$ is cyclic
(Definition \ref{cyclicdef}). We denote by $(g,h): X_0
\rightrightarrows X_{-1}$ a tame correspondence adapted to
$(M,\nabla)$ (Definition \ref{corrdef}) unbranched outside $S$. We
always assume that $X_0$ and $X_{-1}$ are smooth and absolutely
irreducible curves. We assume that the covers $g$ and $h$ are disjoint
(i.e.\ the covers $g,h:X_0 \to X_{-1}$ do not have a common subcover,
or alternatively, there do not exist functions $\phi, \psi_1$ and
$\psi_2$ such that $g=\phi\circ \psi_1$ and $h=\phi\circ\psi_2$ and
$\deg \phi \neq 1$). Denote the common set of singularities of
$(M_g,\nabla_g)$ and $(M_h,\nabla_h)$ by $\mathfrak S$. To the
correspondence $(g,h)$ we associate a tower of curves
\begin{equation}\label{towereq} {\mathcal T}_{g,h}=\left(
\cdots\stackrel{\displaystyle\pi_m}{\longrightarrow}X_m\stackrel{\displaystyle
\pi_{m-1}}{\longrightarrow} X_{m-1}\to \cdots
\stackrel{\displaystyle\pi_{0}}{\longrightarrow}X_0\right),
\end{equation} where $X_m$ is a smooth projective curve and
$\pi_m$ is a cover. For $m \ge 1$, the curve $X_m$ is the
normalization of the curve $X_m'$. The curves $X_m'$ are
defined recursively as follows. We put $X_0'=X_0$. For $m\geq 0$,
we define $X_{m+1}'$ by the fiber product
\[
\xymatrix{
X_{m+1}'\ar[d]_{\pi_{m}}\ar[r]&X_0\ar[d]_h\\
X_{m}'\ar[r]_{g_{m}}&X_{-1},}
\]
where $\pi_m(x_0,\ldots,x_{m+1})=(x_0,\ldots,x_{m})$ and $g_m(x_0,
\ldots, x_{m})=g(x_{m})$. We have
$$X_m'=\{(x_0,\dots,x_m) \in X_0^{m+1} \, | \, h(x_i)=g(x_{i-1}),
\, 1 \le i \le m\}.$$ The cover $\pi_m: X_{m+1}' \to X_m'$ induces
a cover from $X_{m+1}$ to $X_m$ which we again denote by $\pi_m$.
We will also denote by $x_i: X_m \to X_0$ (with $0 \le i \le m$)
the cover defined by $x_i: P \mapsto x_i(P)$.
To obtain a tower of curves, we require that $X_m$ is an absolutely
irreducible curve, for all $m$. Clearly, a necessary condition for
this is that $g$ and $h$ are disjoint. This condition is not
sufficient however. For example if we take $g(t)=t^2+t$ and
$h(t)=1/(t^2+t)$ both defined over a finite field of characteristic
two, then $g$ and $h$ are disjoint, but the corresponding curve $X_2$
is not irreducible (although $X_1$ is).
We now state a sufficient condition for all curves $X_m$ occurring in
the tower ${\mathcal T}_{g,h}$ to be absolutely irreducible. Let $Y$
and $X$ be curves defined over a field $k$ and suppose we are given a
cover $\pi: Y \to X$ defined over $k$. We say a point $P$ of $X$ is
{\sl totally branched} in the cover $\pi$ if there exists a point $Q$
of $Y$ with $\pi(Q)=P$ such that $e(Q|P)=\deg \pi$. The following
lemma is obvious.
\begin{lem}\label{lem:irred} Suppose that the cover $\pi_m: X_m
\to X_{m-1}$ has a totally branched point for any $m \ge 1$. Then all
curves $X_m$ are absolutely irreducible. \end{lem}
The condition that $g$ and $h$ are disjoint is not vital for the
construction of the tower. If $g=\phi\circ
\psi_1$ and $h=\phi\circ\psi_2$, one should replace the
correspondence $(g,h)$ by $(\psi_1, \psi_2)$. Note that if $g$ and $h$ are
disjoint we have $\deg \pi_m=\deg h$. All asymptotically
good towers of function fields found by Garcia--Stichtenoth et al.\
(see e.g.\ \cite{GS1,GS,GS2,GS3}) can be described as a tower $\mathcal
T_{g,h}$ for a suitable correspondence $(g,h)$. For the definition of the tower, we do not need that the
correspondence is adapted to some differential equation. We only
need this afterwards to estimate the number of $\FF_q$-rational
points.
We now state some restrictions we will assume in the rest of this
section.
\begin{assumption}\label{notation}
\begin{itemize}
\item[(a)]{$\mathfrak S=g^{-1}(S)=h^{-1}(S)$.}
\item[(b)]{All curves $X_m$ occurring in the tower ${\mathcal T}_{g,h}$
are absolutely irreducible.}
\item[(c)]{$\deg g=\deg h=:\delta$.}
\end{itemize}
\end{assumption}
We will usually check (b) by using Lemma \ref{lem:irred}. Note
that (c) is a natural restriction, since if $\deg g \neq \deg h$
the tower ${\mathcal T}_{g,h}$ is asymptotically bad \cite{skew}.
We start by estimating the genus $g(\mathcal T_{g,h})$ of a tower
$\mathcal T_{g,h}$. This genus is defined in the following way:
$$g(\mathcal T_{g,h}):=\lim_{m \to \infty} \frac{g(X_m)}{\delta^m}.$$
Here $g(X_m)$ denotes the genus of the curve $X_m$. This limit
exists \cite{GS3}, but may be infinite. A necessary condition for
a tower $\mathcal T$ to be asymptotically good is that $g(\mathcal
T)<\infty$. The following proposition checks this in our
situation.
\begin{prop}\label{genusprop} Let $(M,\nabla)$ be a Fuchsian
differential equation of rank $2$ with set of singularities $S$.
Suppose Assumption \ref{notation} holds. Then for any $m$ and any
point $P$ of $X_m$ we have
\begin{itemize}
\item[i)]{$x_{m-1}(P)\in\mathfrak S \Longleftrightarrow
x_{m}(P)\in\mathfrak S,$}
\item[ii)]{$g({\mathcal T}_{g,h}) \le
g(X_0)+\frac{\# {\mathfrak S}-2}{2}.$}
\end{itemize}
\end{prop}
\proof We extend the constant field to ${\overline{\FF}_q}$,
which does not make a difference since we are only interested in the
genus at this point. We show that the branch locus of the tower
${\mathcal T}_{g,h}$ is contained in $\mathfrak S$. (By branch locus we
mean here the set of points of $X_0$ that are branched in the cover $X_m
\to X_0$, for some $m$.)
Let $P$ be a point of the curve $X_m$. By the recursive definition
of the tower, we have $h(x_m(P))=g(x_{m-1}(P))$. Therefore, if
$x_{m-1}(P) \in \mathfrak S$, we have $x_m(P) \in
h^{-1}g(x_{m-1}(P)) \subset h^{-1}g(\mathfrak S)=\mathfrak S$ by
Assumption (a). Conversely, $x_m(P) \in \mathfrak S$ implies
$x_{m-1}(P) \in \mathfrak S$. This proves the first part of the
proposition.
Recall that the following diagram commutes
\[
\xymatrix{
X_{m}\ar[d]_{\pi_{m-1}}\ar[r]&X_0\ar[d]_h\\
X_{m-1}\ar[r]_{g_{m-1}}&X_{-1}.}
\]
If $P \in X_m$ ramifies in $\pi_{m-1}:X_m \to X_{m-1}$ then
$g(x_{m-1}(P)) \in S$, since we assumed that $h$ is unbranched outside
$S$. We distinguish two cases: $x_m(P) \in \mathfrak S$ and $x_m(P)
\not\in \mathfrak S$. If $x_m(P) \in \mathfrak S$, one obtains from
the first part by induction $x_0(P) \in \mathfrak S$. Now assume
$x_m(P) \not\in \mathfrak S$. Let $e$ be the ramification index of
$x_m(P)$ in the cover $h:X_0 \to X_{-1}$. Since $x_m(P)$ is a regular
point of $(M_h,\nabla_h)$, its local exponents are $0,1$. By
part i) and the assumption $x_m(P) \not\in \mathfrak S$ we have
$x_{m-1}(P) \not\in \mathfrak S$. By considering the local exponents,
we conclude that $x_{m-1}(P)$ has ramification index $e$ in the cover
$g:X_0 \to X_{-1}$. By Abhyankar's lemma, we conclude that the cover
$\pi_0: X_{1} \to X_{0}$ is unbranched at $x_{m-1}(P)$. Consider the
commutative diagram
\[
\xymatrix{
X_{m}\ar[d]_{\pi_{m-1}}\ar[r]^{\psi}&X_1\ar[d]_{\pi_0}\\
X_{m-1}\ar[r]_{x_{m-1}}&X_0,}
\]
with $\psi:X_m \to X_1$ induced by the map $\psi': X_m' \to X_1'$
defined by $\psi(P)=(x_{m-1}(P),x_m(P))$ and $x_{m-1}: X_{m-1} \to
X_0$ defined by $x_{m-1}:Q \mapsto x_{m-1}(Q)$. It follows that
the cover $\pi_{m-1}$ is unramified at $P$, i.e., $x_0(P)$ does
not belong to the branch locus of the tower.
One uses the Riemann--Hurwitz formula for the cover $X_m \to X_0$ to deduce
\[
\frac{g(X_m)-1}{\delta^m} \le g(X_0)-1+\frac{\#\mathfrak S \cdot
(\delta^m-1)}{2\delta^m}.
\]
The proposition follows by letting $m$ tend to infinity. \Endproof
We now investigate the asymptotic behavior of the number of
rational points in the tower $\mathcal T_{g,h}$. A key role is is
played by Proposition \ref{feprop}. Let $\Phi$ be an algebraic
solution of $(M,\nabla)$. Further, let $\Phi_g$ (resp.\ $\Phi_h$)
denote the corresponding solution of $(M_g,\nabla_g)$ (resp.\
$(M_h,\nabla_h)$) (Notation \ref{polysolnot}).
Let $\pi:Y\to X$ be a cover of curves over $k$ and $P$ is a
$k$-rational point of $X$. We say that $P$ is {\sl completely split}
if $P$ is unbranched and every point $Q$ of $Y$ with $\pi(Q)=P$ is
$k$-rational. The following set will turn out to describe a set of
completely split places of $X_0$ in the tower $\mathcal
T_{g,h}$. Define
\begin{equation}\label{Teq}
\mathfrak T:=\{x_0 \in X_0 \, | \, \ord_{x_0}\Phi_g \not\equiv 0
\bmod{p} \makebox{ and } x_0 \not\in \mathfrak S\}.
\end{equation}
Recall that Proposition \ref{degsolprop} implies that $\Ord_{x_0}\Phi_g\equiv 1\bmod{p}$ for $x_0\in \mathfrak{T}$.
The following lemma gives some properties of this set. It will be
useful in the investigation of the number of rational points in the
tower.
\begin{lem}\label{felem} Let $(M,\nabla)$ be a Fuchsian
differential equation of rank $2$ with set of singularities $S$.
Suppose Assumption \ref{notation} holds. Further let $\alpha,\beta
\in X_0$ be such that $h(\beta)=g(\alpha)$. Then
\begin{itemize}
\item[i)]{$\alpha \in \mathfrak T \Longleftrightarrow \beta \in
\mathfrak T,$}
\item[ii)]{$\mathfrak T=\{x_0 \in X_0 \, | \, \ord_{x_0}\Phi_h \not\equiv 0
\bmod{p} \makebox{ and } x_0 \not\in \mathfrak S\}.$}
\end{itemize}
\end{lem}
\proof Suppose $\beta \in \mathfrak T$. By the definition of
$\mathfrak T$ and part i) of Proposition \ref{genusprop}, we have
$\alpha \not\in \mathfrak S$. By Proposition \ref{feprop} we have
$\ord_{\beta} \Phi_h \not\equiv 0 \bmod{p}$. Since
$g(\alpha)=h(\beta)$ and $\alpha\not\in\mathfrak{S}$, we conclude that
$\Phi_h(\beta)=\Phi_g(\alpha)=0$. Proposition \ref{degsolprop} implies
that $\ord_{\alpha} \Phi_g \not\equiv 0 \bmod{p}$. This proves i).
The second part of the lemma follows directly from the (proof of) the
first part.\Endproof
The first part of the above lemma implies that $g(\mathfrak
T)=h(\mathfrak T)$. We write
$$T:=g(\mathfrak T)=h(\mathfrak T).$$
The second part of Lemma \ref{felem} shows that the role of
$g$ and $h$ in the definition of $\mathfrak
T$ can be interchanged.
Given an absolutely irreducible curve $C$ defined over
$\mathbb{F}_q$, we denote by $N_q(C)$ the number of
$\mathbb{F}_q$-rational points of $C$. For a tower $\mathcal
T_{g,h}=(X_0,X_1,\dots)$ defined as above with constant field
$\mathbb{F}_q$ we define the {\sl splitting rate} of the tower
$\mathcal T_{g,h}$ by
$$\nu_q(\mathcal T_{g,h}):=\lim_{m\to \infty }\frac{N_q(X_m)}{\delta^m}.$$
This limit exists \cite{GS3} and is a nonnegative finite number. A
necessary condition for a tower $\mathcal T$ to be asymptotically
good is $\nu_q(\mathcal T)>0$. The following proposition gives an
estimate for $\nu_q(\cal T)$ in our situation.
\begin{prop}\label{countprop} Let $(M,\nabla)$ be a Fuchsian
differential equation of rank $2$ with set of singularities $S$.
Suppose Assumption \ref{notation} holds. Suppose that the constant
field $\mathbb{F}_q$ of the tower $\mathcal T_{g,h}$ is such that
all points of $X_0$ in the set ${\mathfrak T}$ are
defined over $\mathbb{F}_q$. Then
$$\nu_q(\mathcal T_{g,h}) \ge \# \mathfrak T.$$
\end{prop}
\proof Recall that $\delta:=\deg g=\deg h$. Since $ S$
and $T$ are disjoint, for any $\alpha \in T$ there are exactly
$\delta$ points of $X_0$ lying above $\alpha$. Moreover, all these
points of $X_0$ are defined over $\mathbb{F}_q$ by our assumption. Let
$P$ be a point of $X_0$ with $g(P)=\alpha$. Write $h(P)=\beta$. Since
$T=g({\mathfrak T})=h({\mathfrak T})$ it follows that $\beta\in T$.
Now suppose that we have constructed inductively $\delta^{m-1}
\#\mathfrak T$ points of $X_{m-1}$ defined over $\mathbb{F}_q$ and
lying above $\mathfrak T$. Consider the commutative diagram
\[
\xymatrix{
X_{m}\ar[d]_{\pi_{m-1}}\ar[r]^{\psi}&X_1\ar[d]_{x_1}\\
X_{m-1}\ar[r]_{x_{m-1}}&X_0,}
\]
with $\psi:X_m \to X_1$ induced by
the map $\psi': X_m' \to X_1'$ defined by
$\psi'(P)=(x_{m-1}(P),x_m(P))$, the map $x_1:X_1 \to X_0$ is given by
$Q \mapsto x_1(Q)$, and similarly $x_{m-1}: X_{m-1} \to X_0$ is
defined by $R \mapsto x_{m-1}(R)$. Given an $\alpha \in \mathfrak T$,
we can construct $\delta$ points $P$ of $X_1$ defined over
$\mathbb{F}_q$ with $x_1(P)=\alpha$ and $\delta^{m-1}$ points $Q$ of
$X_{m-1}$ also defined over $\mathbb{F}_q$ with
$x_{m-1}(Q)=\alpha$. Given such a $P$ and $Q$, there exists at least
one point $R$ of $X_{m}$ lying above both $P$ and $Q$. Moreover, by
Lemma \ref{felem}, we have $x_m(R) \in \mathfrak T$. It follows that we
have obtained in this way all $\delta^m$ points of $X_m$ lying above
$\alpha$ and that any of these points is defined over
$\mathbb{F}_q$. \Endproof
The field $\mathbb{F}_q$ mentioned in the above proposition is
called the minimal splitting field of the tower $\mathcal
T_{g,h}$. In other words, we have the following definition.
\begin{defn} Given a correspondence $(g,h):X_0 \to X_{-1}$
defining a tower $\mathcal T_{g,h}$, we define the {\sl minimal
splitting field} of this tower to be the smallest field $k$ such
that \begin{itemize} \item[i)]{the correspondence $(g,h):X_0
\rightrightarrows X_{-1}$ is defined over $k$,} \item[ii)]{all
points of $X_1$ in the set $\pi_0^{-1}(\mathfrak T)$ are defined
over $k$.} \end{itemize} \end{defn}
The following theorem gives a sufficient condition for $\mathcal
T_{g,h}$ to be asymptotically good.
\begin{thm}\label{thm:asgood}
Suppose Assumption \ref{notation} holds. Let $\mathbb{F}_q$ be the
minimal splitting field of $\mathcal T_{g,h}$. Suppose that
$\ord_P(\Phi) \not \equiv 0 \bmod{p}$ for some $P \in \PP^1$ and
that $g^{-1}(P) \not \subset \mathfrak S$. Then the tower
$\mathcal T_{g,h}$ is asymptotically good.
\end{thm}
\proof By Proposition \ref{genusprop}, the tower has finite genus.
We will show that the set $\mathfrak T$ is nonempty. Let $Q \in
g^{-1}(P)$ and $Q \not \in \mathfrak S$. Since all ramification in
the cover $g:X_0\to X_{-1}$ is tame, we have $\ord_Q(\Phi_g) \not
\equiv 0 \bmod{p}$. We conclude that $Q \in \mathfrak T$. By
Proposition \ref{countprop} the tower has positive splitting rate.
Hence the tower $\mathcal T_{g,h}$ is asymptotically good.
\Endproof
\begin{thm}\label{thm:asopt}
Suppose Assumption \ref{notation} holds. Let $\mathbb{F}_q$ be the
minimal splitting field of $\mathcal T_{g,h}$. Suppose that
\[
2\#
\mathfrak{T}=(\sqrt{q}-1)(\# \mathfrak S+2g(X_0)-2).
\]
Then the tower $\mathcal T_{g,h}$ is asymptotically optimal.
\end{thm}
\proof This follows immediately from Propositions \ref{genusprop}
and \ref{countprop}. \Endproof
The minimal splitting field is in practice often difficult to
calculate. This is a serious problem in finding asymptotically optimal
towers via the criterion of Theorem \ref{thm:asopt}. Proposition
\ref{zieveprop} is a useful tool to deal with this problem: it
essentially controls the minimal splitting field at the cost of
introducing a new condition on the correspondence
$(g,h):Y\rightrightarrows X$. Namely, we need to suppose that the
correspondence has degree one. In Section
\ref{pullbacksec} we will always make this assumption. In the case of
modular curves (Section \ref{modularsec}) this condition is always
satisfied, see the proof of Lemma \ref{modularzievelem}.
Recall from Section \ref{desec} that if a correspondence
$g,h):Y\rightrightarrows X$ has degree one, then the map $Y\to C$ of
$Y$ onto the curve of correspondence is generically a bijection.
\begin{prop}\label{zieveprop} Let $X$ and $Y$ be smooth and
absolutely irreducible curves defined over $k$, and let
$(g,h):Y \rightrightarrows X$ be a tame correspondence of degree one over $k$.
Let $V \subset X$ be a set of $k$-rational points such that for
any $\alpha, \beta \in Y$ with $h(\beta)=g(\alpha)$ we have
$g(\alpha) \in V \Leftrightarrow g(\beta) \in V$. Let $\alpha \in
Y$ be such that
\begin{itemize}
\item[i)]{$g(\alpha) \in V$.}
\item[ii)]{$g(\alpha)$ is a $k$-rational point of $X$.}
\item[iii)]{$(g(\alpha),h(\alpha))$ is not a singularity of $C$.}
\end{itemize}
Then $\alpha$ is a $k$-rational point of $Y$.
\end{prop}
\proof We first show that $h(\alpha)$ is a $k$-rational point of
$X$. There exists a point $\beta$ such that $h(\alpha)=g(\beta)$.
By the definition of $V$ and $i)$, the point
$h(\alpha)=g(\beta)$ is in $V$ and hence $k$-rational.
Since the map $\phi$ has degree one, it can be inverted for
nonsingular points of $C$. The $k$-rationality of
$(g(\alpha),h(\alpha))$ then implies the $k$-rationality of
$\alpha=\phi^{-1}(g(\alpha),h(\alpha))$. \Endproof
Condition iii) in the above lemma is in practice not a heavy
restriction. Since the number of singularities of $C$ is finite,
they can usually be dealt with by hand in any particular case.
For certain correspondences of degree two this lemma is due to Zieve
(\cite{GS}). We will apply this lemma in the
situation that $V=T(=g(\mathfrak T)=h(\mathfrak T))$. If the
conditions of the above lemma are satisfied, then the points in
the set $\mathfrak T$ are defined over $k$ if the points in the
set $T$ are.
\section{Constructing towers via pull back}\label{pullbacksec}
As before let $(g,h):X_0 \rightrightarrows X_{-1}$ be a correspondence
adapted to a Fuchsian differential equation $(M,\nabla)$, where we
suppose that $g$ and $h$ are disjoint. As always, we suppose that
$(M,\nabla)$ is cyclic. We write $S$ for the set of singularities of
$(M,\nabla)$. Let $f:Y_{-1} \to X_{-1}$ be a (separable) cover of
smooth, absolutely irreducible curves, which is allowed to have wild
ramification and may be ramified outside $S$. We suppose that
all curves and covers are defined over a finite field $k$. Write
$(M_f,\nabla_f)$ for the pull back of $(M,\nabla)$ via $f$ and $S_f$
for the set of singularities of $(M_f,\nabla_f)$. In this section we
make the following additional assumption.
\begin{itemize}
\item[(d)]The correspondence $(g,h):X_0\rightrightarrows X_{-1}$ has
degree one.
\end{itemize}
Recall that we defined the curve of correspondence $C\subset
X_{-1}\times X_{-1}$ by
\begin{equation}\label{Cdefeq}
C:=\{(g(P),h(P)) \, | \, P \makebox{ a point of } X_0\}.
\end{equation}
The curve $C$ is the image of $X_0$ under the map $g*h:X_0 \to X_{-1}
\times X_{-1}$ defined by $(g*h)(P)=(g(P),h(P))$. Assumption (d)
implies that the map $X_0\to C$ has degree one. Denote by $p_1$
(resp. $p_2$) the projections of $C$ onto it first (resp. second)
coordinate. We have the following commutative diagram
\[
\xymatrix{ &&&X_0\ar[d]^{g*h}\ar[dll]_{g}\ar[drr]^{h}\\
&X_{-1}&&C\ar[ll]^{p_1}\ar[rr]_{p_2}&&X_{-1}.}
\]
Let $D\subset Y_{-1}\times Y_{-1}$ be an absolutely irreducible
component of the inverse image of $C$ under $(f,f):Y_{-1} \times
Y_{-1} \to X_{-1} \times X_{-1}$. After extending the field of
definition $k$, we may suppose that $D$ is defined over $k$. We have
the following diagram
\[
\xymatrix{
&X_0\ar[d]^{g*h}&&&D\ar[dlll]_{(f,f)}\ar[dl]^{\tilde{p}_1}\ar[dr]_{\tilde{p}_2}\\
&C\ar[dl]^{p_1}\ar[dr]_>>>>{p_2}&&Y_{-1}\ar[dlll]_<<<<<<<<<{f}&&Y_{-1}\ar[dlll]_f\\
X_{-1}&&X_{-1}.} \] The maps $\tilde{p}_1$ (resp. $\tilde{p}_2$) are
projections of $D$ onto its first (resp. second) coordinate. Recall
that we always suppose that the curve $X_0$ is smooth. Denote by $Y_0$
the normalization of the curve $D$, then we have the following diagram
\[
\xymatrix{ &&&Y_0\ar[d]^{}\ar[dll]_{\tilde{g}}\ar[drr]^{\tilde{h}}\\
&Y_{-1}&&D\ar[ll]^{\tilde{p}_1}\ar[rr]_{\tilde{p}_2}&&Y_{-1}.}
\]
The maps $\tilde{g}$ and $\tilde{h}$ are defined such that the
diagram commutes.
\begin{defn}\label{pullback}
We call $(\tilde{g}, \tilde{h})$ the {\sl pull back} of $(g,h)$
under $f$.
\end{defn}
The following lemma gives a key property of the pull back
correspondence $(\tilde{g},\tilde{h})$.
\begin{lem}\label{adaptlem} The correspondence $(\tilde{g}, \tilde{h})$ is adapted
to $(M_{f},\nabla_f)$. \end{lem}
\proof We have the following (commutative) diagram
\[
\xymatrix{ &&&&Y_0\ar[dlll]\ar[dl]^{\tilde{g}}\ar[dr]_{\tilde{h}}\\
&X_0\ar[dl]^{g}\ar[dr]_>>>>{h}&&Y_{-1}\ar[dlll]_<<<<<<<<<{f}&&Y_{-1}\ar[dlll]_{f}\\
X_{-1}&&X_{-1}.}
\]
Hence the result follows immediately from Lemma \ref{pullbacklem}.
\Endproof
Note that if $\deg g=\deg h$, then $\deg \tilde{g}=\deg
\tilde{h}$. The above lemma motivates that if the tower $\mathcal
T_{g,h}$ is asymptotically good, the tower $\mathcal
T_{\tilde{g},\tilde{h}}$ is a good candidate for being asymptotically
good as well. Recall that we defined (\ref{Teq}) a set $\mathfrak T
\subset X_0$ consisting of completely splitting places of the tower
$\mathcal T_{g,h}$. We denote by $T \subset X_{-1}$ the set $g(\mathfrak
T)=h(\mathfrak T)$.
\begin{thm}\label{thm:pullbackgood} Let $(M,\nabla)$ be a Fuchsian
differential equation of rank $2$ and let $(g,h)$ be a correspondence
adapted to $(M,\nabla)$ all defined over a finite field
$\mathbb{F}_q$. Suppose Assumptions (a), (b), (c), (d) hold. Let $f:
Y_{-1} \to X_{-1}$ be a cover and suppose that Assumption (b)
holds for the pull back correspondence
$(\tilde{g},\tilde{h})$ as well. If the set $\mathfrak T$ is non
empty, then $\mathcal T_{\tilde{g},\tilde{h}}$ is asymptotically good
over some extension field of $\FF_q$. \end{thm}
\proof Our assumptions imply that the ramification locus of the pull
back tower ${\mathcal T}_{\tilde{g}, \tilde{h}}$ is contained in
$f^{-1}(S)$. Therefore Proposition \ref{genusprop} implies that the
genus $g({\mathcal T}_{\tilde{g}, \tilde{h}})$ of the pull back tower
is finite.
We claim that $\nu_q(\mathcal T_{\tilde{g},\tilde{h}})>0$. Denote by
$\phi: Y_0 \to X_0$ the map induced by $f$. Since $\mathfrak T \subset
X_0$ consists of completely splitting points of the tower $T_{g,h}$,
the non empty set $f^{-1}(\mathfrak T) \subset Y_0$ consists of
completely splitting places of the tower $T_{\tilde{g},\tilde{h}}$ if
we extend the constant field suitably. Here we use that ${\mathfrak
T}$ is unbranched in $f$. \Endproof
Before proceeding, we give an example illustrating Theorem
\ref{thm:pullbackgood}.
\begin{exa}\label{goodexa} We consider the correspondence $(g,h):\PP^1
\rightrightarrows \PP^1$ given by $h(t)=t^2$ and $g(t)=4t/(t+1)^2$.
The correspondence $(g,h)$ is adapted to the Gau\ss' hypergeometric
differential operator $L(u)=t(t-1)u''+(2t-1)u'+u/4$, which has the
Deuring polynomial as a solution (Example \ref{GMexa}). We denote the
Deuring polynomial by $\Phi$. Recall that $S=\{0,1,\infty\}$ is the
set of singularities of $L$. The set of singularities of the pull
back differential equation $L_g$ equals ${\mathcal S}=\{0, \pm 1,
\infty\}$. Therefore the correspondence satisfies Assumptions (a) and
(c). One checks that the point $x_0=0$ on $X_0$ is totally branched
in the tower. Therefore Assumption (b) follows from Lemma
\ref{lem:irred}.
The corresponding tower
${\mathcal T}_{g,h}$ is the tower of modular curves $X_0(2^m)$
starting from $m=3$. The tower $\mathcal{T}_{g,h}$ is essentially the
same as a tower considered in \cite{GS}.
The curve of correspondence
$C=\{(g(x),h(x)) \, | \, x \in \PP^1\}$ is given by the equation
\begin{equation}\label{Ceq}
4a(b-2)^2-(a+1)^2b^2=0.
\end{equation}
One checks that $X_0$ is a normalization of $C$. In other words, Assumption (d)
is satisfied. This means that we can apply the pull back construction
to the tower $\mathcal T_{g,h}$. Further note that the point $(-1,2)$
of $C$ is a singularity.
Let $f:\PP^1 \to \PP^1$ be the cover defined by
$t=f(s):=-n(n/(n-1))^{n-1}(s^n-s^{n-1})$ for an integer $n \ge 2$
satisfying $p \not|(n-1)$ and $p\not|n$, where $p$ denotes the
characteristic. In particular $Y_{-1}=\PP^1$. An explicit calculation
shows that the cover $f$ is unbranched outside the set
$\{0,1,\infty\}$. Let $\FF_q$ be the smallest finite field containing
all roots of the polynomial $\Phi(T^m-T^{m-1})$. Using Proposition
\ref{zieveprop} one checks that the pull back tower $\mathcal
T_{\tilde{g},\tilde{h}}$ is asymptotically good over the field
$\FF_q$, for all $n$ for which Assumption (b) holds. We will not
determine the field $\FF_q$ explicitly here. For $n=2$ we obtain an
asymptotically optimal tower which turns out to be the modular tower
$((X_0(2^m))$ starting from $m=4$. For $n>2$ one does not seem to
obtain asymptotically optimal towers. Trivially one sees that $q$
divides $2\cdot n!$.
\end{exa}
Example \ref{goodexa} illustrates how to find asymptotically good
towers via pull back. The only problem is to check Assumption (b) for
the pull back tower. This condition is satisfied if
$\deg(\tilde{g})=\deg(g)$. To obtain asymptotically optimal towers, we
need to impose a condition on the minimal splitting field. This
condition is in practice hard to check.
\begin{thm}\label{thm:pullbackopt} Let $(g,h):X_0\rightrightarrows
X_{-1}$ be a correspondence adapted to a differential equation
$(M,\nabla)$ with singularity set $S$, satisfying Assumptions (a), (b),
(c), (d). Let $f:Y_{-1} \to X_{-1}$ be a tame cover unbranched
outside $S$. Denote by $(\tilde{g},\tilde{h})$ the pull back
correspondence and suppose it satisfies Assumption (b).
If ${\mathcal T}_{g,h}$ is asymptotically optimal and the towers $\mathcal
T_{g,h}$ and $\mathcal T_{\tilde{g},\tilde{h}}$ have the same minimal
splitting field $\FF_q$, then $\mathcal T_{\tilde{g},\tilde{h}}$
is asymptotically optimal also. \end{thm}
\proof By our assumptions $\phi^{-1}(\mathfrak T)$ consists of
completely splitting places of the tower $\mathcal
T_{\tilde{g},\tilde{h}}$. Therefore $\nu_q(\mathcal
T_{\tilde{g},\tilde{h}}) \ge \deg \phi \cdot \# \mathfrak T$ by
Proposition \ref{countprop}.
As usual, denote by $\mathfrak S$ the singularities of the
differential equation $(M_g,\nabla_g)$. The
Riemann-Hurwitz genus formula for the cover $\phi:Y_{0} \to X_{0}$ implies
$$2g(Y_0)-2+\#\phi^{-1}({\mathfrak S})=\deg
\phi\cdot(2g(X_0)-2+\#\mathfrak S).$$ Since $\phi^{-1}({\mathfrak S})$
contains the set of singularities of $(M_{\tilde{g}\circ f},
\nabla_{\tilde{g}\circ f})$, it follows from Proposition \ref{genusprop} that
$g(\mathcal T_{\tilde{g},\tilde{h}}) \le \deg
\phi\cdot(g(X_0)+(\#\mathfrak{S}-2)/2)$. Therefore $\lambda(\mathcal
T_{\tilde{g},\tilde{h}}) \ge \lambda(\mathcal T_{g,h})=q-1$ and we are
done. \Endproof
As a consequence of Theorem \ref{thm:pullbackopt}, we give an
asymptotically optimal tower. We consider again the correspondence
$(g,h):\PP^1 \rightrightarrows \PP^1$ given by
$h(t)=t^2$ and
$g(t)=4t/(t+1)^2$.
Using Proposition \ref{zieveprop} we immediately obtain that the roots
of the Deuring polynomial are squares in $\FF_{p^2}$. In fact
these roots are fourth powers in $\FF_{p^2}$ (see \cite{GS}).
Let $f:\PP^1 \to \PP^1$ be defined by \[
t=f(s)=\frac{16s^2}{(s-1)^4}. \] One checks that
$S_f=f^{-1}\{0,1,\infty\}=\{0,\pm 1, 3\pm\sqrt{2}, \infty\}$. The pull
back differential equation is \[
L_f(v)=v''+\frac{s^4-4s^3+20s^2+8s-1}{s(s^2-1)(s^2-6s+1)}v'+
\frac{16(s^2+1)}{s(s+1)(s-1)^2(s^2-6s+1)}v=0. \] The pull back of $C$
with respect to the map $(f,f)$ has two absolutely irreducible
components of genus $0$ and one of genus $2$. We write $f(A)=a$ and
$f(B)=b$ and use (\ref{Ceq}). The components of genus $0$ of the pull
back of $C$ are then given by the equations
\begin{equation}\label{genus0} -A+A^2+4AB+B^2-AB^2=0
\qquad\makebox{ and }\qquad
1-A+4AB-AB^2+A^2B^2=0. \end{equation}
\noindent One may choose any genus $0$ component $D$ from
(\ref{genus0}) and a coordinate $y$ of its normalization $Y_0$
such that the maps $\phi:Y_0\to X_0$ and $\tilde{g},
\tilde{h}:Y_0\to \PP^1$ are described as follows.
\[
\xymatrix{
&X_0\ar[ld]_g\ar[rd]^h&&&Y_0\ar[lll]_\phi\ar[ld]_{\tilde{g}}\ar[rd]^{\tilde{h}}\\
\PP^1&&\PP^1&\PP^1&&\PP^1,} \] with $\phi(y)=4y^2/(y^2-1)^2$,
$\tilde{h}(y)=y^2$ and $\tilde{g}(y)=-y(y-1)/(y+1)$.
Let ${\mathcal T}_{\tilde{g}, \tilde{h}}=(Y_0,Y_1,\dots,Y_m,\dots)$ be
the tower of curves defined by the correspondence $(\tilde{g},
\tilde{h})$. One checks that $y_0=\infty$ is totally branched in the
tower. Therefore Lemma \ref{lem:irred} implies that the curves $Y_m$
are irreducible for all $m$. Then $Y_{m}$ is given by the equations
\[
y_i^2=-\frac{y_{i-1}(y_{i-1}-1)}{y_{i-1}+1}, \makebox{ with $1 \le
i \le m$}.\]
We write $\Phi_f(y)=\Phi(f(y))$ for the algebraic solution of
$L_f(v)=0$ (Notation \ref{polysolnot}). Using that the correspondence
$(\tilde{g},\tilde{h})$ is adapted to $L_f$ (Lemma \ref{adaptlem}),
one checks that
\[
(y^2-1)^{2p-2}\Phi_f(y^2)=(y^2+1)^{2p-2}\Phi_f\left(\frac{-y(y-1)}{y+1}\right).
\]
This illustrates Proposition \ref{feprop}.
\begin{prop}\label{exaprop} The tower ${\mathcal T}_{\tilde{g},
\tilde{h}}$ is asymptotically optimal if $p\equiv
\pm 1\bmod{8}$. \end{prop}
\proof To apply Theorem \ref{thm:pullbackopt}, we only need to
determine the minimal splitting field of the tower
$\mathcal
T_{\tilde{g},\tilde{h}}$. Using
Proposition \ref{zieveprop} we see that this field is in fact the
splitting field of $\Phi_f(t)$. In other words, we are interested
in the solutions of the equation
\begin{equation} \label{sol}
\frac{16y^2}{(y-1)^4}=\left(\frac{4y}{(y-1)^2}\right)^2=\lambda \
{\makebox{ with }\ \Phi(\lambda)=0.} \end{equation}
We have
already seen that all roots of the Deuring polynomial are squares in
$\FF_{p^2}$. Write $\lambda=\mu^2$. Equation (\ref{sol}) has solutions in
$\FF_{p^2}$ if
and only if $\mu+1$ is a square in $\FF_{p^2}$.
Suppose that $p \equiv \pm 1
\bmod{8}$. We claim that for any root $\lambda$ of $\Phi$ and any
element $\mu$ with $\mu^2=\lambda$ the element $\mu+1$ is a square
in $\FF_{p^2}$. We will prove this claim following the approach by
R{\"u}ck in the appendix of \cite{GS}.
Consider the
elliptic curve $E_{\lambda}$ given by $ Y^2=X(X-1)(X-\lambda).$
Since $\lambda$ is a root of the Deuring polynomial, $E_\lambda$ is
supersingular.
We first suppose that $\lambda \not\in \{-1,2,1/2\}$ and that
$\lambda$ is not a sixth root of unity. It is known that
$\Frob_{p^2}$, the Frobenius automorphism over $\FF_{p^2}$, acts on $E_\lambda$
as multiplication by $\pm p$. This implies that the $x$-coordinate
of any $8$-torsion point of $E_\lambda$ is an element of $\FF_{p^2}$.
Here we use that
$p \equiv \pm 1 \bmod{8}$.
The point $(0,0)$ of $E_{\lambda}$ is a point of order two. For
any point $(a,b)$ satisfying $2(a,b)=(0,0)$ we have
$$a^2=\lambda$$ as can be seen directly from the addition formulas
of $E_{\lambda}$. Hence we may choose $a=-\mu$. For the points
$(c,d)$ satisfying $2(c,d)=(a,b)$ we find in a similar way
$(c-a)^4+4c^2(a-1)^2a=0$. Write $\mu=\nu^2$. Note
that $\nu \in \FF_{p^2}$, since all roots of $\Phi$ are fourth
powers in $\FF_{p^2}$. We obtain
\begin{equation}\label{basiceq}
(c^2+2(-\nu+\mu-\nu \mu)c+\lambda)(c^2+2(\nu+\mu+\nu
\mu)c+\lambda)=0.
\end{equation}
The discriminant of any of these factors is $\mu+1$ up to
multiplication with squares in $\FF_{p^2}$. Since all solutions of
(\ref{basiceq}) are in $\FF_{p^2}$, the claim follows.
If $\lambda \in \{-1,2,1/2\}$, then a direct computation shows
that $\mu+1$ is a square in $\FF_{p^2}$ in our situation. On the
other hand if $\lambda$ is a sixth root of unity, then
$\Frob_{p^6}$, the Frobenius automorphism on $\FF_{p^6}$, acts as
multiplication by $\pm p$ on $E_{\lambda}$. By a similar argument
as above, we conclude that $\sqrt{\mu+1} \in \FF_{p^6}$. On the
other hand, it is obvious that $\sqrt{\mu+1}\in \FF_{p^8}$.
Therefore, $\mu+1$ is a square in $\FF_{p^2}$ in this case as
well.
Theorem \ref{thm:pullbackopt} now implies that the tower $\mathcal
T_{\tilde{g},\tilde{h}}$ is asymptotically optimal. \Endproof
\section{Towers of modular curves}\label{modularsec}
In this section we apply the results of Sections \ref{pointsec} and
\ref{pullbacksec} to towers of modular curves.
Fix an integer $\ell>3$. We do not suppose that $\ell$ is prime.
Write $X_0(\ell^m)$ for the modular curve parameterizing (generalized)
elliptic curves $E$ together with a cyclic isogeny $E\to E'$ of degree
$\ell^m$. For a precise description of the points above $j=\infty$
(the cusps) in terms of generalized elliptic curves we refer to
\cite{DelRap}. The curve $X_0(\ell^m)$ has a natural smooth model
over $\ZZ[1/\ell]$. Denote by $\sigma_m:X_0(\ell^m)\to X_0(\ell^m)$
the Atkin--Lehner involution. It sends an isogeny $E\to E'$ to its
dual isogeny.
We define a correspondence $(g,h):X_0(\ell^2)\rightrightarrows
X_0(\ell)$ as follows. Suppose that $(E_1\to E_2\to E_3)$ corresponds
to point of $X_0(\ell^2)$, i.e.\ $E_1\to E_3$ is a cyclic isogeny of
degree $\ell^2$ and $E_i\to E_{i+1}$ has degree $\ell$. Then $g(E_1\to
E_2\to E_3)=(E_1\to E_2)$ is the standard projection and $h(E_1\to
E_2\to E_3)=(E_2\to E_3)$ is $\sigma_1\circ g\circ \sigma_2$.
Analogous to Example \ref{GMexa}, we obtain a differential equation on
$X_0(\ell^m)$. Fix a prime $p$ relatively prime to $\ell$. We denote
by $X_0(\ell^m)/\FF_p$ the reduction of $X_0(\ell^m)$ to characteristic $p$.
Let $S=\Spec(\FF_p[j, 1/j(j-1728)])$. Write ${\mathcal E}_{\ell^m}$
for the universal elliptic curve on $X_0(\ell^m)$; it exists for
$\ell^m>3$. Let $M_{\ell^m}=H^1_{\dR}({\mathcal E}_{\ell^m}/S)$ be the
de Rham cohomology group and $\nabla:M_{\ell^m}\to \Omega^1_S\otimes
M_{\ell^m}$ the Gau\ss--Manin connection (\cite[Section
7]{Katz}). Then $(M_{\ell^m},\nabla)\in
\MC(X_0(\ell^m))$; its set of singularities $S_{\ell^m}$ is contained in the
inverse image of $j=0,1728, \infty$.
\begin{lem}\label{modularcorrlem}
The correspondence $(g,h):X_0(\ell^2)\rightrightarrows X_0(\ell)$ is
adapted to $(M_\ell, \nabla)$.
\end{lem}
\proof
The pull back of ${\mathcal E}_\ell$ via $g$ is just the
universal elliptic curve ${\mathcal E}_{\ell^2}$. Denote the pull back
of ${\mathcal E}_\ell$ via $h$ by ${\mathcal E}_{\ell^2}'$. The
concrete description of $g$ and $h$ given above implies that there is
an isogeny ${\mathcal E}_{\ell^2}\to {\mathcal E}_{\ell^2}'$. It
induces an isomorphism $H^1_{\dR}({\mathcal E}_{\ell^m}/S)\simeq
H^1_{\dR}({\mathcal E}'_{\ell^m}/S)$ on the de Rham cohomology groups.
This implies the statement of the lemma.
\Endproof
On $X(1)$ we still have a versal family of elliptic curves ${\mathcal
E}_1$. We may choose ${\mathcal E}_1$ such that it is nonsingular
outside $j=0,1728, \infty$. The differential equation corresponding
to $(M_1, \nabla)$ is a hypergeometric
differential equation (i.e.\ a Fuchsian differential equation with
three singularities). Its singularities are $j=0, 1728, \infty$ with
local exponents $ 0, 1/3; 0,1/2; 1/12, 1/12$, respectively. Note
that $(M_{\ell^m}, \nabla)$ is the pull back
of $(M_{1}, \nabla)$ via the natural
projection $(E\to E')\mapsto E$. Denote by $\nu_2(\ell^m)$ (resp.\
$\nu_3(\ell^m)$, resp.\ $\nu_\infty(\ell^m)$) the number of
singularities of $(M_{\ell^m}, \nabla)$
above $j=1728$ (resp.\ $j=0$, resp.\ $j=\infty$). Let $\mu(\ell^m)$
be the degree of $X_0(\ell^m)\to X(1)$. These numbers are computed
in \cite[Proposition 1.43]{Shimura}. Moreover, it is shown in
\cite[Proposition 1.40]{Shimura} that
\begin{equation}\label{genusmodulareq}
g(X_0(\ell^m))=1+\frac{\mu(\ell^m)}{12}-\frac{\nu_2(\ell^m)}{4}-
\frac{\nu_3(\ell^m)}{3}- \frac{\nu_\infty(\ell^m)}{2}.
\end{equation}
The {\sl supersingular polynomial} is defined as
\[
\Phi_1(j)=\prod (j-j(E))\in \FF_p[j],
\]
where the product is taken over the supersingular elliptic curves
$E/\bar{\FF}_p$. Put
\[
\alpha:=\left[\frac{p}{12}\right],\quad
\delta:=\left\{\begin{array}{ll}
0&\mbox{ if }p\equiv 1\bmod{3}\\
1&\mbox{ if }p\equiv 2\bmod{3}
\end{array}\right.
,\quad
\epsilon:=\left\{\begin{array}{ll}
0&\mbox{ if }p\equiv 1\bmod{4}\\
1&\mbox{ if }p\equiv 3\bmod{4} \end{array}.\right. \] There exists
a polynomial $\tilde{\Phi}_1$ of degree $\alpha$ such that
$\Phi_1=j^\delta(j-1728)^\epsilon\tilde{\Phi}_1$. All zeros of
$\tilde{\Phi}_1$ are simple.
\begin{lem}
The polynomial $\Phi_1$ is an algebraic solution of
$(M_{1}, \nabla)$.
\end{lem}
\proof
This is well known. It can for example be checked by direct
verification, or deduced from \cite{Katz84}.
\Endproof
We denote by $\Phi_{\ell^m}$ the induced algebraic solution of
$(M_{\ell^m}, \nabla)\otimes \FF_p$ (Notation
\ref{polysolnot}).
\begin{lem}\label{modularzievelem}
We write
\[
{\mathfrak T}_{\ell}:=\{ x\in X_0(\ell)_{\FF_p}\, |\, \Phi_\ell(x)=0,
\mbox{ and } x\not\in S_\ell\}.
\]
The points of ${\mathfrak T}_{\ell}$ are $\FF_{p^2}$-rational.
\end{lem}
\proof
It is well-known that the roots of $\Phi$ are rational over
$\FF_{p^2}$ (\cite{Silverman}).
For $j_1,j_2\in X(1)$ we write $j_1\sim_\ell j_2$ if there
exists a cyclic isogeny of degree $\ell$ from the elliptic curve with
$j$-invariant $j_1$ to the elliptic curve with $j$-invariant $j_2$.
Define a (singular) curve
\[
{\mathcal C}(\ell)=\{ (j_1,j_2)\in X(1)\times X(1)\, |\, j_1\sim_\ell
j_2\}.
\]
We obtain a commutative diagram
\[
\xymatrix{
&X_0(\ell)\ar[ddl]_{g_0}\ar[ddr]^{h_0}\ar[d]&\\
&{\mathcal C}(\ell)\ar[dl]^{{\rm pr}_1}\ar[dr]_{{\rm pr}_2}&\\
X(1)&&X(1),
}
\]
with $g_0(E_1\to E_2)=j(E_1)$ and $h_0(E_1\to E_2)=j(E_2)$ (compare to
(\ref{Cdefeq})).
If $E_1$ and $E_2$ are elliptic curves without complex multiplication,
there exists at most one isogeny $E_1\to E_2$ of fixed degree
$\ell$. Namely, $\Hom(E_1, E_2)$ is a right $\End(E_2)$-module of
rank one and $\End(E_2)\simeq \ZZ$, since $E_2$ does not have
CM. This implies that the map $X_0(\ell)\to {\mathcal C}(\ell)$ has
degree one and is defined over $\FF_p$. The lemma now follows from
Proposition
\ref{zieveprop}, since the roots of $\Phi$ are $\FF_{p^2}$-rational. \Endproof
\begin{prop}\label{modulartowerprop} Let $\ell>3$ be an integer.
\begin{itemize}
\item[(a)] Let $(g,h):X_0(\ell^2)\rightrightarrows X_0(\ell)$ be the
correspondence defined above. Then the corresponding tower of curves
is isomorphic to ${\mathcal T}_{g,h}=(X_0(\ell^m))$.
\item[(b)] The tower ${\mathcal T}_{g,h}$ is asymptotically optimal.
\end{itemize}
\end{prop}
\proof Part (a) is proved in \cite{Elkies97}. Part (b) follows from
the work of \cite{Ihara}. It is also proved in \cite{TVZ}. We indicate an
alternative proof using our results.
If $\nu_2(\ell^2)=\nu_3(\ell^2)=0$, the proposition follows from
Theorem \ref{thm:asopt}. Otherwise, the estimates for
$g(X_0(\ell^m))$ and $N_{p^2}(X_0(\ell^m))$ given in Section
\ref{pointsec} are not quite good enough. But it is easy to compute
these quantities directly, using the results of
\cite{Shimura}. Namely, one checks that
\[
\lim_{m\to\infty} \frac{\nu_2(\ell^{m+1})}{\delta^m}=\lim_{m\to\infty}
\frac{\nu_3(\ell^{m+1})}{\delta^m}=\lim_{m\to\infty}
\frac{\nu_\infty(\ell^{m+1})}{\delta^m}=0.
\]
Therefore (\ref{genusmodulareq}) implies that the genus of the tower is
\begin{equation}\label{genuseq}
\lim_{m\to\infty}
\frac{g(X_0(\ell^{m+1}))}{\delta^m}=\frac{\mu(\ell)}{12}.
\end{equation}
To estimate the splitting rate in the tower, one needs to count the
points on $X_0(\ell^m)$ above $j=0, 1728$ which are not
singularities. Such points above $j=0$ (resp.\ $j=1728$) are zeros of
the pull back of $\Phi$ if and only if $j=0$ (resp.\ $j=1728$) is
supersingular, i.e.\ $p\equiv 2\bmod{3}$ (resp.\ $p\equiv
3\bmod{4}$). Distinguishing cases acording to the value of
$p\bmod{12}$, one finds that
\begin{equation}\label{splittingeq}
\lim_{m\to \infty} \frac{N_{p^2}(X_0(\ell^{m+1}))}{\delta^m}\geq
\frac{(p-1)\mu(\ell)}{12}.
\end{equation}
Equations (\ref{genuseq}) and (\ref{splittingeq}) imply that the tower
is optimal.
\Endproof
It is easy to see that for every $a\geq 1$ we can consider the tower
defined by the correspondence $(g,h):X_0(\ell^{a+1})\rightrightarrows
X_0(\ell^a)$. This yields the subtower $(X_0(\ell^{a+m}))$ which is of
course again asymptotically optimal.
We now present a variant of this construction. Choose an integer
$\lambda$ relatively prime to $\ell$ and $p$. Consider the pull back
of the correspondence $(g,h):X_0(\ell^{a+1})\rightrightarrows
X_0(\ell^a)$ via the natural projection $X_0(\lambda\ell^a)\to
X_0(\ell^a)$. It is easy to see that the pull back correspondence is
$(\tilde{g}, \tilde{h}):X_0(\lambda\ell^{a+1})\rightrightarrows
X_0(\lambda\ell^{a})$.
\begin{prop}
The tower defined by $(\tilde{g},
\tilde{h}):X_0(\lambda\ell^{a+1})\rightrightarrows
X_0(\lambda\ell^{a})$ is
\[
\cdots\to X_0(\lambda\ell^m)\to \cdots X_0(\lambda\ell^{a+1})\to
X_0(\lambda\ell^{a}).
\]
This tower is asymptotically optimal.
\end{prop}
\proof
This is analogous to the proof of Proposition \ref{modulartowerprop}.
\Endproof
We illustrate in an example how easy it is to compute equations for
modular curves, by using the recursive definition.
\begin{exa}\label{X023exa}
We want to compute equations for the curve $X_0(2\cdot 3^m)$ in
characteristic $p\neq 2,3$. Our method is essentially the same as the
method of Elkies \cite{Elkies97}.
Note that the genus of $X_0(18)$ is zero. For $N=3,6,18$, we write
$L_N(u)=0$ for the differential equation corresponding to
$(H^1_{\dR}({\mathcal E}_N/S), \nabla)$. Using the
description of the cusps in \cite{Shimura}, it is easy to check the
following statements. The differential equation $L_3(u)=0$ has three
singularities. It is no restriction to suppose that these
singularities are $x=0,1,\infty$, where $\infty$ maps to $j=0$ and
$x=0,1$ map to $j=\infty$ with ramification index $1,3,1$,
respectively. The map $X_0(6)\to X_0(3)$ of degree three is totally
branched above $x=\infty$ and has two points $P_0^1, P_0^2$ (resp.\
$P_1^1, P_1^2$ above $x=0$ (resp.\ $x=1$), where $P_\ast^e$ is
ramified of order $e$. Up to normalization, there is a unique such
cover which is given by $x=-27y^2/(y-4)^3$. It follows that the
singularities of $L_6(u)=$ are $S_6=\{0,\infty,-8,1\}$. A look at the
ramification indices of these cusps in $X_0(6)\to X(1)$ tells us that
the Atkin--Lehner involution $\sigma_6$ acts on these points as
$(0,1)(-8, \infty)$, therefore $\sigma_6(y)=-8(x-1)/(x+8)$.
A similar argument shows that the natural projection $g:X_0(18)\to
X_0(6)$ is cyclic of order three and branched at $y=0, \infty$.
Therefore we may suppose that $X_0(18)\to X_0(6)$ is given by
$g(z)=z^3$. The singularities of $L_{18}(u)=0$ are just the
inverse image of $S_6$, i.e.\ $S_{18}=\{0, \infty, -2\zeta_3^i,
\zeta_3^i\}$, where $\zeta_3\in \FF_{p^2}$ is a primitive third
root of unity. The Atkin--Lehner involution is given, up to
normalization, by $\sigma_{18}(z)=-2(z-1)/(z+2)$. We define the
rational function $h(z)=\sigma_6\circ g\circ \sigma_{18}(z)=
z(z^2-2z+4)/(z^2+z+1)$.
This gives the recursive definition for the modular curves
$X_0(2\cdot 3^m)$. \end{exa}
For $\lambda$ relatively prime to $p$, we may define the
congruence subgroup $\Gamma_1(\lambda)\cap \Gamma_0(\ell^a)$. Write
$X_{1,0}(\lambda, \ell^a)_{\CC}$ for the quotient curve of the
completed upper half plane by $\Gamma_1(\lambda)\cap
\Gamma_0(\ell^a)$. It is well known that $X_{1,0}(\lambda,
\ell^a)_{\CC}$ has a model $X_{1,0}(\lambda, \ell^a)_{R}$ over
$R=\ZZ[\zeta_\lambda, 1/\lambda\ell]$, where $\zeta_\lambda$ is a
primitive $\lambda$th root of unity. Write
$\FF_q=\FF_p[\zeta_\lambda]$ and $X_{1,0}(\lambda,
\ell^a)=X_{1,0}(\lambda, \ell^a)_{R}\otimes \FF_q$. Let $f:X_{1,0}(\lambda,
\ell^a)\to X_0(\ell^a)$ be the natural projection.
We can consider the pull back of
$(g,h):X_0(\ell^{a+1})\rightrightarrows X_0(\ell^a)$ via
$f:X_{1,0}(\lambda, \ell^a)\to X_0(\ell^a)$. Write ${\mathcal
T}_{g,h}(f):=(X_{1,0}(\lambda, \ell^{a+n})$ for the corresponding
tower. As in \cite{LMS}, one give a criterion on $p$ for this tower to
be asymptotically optimal. For example, it is easy to show that the
tower of Proposition \ref{exaprop} is isomorphic to $(X_{1,0}(8,
2^{4+m}))$. One can give an alternative proof of the facts on the
minimal splitting field by using this interpretation of the tower.
| 103,533
|
TITLE: How to generate the icosahedral groups $I$ and $I_h$?
QUESTION [1 upvotes]: The icosahedral groups $I$ with 60 elements and $I_h = I \times Z_2$ are also three dimensional point groups. However, ever unlike other point groups, it seems there is rarely reference to give their representation (e.g., matrix representation for each elements). Can any one help me how to generate them?
REPLY [0 votes]: The explicit matrix representations of all 5 irreducible representations of $I$ are given in Hu, Yong, Zhao, and Shu, "The irreducible representation matrices of the icosahedral point groups I and Ih", Superlattices and Microstructures, Volume 3, Issue 4, 1987, pages 391-398, DOI:10.1016/0749-6036(87)90212-6. You can construct the irreps of $I_h$ via the direct product structure with a point inversion.
| 198,540
|
: Lynn, Janice
City: Rockwall
County: Collin
State: TX
Location Notes: Home acreage about 6 miles north of Rockwall.
Accession date: 2013-07-10
Filename: JEL_IMG0593.JPG
Slide Index: PHeliotrope flower underside showing hair
Restrictions: Unrestricted
Collection: Wildflower Center Digital Library
Original Format: Digital
Orientation: Landscape
Shot: Close-up of flower and leaves.
Date Taken: 2013-06-20
NPIN Image Id: 36103
Image VerificationThis image has been verified.
By: JAM
Date: 2013-07-10
From the Image Gallery
Additional resourcesUSDA: Find Heliotropium tenellum images in USDA Plants
Google: Search Google Images for Heliotropium tenellum
| 317,971
|
If scientists have learned anything new about the genetics of rare species in the past three decades, the U.S. Fish and Wildlife Service may not want to hear about it.
In
January, H. Dale Hall, the Service’s Region 2 director,
released a new policy for developing recovery plans for rare
species: Scientists are to use only the genetic information
available at the time a species was first protected under the
Endangered Species Act. To make use of new genetic data, writes
Hall, would be "inappropriate and inconsistent" with the law.
Hall’s decision is based on a 2001 federal court
ruling by Judge Michael Hogan that wild coho salmon are not
genetically distinct from hatchery coho salmon. In response, NOAA
Fisheries, the agency charged with protection of the fish, removed
Oregon’s coastal coho from the endangered species list (HCN,
10/8/01: Coho salmon lose federal protection).
Not
everyone agrees with the new policy: In a March 11 letter to Hall,
Ralph Morgenweck, the Service’s Region 6 director, expressed
concern that the new policy "could run counter to the purpose of
the Endangered Species Act" and "may contradict our direction to
use the best available science."
Jeff Curtis, Western
conservation director for Trout Unlimited in Portland, says of
Hogan’s coho decision: "What we have found is that people use
Hogan to justify doing things that Hogan doesn’t require." He
adds, "Rather than using the bureaucracy to expand (its) areas of
discretion to protect endangered species, (the Service is) trying
to limit (its) areas of discretion."
This isn’t the
first time Hall’s office has limited its own ability to
protect endangered species: In 2003, after a federal court ruled
the U.S. Bureau of Reclamation had "discretion" to release water in
the Middle Rio Grande for the endangered silvery minnow, the
Service reversed its earlier stance and backed the Bureau’s
decision to let the river run dry (HCN, 8/4/03: Truce remains
elusive in Rio Grande water fight).
Topic: Flora & Fauna Department: News Comments: 0
Frozen in time: Endangered species science
Document Actions
- Share this:
- Like
- Tweet
- Tip Jar
- Print this
| 159,591
|
Wow, after just one day and over 500 votes, it seems like you all, rather overwhelmingly, would like to see the AHL tournament go first. Still, just to let you all know what you're voting on, I'm posting the logo pools for both the AHL and ECHL.
AHL
There sure are a lot of red logos up there. But those 29 make up the AHL pool. So vote AHL if you want to vote on these logos first!
ECHL
At least this set is a little more varied in terms of color. These are the 25 logos that make up the ECHL pool. So vote ECHL if you want to vote on these logos first!
We'll get to other leagues in the future including, but not limited to the CHL, OHL, QMJHL, WHL, BCHL, EIHL, IHL, and USHL — just to name a few. It'll be more difficult for me to gather logos for lesser-known leagues to make graphics, so any help you guys might be able to offer would be enormously appreciated. You can email links or images to me at nhllogos@gmail.com.
UPDATE (10/2 3:06 PM): Made some updates to these pools. On the AHL side, I updated the logos for Bridgeport, Lake Erie, Manchester and Manitoba. (I know some of you have told me the wordmark is no longer part of the logo for the Moose, but that's what they use on the web site. I'd be happy to change it, but I'll need a good graphic source. I haven't found an image large enough to make a graphic out of yet. If you can help with that, email me.)
For the ECHL, I added the Gwinnett Gladiators logo which somehow got overlooked when I was transferring files.
Once again, let me know if there's anything else I need to change to get prepared for the new tournament. Thanks everybody!
UPDATE (10/7 10:11 AM): Just updated a few more logos in these two pools. Got rid of the wordmark in the Manitoba logo for the AHL and added better graphics for Elmira, Idaho, Mississippi and Pensacola of the ECH comments:
Its hard to tell because theres no close up but you might be a bit off on the falcons logo,but then again a could be wrong.
why not put the actual lake erie monsters logo on there? in my opinion, that one is alot better than the one on the jerseys.
The falcons logo is incorrect according to there website, the logo site, and NHL 08.
The Falcons logo looks fine to me...
For the Lake Erie monstors u r using the alternate logo. even though it appears on there jersey the m with a hockey puck isnt the logo. its the monster head above water logo, that is ther main logo.
why is the worcester sharks logo the same as the old sj sharks logo?
where's the gwinnet Gladiators??!! (echl)
Thanks for the input everybody. I'm not sure which Falcons logo you guys are referring to, so here are links to large versions of both. You can double check for me.
Springfield Falcons (AHL)
Fresno Falcons (ECHL)
I can't believe I forgot the Gwinnett Gladiators! I know I made the graphic, so it must've just gotten lost somewhere along the way when I was transferring files. I'll make sure it's up later today.
Also, I'll adjust the Lake Erie Monsters to the primary logo instead of the one on the jersey. I guess that one threw me for a bit of a loop.
And according to their web site, the Worcester Sharks' logo is the San Jose Sharks' old logo. I'm just using the version without the wordmark over it.
Thanks for all the help guys! I really appreciate. If you see anything else, please let me know.
the moose's logo is wrong...there is no wordmark on the jersey logo anymore
here is what they currently use
There's no rush brother, let's get to all those leagues eventually!
OHL is defiantly one I want to see!
Hey, cool site. I'll be sending all my readers over here tonight.
for the EcHL you forgot the gulls
you also forgot the Ice piolots
for the EcHL you forgot the gulls
I'm fairly sure the San Diego Gulls folded last summer and therefore don't exist anymore. There is no reference to them on the ECHL's official web site.
you also forgot the Ice piolots
Take another look. The Pensacola Ice Pilots are there. Just below the Gladiators, to the left of the Roadrunners, and above the Wildcatters.
The logo site is incorrect then because it shows that the piolots have a different logo
The mid atlantic hockey league has some really cool logos but only has six teams. You could put them into this but sticking them in the ahl or echl brackets
The logo site is incorrect then because it shows that the piolots have a different logo
Do you mean to say that the Ice Pilots' official web site itself has the wrong logo? Iol
I'm sticking with the one I've got. In the meantime, check out their site.
when I said the logo site I ment chris creamer's sports logo.net
when I said the logo site I ment chris creamer's sports logo.net
Oh, I'm sorry. Yes, that's true. The minor league logos don't seem to be updated accurately on that site so I have to go elsewhere. Apologies for the confusion.
Chris Why don't you use Worcester's Road jersey logo(shield) so that it's something differant and people don't confuse it with the old sharks logo...? just a thought. BTW your site is awesome and your doing a great job.
| 397,607
|
New 2020 Audi SQ5 3.0T Premium
- VIN: WA1B4AFY5L2078602
- Stock: A8143
Basic Info
- Exterior:White
- Drivetrain:quattro
- Transmission:8-Speed Automatic with Tiptronic
- Interior:Black
- Engine:3.0L TFSI
- Fuel Efficiency:18 CITY / 23 HWY
Key Features
- Android Auto
- Apple CarPlay
- Backup Camera
- Blind Spot Monitor
- Bluetooth
- Fog Lights
- Hands-Free Liftgate
- Heated Seats
- Interior Accents
- Keyless Entry
- Leather Seats
- Navigation System
- Power Seats
- Rain Sensing Wipers
- Rear A/C
- Satellite Radio Ready
Description
Vehicle Details
Exterior
- Wheels: 20" x 8" 5-Double-Spoke Star Design -inc: Partly polished
-
- Roof Rack
- Programmable Projector Beam Led Low/High Beam Auto-Leveling Auto High-Beam Daytime Running Lights Preference Setting Headlamps w/Delay-Off
- Front And Rear Fog Lamps
- Perimeter/Approach Lights
- LED Brakelights
Interior
- Sport Bucket Front Seats
- 10-Way Driver Seat
- 10-Way
- Alcantara -inc: 8.3 MMI center screen and MMI all-in-touch w/handwriting-recognition technology
- 10 Speakers
- Regular Amplifier
- Digital Signal Processor
- Streaming Audio
- Window Grid Diversity Antenna
- 2 LCD Monitors In The Front
- Bluetooth Wireless Phone Connectivity SQ5
- Body Style: quattro Sport Utility
- Drivetrain: All Wheel
- EPA Classification: Small Sport Utility Vehicles 4WD
- Passenger Capacity: 5
- Base Curb Weight: 4321
- Fuel Economy Est-Combined: 20
- EPA Fuel Economy Est - City: 18
- EPA Fuel Economy Est - Hwy: 23
-
- Country of Origin: Germany Pkw
Details
- 3.204 Axle Ratio
- Heated Front Seats
- Leather/Alcantara Seating Surfaces
- AM/FM Stereo
- 4-Wheel Disc Brakes
- Air Conditioning
- Electronic Stability Control
- Front Bucket Seats
- Front Center Armrest
- Leather Shift Knob
- Power Liftgate
- Roof rack
- Spoiler
- Tachometer
- ABS brakes
- AM/FM radio
- Adaptive suspension
-
- Audi smartphone interface (Apple CarPlay/Android Auto)
- Driver Air Bag
- Passenger Air Bag
- Front Side Air Bag
- Front Head Air Bag
- Rear Head Air Bag
- Climate Control
- Multi-Zone A/C
- A/C
- Rear A/C
- Security System
- ABS
-
- Power Folding Mirrors
- HD Radio
- Rear Bench Seat
- Passenger Air Bag Sensor
- Blind Spot Monitor
- Headlights-Auto-Leveling
- Hands-Free Liftgate
- Cross-Traffic Alert
- Smart Device Integration
- Requires Subscription
Other Vehicles You May Like
Customer Reviews
This car with Air suspension is like driving a luxury large sedan in the comfort mode. But with more power and better handling with a smaller foot print for parking. Change the driving mode to sport and you have a sports car. Change to allroad or offroad and you you can go on some old country roads at slower speeds. Its like having 3 cars on one.Read more reviews at Cars.com.
| 333,505
|
The bed is not complete without finding the right beddings to go with it. Although the bed sheets, blankets and duvets are essentials for a good night sleep, the mattress is the most significant of the beddings. Therefore finding the ideal mattress to use is vital to ensure you get returns for your investments. You do not want to spend your hard-earned money on something you will not enjoy or have the comfort you may be looking for. However, with a piece of background knowledge on what to expect when purchasing a new bed or buying
Check the Basics
Numerous reasons exist why it is crucial to find the right type of mattress and bed at bedrenaetter.no. First, the mattress should align well with the way you sleep to give you the comfort you need. Whether you are looking for a mattress due to medical recommendations or not, it is crucial to test what you are purchasing. A bad mattress can leave you exhausted and, in some cases, can cause certain musculoskeletal conditions.
The quality of sleep improves when you choose the ideal mattress for you. If you are not sure or aware of what is good for you, your doctor can assess you and recommend the ideal bedding to use. Here are some tips to help you find a suitable mattress for your bed;
Know what is Suitable for you
Every person has their unique needs to the type of bed and mattress good for them. You can have a checklist of what you want in a bed and beddings. If you share the bed with another person, it would be ideal to consider them when choosing the best items. Look at your previous experiences with mattresses and beds to know what to avoid. It will be easy to shop with a list to narrow down the search to what you want.
Compare Items
It is not prudent to purchase the first item you come across. Rather than buying in a hurry, it is beneficial to take the time to compare different products in the market. Maybe the ideal product for your needs is just close by, and you need to search broadly for what you want. The internet can help you in answering the questions you may have. Companies will include descriptions of the products they sell on their websites. Check the platforms to compare merchandises from various brands.
Determine the Firmmess
The mattress firmness is always different for the various brands in the market. Use the description as a guide and do not consider it as the ultimate truth. It is the reason that we advise buyers to try out the bed or mattress they want to buy.
Consider the Size of Bed
Most of the beddings you can buy depending on the bed size. It will be odd to purchase an over or undersized mattress for your bed. Above that, it will be uncomfortable as the former will offer a rugged terrain. The latter leaves spaces on the sides, making it uncomfortable.
| 213,489
|
Aramark Coffee Services
Aramark Refreshment Services supplies more than one billion cups of coffee and 400 million cold drinks annually at approximately 100,000 locations in North America. Professionals from small local businesses to international Fortune 500 companies enjoy using Aramark as a single source for office coffee service, filtered water, brand-name food and beverages, and breakroom essentials.
Coffee and Tea Service - Aramark supplies some of the best coffee beans in the world and a variety of machines for brewing. Coffee brands include ALTERRA, Starbucks, Seattle's Best, Green Mountain, and more. Brewer options start at single serve and move on to bean-to-cup, airport/thermal pot and espresso.
Aramark brings everything you need: creamer, sweetener, napkins and any special requests. The route service representative will even tidy up the breakroom, keeping employees on task. Aramark will bring tea and cocoa as well for the non-coffee drinkers.
Filtered Water and Ice - With Aramark's AquamarkLX water filtration unit, you get limitless clean water without the hassle of jugs or bottles. It taps right into the facility's water supply. The AquamarkLX filters help diminish chemicals, sediment and microorganisms. Everyone in your office will be able to use it to make coffee, soups, drink mixes and much more. Aramark also provides regular service throughout the year to maintain its high quality by changing filters and cleaning the unit.
Marketplace Solutions - Aramark's Marketplace Solutions bring convenient meal and snack options to the workplace. The fresh market offerings can substitute for café services and can be used in partnership with Aramark's other services - cafés, attended retail, catering, and office refreshment service - as a full scale solution. There is a great variety of healthy snack options to choose from, safe and secure payment methods and temperature control to keep food garden-fresh and delicious.
Snacks and Vending - Aramark's innovative equipment brings the latest in technology and reliability. Stock includes a full selection of popular brand-name snacks, juices, teas, and fresh food. A Route Service Representative keeps machines well stocked and in working order. Their RefreshTECHTM inventory management tracks each transaction to learn which snacks are enjoyed most.
Aramark's many offerings lets employers focus on more important things while still providing a refreshment program as an employee benefit.
If you're interested in office coffee machines, beans, and supplies, select below to being customizing your price quote and connect with major coffee vendors.Ready to Compare Office Coffee Services Price Quotes?
| 250,080
|
TITLE: Galois extension and prime number.
QUESTION [4 upvotes]: Let $G$ be a finite group with order $n$, i.e., $|G|=n$. Show that there is a prime number $p\geq n$ and a finite Galois extension $L/K$ with $Gal(L/K)\approx G$ and $[K:\mathbb{Q}]=p!/n$.
Honestly, I have no idea!
REPLY [1 votes]: Consider $G$ as a subgroup of $S_n$, the symmetric group (have $G$ act on itself by left multiplication - $n$ is the order of $G$). The group $S_n$ sits inside of $S_p$, for any prime $p$ larger than or equal to $n$. Now, the inverse Galois problem over $\mathbb Q$ is solved for the symmetric group ( according to https://en.wikipedia.org/wiki/Inverse_Galois_problem, by Hilbert - but see below). So there exists a Galois extension $L/ \mathbb Q$ with group $S_p$. Let $K= L^G$, the fixed field of $G$. Then $p! = [L:{\mathbb Q}] = [L:K] \ [ K:{\mathbb Q}] = n\ [ K:{\mathbb Q}]$. Hence $$G = \mathop{\rm Gal} (L/K )$$ and
$$[K:{\mathbb Q}] = p! /n. $$
For a solution to the inverse Galois problem over $\mathbb Q$ for $S_p$ with $p$ prime $\ge 5$ (courtesy of Brauer), see section 4.10 of Basic Algebra I (surely one of the most discouraging text book titles ever) of Jacobson, Second Edition. The section also refers to a exercise 5, on page 305, which follows a different argument due Tate to solve the inverse Galois problem for $S_n$ over $\mathbb Q$.
| 169,867
|
What size steel building do you need?
A new 20 x 20 steel building could possibly be utilized for a workshop, storage garage, or storage structure, and it’s really the clear span design is perfect for a do-it-yourself assembly. There are so many variables in which have to be looked at while pinpointing the size of the steel building.
Building Use:
Have you ever thought about the end use of the steel building? Do you want to use it for a single function or utilize it for many functions? Planning for a further space with your RV garage or perhaps a home business space? Should your cafe in addition have a substantial private room for receptions or perhaps business conferences? Why don’t you add a lot more room or space inside your barn and give shelter with the hay and also agricultural machines? Think the most effective way that allows you to take benefit from the steel building that will help understanding the ample amount of sq footage.
Get connected to Several Dealers!
Receive multiple free quotes from top Metal Building companies. Compare styles and prices to get the best steel building for you. It’s fast, easy, and free. Start now!
Location of Structure:
Get the existing area suitable for your steel building? Have you ever examined the available area with the property which you attempt to set up your building? What about the size and shape with the buildable portion of the residence. Do you need any kind of easements in your building construction property? In case you’re utilizing it for a commercialized or manufacturing functionality, just how much room is necessary for parking? Responding to all these concerns can offer you a sense of the amount of open area available for a brand-new structure.
Width of Structure:
Should you have decided what amount of sq footage you’ll need for the steel building together with the size and shape of your acreage in the construction site, it becomes a lot easier for you to identify the actual external measurements of the structure. Width is one of vital measurements for most building applications. Whenever unblocked and clear space is undoubtedly critical, the unobstructed area enables you to formulate the room in just about every strategy you choose, having reasonable quantity of space intended for problem less maneuverability. It’s possible to lengthen that steel buildings as much as 150’ minus the inclusion with the structure and support columns. If you happen to do not ever need to have a wide open span configuration as part of your structure, make full use of support columns which can be less expensive plus cost-effective.
Length of Structure:
steel building specifications are generally in 10’ increments. When you’ve got unique blueprints, they may build a structure as documented in your concerns. From the viewpoint of the style and design, you are able to assemble just about any size of building.
Height of Structure:
Decide on the final application of the building truly respectfully as it’s an important component. Eave height is applied to determine the peak of your steel structure.Suppose that it’s essential to assemble a 14’ high garage door to help you squeeze in the Recreational vehicle, you might need a 16’ eave elevation. In the same manner once you include a 2nd floor as well as a mezzanine floor, be prepared for 20’ tall in height side wall membrane. Regarding aircraft hangars, truck structures, and manufacturing facilities, the distance off the ground with the structure is decided by the machines you may be housing and using.
Long Term Expansion:
Ensure you predict whether or not you intend to enhance the framework in the foreseeable future. Luckily for us, putting in framing on the end walls of your steel building needn’t be hard and low-cost, by merely purchasing extra support frames. Make sure you take into consideration also the location of your building on the lot, letting area for virtually any possible future extension.
Personal Spending Plan:
Even with creating a narrow financial budget with regard to the building of the steel building, you’ll need to consider the long term future problems. So why do you want to obtain a smaller sized structure that will likely not meet your company needs and and this is not likely valuable in your immediate future?
So what is the next move in the process?
steel building special offers could differ just a bit from state to state because each building will be individually made with the necessary architectural variations to meet localized strong wind and snowfall load requirements. Fill in the above form right now so that our knowledgeable and professional vendors can help you promptly when deciding on the perfect steel building package deal for ones upcoming building job.
| 19,495
|
Ask Best Of Forum Glossary Members Only Facebook Podcast on Business Books My Library Free Newsletter Making Technology Work For Everyone Loading What Security Software Do You Recommend? Reply Leo April 12, 2016 at 2:42 pm Timing. That's why Leo's advice on staying safe on the internet is so important! Transcript The interactive transcript could not be loaded.
The reason why the changes are flagged by Spybot 2 is that there are also malware programs that disable the notifications so the user does not take note of his security Here, click "Stop" beside the Security Center Service. The system returned: (22) Invalid argument The remote host or network may be down. Loading...
You can take the full course on Experts Exchange at. It does a fine job of detecting malware, does so without adversely impacting system performance, and does so without nagging you for renewals, upgrades, or up-sells. BTW, the efficacy testing I mentioned is actually funded by the banks and used to determine which products are recommended/supplied to their customers. You can right click and have Spybot ''exclude that detection from future scans.'' The second post on this Spybot forum link will explain it better than I canI have had this
Again, thanks a bunch. 0 LVL 97 Overall: Level 97 Windows XP 30 Message Expert Comment by:war1 ID: 147576372005-08-25 Asta, Check your settings in the Security Center. Once a day, once a week? There are three possible alerts: * Firewall Alert me if my computer might be at risk because of my firewall settings Spybot Search And Destroy Post Windows 10 Launch report from a menu, considering criteria only when it is filled… MS Office Office 365 Databases MS Access How to create built-in UI screens with Adobe XD Video by: Bob
Maybe someday that might change, but most likely NOT in my lifetime (I'm now 62), and I've been in the IT-Tech business now for over 20+ years (on both sides of Spybot Disable Live Protection It said to come here, but I specifically noted that these websites that you linked are blocked by my school policy. Equations, Back Color, Alternate Back Color. I'm using WinAntiRansom by WinPatrol.
Do I have to uninstall/disable Windows Defender first (eg by going to Windows services)? Spybot Windows Defender And oh, let's not forget about using an efficient Firewall (although nowadays all OS's come with one). Disruptive posting: Flaming or offending other usersIllegal activities: Promote cracked software, or other illegal contentOffensive: Sexually explicit or offensive languageSpam: Advertisements or commercial links Submit report Cancel report Track this discussion Because Microsucks is THE most commonly used OS and #1 OS in the world, it is also the most commonly attacked OS.
You can ignor the warning. It's not perfect, but no security tool is. Spybot Windows 10 Download Reply Ray Smith September 14, 2016 at 12:39 pm I previously made this comment: "WD’s detection capabilities are broadly similar to the detection capabilities of other products, with usually only a Spybot Live Protection I say should, because most phones nowadays use WPA2 PSK encryption, but I'm not sure if this applies to all phones.
Recommendations Reviews & Affiliate Disclosure MenuExperts Exchange Browse BackBrowse Topics Open Questions Open Projects Solutions Members Articles Videos Courses Contribute Products BackProducts Gigs Live Courses Vendor Services Groups Careers Store Headlines news And, of course, this specialized crimeware is what you really, really want your antivirus program to be able to block. June 21st, 2012Spybot 2 detects registry changes associated with Microsoft Security Center; they are listed as “Windows Security Center”. Take yourself to another level. Spybot Vs Windows Defender Windows 10
What’s the difference between Microsoft Security Essentials and Windows Defender? - Windows Defender is basically Security Essentials by another name… except for the time when it wasn't. So they're never going to be a victim of a phishing-scheme sent to them or click on a link in an E-mail or open an attachment from someone they don't know To fix this issue, please disable Spybot's Security Center Service using the instructions below, as this will prevent Windows from detecting Spybot as an antivirus program. have a peek at these guys If you've ever been stung by this kind of exploit which locks up your computer, and then asks for money, we need another layer.
There is no single, clear, consistent winner. Spybot Windows 10 Cnet It's already installed in Windows 8 and later - there's nothing you have to do. You've got to run the free version scans manually.
But it's time to leave XP. To make your life a little easier, here's a short version that sums it all up. Sorry, there was a problem flagging this post. Spybot 2.5 Windows 10 Download Reply Terry Hoffman April 12, 2016 at 11:15 am I just got a new (used) computer and the store put on windows 10 with Windows defender.
The PC Security Channel [TPSC] 164,947 views 14:53 How to run Malwarebytes and Spybot S&D - Duration: 6:28. Try it. Four Corners Computer Center 81 views 5:49 Spybot Search and Destroy 2.3 review (removal test) - Duration: 15:14. check my blog This is not some policy group but is done by Mcafee.I am so frustrated with Mcafee!
However, when it comes to real-world efficacy testing in relation to specialized financial malware/crimeware – LICAT, Zberp and the numerous ZeuS clones, for example – there’s a much greater disparity. The PC Security Channel [TPSC] 14,095 views 15:14 Malwarebytes Anti-Malware 3.0.5 1299 Premium Crack (Lifetime Free) - Duration: 4:48. She wants to be able to use the laptop all over the house. I have run unprotected except for Win patrol aka Scotty and firewall/router combo.
Sign in to add this video to a playlist. Donate Free Edition Home Edition Pro Edition Download Free Software Tools Donate sbNet Corporate Technician Evaluation Licenses Becoming a Reseller About Contact Privacy Policy Terms & Conditions of Sale Site Map Doing so can cause conflicts, system instability and is really a waste of both computer resources and money. The license covers 2 computers.
About Press Copyright Creators Advertise Developers +YouTube Terms Privacy Policy & Safety Send feedback Test new features Loading...
| 347,796
|
No photos available yet
The seller's agent hasn't uploaded any photos — check back soon
$400,000$280/sf
Sold10 years ago
$400,000$280/sf
Sold10 years ago
- Home, 4 beds 2 baths, 1,427 sf, 7,840 sf lot
- HOA dues: $125/mo
Features
- - Built 1973
- - Heating: forced air
- - View: local/neighborhood
- - Building: attached
- - Garage/parking: detached
- - Kitchen: dishwasher
- - Garbage disposal
- - Oven range
- - Laundry: in laundry room
- - Foundation: concrete slab
- - Water: heater - gas
- - Public
- - Style: contemporary
- - Patio(s)
- - Patio(s)/deck(s) - covered
- -
| 332,037
|
After a decade with Nike Golf, Mike McKennon started his own company, McKennon Golf Bag Company, making leather golf bags in his basement. Now he's making them for presidents.
"We had set out to make something very presidential, but not too gaudy," he said in keeping with the company philsophy. "It's a simple navy blue bag with white accents. We've got the presidential seal on the headovers and bottom. And, subtly, on the side we put Golf Bag One."
McKennon said he was able to get it in President Obama's hands, though he would not disclose how he did so.
He has made one for Davis Love III, the next U.S. Ryder Cup captain, who used it in the Par 3 Contest at the Masters this year. "I made him a North Carolina bag," McKennon said. "Lucas Glover saw it and called and wanted a Clemson one."
McKennon's philsophy evolved from the statement at the top of his website home page: "As we developed the McKennon Player's Bag we considered every single feature of the modern golf bag. We didn't use a damn one of them."
"The whole premise of our company," he said, "is simplicity and intuitive golf bags. We don't have to have 13 zippers. You know where things are. We don't clutter it up."
The Player's Bag is handcrafted from aniline leather, sells for $799 and comes with either one or two zippered pockets.
In addition, McKennon purchased Jones Sporting Goods, maker of the Jones Bag that was so popular 30 and 40 years ago. "This is the only bag in golf where you can look at it and go, 'I remember that.' I carried a Jones bag until I was probably 25. It's cool to see it come back to life."
McKennon no longer works from his basement. He now has a 3,200-square foot shop and recently hired a fifth employee. "We're trying to turn everything upside down," he said. "We're going to go head-to-head with the bigger companies."
-- John Strege
| 62,263
|
Betty Moone
Madrid, Spain
Hi
My name is Betty and I am the Community Manager of driveme.es, which is a luxury chauffeur driven car & vehicle company based in Madrid (Spain).
We also have the web site driveme.tours for our new Private Tours business unit.
We hope you like it!
| 334,459
|
\begin{document}
\begin{abstract}
In this article we prove certain results comparing rationality of algebraic cycles over the function field of a quadric and over the base field.
Those results have already been proved by Alexander Vishik in the case of characteristic $0$, which allowed him to work with algebraic cobordism theory. Our proofs use the modulo $2$ Steenrod operations in the Chow theory and work in any characteristic $\neq 2$.
\end{abstract}
\maketitle
\medskip
\medskip
In characteristic $0$, the results of this note (Theorem 1.1 and 2.4 and Proposition 2.1) have been proved several years ago by Alexander Vishik
in [3] (exact references are given right before each statement) with the help of the algebraic cobordism theory and especially \textit{symmetric operations} of [4]. In fact, putting aside characteristic, the original versions are stronger. Indeed, an exponent $2$ element appears in our conclusions, while the use of symmetric operations in the algebraic cobordism theory allows to obtain results without an exponent $2$ element (see Remark on page 370 of [3]).
In a way, most results of this note are generalizations of some results proved by Nikita Karpenko in [2].
Our proofs are, to a great extent, inspired by the proofs of [2].
In our proofs, the base field is allowed to be of any characteristic different from $2$ because the Landweber-Novikov operations used in [3, Remark after proof of Theorem 3.1] are replaced here by the Steenrod operations on the modulo $2$ Chow groups.
We refer to [3] and [2] for an introduction into the subject. Notation is introduced in the beginning of Section 1.
\section{Main Result}
Let $F$ be a field of characteristic $\neq 2$, $Q$ a smooth projective quadric over $F$ of dimension $n\geq 0$, $Y$ a smooth quasi-projective
$F$-variety (a \textit{variety} is a separated scheme of finite type over a field).
The function field $F(Q)$ is defined if $n\geq 1$ or if $Q$ is anisotropic. In the case of $n=0$ and isotropic $Q$ we have
$Q= Spec F\coprod Spec F$ and we set $F(Q):=F$.
We write $CH(Y)$ for the integral Chow group of $Y$ (see [1, Chapter X]) and we write $Ch(Y)$ for $CH(Y)$ modulo $2$.
We write $\overline{Y}:=Y_{\overline{F}}$ where $\overline{F}$ is an algebraic closure of $F$.
Let $X$ be a geometrically integral variety over $F$. An element $\overline{y}$ of $Ch(\overline{Y})$ (or of $CH(\overline{Y})$)
is \textit{$F(X)$-rational} if its image $\overline{y}_{\overline{F}(X)}$ under $Ch(\overline{Y})\rightarrow Ch(Y_{\overline{F}(X)})$
(resp. $CH(\overline{Y})\rightarrow CH(Y_{\overline{F}(X)})$) is in the image of $Ch(Y_{F(X)})\rightarrow Ch(Y_{\overline{F}(X)})$
(resp. $CH(Y_{F(X)})\rightarrow CH(Y_{\overline{F}(X)})$). Finally, an element $\overline{y}$ of $Ch(\overline{Y})$ (or of $CH(\overline{Y})$)
is called \textit{rational} if it is in the image of $Ch(Y)\rightarrow Ch(\overline{Y})$ (resp. $CH(Y)\rightarrow CH(\overline{Y})$).
\medskip
In a way, the following result is a generalization of [3, Theorem 3.1(1)]. Indeed, the use of the Steenrod operations on the modulo $2$ Chow groups allows to obtain a valid result in any characteristic different from $2$. Nevertheless, an exponent $2$ element appears in our conclusion while it is not the case in [3, Theorem 3.1(1)]. In addition, this result is also a generalization of [2, Theorem 2.1] in the sense that it
allows a larger codimension for the considered cycle.
\begin{thm}
\textit{Assume that $m<n/2+j$. Let $\overline{y}$ be an $F(Q)$-rational element of $Ch^m(\overline{Y})$.
Then $S^{j}(\overline{y})$ is the sum of a rational element and the class modulo $2$ of an integral element of exponent $2$.}
\end{thm}
\begin{proof}
We assume that $m\geq 0$ in the proof. We also assume that $j \leq m$ (otherwise we get $S^{j}(\overline{y})=0$). The element $\overline{y}$ being $F(Q)$-rational, there exists $y \in Ch^m(Y_{F(Q)})$ mapped to $\overline{y}_{\overline{F}(Q)}$ under the homomorphism
\[Ch^m(Y_{F(Q)})\rightarrow Ch^m(Y_{\overline{F}(Q)}).\]
Let us fix an element $x \in Ch^m(Q\times Y)$ mapped to $y\;\;mod\;2\;$ under the surjection
\[Ch^m(Q\times Y)\twoheadrightarrow Ch^m(Y_{F(Q)}).\]
Since over $\overline{F}$ the variety $Q$ becomes cellular, the image $\overline{x}\in Ch^m(\overline{Q}\times \overline{Y})$
of $x$ decomposes as
\[\overline{x}=h^0\times y^m +\cdot \cdot \cdot
+h^{[\frac{n}{2}]} \times y^{m-[\frac{n}{2}]}+l_{[\frac{n}{2}]}\times z^{m+[\frac{n}{2}]-n}+
\cdot \cdot \cdot +l_{[\frac{n}{2}]-j}\times z^{m+[\frac{n}{2}]-j-n}\]
with some $y^i \in Ch^i(\overline{Y})$ and some $z^i \in Ch^i(\overline{Y})$, where $y^m=\overline{y}$, and where $h^i \in Ch^i(\overline{Q})$ is the $i$th power of the hyperplane section class while $l_i \in Ch_i(\overline{Q})$ is the class of an $i$-dimensional
subspace of $\mathbb{P}(W)$, where $W$ is a maximal totally isotropic subspace associated with the
quadric $\overline{Q}$ (see [1, \S68]).
\medskip
For every $i=0,...,m$, let $s^i$ be the image in $CH^{m+i}(\overline{Q}\times \overline{Y})$ of an element in $CH^{m+i}(Q\times Y)$
representing $S^i(x)\in Ch^{m+i}(Q\times Y)$. We also set $s^i:=0$ for $i>m$ as well as for $i<0$.
\medskip
The integer $n$ can be uniquely written in the form $n=2^t-1+s$, where $t$ is a non-negative integer and $0 \leq s < 2^t$.
Let us denote $2^t-1$ as $d$. Since $d \leq n$, we can fix a smooth subquadric $P$ of $Q$ of dimension $d$; we write $in$
for the imbedding
\[(P\hookrightarrow Q)\times id_Y: P\times Y \hookrightarrow Q\times Y.\]
\begin{lemme}
\textit{For any integer $r$, one has
\[S^r pr_{\ast}in^{\ast}x=\sum_{i=0}^r pr_{\ast}(c_i(-T_P)\cdot in^{\ast}S^{r-i}(x))\;\;\;\;\;in\;\;Ch^{r+m-d}(Y)\]
(where $T_P$ is the tangent bundle of $P$, $c_i$ are the Chern classes, et $pr$ is the projection
$P\times Y \rightarrow Y$).}
\end{lemme}
\begin{proof}
The morphism $pr:P\times Y \rightarrow Y$ is a smooth projective morphism between smooth schemes. Thus, for any integer $r$,
we have by [1, Proposition 61.10],
\[S^r\circ pr_{\ast}=\sum_{i=0}^r pr_{\ast}(c_i(-T_{pr})\cdot S^{r-i})\]
where $T_{pr}$ is the relative tangent bundle of $pr$ over $P\times Y$. Furthermore, since $pr$ is the projection $P\times Y \rightarrow Y$,
one has $T_{pr}=T_P$. Hence, we get
\[S^r pr_{\ast}in^{\ast}x=\sum_{i=0}^r pr_{\ast}(c_i(-T_P)\cdot S^{r-i}(in^{\ast}x)).\]
Finally, since $in:P\times Y \hookrightarrow Q\times Y$ is a morphism between smooth schemes, the Steenrod operations of cohomological type
commute with $in^{\ast}$ (see [1, Theorem 61.9]), we are done.
\end{proof}
\medskip
We apply Lemma 1.2 taking $r=d+j$. Since $pr_{\ast}in^{\ast}x \in Ch^{m-d}(Y)$ and $m-d<d+j$ (indeed, $m-d<n/2+j-d$ by assumption,
and $n/2<2d$ thanks to our choice of $d$), we have $S^{d+j} pr_{\ast}in^{\ast}x=0$.
Hence, we have by Lemma 1.2,
\[\sum_{i=0}^{d+j} pr_{\ast}(c_i(-T_P)\cdot in^{\ast}S^{d+j-i}(x))=0\;\;\;\;\;\text{in}\;\;Ch^{m+j}(Y).\]
In addition, for any $i=0,...,d$, by [1, Lemma 78.1] we have $c_i(-T_P)={-d-2 \choose i}\cdot h^i$, where $h^i \in Ch^i(P)$
is the $i$th power of the hyperplane section class, and where the binomial coefficient is considered modulo $2$.
Furthermore, for any $i=0,...,d$, the binomial coefficient ${-d-2 \choose i}={d+i+1 \choose i}$ is odd (because $d$ is a power
of $2$ minus 1, cf. [1, Lemma 78.6]). Moreover, for $i>d$, we have $c_i(-T_P)=0$ in $CH^i(P)$ by definition of Chern classes.
Thus, we get
\[\sum_{i=0}^{d} pr_{\ast}(h^i\cdot in^{\ast}S^{d+j-i}(x))=0\;\;\;\;\;\text{in}\;\;Ch^{m+j}(Y).\]
Therefore, the element
\[\sum_{i=0}^{d} pr_{\ast}(h^i\cdot in^{\ast}s^{d+j-i}) \in CH^{m+j}(\overline{Y})\]
is twice a rational element.
\medskip
Furthermore, for any $i=0,...,d$, we have
\[pr_{\ast}(h^i\cdot in^{\ast}s^{d+j-i})=pr_{\ast}(in_{\ast}(h^i\cdot in^{\ast}s^{d+j-i}))\]
(the first $pr$ is the projection $P\times Y \rightarrow Y$ while the second $pr$ is the projection
$Q\times Y \rightarrow Y$). Since $in$ is a proper morphism between smooth schemes, we have by [1, Proposition 56.9],
\[in_{\ast}h^i\cdot in^{\ast}s^{d+j-i}=in_{\ast}h^i\cdot s^{d+j-i}=h^{n-d+i}\cdot s^{d+j-i}\]
and we finally get
\[pr_{\ast}(h^i\cdot in^{\ast}s^{d+j-i})=pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i}).\]
\medskip
Hence, we get that the element
\[\sum_{i=0}^{d}pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\in CH^{m+j}(\overline{Y})\]
is twice a rational element.
\medskip
We would like to compute the sum obtained modulo $4$. Since $s^{d+j-i}=0$ if $d+j-i>m$, the $i$th summand is $0$ for any
$i<d+j-m$ ($(j-m) \leq 0$ by assumption). Otherwise -- if $i \geq d+j-m$ -- the factor $h^{n-d+i}$ is divisible by $2$ (indeed, we have
$h^{n-d+i}=2l_{d-i}$ because $n-d+i\geq n+j-m>n/2$) and in order to compute the $i$th summand modulo $4$ it suffices to compute
$s^{d+j-i}$ modulo $2$, that is, to compute $S^{d+j-i}(\overline{x})$.
We recall that
\[\overline{x}=h^0\times y^m +\cdot \cdot \cdot+
h^{[\frac{n}{2}]} \times y^{m-[\frac{n}{2}]}+l_{[\frac{n}{2}]}\times z^{m+[\frac{n}{2}]-n}+
\cdot \cdot \cdot +l_{[\frac{n}{2}]-j}\times z^{m+[\frac{n}{2}]-j-n}.\]
Therefore, we have
\[S^{d+j-i}(\overline{x})=\sum_{k=0}^{[\frac{n}{2}]} S^{d+j-i}(h^k \times y^{m-k})+\sum_{k=0}^{j} S^{d+j-i}(l_{[\frac{n}{2}]-k}
\times z^{m+[\frac{n}{2}]-k-n}).\]
And we set
\[A_i:=\sum_{k=0}^{[\frac{n}{2}]} S^{d+j-i}(h^k \times y^{m-k})\;\;and\;\;B_i:=\sum_{k=0}^{j} S^{d+j-i}(l_{[\frac{n}{2}]-k}
\times z^{m+[\frac{n}{2}]-k-n}).\]
\medskip
For any $k=0,...,[\frac{n}{2}]$, we have by [1, Theorem 61.14],
\[S^{d+j-i}(h^k \times y^{m-k})=\sum_{l=0}^{d+j-i}S^{d+j-i-l}(h^k)\times S^l(y^{m-k}).\]
Moreover, for any $l=0,...,d+j-i$, we have by [1, Corollary 78.5],
\[S^{d+j-i-l}(h^k)={k \choose d+j-i-l}h^{d+j+k-i-l}.\]
Thus, choosing an integral representative $\varepsilon_{k,l}\in CH^{m-k+l}(\overline{Y})$ of $S^l(y^{m-k})$
(we choose $\varepsilon_{k,l}=0$ if $l>m-k$), we get that the element
\[\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(h^{d+j+k-i-l}\times \varepsilon_{k,l}) \in CH^{m+d+j-i}(\overline{Q}\times \overline{Y})\]
is an integral representative of $A_i$.
\medskip
Therefore, for any $i \geq d+j-m$, choosing an integral representative $\tilde{B_i}$ of $B_i$, there exists $\gamma_i \in CH^{m+d+j-i}(\overline{Q}\times \overline{Y})$ such that
\[s^{d+j-i}=\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(h^{d+j+k-i-l}\times \varepsilon_{k,l})
+\tilde{B_i}+ 2\gamma_i.\]
Hence, according to the multiplication rules in the ring $CH(\overline{Q})$ described in [1, Proposition 68.1],
for any $i \geq d+j-m$, we have
\[h^{n-d+i}\cdot s^{d+j-i}=2\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(l_{l-j-k}\times \varepsilon_{k,l})
+h^{n-d+i}\cdot \tilde{B_i}+ 4l_{d-i}\cdot \gamma_i. \]
If $k \leq d-i$, one has $j+k\leq d+j-i$, and for any $0\leq l \leq d+j-i$, we have by dimensional reasons,
\[pr_{\ast}(l_{l-j-k}\times \varepsilon_{k,l})=\left\{\begin{array}{rcl} &\varepsilon_{k,l} & \text{if}\,\,l=j+k\\
&0&\text{otherwise.}
\end{array}\right.\]
Otherwise $k>d-i$, and $pr_{\ast}(l_{l-j-k}\times \varepsilon_{k,l})=0$ for any $0\leq l \leq d+j-i$.
Moreover, for $k>d-i$, one has $j+k>j+d-i\geq m >m-k$, therefore $\varepsilon_{k,j+k}=0$.
Thus we deduce the identity
\[pr_{\ast}\left(2\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(l_{l-j-k}\times \varepsilon_{k,l})\right)=
2\sum_{k=0}^{[\frac{n}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k}.\]
Then,
\begin{multline*}
\sum_{i=d+j-m}^d pr_{\ast}\left(2\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(l_{l-j-k}\times \varepsilon_{k,l})\right)\\
=2\sum_{i=d+j-m}^d \sum_{k=0}^{[\frac{n}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k}.
\end{multline*}
In the latest expression, for every $k=0,...,[\frac{n}{2}]$, the total coefficient near $\varepsilon_{k,j+k}$ is
\[2\sum_{i=d+j-m}^{d}{k \choose d-i-k}=2\sum_{i=d-2k}^{d-k}{k \choose d-i-k}=2\sum_{s=0}^{k}{k \choose s}=2^{k+1},\]
which is divisible by $4$ for $k \geq 1$.
\medskip
Therefore, the cycle $\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\in CH^{m+j}(\overline{Y})$ is congruent modulo 4 to
\[2\varepsilon_{0,j}+\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i}).\]
Thus, the cycle $2\varepsilon_{0,j}+\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i})$ is congruent modulo $4$ to twice a rational element.
\medskip
Finally, the following lemma will lead to the conclusion.
\begin{lemme}
\textit{For any $d+j-m\leq i \leq d$, one can choose an integral representative $\tilde{B_i}$ of $B_i$ so that
\[pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i})=0.\]}
\end{lemme}
\begin{proof}
We recall that $B_i:=\sum_{k=0}^{j} S^{d+j-i}(l_{[\frac{n}{2}]-k}\times z^{m+[\frac{n}{2}]-k-n})$.
For any $k=0,...,j$, we have by [1, Theorem 61.14],
\[S^{d+j-i}(l_{[\frac{n}{2}]-k} \times z^{m+[\frac{n}{2}]-k-n})=\sum_{l=0}^{d+j-i}S^{d+j-i-l}(l_{[\frac{n}{2}]-k})\times S^l(z^{m+[\frac{n}{2}]-k-n}).\]
And for any $l=0,...,d+j-i$, we have by [1, Corollary 78.5],
\[S^{d+j-i-l}(l_{[\frac{n}{2}]-k})={n+1-[\frac{n}{2}]+k \choose d+j-i-l}l_{[\frac{n}{2}]-k-d-j+i+l}.\]
Thus, choosing an integral representative $\delta_{k,l}\in CH^{m-k+l}(\overline{Y})$ of $S^l(z^{m+[\frac{n}{2}]-k-n})$
(we choose $\delta_{k,l}=0$ if $l>m+[\frac{n}{2}]-k-n$), we get that the element
\[\sum_{k=0}^{j} \sum_{l=0}^{d+j-i}{n+1-[\frac{n}{2}]+k \choose d+j-i-l}(l_{[\frac{n}{2}]-k-d-j+i+l}\times \delta_{k,l}) \in CH^{m+d+j-i}(\overline{Q}\times \overline{Y})\]
is an integral representative of $B_i$. Let us note it $\tilde{B_i}$.
\medskip
Hence, we have
\[h^{n-d+i}\cdot \tilde{B_i}=\sum_{k=0}^{j} \sum_{l=0}^{d+j-i}{n+1-[\frac{n}{2}]+k \choose d+j-i-l}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l}).\]
Moreover, we have
\[pr_{\ast}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l})\neq 0\Longrightarrow l=j+k+n-[\frac{n}{2}].\]
Furthermore, for any $0\leq k \leq j$, we have $d+j-i \leq m <j+\frac{n}{2}\leq j+n-[\frac{n}{2}]\leq j+k+n-[\frac{n}{2}]$.
Thus, for any $0 \leq l\leq d+j-i$ and for any $0\leq k \leq j$, we have $pr_{\ast}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l})=0$.
It follows that $pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i})=0$ and we are done.
\end{proof}
We deduce from Lemma 1.3 that the cycle $2\varepsilon_{0,j} \in CH^{m+j}(\overline{Y})$ is congruent modulo $4$ to twice a rational cycle.
Therefore, there exist a cycle $\gamma \in CH^{m+j}(\overline{Y})$ and a rational cycle $\alpha \in CH^{m+j}(\overline{Y})$ so that
\[2\varepsilon_{0,j}=2\alpha + 4\gamma,\]
hence, there exists an exponent $2$ element $\delta \in CH^{m+j}(\overline{Y})$ so that
\[\varepsilon_{0,j}=\alpha + 2\gamma + \delta.\]
\medskip
Finally, since $\varepsilon_{0,j}$ is an integral representative of $S^{j}(\overline{y})$, we get that
$S^{j}(\overline{y})$ is the sum of a rational element and the class modulo $2$ of an integral element of exponent $2$.
We are done with the proof of Theorem 1.1.
\end{proof}
\section{Other results}
In this section we continue to use notation introduced in the beginning of Section~1.
In the same way as before, the following proposition is a generalization of [3, Proposition 3.3(2)]
(although, putting aside characteristic, our proposition is still weaker than the original version
in the sense that an exponent $2$ element appears in the conclusion).
\begin{prop}
\textit{Let $x \in Ch^m(Q\times Y)$ be some element, and $y^i, z^i \in Ch^i(\overline{Y})$ be the coordinates of $\overline{x}$ as
in the beginning of proof of Theorem 1.1.
Assume that $m=[\frac{n+1}{2}]+j$ and that $n\geq 1$.}
\textit{Then $S^j(y^m)+y^m\cdot z^j$ differs from a rational element by the class of an exponent $2$ element of
$CH^{m+j}(\overline{Y})$.}
\end{prop}
\begin{proof}
The image $\overline{x}\in Ch^m(\overline{Q}\times \overline{Y})$ of $x$ decomposes as
\[\overline{x}=h^0\times y^m +\cdot \cdot \cdot
+h^{[\frac{n}{2}]} \times y^{m-[\frac{n}{2}]}+l_{[\frac{n}{2}]}\times z^{m+[\frac{n}{2}]-n}+
\cdot \cdot \cdot +l_{[\frac{n}{2}]-j-1}\times z^{m+[\frac{n}{2}]-j-n}.\]
\medskip
Let $\textbf{x} \in CH^m(Q \times Y)$ be an integral representative of $x$.
The image $\overline{\textbf{x}} \in CH^m(\overline{Q} \times \overline{Y})$ decomposes as
\[\overline{\textbf{x}}=h^0\times \textbf{y}^m +\cdot \cdot \cdot
+h^{[\frac{n}{2}]} \times \textbf{y}^{m-[\frac{n}{2}]}+l_{[\frac{n}{2}]}\times \textbf{z}^{m+[\frac{n}{2}]-n}+
\cdot \cdot \cdot +l_{[\frac{n}{2}]-j-1}\times \textbf{z}^{m+[\frac{n}{2}]-j-n}\]
where the elements $\textbf{y}^i \in CH^i(\overline{Y})$ (resp. $\textbf{z}^i \in CH^i(\overline{Y})$) are some integral representatives of
the elements $y^i$ (resp. $z^i$).
\medskip
For every $i=0,...,m-1$, let $s^i$ be the image in $CH^{m+i}(\overline{Q}\times \overline{Y})$ of an element in $CH^{m+i}(Q\times Y)$
representing $S^i(x)\in Ch^{m+i}(Q\times Y)$. We also set $s^i:=0$ for $i>m$ as well as for $i<0$. Finally, we set $s^0:=\overline{\textbf{x}}$
and $s^m:={(s^0)}^2$. Therefore, for any integer $i$, $s^i$ is the image in $CH^{m+i}(\overline{Q} \times \overline{Y})$ of an integral representative of $S^i(x)$.
\medskip
The integer $n$ can be uniquely written in the form $n=2^t-1+s$, where $t$ is a non-negative integer and $0 \leq s < 2^t$.
Let us denote $2^t-1$ as $d$.
\medskip
We would like to use again Lemma 1.2 to get that the sum
\[\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\in CH^{m+j}(\overline{Y})\]
is twice a rational element. To do this, it suffices to check that $m-d<d+j$.
Then the same reasoning as the one used during the proof of Theorem 1.1 gives us the desired result.
We have $m-d=[\frac{n+1}{2}]+j-d=d+j+([\frac{n+1}{2}]-2d)$, and since our choice of $d$ and the assumption $n\geq 1$,
one can easily check that $2d>[\frac{n+1}{2}]$. Thus we do get that the sum
\[\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\in CH^{m+j}(\overline{Y})\]
is twice a rational element. We would like to compute that sum modulo $4$.
\medskip
For any $i\geq d+j-m$, the factor $s^{d+j-i}$ present in the $i$th summand is congruent modulo $2$ to
$S^{d+j-i}(\overline{x})$, which is represented by $\tilde{A_i}+ \tilde{B_i}$, where
\[\tilde{A_i}:=\sum_{k=0}^{[\frac{n}{2}]} \sum_{l=0}^{d+j-i}{k \choose d+j-i-l}(h^{d+j+k-i-l}\times \varepsilon_{k,l})\]
and
\[\tilde{B_i}:=\sum_{k=0}^{j} \sum_{l=0}^{d+j-i}{n+1-[\frac{n}{2}]+k \choose d+j-i-l}(l_{[\frac{n}{2}]-k-d-j+i+l}\times \delta_{k,l})\]
where $\varepsilon_{k,l}\in CH^{m-k+l}(\overline{Y})$ (resp. $\delta_{k,l}\in CH^{m-k+l}(\overline{Y})$) is an integral representative of $S^l(y^{m-k})$ (resp. of $S^l(z^{m+[\frac{n}{2}]-k-n})$),
and we choose $\varepsilon_{k,l}=0$ if $l>m-k$ (resp. $\delta_{k,l}=0$ if $l>m+[\frac{n}{2}]-k-n$).
Finally, in the case of even $m-j$ , we choose $\varepsilon_{\frac{m-j}{2},\frac{m+j}{2}}=(\textbf{y}^{\frac{m+j}{2}})^2$.
\medskip
Furthermore, for any $i\geq d+j-m$, we have
\[h^{n-d+i}\cdot \tilde{B_i}=\sum_{k=0}^{j} \sum_{l=0}^{d+j-i}{n+1-[\frac{n}{2}]+k \choose d+j-i-l}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l}).\]
And we have
\[pr_{\ast}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l})\neq 0\Longrightarrow l=j+k+n-[\frac{n}{2}].\]
On the one hand, for any $i>d+j-m$, we have $d+j-i<m=n-[\frac{n}{2}]+j \leq j+k+n-[\frac{n}{2}]$.
Hence, for any $0 \leq l\leq d+j-i$ and for any $0\leq k \leq j$, we have $pr_{\ast}(l_{[\frac{n}{2}]-k-n-j+l}\times \delta_{k,l})=0$.
Then, for any $i>d+j-m$, we get that $pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i})=0$.
\medskip
On the other hand, for $i=d+j-m$, we have $d+j-i=j+n-[n/2]$ and
\[l=j+k+n-[\frac{n}{2}] \Longleftrightarrow k=0\;\;and\;\;l=d+j-i.\]
Thus, we have
\[pr_{\ast}(h^{n+j-m}\cdot \tilde{B}_{d+j-m})=\delta_{0,m}.\]
Since $m>m+[n/2]-n$, we get that $\delta_{0,m}=0$.
\medskip
Therefore, for any $i\geq d+j-m$, we have
\[pr_{\ast}(h^{n-d+i}\cdot \tilde{B_i})=0.\]
\medskip
Then, for any $i>d+j-m$, the cycle $h^{n-d+i}$ is divisible by $2$. Hence, according to the multiplication rules in the ring $CH(\overline{Q})$ described in [1, Proposition 68.1] and by doing the same computations as those done during the proof of Theorem 1.1, for any $i>d+j-m$, we get the congruence
\[pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\equiv 2\sum_{k=0}^{[\frac{n}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k} \;\;(mod\;4).\]
Moreover, since $d-i-k\leq k$ if and only if $k\leq [\frac{m-j}{2}]$, for any $i>d+j-m$, we have the congruence
\begin{equation} pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\equiv 2\sum_{k=0}^{[\frac{m-j}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k} \;\;(mod\;4). \end{equation}
Now, we would like to study the $(d+j-m)$th summand, that is to say the cycle $pr_{\ast}(h^{n+j-m}\cdot s^m)$ modulo $4$.
That is the purpose of the following lemma.
\medskip
\begin{lemme}
\textit{One has
\[pr_{\ast}(h^{n+j-m}\cdot s^m)\equiv \left\{\begin{array}{rcl} &2\varepsilon_{\frac{m-j}{2},\frac{m+j}{2}}+2\textbf{y}^m\cdot \textbf{z}^j
\;\;(mod\;4) & if\;m-j\;is\;even\\
&2\textbf{y}^m\cdot \textbf{z}^j \;\;(mod\;4) &if\;m-j\;is\;odd.
\end{array}\right.\]}
\end{lemme}
\begin{proof}
We recall that $s^m=(\overline{\textbf{x}})^2$. Thus, we have
\[h^{n+j-m}\cdot s^m=h^{n+j-m}\cdot (A+B+C)\]
where
\[A:=\sum_{0\leq i,l \leq [\frac{n}{2}]} h^{i+l}\times (\textbf{y}^{m-i}\cdot \textbf{y}^{m-l}),\]
\[B:=\sum_{0\leq i,l \leq j} (l_{[\frac{n}{2}]-i}\cdot l_{[\frac{n}{2}]-l})\times (\textbf{z}^{j-i}\cdot \textbf{z}^{j-l})\]
and
\[C:=2\sum_{i=0}^{[\frac{n}{2}]} h^i\times \textbf{y}^{m-i} \cdot \sum_{l=0}^j l_{[\frac{n}{2}]-l}\times \textbf{z}^{j-l}.\]
First of all, we have
\[h^{n+j-m}\cdot A=\sum_{0\leq i,l \leq [\frac{n}{2}]} h^{n+j-m+i+l}\times (\textbf{y}^{m-i}\cdot \textbf{y}^{m-l}).\]
Now we have $m=[\frac{n+1}{2}]+j$, so $n+j-m+i+l=[\frac{n}{2}]+i+l$. Thus, if $i\geq 1$ or $l\geq 1$, we have
$n+j-m+i+l>[\frac{n}{2}]$, and in this case we have $h^{n+j-m+i+l}=2l_{m-i-l-j}$. Therefore,
the cycle $h^{n+j-m}\cdot A$ is equal to
\[h^{n+j-m}\times (\textbf{y}^m)^2 + 4\sum_{1\leq i,l \leq [\frac{n}{2}]}l_{m-i-l-j} \times (\textbf{y}^{m-i}\cdot \textbf{y}^{m-l})
+2\sum_{i=1}^{[\frac{n}{2}]} l_{m-j-2i}\times (\textbf{y}^{m-i})^2.\]
\medskip
Then, since $n\geq 1$, we have $n+j-m \neq n$. It follows that $pr_{\ast}(h^{n+j-m}\times (\textbf{y}^m)^2)=0$.
\medskip
Furthermore, we have
\[pr_{\ast}(\sum_{i=1}^{[\frac{n}{2}]} l_{m-j-2i}\times (\textbf{y}^{m-i})^2)=\left\{\begin{array}{rcl} &(\textbf{y}^{\frac{m+j}{2}})^2 & \text{if $m-j$ is even}\\
&0& \text{if $m-j$ is odd.}
\end{array}\right.\]
Therefore, $pr_{\ast}(h^{n+j-m}\cdot A)$ is congruent modulo $4$ to $2\varepsilon_{\frac{m-j}{2},\frac{m+j}{2}}$ if $m-j$ is even, and to $0$ if $m-j$ is odd.
\medskip
Then, by dimensional reasons, we have $l_{[\frac{n}{2}]-i}\cdot l_{[\frac{n}{2}]-l}=0$ if $i\geq 1$ or if $l\geq 1$.
Hence, we have $B=(l_{[\frac{n}{2}]}\cdot l_{[\frac{n}{2}]})\times (\textbf{z}^j)^2$. It follows that
\[h^{n+j-m}\cdot B=(l_0 \cdot l_{[\frac{n}{2}]})\times (\textbf{z}^j)^2\]
and $l_0 \cdot l_{[\frac{n}{2}]}=0$ by dimensional reasons. Therefore, we get that $h^{n+j-m}\cdot B=0$.
\medskip
Finally, we have
\[h^{n+j-m}\cdot C=2\sum_{i=0}^{[\frac{n}{2}]} h^{n+j-m+i}\times \textbf{y}^{m-i} \cdot \sum_{l=0}^j l_{[\frac{n}{2}]-l}\times \textbf{z}^{j-l}.\]
Now for any $i\geq 1$, we have $n+j-m+i>[\frac{n}{2}]$, and in this case the cycle $h^{n+j-m+i}$ is divisible by $2$.
Thus, the element $h^{n+j-m}\cdot C$ is congruent modulo $4$ to
\[2\sum_{l=0}^j (h^{[\frac{n}{2}]}\cdot l_{[\frac{n}{2}]-l})\times (\textbf{y}^{m} \cdot \textbf{z}^{j-l}),\]
and, by dimensional reasons, in the latest sum, each summand is $0$ except the one corresponding to $l=0$.
Therefore, the cycle $h^{n+j-m}\cdot C$ is congruent modulo $4$ to $2l_0 \times (\textbf{y}^{m} \cdot \textbf{z}^{j})$.
It follows that $pr_{\ast}(h^{n+j-m}\cdot C)$ is congruent modulo $4$ to $2\textbf{y}^{m} \cdot \textbf{z}^{j}$.
We are done.
\end{proof}
By the congruence (1) and Lemma 2.2, we deduce that the cycle
\[\sum_{i=d+j-m}^{d}pr_{\ast}(h^{n-d+i}\cdot s^{d+j-i})\]
is congruent modulo $4$ to
\[2\sum_{i=d+j-m}^d \sum_{k=0}^{[\frac{m-j}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k} + 2\textbf{y}^m\cdot \textbf{z}^j.\]
It follows that the cycle
\[2\sum_{i=d+j-m}^d \sum_{k=0}^{[\frac{m-j}{2}]}{k \choose d-i-k}\varepsilon_{k,j+k} + 2\textbf{y}^m\cdot \textbf{z}^j\]
is congruent modulo $4$ to twice a rational element $\alpha \in CH^{m+j}(\overline{Y})$.
Then, we finish as in the proof of Theorem 1.1. For every $k=0,...,[(m-j)/2]$, the total coefficient near $\epsilon_{k,j+k}$ is $2^{k+1}$, which is divisible by $4$ for $k \geq 1$.
Therefore, there exists a cycle $\gamma \in CH^{m+j}(\overline{Y})$ such that
\[2\varepsilon_{0,j}+2\textbf{y}^m\cdot \textbf{z}^j=2\alpha + 4\gamma,\]
hence, there exists an exponent $2$ element $\delta \in CH^{m+j}(\overline{Y})$ so that
\[\varepsilon_{0,j}+\textbf{y}^m\cdot \textbf{z}^j=\alpha + 2\gamma + \delta.\]
\medskip
Finally, since $\varepsilon_{0,j}$ is an integral representative of $S^{j}(y^m)$ and $\textbf{y}^m$ (resp. $\textbf{z}^j$)
is an integral representative of $y^m$ (resp. of $z^j$), we get that $S^j(y^m)~+~y^m\cdot ~z^j$
differs from a rational element by the class of an exponent $2$ element of
$CH^{m+j}(\overline{Y})$.
We are done with the proof of Proposition 2.1.
\end{proof}
\medskip
\begin{rem}
In the case of $j=0$, and if we make the extra assumption that the image of $x$ under the composition
\[Ch^m(Q \times Y) \rightarrow Ch^m(Q_{F(Y)}) \rightarrow Ch^m(Q_{\overline{F}(Y)}) \rightarrow Ch^m(\overline{Q})\]
(the last passage is given by the inverse of the change of field isomorphism)
is rational, then we get the stronger result that the cycle $y^m$ differs from a rational element by the class of an exponent $2$ element of
$CH^{m}(\overline{Y})$. That is the object of [2, Proposition 4.1].
\end{rem}
\medskip
Finally, the following theorem is a consequence of Proposition 2.1. In the same way as before, it is a generalization of [3, Theorem 3.1(2)].
For a variety $X$, we write $r$ for the restriction map $Ch^{\ast}(X)\rightarrow Ch^{\ast}(\overline{X})$.
\begin{thm}
\textit{Assume that $m=[\frac{n+1}{2}]+j$ and $n\geq 1$. Let $\overline{y}$ be an $F(Q)$-rational element of $Ch^m(\overline{Y})$.
Then there exists a rational element $z \in Ch^j(\overline{Y})$ such that $S^{j}(\overline{y})+\overline{y}\cdot z$ is the sum of a rational element and the class modulo $2$ of an integral element of exponent $2$.}
\end{thm}
\begin{proof}
The element $\overline{y}$ being $F(Q)$-rational, there exists $x \in Ch^m(Q\times Y)$ mapped to $\overline{y}_{\overline{F}(Q)}$ under the composition
\[Ch^m(Q \times Y) \rightarrow Ch^m(Y_{F(Q)}) \rightarrow Ch^m(Y_{\overline{F}(Q)}).\]
Moreover, the image $\overline{x}\in Ch^m(\overline{Q}\times \overline{Y})$ of $x$ decomposes as
\[\overline{x}=h^0\times y^m +\cdot \cdot \cdot
+h^{[\frac{n}{2}]} \times y^{m-[\frac{n}{2}]}+l_{[\frac{n}{2}]}\times z^{m+[\frac{n}{2}]-n}+
\cdot \cdot \cdot +l_{[\frac{n}{2}]-j-1}\times z^{m+[\frac{n}{2}]-j-n}\]
with some $y^i \in Ch^i(\overline{Y})$, and some $z^i \in Ch^i(\overline{Y})$, and where $y^m=\overline{y}$.
Thus, by Proposition 2.1, the cycle $S^{j}(\overline{y})+\overline{y}\cdot z^j$
is the sum of a rational element and the class an element of exponent $2$.
\medskip
Finally, we have by [1, Proposition 49.20],
\[r\circ(pr)_{\ast}(x\cdot h^{[\frac{n}{2}]})=pr_{\ast}(\overline{x}\cdot h^{[\frac{n}{2}]})=z^j.\]
We are done with the proof of Theorem 2.4.
\end{proof}
| 178,169
|
- Track your orders
- Save your details for express checkout
Note: Due to unprecedented demand, lead time for NZ manufactured items made-to-order is now approx. 13 weeks.
**Sale Price applies to NZ Stock only.**
Add a refined, textural statement to your space with the Zahra Feather Bedside. A sleek matt black case grounds the silhouette while feature natural bone inlay drawers add tangible interest and visual elevation.
SIDE-ZAH-FTH
Browse By Category, Bedsides and Dressers, Zahra, New Master Codes 2021, Zahra
| 150,297
|
TITLE: Probability Unknown Distribution
QUESTION [0 upvotes]: Suppose that the mean for some data is 75 mm and the standard deviation is 10 mm. The distribution for this data is not necessarily normal.
Is it possible to determine the probability of the distance being greater than 73 mm?
REPLY [1 votes]: Expanding on my comment, let us consider distributions which take only two values and have mean 75mm and standard deviation 10mm. Every such distribution will be a shifted and resealed Bernoulli distribution.
Let $X_p$ be the random variable that takes the value $1$ with probability $p$ and the value $0$ with probability $q=1-p$. The mean is $p$ and the standard deviation is $\sqrt{pq}$. Therefore, we must scale by $\frac{10}{\sqrt{pq}}$ to correct the standard deviation, and since $\frac{10X_p}{\sqrt{pq}}$ has mean $10\sqrt{p/q}$, our desired distribution is the distribution of $Y_p=\frac{10X_p}{\sqrt{pq}}+75-10\sqrt{p/q}$
What is the probability that $Y_p\geq 73$? Since $(75-73)/10=1/5$, it will be the same as the probability that $X_p\geq p-\frac{\sqrt{pq}}{5}$. This will be $p$ if $p-\frac{\sqrt{pq}}{5}>0$, and $1$ otherwise. Since $p-\frac{\sqrt{pq}}{5}$ has a root at $p=1/26$, we have that the probability for this simple family of distributions lies in $(1/26,1]$.
| 115,635
|
tag:blogger.com,1999:blog-51351272399259464912009-04-29T09:42:42.316-04:00The Authority - CT Authors & Publishers Assoc.The Authority is the voice of the Connecticut Authors and Publishers Association, a group of about 150 authors who share information about writing and the world of publishing and marketing books and work to support one another by helping new writers to succeed and established writers to find new marketing avenues."The Authority" CT Authors & Publishers Assoc. Blogs<div style="text-align: center;"><span style="font-size:180%;">Every Author needs a Book Blog</span> <br /></div> <br /><div style="text-align: center;">By Peggy Gaffney <br /></div> <br /> <br />Now. <br /> <br /. <br /> <br /. <br /> <br /. <br /> <br /. <br /> <br />The next thing I did was go to “Write” and clicked tags. This is where I put all the words that I thought that people would type if they were looking for my topic. <br /> <br /. <br /> <br /. <br /> <br /> <br /><span style="font-size:85%;">Peggy Gaffney is the author of The Crafty Dog Knits series published by Kanine Knits <br /> <br />and Do It Yourself: Publishing Nonfiction In Your Spare Time <br /> <br /></span><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. Welcomes Writers<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5215202426314951378" border="0" /></a><br /><p class="MsoNormal"><span style="font-size:130%;">CAPA Welcomes Writer</span><span style="font-size:130%;">s to</span><span style="font-size:130%;"> 5<sup>th</sup></span><span style="font-size:16;"><span style="font-size:130%;"> annual CAPA-University</span><o:p></o:p></span></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">By Peggy Gaffney</p> <p class="MsoNormal"><o:p> </o:p>Again this year the Connecticut Authors and<span style=""> </span>Publishers Association welcomes writers , and those interested in all aspects of the book world to its professional development program, CAPA-U, which celebrates its fifth year. </p> <p class="MsoNormal">This year for the first time, all participants who have written books are offered an opportunity to market them in the CAPA-U Bookstore. This gives attendees a chance to <span style=""> </span>show everyone their talent. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">Another major addition to the program is the chance for beginning writers to meet with top notch editors and learn how to get their books ready to be published. <a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5215203138396353586" border="0" /></a>This guidance is invaluable. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">The keynote speaker, celebrated author Wally Lamb, entertains and inspires. His award winning books <i style="">I Know This Much Is True</i> and <i style="">She’s Com</i><i style="">e Undone</i> have been NY Times Best Book of the Year choice and Oprah Book Club selections. <span style=""> </span></p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">An outstanding team of agents<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 221px; height: 128px;" src="" alt="" id="BLOGGER_PHOTO_ID_5215202720797812962" border="0" /></a>, marketers and editors are on hand to help guide both novice and experienced writers through the intricate maze of the publishing world. These one-on-one sessions are supplemented with professional development workshops covering every topic of interest to writers inclu<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5215203823018593474" border="0" /></a>ding: fiction, non-fiction, memoir, the op-ed page, sports writing, writing book proposals, working with an agent, publishing, self-publishing, creating a buzz on Amazon, marketing, and even a panel that combines the agent, the writer and the publisher to answer the participants questions on how to travel the road to becoming a published author. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">With a delicious breakfast and lunch provided by the chefs at the > and the chance to explore the Bookstore to find new authors to read and new books by favorite authors, it makes the day one that is not to be missed.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5215204754892286882" border="0" /></a></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. of dog portrait knitting books speaks<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5180620610702567570" border="0" /></a><br /><span style="font-size:180%;">Peggy Gaffney speaks on Self-Publishing</span> <p class="MsoNormal">By Carol Keeney</p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal"><span style=""> </span><span style=""> </span>Peggy Gaffney, who will be giving a workshop at CAPA-U on self-publishing in May, gave us a preview in her talk to CAPA members at the February meeting. She published her first book in 2005 but the seeds of her efforts were sown throughout her life. She has enjoyed a career of teaching and hobbies like knitting and raising Samoyed dogs. A few years before retiring Peggy decided to start a business which would serve as a source of future income. She began knitting sweaters which portrayed the Samoyed dogs she loved. When she presented her samples at popular dog shows, she was pleased to accept many orders for sweaters depicting breeds of all types. Peggy’s knitting business began to soar.</p> <p class="MsoNormal"><span style=""> </span><span style=""> </span>Peggy noticed that some of her customers were interested in purchasing patterns to knit these sweaters themselves. At first she resisted sharing her patterns with them and then a light went on in her head. Why not publish these patterns and offer them for sale to the general public. Peggy drew her own pictures from photos she had collected through the years. Each one of them offered a visual example of the personality of the breed.<span style=""> </span>She then calculated how to set<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5180618123916503138" border="0" /></a> up a graph for each design.</p> <p class="MsoNormal" style="text-indent: 0.5in;">The charm of Peggy Gaffney’s story is that she could understand the world of dogs and how people within it relate to the love of their particular pet. She enjoyed hearing anecdotes about them many of which she includes in her books. She told the CAPA audience one story of a the dog that at the last minute in her run around the agility course, momentarily ran off her path to a first place win to grab a bite of a sandwich from a bystander. She collects stories from her network of friends who showed dogs and include them in knitting books. The stories of the breed’s activities and the fact that her designs based on those activities of a single breed within each book makes them unique. </p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy has since written and published so far in this series a total of three books. The first deals with her own breed which she has bred a<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5180616234130892850" border="0" /></a>nd shown for 40 years, the Samoyed. The second deals with the Labrador Retriever and the third with the Golden Retriever. Peggy follows a similar construction when writing each book. A third of the book contains 25 basic knitting patterns for various projects. Another third deals with anecdotes about the particular breed, and the other third is made up of charts showing all of the dog’s activities converted to graphic patterns to be knit for that particular breed. Keeping this same basic theme for layout in every book saves time and effort.</p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy self-publishes all her books so that she can get the most profit from her efforts. ISBN numbers are gotten from a site for “Bowker.” Her photos are taken by herself or friends such as a field photographer who agreed to do the photos as illustrator for the retriever books in order to get illustrator credit for her portfolio and so that more people would see and appreciate her work.<span style=""> </span></p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy lays out her own manuscripts, covers, and converts them to pdf’s. She then gets in touch with her contact, Bobbi Rodriguez, at the book printer Fiddler Doubleday.<span style=""> </span>Bobbi quotes a price and sets a printing schedule for the work and emails it back to P<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5180616835426314306" border="0" /></a>eggy for approval. Once Peggy gives her final approval, Fidlar Doubleday goes ahead and prints the book. Within eighteen days Peggy receives a shipment of books which she stores in her own home. Peggy does her own fulfillment.<span style=""> </span></p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy’s books can be purchased on Amazon or from her Kanine Knits website. Peggy has gotten her books on Amazon through the Amazon Advantage Program. She also had a distributor specializing in craft books that handles making her books available to knitti<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer;" src="" alt="" id="BLOGGER_PHOTO_ID_5180618845471008882" border="0" /></a>ng shops. </p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy networks on the computer through Yahoo. <span style=""> </span>She has constructed a her own Kanine Knits website, a Kanine Knits Blog, a My Space Page as well as setting up a page on the new social networking site for knitters, Ravelry. <span style=""> </span>If you Google Peggy Gaffney, you will appreciate some of the results of her efforts. Congratulations, Peggy! </p> <p class="MsoNormal" style="text-indent: 0.5in;">Peggy Gaffney is the author of the Crafty Dog Knits series: The Crafty Samoyed Knits, The Crafty Labrador Retriever Knits, The Crafty Golden Retriever Knits, and the soon to be published Crafty Bernese Mt. Dog Knits. She also will soon be bringing out a book on self-publishing and one on dog obedience and square dancing with dogs..</p><br /><p class="MsoNormal" style="text-indent: 0.5in;"></p><span style="font-style: italic;">Carol Keeney is the author of "Brand New Teacher" a guide book for educators.</span><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. Author's Success<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142057681881377938" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoNormal" style="TEXT-ALIGN: center" align="center"><b><span style="font-family:Arial;font-size:14;">Getting My First Children’s Picture Book Published<?xml:namespace prefix = o /><o:p></o:p></span></b></p><p class="MsoNormal" style="TEXT-ALIGN: center" align="center"><b><span style="font-family:Arial;font-size:14;">By<o:p></o:p></span></b></p><p class="MsoNormal" style="TEXT-ALIGN: center" align="center"><b><span style="font-family:Arial;font-size:14;">Doreen Tango <?xml:namespace prefix = st1 /><st1:city<st1:placeHampton</st1:place></st1:city><o:p></o:p></span></b></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>I am officially a children’s book author. My children’s picture book titled <u style="FONT-WEIGHT: bold; FONT-STYLE: italic">I Like Gum</u> was recently released by Shenanigan Books. I have been asked to share my publishing experience with fellow CAPA members. I hope my story will provide some helpful insights.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>In September 2005, I had written a series of four children’s picture book manuscripts. I was an agent-less, first-time author with absolutely no knowledge of the publishing industry. What next? I went to my local library and asked for assistance. With the resourceful help of the Reference Librarian, I discovered an invaluable guide to the children’s publishing world titled The Children’s Writer’s & Illustrator’s Market. <o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>The book contained lists of children’s book publishers and agents. There is also information regarding the manuscript submission package. Not only did I learn exactly what I should include with my submission, I also discovered what <i>not</i> to include. <o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>To my relief, I learned publishers want <i>text only</i> manuscripts from authors who are not illustrators. It is the job of the publishing house to match the author’s text with an artist’s illustrations. If you decide to submit your friend’s illustrations with your manuscript, you run the risk of a visual rejection before your story gets reviewed.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>I heard the term “query” letter for the first time. I learned the significance of composing a strong query letter and the ramifications of writing a weak one. An interesting, well-written piece might pique an editor’s interest. A trite, boring letter will not produce an editor’s request for the manuscript.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>I reviewed each publisher’s manuscript submission policies and followed each set of guidelines to the letter. I did not want my masterpiece ending up in a recycling bin without ever having been read. Some publishers required only a query letter. Others wanted the entire manuscript. Some requested electronic submissions; while, others preferred a hard copy in the mail. The manuscript had to be ether single spaced or double spaced, depending on the guidelines. I created a database listing all the publishers to whom I had sent my manuscript (I didn’t want to send my story to the same house twice).<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Be diligent, but remain objective. Do not waste a publisher’s time or yours. Many publishers clearly state they will NOT accept unsolicited manuscripts. Some houses reject multiple submissions. They want exclusive submissions sent only to them. Others accept manuscripts during a stated timeframe. Still others have an open-ended submissions policy. Following all the guidelines is meaningless if your story does not match the publisher’s needs. Do not submit your manuscript if your story is exactly what the publisher is <i>not </i>looking for at the present time.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>As far as securing a literary agent, I found it to be a tougher proposition than finding a publisher. Agents are extremely selective. An unpublished, first time author is a big risk. Most agents are looking for established authors. I found myself identifying with the lament of the college graduate: “I can’t get a job without experience, but how do I get experience if no one will hire me?”<span style="font-size:+0;"> </span>(Needless to say, I secured my publishing contract for <u>I Like Gum</u> with no literary agent.)<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Even more information regarding publishers, agents and manuscript submissions can be researched on the Internet. I googled “children’s book publishers,” “children’s literary agents” and “query letters.” <span style="font-size:+0;"></span>The Internet became my most utilized resource. Be sure you are reviewing current websites. I discovered some websites had not been updated in years. An out-of-date website is of no benefit.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Spread the word. Talk about your book to everyone. I began scouring local newspapers for book events. I attended author singings, readings and book fairs. I asked questions constantly and found authors graciously shared helpful information. <o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Check out publishing house catalogs to learn what individual publishers are producing at the present time. Visit bookstores to see what books are “hot.” I sat in the Barnes & Noble in <st1:city<st1:placeCanton</st1:place></st1:city> one afternoon and just watched the reactions of children. They are my audience, so I listen and learn from them.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Join organizations. I heard about the Society of Children’s Book Writers and Illustrators (SCBWI) and immediately signed up. The newsletter provides a wealth of knowledge. There are two SCBWI conventions each year that offer workshops, guest lecturers and manuscript critiquing. There is a <st1:placeSouthern New England</st1:place> chapter that offers regional events.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Enter writing contests. Not only is the exposure beneficial, the financial rewards are a nice addition. The Internet lists numerous writing contests. Scholastic and Women’s Day Magazine jointly sponsor an annual writing contest specifically for children’s book manuscripts. The Tassy Walden Awards: New Voices in Children’s Literature sponsored by The Shoreline Arts Alliance is a competition open to writers and illustrators in the state of CT.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;">Despite all of the resources outlined above, getting a children’s book published is not for the faint of heart. It will take every ounce of tenacity, persistence and perseverance you can muster. From September 2005 until August 2006, I worked tirelessly to find a publisher and/or an agent. My manuscript was rejected by 82 publishers and 64 agents. Instead of feeling discouraged, I decided to submit another story.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;">After spending a weekend with my five-year old niece, I came up with the <u>I Like Gum</u> concept. I worked on the story for a weekend. I decided to submit my manuscript to the one publisher who had sent me the nicest and most constructive rejection letter. Five days later I received an email. Shenanigan Books wanted to publish <u>I Like Gum</u>. I signed the contract in September 2006. My book was released in September 2007. The entire process, from conception to publication, lasted one year.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-family:Arial;"><o:p></o:p>Currently, my publisher and I are working out the details of my second book. It has been an exhilarating, joyful and at times frustrating process. But, I am loving every mile of this ride. I hope you, too, get to savor your first publishing experience, with the anticipation of many more in the future.<o:p></o:p></span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. Authors Do the Big E<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142041618703690818" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoNormal" style="TEXT-ALIGN: center"><span style="font-size:180%;">CAPA at the Big E</span><span style="font-size:+0;"><span style="font-size:180%;"> </span></span><br /></p><p class="MsoNormal" style="TEXT-ALIGN: center">by Peggy Gaffney</p><p class="MsoNormal"><?xml:namespace prefix = o /><o:p></o:p></p><p class="MsoNormal">This was the first experience for CAPA member to market their books through a booth at the Big E and I thought I’d share my experience. </p><p class="MsoNormal"><o:p></o:p>On Tuesday I was scheduled for the afternoon-evening session and when Debbie Kilday and I arrived, the booth was busy so we explored before it was our time to take over. Once we got going, there was a flow of people interested in the books, CAPA and the whole concept of authorship. People were very friendly and they for the most part were looking for fiction (primarily mysteries), history, travel, and children’s picture books. Though they were fascinated with my knitting books, they weren’t selling. </p><p class="MsoNormal"><o:p></o:p>The six hour session went relatively quickly because people stopped to talk and check out the books. Everyone who showed any interest was given a contact list with the names and books of the authors involved and their websites or email.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142042675265645650" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a> </p><p class="MsoNormal"><o:p></o:p>The only real surprise for me is the people who when asked if they’d like to know more about the books, told me point blank that they didn’t read books or that they hated to read. This concept was new to me and I feel really sorry for them. Luckily there weren’t many of them.</p><p class="MsoNormal"><o:p></o:p>Driving home from the Big E at the end of the session was a challenge in that they are repaving I91 and what should have been a 50 minute trip became a 90 minute stop and go session.</p><p class="MsoNormal"><o:p></o:p>However, I was up again the next morning to do a double shift at the CAPA booth. I brought my knitting so I wouldn’t get bored, and shared the morning session with Elizabeth Faragher. <?xml:namespace prefix = st1 /><st1:city<st1:placeElizabeth</st1:place></st1:city> was the “featured author” first and did a land office business with her beautifully illustrated children’s book “Off to the Fair.”<span style="font-size:+0;"> </span>It was the perfect sell for the day because people connected the fair experience. The crowds were constant and the booth was full of browsers all day. </p><p class="MsoNormal">This was Connecticut Day and there were many people from the state walking through. The interest in CAPA was very high. It seams that everyone has a book inside him just waiting for someone to show him how to get it published. I chatted with people from my old home town and my present one.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142042980208323682" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a></p><p class="MsoNormal"><o:p></o:p>When it came my time to be “featured author” I was glad to sit down. I got out my knitting and my sweaters and books and began talking to everyone who stopped to look. There was a lot of interest in books coming in the future and the pattern they could order off the website. I even had a bunch of people interested in ordering custom knit sweaters. However, I discovered as the day progressed, that though there was a lot of interest in the topic of my books, there were very few knitters visiting the Big E that day. However, as the day continued, my pitch changed to having them pass on my card to friends and relatives who were much more my target market. I gave out well over a hundred cards that day.</p><p class="MsoNormal"><o:p></o:p>The high points of the day came later. <st1:city<st1:placeElizabeth</st1:place></st1:city> had finished her shift a<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142043246496296050" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a>nd Jan Mann had taken over. Miss <st1:state<st1:placeConnecticut</st1:place></st1:state> came through with her publicist and explored the booth checking out the children’s books. Then about four o’clock, Governor Rell and her group came through. She loved the booth and my books. One of her aids took a few photos of me with the Governor and Jan got a chance to talk with her as well. </p><p class="MsoNormal"><o:p></o:p>Jan’s session as “featured author” ended the day and she sold a number of books to people interested in the concept of Cruising Connecticut with a Picnic Basket. I was able to get shots of people visiting the booth while I was shar<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5142043869266553986" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a>ing more information about the group. </p><p class="MsoNormal"><o:p></o:p>All in all, I’d say it was a good experience. Lots of positive contacts were made. The word got out that there is a large group of authors available for talks in Connecticut, and as the manager of the building said, CAPA give’s the Connecticut Building “a touch of class. “</p><p class="MsoNormal"><o:p></o:p>At the end of the day I was left feeling that it had been worthwhile. As for the booth as a place to make money, I’d say it depends on the type of book you sell. General fiction, non-fiction and children’s books are the best sell for this crowd. </p><p class="MsoNormal"><o:p></o:p>The trip home took 2 hours and 10 minutes, so I hope if we do this next year, they won’t feel the need to repave in September. </p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. marketing<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5092321889625292338" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoNormal" style="TEXT-ALIGN: center"><span style="FONT-WEIGHT: bold">To Market....... To Market</span></p><div style="TEXT-ALIGN: center">by Cynthia S. Bercowetz<br /><br /></div><p class="MsoNormal">When I was in sixth grade, I wrote the following in a car contest: "If a strong and sturdy car is your delight, come to Shapiro's and be treated right." I won! I must have had ideas about promoting at that young age.<br />I have had many ideas on promoting and marketing my books. They may not be the traditonal ways but they have proven successful.<br />My biggest promotion is a book party at my home. I usually have 75 to 100 friends, town officials and anyone interested in attending a party. I have had an orchestra for the past two books.<br />For my second book, <span style="FONT-WEIGHT: bold; FONT-STYLE: italic">"Unforgettable Recipes and Savvy Consumer Tips"</span>, I had local cooks bring samples of their recipes in the books to the book party. The introduction of the book is by a well-known chef and he attended too with his recipes that made a hit.<br />For my third book,<span style="FONT-WEIGHT: bold; FONT-STYLE: italic">"Grandpa Herman's Pettng Zoo",</span> a true story about animals at a former petting zoo at COPACO in <?xml:namespace prefix = st1 /><st1:city<st1:placeBloomfield</st1:place></st1:city>, I will have a local farm bring over some of the animals that were similar to the ones in the book. They will be outside on our lawn.<br />I tuck a book at my dentist's office and other medical offices I visit. It pays off. P<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5092322052834049602" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a>atients have called to purchase books.<br />In <st1:city<st1:placeNaples</st1:place></st1:city>, Fl., I had a book party at the Elks Lodge. The chef at the Lodge has recipes in the book. It was most successful.<br />I had a book signing at a book store in <st1:city<st1:placeNaples</st1:place></st1:city> that did not have a high attendance. I went across the way to a pizza place for lunch. Customers asked why I was dressed up and the others were very casual. I told them about the book signing and many in line wanted to buy my books. So, there I was book signing at the pizza place.<br />I also like to help other authors to promote their books. I have a TV show on Channel Five and interview guests. For further information, contact me at consumreye@aol.com.<br />Cynthia Susan Bercowetz.<br /><br /><br /><span style="FONT-STYLE: italic">Unforgettable Recipes and Savvy Consumer Tips<br />ISBN: 0-9708430-7-0</span> <span style="FONT-STYLE: italic">Paperback $14.95</span> <span style="FONT-STYLE: italic">Contact me for an Autographed copy</span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc.<p class="MsoNormal" style="text-align: center;" align="center"><span style="font-size:180%;">Podcasts – Radio for the Internet?</span></p> <p class="MsoNormal" style="text-align: center;" align="center">By Peggy Gaffney</p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">What is a podcast? Well they are in many ways similar to a radio program created by non-professionals to be shared with people of like interests. However, this isn’t Fibber McGee & Molly. These are usually talk broadcasts that could be interviews, discussions, lectures, readings or just someone chatting about something that interests him.</p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">So what value do they have to writers? How many of us can afford to take classes and continue our education on a daily basis? How many can afford the time to attend workshops to hear ideas in your field? If the answer is that you don’t have time or the money, the podcast might provide an answer. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">How do you get a podcast and what does it cost? There are many ways to receive podcasts and more growing every day with iPods, interactive cell phones and a fascinating collection of pricey gadgets. However, the cheapest and probably most practical way won’t cost a thing. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">On your computer you access the internet and go to the site for iTunes. You can follow the directions to download the iTunes program into your computer. Once this is done, you click on the iTunes Store and find the word podcasts and click on that. When it shows you some of the thousands of podcasts on every subject, go to the upper right hand corner and you’ll find a search section. If you type in “writers, writing” up will come a long list of shows on these topics. Most of these are free and can be had by just clicking on the subscribe button and downloading them. Cost for most of them is nothing.</p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">A sample of the shows available include: Grammar Girl’s Quick and Dirty tips for Better Writing, Writers on Writing, AmericanWriters.com – Creative Writing Podcast, I Should Be Writing, The Writing Show, Holly Lisle On Writing, WrimoRadio: The Official National Novel Writing Month Podcast, Writers Talking, Write Away: Podcast of the Texas A&M University Writing Center, “The Kissy Bits” Romance Writing Without Cooties, Odyssey: SF/F Writing Workshop Podcast, Scottish Writers’ Podcast, Writers Voice, Behind the Black Mask: Mystery Writer Revealed, Tech Writers Voices: Podcasts on Technical Writing, Litopia Writers’ Podcast, Writers in the Sky, Arthur Slade.com, Writing for Young Adults, Tips for Media Writers, Kelly Writers House Podcast, Mom Writers’ Talk Radio, Little Red’s Writing Hood, Confessions of a Struggling Writer,</p> <p class="MsoNormal">Mark Moxon - Travel Writer, Novel Writing with Brian Jepesen, Lit Law for Writers on the Go!, Femslash Writer’s Corner, The Kwantien Writers’ Guild and Writer’s Voice.</p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">Now these are only a sample of writing podcasts and each might have 20 to 50 episodes so the wealth of information is mind boggling. Also these are just ones on writing, there is also marketing, self publishing and speaking to say nothing of the podcasts on your specific genre. For example, I have been listening to a number of podcasts and one appealed to me to the point of my sending the moderator copies of my books along with the online information to point her to my website and a nice note about how much I enjoyed her weekly broadcasts. The next thing I know, she is reviewing (very favorably) my books on her show and telling everyone to go to her webpage where my contact information will be listed. She has a listening audience of a half-million. I was delighted. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">The best part of the podcast world for me is that these podcasts can play on my computer any time I want and they don’t interfere with other programs. So as I work on sizing photos for my next book, I can at the same time listing to a half hour discussion of yarn and knitting or writing mysteries or marketing self published books. </p> <p class="MsoNormal"><o:p> </o:p></p> <p class="MsoNormal">So give podcasts a try. Listen and learn. Maybe some day you’ll want to do your own. If you think you might, there are even podcasts telling you how to do that as well. It may not be Fibber McGee but its “radio” for a new age and something we can all enjoy.</p> <p class="MsoNormal"><span style=""> </span></p> <p class="MsoNormal"><i style=""><span style="font-size:85%;">Peggy Gaffney is the author of the knitting book series <span style="font-weight: bold;">The Crafty Dog Knits</span> whose books cover the dog breeds Samoyed, Labrador Retriever & Golden Retriever. Shes is also the designer of a large collection of knitting patterns issued under the label <span style="font-weight: bold;">Kanine Knits. </span>Visit. </span><o:p></o:p></i></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. is going to the Big E<div style="TEXT-ALIGN: center"><span style="FONT-WEIGHT: bold; FONT-STYLE: italicfont-size:180%;" >Market To Thousands * Up Close and Personal<br /></span><br /><span style="FONT-WEIGHT: bold; FONT-STYLE: italicfont-size:180%;" >The CAPA Big E Bookstore</span><br /></div><br />Top 3 reasons why every Connecticut author should be part of the CAPA Big E Bookstore<br />1. You will have your books on display for 17 days.<br />2. You will have a quality signing opportunity at the largest fair in the Northeast.<br />3. You will have a great time while being part of an exciting CAPA event.<br /><br />CAPA members have a unique opportunity to meet the public and market their books. The CAPA Bookstore will be located in the Connecticut Building for the 17 days in September with hundreds of thousands of people walking through and getting a chance to meet the authors and buy their books. You can choose the day to meet your public, but your books will be on display and available for sale throughout the entire 17 day event.<br /><br />Applications were mailed to members in their latest copy of The Authority. Applications MUST be returned to Jan Mann by July 14. Anyone needing more information may e-mail her at jan@cruisingconnecticut.com.<br /><br />This is a perfect opportunity to do a booksigning to record crowds and get your titles known by the thousands of potential buyers attending the fair. <span style="font-size:130%;"><span style="FONT-WEIGHT: bold; COLOR: rgb(255,0,0)">Don't let this chance pass you by!</span><br /></span><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. & Marketing "offshoot" books<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5076732625044865682" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoSubtitle" style="TEXT-ALIGN: center"><span style="font-size:180%;">Riding Pop-Culture Whales<br /></span><span style="font-size:12;"><span style="font-size:130%;">The Adventures of a Book Barnacle</span><?xml:namespace prefix = o /><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: center"><o:p></o:p><span style="font-size:100%;">By Robert Trexler<span style="font-size:+0;"> </span><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;"><o:p></o:p></span><span style="font-size:100%;">A reporter from the <i>Wall Street Journal</i> called about <i>Unlocking Harry Potter: Five Keys for the Serious Reader</i> (Zossima Press). When the article appeared on May 10, 2007 the title was “<i>Last Hurrah for Harry Offshoots?</i>” with the </span><span style="font-size:100%;">subtitle “<i>As series draws to a close, market for related books may well spike, then fade</i>.” He was fair and accurate, laying the background of the “off-shoot” books with a statistic from R.R. Bowker that over 190 <i>Harry Potter</i> related titles are in print. He captured the phenomenon of “offshoot” books in this picturesque sentence: “Much like George Lucas’s “Star Wars” films and Dan Brown’s “The Da Vinci Code,” the <i>Harry Potter</i> books are whales to which many barnacles have attached themselves.”</span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5076732749598917282" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a><span style="font-size:100%;"> <o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;">John Granger, a classicist asked Robert Trexler, editor of a bi-monthly literary publication <i>CSL: The Bulletin of the New York C.S. Lewis </i>to join the effort to p</span><span style="font-size:100%;">roduce and market his first book, <i>The Hidden Key to Harry Potter</i>, that eventually sold 5,000 copies in about one year. This led to a book deal with Tyndale Publishers with a book titled <i>Looking for God in Harry Potter</i> (2004). That book sold over 50,000 copies and is in its second updated edition. In April 2006 John was offered a book deal with Putnam-Penguin for another <i>Harry Potter</i> related book but John and Robert formed a business partnership and determined that they could do at least as well and wanted to retain creative control of the content and marketing .<o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;">It was Robert’s job to find the appropriate business/marketing plan and set the pieces in place to make everything happen. It was John’s job to write another book and edit a series of <i>Harry Potter</i> essays that became their first book, <i>Who Killed Albus Dumbledore?</i> (Nov. 2007). It was also John’s job to be the “personality” who could put his written ideas across to the media and to live audiences. <i>Unlocking Harry Potter</i>, took a while longer to be published (March 2007). But it was a fortunate delay, because just as we were going to press a Rowling quotation appeared from an interview with <i>The Herald</i>, a newspaper in</span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5076732964347282098" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a><span style="font-size:100%;"> <?xml:namespace prefix = st1 /><st1:place<st1:country-regionEngland</st1:country-region></st1:place>, where she stated: “To invent this wizarding world I’ve learned a ridiculous amount about alchemy . . . to set the parameters and establish the stories internal logic.” It was just the marketing “hook” we needed. Our book is the only book of the 190 spin-offs that explores the literary alchemy connection in the <i>Potter</i> books at length and in depth. <i>Unlocking</i> contains 70 pages covering this “key” to understanding her fantasy series. <o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;">As the saying goes, “Advertising is what you pay for, and publicity is what you pray for.” Well, our prayers were answered.<span style="font-size:+0;"> </span>A week after the <i>WSJ</i> article they were contacted by a Warner Brothers movie studio producer developing a TV program to be aired the week before the fifth <i>Harry Potter</i> movie is released (July 15, 2007). On May 18<sup>th</sup> John was flown to <st1:place<st1:cityBurbank</st1:city>, <st1:stateCalifornia</st1:state></st1:place> as one of three Harry Potter experts for the program. There is a possibility that the program will also be one of the “extra features” when the CD of movie version of <i>Order of the Phoenix </i>appears.<o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoSubtitle" style="TEXT-ALIGN: left" align="left"><span style="font-size:100%;">What distinguishes their attempt from the get-rich-quick approach to barnacle book writing is that they started with a conviction of the importance of our message – and knew that it offered a unique viewpoint. It took five years to get the “call from <st1:cityHollywood</st1:city>,” but appealing to <st1:city<st1:placeHollywood</st1:place></st1:city> was not foremost in our thinking with the first book in 2002. However</span><span style="font-size:100%;"> being a barnacle on a whale is better than being a pebble on the beach.</span><span style="font-size:12;"><o:p></o:p></span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc.<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5073873005754417794" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoNormal" style="TEXT-ALIGN: center" align="center"><span style="font-size:130%;">Jan Mann Speaks at CAPA-SE</span></p><p class="MsoNormal" style="TEXT-ALIGN: center" align="center">by John Benjamin Sciarra</p><p class="MsoNormal"><?xml:namespace prefix = o /><o:p></o:p><?xml:namespace prefix = st1 /><st1:cityGroton</st1:city> — <st1:place<st1:stateConnecticut</st1:state></st1:place> author Jan Mann encouraged fledgling writers to “pursue their passion,” at the Groton Library Monday evening, April 16<sup>th</sup>. Speaking to the satellite chapter of CAPA, the Southeast group enjoyed Mann’s presentation immensely as she detailed her 22 year journey to published author.</p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal">Mann’s book, <span style="FONT-WEIGHT: bold; FONT-STYLE: italic">“Cruising Connecticut in a Picnic Basket”</span>.</p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal" <st1:place<st1:stateMichigan</st1:state></st1:place> called “Edwards Brothers” delivered her books in one week and the results met Mann’s demanding standards of perfection and">Whatever it takes, Jan Mann has it. Through dogged determination and unyielding hard work, Jan is going to put <st1:place<st1:stateConnecticut</st1:state></st1:place> on the map of places to visit and cut a path for herself in the world of publishing. We, too, can have that kind of success if we apply ourselves in the same Mann—er.</p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. to a Niche Market - Teachers<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5070485572801012642" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" /></a><br /><p class="MsoNormal" style="TEXT-ALIGN: center"><span style="font-size:10;"><span style="FONT-WEIGHT: bold">On Marketing Brand New Teacher</span><?xml:namespace prefix = o /><o:p></o:p></span></p><p class="MsoNormal" style="TEXT-ALIGN: center"><span style="font-size:85%;"><span style="FONT-WEIGHT: bold; FONT-STYLE: italic">by Carol and Joe Keeney</span><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;">In the latter part of 2006, the same year our book was published, we started receiving checks from our distributor. While the checks are not big, $300 or less, the feeling that we were doing something right was beneficial to us. Before listing our marketing adventures that brought us this far monetarily, it would be to your advantage to know something about us: Our book is of the academic variety and we had no experience in promoting it; and early efforts to advance our book failed, it was like pouring sweat into a black box that remained perpetually empty; and as a marketing compass, we used The <i>Self Publishing Manual</i>, a great book, however, it catered more toward the general-audience-type book. <o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;">We did not get it <i>Brand New Teacher: How to Guide and Teach the Early Grades Using Scripts</i> had to be sold differently than other books. The worm turned, so to speak, when we read Brian Jud’s “Beyond the Bookstore;” We learned the meaning of the word <i>niche</i> and it gave us a roadmap for reorienting our selling efforts. Below is a list of efforts that should have targeted academic education from the beginning:<o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;"><o:p></o:p></span></p><ul style="MARGIN-TOP: 0in" type="disc"><li class="MsoNormal"><span style="font-size:100%;">Mailed pre-published copies to reviewers such as Publisher’s weekly. <o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Mailed post-publication books to reviewers and wholesalers. <o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Contracted with the distributor Atlas Books, who promised to move our books to booksellers and wholesalers.<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Participated in PMA mailing programs to reviewers and schools. <o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Sent press releases with publicity kit to major newspapers across the country<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Carol went into chat forums on the internet to help new teachers<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We set up a web site <a href=""></a> for the book and use Link Metro for link exchanges; <o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Carol made contact with a school principal enabling her to sell books to <?xml:namespace prefix = st1 /><st1:city<st1:placeHartford</st1:place></st1:city> schools and do a seminar for new teachers.<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Carol got a review in a local newspaper.<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We participated in the CT library association through CAPA<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We sent a book to the First Lady Nancy Bush because of her interest in education.<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We placed an ad in New York Teacher, the union newspaper for the New York City Schools at the start of the new school year<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">CAPA member <st1:personnameDennis Schleicher</st1:personname> helped us with Amazon to increase our ranking and sales.<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">Carol did a book signing at Barnes and Noble in <st1:city<st1:placeNorwalk</st1:place></st1:city><o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We visited bookstores and got some of them to stock our books. <o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We have approached a School Supply Company to include us in their catalog<o:p></o:p></span></li><li class="MsoNormal"><span style="font-size:100%;">We contacted wholesalers more geared toward academic education.<o:p></o:p></span></li></ul><p class="MsoNormal"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;">We got a handful of review requests from our mailings; the ad we had in <i>New York Teacher</i> sold about 50 books; orders through the distributor came as a direct result of our personal contact with wholesalers; Carol, through her Internet forums, drove some traffic to Amazon; ditto for Dennis and his Amazon positioning strategies; Carol’s seminar sold 15 books to the Hartford school system. As you can see, our cash producing activities were spotty; it did however inspire us to do more. <o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;">Doing-more will always be our biggest challenge. Carol likes to say, “We say to the universe give-me, give-me…and in return…the universe says give-me, give-me, back.” What this means is that the universe wants its share first. Our early non-rewarded efforts are how we paid our share to the universe. The other side of the coin is about the reward. From our experience the universe pays back synergistically. It gives us more than what we put in. We know it because the efforts we put forth are less than the size of our checks. The only answer we have for this is that when we do something it eventually attracts something else in the universe. The other moral of the story is to keep-doing.<o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;"><o:p></o:p></span></p><p class="MsoNormal"><span style="font-size:100%;">How we get enthused to market our book is through reading. In addition to <st1:personnameBrian Jud</st1:personname>’s inspiring book, John Kremer’s <i>One Thousand Ways to Market Your Book</i> is filled with ideas that have a tendency to motivate us. Our CAPA membership, however, is our number one source for getting new ideas and inspiration; we hardly miss a meeting. Currently, we are exhibiting our book at the New England College Bookstore show, and Carol is planning a seminar in <st1:city<st1:placeRidgefield</st1:place></st1:city> for new teachers, all of this because of our association with CAPA. Thank you CAPA.</span></p><p class="MsoNormal"></p><p><span style="font-size:78%;"><i>Carol Keeney has thirty-years experience teaching early childhood grades. Many years were spent as a first grade teacher in the <st1:city<st1:placeNew York City</st1:place></st1:city> school system. While tenured with the school system, Carol helped develop a science curriculum at the request of her school district. She spent many years as an adjunct professor at the <st1:placetypeCollege</st1:placetype> of <st1:placenameNew Rochelle</st1:placename> in <st1:state<st1:placeNew York</st1:place></st1:state>, and while tenured, has critiqued peer professors at the request of the college. She is currently working as an Adjunct Professor at <st1:placenameNorwalk</st1:placename> <st1:placetypeCommunity College</st1:placetype> in <st1:place<st1:cityNorwalk</st1:city>, <st1:stateConnecticut</st1:state></st1:place>. Carol holds an MS degree in Education from <st1:place<st1:placenameSt. John's</st1:placename> <st1:placetypeUniversity</st1:placetype></st1:place>. She was nominated and listed in the </i><em><span style="FONT-STYLE: normal">Who's Who of American Teachers</span></em></span><i><span style="font-size:8;"><span style="font-size:78%;">.She has developed and taught many courses. Her favorites are Methods of Teaching Early childhood Education, Methods of Teaching <st1:cityReading</st1:city> to <st1:city<st1:placeNormal</st1:place></st1:city> </span><span style="font-size:78%;">and Special Children and Methods of Teaching Creative Arts in the Classroom.</span> <o:p></o:p></span></i></p><p class="MsoNormal"><br /><span style="font-size:10;"><o:p></o:p></span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. tip from Brian Jud<div style="TEXT-ALIGN: center"><span style="FONT-STYLE: italic"><span style="FONT-WEIGHT: bold"><span style="font-size:130%;">Give them what they want!<br /><br /></span></span></span><div style="TEXT-ALIGN: justify"><p class="MsoSubtitle" style="MARGIN: 0in -0.25in 0pt -9pt; TEXT-INDENT: 9pt; TEXT-ALIGN: left" align="left"><span style="FONT-WEIGHT: normal;font-size:12;" ><span style="font-size:0;"></span>Buyers want to buy helpful information, not necessarily <i>books</i>. This gives you the flexibility to customize the form in which the information is delivered. It may be a comb-bound or spiral-bound manual that lies flat when used as a workbook during your seminars. Or, it may be a 3-ring binder allowing people to add or change pages easily. You may choose to serve the needs of you potential customers with a video program, DVD, CD or saddle-stitched booklet.<?xml:namespace prefix = o /><o:p></o:p></span></p><br /><span style="FONT-STYLE: italic"><span style="FONT-WEIGHT: bold"></span></span></div></div><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. University<div style="TEXT-ALIGN: center"><span style="font-size:130%;"><span style="FONT-WEIGHT: bold">4</span><sup style="FONT-WEIGHT: bold">th</sup><span style="FONT-WEIGHT: bold"> Annual CAPA University a Success<br /><span style="font-size:85%;">by Peggy Gaffney</span><br /></span></span></div><p class="MsoNormal"></p><p class="MsoNormal"><?xml:namespace prefix = o /><o:p></o:p></p><p class="MsoNormal">On May 12 authors and potential authors gathered at the <?xml:namespace prefix = st1 />> for an exceptional professional development day. The sellout crowd had the rare opportunity to hear from keyn<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5068970420008133506" style="FLOAT: left; MARGIN: 0pt 10px 10px 0pt; CURSOR: pointer" alt="" src="" border="0" /></a>ote speaker Marcella A. Smith, Director of Small Press and Vendor Relations for Barnes & Noble.<span style="font-size:+0;"> </span>Her talk and the questions she answered in the follow-up session dealt with the specific nuts and bolts of getting your book onto the shelves of Barnes & Noble. </p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal".<span style="font-size:+0;"> </span></p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal"<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img id="BLOGGER_PHOTO_ID_5068970673411203986" style="FLOAT: right; MARGIN: 0pt 0pt 10px 10px; CURSOR: pointer" alt="" src="" border="0" />< <st1:personnameBrian Jud</st1:personname> from Book Marketing Works. <span style="font-size:+0;"></span></p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal". </p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal">With all these sessions plus good food and good company, the day was declared a success by all. </p><p class="MsoNormal"><o:p></o:p></p><p class="MsoNormal"><span style="font-size:78%;">Peggy Gaffney serves as co-editor of <span style="FONT-WEIGHT: bold; FONT-STYLE: italic">The Authority</span>.<br /></span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. Articles<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href=""><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 174px; height: 225px;" src="" alt="" id="BLOGGER_PHOTO_ID_5076735180550406850" border="0" /></a><br /><p class="MsoNormal" style="text-align: center;" align="center"><span style="font-size:130%;">Marketing Your Food Articles, and More</span></p> <p class="MsoNormal" style="text-indent: 0.5in;"><o:p> </o:p></p> <p class="MsoNormal"><o:p> </o:p></p> <div style="text-align: center;">By Richard Moriarty<br /></div> <p class="MsoNormal">;"><br /><o:p></o:p>After twelve years as a food columnist I still believe that marketing your articles often relies on a pinch of this and a dash of that, and a willingness to try different approaches.<br /><br />For many years I toiled in noisy, hot, often cramped restaurant kitchens, selling the fruits of my labor to happy and satisfied customers.<span style=""> </span>Hungry patrons paid me for doing what I loved to do most – cook.<span style=""> </span>But, deep down inside my creative chef’s soul I longed for something more than culinary perfection.<br /><br />I wanted to write about it.<span style=""> </span>But I didn’t have a clue how to get started.<br /><br />My good friend, Mike Covello, shared a similar goal.<span style=""> </span>He wanted to write about cars, but couldn’t get the editor of the local daily newspaper to respond to his numerous queries.<span style=""> </span>One night we brainstormed how he could market his idea.<br /><br />Michael was selling insurance at the time, and one of his customers was a large automobile dealer that spent lots of advertising dollars.<span style=""> </span>I suggested he ask the business owner to contact the newspaper and mention Michael as someone who might make a good car reviewer.<span style=""><br /></span><br />A letter from the editor arrived in Mike’s mailbox soon after, and he’s been writing about cars ever since.<span style=""> </span>And, we had learned an important lesson about marketing; business connections can be very valuable.<br /><br />I wanted to know more about the process of marketing my food writing, so I subscribed to two or three writer’s magazines.<span style=""> </span>Their how-to articles about researching, writing and getting published were always interesting and would prove invaluable.<span style=""><br /></span><br />Shortly after my friend’s success an article in <i>the Writer</i> caught my attention.<span style=""> </span>The author explained that newspaper editors were notoriously overworked. They often would not consider hiring new writers simply because the initial correspondence took too much time and effort.<span style=""> </span>The author said that instead of sending a SASE (self addressed stamped envelope) with your sample articles and query letter, to send a self addressed stamped postcard.<span style=""> </span>On the back of the card put two boxes, “I’m interested,” or, “I’m not interested,” and ask the editor to check one.<br /><br />I took that advice, and the postcard arrived in my mailbox a week later.<span style=""> </span>The first box was checked, and I’ve been writing a food column for that newspaper for over twelve years now.<br /><br />When I moved from the kitchen to the classroom I was presented with even more opportunities to market my writing.<span style=""> </span>My employer, the Center for Culinary Arts, in Cromwell CT, was a brand new post-secondary training facility.<span style=""> </span>I was fortunate to be hired to develop the curriculum and head the educational department.<span style=""> </span>Being the new school on the block, the owners felt they had to earn name recognition.<span style=""> </span>I showed my boss some of my food columns and mentioned that I thought the school might benefit from sponsoring cooking articles in area newspapers.<br /><br />Not long after I had planted that seed, the school’s public relations firm worked out a deal with a central Connecticut newspaper chain whereby the newspaper got free articles in exchange for advertising, and the school paid my writer’s fee.<span style=""> </span>Best of all, my column would run in four to five newspapers on a weekly basis.<br /><br />Business connections would eventually play another role in helping to market my writing.<br /><br />Part of my school job was to choose the textbooks that would be used for each course.<span style=""> </span>Over the years I met and communicated regularly with many publisher’s representatives and learned a lot about the textbook segment of the market.<span style=""> </span>One day an editor at one of the largest textbook publishers, Pearson Prentice Hall, called with an offer I couldn’t refuse.<br /><br />Would I be interested in proofreading, for content, a new culinary arts textbook that was in the process of being written?<span style=""> </span>The fee was small, but I thought the possibilities seemed endless.<span style=""> </span>I didn’t think twice before saying, “Yes.”<br /><br />The book turned out to be over 1000 pages.<span style=""> </span>Twelve months and countless rereads later, the project was done.<span style=""> </span>I received my check and a polite thank you.<span style=""> </span>And I thought that was the end of it.<span style=""> </span>But I kept in touch with the publisher.<br /><br />Two months later the editor at Prentice Hall called.<span style=""> </span>Would I be interested in writing the Teacher’s Manual and Student Study Guide that would go with the textbook?<span style=""> </span>The Teacher’s Manual paid a flat fee, but the Study Guide came with the promise of future royalties.<span style=""> </span>Where do I sign?<br /><br />When I received my first writing assignment from my hometown newspaper, I had no idea what doors it would eventually open for me.<span style=""> </span>I’m not sure that the postcard did the trick or not, but it showed that I understood the editor’s situation, and cared about his valuable time.<span style=""> </span>Likewise, when I began my teaching career I never dreamed that it would lead to my having a byline in four more newspapers and a book contract.<br /><br />Marketing our articles is usually the last thing we want to do, and probably the first thing we should be thinking about.<span style=""> </span>I’ve learned that the art of selling my writing is about keeping my eyes and my business contacts open.<span style=""> </span>Also, a pinch of advice and a dash of luck don’t hurt.<br /><br /><span style=";font-family:Arial,Helvetica,sans-serif;font-size:85%;" ><span style=";font-family:Arial,Helvetica,sans-serif;font-size:85%;" ><em><strong>About the author:</strong> Richard Moriarty is a graduate of the Culinary Institute of America and teaches professional chef training classes at the Center for Culinary Arts in Cromwell, CT. He is a food columnist and a restaurant consultant specializing in kitchen design. Chef Moriarty also offers personalized cooking classes in the privacy of clients' homes. Chef Moriarty can be reached online at <a href="mailto:remoriarty@snet.net">remoriarty@snet.net</a>.</em></span></span><br /><span style=""> </span></p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. Articles<p class="MsoNormal" style="text-align: justify; line-height: 200%;"> </p><p class="MsoNormal" style="line-height: 200%; text-align: center;"><i style=""><span style="line-height: 200%;font-size:14;color:black;" ><span style="font-weight: bold;">Maximizing Markets</span><o:p></o:p></span></i><span style="color:black;"><span style=""> <span style="font-weight: bold;"> </span></span></span><span style="line-height: 200%; font-weight: bold;font-family:Arial;font-size:10;color:black;" >By Patricia D'Ascoli, CAPA Member</span><span style="color:black;"><o:p></o:p></span><br /></p><div style="text-align: justify;"> </div><div style="text-align: justify;"> </div><div style="text-align: justify;"> </div><span style="color:black;"></span><span style="color:black;"><span style=""></span></span>Freelance writing can be a rewarding way to make a living, but it will rarely make you rich.<span style=""> </span>In order to generate a steady flow of income from this endeavor, a writer really needs to be creative in finding a variety of markets for his or her work.<span style=""> </span>This entails making the most out of every story idea that comes to mind - looking for as many publications as possible that might be interested in publishing a particular story.<span style=""><br /><br /></span>Sometimes all it takes is a quick rewrite and a story can be regenerated for use in a second or even a third market.<span style=""> </span>Occasionally, you can simultaneously write several versions of the same story if you know in advance which publications will be purchasing the piece.<span style=""> </span>And if you own the rights to a story, some publications will pay to <em><span style="color:black;">reprint </span></em>stories that have appeared elsewhere.<span style=""> </span><br /><div> </div><div style="text-align: justify;"> </div><p class="MsoNormal" style="text-indent: 0.5in; text-align: justify;"><b style=""> Getting Started:<span style=""> </span></b>Choose a topic that will sustain your interest, because it’s a lot more fun to write about something you like, especially when you are not getting paid very much to do it!<span style=""> </span>When I embarked on freelance writing, I decided I wanted to write about authors and the books they write, not only because I love reading, but also because I am fascinated by the book writing process.<span style=""> </span>I chose <st1:stateConnecticut</st1:state> authors to give it a unique perspective, and since there are so many different <st1:place<st1:stateConnecticut</st1:state></st1:place> authors, I haven’t run out of material yet.</p><div style="text-align: justify;"> </div><p class="MsoNormal" style="text-indent: 0.5in; text-align: justify;"><b style="">What’s the angle: </b><span style=""> </span>Whether your nonfiction piece is a <i style="">profile </i>of a person or an entity, chances are it will entail doing research and interviews.<span style=""> </span>My angle is this: when a local author publishes a new book that’s worthy of a story!<span style=""> </span>Alternatively, I might write a review of the new book, but reviewing the book is usually part of the research process for writing about an author as well.<span style=""> </span>These are the feature articles I generally write for newspapers in <st1:place<st1:stateConnecticut</st1:state></st1:place>.</p><div style="text-align: justify;"> </div><p class="MsoNormal" style="text-indent: 0.5in; text-align: justify;"><b style="">Who cares:<span style=""> </span></b>Who is interested in your story?<span style=""> </span>Publications related to your story’s <u>subject matter</u> are, so think local as well as global.<span style=""> </span>When an author publishes a new book, the first market to consider is the local newspaper that covers the town in which the author resides.<span style=""> </span>If this is an author’s first published book, sometimes the newspaper from the author’s hometown might also be interested in a story.<span style=""> </span>As far as <i style="">subject matter </i>goes, any publication that includes articles about books and authors is fair game, as are markets dealing with the specific <i style="">type </i>of book the author has published.</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">Here is an example of the myriad possibilities associated with one <st1:state<st1:placeConnecticut</st1:place></st1:state> author I recently profiled.<span style=""> </span>The author is a woman who published an inspirational book for Catholic mothers.<span style=""> </span>This is a sampling of possible markets for this one story:</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">1.<span style=""> </span><b style="">Local newspaper</b> in the town where the author currently lives</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">2.<span style=""> </span><b style="">Local newspaper</b> in the town where the author grew up</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">3.<span style=""> </span><b style="">Alumni magazine</b> for the college(s) the author attended</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">4.<span style=""> </span><b style="">Religious (Catholic)</b> newspapers, magazines and websites</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">5.<span style=""> </span><b style="">Women’s</b> magazines (print and online) and websites</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">6.<span style=""> </span><b style="">Parents </b>magazines (print and online) and websites</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal">7.<span style=""> </span><b style="">Writing</b> magazines (print and online) and websites</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal"><span style=""> </span><b style="">How to find markets: </b>Both Writers Market and <a href=""></a> are great sources for markets - both print and online.<span style=""> </span>You don’t need both of these; choose the format that works best for you.<span style=""> </span>Each contains a listing of publications with the associated writers’ guidelines.<span style=""> </span>It is always a good idea to review the guidelines and read a sample of the publication before sending a query.<span style=""> </span>If you have already written the story for a local newspaper, you might want to include a copy of that along with the query to give the editor an idea of your style and especially how you have successfully handled the specific topic already!</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal"><span style=""> </span><b style="">Make your own market: </b>If you are passionate about a subject and there is enough material about that subject to last indefinitely, you might also want to consider writing and publishing your own newsletter in addition to freelancing for other publications.<span style=""> </span>Sometimes the newsletter can give you credibility when you are trying to break into a new market.<span style=""> </span>I started publishing <i style="">Connecticut Muse</i> at the same time I began querying newspapers and magazines on the subject of <st1:stateConnecticut</st1:state> authors and their books, and now I regularly write about <st1:state<st1:placeConnecticut</st1:place></st1:state> authors for a variety of newspapers in addition to publishing <i style="">Connecticut Muse</i>.</p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal"><span style=""> </span>I would be happy to help you get started on finding markets for your own special topic.<span style=""> </span>Feel free to email me at <a href="mailto:pfdasc@aol.com">pfdasc@aol.com</a> or call me at 860-354-6488.<span style=""> </span></p><div style="text-align: justify;"> </div><p style="text-align: justify;" class="MsoNormal"> </p><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc. CAPA Writing Contest<div style="TEXT-ALIGN: center"><p class="MsoNormal"><span style="font-size:180%;"><strong><span lang="EN"><br />Deadline – May 31<sup>st<br /></sup><br />Annual Writing Contest</span></strong></span><?xml:namespace prefix = o /><o:p></o:p></p><div style="TEXT-ALIGN: justify"><p class="MsoNormal"><span lang="EN">The Annual <span style="FONT-WEIGHT: bold; FONT-STYLE: italic">Connecticut Authors & Publishers Association</span> Writing Contest closes entries on May 31<sup>st</sup>.<span style="font-size:+0;"> </span>Entries are accepted in Poetry, Short Story, Personal Essay and Children Stories.<span style="font-size:+0;"> </span><o:p></o:p></span></p><div style="TEXT-ALIGN: center"></div><p class="MsoNormal" style="TEXT-ALIGN: center"><span lang="EN"><span style="font-size:130%;"><span style="FONT-WEIGHT: bold">Four easy steps</span></span><o:p></o:p></span></p><ul style="MARGIN-TOP: 0in" type="disc"><li><span lang="EN"></span><span lang="EN" style="FONT-WEIGHT: bold">Visit </span></li><li class="MsoNormal" style="FONT-WEIGHT: bold"><span lang="EN">Review the rules<o:p></o:p></span></li><li class="MsoNormal" style="FONT-WEIGHT: bold"><span lang="EN">Print the entry form <o:p></o:p></span></li><li class="MsoNormal"><span lang="EN"><span style="FONT-WEIGHT: bold">Mail in your entries.</span> <o:p></o:p></span></li></ul><p class="MsoNormal"><span lang="EN"><o:p></o:p><br /></span></p></div></div><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc.<div style="text-align: center;"><span style="font-size:130%;"><span style="font-weight: bold;"><span style="font-family:times new roman;"><span style="font-size:180%;">"The Authority"<br /></span></span></span></span><div style="text-align: justify;"><span style="font-size:130%;"><span style="font-size:100%;"> <span style="font-size:85%;">The Authority is the voice of the Connecticut Authors and Publishers Association which is a group of approximately 150 authors who meet on a monthly basis to share information about writing and the world of publishing and marketing books. They also work to support one another by helping new writers to succeed and established writers to find new marketing avenues.<br /><br /> Each month a speaker shares his expertise in an effort to help members improve their craft. CAPA is open to anyone who has a love for the world of writing and an interest in improving their skills.<br /><br />This Blog will showcase </span></span></span><span style="font-size:130%;"><span style="font-size:100%;"><span style="font-size:85%;">both </span></span></span><span style="font-size:130%;"><span style="font-size:100%;"><span style="font-size:85%;">some writings by members on various topics and stories about CAPA members. We hope that you will enjoy "The Authority."<br /></span></span></span></div></div><div class="blogger-post-footer"><img width='1' height='1' src=''/></div>"The Authority" CT Authors & Publishers Assoc.
| 335,472
|
- Shopping Bag ( 0 items )
- Spend $25, Get FREE SHIPPING
From BN.com
Used & New From our Trusted Marketplace Sellers
Usually ships in 1-2 business days
From BN.com
How Textbook Rentals Work
From BN.comfor 180 Days How NOOK Study eTextbook rentals work
NOOK Study eTextbooks from Barnes & Noble are read with the NOOK Study eReader for your PC and Mac.
How NOOK Study eTextbooks work. is professor of Psychology at the University of Kansas, Lawrence. Wrightsman is an author or editor of ten other books relevant to the legal system, including PSYCHOLOGY AND THE LEGAL SYSTEM, FIFTH EDITION, THE AMERICAN JURY ON TRIAL, and JUDICIAL DECISION MAKING: IS PSYCHOLOGY RELEVANT? He was invited to contribute the entry on the law and psychology for the ENCYCLOPEDIA OF PSYCHOLOGY, sponsored by the American Psychological Association and published by Oxford University Press. His research topics include jury selection procedures, reactions to police interrogations, and the impact of judicial instructions. He.
| 411,815
|
KOLKATA: The 180-lot sale of 20th Century Decorative Arts & Design at Sotheby's Paris on May 22 will offer a large range of works selected for their quality, prestigious provenance and importance to the history of Decorative Arts since 1900. Emile Galle, Louis Majorelle, Rembrandt Bugatti, Jacques-Emile Ruhlmann, Jean Royere, Dominique, Diego Giacometti and Jean Prouve are among the celebrated designers featured, while the sale also includes furniture by such major 20th century creators as Alberto Giacometti, Claude & Francois-Xavier Lalanne, Antonio Saura, Alexander Calder, Philolaos and Arman.
| 164,184
|
Apply for Impact Factor
Request Impact Factor for Your Scholarly Journal to Increase Excellence.
Journal Aim and Scope:
Journal of computer intelligence and innovation is a peer reviewed multidisciplinary, monthly and open access international journal publishing original and superb articles covering an extensive variety of subjects in computer science and engineering. JOCII aims to contribute to the scientific research, so as to promote research in the field of computer science.
Scholar Article Impact Factor does not guarantee any accuracy of journal information in this page. Please contact journal administration or editor through their website for any query or specific information, journal status and latest updates. Thank you
| 193,014
|
Vodafone App Hunt the Hearts & Win Free Data upto 1.7 GB . Applicable only on MyVodafone App .
Contest Details :-
Get FREE 100 MB Data on Finding 1 Heart
Get FREE 500 MB Data more on Finding 4 Hearts.
Get FREE 1.7 GB Data more on Finding 8 Hearts.
Locations for MyVodafone App Hunt:-
Profile > Add Account > bottom right.
Home page > Bottom Menu > Amazing offers for you > bottom right.
Roaming > Select ‘With in India”> bottom right.
Menu > Shop > Search > bottom right.
Balance Transfer > Enter Vodafone number > Get on last page.
Banner of Switch to 4g.
Internet page > Buy Data.
How to Get Participate?
Download MyVodafone or update (working on latest version)
Register/Login
Click Banner Search for love & Start Hunting!
Cupid has hidden 8 love envelopes in the MyVodafone App.
Search all screens of the App. Tapping on the Envelope will reveal your gift
Some heart envelopes will appear only on special days. Keep your eyes open
You will receive data within 24.
| 320,868
|
Final Results (PDF)LONG BEACH, Calif. - The weekend at the Beach Invitational for DePaul track & field ended with multiple Blue Demons recording personal bests on Saturday.DePaul’s 4x100 teams posted big performances on Saturday. On the women’s side, Jessica Watkins, Alexia Harvey, Laura Edwards and Shayna Nwokenkwo finished with a time of 46.86 to finish fifth in its section and sixth overall. The men’s team of Adam Nelson, Isaac Walker, Chaz Bottoms and Sebastian Feyersinger debuted in the event with a time of 41.72 to place eighth.Feyersinger dropped his season-best time in the men’s 400m by crossing the finish line at 48.10. With the performance, the junior placed sixth overall out of 86 total runners. In the 200m, Feyersinger fought off the wind to finish at 21.89 for 20th overall. Harvey and Ariel Davis each recorded season-best times in the 100m hurdles with Harvey leading the way at 14.08. Harvey took 26th overall out of 91 competitors, while Davis claimed 30th after finishing at 14.24.Freshman Nelson had two personal bests in the men’s 100m and 200m events. He posted his best time in the 100m after sprinting across the line with a time of 10.87. Nelson finished 44th overall out of the 96 men’s sprinters. Later in the afternoon, the freshman registered a time of 22.26 in the 200m.Watkins and Nwokenkwo each broke the 12-second mark in the women’s 100m dash. Watkins produced a PR of 11.97 in the event and tied for 24th overall. Nwokenkwo was a hundredth of a second behind to take 27th at 11.98. Watkins recorded another personal best in the 200m when she completed the event in 25.01. The men’s 4x400 relay also made its season debut on Saturday. The team of Mac Melto, Bottoms, Jarvis Hill and Feyersinger finished as the runner-up with a time of 3:14.94.Xavier Taylor marked a personal record in the men’s long jump by hitting 6.65 meters on his best attempt. The freshman was .07 meters behind teammate Brian Mada in the event. Emily Eller placed 26th in a field that featured 68 throwers in the women's shot put. The sophomore recorded a best toss of 13.92 meters, her second-best throw of the season. In the men’s shot put, Isaac Jimenez cracked 15 meters for the first time in his collegiate career after tossing a distance of 15.09 meters. Jimenez outthrew 25 other competitors in the event. The Blue Demons return to Illinois for their next meet at the Illinois Twilight. The one-day meet hosted by Illinois is scheduled for April 22.
| 120,935
|
UH Manoa receives approval for three bachelor of science degreesUniversity of Hawaiʻi
Carolyn Tanaka, (808) 956-8109
Associate Vice President, External Affairs and University Relations
Associate Vice President, External Affairs and University Relations
Posted: Nov 20, 2009
HONOLULU, Hawai’i – At its monthly meeting held yesterday at Leeward Community College, the University of Hawai‘i Board of Regents (BOR) approved a bachelor of science degree in Computer Science to be administered by the College of Engineering at UH Mānoa. The BOR also granted established status of the bachelor of science degrees in Tropical Plant and Soil Science and Natural Resources and Environmental Management administered by UH Mānoa’s College of Tropical Agriculture and Human Resources.
The College of Engineering will offer a bachelor of science degree in Computer Engineering effective fall 2010. The program will train a workforce that can design, analyze and integrate hardware and software of computers. The BS in Computer Engineering will be the first and only program in the state, providing an affordable computer engineering education to students who wish to remain in Hawaiʻi.
The College of Tropical Agriculture and Human Resources (CTAHR) has been granted established status of a bachelor of science degree in Tropical Plant and Soil Sciences, effective fall 2009. Since its provisional approval in June 2001, more than 50 students have graduated from the program, and there are currently 45 majors. The program emphasizes modern plant production strategies and the adaptation and application of biotechnology that is environmentally and economically sustainable.
CTAHR’s bachelor of science degree in Natural Resources and Environmental Management was also changed from provisional to established status, effective fall 2009. The program has graduated 66 students since its provisional approval in June 2001. This interdisciplinary science degree, which currently has about 50 students enrolled in the program, emphasizes tropical island ecology and natural resources. The program provides students with scientific knowledge on the physical, chemical, biological, economic, social, and policy elements of natural resources management.
The offering of these degrees directly support the UH System’s strategic plan by increasing the educational capital in the state, developing a globally competitive workforce and diversifying the economy.
| 181,758
|
Mail us 24/7 on info@baunat.com
0.305-0030S
Suited for diamond engagement ringCertified round diamond, 1 x 0.30 Ct, of high quality
18 Kt yellow gold
Measurements:
Width ring: 2.40 mm
Thickness metal: 1.40 mm
Thickness prongs: 1.20 mm
Height ring: 6.40.30 carat solitaire diamond ring in yellow gold
Our customers can share their personal experience with BAUNAT by writing a review using Trustpilot. The original reviews for this jewel have been translated for your convenience.
Great experience
Very good experience buying a diamond jewel. We are really in love with this diamond ring.
Outstanding service and quality!!
Great service
The diamond ring that I ordered for my fiancée is beautiful. I was frequently updated on the progress of my order by email and it arrived on time. Very satisfied!
Inspirational insights
The evolution of the wedding ring
A wedding ring as symbol for the commitment two people make to each other is timeless. Do you want a unique wedding ring? Look in the collection of BAUNAT.
Buying diamonds: 5 things you did not know about diamonds
Diamond inspires. A lot has been written about diamonds but there certainly are things you did not know..
| 209,240
|
In part one of Crystal Energy we discussed the fundamental principles at work when we bring crystals into our energy field. We talked about the importance of engaging our intuition to choose specific crystals to work with on any given issue, be it physical, emotional or spiritual.
Now, let’s have a look at some other interesting elements that make crystals, minerals and stones a truly magical andpractical process that can be utilized for the benefit of us all!
I think it bears repeating that crystals and minerals are electromagnetic in nature and that what they do is find resonance with whatever comes into their own EME (electromagnetic energy) field. In this piece what we are concerned with is how they interact with us. Keep in mind that we are literally electromagnetic in nature! Therefore one can reason that crystals most certainly have the ability to find resonance with the biological system and quite literally – they do!
But just what are they finding resonance with? Let’s start with a baseline assumption that we as human beings have everything (literally everything) already within us, physiologically, psychologically and spiritually – to solve, heal, attract or repel. This includes the ability to manifest circumstances on the physical plane to healing our bodies of dis-ease, and everything in between. There are steps we can take that will help us to activate and utilize each and every one of these qualities. Due to a number of factors over the course of our development however, most of us lost easy access to many of these faculties, save for a handful of spiritual adepts who are said to have the ability to call up many of these qualities, including the act of materializing physical objects – at will!
So it’s safe to assume that these qualities within us have gone largely dormant. Remember that our brain only processes a tiny, tiny fraction of information on a conscious level – 5-9 bits, roughly. What crystals represent in this case are tools to help us activate some of this “unconscious” information. I want to stress that we are an energy source in and of ourselves and crystals represent the medium – the tool to help us activate and in some cases, restructure that energy. Due to the fact that each crystal (according to its internal crystalline/molecular structure) corresponds to different frequency sets within the electromagnetic spectrum and again, going with the assumption that we carry these same frequencies, the crystals tune themselves according to the corresponding area that needs addressing. This will happen by default. The given area will be activated within the recipient and the physical law of resonance kicks in. Now the issue is brought to the fore and ready to be dealt with.
However, the critical factor is that we then use our mental and emotional energies to shape the results we are trying to achieve (thoughts and feelings are also electromagnetic in nature) to bring about the solution. It is really you who are doing the healing, the solving, the creating – the crystal is simply the conduit to help you activate the process!
Just what does clearing a crystal mean?
Our modern texts talk about the myriad ways in which we can clear or cleanse crystals prior to working with them. Whether putting them in salt or salt water, in the earth (where most of them come from), in sunlight or moonlight or running them through sage smoke (also known as smudging), I suppose many of these can be somewhat effective. However, here’s what I’ve found to be the most effective and the least talked about in my own research. Sound Vibration! Yes, sound is probably the most effective, quickest and strongest way to get a crystal back to its base resonant frequency. And that’s what “clearing” means. It has been said, although still somewhat ambiguous, that crystals do not absorb or hold an energy pattern – rather, energy will pass through crystals and the crystal will either vibrate with the energy field it comes into contact with, or it won’t. In using sound vibration, what you will do is break the current pattern the crystal is vibrating with, much like you would do if you were to stop a pendulum from oscillating if you were to simply put your hand on it to stop it from moving. That’s it! The louder the sound – whether it is coming from striking a loud key of C on the piano or putting your crystal in front of a speaker playing loud music or emitting a loud sound from your mouth, the vibration being emitted from the sound (we’ve all felt the vibrations of loud sounds in our own body) will interrupt the crystal’s current pattern of oscillation and reset it to its baseline frequency. This is a most effective (and quick) way to clear a crystal, from my perspective. However, again, I encourage you to experiment on your own!
Clearing vs. Charging…
Here’s a little anecdote that I thought was worth sharing. I have a wonderful crystal that is called ametrine (a combination of amethyst and citrine). I have used this stone to help with mild cases of insomnia, where I would literally lean the stone against the small of my back while in bed, and in most every case of using it in this way, I would be fast asleep within 10 minutes!! Now, I must say that I do not know where I got the idea to utilize the ametrine in this way – I’ll have to assume, I was simply drawn to do so! Nonetheless, one night I decided to cleanse the stone under the light of the full moon. I’d never done this before and had never even cleared this particular crystal by any method. After leaving it in full moonlight for the night, the following night I again placed the crystal in its usual position and what happened?? I was awake all night, tossing and turning as if having three cups of coffee just prior to bed.
What was going on here? This again, was purely experimental. And this is why I do advocate trying out these things on your own, without the aid of a book, so you can determine whether the effects you are experiencing are clearly yours or assumptions based on what you’ve read or heard.
It seems, that what happened is that by receiving strong light energy (which is information) from the moon, that rather than clear the ametrine, it charged it – somehow reversing the effect its native attributes had on my own system. Needless to say, that crystal will never see the light of the moon again!!
Other Experiments
Using a pendulum to determine consistency…
Here’s another inclination I had, simply to demonstrate to myself the stability and inherent patterns of certain crystals. We spoke of a pendulum as another tool to demonstrate oscillation. Many use pendulums (some are made of crystals) to “dowse” or divine answers to certain questions or problems. What I have done is taken a given crystal or set of different crystals and one at a time I will place the pendulum about 2 inches above the crystal and simply watch the direction or intensity that the pendulum will move. I found it fascinating that each crystal emitted its own unique movement associated with the pendulum – whether the movement was an oscillation (back and forth), clockwise or counterclockwise motion. Some movements were rather subtle, others clearly pronounced, depending on the crystal. What made this little experiment even more intriguing was the fact that each time I came back to a given crystal, whether an hour, a day or a week later, the same fundamental movement occurred proving to me that crystals do carry their own unique signature!
Feeling Sensations…
It is said that our feet are perhaps the most sensitive part of our bodies. In Chinese medicine, the acupressure points are used on the feet to identify and treat many ailments. Reflexology has become a popular method in using as a subtle energetic modality. Again, purely experimental, what I have done is taken a crystal, specifically a clear quartz crystal wand, and while holding the wand with its larger end facing the soles of my feet, I simply scan different parts of the foot, without the wand actually touching the foot. As I move the wand back and forth, I can feel certain subtle pulses, vibrations or “tickles” depending on where the wand is located on the foot. Sometimes I have been able to pick up a vibration in another part of my body while holding the wand at a certain area, say the big toe or heel, for instance. To test the accuracy of the sensation, I will simply run the wand back and forth in the area where the vibration is concentrated and interestingly I can feel the energy moving back and forth in perfect accord with the movement of the wand! This simply demonstrated to me that we are clearly having a definite electromagnetic interaction with our crystals.
Make Room for the “Mystery”…
In part one of this article, I mentioned my reverence for the scientific method relative to how and why crystals work, although one should leave room for the mystery at the other end of the spectrum. I have admittedly had a handful of what I would call anomalous experiences with several of my crystals that elude me ‘til this day.
The Crystal and “The Vision”
On one occasion, I had been working with a medium sized *pure citrine crystal while gazing into its structure, noticing the beautiful rainbows and inclusions it had throughout. I found myself pulling the crystal closer to my eyes while being more and more entranced with its beauty. All of a sudden a three dimensional image of my mother (who passed over a decade ago), formed right in front of my eyes within the crystal! The features were unmistakable. I could even see expression on her face as if she were looking right at me! The image was so distinct, it startled me! I had never had an experience quite like this and certainly not with a crystal! It’s as if the crystal became a holographic plate, imprinting what appeared to be a 3D image of my mother’s face within its structure. I later went back to the crystal to see if I could “find her” again, but to no avail! This, I suppose was a once in a lifetime occurrence. I do not know what prompted the experience but I do know that it was an authentic albeit mysterious one!
*Note that some citrines are actually heat treated amethyst and not citrine in the truest sense.
Shakti and the Jungle
On another occasion I had been simply spending time with some of my crystals and stones that I had placed around one of the rooms in my home. Periodically, I like to pick up different crystals and try to sense if there’s anything I should know about them, energetically or otherwise.
Upon picking up a large jasper stone, I heard (telepathically), although loud and emphatic, “SHAKTI!”Again, I was taken aback at the spontaneity of this message. I did not expect to get such a definitive communication. But it was just one word and it was clear – Shakti!
Derived from Sanskrit,. (source: Wikipedia).
After hearing the word shakti and knowing somewhat of its meaning, I then went to my crystal reference book (A great time to use books is after you have a direct experience with a crystal to see if it correlates), and among one of the attributes listed in this particular book for jasper: feminine energy – SHAKTI!Enough said!
On that same day, I took a specimen of adamite (a zinc arsenate mineral with an orthorhombic crystalline structure), and held it in my hand, eyes closed. I focused on an image of a candle flame in my mind’s eye – I did this simply to clear my mind of any conscious thoughts that might creep in. Again, out of the blue, I found myself viewing a scene. First I saw what looked to be a barren desert, but then the scene shifted quickly to that of a “jungle-like” environment complete with an image of a leopard! I was floored by the distinctiveness of the scene and even more intrigued by how the images morphed from one scene to another, like it was trying to tell me a story! Off I went to my book source to see where deposits of adamite have been mined in abundance over the years. Lo and behold, Namibia, a country in southern Africa was listed as having among one of the most abundant sources of adamite! Namibia is known for its deserts and vast jungles and apparently I was witnessing this landscape first-hand!
With both of these truly fascinating and mysterious first-hand experiences, it’s interesting that in each case, I did not ask for a specific piece of information, like “where are you from” or “are you masculine or feminine.” Rather, I simply brought the crystals into my energy field and let them do the communicating…and they did so in their own unique way!
Here is a technique that I have personally cultivated and feel to be very effective in engaging my crystals for whatever purpose I may have.
Connecting through the Rainbow Ray…
I have a personal affinity for rainbows. Whether in the sky or in a crystal, I find them intriguing and yes, somewhat magical. Whenever I look for crystals to add to my collection I’m especially drawn to those with little rainbows within their structure. When you see a rainbow in a crystal what you are seeing is refracted light bouncing off of the crystal’s inclusions. As we know, light equals energy and energy equals information, therefore it always made sense to me that a key way to engage a crystal is to tune into the information being emitted from its rainbow ray!
Here’s what I do. Prior to working with a rainbow crystal (many of which are clear quartz), I will locate the rainbow within its structure. I put the crystal within close proximity to a light source that will illuminate the refractions and show the rainbow. Once located, I then gaze at the rainbow, taking in all of the colors in the spectrum (blue, green, red, yellow, etc.). While focusing intently, I call attention to my heart center (our largest electromagnetic processor of energy) and begin “feeling” the rainbow within my heart. It is absolutely necessary to allow yourself to physically feel within your heart that sensation we get when we have a powerful emotion, except in this case the emotion is directed toward the rainbow. Eventually I can feel a connection being made between my own heart and that of the crystal – this is a very real energetic resonance that is taking place. I will then imagine a thin string of pulsating light connecting the crystal with my heart which ensures the two are in resonance until I finish my session. This helps tremendously when using crystals in meditation or just sitting with them or even carrying them with you during the day. You have made the energetic connection and are now poised to benefit from whatever it is the crystal might have to offer at any given time.
As you begin (or continue) your journey with crystals, minerals and stones, I encourage you to work with them consciously and subjectively. In fact, here’s a little experiment you can try on your own (especially if you are new to working with crystals). Get 4 or 5 different crystals, say, an amethyst, citrine, carnelian, rose quartz and smoky quartz. Do not look up anywhere the metaphysical meanings associated with the crystal. The key is to try and determine how best to utilize each different crystal for you! Allow 24-48 hours minimum to work with each crystal. You can meditate with the crystal, holding it preferably in your left hand – which is connected to the right hemisphere of the brain – or simply put it in your pillow case and sleep with it at night. Remember to engage the heart center with feeling as much as possible. Within the 24-48 hour period, try to pick up any impressions you get from the crystal and jot them down. Even if you are not with the crystal for the entire period, you will still get impressions from it as you will be influenced somewhat by its energy field. Do this with each of your crystals. Once you feel confident you’ve received specific impressions (physical sensations, emotions, inspirations, memories, etc.) you can find several resources either online or in a book that will describe what the crystal is said to represent. These may correspond with your personal experience or they may not. Either way, you will have determined how best to engage the crystal for your personal use!
As has been acknowledged, not everyone is going to be drawn to crystals for their metaphysical properties. This is a very personal and powerful journey. That is what makes it so special. Is it not natural to look at these absolutely exquisite pieces of Mother Earth and ponder her beauty, wisdom and potential? This earth is rich with flora and fauna and crystals are a central part of her landscape, both inner and outer. Most importantly, in realizing this we also come to realize that the “inner and the outer” are but reflections of one fundamental landscape. If we peer deeply into a crystal with reverence and appreciation we may begin to understand and appreciate some of the deepest and most exquisite aspects that we call our-selves! Enjoy the journey!
Recommended resources to further enhance your journey through the mineral kingdom:
Interview from 1984 – Crystal Applications in Technology and for Self, (Inc. the work of Marcel Vogel)
The Book of Stones – Who They Are & What They Teach, Robert Simmons and Naisha Ahsian (This is an excellent reference book with hundreds of beautiful color photos and descriptions of crystals, minerals and stones.
Vibrational Medicine – The #1 Handbook of Subtle-Energy Therapies, Richard Gerber, M.D. (This book covers a range of energetic modalities but has a significant section on crystals)
*The Magic of Precious Stones, Mellie Uyldert – The hidden powers of gems, their influence on man, and their connection with magic, astrology, healing and religion.
*This title may be a bit difficult to find. Check your local and online book resources, especially those who carry out of print, or used books.
| 35,252
|
\begin{document}
\title{On symmetric CR geometries of hypersurface type}
\author{Jan Gregorovi\v c and Lenka Zalabov\' a}
\address{J.G. Faculty of Mathematics, University of Vienna, Oskar Morgenstern Platz 1, 1090 Wien, Austria; L.Z. Institute of Mathematics, Faculty of Science, University of South Bohemia, Brani\v sovsk\' a 1760, \v Cesk\' e Bud\v ejovice, 370 05, Czech Republic and Department of Mathematics and Statistics, Faculty of Science, Masaryk University, Kotl\' a\v rsk\' a 2, Brno, 611 37, Czech Republic}
\email{jan.gregorovic@seznam.cz, lzalabova@gmail.com}
\thanks{First author supported by the project P29468 of the Austrian Science Fund (FWF). Second author supported by the grant 17-01171S of the Czech Science Foundation (GA\v CR)}
\keywords{CR geometry, homogeneous manifold, Webster metric}
\subjclass[2010]{32V05, 32V30, 53C30}
\maketitle
\begin{abstract}
We study non--degenerate CR geometries of hypersurface type that are symmetric in the sense that, at each point, there is a CR transformation reversing the CR distribution at that point. We show that such geometries are either flat or homogeneous. We show that non--flat non--degenerate symmetric CR geometries of hypersurface type are covered by CR geometries with a compatible pseudo--Riemannian metric preserved by all symmetries. We construct examples of simply connected flat non--degenerate symmetric CR geometries of hypersurface type that do not carry a pseudo--Riemannian metric compatible with the symmetries.
\end{abstract}
\section{Introduction} \label{prvni-cast}
In \cite{KZ}, Kaup and Zaitsev generalized Riemannian symmetric spaces to the setting of CR geometries, i.e., smooth manifolds with so--called CR distribution endowed with a complex structure. They consider a Riemannian metric compatible with CR geometry in the sense that the Riemannian metric is preserved by the complex multiplication on the CR distribution. Such manifold is symmetric in the sense of \cite{KZ} if, at each point, there is an isometric CR transformation that preserves the point and which, at that point, acts as $-\id$ on the CR distribution \cite[Definition 3.5.]{KZ}. They show that such isometric CR transformations are uniquely determined by the tangent action on the CR distribution \cite[Theorem 3.3]{KZ}. They also show that such CR geometries are homogeneous \cite[Proposition 3.6]{KZ}. In fact, these CR geometries may be considered as reflexion spaces in the sense of \cite{L1}. In \cite{AMN}, the authors study these CR geometries in the setting of so--called CR algebras.
We studied in \cite{my-sigma} filtered geometric structures that carry an automorphism at each point that fixes the point and acts as $-$id on a distinguished part of the filtration at the point. Let us point out that the non--degenerate CR geometries of hypersurface type, i.e., those with CR distribution of codimension $1$, are among these geometries. We answered the question whether these filtered geometries are homogeneous and can be considered as reflexion spaces. However, our result \cite[Theorem 5.7.]{my-sigma} holds under weaker conditions than the result of \cite{KZ} for non--degenerate CR geometries of hypersurface type. In particular, the sufficient condition for such non--degenerate CR geometry of hypersurface type to be homogeneous is that it is non--flat at one point.
In this paper, we study the case of non--degenerate CR geometries of hypersurface type in more detailed way. We consider point preserving CR transformations which, at that point, induce $-$id on the CR distribution. We say that a non--degenerate CR geometry of hypersurface type is symmetric (in our sense) if there exists a symmetry at each point, see Definition \ref{def1}. In particular, our definition does not require the existence of a metric compatible with the CR geometry. We adapt and significantly improve general results of \cite{my-TG,my-dga,my-sigma} for our particular class of CR geometries. Let us emphasize that every non--degenerate CR geometry of hypersurface type that is symmetric in the sense of \cite{KZ} is symmetric (in our sense).
Let us say that \cite[Theorem 5.7.]{my-sigma} is formulated in the general setting of parabolic geometries. We provide here the particular results of this theorem for CR geometries. We also provide new direct proofs, because we will need the presented ideas to explain new results, see Lemmas \ref{dve-symetrie}, \ref{involutivity} and Propositions \ref{hladky-sys}, \ref{thm1}. This allows us to compare our results with results of \cite{KZ} and \cite{AMN}.
We prove in Theorem \ref{thm3} that non--flat non--degenerate CR geometries of hypersurface type that are symmetric (in our sense) are covered by symmetric non--degenerate CR geometries of hypersurface type that carry a pseudo--Riemannian metric compatible with the CR geometry that is preserved by all our symmetries. In the Riemannian signature, these coverings are symmetric in the sense of \cite{KZ}, see Theorem \ref{thm4}. Moreover, we show in Theorem \ref{embed} that it is always possible to embed the CR geometry on these coverings into a complex manifold. In the Riemannian signature, this embedding is provided by a different construction than the one in \cite[Proposition 7.3]{KZ}.
Finally, we construct examples of non--homogeneous symmetric (in our sense) flat non--degenerate CR geometries of hypersurface type. These examples do not admit a pseudo--Riemannian metric that would be preserved by some symmetry at each point and in particular, they are not symmetric in the sense of \cite{KZ}. We also discuss examples of homogeneous CR geometries on orbits of real forms in complex flag manifolds. In particular, we show that there are homogeneous CR geometries which are locally symmetric but not globally symmetric. In fact, Theorem \ref{orbit} provides complete description of all possible cases.
\section{CR geometries of hypersurface type}
\subsection{CR geometries}
Let $M$ be a smooth manifold of dimension $2n+1$ for $n>1$ together with a distribution $\mathcal{H} \subset TM$ of dimension $2n$ and a \emph{complex structure} $J$ on $\mathcal{H}$, i.e., $J:\mathcal{H} \to \mathcal{H}$ is an endomorphism with the property that $J^2=-\id$. The triple $(M,\mathcal{H},J)$ is called a \emph{CR geometry of hypersurface type} if the $i$--eigenspace $\mathcal{H}^{1,0}$ of $J$ in the complexification of $\mathcal{H}$ is integrable, i.e., $[\mathcal{H}^{1,0},\mathcal{H}^{1,0}] \subset \mathcal{H}^{1,0}$. CR geometry $(M,\mathcal{H},J)$ is called \emph{non--degenerate} if $\mathcal{H}$ is completely non--integrable.
On $\mathcal{H}$ there exists a symmetric bilinear form $h$ with values in the line bundle $TM/\mathcal{H}$ given by $h(\xi,\eta)= \frac12 \pi([\xi,J \eta])$ for all $\xi,\eta \in \Gamma(\mathcal{H})$, where $\pi: TM\to TM/\mathcal{H}$ is a natural projection. Let us recall that $h$ is the real part of the Levi form $\tilde h$ of $(M,\mathcal{H},J)$, while the imaginary part of the Levi form is the map given by $\frac12 \pi([\xi,\eta])$. We assume that $M$ is orientable and denote by $(p,q)$ the signature of the Levi form, where our convention is $p\leq q$, $p+q=n$. Then the signature of $h$ is $(2p,2q)$.
The homogeneous space $PSU(p+1,q+1)/P$ is usually called the \emph{standard model} of non--degenerate CR geometry of hypersurface type of signature $(p,q)$, where the group $PSU(p+1,q+1)$ is the projectivization of the group of matrices preserving the pseudo--Hermitian form
$$m((u_0,\dots,u_{n+1}), (v_0,\dots,v_{n+1}))=u_0\overline{v_{n+1}}+u_{n+1} \overline{v_0}+\sum_{k=1}^p u_k\overline{v_k}-\sum_{k=p+1}^n u_k\overline{v_k}$$
on $\C^{n+2}$ and $P$ is the stabilizer of the complex line generated by the first basis vector in the standard basis of $\C^{n+2}$. The standard model $PSU(p+1,q+1)/P$ is a smooth real hypersurface in $\C P^{n+1}$ that can be also viewed as the projectivization of the null cone of $m$ in $\C^{n+2}$.
In the rest of the paper, by a CR geometry we mean a non--degenerate CR geometry of hypersurface type of signature $(p,q)$ for $p\leq q$. Such CR geometries can be equivalently described as parabolic geometries modeled on standard models $PSU(p+1,q+1)/P$. This description can be found in \cite[Section 4.2.4]{parabook}. We only use several consequences of this description later in the text.
\subsection{Distinguished connections}
There exist many admissible connections, i.e., connections preserving $\mathcal{H}$ and $J$, on CR geometries.
In particular, there are several distinguished classes of admissible connections given by a particular normalization condition on the torsion of admissible connections in the class. The most common class is the class of Webster--Tanaka connections \cite [Section 5.2.12]{parabook}. Another important class is the class of Weyl connections \cite[Sections 5.1.2 and 5.2.13]{parabook}. In this paper, we consider the class of Weyl connections, because in our proofs we use relations between CR transformations and geodesic transformations of normal Weyl connections \cite[Section 5.1.12]{parabook}.
In fact, Webster--Tanaka connections and Weyl connections induce the same class of distinguished partial connections $\na$ on $\mathcal{H}$.
Such two distinguished partial connections $\na$ and $\hat \na$ are related by the formula
\begin{align} \label{rozdil}
\begin{split}
\hat \na_{\xi}(\eta)&=\na_{\xi}(\eta)+
F(\xi)\eta+F(\eta)\xi-\tilde h(\xi, \eta)\tilde h^{-1}(F), \\
\end{split}
\end{align}
where $\xi,\eta$ are vector fields on $\mathcal{H}$ and $F$ is a one--form in $\mathcal{H}^*$.
Here $\tilde h^{-1}$ is the inverse of the Levi form $\tilde h$. We will write shortly $\hat \na=\na + F$ instead of the entire formula $(\ref{rozdil})$.
Each Weyl connection $D$ is associated with particular decompositions $TM \simeq \mathcal{H} \oplus \ell$ and $T^*M\simeq \mathcal{H}^* \oplus \ell^*$ that are preserved by $D$, where $\ell$ is a one--dimensional distribution complementary to $\mathcal{H}$. In fact, one-form $F$ in $\mathcal{H}^*$ from the formula (\ref{rozdil}) also describes the change of the decompositions of $TM$ and $T^*M$ associated with $D$ and $\hat D$. The precise formula for the change of the decompositions can be easily computed using \cite[Section 5.1.5]{parabook}. In general, arbitrary two Weyl connections $D$ and $\hat D$ are related by a suitable
action of a one--form $\U=\U_1+\U_2$ in $T^*M=\mathcal{H}^* \oplus \ell^*$, where we consider the decomposition associated with the Weyl connection $D$. We write shortly $\hat D=D+ \U_1+\U_2$ instead of the explicit formula for the change, which is complicated and can be computed using \cite[Section 5.1.6]{parabook}. Let us emphasize that $\U_1$ coincides with $F$ from the formula (\ref{rozdil}) for the corresponding partial connections $\na$, $\hat \na$ determined by $D$, $\hat D$.
Let us finally point out that admissible connections provide the fundamental invariant $W$ of CR geometries which is known as \emph{Chern--Moser tensor} or \emph{Weyl tensor} and coincides with the totally trace--free part of the curvature of arbitrary Weyl or Webster--Tanaka connection. Vanishing of this invariant implies that CR geometry is \emph{flat}, meaning that CR geometry is locally equivalent to the standard model $PSU(p+1,q+1)/P$.
\section{Symmetries of CR geometries}
\subsection{Definition of symmetries}\label{sec31}
A \emph{CR transformation} of CR geometry $(M,\mathcal{H},J)$ is a diffeomorphism of $M$ such that the tangent map preserves the \emph{CR distribution} $\mathcal{H}$ and its restriction to $\mathcal{H}$ is complex linear. We study the following CR transformations.
\begin{def*}\label{def1}
A \emph{symmetry} at $x \in M$ on a CR manifold $(M,\mathcal{H},J)$ is a CR transformation $S_x$ of $M$ such that:
\begin{enumerate}
\item $S_x(x)=x$,
\item $T_xS_x=-$id on $\mathcal{H}$.
\end{enumerate}
We say that CR geometry is \emph{symmetric} if there exists a symmetry at each point $x \in M$. A \emph{system of symmetries} on $M$ is a choice of a symmetry $S_x$ at each $x \in M$. We call the system \emph{smooth}, if the map $S: M \times M \to M$ given by $S(x,y)=S_x(y)$ is smooth in both variables.
\end{def*}
Let us show that the standard model $PSU(p+1,q+1)/P$ is symmetric. The Lie group $PSU(p+1,q+1)$ is the group of all CR transformations of the standard model $PSU(p+1,q+1)/P$, where we consider left action.
Direct computation gives that all symmetries of the standard model $PSU(p+1,q+1)/P$ at the origin $eP$ are represented by $(1,n,1)$--block matrices of the form
\begin{align}\label{symform}
s_{Z,z}=\left(
\begin{matrix}
-1&-Z& iz+\frac12ZIZ^* \\ 0&E&-IZ^* \\ 0&0&-1
\end{matrix}
\right)
,\end{align}
where $Z \in \C^{n*}$, $z \in \R^*$ are arbitrary, $E$ is the identity matrix of the rank $n$ and $I$ is the diagonal matrix with the first $p$ entries equal to $1$ and the remaining $q$ entries equal to $-1$.
\begin{lemma*} \label{standard}
There exists an infinite number of symmetries at each point $kP$ of $PSU(p+1,q+1)/P$ given by matrices of the form $ks_{Z,z}k^{-1}$ for all $Z \in \C^{n*}$ and $z \in \R^*$. In particular:
\begin{enumerate}
\item There exists an infinite number of involutive symmetries at each point characterized by the condition $z=0$. For each such symmetry, there is a different metric preserved by this symmetry compatible with the CR geometry.
\item There exists an infinite number of non--involutive symmetries at each point characterized by the condition $z\neq 0$. They do not preserve any metric compatible with the CR geometry.
\end{enumerate}
\end{lemma*}
\begin{proof}
Each element $s_{Z,z}$ satisfying $z=0$ is conjugated to an element of maximal compact subgroup of $PSU(p+1,q+1)$, and thus preserves the corresponding metric. Each element $s_{Z,z}$ satisfying $z\neq 0$ is contained in a different orbit with respect to the conjugation than the maximal compact subgroup of $PSU(p+1,q+1)$, and thus it cannot preserve any compatible metric.
\end{proof}
The standard model $PSU(p+1,q+1)/P$ is endowed with a pseudo--Riemannian metric compatible with the CR geometry, because the maximal compact subgroup of $PSU(p+1,q+1)$ acts transitively on the standard model. Moreover, there is exactly one involutive symmetry at each point of this model that is contained in the maximal compact subgroup. These symmetries preserve the corresponding pseudo--Riemannian metric and form a smooth system. This means that in the Riemannian signature, the standard model $PSU(1,n+1)/P$ is symmetric in the sense of \cite{KZ}.
On flat CR geometries, the set of symmetries is locally identical with the one on the standard model. However, these symmetries may not be defined globally. This means that on flat CR geometries, there locally always exists a pseudo--Riemannian metric compatible with the CR geometry preserved by some symmetry at each point. We show in Example \ref{exam1} that such pseudo--Riemannian metric compatible with CR geometry does not have to exist globally.
\subsection{Involutive and non--involutive symmetries}
Suppose that there is a symmetry $S_x$ at $x$ on CR geometry $(M,\mathcal{H},J)$. If $D$ is a Weyl connection, then $S_x^*D$ is a Weyl connection, too. Therefore, there is a one--form $\U_1+\U_2 \in \mathcal{H}^*\oplus \ell^*$ such that
\begin{align} \label{zmena}
S_x^*D=D+ \U_1+\U_2.
\end{align}
\begin{lemma*} \label{dve-symetrie}
Suppose $S_x$ is a symmetry at $x \in M$. Let $D$ be an arbitrary Weyl connection and let $\U_1+\U_2 \in \mathcal{H}^*\oplus \ell^*$ be the one--form from the formula $(\ref{zmena})$. Then the following claims are equivalent:
\begin{enumerate}
\item the symmetry $S_x$ is involutive,
\item $\U_2(x)=0$, and
\item the diffeomorphism $S_x$ is linear in the normal coordinates given by the normal Weyl connection $\bar D$ at $x$ that is uniquely determined by the property that $\bar D$ coincides with the Weyl connection $D+{1 \over 2} \U_1$ at $x$.
\end{enumerate}
Moreover, the partial connection $\nabla^{S_x}$ induced by the Weyl connection $D^{S_x}:=D+{1 \over 2} \U_1$ does not depend on the choice of $D$ at $x$ and satisfies
\begin{itemize}
\item $S_x^*(\nabla^{S_x})=\nabla^{S_x}$ at $x$, and
\item $\na^{S_x} W(x)=0$.
\end{itemize}
\end{lemma*}
\begin{proof}
Iterating the formula $(\ref{zmena})$ we compute $$S_x^*S_x^*D=D+ \U_1+ S_x^*(\U_1)+\U_2+S_x^*(\U_2).$$ The component of the (dual) action of $T_xS_x$ on $T_x^*M$
preserving the decomposition $T_x^*M=\mathcal{H}^*(x)\oplus \ell^*(x)$ is $-\id\oplus \id$, and the component that maps $\mathcal{H}^*(x)$ into $\ell^*(x)$ depends linearly on $\U_1$ and is antisymmetric as a map $\mathcal{H}^*(x)\otimes \mathcal{H}^*(x)\to \ell^*(x)$. Therefore, $ S_x^*(\U_1)(x)=-\U_1(x)$ and $S_x^*(\U_2)(x)=\U_2(x)$.
If the symmetry $S_x$ is involutive, i.e., $S_x^2=\id$, then $$0=\U_2(x)+S_x^*(\U_2)(x)=2\U_2(x)$$ and thus $\U_2(x)=0$.
If $\U_2(x)=0$, then the normal Weyl connection $\bar D$ that coincides with the Weyl connection $D+{1 \over 2}\U_1$ at $x$ satisfies $$S_x^*(D+{1 \over 2}\U_1)=D+\U_1+S_x^*({1 \over 2}\U_1).$$ At the point $x$, we get $$\U_1(x)+S_x^*({1 \over 2}\U_1)(x)=\U_1(x)-{1 \over 2}\U_1(x)={1 \over 2}\U_1(x)$$ and thus $S_x^*\bar D=\bar D$ follows from the normality \cite[Section 5.1.12]{parabook}. Thus $S_x$ is an affine map, which is linear in the normal coordinates.
If the symmetry $S_x$ at $x$ is linear in the normal coordinates of a Weyl connection, then its (dual) tangent action preserves the decomposition $T_x^*M=\mathcal{H}^*(x)\oplus \ell^*(x)$ and therefore $(T_xS_x)^2=\id$. Then it follows from the linearity that $S_x$ is involutive.
Finally, the last claim follows, because $(\na^{S_x} W)(x)$ is a tensor of type $\otimes^4 \mathcal{H}_x^* \otimes \mathcal{H}_x$ invariant with respect to $S_x$.
\end{proof}
\begin{lemma*} \label{involutivity}
Suppose that there is a symmetry $S_x$ at $x \in M$.
Let $D$ be an arbitrary Weyl connection and let $\U_1+\U_2 \in \mathcal{H}^* \oplus \ell^*$ be the one--form from the formula $(\ref{zmena})$.
If $W(x)\neq 0$, then $\U_2(x)=0$ and the symmetry $S_x$ is involutive.
\end{lemma*}
\begin{proof}
Consider the covariant derivative of $W$ with respect to $D+{1 \over 2}\U_1$ in the direction $\ell$ and compute $S_x^*(D+{1 \over 2}\U_1)_{r}W(x)$ for $r\in \ell(x)$. We know that $W(x)$ is $S_x$--invariant and thus
$$S_x^*(D+{1 \over 2}\U_1)_{r}W(x)=(D+{1 \over 2}\U_1)_{S_x^*(r)}W(x)=(D+{1 \over 2}\U_1)_{r}W(x).$$
On the other hand, it generally holds that $S_x^*(D+{1 \over 2}\U_1)=D+{1 \over 2}\U_1+\U_2$ and $\U_2(x)=a\theta(x)$ for a
covector $\theta \in \ell^*(x)$ such that $\theta(r)=1$. Then $$S_x^*(D+{1 \over 2}\U_1)_{r}W(x)=(D+{1 \over 2}\U_1)_{r}W(x)+2aW(x)$$ and thus $a=0$ which implies $\U_2(x)=0$.
\end{proof}
\subsection{Smooth systems of involutive symmetries}
Let us show that the assumption on $W$ to be nowhere vanishing not only implies that all symmetries are involutive, but also that there is at most one symmetry at each point of $M$ and that these symmetries change smoothly along $M$.
\begin{prop*} \label{hladky-sys}
Suppose that $(M,\mathcal{H},J)$ is a symmetric CR geometry such that $W(x)\neq 0$ for all $x\in M$. Then
\begin{enumerate}
\item there is a unique symmetry $S_x$ at each $x\in M$,
\item the map $S: x\mapsto S_x$ is smooth, and
\item $S_x \circ S_y \circ S_x=S_{S_x(y)}$ holds for all $x,y\in M$.
\end{enumerate}
In particular, $(M,S)$ is a reflexion space, i.e., $S: M\times M\to M$ is a smooth map that for all $x,y,z\in M$ satisfies that
\begin{itemize}
\item $S(x,x)=x,$
\item $S(x,S(x,y))=y,$ and
\item $S(x,S(y,z))=S(S(x,y),S(x,z))$.
\end{itemize}
\end{prop*}
\begin{proof}
We show that if there are two different symmetries at $x$ on CR geometry $(M,\mathcal{H},J)$, then $W$ vanishes at $x$.
Consider two different symmetries $S_x$ and $S'_x$ at $x$ (both must be involutive).
We know from Lemma \ref{dve-symetrie} that $\na^{S_x} W(x)=0$ and $\na^{S_x'} W(x)=0$ hold for partial connections $\na^{S_x},\na^{S_x'}$. These partial connections are different (at $x$) due to the claim (3) of Lemma \ref{dve-symetrie}, i.e., $\na^{S_x'}=\na^{S_x}+F$ holds according to the formula (\ref{rozdil}) for $F(x)\neq 0$. This means that the linear map $\mathcal{H}_x\to \mathcal{H}_x$ given by
\begin{align} \label{alg}
\eta\mapsto (F(\xi)\eta+F(\eta)\xi-\tilde h(\xi, \eta)\tilde h^{-1}(F))(x)
\end{align}
defines a non--zero element $\xi(F)(x)$ of a Lie algebra $\frak{csu}(p,q)$ for each $\xi \in \mathcal{H}_x$, where we identify $\frak{csu}(p,q)$ with $$\{X\in \frak{gl}(\mathcal{H}_x) : [X,J_x]=0, h_x(X(\xi),\nu)+h_x(\xi,X(\nu))=a\cdot h_x(\xi,\nu), a\in \mathbb{R}\}.$$
Moreover, the element $\xi(F)(x)$ of $\frak{csu}(p,q)$ has to act trivially on $W(x)$ for all vectors $\xi$. Let us denote by $\Ann(W_x)$ the set of all $A\in \frak{csu}(p,q)$ such that $A$ acts trivially on $W(x)$. Then we get
$$F(x)\in \Ann(W_x)^{(1)}:=\{F : \xi(F)(x)\in \Ann(W_x)\ {\rm for\ all} \ \xi\in \mathcal{H}_x\}.$$
The result of \cite{KT} states that if $W(x)$ is non--trivial, then $\Ann(W_x)^{(1)}=0,$ and thus $\xi(F)(x)=0$ for all $\xi \in \mathcal{H}_x$. Since $\xi(-)(x): \mathcal{H}_x^*\to \frak{csu}(p,q)$ is a linear map at each $x \in M$, this implies $F(x)=0$, which is a contradiction. This proves the uniqueness of symmetries at $x$ in the case $W(x) \neq 0$.
Since $S_x \circ S_y \circ S_x$ is a symmetry at $S_x(y)$, the condition $S_x \circ S_y \circ S_x=S_{S_x(y)}$ trivially follows from the uniqueness of symmetries. Thus it remains to prove the smoothness of $S$.
Let us fix a partial Weyl connection $\nabla$. For each $y\in M$, there is $F(y)$ such that $(\nabla^{S_y}-\nabla)(y)=F(y)$ by the formula (\ref{rozdil}), which is well--defined due to the uniqueness of $\nabla^{S_y}$ at $y$. Thus $\nabla W(y)$ is given by the algebraic action (\ref{alg}) of $\xi(F(y))$ on $W(y)$ for each $\xi\in \mathcal{H}_y$. Since $\nabla W(y)$ is smooth, the image of $\xi(F(y))$ in $\frak{csu}(p,q)$ depends smoothly on $y$ for each $\xi\in \mathcal{H}_y$. Since the kernel of the action coincides with $\Ann(W_y)^{(1)}$, we conclude that $F(y)$ depends smoothly on $y$.
Let $D$ be an arbitrary Weyl connection inducing the partial Weyl connection $\nabla$. Then $S_y$ is linear in the normal coordinates of the normal Weyl connection $\bar D$ constructed for $D+\frac12F(y)$ due to the claim (3) of Lemma \ref{dve-symetrie}. Since $\bar D$ depends smoothly on $y$, we get that $S$ is smooth.
It clearly holds that $S_x(x)=x$ and $ S_x^2=\id$ for all $x \in M$. We have proved that $S$ is smooth and satisfies $S_x \circ S_y=S_{S_x(y)} \circ S_x^{-1}=S_{S_x(y)} \circ S_x$ for all $x,y \in M$. Thus it follows that $(M,S)$ satisfies the conditions of the reflexion space.
\end{proof}
Proposition \ref{hladky-sys} has the following consequence.
\begin{prop*}\label{thm1}
Suppose that $(M,\mathcal{H},J)$ is a symmetric CR geometry. Then either
\begin{enumerate}
\item $W=0$ and the CR geometry is locally equivalent to the standard model, or
\item $W\neq 0$ and the group generated by symmetries is a Lie group that acts transitively on $M$, i.e., CR geometry is homogeneous. In particular, the reflexion space $(M,S)$ from the Proposition \ref{hladky-sys} is a homogeneous reflexion space.
\end{enumerate}
\end{prop*}
\begin{proof}
Suppose that $U \subset M$ consists of all points with non--trivial $W$. It suffices to prove that the group generated by symmetries at points in $U$ acts transitively on $U$ to obtain the claim of the Theorem, because then $W$ is constant on $U$ due to the homogeneity. The fact that the group generated by symmetries on a reflexion space is a Lie group can be found in \cite{L1}.
Let $c(t)$ be a curve in $U$ such that $c(0)=x$ and ${d \over dt}|_{t=0}c(t)=X\in \mathcal{H}_x$. Then ${d \over dt}|_{t=0}S_{c(t)}(x)$ is tangent to the orbit of the action of the group generated by symmetries at points in $U$. Differentiation of the equality $c(t)=S_{c(t)}c(t)$ gives
$$X={d \over dt}|_{t=0}S_{c(t)}(c(t))={d \over dt}|_{t=0}S_{c(t)}(x)+T_xS_x.X,$$ and we get
$${d \over dt}|_{t=0}S_{c(t)}(x)=X-T_xS_x.X=2X.$$
Thus at all $x\in U$, the CR distribution $\mathcal{H}$ is tangent to the orbit of the group generated by symmetries at points in $U$. Therefore the group generated by symmetries at points in $U$ acts transitively on $U$.
\end{proof}
Flat symmetric CR geometries do not have to be homogeneous. We construct an explicit example in Section \ref{sec6}.
\section{Non--flat symmetric CR geometries}
\subsection{Homogeneous CR geometries and their symmetries}
There are several possible ways how to describe a homogeneous CR geometry. We will use the description from \cite[Section 1.5.15]{parabook} that is closely tight with the setting of Cartan geometries, but as we show in this section, it can be treated independently of the general theory. We need only to recall that the Lie algebra $\frak{su}(p+1,q+1)$ of $PSU(p+1,q+1)$ consists of the $(1,n,1)$--block matrices
$$\left(
\begin{smallmatrix}
a&Z&iz \\ X&A&-IZ^* \\ ix&-X^*I&-\bar a
\end{smallmatrix}
\right),$$
where $\mathfrak{csu}(p,q)=\{(a,A):a \in \C,\ A \in \frak{u}(n),\ a+tr(A)-\bar a=0 \}$, $X \in \C^n $, $Z \in \C^{n*} $, $x \in \R $ and $z \in \R^*$. This means that we have the following decomposition
$$\frak{su}(p+1,q+1)=\R \oplus \C^{n}\oplus\mathfrak{csu}(p,q)\oplus \C^{n*}\oplus\R^*.$$
The Lie algebra $\fp$ of $P$ corresponds to $(1,n,1)$--block upper triangular part and decomposes as $\fp=\mathfrak{csu}(p,q)\oplus \C^{n*}\oplus\R^*.$ In fact, $P\cong CSU(p,q)\exp(\C^{n*}\oplus\R^*)$, where $CSU(p,q)$ consists of all elements of $P$ preserving the above decomposition.
\begin{lemma*}\label{lem4}
Let $K$ be an arbitrary transitive Lie group of CR transformations of a homogeneous CR geometry $(M,\mathcal{H},J)$ and let $L \subset K$ be the stabilizer of a point. Then there is a pair of maps $(\alpha,i)$ such that $i$ is an injective Lie group homomorphism $i:L \rightarrow P$ and $\alpha$ is a linear map $\alpha: \mathfrak{k}\to \mathfrak{\frak{su}}(p+1,q+1)$ satisfying the following conditions:
\begin{enumerate}
\item $\alpha: \fk \to \mathfrak{su}(p+1,q+1)$ is a linear map extending $T_ei:\mathfrak{l}\to \fp$,
\item $\alpha$ induces an isomorphism $\underline \alpha: \mathfrak{k}/\mathfrak{l} \rightarrow \mathfrak{su}(p+1,q+1)/\fp$ of vector spaces,
\item $\Ad(i(l))\circ \alpha=\alpha \circ \Ad(l)$ holds for all $l\in L$,
\item the linear map $\wedge^2 \fk\to \frak{su}(p+1,q+1)$ given by the formula $[\alpha(X),\alpha(Y)]-\alpha([X,Y])$ for all $X,Y\in \fk$ has values in $\fp$ and defines a $K$--invariant two--form $\kappa$ with values in $K\times_{\Ad\circ i} \fp$,
\item the component of $\kappa$ in $K\times_{\underline{\Ad}\circ i} \mathfrak{csu}(p,q)$ is a tensor that coincides with $W$, where $\underline{\Ad}$ is the induced action of $P$ on $\mathfrak{csu}(p,q)\cong \fp/(\C^{n*}\oplus\R^*).$
\end{enumerate}
Conversely, suppose that $(\alpha,i)$ is such pair of maps from $(K,L)$ to $(PSU(p+1,q+1),P)$. Then there is $K$--homogeneous CR geometry $(K/L,\mathcal{H},J)$ satisfying $\mathcal{H}_{eL}=\alpha^{-1}(\C^{n}\oplus \fp)/\fl$ and $J_{eL}=\underline{\alpha}^*(J)$, where $J$ is the complex structure on $\C^n$.
\end{lemma*}
A pair $(\alpha,i)$ satisfying the conditions (1)--(3) of Lemma \ref{lem4} is usually called an \emph{extension} of $(K,L)$ to $(PSU(p+1,q+1),P)$. The two--form $\kappa$ from the condition (4) is the curvature of the Cartan connection given by the extension $(\alpha,i)$. Finally, the condition (5) is the normalization condition on the curvature $\kappa$ that can be also expressed as $\partial^*\kappa=0$, where $\partial^*$ is the Kostant's co--differential \cite[Section 3.1.11]{parabook}.
\begin{proof}
It is shown in \cite[Section 1.5.15]{parabook} that each homogeneous Cartan (and thus parabolic) geometry can be described by a particular extension and that each extension determines a homogeneous Cartan geometry. The formula for $\kappa$ in the condition (4) is obtained from \cite[Section 1.5.16]{parabook}. Therefore, it follows from the description of CR geometries in \cite[Section 4.2.4]{parabook} that the conditions (4) and (5) on the curvature $\kappa$ have to be satisfied.
\end{proof}
\begin{def*}
The pair $(\alpha,i)$ from Lemma \ref{lem4} is called the \emph{normal extension} of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing the homogeneous CR geometry $(M,\mathcal{H},J)$.
\end{def*}
Examples of normal extensions describing certain homogeneous CR geometries and the explicit formula from the condition (5) of Lemma \ref{lem4} can be found in \cite{HG-dga}.
It is clear from the second part of Lemma \ref{lem4} that only the maps $i$ and $\underline{\alpha}$ are sufficient to determine CR geometry. This means that there are many normal extensions $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing the same CR geometry. The other parts of $\alpha$ are completely determined by the condition (5) from Lemma \ref{lem4} and carry the information about Weyl connections. The remaining freedom (for fixed $i$) is in the choice of a complex basis of $\underline{\alpha}^{-1}(\C^n)$. In general, if $h\in P$, then the pair $(\Ad(h)\circ \alpha,conj(h)\circ i)$ is also a normal extension of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing the same CR geometry as the normal extension $(\alpha,i)$.
Let us summarize the results characterizing symmetric non--flat homogeneous CR geometries following from \cite{GZ-Lie,my-dga}.
\begin{prop*}\label{existence-2}
Let $K$ be the Lie group of all CR transformations of a non--flat homogeneous CR geometry $(M,\mathcal{H},J)$. Then the following is equivalent:
\begin{enumerate}
\item There is a (unique) symmetry at each point.
\item There is $s\in L$ such that the triple $(K,L,s)$ is a (non--prime) homogeneous reflexion space, i.e.,
\begin{itemize}
\item $s$ commutes with all elements of $L$,
\item $s^2=e$, where $e$ is the identity element of $L$, and
\item all symmetries are of the form $S_{kL}=ksk^{-1}$ for $k\in K$.
\end{itemize}
\item There is a normal extension $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing $(M,\mathcal{H},J)$ such that $i(L)\subset CSU(p,q)$ and $s_{0,0}\in i(L)$ (see the formula (\ref{symform})).
\item For each normal extension $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing $(M,\mathcal{H},J)$, there is a (unique) $Z\in \C^{n*}$ such that $\Ad(\exp (Z))\alpha(\fk)$ is preserved by $\Ad(s_{0,0})$, and the Lie algebra automorphism of $\fk$ given by $\Ad(s_{0,0})$ defines an automorphism of the Lie group $K$.
\end{enumerate}
\end{prop*}
The condition (3) of Proposition \ref{existence-2} immediately implies that there are $K$--invariant Weyl connections on a symmetric non--flat CR geometry $(M,\mathcal{H},J)$. According to \cite[Proposition 1.4.8]{parabook}, a $K$--invariant connection on $T(K/L)$ can be described by a map $\gamma: \fk\to \frak{gl}(\fk/\fl)$ such that
\begin{itemize}
\item $\gamma|_\fl=\underline{\ad}$, and
\item $\gamma(\Ad(h)(X))=\underline{\Ad}(h)\circ \gamma(X)\circ \underline{\Ad}(h)^{-1}$
\end{itemize}
hold for all $X\in \fk$ and $h\in L$, where $\underline{\Ad}: L\to Gl(\fk/\fl)$ is induced by the adjoint representation.
\begin{prop*}\label{connexe}
Let $K$ be the Lie group of all CR transformations of a non--flat symmetric CR geometry $(M,\mathcal{H},J)$. Let $(\alpha,i)$ be a normal extension of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing $(M,\mathcal{H},J)$ such that $i(L)\subset CSU(p,q)$ and $s_{0,0}\in i(L)$. Then $\gamma:=\underline{\alpha}^*(\underline{\ad}\circ r_0)$ describes a $K$--invariant Weyl connection, where $r_0: \mathfrak{su}(p+1,q+1)\to \mathfrak{csu}(p,q)$ is the projection along $\R \oplus \C^{n}\oplus \C^{n*}\oplus\R^*$.
In particular, there is a bijection between the set of $K$--invariant Weyl connections on $M$ and the set of $z\in \R^*$ such that $conj(\exp(z))\circ i(L)\subset CSU(p,q)$ holds for the extension $(\alpha,i)$.
\end{prop*}
\begin{proof}
We proved the existence of $K$--invariant Weyl connections on non--flat symmetric CR geometries in \cite{my-dga}. Therefore it is enough to check that they can be described by the functions $\gamma$. Since $i(L)\subset CSU(p,q)$, the projection $r_0$ is $i(L)$--equivariant and $\gamma|_\fl=\underline{\ad}$ holds. Therefore each $\gamma$ describes a $K$--invariant connection. The fact that this is a Weyl connection follows directly from the condition (5) in Lemma \ref{lem4}.
It is clear that the one--form $\U_1+\U_2$ measuring the `difference' between two $K$--invariant Weyl connections is given by an $i(L)$--invariant element of $\C^{n*} \oplus \R^*$. Since $s_{0,0}\in i(L)$, it has to be an element of $\R^*$. It is clear that $z\in \R^*$ is $i(L)$--invariant element if and only if $conj(\exp(z))\circ i(L)\subset CSU(p,q)$ holds.
\end{proof}
\subsection{
Groups generated by symmetries}\label{sec4.2}
The following Theorem significantly improves the characterization of non--flat symmetric homogeneous CR geometries given by Propositions \ref{thm1} and \ref{existence-2}.
\begin{thm*} \label{thm2}
Let $K$ be the Lie group generated by all symmetries of a non--flat symmetric CR geometry $(M,\mathcal{H},J)$. Let $(\alpha,i)$ be a normal extension of $(K,L)$ to $(PSU(p+1,q+1),P)$ describing the CR geometry that satisfies $i(s)=s_{0,0}$ and $i(L)\subset CSU(p,q)$. Denote by $\fh$ the $1$--eigenspace of $s$ in $\fk$ and by $\fm$ the $-1$--eigenspace of $s$ in $\fk$. Then:
\begin{enumerate}
\item The following conditions hold
\begin{itemize}
\item $\alpha(\fl)\subset \frak{u}(p,q)$,
\item $\alpha(\fm)\subset \C^{n}\oplus \C^{n*}$, and
\item $\alpha(\fh)\subset \R \oplus\mathfrak{csu}(p,q)\oplus \R^*$ is a Lie subalgebra.
\end{itemize}
\item There is a basis of $\fh/\fl\oplus \fm$ such that for a vector in $\fh/\fl\oplus \fm$ with coordinates $(x,X)$ holds
$$\alpha((x,X)+\fl)=\Ad(\exp(z))\circ \left(
\begin{matrix}
aix&\Rho_1(X)& \Rho_2(x)i \\ X&Ax&-I\Rho_1(X)^* \\ xi&-IX^*&aix
\end{matrix}
\right)+\alpha(\fl),$$
where $z\in \R^*$, $\Rho_1:\C^{n}\to \C^{n*}$, $\Rho_2: \R\to \R^*$
and $(a,A)\in \frak{u}(p,q)$ normalizes $\alpha(\fl)$.
\item The maps $\Rho_1,\Rho_2$ and the matrix $(a,A)$ are completely determined by the condition (5) from Lemma \ref{lem4}.
\end{enumerate}
\end{thm*}
\begin{proof}
We know from Proposition \ref{existence-2} that there exists a normal extension $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ satisfying our assumptions.
Consider the canonical decomposition $\fk = \fh \oplus \fm$, where $\fh$ is $1$--eigenspace of $s$ and $\fm$ is $-1$--eigenspace of $s$. Then $\alpha(\fm)\subset \C^{n}\oplus \C^{n*}$ and $\alpha(\fh)\subset \R \oplus\mathfrak{csu}(p,q)\oplus \R^*$ follow from the assumption $i(s)=s_{0,0}$ and $\alpha(\fh)$ is a Lie subalgebra, because $dim(\fh/\fl)=1$. We can identify $\fm$ with $\C^{n}$ via $\alpha$, because the restriction of $\alpha$ to the map $\fm\to \C^{n}$ is injective. Indeed, if the restriction is not injective, then the elements in its kernel would be another symmetries at $eL$, but we know that there is only one symmetry. This identification uniquely determines the map $i: L\to CSU(p,q)$.
Further, $[\fm,\fm] \subset \fh$ holds and we have the corresponding symmetric space $K/H^0$, where $H^0$ is the connected component of identity of the fixed point set of the conjugation by $s$. Therefore $\exp([X,Y]) \in H^0$ for each $X,Y \in \fm$. The map $\Ad: H^0\to GL(\fm)$ can be restricted to the connected component of identity $L^0$ of $L$ and the restriction coincides with $i$. Therefore, it suffices to show that the element $\ad([X,Y])\in \frak{gl}(\fm)$ belongs to $\frak{sl}(\fm)$ for all $X,Y\in \fm$. But we have
$\ad([X,Y])=\ad(X) \circ \ad(Y) - \ad(Y) \circ \ad(X)$ and the trace equals to
$$
tr(\ad([X,Y]))=tr(\ad(X) \circ \ad(Y) - \ad(Y) \circ \ad(X))=B(X,Y)-B(Y,X)
,$$
where $B$ denotes the Killing form, which is symmetric. Therefore $i(L^0) \subset U(p,q)$ and $T_ei(\fl)\subset \frak{u}(p,q)$. In particular, the claim (1) holds. The map $\alpha$ can be expressed as in the claim (2), because there is always $z\in \R^*$ such that the extension $(\Ad(\exp(-z))\circ \alpha,conj(\exp(-z))\circ i)$ satisfies
$$\Ad(\exp(-z))\circ \alpha((x,0)+\fl)= \left(
\begin{matrix}
aix&0& \Rho_2(x)i \\ 0&Ax&0 \\ xi&0&aix
\end{matrix}
\right)+\Ad(\exp(-z))\circ \alpha(\fl).$$
Since the CR geometry $(M,\mathcal{H},J)$ does not depend on parts $\Rho_1,\Rho_2$ and $(a,A)$ of $\alpha$, these parts are completely determined by the condition (5) from Lemma \ref{lem4}.
\end{proof}
Let us remark that although the Lie algebra homomorphism $i$ is uniquely determined by the isomorphism $\fm\cong \C^{n}$ given by $\alpha$, the converse is not true. See \cite{HG-dga} for examples of non--equivalent CR geometries described by extensions with the same Lie group homomorphism $i$.
Let us further remark that we are not aware of any example of an extension $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ where $z\in \R^*$ from the claim (2) of Theorem \ref{thm2} does not correspond to an invariant Weyl connection. The main reason for this is the following result.
\begin{prop*}\label{prop5}
Suppose that $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$. Then $i(L)\subset U(p,q)$ and there is a bijection between $ \R^*$ and the set of $K$--invariant Weyl connections. In particular, there is a unique $K$--invariant Weyl connection corresponding to the normal extension $(\alpha,i)$ satisfying $i(L)\subset U(p,q)$, $i(s)=s_{0,0}$ and
\begin{align} \label{form-prop5}
\alpha((x,X)+\fl)=\left(
\begin{matrix}
aix&\Rho_1(X)& \Rho_2(x)i \\ X&Ax&-I\Rho_1(X)^* \\ xi&-IX^*&aix
\end{matrix}
\right)+\alpha(\fl).
\end{align}
This particularly holds when the transitive group $K$ is semisimple.
\end{prop*}
\begin{proof}
If $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$, then $i(L)\subset U(p,q)$ holds and the claim follows. It follows from the classification of semisimple symmetric spaces that $H$ is reductive and there is a complement to $\fl$ in the center of $\fh$. Consequently $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$.
\end{proof}
\subsection{Relations to CR algebras}
We explain here relations between our concept and the concept of CR algebras introduced in \cite{AMN}. We denote here by $\frak{n}_\C$ the complexification of a Lie algebra $\frak{n}$.
Let $(\alpha,i)$ be an extension of $(K,L)$ to $(PSU(p+1,q+1),P)$. We complexify the linear map $\alpha$ to obtain a map
$$\alpha_{\mathbb{C}}: \fk_{\mathbb{C}}\to \frak{sl}(n+2,\mathbb{C}).$$
The Lie algebra $\frak{sl}(n+2,\mathbb{C})$ decomposes as
$$\frak{sl}(n+2,\mathbb{C})=\mathbb{C}\oplus (\mathbb{C}^n\oplus \mathbb{C}^{n*})\oplus (\frak{gl}(n,\mathbb{C})\oplus \mathbb{C})\oplus (\mathbb{C}^{n*}\oplus \mathbb{C}^n) \oplus \mathbb{C},$$ where $\fp_{\mathbb{C}}= (\frak{gl}(n,\mathbb{C})\oplus \mathbb{C})\oplus (\mathbb{C}^{n*}\oplus \mathbb{C}^n) \oplus \mathbb{C}$. The subspace $\mathbb{C}^{n*}\oplus \fp_{\mathbb{C}}$ is a parabolic subalgebra of $\frak{sl}(n+2,\mathbb{C})$ that satisfies $$\mathcal{H}^{01}_{eL}=\alpha^{-1}_{\mathbb{C}}(\mathbb{C}^{n*}\oplus \fp_{\mathbb{C}})/\fl_{\mathbb{C}}.$$
Therefore, the preimage $\fq$ of $\mathcal{H}^{01}_{eL}$ in $\fk_{\mathbb{C}}$ is a Lie subalgebra of the form $$\fq=\alpha^{-1}_{\mathbb{C}}(\mathbb{C}^{n*}\oplus \fp_{\mathbb{C}}).$$
The pair $(\fk,\fq)$ satisfies conditions of \emph{CR algebra} from \cite[Section 1.2.]{AMN}.
It is proved in \cite{AMN} that this is the minimal set of data describing CR geometry on the homogeneous space $K/L$. However, CR algebras do not provide as much information as the extension $(\alpha,i)$.
In particular, we cannot obtain directly the curvature $\kappa$ of the corresponding Cartan connection from the CR algebra. Therefore, it is not easy to distinguish whether two CR algebras correspond to equivalent CR geometries.
There are conditions in \cite[Section 1.4]{AMN} that characterize CR algebras of CR geometries that are symmetric in the sense of \cite{KZ}.
One of these conditions ensures that there is a Riemannian metric compatible with CR geometry. Other conditions are analogous to the condition (4) of Proposition \ref{existence-2} which says that the Lie algebra automorphism of $\fk$ given by $\Ad(s_{0,0})$ defines an automorphism of the Lie group $K$.
There is the following method to check whether CR geometries corresponding to CR algebras $(\fk,\fq)$ are symmetric (in our sense) and to construct the normal extensions $(\alpha,i)$ that describe them.
\\[1mm]
$(1)$ We consider $\fl=\fk \cap \fq\cap \bar{\fq}$ and $\mathcal{H}_{eL}=\fk/\fl \cap (\fq+\bar{\fq})/(\fq\cap \bar{\fq})$, where $\bar{\fq}$ is the subalgebra conjugated to $\fq$ in $\fk_{\mathbb{C}}$.
\\[1mm]
$(2)$ We choose a complex basis of $\mathcal{H}_{eL}$. This choice defines a Lie algebra homomorphism $\fl\to \frak{csu}(p,q)$ and the following facts hold:
\\$(2a)$
If this homomorphism is not injective, then CR geometry is flat (we will discuss this situation later).
\\$(2b)$
If this homomorphism is injective and the CR geometry is symmetric, then it coincides with the restriction of $\alpha$ to $\fl$ for some normal extension $(\alpha, i)$ describing the CR geometry.
\\$(2c)$
If this homomorphism is injective and the CR geometry is not symmetric, then the homomorphism corresponds only to associated graded map corresponding to restriction of $\alpha$ to $\fl$ for some normal extension $(\alpha, i)$ describing the CR geometry.
\\[1mm]
$(3)$ Each choice of representatives (in $\fk$) of the complex basis of $\mathcal{H}_{eL}$ from (2) together with a choice of an element of $\fk$ complementary to $\mathcal{H}_{eL}$ allows us to define
\\$(3a)$ a linear map $\alpha$ of the form $(\ref{form-prop5})$ from Proposition \ref{prop5} for (at this point) unknown linear maps $a,A,\Rho_1,\Rho_2$,
\\$(3b)$ a linear map $\tau: \wedge^2 \fk\to \frak{su}(p+1,q+1)$ given for all $X,Y\in \fk$ by the formula $$\tau(X,Y):=[\alpha(X),\alpha(Y)]-\alpha([X,Y]),$$
$(3c)$ a linear map $\nu: \fk\to \fk$ such that $\nu$ equals to
\begin{itemize}
\item
$-\id$ on the representatives (in $\fk$) of complex basis of $\mathcal{H}_{eL}$, and
\item $\id$ on the element of $\fk$ complementary to $\mathcal{H}_{eL}$ and on $\fl$.
\end{itemize}
Moreover, we consider only the choices that satisfy the equivalent conditions of the following statement.
\begin{prop*}
The map $\nu$ is a Lie algebra automorphism of $\fk$ if and only if the components $$(\mathbb{R}\oplus \alpha(\fl))\otimes \mathbb{C}^n\to \mathbb{R}\oplus \frak{csu}(p,q)\oplus \mathbb{R}^*, \ \ \ \mathbb{C}^n\otimes \mathbb{C}^n\to \mathbb{C}^{n}\oplus \mathbb{C}^{n*}$$ of $\tau$ vanish for all linear maps $a,A,\Rho_1,\Rho_2$.
\end{prop*}
\begin{proof}
A consequence of the formula for $\tau$ is that $\nu$ is a Lie algebra automorphism of $\fk$ if and only if $$\Ad(s_{0,0})\tau(\nu(X),\nu(Y))=\tau(X,Y),\ \ \ \Ad(s_{0,0})[\alpha(\nu(X)),\alpha(\nu(X))]=[\alpha(X),\alpha(Y)]$$ hold for all $X,Y\in \fk$. If $\alpha$ is of the form $(\ref{form-prop5})$, then $$\Ad(s_{0,0})[\alpha(\nu(X)),\alpha(\nu(X))]=[\alpha(X),\alpha(Y)]$$ holds for all $X,Y\in \fk$ and all linear maps $a,A,\Rho_1,\Rho_2$, and $$\Ad(s_{0,0})\tau(\nu(X),\nu(Y))=\tau(X,Y)$$ holds for all $X,Y\in \fk$ if and only if the claimed components vanish.
\end{proof}
\noindent
$(4)$ There are the following possibilities for the choice in the step $(3)$.
\\ $(4a)$
If there is no choice such that $\nu$ is a Lie algebra automorphism of $\fk$, then the CR geometry corresponding to CR algebra $(\fk,\fq)$ is not symmetric.
\\$(4b)$ If there is a choice such that $\nu$ is a Lie algebra automorphism of $\fk$, then the CR geometry corresponding to CR algebra $(\fk,\fq)$ is symmetric if and only if $\nu$ induces Lie group automorphism of $K$ and $L$ is contained in fixed point set of $\nu$.
\\[1mm]
$(5)$
We require from now that CR geometry corresponding to CR algebra $(\fk,\fq)$ is symmetric. The remaining step is to determine the choice of an element of $\fk$ complementary to $\mathcal{H}_{eL}$ and $i:L\to P$ such that $(\alpha, i)$ is a normal extension describing the CR geometry.
We know that there is a choice such that $(\Ad(\exp(z))\circ \alpha,i')$ is an extension for some $z\in \mathbb{R}^*$, where the Lie group homomorphisms $i': L\to P$ is induced by (adjoint) action of $L$ on $\mathcal{H}_{eL}$ and $\alpha(\fl)$. Thus it suffices to check the vanishing of components $$\alpha(\fl)\otimes \mathbb{R}\to \frak{su}(p+1,q+1), \ \ \ \mathbb{C}^n\otimes \mathbb{C}^n\to \mathbb{R}$$ of $\tau$ for all linear maps $a,A,\Rho_1,\Rho_2$.
The condition (5) of Lemma \ref{lem4} provides linear equations that determine uniquely the linear maps $a,A,\Rho_1,\Rho_2$ for which the extension $(\alpha,i)$ is normal.
\subsection{Example of non--flat symmetric CR geometries}
Consider a Lie group $E(2)=\R^2 \oplus \frak{so}(2)$ of isometries of Euclidean plane. There is the following normal extension $(\alpha,i)$ of $(E(2),\{\id\})$ to $(PSU(1,2),P)$ of the form
\begin{align} \label{neE2}
&\alpha
\left( \begin {matrix}
0&0&0\\
{x\over 2}&0&-X_1
\\ X_2 &X_1&0\end{matrix} \right)= \left( \begin{matrix}
{ix \over 16} & -\frac{5}{16}X_1-\frac{3i}{16}X_2 &-{\frac {15ix}{256}}\\
X_1+iX_2&-{ix \over 8}& \frac{5}{16}X_1-\frac{3i}{16}X_2 \\ ix&-X_1+iX_2& {ix \over 16}\end{matrix} \right),
\end{align}
where the choice of the basis of the Lie algebra of $\R^2 \oplus \frak{so}(2)=\langle x,X_2\rangle\oplus \langle X_1\rangle$ reflects the convention from Section \ref{sec4.2}, i.e., $(x,(X_1,X_2))$ are the distinguished coordinates from Theorem \ref{thm2}.
Indeed, since $i$ is trivial and
\begin{align*}
&\tau\left((x,(X_1,X_2)),(y,(Y_1,Y_2))\right)= \\
&\left( \begin{matrix}
0& \frac {3i}{32} yX_1-\frac {3}{32}yX_2
-\frac {3i}{32} xY_1+\frac {3}{32} xY_2&0\\ 0&0& \frac {3i}{32} yX_1+\frac {3}{32}yX_2
-\frac {3i}{32} xY_1-\frac {3}{32} xY_2\\
0&0& 0\end{matrix} \right)
\end{align*}
holds for the linear map $\tau$ determining the curvature $\kappa$,
it follows that $(\alpha,i)$ is a normal extension describing a non--flat symmetric CR geometry.
In fact, any linear invertible linear map
$B: \mathbb{R}^2\oplus \frak{so}(2)\to \mathbb{R}\oplus \mathbb{C}$ defines a CR algebra $(\fk,\fq)$ for $$\fq=B^{-1}_{\mathbb{C}}(\mathbb{C}^*\oplus \fp_{\mathbb{C}})$$
and we ask the following question: Which maps $B$ correspond to non--equivalent non--degenerate symmetric CR geometries of hypersurface type on the Lie group $E(2)$ of isometries of Euclidean plane?
We give the answer to this question (using the algorithm from previous section and \cite[Lemma 3.5]{HG-dga}) in the following statement.
\begin{prop*} \label{sedm}
The normal extension $(\alpha, i)$ of the form $(\ref{neE2})$ describes the unique (up to equivalence) non--degenerate symmetric CR geometry of hypersurface type on the Lie group $E(2)$.
\end{prop*}
\begin{proof}
Consider an invertible linear map $B: \mathbb{R}^2\oplus \frak{so}(2)\to \mathbb{R}\otimes \mathbb{C}^2$.
The construction of the objects from the algorithm is clear in this case.
We need to find for which maps $B$ the components
$$\mathbb{R}\otimes \mathbb{C}\to \mathbb{R}\oplus \frak{csu}(1)\oplus \mathbb{R}^*, \ \ \ \mathbb{C}\otimes \mathbb{C}\to \mathbb{C}\oplus \mathbb{C}^*,\ \ \ \ \mathbb{C}\otimes \mathbb{C}\to \mathbb{R}$$
of $\tau$ vanish for all linear maps $a,A,\Rho_1,\Rho_2$. In fact, this provides three equations on the entries of the matrix $B$ that can be solved explicitly. In the standard bases of $\mathbb{R}^2\oplus \frak{so}(2)$ and $\mathbb{R}\otimes \mathbb{C}$, the inverses of matrices $B$ that satisfy these equations define the following subvariety:
\begin{align} \label{var}
\left( \begin{matrix}
{p_1p_2 -p_3p_4 \over 2} & p_5 & {p_5p_3 -2p_6 \over 2}\\
p_6&p_4&p_2\\
0&p_1&p_3
\end{matrix} \right).
\end{align}
Thus it remains to check the action of morphisms from \cite[Lemma 3.5]{HG-dga} that determine which extensions define equivalent CR geometries. In particular, there are
\begin{itemize}
\item four--dimensional Lie group of derivations of $\mathbb{R}^2\oplus \frak{so}(2)$ that in addition contains the homothethies, and
\item two--dimensional Lie subgroup that forms center of $CSU(p,q)$.
\end{itemize}
We compute that the induced action of these morphisms on the six--dimensional variety $(\ref{var})$ is transitive and the matrix
$$
\left( \begin{matrix} {1 \over 2}&0&0\\
0&0&1\\
0&1&0
\end{matrix} \right)
$$
corresponds to the extension $(\ref{neE2})$.
\end{proof}
\section{Metrizability and CR embeddings}
In this section, we always consider the $K$--invariant Weyl connection $D$ corresponding to a normal extension $(\alpha,i)$ describing a homogeneous CR geometry $(M,\mathcal{H},J)$ that satisfies $i(L)\subset U(p,q)$, $i(s)=s_{0,0}$ and
$$\alpha((x,X)+\fl)=\left(
\begin{matrix}
aix&\Rho_1(X)& \Rho_2(x) \\ X&Ax&-I\Rho_1(X)^* \\ x&-IX^*&aix
\end{matrix}
\right)+\alpha(\fl).$$ Moreover, we always assume $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$, where $L^0$ is the component of identity of $L$. This gives almost no restriction, because this condition is always satisfied on the symmetric CR geometry on the covering $K^0/L^0\to K/L$.
\subsection{Distinguished metrics compatible with the CR geometry}
The symmetric bilinear form $h$ generally does not define a pseudo--Riemannian metric on $\mathcal{H}$, because there is no natural way, how to measure the length of elements of $TM/\mathcal{H}$. The situation is different, if there is a Weyl connection preserving not only the decomposition $\mathcal{H} \oplus \ell$, but also a non--zero vector field $r$ in $\ell$. Such Weyl connection is called \emph{exact} and the vector field $r$ is called the \emph{Reed field}. Equivalently, each exact Weyl connection corresponds to the \emph{contact form} $\theta$ that annihilates $\mathcal{H}$ and satisfies $\theta(r)=1$ for the Reeb field $r$.
If there is an exact Weyl connection, then $\theta\circ h$ is a pseudo--Riemannian metric on $\mathcal{H}$. This metric is compatible with the CR--structure, because the form $h$ satisfies $h(J\xi,J\nu)=h(\xi,\nu)$ for all sections $\xi,\nu$ of $\mathcal{H}$. The exact Weyl connection preserves this metric and the Reeb field can be used to construct a pseudo--Riemannian metric on $TM$, for which the connection is a metric connection. This metric is usually called a \emph{Webster metric}. However, the Webster metric neither has to exist nor has to be compatible with the symmetries. Therefore, if we want to find a metric compatible with the CR geometry that is preserved by all symmetries, we need to show that the distinguished Weyl connection $D$ is exact.
\begin{thm*} \label{thm3}
Let $K$ be the Lie group generated by all symmetries of a non--flat symmetric CR geometry $(M,\mathcal{H},J)$. Suppose that $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$. The distinguished Weyl connection $D$ is exact and furthermore, there exists
\begin{itemize}
\item a $K$--invariant contact form $\theta$,
\item a $K$--invariant pseudo--Riemannian metric $\bar g:= \theta\circ h$ on $\mathcal{H}$, and
\item a $K$--invariant Webster metric $g:=\theta\circ h+\theta\otimes \theta$ on $TM$
\end{itemize}
such that
\begin{enumerate}
\item $D\bar g=0, Dg=0$,
\item $g|_{\mathcal{H}}=\bar g$ and the Reeb field of $D$ is orthogonal to $\mathcal{H}$ and has length $1$,
\item choosing the Reeb field of $D$ as a trivialization of $TM/\mathcal{H}\otimes \C$, the pseudo--Riemannian metric $\bar g$ on $\mathcal{H}$ coincides with the real part of the Levi form up to a constant multiple,
\item the symmetry at $x$ is linear in geodesic coordinates of $D$ at $x$, reverses the directions of $\mathcal{H}_x$ and preserves the direction of the Reeb field of $D$ at $x$.
\end{enumerate}
\end{thm*}
\begin{proof}
The image of $\alpha$ is contained in $\R \oplus\C^n \oplus \frak{u}(p,q) \oplus\C^{n*} \oplus \R^*$ and thus $\gamma$ describing the corresponding $K$--invariant Weyl connection has values in $\underline{\ad}(\frak{u}(p,q))$. Furthermore, the assumption $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$ implies that $i(L)\subset U(p,q)$ and therefore the maps $\underline{\ad}^{-1}\circ \gamma$ and $i$ satisfy all the conditions of \cite[Theorem 1.4.5]{parabook}. This means that the Weyl connection $D$ is an associated connection to a $K$--invariant principal connection on the bundle $K\times_{i(L)}U(p,q)\to K/L$. Therefore it is an exact Weyl connection, because its holonomy is contained in $U(p,q)$. The remaining claims then follow from general theory.
\end{proof}
In the Riemannian signature, Theorem \ref{thm3} particularly allows to compare symmetric CR geometries (in our sense) with the symmetric CR geometries in the sense of \cite{KZ}, because we have found a metric compatible with CR geometry that is preserved by all symmetries.
\begin{thm*} \label{thm4}
Suppose that $p=0$. Then each non--flat symmetric CR geometry is covered by a symmetric CR geometry in the sense of \cite{KZ}, where the covering is a CR map that intertwines the symmetries.
\end{thm*}
\subsection{CR embeddings}
Consider the fiber bundle $K\times_{i} CSU(p,q)/U(p,q)\to K/L$. If $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$ holds, then this bundle is trivial, i.e.,
$$K\times_{i} CSU(p,q)/U(p,q)=K/L\times \mathbb{R}.$$
Let us prove the following statement:
\begin{thm*}\label{embed}
Let $K$ be the Lie group generated by all symmetries of a non--flat symmetric CR geometry $(M,\mathcal{H},J)$. Suppose that $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$. Then:
\begin{enumerate}
\item the manifold $K/L\times \mathbb{R}$ is a complex manifold, and
\item the inclusion $K/L\to K/L\times \mathbb{R}$ given as a zero section is a CR embedding.
\end{enumerate}
\end{thm*}
\begin{proof}
We need some more details from the theory of Cartan geometries from \cite[Sections 1.5.13 and 3.1.2]{parabook} to proceed with the proof. First, there is a natural complement of $\frak{u}(p,q)$ in $\frak{csu}(p,q)$ given by so--called grading element, which is the unique element $Z\in \frak{csu}(p,q)$ acting by $-2$ on $\mathbb{R}$, $-1$ on $\C^n$, $0$ on $\frak{csu}(p,q)$, $1$ on $\C^{n*}$ and $2$ on $\R^*$. Furthermore, there is a Cartan connection on $K/L\times \mathbb{R}$ induced by CR geometry, where we identify $\mathbb{R}$ (via $\exp$) with the multiples of the grading element $Z$. Then the Weyl connection $D$ provides a reduction of this Cartan connection to $U(p,q)$, which allows us to identify the tangent space of $K/L\times \mathbb{R}$ with the fiber bundle $(K\times \mathbb{R})\times_{i} (\R \oplus\C^n\oplus \frak{csu}(p,q)/\frak{u}(p,q))$. We can extend the complex structure on $\C^n$ to $\R \oplus\C^n\oplus \frak{csu}(p,q)/\frak{u}(p,q)$ by declaring $\R$ to be the imaginary part of $\mathbb{C}$ and the multiples of the grading element in $ \frak{csu}(p,q)/\frak{u}(p,q)$ to form the real part of $\mathbb{C}$. This definition is clearly $U(p,q)$--invariant (and thus $K$--invariant) and defines an almost complex structure $J$ on $K\times_{i} CSU(p,q)/U(p,q)$.
Let us compute the Nijenhuis tensor $[\xi,\eta]-[J\xi,J\eta]+J([J\xi,\eta]+[\xi,J\eta])$ of $J$ for $\xi,\eta\in T(K/L\times \mathbb{R})$. For each $x\in K/L\times \mathbb{R}$, there are vector fields $\tilde \xi, \tilde \eta$ such that $\tilde \xi(x)=\xi(x), \tilde \eta(x)=\eta(x)$ and that the element $[\tilde \xi,\tilde \eta](x)$ is identified with the element $$[X,Y]-[\alpha(X+\fh),\alpha(Y+\fh)]+\alpha([X+\fh,Y+\fh]) \mod \frak{u}(p,q)\oplus\C^{n*} \oplus \R^*,$$
where $\xi(x), \eta(x)$ are identified with $X,Y\in \R \oplus\C^n\oplus \frak{csu}(p,q)/\frak{u}(p,q)$. This identification can be obtained using the technique analogous to \cite[Proposition 3.1.8]{parabook} for $T(K/L\times \mathbb{R})$ instead of $T(K/L)$. Indeed, the Cartan connection in the background remains the same and we only need to restrict ourselves to normal Weyl connections that coincide with $D$ at $x$ and project the results given by the Cartan connection to $T(K/L\times \mathbb{R})$ instead of $T(K/L)$.
However, $$[X,Y]-[\alpha(X+\fh),\alpha(Y+\fh)]+\alpha([X+\fh,Y+\fh])=[X,Y] \mod \frak{u}(p,q)\oplus\C^{n*} \oplus \R^*$$
due to the condition (5) from Lemma \ref{lem4}. Therefore we have $$([\xi,\eta]-[J\xi,J\eta]+J([J\xi,\eta]+[\xi,J\eta]))(x)=[X,Y]-[JX,JY]+J([JX,Y]+[X,JY]).$$
Let us now discuss possible values of this expression for all possible incomes:
\begin{itemize}
\item For $X,Y\in \C^n$ we have $[X,Y]-[JX,JY]+J([JX,Y]+[X,JY])=0$.
\item For $X\in \C^n$ and $Y=JZ\in \R$ we have $[X,Y]-[JX,JY]+J([JX,Y]+[X,JY])=[JX,Z]-J([X,Z])=0$.
\item For $X\in \C^n$ and $Y=Z$ we have $[X,Y]-[JX,JY]+J([JX,Y]+[X,JY])=[X,Z]+J([JX,Z])=0$.
\item For $X=JZ\in \R$ and $Y=Z$ we have $[X,Y]-[JX,JY]+J([JX,Y]+[X,JY])=[JZ,Z]+[Z,JZ]=0$.
\end{itemize}
The remaining possibilities vanish trivially.
Thus the complex structure is integrable. Then the zero section is a CR embedding, because it is a closed orbit.
\end{proof}
In holomorphic coordinates on $U\subset K/L\times \mathbb{R}$, the hypersurface $K/L\cap U \subset \C^{n+1}$ may be described as a zero set of a function $F: U\to \mathbb{R}$. Theorem \ref{embed} and Lemma \ref{dve-symetrie} provide distinguished holomorphic coordinates, in which the function $F$ has a specific form.
\begin{cor*}
Let $K$ be the Lie group generated by all symmetries of a non--flat symmetric CR geometry $(M,\mathcal{H},J)$. Suppose that $\Ad(L^0)|_{\fh/\fl}=\Ad(L)|_{\fh/\fl}$. Then for every point $x\in M$, there is a holomorphic coordinate system on $U\subset K/L\times \mathbb{R}$ centred at $x$ such that the function $F(z,w)$ defining $M$ satisfies $F(z,w)=F(-z,w)$.
\end{cor*}
\section{Locally flat CR symmetric spaces} \label{sec6}
Locally flat CR geometries are always locally symmetric (in our sense). Therefore, the following question appear:
Which local symmetries are globally defined?
The answer depends on the topology of the manifold. We show on series of examples that various situations are possible. There are two sources of examples that we study here that are related to flag manifolds. The first series of examples follows the construction from \cite{ja-CEJM,conf} that we apply to CR geometries. The second series of examples involves CR geometries on orbits of real forms in flag manifolds from \cite{AMN}.
\subsection{Non--homogeneous symmetric CR geometries}
Let us apply the construction from \cite{ja-CEJM,conf} to CR geometries. We start with the standard model $PSU(p+1,q+1)/P$. Consider the CR manifold $M:=PSU(p+1,q+1)/P - \{\langle u \rangle,\langle v \rangle\}$, where $u,v \in \C^{n+2}$ are arbitrary non--zero null vectors of $m$. The group $K(u,v)$ of CR transformations of the flat CR geometry on $M$ has two connected components. The identity component of $K(u,v)$ is the intersection of stabilizers of $\langle u \rangle$ and $\langle v \rangle$. The other component contains the elements that swap $\langle u \rangle$ and $\langle v \rangle$. We check whether there is a symmetry at each $K(u,v)$--orbit on $M$. Let us emphasize that if all symmetries at one point of a $K(u,v)$--orbit preserve or swap the points $\langle u\rangle$ and $\langle v\rangle$ then all symmetries at all points of the whole orbit have the same property.
The orbits of the action of $K(u,v)$ on $M$ are characterized by the fact that the action preserves
\begin{itemize}
\item the subspace $\langle u,v\rangle$, and
\item the (non)--isotropy with respect to the Hermitian form $m$.
\end{itemize}
Moreover, the action of $K(u,v)$ on $\langle u,v\rangle$ depends on whether $\langle u,v\rangle$ is isotropic subspace or not.
\begin{exam} \label{exam1}
Assume that $p,q > 1$, i.e., not the Riemannian signature. Consider the CR manifold $M=PSU(p+1,q+1)/P - \{\langle u \rangle,\langle v \rangle\}$ for arbitrary non--zero null vectors $u,v \in \C^{n+2}$ isotropic with respect to $m$, i.e., $m(u,v)=0$. Then $\langle u,v\rangle -\{\langle u\rangle ,\langle v \rangle \}$ consists of a single orbit of $K(u,v)$. Furthermore, $K(u,v)$--orbits of points $\langle x\rangle $ such that $x \notin \langle u,v\rangle-\{\langle u\rangle ,\langle v \rangle \}$ depend only on the (non)--isotropy of $x$ with respect to $u, v$.
We show that there exist symmetries at all points of each orbit of $K(u,v)$. Instead of fixing $\langle u\rangle,\langle v \rangle$ and discussing symmetries at various points $\langle x \rangle$, we fix the point $\langle x \rangle$ as the point $\langle e_0 \rangle$ given by the first vector of the standard basis $e_0,\dots,e_{n+1}$ of $\C^{n+2}$ and we choose admissible $\langle u \rangle$ and $\langle v \rangle$ such that $\langle e_0 \rangle$ lies in the correct orbit. Then we find all symmetries at $\langle e_0 \rangle$. Let us recall that all symmetries of the standard model at the origin $\langle e_0 \rangle$ are of the form
\begin{align*}
s_{Z,z}=\left(
\begin{matrix}
-1&-Z& iz+\frac12ZIZ^* \\ 0&E&-IZ^* \\ 0&0&-1
\end{matrix}
\right)
,\end{align*}
where $Z=(z_1, \dots, z_n) \in \C^{n*}$ and $z \in \R^*$ are arbitrary, and involutive are those satisfying $z=0$.
\\[1mm]
(1) Let us start with the orbit corresponding to the case $m(e_0,u)\neq 0$ and $m(e_0,v)\neq 0$. We choose $u=ie_0+\sqrt{2}e_1-ie_{n+1}$ and $v=ie_0-\sqrt{2}e_{n}+ie_{n+1}$. Direct computation gives that there is exactly one symmetry $s_{Z,z}$, where $Z=(-i\sqrt{2},0, \dots, 0,-i\sqrt{2})$ and $z=0$. This symmetry is involutive and swaps $\langle u \rangle$ and $\langle v \rangle$. There is no symmetry preserving them.
\\[1mm]
(2) Let us now consider the orbit for the case $m(e_0,u)=0$ and $m(e_0,v)\neq 0$ (which is the same orbit as the orbit for the case $m(e_0,u)\neq 0$ and $m(e_0,v)=0$). We choose $u=e_1+e_{n}$ and $v=ie_{n+1}$. Direct computation gives that there is exactly one symmetry $s_{Z,z}$, where $Z=(0, \dots,0)$ and $z=0$. This symmetry is involutive and preserves $\langle u \rangle$ and $\langle v \rangle$. There is no symmetry swapping them.
\\[1mm]
(3) The next possibility is the orbit for the case $m(e_0,u)=m(e_0,v)=0$ and $e_0 \in \langle u,v \rangle$. We choose $u=e_{1}+e_{n}$ and $v=e_0+e_{1}+e_{n}$. Computation gives that there are (many) symmetries $s_{Z,z}$, where $Z=(z_1, \dots,z_{n})$ with components $z_k=a_k+ib_k$ for $k=1, \dots, n$ satisfies $a_1+a_{n}+1=0$ and $b_1+b_{n}=0$, and $a_k,b_k$ for $k=2, \dots, n-1$ and $z$ are arbitrary.
All these symmetries swap $\langle u \rangle$ and $\langle v \rangle$, and there are no symmetries preserving them. In particular, there are also non--involutive symmetries for $z\neq 0$.
\\[1mm]
In fact, this covers all possible orbits for the case $p=1$ or $q=1$, i.e., the Lorentzian signature.
In the other cases, there is one more orbit.
\\[1mm]
(4) Consider the orbit for the case $m(u,e_0)=m(v,e_0)=0$ and $e_0 \notin \langle u,v \rangle$. We choose $u=e_1+e_{n}$ and $v=e_2+e_{n-1}$. Computation gives that there are (many) symmetries $s_{Z,z}$, where $Z=(z_1, \dots,z_{n})$ satisfies $a_1+a_n=0$, $b_1+b_n=0$, $a_2+a_{n-1}=0$ and $b_2+b_{n-1}=0$ and $a_k,b_k$ for $k=3, \dots, n-2$ and $z$ are arbitrary. All these symmetries preserve $\langle u \rangle$ and $\langle v \rangle$ and there are no symmetries swapping them. In particular, there are also non--involutive symmetries for $z\neq 0$.
\\[1mm]
Altogether, symmetries at different orbits behave differently. Therefore, there is no smooth system of symmetries. In particular, there is no pseudo--Riemannian metric compatible with the CR geometry that would be preserved by some symmetry at every point.
\hspace{\fill} $\Diamond$
\end{exam}
Let us show that this principle does not work if we remove two points corresponding to non--isotropic vectors.
\begin{exam}
Consider the manifold $M=PSU(p+1,q+1)/P - \{\langle u \rangle,\langle v\rangle\}$ for arbitrary non--zero null vectors $u,v \in \C^{n+2}$ that are non--isotropic for $m$, i.e. $m(u,v) \neq 0$.
We choose $u=e_{n+1}$ and $v=e_0+\sqrt{2}e_1+(1+i)e_{n}$. Computation gives that there is no symmetry at $\langle e_0\rangle$ preserving or swapping $\langle u \rangle$ and $\langle v\rangle$. Let us remark that the component of identity of $K(u,v)$ is isomorphic to the group $CSU(p,q)$ and $K(u,v)$ does not act transitively on $\langle u,v\rangle-\{\langle u\rangle ,\langle v \rangle \}$.
\hspace{\fill} $\Diamond$
\end{exam}
\subsection{Flat homogeneous symmetric CR geometries and orbits of real forms in complex flag manifolds}
It follows from Lemma \ref{lem4} that an extension $(\alpha,i)$ of $(K,L)$ to $(PSU(p+1,q+1),P)$ corresponds to flat CR geometry if and only if $\alpha$ is a Lie algebra homomorphisms. Therefore, we can present examples of extensions describing flat homogeneous symmetric CR geometries just by specifying the Lie subalgebra of $\frak{su}(p+1,q+1)$ that coincides with the image of $\alpha$. In general, the group $K$ does not have to contain symmetries. Moreover, symmetries do not have to preserve $\alpha(\fk)$. This is satisfied if $K$ is the group generated by symmetries or the full group of CR automorphisms.
\begin{exam}
Consider the orbits of $PSp(1,1)$ on $\mathbb{C}P^4$ given by inclusion $PSp(1,1)\subset PSp(4,\mathbb{C})\subset PGl(4,\mathbb{C})$. Due to the isomorphisms $PSp(1,1)\cong PO(1,4)$, these orbits can also be interpreted as orbits in the flag manifold of 2--planes in quadric in $\mathbb{C}P^5$. There is a normal extension given by identifying the following Lie subalgebra of $\frak{su}(2,2)$ with the image of $\alpha(\frak{sp}(1,1)):$
$$
\left( \begin{matrix} l_1+il_2&i X_2+i
l_5- X_1+l_4&iX_4+il_5-X_3+
l_4&i \left(l_3+x \right) \\ iX_2+X_1&-{i \over 2} \left( 2l_2+2x+ l_3 \right) &-{il_3 \over 2
}-l_1&iX_2+il_5+X_1-l_4
\\ iX_4+ X_3&{il_3 \over 2}-l_1
&-{i \over 2} \left( 2l_2-2x-l_3 \right) &-iX_4-i
l_5-X_3+l_4\\ ix&iX_2-X_1&X_3-iX_4&- l_1+il_2\end{matrix}
\right),
$$
where $l_i$--entries generate the Lie algebra of the stabilizer $L=CSO(2)\rtimes S^2\mathbb{R}^2$ of a point in the minimal orbit. Precisely, $\langle l_1,l_2\rangle=\frak{cso}(2)$ and $\langle l_3,l_4,l_5\rangle=S^2\mathbb{R}^2$.
\end{exam}
\begin{exam}
Consider the orbits of $PSp(4,\mathbb{R})$ on $\mathbb{C}P^4$ given by inclusion $PSp(4,\mathbb{R})\subset PSp(4,\mathbb{C})\subset PGl(4,\mathbb{C})$. Due to the isomorphisms $PSp(4,\mathbb{R})\cong PO(2,3)$, these orbits can again be interpreted as orbits in the flag manifold of 2--planes in quadric in $\mathbb{C}P^5$. There is a normal extension given by identifying the following Lie subalgebra of $\frak{su}(2,2)$ with the image of $\alpha(\frak{sp}(n+2,\mathbb{R})):$
$$
\left( \begin {matrix} l_1+i l_2&X_1-i
X_2+l_4+il_5&X_3-i X_4-l_4-i
l_5&i \left(l_3+x \right) \\ i X_2+ X_1&-{i \over 2} \left( 2 l_2-2x- l_3 \right) &
l_1-{il_3 \over 2}&-iX_2+i l_5- X_1- l_4
\\ iX_4+ X_3& l_1+{il_3 \over 2}
&-{i \over 2} \left( 2l_2+2x+l_3 \right) &iX_4+il_5+X_3-l_4\\ ix&iX_2-X_1&X_3-iX_4&- l_1+il_2\end{matrix}
\right),
$$
where $l_i$--entries generate the Lie algebra of the stabilizer $L=CSO(2)\rtimes S^2\mathbb{R}^2$ of a point in $5$--dimensional orbit (which is not minimal). Precisely, $\langle l_1,l_2\rangle=\frak{cso}(2)$ and $\langle l_3,l_4,l_5\rangle=S^2\mathbb{R}^2$.
\end{exam}
In both examples, $\fk$ is simple and $\fq$ is a parabolic subgalgebra of $\fk_\mathbb{C}$. In \cite{AMN}, the authors discuss which CR algebras $(\fk,\fq)$ for simple Lie algebras $\fk$ and parabolic subalgebras $\fq$ of the complexification of $\fk$ are symmetric. In fact, they correspond to orbits of real forms in complex flag varieties. Therefore, symmetric CR algebras of these types generalize bounded symmetric domains.
We show that if CR algebra $(\fk,\fq)$ for a simple Lie algebra $\fk$ and a parabolic subalgebra $\fq$ of the complexification $\fk_{\mathbb{C}}$ of $\fk$ corresponds to non--degenerate symmetric CR geometry of hypersurface type, then the geometry is necessarily flat. Therefore, we can use the results of \cite{O} to classify all possible cases.
\begin{prop*}
Let $(\fk,\fq)$ be a CR algebra such that $\fk$ is simple and $\fq$ is a parabolic subalgebra of $\fk_{\mathbb{C}}$ and the corresponding CR geometry is non--degenerate and of hypersurface type. Then the following statements hold:
\begin{enumerate}
\item If the CR geometry is symmetric, then the CR geometry is flat.
\item If the CR geometry is flat, then it corresponds to one of the following possibilities:
\begin{enumerate}
\item $\fk=\frak{su}(p+1,q+1)$ and $\fl=\fp$,
\item $\fk=\frak{sp}({p+1\over 2},{q+1\over 2})$ and $\fl=\frak{co}(2)\oplus \frak{sp}({p-1 \over 2},{q-1\over 2}) \oplus \mathbb{R}^{2}\otimes\mathbb{R}^{n-2*}\oplus S^2\mathbb{R}^2$,
\item $\fk=\frak{sp}(n+2,\mathbb{R})$ and $\fl=\frak{co}(2)\oplus \frak{sp}(n-2,\mathbb{R}) \oplus \mathbb{R}^{2}\otimes\mathbb{R}^{n-2*}\oplus S^2\mathbb{R}^2$, where $\mathbb{R}^{2}\otimes\mathbb{R}^{n-2*}\oplus S^2\mathbb{R}^2$ is the positive part of the parabolic subalgebra corresponding to the stabilizer of a Lagrangian 2--plane in $\mathbb{R}^{n+2}$.
\end{enumerate}
\item If the CR geometry is flat and $\fk_{\mathbb{C}}=\frak{sp}(n+2,\mathbb{C})$ is the full Lie algebra of complete infinitesimal automorphism and $n>2$, then the corresponding CR geometry is not symmetric.
\item If the CR geometry is flat and corresponds to $(2n+1)$--dimensional orbit of the real form of $\frak{sp}(n+2,\mathbb{C})$ in $\mathbb{C}P^{n+2}$, then the corresponding CR geometry is symmetric if and only if $n=2$ or the orbit is minimal, i.e., if $\fk\neq \frak{sp}(n+2,\mathbb{R})$.
\end{enumerate}
\end{prop*}
\begin{proof}
If such symmetric CR geometry is non--flat, then $K$ has to be generated by symmetries and it follows from \cite[Theorem 3.1]{HG-dga} that the complexification of $\fl$ does not contain a Cartan subalgebra of $\fk_{\mathbb{C}}$.
On the other hand, if $\fk$ is simple and $\fq$ is a parabolic subalgebra of $\fk_{\mathbb{C}}$, then $\fq\cap \bar{\fq}$ contains a Cartan subalgebra of $\fk_{\mathbb{C}}$. This is a contradiction and therefore, the claim (1) holds.
If such symmetric CR geometry is flat, then $\fk_{\mathbb{C}}$ is isomorphic to a Lie subalgebra of $\frak{sl}(n+2,\mathbb{C})$, $\fq=\fk_\mathbb{C}\cap (\mathbb{C}^*\oplus \fp_{\mathbb{C}})$ is a parabolic subalgebra of $\fk_\mathbb{C}$ and $\fk_{\mathbb{C}}/\fq=\frak{sl}(n+2,\mathbb{C})/ (\mathbb{C}^*\oplus \fp_{\mathbb{C}}).$ All such cases are classified in \cite{O} and it follows that $\fk_{\mathbb{C}}=\frak{sl}(n+2,\mathbb{C})$ or $\fk_{\mathbb{C}}=\frak{sp}(2n+2,\mathbb{C})$. The first case corresponds to the standard model. The remaining cases correspond to the symmetric pair $(\frak{sl}(n+2,\mathbb{C}), \frak{sp}(2n+2,\mathbb{C}))$. Real forms of this symmetric pair are well--known and correspond to suitable inclusions $\frak{sp}({p+1\over 2},{q+1\over 2})\subset \frak{su}(p+1,q+1)$ or $\frak{sp}(n+2,\mathbb{R})\subset \frak{su}(n+1,n+1)$. If such inclusion provides an extension, then it is unique (up to equivalence). Therefore, it suffices to show that the cases in the claim (2) correspond to non--degenerate CR geometries of hypersurface type. This follows from the fact that $\frak{co}(2)\cong \mathbb{C}$ defines a complex structure on the whole $\fk/\fl$ with the exception of the trace part of $(S^2\mathbb{R}^2)^*$.
If
$\fk_{\mathbb{C}}=\frak{sp}(n+2,\mathbb{C})$ is the full Lie algebra of complete infinitesimal automorphism and the corresponding CR geometry is symmetric, then $\Ad(s_{0,0})$ induces an involution of $\frak{sp}(n+2,\mathbb{R})$. It follows from the description of $\fl$ that the stabilizer has to have the form $\frak{gl}(2,\mathbb{C})\oplus \frak{sp}(n-2,\mathbb{C})$. Therefore the claim (3) follows from the fact that this stabilizer does not appear in the classification of simple symmetric spaces if $n>2$.
Since $\frak{sp}(n+2,\mathbb{C})$ is maximal subalgebra of $\frak{sl}(n+2,\mathbb{C})$, the only possibility for the orbit to be symmetric is to be equivalent to standard model which is compact. Since the orbit is compact if only if the orbit is minimal, the claim (4) follows. It follows from \cite{AMN} that the orbit is minimal if and only if $\fk\neq \frak{sp}(n+2,\mathbb{R})$.
\end{proof}
| 129,445
|
TITLE: Examples and further results about the order of the product of two elements in a group
QUESTION [49 upvotes]: Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?
Wikipedia says "not much":
There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.
On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:
What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?
Wikipedia then states:
If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$
which is easy to prove, but not very effective. So:
What are some similar results about the order of a product, perhaps with some additional hypotheses?
REPLY [0 votes]: A different sort of example of elements of finite order having an infinite order product:
Consider the permutation $f$ on $\mathbb Z^+$ that sends $2i+1$ to $2i+2$ and $2i+2$ to $2i+1$ for every non-negative integer $i$.
Now consider the permutation $g$ also on $\mathbb Z^+$ that sends $1$ to $1$, and sends $2i+2$ to $2i+3$ and $2i+3$ to $2i+2$ for every non-negative integer $i$.
Note both these permutations have order $2$, however $g \circ f$ sends every odd integer $k$ to $k+2$ and is thus of infinite order.
| 120,242
|
It happened around 3:15pm in the 4200 block of Lavender at Cavalcade in northeast Houston.
Police said two cousins were moving a camper trailer. One was standing behind the trailer and the other was driving, pulling the trailer.
The driver did not realize his cousin was there and backed out, pinning him between the guardrail of another truck and the trailer.
The injured cousin -- a 65-year-old man -- was rushed to an area hospital in critical condition where he later died. His identity is pending verification by the Harris County Institute of Forensic Sciences.
Police said they consider this a tragic accident and no charges are expected.
| 165,333
|
This site uses different types of cookies, including analytics and functional cookies (its own and from other sites). To change your cookie settings or find out more, click here. If you continue browsing our website, you accept these cookies.
If you're anything like me, my LinkedIn is constantly full of empowering and thoughtful articles shared by peers who inspire me. From articles highlighting the need for equal educational resources for all students, to reading how @tessaenns motivates her team to "Alteryx Everything," it's clear that the world around me is full of like-minded thinkers who are passionate about making a positive difference.
The Women of Analytics initiative at Alteryx is just one way we're working to empower the amazing women driving the analytics industry, and I want to call out some of the other incredible organizations who are providing tools to women and girls to make sure they succeed.
Organizations that immediately come to mind are Girls who Code, working to close the gender gap in the tech space, and ForbesWoman, facilitating conversations with successful women around the world.
What organizations have inspired you in your career? Have you volunteered, mentored, or had a mentor help you get to where you are today? Share your story with us, shout out other amazing organizations, and help to inspire others by commenting below!
Maddie Johannsen
Customer Advocacy Coordinator, Alteryx
We'd love to highlight your experience and achievements in an upcoming blog spotlight! Fill out our Women of Analytics Survey for the opportunity to share your story with the Alteryx Community. The survey should take less than 15-20 minutes to complete. We also plan to leverage survey submissions to identify potential panelists for upcoming events.
Be sure to click the "Join the Group" button on the Women of Analytics page to start participating today and automatically subscribe to receive notifications when we announce new events. Want more control? You can personalize your subscriptions and notifications via My Settings.
There are many organizations who push to create equality for women. I love the work of Women who Code. I appreciate Girl Scouts, and all they do for young women.
For me, education made a huge difference in my outlook. My university experience opened my eyes. In Business School, our Business Stats program was taught by a woman. In her, I could suddenly see what sort of woman I wanted to be. When I was in school, I was judged by my work, and not as much by my gender. I saw women taking leadership roles. Women didn't have to be glorified secretaries. They were free to spread their wings and soar.
Personally, I try to provide that example on a regular basis to the women around me. I write a regular blog as part of my job. I make sure that I feature women equally. I need to see that women are succeeding. I need to know that my experiences are "normal'. So, I write the blog. I provide to the world the things that I need to help me keep pushing. Here are two recent blog posts about women doing really cool things in analytics.
Thanks for posting this @MaddieJ!
An organization that comes to mind for me, is UN Women. whose purpose is to support global standards for achieving gender equality -- They work with governments and civil society to design laws, policies, and programs/services to ensure that standards are effectively implemented and truly benefit women and girls worldwide.
UN Women initiated the HeForShe movement, inviting people around the world to stand together to create a bold, visible force for gender equality.
I also wanted to point out Twitter's effort to publicize their initiatives to build a diverse and inclusive workforce: Building a more inclusive Twitter in 2016
While I haven't seen an updated version of this for 2017, and just recently discovered their VP of Diversity and Inclusion (who published the blog) is no longer with the company, I still think it's a good example of a company trying to do the right thing. I'd love to see more businesses openly share EEO reports and progress towards diversity goals. It's pretty inspiring stuff ☺️
Leah Knowles
Community Engagement Programs Manager
Alteryx
I have found some great resources through Google's Women Techmakers program. They aren't really active on LinkedIn, but their twitter feed (@WomenTechmakers) is great!
I've been inspired by HBR (Harvard Business Review) podcasts and publications for years. They started a podcast earlier in 2018 called HBR Women at Work and it is fantastic.
I've also been inspired by the work that Alteryx and Tableau are doing with Women of Analytics and Data+Women.
There is a non-profit called Women In Data that is also focused on diversity in data science.
There are so many great resources out there!
| 350,546
|
Renderings Released of High Rise on Queens Boulevard Containing Regal Cinemas
The majority of the space, 95,000 square feet, will be dedicated to medical offices. The facilities have been designed for .....
New Renderings for Crown Heights Bedford Union Armory Redevelopment Have Been Released
Greenwood Gas Station to Be Replaced by 150-Unit Residential Building
Renderings Revealed for 1 Flatbush Avenue, Downtown Brooklyn
Reveal for ODA’s Tetris-Inspired 101 West 14th Street in Greenwich Village
Another Tall Tower Is Headed for Long Island City
MOST RECENT
New Renderings for Temple Israel’s Expansion at 112 East 75th Street on the UES
No Sign Can Replace ‘Watchtower,’ Because It Was Illegal All Along, City Claims
- New Tower Planned for Brooklyn Draws Community Ire
- Lottery Launches for 250+ Mixed-Income Apartments in East Tremont
- 270 Park Avenue Will Become the Tallest Buildings Ever Demolished
- Demolition Permits Filed for Stand-Alone Garage on Thomson Avenue
- New Renderings of Domino Sugar Factory’s Waterfront Park and Esplanade
- Renderings Revealed for 3254, 3258 Parkside Place
- The Most Affordable Neighborhoods in NYC: A 2018 Buyer’s Guide
New Historic District More Than Doubles the Landmarked Buildings in Boerum Hill
Neighbors Worry Footbridge From Montague St. to Brooklyn Bridge Park Will Bring Circus to Their Door
New Renderings for The Lower East Side’s Next Skyscraper at 259 Clinton Street
New Hunters Point Towers Will Be 55 and 32 Stories, Construction to Begin Next Year
Jehovah’s Witnesses’ DUMBO Hotel Will Become Housing for Formerly Homeless
Permits Filed to Build Rooftop Soccer Field Atop Three-Story Building in Red Hook
Resident Concerned Over Cracks Caused by Excavation Work in Lot Next Door
Coney Island Shore Theater to Become Seven-Story Hotel
Plans by the NYC Department of Transportation (DOT) have been submitted to the Landmark Preservation Commission calling for alterations of the Brooklyn Bridge. .....<<
| 205,779
|
Search
You searched for: Political Geography United States Remove constraint Political Geography: United States Topic Foreign Policy Remove constraint Topic: Foreign
2. The Pathologies of Power: Fear, Honor, Glory, and Hubris in U.S. Foreign Policy, Christopher J. Fettweis
- Publication Date: 01-2015
- Content Type: Journal Article
- Journal: Political Science Quarterly
- Institution: Academy of Political Science
- Abstract: Books about improving U.S. foreign policy are a dime a dozen. But in The Pathologies of Power, Christopher Fettweis offers an unusual take on what he sees as the subpar foreign policy performance of the planet's sole superpower. Fettweis claims that U.S. foreign policy is driven by four pathological beliefs—fear, honor, glory, and hubris—that lead to poor policymaking. The book devotes a chapter to each of the beliefs that Fettweis contends account for foreign policy disasters like the Iraq war and the Vietnam war. - See more at:
- Topic: Foreign Policy
- Political Geography: United States, America
- Author: Edward Rhodes
- Publication Date: 01-2015
- Content Type: Journal Article
- Journal: Political Science Quarterly
- Institution: Academy of Political Science
- Abstract: “History,” Winston Churchill is reported to have observed, “is written by the vic¬tors.” The losers, if they are lucky enough to avoid vilification, are airbrushed out. When it comes to our understanding of American foreign policies of the first four decades of the twentieth century, the history-writing victors have, for the most part, been liberal internationalists. Democrats and Republicans alike, in the wake of the Second World War, concluded that the task of making the world safe for America demanded active, global U.S. politico-military engagement. In the name of liberal international institutions, Washington's “Farewell” injunctions against entangling alliances would be consigned to the waste bin of quaint anachronisms.- See more at:
- Topic: Foreign Policy, Education, War
- Political Geography: United States, Washington
-
- Publication Date: 03-2015
- Content Type: Journal Article
- Journal: Journal of Military and Strategic Studies
- Institution: Centre for Military and Strategic Studies
- Abstract: In America and the Rogue States, Thomas Henriksen lays out the relationships that exist, and have existed, between America and the states that made up George W. Bush's 'Axis of Evil.' Henriksen outlines the history of the interactions between the United States and North Korea, pre-invasion Iraq, and Iran, and through this draws out a number of themes. He also shows that the ways the relationships have played out are highly situational and there is no one-size-fits-all solution. In the last chapter, Henriksen explores American relationships with a number of states that were either once considered rogue or could become rogue, like Libya, Syria, and Cuba, referring to them as either “lesser rogues” or “troublesome states.” These states have remained “a puzzle for US foreign policy” (1) and are characterized by three things: autocratic governance, sponsorship of terrorism, and pursuit of weapons of mass destruction (WMD). There is no clear definition provided by Henriksen for what can be considered a rogue state, making it difficult to judge what other states, if any, could be considered rogue. Henriksen seems to arbitrarily decide who is rogue and who is not: Cuba is a rogue state, while Myanmar is merely troublesome. Instead of synthesizing a clear definition of the term, something that could then be applied to other states in order to judge their 'rogueness,' Henriksen uses the Bush administration's criteria (the term itself was coined by President Bill Clinton in a 1994 speech in Brussels), which was outlined in the National Security Strategy of 2002 (NSS-2002). These were “brutality toward their own people; contempt for international law; determination to acquire weapons of mass destruction (WMD); advanced military technology; sponsorship of terrorism; rejection of human rights values; and hatred for the United States and 'everything it stands for'”. The use of the NSS-2002 definition allows for the 'Axis of Evil' to fit neatly into the term, which constitutes a problem of tautology, at least for the Bush administration. Further compounding this was that, according to Henriksen at least, the administration was set on going to war in Iraq prior to assuming office. This creates a situation in which it is hard to determine whether the idea of rogue states was created to justify this desire, or it informed the desire prior to the administration taking office.
- Topic: Foreign Policy, Terrorism
- Political Geography: United States, America, North Korea, Libya
- Author:
-
| 45,502
|
\begin{document}
\maketitle
\begin{abstract}
The main aim of this paper is to characterize ideals $I$ in the power series ring $R=K[[x_1,\ldots,x_s]]$ that are finitely determined up to contact equivalence by proving that this is the case if and only if $I$ is an isolated complete intersection singularity, provided dim$(R/I) > 0$ and $K$ is an infinite field (of arbitrary characteristic). Here two ideals $I$ and $J$ are contact equivalent if
the local $K$--algebras $R/I$ and $R/J$ are isomorphic. If $I$ is minimally generated by $a_1,\ldots,a_m$, we call $I${ \em finitely contact determined} if it is contact equivalent to any ideal $J$ that can be generated by $b_1,\ldots,b_m$ with $a_i - b_i \in \langle x_1,\ldots,x_s\rangle^k$ for some integer $k$.
We give also computable and semicontinuous determinacy bounds.
The above result is proved by considering left--right equivalence on the ring
$M_{m,n}$ of $m\times n$ matrices $A$ with entries in $R$ and we show that the Fitting ideals of a finitely determined matrix in $M_{m,n}$ have maximal height, a result of independent interest. The case of ideals is treated by considering 1-column matrices. Fitting ideals together with a special construction are used to prove the characterization of finite determinacy for ideals in $R$.
Some results of this paper are known in characteristic 0, but they need new (and more sophisticated) arguments in positive characteristic partly because the tangent space to the orbit of the left-right group cannot be described in the classical way. In addition we point out several other oddities, including the concept of specialization for power series, where the classical approach (due to Krull) does not work anymore. We include some open problems and a conjecture.
\end{abstract}
\section{Introduction}\label{introduction}
Throughout this paper let $K$ denote a field of arbitrary characteristic and
\[R:=K[[{\bf{x}}]]=K[[x_1,\ldots, x_s]]\]
the formal power series ring over $K$ in $s$ variables with maximal ideal $\mathfrak{m} = \langle x_1, \ldots , x_s \rangle$. We denote by
\[M_{m,n}:=Mat(m,n, R)\]
the set of all $m\times n$ matrices ($m$ rows and $n$ columns) with entries in $R$.
Two matrices $A, B \in M_{m,n}$ are called {\it left--right equivalent}, denoted $A\mathop\sim\limits^{G} B$, if they belong to the same orbit of the left--right group
\[G:=(GL(m,R)\times GL(n,R)^{op}) \rtimes \mathcal{R},\]
\noindent where $G^{op}$ is the opposite group of a group $G$ and
$\mathcal{R}:=Aut(R)$. The group $G$ acts on $M_{m,n}$ by
\[(U, V, \phi, A)\mapsto U\cdot\phi(A)\cdot V,\]
with $U\in GL(m,R), \ V\in GL(n,R), \ \phi\in Aut(R)$ and if $A=[a_{ij}({\bf x})]$ then
$\phi(A)=[a_{ij}\left(\phi({\bf x})\right)]$ with $\phi({\bf x})=\left(\phi(x_1),\ldots,\phi(x_s) \right)$.
A matrix $A\in M_{m,n}$ is said to be {\it $G$ $k$--determined} if for each matrix $B\in M_{m,n}$ with $B-A\in \mathfrak{m}^{k+1} M_{m,n}$ we have $B\mathop\sim\limits^{G} A$, $\mathfrak{m} = \langle x_1,\ldots,x_s\rangle$ the maximal ideal of $R$.
$A$ is called {\it finitely $G$--determined} if there exists a positive integer $k$ such that it is $G$ $k$--determined.
Finite determinacy for analytic map-germs and ideals with respect to various equivalence relations has been intensively studied by e.g. \cite{Hi65}, \cite{Tou68}, \cite{Mat68}, \cite{Gaf79}, \cite{Ple80}, \cite{Dam81}, \cite{Wal81}, \cite{BdPW87}, \cite{CS97}, and many more, mainly with the aim to find sufficient conditions for finite determinacy.
In \cite{BK16}, the authors study finite determinacy for matrices with entries in the convergent power series ring $\C\{{\bf x}\}$, without giving determinacy bounds. In \cite {Pha16} and \cite {GP16} the authors started the study of finite determinacy for matrices of power series over fields of arbitrary characteristic. The case of one power series, i.e. $m=n=1$, is classical over the complex and real numbers. It was treated over a field of arbitrary characteristic in \cite {GK90} for contact equivalence and in \cite {BGM12} for right and contact equivalence.
Let us give a rough overview of the main results, for more detailed statements we refer to the main body of the paper.
For $A\in \mathfrak{m}\cdot M_{m,n}$ we define the following submodules of $M_{m,n}$,
\begin{align*}
\tilde T_A(GA)&:=\langle E_{m, pq}\cdot A\rangle +\langle A\cdot E_{n, hl}\rangle+ \mathfrak{m}\cdot\left\langle\frac{\partial A}{\partial x_\nu}\right\rangle \hskip 6pt{\text{resp.}}\\
\tilde T^e_A(GA)&:=\langle E_{m, pq}\cdot A\rangle +\langle A\cdot E_{n, hl}\rangle+ \left\langle\frac{\partial A}{\partial x_\nu}\right\rangle,
\end{align*}
and call $\tilde T_A(GA)$ resp. $\tilde T^e_A(GA)$ the {\it tangent image} resp. {\em extended tangent image} at $A$ to the orbit $GA$.
Here $\langle E_{m, pq}\cdot A\rangle$ is the $R$-submodule generated by $E_{m, pq}\cdot A$, $p,q=1,\ldots,m$, with $E_{m,pq}$ the $(p,q)$-th canonical matrix of $Mat(m,m,R)$ (1 at place $(p,q)$ and 0 else) and $\left\langle\frac{\partial A}{\partial x_\nu}\right\rangle$ is the $R$-submodule generated by the matrices $\frac{\partial A}{\partial x_\nu} =\left [\frac{\partial a_{ij}}{\partial x_\nu} ({\bf x})\right], \nu = 1, \ldots, s$. The submodules $\tilde T_A(GA)$ and $\tilde T^e_A(GA)$ were introduced in \cite{GP16} and it is proved in \cite[Proposition 2.5]{GP16} that $\tilde T_A(GA)$ is the image of the tangent map of the orbit map $G \to GA$. Moreover, the following is proved in the same paper, with $\order(A)$ the minimum of the orders of the entries of $A$.
\begin{Theorem}\rm(\cite[Theorem 3.2]{GP16})\label{GP16}
\begin{enumerate}
\item Let $A=[a_{ij}]\in \mathfrak{m}\cdot M_{m,n}$. If there is an integer $k\ge 0$ such that
\begin{align*}
\mathfrak{m}^{k+2}\cdot M_{m,n}\subset \mathfrak{m}\cdot \tilde T_A(GA)\tag{1.1}\label{main},
\end{align*}
then $A$ is $G$ $(2k-\order(A)+2)$-determined. Moreover, \eqref{main} holds
for some $k$ iff $\tilde T_A(GA)$ (equivalently $\tilde T^e_A(GA)$) is of finite $K-$codimension in $M_{m,n}$.
\item If $\characteristic(K)= 0$ then the condition \eqref{main} for some $k$ is {\em equivalent} to $A$ being finitely $G$--determined. \end{enumerate}
\end{Theorem}
\begin{Remark} \rm
Condition \eqref{main} gives a {\em sufficient} criterion for finite determinacy of $A$ in any characteristic. We do not know whether, for arbitrary $m$ and $n$, the finite codimension of the tangent image $\tilde T_A(GA)$ in $M_{m,n}$ is necessary in positive characteristic. On the other hand, the finite codimension of the tangent space $T_A(GA) \subset M_{m,n}$ is easily seen to be necessary in any characteristic (cf. \cite [Lemma 2.11] {GP16}).
If $\characteristic(K)= 0$ then $\tilde T_A(GA)$ coincides $T_A(GA)$ (\cite[Lemma 2.8] {GP16}) and hence condition \eqref{main} for some $k$ is necessary and sufficient.
In general we have $\tilde T_A(GA)\subset T_A(GA)$, and it is important to notice that the inclusion can be strict in positive characteristic (see Example 2.9 in \cite{GP16}).
\end{Remark}
\begin {Conjecture} \rm
Statement {\it 2.} of Theorem \ref{GP16} holds also in positive characteristic.
\end {Conjecture}
One of the main aims of this paper is to prove the conjecture for
matrices in $M_{m,1}$ (c.f. Theorem \ref{column matrix}, where we give also determinacy bounds).
\begin{Theorem} \label{th1.2}
{\it For $A \in \mathfrak{m} \cdot Mat(m,1,R)$ with $K$ infinite, the following are equivalent:
\begin{enumerate}
\item $A$ is finitely left--right determined.
\item $\dim_K\left(M_{m,1}\big/\tilde T_A(GA)\right) <\infty.$
\end{enumerate} }
$K$ infinite is only needed for (1) $\Rightarrow$ (2) and for $m<s$ (= number of variables).
\end{Theorem}
Theorem \ref{th1.2} was proved by Mather in \cite[Theorem 3.5] {Mat68} for real and complex analytic maps and in \cite{CS97} in a slightly more general setting for char($K$) = 0, using classical methods.
The proof of the interesting direction {\em 1.} $\Rightarrow$ {\em 2.} for $\characteristic(K)> 0$ resisted previous attempts and is more involved as one might think. It requires a special deformation (unknown to us for arbitrary $m,n$) and the above mentioned result about Fitting ideals. \\
Consider now the ideals $I$ resp. $J$ generated by the entries of $A$ resp. $B \in M_{m,1}$. Then $A$ is left--right equivalent to $B$ iff $I$ and $J$ are {\it contact equivalent} (c.f. Proposition \ref{contact equivalence}).
Recall that $R/I$ is called a {\it complete intersection} if $\dim(R/I) = s - \mng(I)$ with $\mng(I)$ the minimal number of generators of $I \subset R$. The complete intersection $R/I$ is called an {\it isolated complete intersection singularity (ICIS)} if the ideal of the singular locus,
$I+I_m\left(\left[\frac{\partial f_i}{\partial x_j}\right]\right)$, contains a power of the maximal ideal, where $\{f_1,\ldots, f_m\}$ is a minimal set of generators of $I$ and $I_t(A)$ denotes the ideal of $t\times t$ minors of the matrix $A$.
Theorem \ref{th1.2} implies
\begin {Theorem} {\label{theo1.5}}
Let $I$ be a proper ideal of $R$.
\begin {enumerate}
\item If $\dim(R/I) = 0$ then $I$ is finitely contact determined.
\item If $\dim(R/I) >0$ and $K$ is infinite, then $I$ is finitely contact determined if and only if $R/I$ is an isolated complete intersection singularity.
\end {enumerate}
\end {Theorem}
See Theorem \ref{geomchar} for a more detailed statement which contains also the determinacy bound $(2\tau(I)-\order(I)+2)$, where $\tau(I)$ is the Tjurina number of $I$.
Finite determinacy of $I$ implies that $R/I$ is {\em algebraic}, i.e. $R/I \cong R/J$ where $J$ is generated by polynomials. But it is much stronger since by a result of Artin (\cite[Theorem 3.8]{Ar69}) every isolated singularity $R/I$ (not necessarily an {\em ICIS}) is algebraic.
\begin {Problem} \rm
The assumption that $K$ is infinite in 2. is only needed to show that a finitely determined $I$ defines an $ICIS$. It is due to our method of proof but we do not know whether it is necessary. For hypersufaces however we show in Theorem \ref{hypersurface} that it is not necessary.
\end {Problem}
To prove our results we derive a necessary condition for finite $G$-determinacy for matrices in section \ref{necessary}, Theorem \ref{height}, by showing that the Fitting ideals of a finitely $G$-determined matrix have maximal height. For this we use the specialization of ideals depending on parameters, which was introduced by W. Krull and then extended and systematically studied by D.V. Nhi and N.V. Trung for finitely generated modules over polynomial rings and localizations thereof (cf. \cite{NT99}, \cite{NT00}).
\begin {Problem} \rm
A satisfactory theory for specialization of ideals in power series rings depending on parameters has not yet been developed.
We show in Example \ref{osgood} that a straightforward generalization of specialization from (localization of) polynomial rings to power series rings does not work. In Remark \ref{specialization} we propose an approach which is reasonable for uncountable fields $K$ e.g. for $\R, \C$. For a concrete open problem see Problem \ref{pr2.6}.
\end {Problem}
In section \ref {special case} we study $G$--equivalence for 1-column matrices and use the results of section \ref{necessary} to prove Theorem \ref{th1.2}.
We need and prove a semicontinuity result for modules over a power series ring depending on parameters (Proposition \ref{semi-continuity}) which should be well known, but for which we could not find a reference.
In section \ref {complete intersections} we apply the results of section \ref {special case} to contact equivalence for ideals and prove Theorem \ref{geomchar}. Finally we prove in Theorem \ref{hypersurface} that a power series $f \in K[[{\bf x}]]$ is finitely contact (resp. right) determined iff the Tjurina number (resp. the Milnor number) of $f$ is finite, also in positive characteristic.
\section{A necessary finite determinacy criterion by Fitting ideals}\label{necessary}
In this section, we establish a necessary condition for finite $G$-determinacy of matrices in $M_{m,n}$.
Without loss of generality, we assume that $n\le m$.
For a matrix $A\in Mat(m,n, P)$, $P$ a commutative Noetherian ring, and an integer $t$, let $I_t(A)$ denote the ideal of $P$ generated by all $t\times t$ minors of $A$, also called the ($m-t$)-th Fitting ideal of the cokernel of the map
$A: R^n \to R^m$.
Let ${\bf u}= (u_1,\ldots, u_r)$ be a new set of indeterminates. For an ideal $I \subset K({\bf u})[{\bf x}]$ and $a\in K^r$ the ideal
$$I_a:=\ \{f(a,{\bf x})\ | \ f({\bf u},{\bf x})\in I\cap K[{\bf u}][{\bf x}] \},$$
is called the {\em specialization} of $I$ and for an ideal $J \subset K[{\bf u},{\bf x}]$,
$$J^e :=J\cdot K({\bf u})[{\bf x}]$$
denotes the {\em extension} of $J$.
We say that a property holds for {\em generic} $a \in K^r$ if there exists a non-empty Zariski open set $U\subset K^r$ such that the considered property holds for all $a\in U$.
\begin{Lemma}\label{height of determinantal ideal}
Let $K$ be infinite and $M=[g_{ij}({\bf u},{\bf x})]\in Mat(m, n, K[{\bf u}, {\bf x}])$. For $a\in K^r$ set $M_a=[ g_{ij}(a,{\bf x})]\in Mat(m, n,K[{\bf x}])$.
Then for generic $a\in K^r$ and for all $t=1,\ldots, n$ the following holds:
\begin{enumerate}
\item [(i)] $I_t(M_a)=(I_t(M)^e)_a,$
\item [(ii)] if $I_t(M_a)$ is a proper ideal then $I_t(M)^e$ is proper too,
\item [(iii)] if $I_t(M_a)$ is proper then $\height \left(I_t(M_a)\right)=\height (I_t(M)^e), \, \depth \left(I_t(M_a)\right)= \depth (I_t(M)^e).$
\end{enumerate}
\end{Lemma}
\begin{proof}
$(i)$ For all $a\in K^r$, $I_t(M_a)$ is the ideal of $K[{\bf x}]$ generated by $d_1^{(t)}(a, {\bf x}),\ldots, d_{l_t}^{(t)}(a, {\bf x})$, where $d_1^{(t)}({\bf u}, {\bf x}), \ldots, d_{l_t}^{(t)}({\bf u}, {\bf x})$ are the $t\times t$ minors of $M$. On the other hand, for each generator $f^{(t)}({\bf u},{\bf x})$ of the ideal $I_t(M)^e\cap K[{\bf u},{\bf x}]$ of $K[{\bf u},{\bf x}]$, there is a polynomial $b^{(t)}({\bf u})\in K[{\bf u}]\setminus \{0\}$ such that $b^{(t)}\cdot f^{(t)}\in I_t(M)$. Therefore, for $a\in K^r$, which is not a zero of any $b^{(t)}({\bf u})$, we have that $(I_t(M)^e)_a$ is generated by $d_1^{(t)}(a, {\bf x}), \ldots, d_{l_t}^{(t)}(a, {\bf x})$. Hence, the first assertion holds. \\
$(ii)$ If $I_t(M)^e$ is not proper then $K[{\bf x}]=(I_t(M)^e)_a= I_t(M_a)$ by
$(i)$, contradicting the assumption.\\
$(iii)$ The other statements follow from $(i), (ii)$ and \cite[Theorem 3.4 (ii) and Corollary 4.4]{NT99}.
\end{proof}
Specialization and extension is also used in the proof of the following theorem which shows that we can modify a matrix with polynomial entries by adding polynomials of arbitrary high order such that the Fitting ideals of the modified matrix have maximal height. The proof was communicated to the authors by Ng{\^o}~Vi\d{\^e}t Trung in \cite{Tru15} for $t=n$. Using his arguments, we prove for arbitrary $t$:
\begin{Theorem}\label{generic determinantal ideals}
Let $A=[f_{ij}]\in Mat(m, n, K[{\bf x}])$, $f_{ij}\in \langle x_1,\ldots, x_s\rangle\cdot K[{\bf x}]$ with $K$ infinite. For $N\ge 1$ let $B=[g_{ij}]\in Mat(m, n, K[{\bf x}])$ be the matrix with entries of the form
$$g_{ij}=\sum\limits_{k=1}^s a_{ijk}x_k^N, \ a_{ijk}\in K.$$
Then, for $t\in \{1,\ldots, n\}$ and generic $(a_{ijk})$ in $K^{mns}$, we have with $m_t:=(m-t+1)(n-t+1)$
$$\height (I_t(A+B))=\min\{s,m_t\}.$$
If $m_t < s$ then $K[{\bf x}] / I_t(A+B)$ is Cohen-Macaulay and moreover, if $K$ is algebraically closed, then $I_t(A+B)$ is a prime ideal.
\end{Theorem}
\begin{proof}
For $i=1,\ldots, m$, $j=1,\ldots, n$, let
$$F_{ij}=\sum\limits_{k=1}^s u_{ijk}x_k^N+f_{ij},$$
where ${\bf u}=\{u_{ijk}\mid i=1,\ldots, m, j=1,\ldots,n, k=1,\ldots, s\}$ is a set of new indeterminates. Set $S=K[{\bf u}, {\bf x}]$ and let
$$M=[F_{ij}]\in Mat(m, n, S).$$
The main work is done to prove the following
\vskip 4pt
{\bf Claim:} For all $t=1,\ldots, n$, $I_t(M)^e$ is a proper ideal of $K({\bf u})[{\bf x}]$ and
$$\height (I_t(M)^e)=\min\{s, m_t\}.$$
Indeed, the first assertion of the claim follows from Lemma \ref{height of determinantal ideal}. Now we prove the second statement. Fix $t\in\{1,\ldots, n\}$. For every $k=1,\ldots, s$ and for all $i, j$
we have that in $S\left[\frac{1}{x_k}\right]$ $$\frac{1}{x_k^N}F_{ij}=u_{ijk}+\frac{1}{x_k^N}\left(\sum\limits_{h\ne k} u_{ijh}x_h^N+f_{ij}\right).$$
Therefore, the elements $\frac{1}{x_k^N}F_{ij}$, $k=1,\ldots, s$, $i=1,\ldots,m$, $j=1,\ldots, n$ are algebraically independent over $K\left[{\bf u'}, {\bf x}, \frac{1}{x_k}\right]$, where ${\bf u'}={\bf u}\smallsetminus\{u_{ijk} \, |\, i=1,\ldots, m, j=1,\ldots, n\}$. This means that $\frac{1}{x_k^N}\cdot M$ is a generic matrix over the ring $K\left[{\bf u'}, {\bf x}, \frac{1}{x_k}\right]$ (generic in the sense that the entries are indeterminates, not to be confused with generic points). Note that
\[K[{\bf u'}] [{\bf x}]\left[\frac{1}{x_k}\right]\left[\frac{1}{x_k^N} F_{ij} | \ i= 1,\ldots ,m, \ j=1,\ldots, n\right] = S\left[\frac{1}{x_k}\right].\]
It is well known that the determinantal ideals of a generic matrix are prime and have maximal height (cf. e.g. \cite[(2.13) and (5.18)]{BV88}).
Hence, $I_t\left(\frac{1}{x_k^N}\cdot M\right)$, the ideal of $S\left[\frac{1}{x_k}\right]$ generated by all $t\times t$ minors of $\frac{1}{x_k^N}\cdot M$, is a prime ideal of the height $m_t$.
Since
$I_t(M)\cdot S\left[\frac{1}{x_k}\right]=I_t\left(\frac{1}{x_k^N}\cdot M\right),$ we have that
$I_t(M)\cdot S\left[\frac{1}{x_k}\right]$ is a prime ideal of height $m_t$.
This implies that $I_t(M)$ has a prime component, say $P^{(t)}_k$, which does not contain $x_k$, and all other associated primes must contain $x_k$.
Let now $k'\in\{1,\ldots, s\}$ and $k'\ne k$. By a similar argument, $I_t(M)\cdot S\left[\frac{1}{x_k\cdot x_{k'}}\right]$ is a prime ideal of $S\left[\frac{1}{x_k\cdot x_{k'}}\right]$ . It follows that
$I_t(M)$ has a prime component, say $P^{(t)}_{k,k'}$, which does not contain $x_k$ and $x_{k'}$, and all other associated primes must contain the product $x_k\cdot x_{k'}.$ Therefore, $P^{(t)}_k= P^{(t)}_{k'}$ for all $k\ne {k'}$. Let $P_t$ denote this prime component. Then $P_t$ does not contain any $x_k$, $k=1,\ldots, s$, and all other associated primes of $I_t(M)$ must contain all $x_1,\ldots, x_s.$ Let $Q_t$ be the intersection of all primary components of $I_t(M)$ whose associated primes contain $x_1,\ldots, x_s.$ Note that $Q_t=S$ if such components do not exit. Then
$$I_t(M)=P_t\cap Q_t.$$
Moreover, since $P_t$ is the only associated prime of $I_t(M)$ which does not contain any $x_k$,
$$\height (P_t)=\height \left(I_t(M)\cdot S\left[\frac{1}{x_k}\right]\right)=m_t.$$
We have
$$I_t(M)^e=\left(P_t\right)^e\cap (Q_t)^e.$$
Let first $t$ be such that $s\le m_t.$ We consider two cases:
{\textit {Case 1:}} $\left(P_t\right)^e$ is strictly contained in $ K({\bf u})[{\bf x}]$. Then
$$\height \left(\left(P_t\right)^e\right)=\height (P_t)=m_t\ge s,$$
so that $\height \left(\left(P_t\right)^e\right)=s$.
On the other hand, since all other associated primes of $I_t(M)$ contain $x_1,\ldots, x_s$ if they exist, all associated primes of $\left(Q_t\right)^e$ have the height $s$. Hence, in this case $\height (I_t(M)^e)=s.$
\vskip 5pt
{\textit {Case 2:}} $(P_t)^e=K({\bf u})[{\bf x}]$. Then
$I_t(M)^e=(Q_t)^e$
so that $\height (I_t(M)^e)=s.$
\vskip 5pt
\noindent Let now $t$ be such that $s>m_t$. In this case $P_t$ has the least height among the associated primes of $I_t(M)$. Hence, $\height (I_t(M))=\height (P_t)=m_t$. By the generic perfection \cite{HE71}, $I_t(M)$ is a perfect ideal of $S$, and $S/I_t(M)$ is a Cohen-Macaulay ring. Hence,
all associated primes of $I_t(M)$ have the same height \cite{HE71} and $Q_t$ does not exist, showing that $I_t(M)=P_t$. Hence $I_t(M)$ is a prime ideal and
$$\height (I_t(M)^e)=\height( P_t)=m_t.$$
This finishes the claim.
Now let $r=mns$ be the number of the new indeterminates ${\bf u}=u_{ijk}$. By Lemma \ref{height of determinantal ideal} we get for generic $a\in K^r$,
$\height (I_t(M_a))=\height (I_t(M)^e)=\min\{s, m_t\}$,
and if $m_t<s$ then
$K[ {\bf x}] / I_t(M_a)$ is Cohen-Macaulay.
Let $K$ be algebraically closed and $t$ such that $m_t<s$. To see that $I_t(M_a)$ is a prime ideal for generic $a$ note that $I_t(M)^e$ is a prime ideal in $K(\bf u) [\bf x]$. In fact, let $fg \in I_t(M)^e$
with $f, g \in K(\bf u)[\bf x]$. Clearing denominators, we may assume $f, g \in K[\bf u, \bf x]$. There exists $0 \neq h(u) \in K[\bf u]$ s.t. $hfg \in I_t(M)$. Since $h(u) \notin I_t(M)$ (otherwise $1 \in I_t(M)^e$, a contradiction) we get $fg \in I_t(M)$ and hence $f$ or $g$ is in $I_t(M)$, showing that $I_t(M)^e$ is prime. Now we get the result from \cite[Proposition 3.5 ]{NT99}. \end{proof}
\begin{Problem}\rm {\label{pr2.6}}
Does an analogous statement as in Theorem \ref{generic determinantal ideals} hold for matrices with entries in $K[[\bf x]]$ instead of $K[\bf x]$? Using Remark \ref {specialization} with a new definition of specialization in power series rings, we expect this for $K$ uncountable and $a$ outside the union of countably many nowhere dense subvarieties.
\end {Problem}\rm
The above proof does not work for power series, since the straightforward definition of specialization $I_a:=\{f(a,{\bf x})\mid f({\bf u},{\bf x})\in I\cap K[{\bf u}][[{\bf x}]]\}$ for an ideal $I$ in $K({\bf u})[[{\bf x}]]$ may be 0 for all $a$ even if $I \neq 0$, as the following example shows. This example is due to Osgood and was also used by Gabrielov in his counter example to the nested approximation theorem in the analytic case (cf. \cite {Ro13}):
\begin {Example} \label{osgood} \rm{
Consider the morphism
\[ \hat{\varphi} : \C[[u,x_1, x_2]] \to \C[[y_1,y_2]], \ u \mapsto y_1, \, x_1 \mapsto y_1y_2, \, x_2 \mapsto y_1y_2 \cdot \exp(y_2),\]
and let $\varphi : \C(u)[[x_1, x_2]] \to \C[[y_1,y_2]]$ be given by the same assignment.
It is shown in \cite {Os16} that $\ker (\hat{\varphi}) = 0$. However, $I:= \ker(\varphi) \neq 0$ since it contains $x_2 - x_1 \cdot \exp(x_1/u)$, while $I \cap \C[u][[x_1, x_2]] \subset \ker (\hat{\varphi})=0$} and hence $I_a =0$ for all $a \in \C$.
\end {Example}
\begin{Remark} \label {specialization} \rm
The problem is, that elements in the ring $K({\bf u})[[{\bf x}]]$ may have infinitely many denominators. A reasonable definition for a specialization in this ring is
$$I_a := \{f(a,{\bf x})\ | \ f({\bf u},{\bf x})\in I \text {, no denominator of } f \text { vanishes at } a\}. $$
With this definition, which can be easily extended to finitely generated submodules of $(K({\bf u})[[{\bf x}]])^p$, many properties of $I$ hold also for $I_a$ (e.g. the Hilbert-Samuel functions coincide) if $a$ is contained in the complement of countably many closed proper subvarieties of $K^r$. For this to be useful we must assume that $K$ is uncountable.
We do not pursue this here, since we need only the specialization for ideals $I \subset K({\bf u})[{\bf x}]$.
\end{Remark}
\medskip
The following theorem provides a necessary condition for finite determinacy for matrices with entries in $R = K[[x_1, \ldots, x_s]]$ with respect to the group $G$.
\begin{Theorem}{\label{height}}
Let $A=[a_{ij}]\in \mathfrak{m}\cdot Mat(m,n,R)$ be finitely $G$-determined. Then the following holds:
\begin{enumerate}
\item\label{height is expected} Let $K$ be infinite. Then we have for $t\in\{1, \ldots, n\}$ with $m_t:=(m-t+1)(n-t+1)$
\[\height (I_t(A))=\min\{ s, m_t\}.\]
\item\label{height 2} For any $K$ the following holds.
(i) If $ s \geq mn$ then $\height(I_1(A))=mn$, i.e. $\{a_{ij}\}$ is an $R$-sequence.
(ii) If $s\le mn$ then $I_1(A)\supset \mathfrak{m}^k$ for some positive integer $k$, i.e. the entries of $A$ generate an $\mathfrak{m}$-primary ideal in $R$.
\end{enumerate}
\end{Theorem}
\begin{proof}
1. Assume that $A$ is $G$ $l$-determined and let $N \geq l+1$. Applying Theorem \ref{generic determinantal ideals} to $A_0=jet_l(A)$ (the truncation of the entries of $A$ after degree $l$ such that $A_0$ has polynomial entries and is $G$-equivalent to A), there is a matrix $B=[g_{ij}]\in Mat(m, n, K[{\bf x}])$ with entries of the form
$$g_{ij}=\sum\limits_{k=1}^s c_{ijk}x_k^{N},\hskip 5pt c_{ijk}\in K,$$
such that for all $t\in \{1,\ldots, n\}$ we obtain
$$\height I_t(A_0+B)=\min\{s, m_t\}.$$
The same equality holds
for the extension of the ideal $I_t(A_0+B)$ in $R$ since the morphism $K[{\bf x}]\hookrightarrow R$ is flat.
Now, since $A\mathop\sim\limits^{G}A_0\mathop\sim\limits^{G}A_0+B$, we have
$\height I_t(A)=\height I_t(A_0)=\height I_t(A_0+B)$, since Fitting ideals are invariant under $G$-equivalence.
2. This holds for any $K$ since
we do not need Theorem \ref{generic determinantal ideals} (where $K$ infinite was used) to show that
\[\height(I_1(A))=\min\{s, mn\}.\]
In fact, in case (i) we choose the entries of the matrix $B=[g_{ij}]$ to be $x_1^{N}, \ldots, x_{mn}^{N}$ and in case (ii) we choose the
first $s$ entries of the matrix $B$ to be $x_1^{N}, \ldots, x_s^{N}$ and the remaining entries 0, with $N$ sufficiently big. Then $\height \left(I_1(A_0+B)\right) = \min\{s, mn\}$, which can be seen for arbitrary $K$ as follows. Choose a global degree ordering on the variables, such that $x_1^{N}, \ldots, x_{mn}^{N}$ in case (i) and $x_1^{N}, \ldots, x_s^{N}$ in case (ii) generate the leading ideal of $I_1(A_0+B)$. By \cite [Corollary 5.3.14]{GP07} the dimension of $K[{\bf x}]/I_1(A_0+B)$ is $s-mn$ in case (i) and $0$ in case (ii). In case (i) the entries of $A_0+B$ are a regular sequence and generate a complete intersection. This is unmixed and therefore $\dim(R/I_1(A_0+B))=s-mn$. In case (ii) we get $\dim(R/I_1(A_0+B))=0$.
Since $A\mathop\sim\limits^{G}A_0+B$ we get $\height \left(I_1(A)\right)=\min\{mn,s\}$ and hence the result.
\end{proof}
\begin{Remark}\rm
The above necessary condition is of course not sufficient. For example, in any characteristic $f=x^k\in K[[x,y]]$ is not finitely contact determined by Theorem \ref{hypersurface} but $\height\left(\langle f\rangle\right)=1.$
\end{Remark}
\section{A finite determinacy criterion for column matrices}\label{special case}
Theorem \ref{height} shows that finite $G-$determinacy of matrices in $M_{m,n}$ is rather restrictive. A criterion which is at the same time necessary and sufficient for finite $G$-determinacy for arbitrary $m, n$ in positive characteristic is unknown to us.
In this section we prove such a criterion for 1-column matrices in Theorem \ref{column matrix}.
For a matrix $A=[a_1\hskip 4pt a_2\hskip 4pt\ldots\hskip 4pt a_m]^T\in M_{m,1},$
we denote by
$Jac(A):=\left[\frac{\partial a_i}{\partial x_j}\right]\in M_{m,s}$
the Jacobian matrix of the vector $(a_1,\ldots, a_m)\in R^m$ and call it the Jacobian matrix of $A$. The extended tangent image has then the following concrete description as submodule of $R^m = M_{m,1}$:
\[\tilde T_A^e\left(GA\right) = IR^m+\left(\frac{\partial a_i}{\partial x_j}\right)\cdot R^s,\]
where $I=\langle a_1,\ldots, a_m\rangle $ is the ideal in $R$ generated by the entries of $A$ and $\left(\frac{\partial a_i}{\partial x_j}\right)\cdot R^s$ is the $R$-submodule of $R^m$ generated by the columns of $Jac(A)$.
\begin{Lemma}\label{isolated and presentation matrix}
Let $A=[a_1\hskip 4pt a_2\hskip 4pt\ldots\hskip 4pt a_m]^T\in Mat(m, 1, \mathfrak{m})$ be such that $ m\le s$, where $s$ is the number of variables. Let $\Theta_{(G,A)}$ be a presentation matrix of $M_{m,1}/\tilde T^e_A(GA)$, where $\tilde T^e_A(GA)$ is the extended tangent image at $A$ to the orbit $GA$. Then
\[\sqrt{I_1(A)+I_m(Jac(A))}= \sqrt{I_m\left(\Theta_{(G,A)}\right)}
= \sqrt{\Ann_R(M_{m,1}/\tilde T^e_A(GA))}.\]
In particular, $\dim_K (M_{m,1}/\tilde T^e_A(GA)) < \infty$ iff $\dim_K\big(R/I_1(A)+I_m(Jac(A))\big)< \infty$.
\end{Lemma}
\begin{proof}
A presentation matrix of $M_{m,1}/\tilde T^e_A(GA)$ is the following two-block matrix of size $m\times( m^2+s)$
\[\Theta_{(G,A)}=
\left[ {\begin{array}{*{20}c}
{\frac{{\partial a_{1} }}{{\partial x_1 }}} & {\frac{{\partial a_{1} }}{{\partial x_2 }}} & {\ldots} & {\frac{{\partial a_{1} }}{{\partial x_s }}} \\
{\frac{{\partial a_{2} }}{{\partial x_1 }}} & {\frac{{\partial a_{2} }}{{\partial x_2 }}} & {\ldots} & {\frac{{\partial a_{2} }}{{\partial x_s }}} \\
{\vdots} & {\vdots} & {\ldots} & {\vdots} \\
{\frac{{\partial a_{m} }}{{\partial x_1 }}} & {\frac{{\partial a_{m} }}{{\partial x_2}}} & {\ldots} & {\frac{{\partial a_{m} }}{{\partial x_s }}} \\
\end{array}} \right.\left| {\begin{array}{*{20}c}
\vec{a} & \vec{0} & \vec{0} & {\ldots} & \vec{0}\\
\vec{0} & \vec{a} & \vec{0} & {\ldots} & \vec{0} \\
{\vdots} & {\vdots} & {\vdots} & {\ldots} & {\vdots} \\
\vec{0} & \vec{0} & \vec{0} & {\ldots} & \vec{a} \\
\end{array}} \right],
\]
where $\vec{a}:=(a_1,\ldots, a_m)$ and $\vec{0}:=(0,\ldots,0)\in K^m$. Since $m\le s$, it is obvious that
\[I_m\left(\Theta_{(G,A)}\right)\subset \langle a_1,\ldots, a_m\rangle + I_m(Jac(A)).\]
On the other hand we have the inclusion
\[\sqrt{\langle a_1,\ldots, a_m\rangle+I_m(Jac(A))}\subset
\sqrt{\sqrt{\langle a_1,\ldots, a_m\rangle}+\sqrt{I_m(Jac(A))}}\subset\sqrt {I_m(\Theta_{(G,A)})}.\]
since $\langle a_1,\ldots, a_m\rangle^m$ is the ideal generated by $m\times m$ minors of the right-hand block of $\Theta_{(G,A)}$ and $Jac(A)$ is a block of $\Theta_{(G,A)}$. The rest is well-known (see also \cite[Proposition 4.2]{GP16}).
\end{proof}
We show now that there exist finitely $G$-determined matrices in $Mat(m,1, R)$ with entries of arbitrary high order.
\begin{Proposition}\label{example isolated}
Let $R=K[[x_1, \ldots, x_s]]$ and $M_{m,1}=Mat(m,1, R)$ with $m \leq s$. Let $\characteristic(K)=p \geq 0$, $N\ge 2$ an integer and if $p>0$ let $p\nmid N$. Assume there are $c_{ij} \in K, i=1,\ldots, m, j=1,\ldots, s$, such that no maximal minor $m_1, \ldots, m_r$ of the $m\times s$ matrix $[c_{ij}]_{i=1,\ldots, m, j=1,\ldots, s}$ vanishes (which is always possible if $K$ is infinite or if $K$ is arbitrary and $m=1$). Set
\[ f_i := c_{i1}x^N_1+\cdots+c_{is}x^N_s, \ \ i=1,\ldots, m. \]
Then $A :=\left[f_1 \dots f_m\right]^T$ is finitely $G$-determined.
\end{Proposition}
\begin{proof}
Let $J$ be the ideal of $K[{\bf x}]$ generated by $f_1, f_2,\ldots, f_m$ and all $m\times m$ minors of $Jac(A)$. We claim that $V(J)=\{0\}$ in $\bar{K}^s$, with $\bar{K}$ an algebraic closure of $K$. Indeed, let ${\bf a}=(a_1,a_2,\ldots,a_s)\in V(J)$. Then at least $s-m+1$ components of ${\bf a}$ must be zero, since ${\bf a}$ is a zero of all products $x^{N-1}_{j_{1}} \cdot x^{N-1}_{j_{2}}\cdots x^{N-1}_{j_{m}}$, where $j_i\in\{1,2,\ldots,s\}$ for all $i=1,\ldots,m$ and $j_i\ne j_k$ for all $i\ne k$, and $m \leq s$. Without loss of generality we assume that the last $s-k$ components of ${\bf a}$ are zero for some $k\le m-1$. Then $f_i({\bf a}) = c_{i1}a_1^N+c_{i2}a_2^N+\cdots +c_{ik}a_k^N.$
Consider the homogeneous system of $m$ linear equations in $k$ variables $y_1,\ldots,y_k$
$$(H): \sum\limits_{j=1}^{k}c_{ij}y_j=0, \hskip 10pt i=1,\ldots,m.$$
Since ${\bf a}$ is also a zero of $f_1,\ldots, f_m$, it follows that $y_j=a^N_j$, $j=1,\ldots,k$, is a solution of $(H)$.
By the choice of $c_{ij}$ there must be a non-zero $k\times k$ sub-determinant of the coefficient matrix of $(H)$. This implies that $(H)$ has only the trivial solution. Therefore, ${\bf a}={\bf 0}$ and the claim follows. As a consequence, $\dim(K[{\bf x}]/J)=0$ and hence $\dim(K[[{\bf x}]]/J\cdot K[[{\bf x}]])=0$. Applying Lemma \ref{isolated and presentation matrix} and \cite[Proposition 4.2.5]{GP16}, $A$ is finitely $G$-determined.
\end{proof}
\begin{Example}\rm
$K^{ms}\smallsetminus V(m_1\cdot m_2\cdots m_r) =\emptyset$ may happen for finite $K$. If $K=\{0,1\}, m=2$ and $s=4$, then it is easy to see that at least one of the six 2-minors of $[c_{ij}]_{i=1,\ldots, 2, j=1,\ldots, 4}$ is 0 for any choice of $c_{ij} \in K$. However, if $m=1$ we can choose $c_{1j} =1$ for all $j$ and then $A=[f_1]$ is finitely $G-$determined for any $K$.
\end{Example}
We need the semi-continuity of the $K$-dimension of a 1-parameter family of finitely generated modules over a power series ring. This is well known for complex analytic power series by the finite coherence theorem. But since we could not find a reference for our situation, we give a proof here.
Let $P=K[t][[{\bf x}]]$,
${\bf x}=(x_1,\ldots,x_s)$, $K$ an arbitrary field, and $M$ a finitely generated $P$-module. For
$t_0\in K = \A^1$, set
\begin{align*}
M(t_0):&=M\mathop\otimes\limits_{K[t]}\ (K[t]/\langle t-t_0\rangle)
\cong M/\langle t-t_0\rangle\cdot M,\\
& \mathfrak{m}_{t_0}:=\langle x_1,\ldots, x_s, t-t_0\rangle\subset P.
\end{align*}
We remark that $M(t_0) \cong M_{\mathfrak{m}_{t_0}}\big/\langle t-t_0\rangle\cdot M_{\mathfrak{m}_{t_0}}$.
\begin{Proposition}\label{semi-continuity}
With the above notations, for any $o \in \A^1$ there is an open neighborhood $U$ of $o$ such that for all $t_0\in U$, we have
\[\dim_KM(t_0)\le\dim_KM(o).\]
\end{Proposition}
\begin{proof}
Without loss of generality we may assume $o=0$ and that $\dim_K M(0)<\infty.$ Then $M$ is quasi-finite but in general not finite over $K[t]$. However, we show that the restriction of $M$ to some open subset of $\A^1$ is in fact finitely generated over $K[t]$ (\textit {Step 4}), so that we can apply the semicontinuity of the rank of the presentation matrix of $M$ as $K[t]$--module.
The first steps in the proof are used to show that we can reduce to this case. We may assume that $K$ is infinite, since otherwise $\{0\}$ is open and the statement is trivially true.
Consider a primary decomposition of $\annihilator_P(M)$,
\[\annihilator_P(M)=\mathop\cap\limits_{i=1}^rQ_i\subset P.\]
For all $i=1,\ldots, r$ let $\bar{Q_i}$ denote the image of $Q_i$ under the morphism $P\to P/\langle x_1,\ldots, x_s\rangle\cong K[t]$.
\noindent{\bf\textit {Case 1:}} $\variety(\langle x_1,\ldots, x_s\rangle)\not\subset \variety(Q_i)$ for all $i=1,\ldots,r$ in $\spectrum(P)$, i.e. $Q_i\not\subset\langle x_1,\ldots, x_s\rangle$ for all $i=1,\ldots,r$. In this case $\bar{Q_i}=\langle f_i\rangle\subset K[t]$ for some $f_i \neq 0$. Since
\[U:=\A^1\smallsetminus\mathop\cup\limits_{i=1}^r\variety(f_i)\ne\emptyset,\]
for all $t_0\in U$ we have
$M_{\mathfrak{m}_{t_0}}=0$ and hence $\dim_K M(t_0)=0\le \dim_KM(0).$
\noindent{\bf\textit {Case 2:}} $\variety(\langle x_1,\ldots, x_s\rangle)\subset \variety(Q_i)$ for some $i\in\{1,\ldots,r\}$, i.e. $Q_i\subset\langle x_1,\ldots, x_s\rangle$.
\textit{Step 1:} For $Q_i\subset\langle x_1,\ldots, x_s\rangle$ we have that $Q_i$ is unique and
\begin{align*}
\sqrt{Q_i}=\langle x_1,\ldots, x_s\rangle. \tag{3.1}\label{21}
\end{align*}
Indeed, we have
$\dim_K M_{\mathfrak{m}_0}/\langle t\rangle\cdot M_{\mathfrak{m}_0}=\dim_KM(0)<\infty$, and $M_{\mathfrak{m}_0}$ is finitely generated over the local ring $P_{\mathfrak{m}_0}$. Therefore, by Krull's principal ideal theorem (\cite [Theorem B.2.1]{GLS07}) $\dim (M_{\mathfrak{m}_0})\le 1,$ which implies
$\dim P_{\mathfrak{m}_0}/(\sqrt{ Q_{i}})_{\mathfrak{m}_0}=1.$
Therefore
\[\sqrt{ Q_{i}}=\langle x_1,\ldots, x_s\rangle\]
and $\sqrt{ Q_{i}}$ is a minimal associated prime of $\annihilator_P M$. Hence $Q_i$ is unique.
{\textit {Step 2:}} Let, by \textit{Step 1}, $Q$
be the only primary component of $\annihilator_PM$ contained in $\langle x_1,\ldots, x_s\rangle$. We set
$$\bar M := M/Q \cdot M$$
and $\bar M(t_0):= \bar M\mathop\otimes\limits_{K[t]}K[t]/\langle t-t_0\rangle$ for $t_0\in K$.
Then, for $t_0=0$ we have
\begin{align*}
\dim_K \bar M(0)\le \dim_K M(0)\tag{3.2}\label{2}.
\end{align*}
{\textit{Step 3:}} Set
$W:=\variety(\bar Q)\smallsetminus (\mathop\cup\limits_{Q_i\ne Q}\variety(\bar{Q_i})) =\A^1\smallsetminus (\mathop\cup\limits_{Q_i\ne Q}\variety(\bar{Q_i}))\ne\emptyset.$
We claim that for $t_0\in W$
\begin{align*}
\dim_K\bar M(t_0)=\dim_KM(t_0)\tag{3.3}\label{4}.
\end{align*}
In fact, since $\mathfrak{m}_{t_0}\in \variety(Q)\smallsetminus \mathop\cup\limits_{Q_i\ne Q}\variety(Q_i)$,
it follows that $(Q\cdot M)_{\mathfrak{m}_{t_0}}=0.$
Hence, there is a $P$-module homomorphism
$\varphi: \bar M_{\mathfrak{m}_{t_0}}\cong M_{\mathfrak{m}_{t_0}}/(Q\cdot M)_{\mathfrak{m}_{t_0}}=M_{\mathfrak{m}_{t_0}}$ with
$\varphi\left(\langle t-t_0\rangle\cdot \bar M_{\mathfrak{m}_{t_0}}\right)=\langle t-t_0\rangle\cdot M_{\mathfrak{m}_{t_0}}$, implying the claim.
{\textit{Step 4:}}
By \eqref {21} $\sqrt{Q}=\langle x_1,\ldots, x_s\rangle$, hence $P/Q$ is a finitely generated $K[t]$-module. Thus $M/Q\cdot M$ is a finitely generated $K[t]$-module, having a presentation
\[K[t]^m\stackrel{\psi(t)}\longrightarrow K[t]^n\to \bar M\to 0.\]
This implies for $t_0\in K$ the exact sequence
$K^m\stackrel{\psi(t_0)}\longrightarrow K^n\to \bar M(t_0)\to 0,$
which yields
$\dim_K\bar M(t_0)=n-\rank \psi(t_0).$
Since $\rank \psi(t)$ is lower semi-continuous on $\A^1$, there is an open neighborhood $V$ of $0$ in $\A^1$ such that for all $t_0\in V$ we get the inequality
\begin{align*}
\dim_K\bar M(t_0)\le \dim_K\bar M(0).\tag{3.4}\label{5}
\end{align*}
With $U=W\cap V$ the assertion of the proposition follows from \eqref{4}, \eqref{5}, and \eqref{2}.
\end{proof}
We prove now our main result of this section.
\begin{Theorem}\label{column matrix}
Let $A=[f_1 \ldots f_m]^T\in \mathfrak{m}\cdot M_{m,1}$, $m \geq 1$. For $1<m<s$ assume $K$ to be infinite. Then the following are equivalent:
\begin{enumerate}
\item $A$ is finitely $G$-determined.
\item $\dim_K\left(M_{m,1}\big/\tilde T^e_A(GA)\right)=:d_e<\infty.$
\end{enumerate}
$K$ infinite is not needed for 2 $\ \Rightarrow$ 1. Moreover, if condition 2 is satisfied then $A$ is $G$ $(2d_e-\order(A)+2)$-determined.
\end{Theorem}
\begin{proof}
2. $\Rightarrow$ 1. is a consequence of Theorem \ref{GP16}, as well as 1. $ \Rightarrow$ 2. for char($K$)= 0.
Let $\characteristic(K)=p>0$ and assume that $A$ is $G$ $k$-determined.\\
{\bf{\textit {Case 1: $m < s$}}}.
By finite determinacy we may assume that $A=[f_1 \ldots f_m]^T$ is a matrix of polynomials. Let $N\in \N$ be such that $N>k$ and $p\nmid N$. Let $B=[g_1 \ldots g_m]^T\in Mat(m,1,\mathfrak{m})$, where for $i=1,\ldots,m$,
\[g_i=c_{i1}x^N_1+c_{i2}x^N_2+\cdots+c_{is}x^N_s\]
and $c_{ij}\in K$ as in Proposition \ref{example isolated} (which is possible by assumption). Consider
\[B_t=B+tA=[g_1+tf_1\hskip 5pt\ldots\hskip5pt g_m+tf_m]^T\in Mat(m,1, K[t][{\bf x}]).\]
Let $Q_t$ be the ideal of $K[t][[{\bf x}]]$ generated by the entries of $B_t$ and all $m\times m$ minors of the matrix $\left[\frac{\partial(g_i+tf_i)}{\partial x_j}\right]$. Then by Proposition \ref{semi-continuity}, there exists a neighborhood $U\subset \A^1_K$ of $0$ such that for all $t_0\in U$,
\[\dim_K\left(K[[{\bf x}]]/Q_{t_0}\right)\le\dim_K\left(K[[{\bf x}]]/Q_0\right)<\infty,\]
where the second inequality follows from the proof of Proposition \ref{example isolated}.
By Lemma \ref{isolated and presentation matrix} we get that also
$\dim_K\left(M_{m,1}\Big/\tilde T^e_{B_{t_0}}(GB_{t_0})\right)<\infty$.
Let $t_0\in U$, $t_0\ne 0$. Since $B_{t_0}\mathop\sim\limits^{G}t_0A\mathop\sim\limits^{G}A$ we obtain
\[\dim_K\left(M_{m,1}\Big/\tilde T^e_{ A}(G A)\right)=\dim_K\left(M_{m,1}\Big/\tilde T^e_{B_{t_0}}(GB_{t_0})\right)<\infty.\]
{\bf{\textit {Case 2: $m \ge s$}}}. By Theorem \ref{height}.\ref{height 2} (ii), $I=\langle f_1,\ldots, f_m\rangle $ is $\mathfrak m$-primary and since $IR^m \subset \tilde T^e_{ A}(G A)$ the claim follows.
\end{proof}
The following corollary shows a criterion for finite $G$-determinacy of a column matrix, which does not use $R-$modules but only ideals in $R =K[[x_1,\ldots,x_s]]$. In addition, another determinacy bound is provided.
\begin{Corollary}\label{main corollary}
Let $A=[a_1 \ldots a_m]^T\in Mat(m, 1, \mathfrak{m})$.
\begin{enumerate}
\item\label{main 1} If $m\ge s$
then $A$ is finitely $G$-determined if and only if there is some integer $k\ge 0$ such that
\begin{align*}
\mathfrak{m}\cdot\langle a_1,\ldots, a_m\rangle\supset \mathfrak{m}^{k+2},
\end{align*}
i.e. $I_1(A)$ is primary. $A$ is then $(2k-\order(A)+2)$-determined.
\item \label{main 2} Let $m\le s$ and for $m >1 $ assume that $K$ is infinite. Then $A$ is finitely $G$-determined if and only if there is some integer $k\ge 0$ such that
\begin{align*}
\langle a_1,\ldots, a_m\rangle + I_m(Jac(A))\supset \mathfrak{m}^k.
\end{align*}
Furthermore, $A$ is then $G$ $(2km-\order(A)+2)$-determined.
\end{enumerate}
\end{Corollary}
\begin{proof}
1. If $A$ is finitely $G$-determined, Theorem \ref{height}.\ref{height 2}\ (ii) implies the inclusion $ ``\supset"$. If $``\supset"$ holds, we have
\[\mathfrak{m}^{k+2}\cdot M_{m,1}\subset \mathfrak{m}\cdot I_1(A)\cdot M_{m,1}+\mathfrak{m}^{2}\cdot\left\langle\frac{\partial A}{\partial x_1},\ldots,\frac{\partial A}{\partial x_s} \right\rangle= \mathfrak{m}\cdot \tilde T_A(GA).\]
By Theorem \ref{GP16}, $A$ is $G$ $(2k-\order(A)+2)$-determined.
2. The characterization of finite determinacy follows from Theorem \ref{column matrix} and Lemma \ref{isolated and presentation matrix}.
For the determinacy bound, we have $$\mathfrak{m}^{km}\subset\langle a_1,\ldots, a_m\rangle^m + I_m(Jac(A))\subset I_m(\Theta_{(G,A)}),$$
where the second inclusion follows from the proof of Lemma \ref{isolated and presentation matrix}. By \cite[Proposition 4.2]{GP16} we obtain the result.
\end{proof}
\section{Contact equivalence of ideals} \label{complete intersections}
By associating to a matrix $A=[a_1 \ldots a_m]^T\in M_{m,1}$ the ideal $I=\langle a_1,\ldots, a_m\rangle\subset R$ generated by the entries of $A$, the results of the previous section about $G$-equivalence for matrices in $M_{m,1}$ can easily be transferred to contact equivalence for ideals in $R$. The minimal number of generators of $I$ is denoted by $\mng(I)$.
Recall that two ideals $I$ and $J$ of $R$ are called {\textit {contact equivalent}}, denoted $I\mathop \sim \limits^c J$, if $R/I \cong R/J$ as local $K$-algebras.
\begin{Definition} \label{determined}
Let $I$ be a proper ideal of $R$ and $a_1, \ldots, a_m$ a minimal set of generators of $I$.
$I$ is called {\bf{contact $k$-determined}} if for every ideal $J$ of $R$ that can be generated by $m$ elements $b_1, \ldots, b_m$ with $b_i-a_i\in\mathfrak{m}^{k+1}$ for $i=1,\ldots,m$, we have $I\mathop\sim\limits^c J$.
$I$ is called {\bf {finitely contact determined}} if $I$ is contact $k$-determined for some $k$.
\end{Definition}
The notion of contact determinacy is independent of the minimal set of generators of $I$ and hence an intrinsic property of $I$. In fact, let $I$ be contact $k$-determined w.r.t. $a_1, \ldots, a_m$ and let
$a'_1, \ldots, a'_m$ be another minimal set of generators of $I$ and denote by $A$ resp. $A'$ are the corresponding column matrices. If an ideal $J$ can be generated by $b_1, \ldots, b_m$, satisfying $b_i - a'_i \in\mathfrak{m}^{k+1}$, we have to show that $I\mathop\sim\limits^c J$:
By Lemma \ref {Mather 1} below, $A=U \cdot A'$ with $U\in GL(m, R)$. Let $b'_i$ be the entries of $B' = U \cdot B$, $B$ the matrix with entries $b_i$ then $ B'-A = U\cdot (B - A') \in \mathfrak{m}^{k+1}M_{m,1}$ and $J$ is generated by $b'_1, \ldots, b'_m$ with $b'_i - a_i \in\mathfrak{m}^{k+1}$. Hence $I\mathop\sim\limits^c J$ and $I$ is contact $k$-determined w.r.t. $a'_1, \ldots, a'_m$.
The following lemma is proved in \cite[2.3 Proposition]{Mat68} for map-germs, but the proof works also in our more general situation.
\begin{Lemma}\label{Mather 1}
Let $I$ be an ideal of a unital Noetherian local ring $Q$ generated by two lists of $m$ elements, say $\langle a_1,\ldots, a_m\rangle$ and $\langle b_1\ldots, b_m\rangle$. Let $A=[a_1\hskip 3pt a_2\hskip 3pt\ldots\hskip 3pt a_m]^T$ and $B=[b_1\hskip 3pt b_2\hskip 3pt\ldots\hskip 3pt b_m]^T$ be the corresponding column matrices. Then there is an invertible matrix $U\in GL(m, Q)$ such that $B=U\cdot A$.
\end{Lemma}
Note that for 1--column matrices
left-right equivalence (i.e., $G$-equivalence) and left equivalence coincide.
\begin{Proposition}{\label{contact equivalence}}
Let $I$ and $J$ be proper ideals of $R$ given by the same number of $m$ generators
and let $A$ resp. $B$ be the corresponding matrices in $M_{m,1}$.
Then the following are equivalent:
\begin{enumerate}
\item $I\mathop \sim \limits^c J$.
\item There is an automorphism $\phi\in Aut(R)$ such that $\phi(I)=J$.
\item There are $\phi\in Aut(R)$ and $U\in GL(m,R)$ such that $B=U\cdot \phi(A)$.
\item $A\mathop\sim\limits^G B$.
\end{enumerate}
Moreover, if $m = \mng(I)$, then $I$ is contact $k$-determined iff $A$ is $G$ $k$-determined.
\end{Proposition}
\begin{proof}
The equivalence of 1. and 2. follows from the lifting lemma \cite[Lemma 1.23]{GLS07}, that of 2. and 3. from Lemma \ref{Mather 1}. The equivalence of 3. and 4. is obvious. The last statement follows from the equivalence of 1. and 4. and from the definition of determinacy.
\end{proof}
\begin{Definition}\label{definition tjurina}
Let $I$ be a proper ideal of $R$ with $\mng(I)=m$ and $A\in M_{m,1}$ the column matrix corresponding to a minimal set of generators of $I$. We define
\[T_I:=M_{m,1}{\Big/}\left(I\cdot M_{m,1}+\left\langle\frac{\partial A}{\partial x_1},\ldots,\frac{\partial A}{\partial x_s} \right\rangle\right)\]
and set $\tau(I):=\dim_KT_I.$
\end{Definition}
\begin{Remark}\label{Tjurina}\rm
We have
$T_I=M_{m,1}\big/\tilde T_A^e\left(GA\right) = R^m\Big/IR^m+\left(\frac{\partial a_i}{\partial x_j}\right)\cdot R^s.$
By Proposition \ref{semi-continuity} $\tau(I)$ is semicontinuous if we perturb the entries of $A$.
For a complete intersection the $R/I$-module $T_I$ is called the {\em Tjurina module of I} and $\tau(I)=\dim_KT_I$ the {\em Tjurina number of I} (see \cite[Theorem 1.16 and Definition 1.19]{GLS07} for the complex analytic case). The support of $T_I$ is then the singular locus of $I$.
\end{Remark}
\begin{Theorem}{\label{geomchar}}
Let $I$ be a proper ideal of $R$.
\begin {enumerate}
\item If $\dim(R/I) = 0$ then $I$ is finitely contact determined.
\item If $\dim(R/I) >0$ and $K$ is infinite, then the following are equivalent:\\
(i) $I$ is finitely contact determined. \\
(ii) $\tau(I)<\infty$. \\
(iii) $R/I$ is an isolated complete intersection singularity. \\
If one of these condition is satisfied then $I$ is contact $(2\tau(I)-\order(I)+2)$-determined.
Moreover, if $J$ is an ideal such that $R/J$ is a deformation of $R/I$, then $J$ is contact $(2\tau(I)+1)$-determined resp. $2\tau(I)$-determined if $\order(J) \geq 2$.
\end{enumerate}
\end{Theorem}
\begin{proof}
Let $A$ be the column matrix corresponding to a minimal set of $m$ generators of $I$.
1. If $\dim(R/I) = 0$ then some power of $\mathfrak m$ is contained in $I$ and $m = \mng(I) \geq s$. Then $A$ is finitely $G$-determined by Corollary \ref{main corollary}.\ref{main 1} and hence $I$ is finitely contact determined by Proposition \ref{contact equivalence}.
2. The equivalence of (i) and (ii) as well as the determinacy bound follows from Theorem \ref{column matrix} without the assumption $\dim(R/I)>0$.
We prove the equivalence of (i) and (iii). If $I$ is finitely determined then $m < s$ by Theorem \ref{height}.\ref{height 2} (ii). By Theorem \ref{height}.\ref{height 2} (i), we get $\height(I_1(A))=m$, i.e. $\dim R/I = s-m$ and thus $R/I$ is a complete intersection. The isolatedness of $R/I$ and the converse direction (iii) $\Rightarrow$ (i) are direct consequences of Corollary \ref{main corollary}.\ref{main 2}.
If $R/J$ is a deformation of $R/I$, $J$ defines an ICIS of the same dimension and hence $J$ is contact $(2\tau(J)+1)$-determined. By semicontinuity of $\tau$ (Remark \ref{Tjurina}) $\tau(I) \ge \tau(J)$, implying that $J$ is $(2\tau(I)+1)$-determined, resp. $2\tau(I)$-determined if ord($J$) $\geq 2$.
\end{proof}
\begin {Problem}\rm
Note that (ii) $\Leftrightarrow$ (iii) $\Rightarrow$ (i) in Theorem \ref{geomchar} holds for any $K$, as well as (i) $\Rightarrow$ (iii) for hypersurfaces by Theorem \ref{hypersurface}. However, we do not know whether (i) $\Rightarrow$ (iii) holds for finite $K$ and a complete intersection which is not a hypersurface.
\end {Problem}
We complete the section with the hypersurface case, i.e. ideals generated by one element $f\in R=K[[{\bf x}]]$.
In addition to contact equivalence we consider also {\em right equivalence} for $f\in R$, with $f$ being right equivalent to $g$ if $\phi(f) = g$ for some $\phi \in Aut(K[[{\bf x}]])$. $f$ is {\em right k-determined} if any $g$ with $f-g \in \mathfrak m^{k+1}$ is right equivalent to $f$.
Then the Tjurina number is
$$\tau(f)=\dim_K K[[{\bf x}]]/\langle f, {\partial f}/{\partial x_1}, \ldots, {\partial f}/{\partial x_s}\rangle$$
and the Milnor number is
$$\mu(f)=\dim_K K[[{\bf x}]]/\langle {\partial f}/{\partial x_1}, \ldots, {\partial f}/{\partial x_s}\rangle.$$
\begin{Theorem}\label{hypersurface}
Let $K$ be any field and $f\in \mathfrak{m} \subset R$.
\begin{enumerate}
\item $f$ is finitely contact determined iff $\tau(f)$ is finite.
\item $f$ is finitely right determined iff $\mu(f)$ is finite.
\end{enumerate}
Moreover, if $\tau(f) < \infty$ (resp. $\mu(f) < \infty$), then $f$ is contact $(2\tau(f)-\order(f)+2)$--determined
(resp. right $(2\mu(f)-\order(f)+2)$--determined).
\end{Theorem}
\begin{proof}
Since the statement is obviously true for $f=0$ we may assume that $f \neq 0$. For contact equivalence, the statement follows from Theorem \ref{column matrix} with $m=1$ (for this case it was not assumed that $K$ is infinite). For right equivalence, if $ \mu(f) < \infty$, then $f $ is finitely right determined by \cite[Theorem 3.2]{GP16} for any $K$. The proof of the converse goes as for contact equivalence.
\end{proof}
\begin{Remark}\rm
Theorem \ref{hypersurface} was already stated in \cite[Theorem 5]{BGM12}. However, the proof of the direction that finite determinacy implies the finiteness of $\tau$ resp. $\mu$ in \cite {BGM12} contains a gap
since it assumes the orbit map $G^{(k)}\to G^{(k)}jet_k(f)$ to be separable. Separability of the orbit map holds always in characteristic 0 and quite often in positive characteristic, but not always (see \cite[Example 2.9]{GP16}).
\end{Remark}
{\bf Acknowledgement:} We thank Andr\'e Galligo for suggesting Gabrielov's example and Ng{\^o}~V. Trung for useful discussions, in particular for providing a proof of Theorem \ref {generic determinantal ideals}. The second
author would like to thank the Abdus Salam International Centre for Theoretical Physics (ICTP) for support and the department of mathematics of the University of Kaiserslautern for its hospitality.
\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]{
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
| 188,063
|
The hack works efficiently for both Android and iOS version of the game. The addictive mega- hit Temple Run is now out for the Amazon Appstore! Fire and water prove opposites attract when they team up to explore the ancient Forest Temple. Stone Stephen Brown the program centered on a temple that was " filled with lost treasures protected by mysterious Mayan temple guards".
It was released on the App Store on January 16 on Google Play on January 24 on Windows Phone 8 on December 20. We were asked to build a working hack tool for Temple Run 2 so after several weeks of hard work we managed to advance from beta testing so we release today Temple Run 2 Hack v1.
Temple run game free version. Temple run game free version. Read reviews see screenshots, compare customer ratings learn more about Temple Run. Temple Run Online - hints tips, cheats news about the smash hit game!
Navigate perilous cliffs mines , zip lines forests as you try to escape with the cursed idol. Comment by therockstar95 the black temple is the home 2 the ceint illidan i am sure no1 in the game will be able 2 kill him I MEAN IT no playewr is good enough i. Temple Run 2 is an endless running video game developed and published by Imangi Studios. This game continues the main idea from the old Temple Run, so you objective.
Fireboy Watergirl: The Forest Temple is Safe, game android apk, mod apk, apk mod, kumpulan game android, Cool to play , game android terbaik, download game android, download permainan gratis, apk game download apk mod. All your friends are playing it - can you beat their high scores? Welcome to Share- Games!
We have thousands of addicting pc games for you to download we add new downloadable games every day! Created by David G.
Now get more of the exhilarating running, jumping, turning and sliding you love in Temple Run 2!
| 392,235
|
All Total Rewards members immediately enjoy Gold status benefits at all our casinos.
Already a Total Rewards member? SIGN IN now to view your account and benefits.
TIER CRITERIA
You may also qualify for a Tier Upgrade based on daily play;
requirements vary by location.
All benefits are subject to program Rules and Regulations.
| 107,074
|
University of Nevada, Reno
You are here: Nevada Home > Nevada Today > News @ Nevada > 2007 News Stories > February > National Student Exchange gives students new insight
Media professionals interested in reporting on university-related stories are encouraged to visit the media newsroom.
February 20, 2007
Studying abroad is beyond the budget of some University students. National Student Exchange, a program coordinated within the University's transfer center, offers an alternative for students who wish to gain a similar experience for a fraction of the cost. NSE allows students to choose from 190 colleges in the United States, Puerto Rico and Guam for semester and year-long exchange programs. Travel and housing expenses, in-state tuition and a $115 application fee are the only costs for the student.
"It's a really great way for students to connect with different cultures and communities," says Jeri Russell, transfer center Coordinator. "I like to challenge students to do something completely different than what they know."
National Student Exchange has been an option for students nationally for 38 years and has been working with the University for more than 30 years. Seven University students are currently attending another school through NSE. Some of their destinations include Oregon, Illinois, New York, California and Hawaii.
"We have an even exchange program," Russell says. "That means that for every student that leaves, we have one that comes here."
Kyle Eisenreich, a junior who studied at Stony Brook University in Long Island through NSE said his experience helped broaden his view of his world.
"Before I left, Nevada was all that I knew," Eisenreich said. "It was great to experience a different campus with a different culture without having to leave the country."
Eisenreich said he would recommend studying with NSE to anyone.
"It's sad that a lot of people don't know about [the program]," Eisenreich said.
Russell says out-of-state students decide to study at the University for a variety of reasons.
"We had an engineering student from another university come here because we offer a different background in engineering," Russell says. "We have also had students come here just to test it out."
Students are required to meet with their advisor to determine which classes they will take at their host school before they leave. All classes are required to be approved by the advisor and the transfer center to ensure the classes will transfer to fulfill University requirements.
"This is just to make sure there are no surprises when the students get back," Russell says.
Applications for exchanges during the Fall and Spring semesters are due by Feb. 28. Each completed application must include a written recommendation from the student's advisor, an unofficial copy of the student's University transcript, a check or money order for the $115 application fee and a form explaining any special financial or health circumstances. The program is only available for Nevada residents.
| 99,110
|
Re: Sean Hannity Says 98% of Republican Men Would Allow Bush To Rape Their Ass
- From: Deaf Power <deaf@xxxxxxxxx>
- Date: Thu, 01 Nov 2007 22:37:37 -0400
On Thu, 01 Nov 2007 22:00:29 -0400, Horatio Fudruckerton
<nospam@xxxxxxxxxxx> wrote:
On Wed, 31 Oct 2007 23:45:54 -0400, Clint Hunter <ciceroi@xxxxxxxxxx>
wrote:
Another Republican Sex Criminal Busted for Diddling a Boy..
<...>
Donald's email address still on Brown County GOP website: mailto:Don0...
@aol.com
---
Do you like male masturbation videos? Do you not like paying for
outrageously expensive professional male masturbation videos? If
you answered yes to either of those questions, then this posting
is for you.
A few years ago, I ordered a couple of amateur videos from three
women off the Internet. Their video was nice enough, but many of
the scenes were of very poor quality due to lighting and camera
angles. I decided I could make some videos of myself that were
at least that good. I needed some extra money anyway and I
wanted to play with my new video and computer equipment so I
decided to make some videos of myself and sell copies online.
I haven't been selling them for several months now, but, when an
editor of a gay men's magazine emailed me asking if he could
review and feature my videos in his publication, I decided to
start selling them again for a while.
I have a web site up that explains what's in the videos and how
long each one runs. There are pictures there that I captured
from the videos also so you can get an idea of what I look like
and whether or not you like what you'd see in the videos. This
web site is 100% FREE... no passwords, adult check nonsense, no
money to me or to anyone else at all to browse it. There are
plenty of products for parents to use to screen what their kids
see online and I believe strongly in the First Amendment so
look at the site all you want without spending a dime.
If you do decide you'd like to buy my videos after looking at
my pictures at the web site, there is an online order form you
can use. I'm selling copies as a hobby, but I do incur expenses
by providing them for others to view so I can't afford to just
give them away. Unlike professional male masturbation videos
I've seen for sale that sell for $60, mine are only $20 tops...
cheaper if you buy more than one.
So stop by my web site, look around at my pictures, and, if
you'd like, order my videos.
Are you pissed because you paid for your videos and didn't get them ?
It's the Republican style. They're all pedophiles like you!
--
.
- Prev by Date: Re: Screwing boys in the ass OKAY for Republicans
- Next by Date: Bush's America... A Nazi Torture Country.
- Previous by thread: Re: Screwing boys in the ass OKAY for Republicans
- Next by thread: Bush's America... A Nazi Torture Country.
- Index(es):
| 163,917
|
TITLE: How to find an orbit in implicit form for a first order non-linear system of differential equations?
QUESTION [0 upvotes]: How to find an orbit in implicit form for a first order non-linear system of differential equations? Say $x'= x - xy$, $y'= y - 2xy$ is our system. How do we find an orbit of it in an implicit form?
REPLY [0 votes]: Assume that $(x(t),y(t))$ belongs to the curve of equation $$H(x)=K(y),$$ for every $t$, that is, that, for every $t$, $$H(x(t))=K(y(t)).$$Then $$\frac{\mathrm d}{\mathrm dt}H(x(t))=\frac{\mathrm d}{\mathrm dt}K(y(t)),$$ that is, $$H'(x(t))x'(t)=K'(y(t))y'(t).$$ Thus, a sufficient conditio, is that, for every $(x,y)$, $$(x-xy)H'(x)=(y-2xy)K'(y),$$ that is, $$\frac{x}{1-2x}H'(x)=\frac{y}{1-y}K'(y).$$ The LHS does not depend on $y$ and the RHS does not depend on $x$, hence both are constant, without loss of generality, $$H'(x)=\frac{1-2x}x,\qquad K'(y)=\frac{1-y}y,$$ that is, $$H(x)=\log x-2x+h,\qquad K(y)=\log y-y+k.$$ Finally, the orbits are defined by the condition that $(\log y-y)-(\log x-2x)$ is constant or, equivalently, by the condition that, for some constant $C$, $$y\mathrm e^{-y}=Cx\mathrm e^{-2x}.$$
| 26,257
|
TITLE: breadth of join-semilattice
QUESTION [2 upvotes]: On page 16, Lattice Theory: Foundation, George Grätze(2011),
1.19 Show that a join-semilattice $(L; \lor)$ has breadth at most $n$, for a
positive integer n, iff for every nonempty finite subset $X$ of $L$, there
exists a nonempty $Y \subseteq X$ with at most $n$ elements such that $$\bigvee X = \bigvee Y$$
I don't know in which way we shall establish this equivalence by appealing to the definition of breadth, which is
Let $n$ be a positive integer. We say that an order $P$ has breadth at most $n$
if for all elements $x_0, x_1, x_2,……x_n, y_0, y_1,……y_n$,in $P$, if $x_i \leq y_j$ for all $i \neq j$ in $\{0, 1, ……,n\}$, then there exists $ i \in\{0, 1, ……,n\}$,such that $x_i \leq y_i$. The breadth
of $P$, in notation, breadth$(P)$, is the least positive integer $n$ such that $P$ has
breadth at most $n$ if such an $n$ exists.
Observe that this definition of breadth is selfdual. For an equivalent
denition for a join-semilattice (or for a lattice), see Exercise 1.19
Besides the aforementioned exercise, I have difficulty in visualizing this concept, breadth of an order. I don't know the meaning of selfdual, either.
REPLY [1 votes]: $\newcommand{\br}{\operatorname{breadth}}$Let’s start by assuming that for every non-empty finite $X\subseteq L$ there is a non-empty $Y\subseteq X$ such that $|Y|\le n$ and $\bigvee Y=\bigvee X$. We want to show that $\br(L)\le n$, so suppose that we have elements $x_0,\dots,x_n,y_0,\dots,y_n\in L$ such that $x_i\le y_j$ whenever $i,j\in\{0,\dots,n\}$ and $i\ne j$. We want to show that there is a $k\in\{0,\dots,n\}$ such that $x_k\le y_k$.
Let $X=\{x_0,\dots,x_n\}$; by hypothesis there is a non-empty $Z\subseteq X$ such that $|Z|\le n$ and $\bigvee Z=\bigvee X$. Since $|Z|\le n$, there is a $k\in\{0,\dots,n\}$ such that $x_k\notin Z$. For each $x_i\in Z$ we have $i\ne k$ and therefore $x_i\le y_k$, so $x_k\le\bigvee X=\bigvee Z\le y_k$, and we’re done: $x_k\le y_k$.
You can prove the other direction by induction on $|X|$. Assume that $\br(L)\le n$. It’s certainly true that if $|X|\le n$, then there is a non-empty $Y\subseteq X$ such that $|Y|\le n$ and $\bigvee Y=\bigvee X$: just take $Y=X$.
Now suppose that $|X|=n+1$, and let $X=\{x_0,\dots,x_n\}$. For $j\in\{0,\dots,n\}$ let $$y_j=\bigvee \left(X\setminus\{x_j\}\right)\;;$$ clearly this definition ensures that $x_i\le y_j$ whenever $i,j\in\{0,\dots,n\}$ and $i\ne j$. Since $\br(L)\le n$, there is a $k\in\{0,\dots,n\}$ such that $x_k\le y_k$ for some $k\in\{0,\dots,n\}$. Then $$x_k\le\bigvee\left(X\setminus\{x_k\}\right)\;,$$ so $$\bigvee X=\bigvee\left(X\setminus\{x_k\}\right)\;,$$ and we can take $Y=X\setminus\{x_k\}$.
Now suppose that $m>n+1$, and for each non-empty $X\subseteq L$ with $|X|<m$ there is a non-empty $Y\subseteq X$ such that $|Y|\le n$ and $\bigvee Y=\bigvee X$. Let $X$ be a subset of $L$ of cardinality $m$; we show that there is a non-empty $Y\subseteq X$ such that $|Y|\le n$ and $\bigvee Y=\bigvee X$. Fix $x\in X$, and let $X_0=X\setminus\{x\}$. By the induction hypothesis there is an $X_1\subseteq X_0$ such that $|X_1|\le n$ and $\bigvee X_1=\bigvee X_0$. Let $X_2=X_1\cup\{x\}$. Clearly $\bigvee X_2=\bigvee X$, so if $|X_2|\le n$, we’re done. Otherwise, $|X_2|=n+1$, and we just proved the result for that case. The theorem now follows by induction.
To say that the definition of breadth is selfdual is to say that replacing the definition by its dual does not change the concept. The definition of breadth:
An order $P$ has breadth at most $n$ if for all elements $x_0,\dots,x_n,y_n,\dots,y_n\in P$, if $x_i\le y_j$ for all $i,j\in\{0,\dots,n\}$ with $i\ne j$, then there is a $k\in\{0,\dots,n\}$ such that $x_k\le y_k$.
The dual definition:
An order $P$ has dual-breadth at most $n$ if for all elements $x_0,\dots,x_n,y_n,\dots,y_n\in P$, if $x_i\ge y_j$ for all $i,j\in\{0,\dots,n\}$ with $i\ne j$, then there is a $k\in\{0,\dots,n\}$ such that $x_k\ge y_k$.
If you interchange $x_i$ and $y_i$ for each $i\in\{0,\dots,n\}$, you transform each of these definitions into the other. Thus, they actually say the same thing and define the same concept: what I called dual-breadth is just breadth.
| 1,784
|
London:
A bright two bedroom apartment in the heart of Marylebone and minutes from the vibrant Marylebone High Street.
Large one double bedroom ground and first floor flat set within moments to Baker Street and Marylebone Train Stations. Large reception room, fully fitted kitchen and modern bathroom.
Beautifully presented four bedroom flat the third floor with lift in this portered red brick mansion block in the heart of Marylebone.
One bedroom apartment situated in a well respected, secure period mansion block off Baker Street.
Managed
by KFH
Beautifully presented two double bedroom penthouse in this Georgian Conversion in Marylebone.
A third floor lateral apartment in a red brick block facing Bickenhall Street.
An extremely spacious and airy one double bedroom flat which has recently undergone full refurbishment.
Thank you for your hard work getting the right price for our house.
Abraham
| 133,037
|
TITLE: Why Laplace-Beltrami operator is so popular for 3D shape analysis.?
QUESTION [1 upvotes]: Apart from providing orthogonal basis in form of eigen functions what is the reason that Laplace-Beltrami operator is so popular in shape and point cloud processing.
REPLY [1 votes]: Because it provides many useful properties for shape analysis. If you view the 3D shape as "samplings" of a Riemannian manifold, then you get an associated Laplace-Beltrami operator (LBO). Knowing this operator, or an approximation of it, lets you do many interesting things, known as "spectral shape theory" and "diffusion geometry".
For instance, if you compute the eigenvalues and eigenfunctions of the LBO, you can get a "signature" of the shape. This has many powerful properties, such as being isometry invariant and "similarity" (meaning small perturbations in the shape lead to small perturbations in the eigenvalues).
(See for instance: Reuter, Martin, Franz-Erich Wolter, and Niklas Peinecke. "Laplace–Beltrami spectra as ‘Shape-DNA’of surfaces and solids." Computer-Aided Design 38.4 (2006): 342-366.)
Any of the "good" discrete LBOs approximate the continuous one of the underlying manifold (or converge in the point cloud case).
Another popular signature is the "autodiffusion" heat kernel, which is directly derived from the LBO eigenvalues and eigenfunctions. It can act as a local and multiscale signature, so it can work well even on partial shapes.
See: Sun, Jian, Maks Ovsjanikov, and Leonidas Guibas. "A Concise and Provably Informative Multi‐Scale Signature Based on Heat Diffusion." Computer graphics forum. Vol. 28. No. 5. Blackwell Publishing Ltd, 2009.
Thus, spectral descriptors let you do shape retrieval and matching, even for partially missing or articulated shapes.
If you are computationally oriented, spectral shape analysis has been used for fairly sophisticated computer vision and machine learning tasks; for instance:
Masci, Jonathan, et al. "Geodesic convolutional neural networks on riemannian manifolds." Proceedings of the IEEE International Conference on Computer Vision Workshops. 2015.
Litman, Roee, and Alexander M. Bronstein. "Learning spectral descriptors for deformable shape correspondence." IEEE transactions on pattern analysis and machine intelligence 36.1 (2014): 171-180.
| 12,721
|
TITLE: Vector Space - how to visualize it for understanding?
QUESTION [3 upvotes]: I read on Wikipedia about vector spaces, but I don't understand them in a way that I can visualize the vector spaces in my head. During the process of understanding, I had several concepts in my head and I am at a point now, where I am totally confused. Maybe I am in a dead end as well. I have drawn four of these concepts, so you can imagine what happened in my head.
Pictures: my approaches for vector spaces
Picture A
$ \vec{r} $ is the vector space, which means the space is linear on the line of the vector. $ \vec{r} $ contains infinite vectors like $ \vec{a} $, $ \vec{b} $ and $ \vec{c} $. The last three vectors only exist in $ \vec{r} $ or vector spaces which are bigger or equal to themselves. An orthogonal vector of $ \vec{b} $ is not a part of $ \vec{r} $.
Picture B
The vector space is an area where one or multiple vectors like $ \vec{r} $ and $ \vec{m} $ exist. The space is infinite, which doesn't make much sense to define a space. But it is a space. In the picture it is the striped zone of the diagram.
Picture C
$ \vec{r} $ can be build by the linear combination of $ \vec{a} $ + $ \vec{b} $, $ \vec{c} $ + $ \vec{d} $ or any other combination of two vectors within the red striped zone. But what is with combinations outside of the red striped zone? Here it destroys my concept probably.
Picture D
$ \vec{r} $ is the shortest vector to the target point. $ \vec{a} $, $ \vec{b} $, $ \vec{c} $ and $ \vec{d} $ are one linear combination of multiple possible linear combinations to the target. Is the red striped area the vector space or red and yellow together?
Is one of my concepts the right concept of vector spaces?
I really appreciate your inputs and hope to get a explanation which my brain can visualize. Maybe you could draw it?
REPLY [1 votes]: A, B are reasonable pictures.
C, D are not.
the vector space is the set of all linear combinations of some set of basis vectors.
That means that the vector space is never bounded in the way you have it pictured in these two pictures. If $a$ and $b$ are in your space, so is $2a$, and $2b$ and $a+b$ and $\frac 12a + 3b$ etc. And in D) $r, c, d$ can each be described as some combination of $a,b$
| 32,084
|
Abstract
Localising activity in the human midbrain with conventional functional MRI (fMRI) is challenging because the midbrain nuclei are small and located in an area that is prone to physiological artefacts. Here we present a replicable and automated method to improve the detection and localisation of midbrain fMRI signals. We designed a visual fMRI task that was predicted would activate the superior colliculi (SC) bilaterally. A limited number of coronal slices were scanned, orientated along the long axis of the brainstem, whilst simultaneously recording cardiac and respiratory traces. A novel anatomical registration pathway was used to optimise the localisation of the small midbrain nuclei in stereotactic space. Two additional structural scans were used to improve registration between functional and structural T1-weighted images: an echo-planar image (EPI) that matched the functional data but had whole-brain coverage, and a whole-brain T2-weighted image. This pathway was compared to conventional registration pathways, and was shown to significantly improve midbrain registration. To reduce the physiological artefacts in the functional data, we estimated and removed structured noise using a modified version of a previously described physiological noise model (PNM). Whereas a conventional analysis revealed only unilateral SC activity, the PNM analysis revealed the predicted bilateral activity. We demonstrate that these methods improve the measurement of a biologically plausible fMRI signal. Moreover they could be used to investigate the function of other midbrain nuclei.
Structured keywords
- Brain and Behaviour
| 373,261
|
TITLE: What are some examples of these kinds of commutative semirings?
QUESTION [1 upvotes]: What are some examples of commutative semirings such that the following hold?
Multiplication is idempotent i.e. we have $xx=x$ for all elements $x$.
Addition is not idempotent i.e. there is at least one element $x$ with $x+x \neq x$.
There is at least one element $x$ with $x+x \neq 0$.
I cannot think of any examples, with or without commutativity.
Discussion. In ring theory, it is a theorem that if multiplication is idempotent, then $x+x=0$ for all elements $x$. However, in semiring theory, this isn't necessarily the case; take, in particular, any non-trivial distributive lattice. Then it is certainly the case that $x+x \neq 0$ for all $x,$ except $x=0$ of course. In fact, we have that $x+x=x$ for all elements $x$.
REPLY [2 votes]: The initial semiring is $\mathbb{N}$ (with the usual operations). Hence, the initial boolean semiring is $R = \mathbb{N} / \sim$, where $\sim$ is the smallest congruence relation such that $n \sim n^2$ for all $n \in \mathbb{N}$. Hence, we also have $n_1^2 + \dotsc + n_s^2 \sim n_1 + \dotsc + n_s$. The elements of $R$ are therefore
$$[0],[1],[2],[3],[4]=[2],[5]=[3],[6]=[2],[7]=[3],\dotsc$$
We find that $R$ is actually finite, it has exactly $4$ elements. This ring satisfies your requirements.
Another example is the free boolean semiring on one generator $x$. It is given by $\{a+bx : a,b \in R\}$ modulo the relation $a+bx \sim (a+bx)^2 = (a+b)+(2ab)x$. This semiring is also finite and its elements may be listed explicitly, if one wants to.
| 150,403
|
TITLE: How do you complexify vector spaces?
QUESTION [1 upvotes]: Suppose you have a linear map $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by $$Tv = \begin{bmatrix}
a &b \\
c & d
\end{bmatrix}v$$
So, if I understand correctly complexification (of the domain and the target space and the map) means $\tilde{T}: \mathbb{R}^2 \otimes \mathbb{C} \to \mathbb{R}^2 \otimes \mathbb{C}$ where $$\tilde{T}(v \otimes z) = T(v) \otimes z$$ as I can not think of any other way to define a map from $\mathbb{R}^2 \otimes \mathbb{C} \to \mathbb{R}^2 \otimes \mathbb{C}$. This will now be linear map from a 4 dimensional space to a 4 dimensional space. And if we use the standard tensor product basis then $\tilde T$ is just a block diagonal matrix with each block identical to T. This seems pretty pointless.
Now if we look at a map $f: \mathbb{C} \to \mathbb{C}$ which is given by $f = u+iv$ or $f = (u,v)$ thinking of $\mathbb{C}$ as $\mathbb{R}^2$. Then the derivative $f^*: T_p \mathbb{R}^2 \to T_{f(p)} \mathbb{R}^2$ is given by $$\begin{bmatrix}
u_x &u_y \\
v_x & v_y
\end{bmatrix}$$ if we let the basis of both tangent spaces to be $\partial_x, \partial_y$. Now, if we play the same game and complexify these tangent spaces and $f^*$ and write it in terms of the standard tensor product basis then we should get a block diagonal matrix with each block identical to that of $f^*$. Now if we write the matrix corresponding to the complexification with respect to the basis $$\partial_z = \partial_x \otimes1 - \partial_y \otimes i \\ \partial_{\bar z} = \partial_x \otimes1 + \partial_y \otimes i$$ (what are the other two basis vectors) why do we get the following matrix?
$$\begin{bmatrix}
f_z &f_{\bar z} \\
\bar f_z & \bar f _{\bar z}
\end{bmatrix}$$ i.e. a 2 by 2 and not a 4 by 4 matrix.
REPLY [1 votes]: $\newcommand{\dd}{\partial}\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Cpx}{\Number{C}}\newcommand{\Reals}{\Number{R}}$If $B = (v_{1}, \dots, v_{n})$ is an ordered basis of a real vector space $V$, then
$B = (v_{1}, \dots, v_{n})$ is an ordered complex basis of the complexification $V \otimes \Cpx$;
$B_{\Cpx} := (v_{1}, \dots, v_{n}, iv_{1}, \dots, iv_{n})$ is an ordered real basis of $V \otimes \Cpx$. Note that we're suggestively writing $v_{k} \otimes 1 = v_{k}$ and $v_{k} \otimes i = iv_{k}$. (Caution: This is not the only ordering used in the literature.)
Correspondingly, if the linear operator $T:V \to V$ has (real $n \times n$) matrix $A$, then the induced complex-linear operator $T$ on $V \otimes \Cpx$
(also) has matrix $A$ with respect to $B$;
(as you say) has block diagonal matrix $A \oplus A$ with respect to $B_{\Cpx}$.
Particularly, if we choose $(\dd_{x}, \dd_{y})$ as a basis of $\Reals^{2}$, then $(\dd_{x}, \dd_{y})$ is a complex basis of $\Reals^{2} \otimes \Cpx = \Cpx^{2}$, and $(\dd_{x}, \dd_{y}, i \dd_{x}, i \dd_{y})$ is a real basis of $\Reals^{2} \otimes \Cpx = \Cpx^{2}$. As Brevan Ellefsen notes in the comments, there is a straightforward change of basis calculation, using the complex bases $(\dd_{x}, \dd_{y})$ and
$$
\dd_{z} = \tfrac{1}{2}(\dd_{x} - i \dd_{y}),\qquad
\dd_{\bar{z}} = \tfrac{1}{2}(\dd_{x} + i \dd_{y}).
$$
(The factors of $\frac{1}{2}$ aren't essential for the calculation, but are needed to make these vectors dual to the complex $1$-forms $dz = dx + i\, dy$ and $d\bar{z} = dx - i\, dy$.)
| 206,375
|
Column 425
Mr. Dalyell : To ask the Secretary of State for Foreign and Commonwealth Affairs, pursuant to the letter from the Minister for Overseas Development of 18 February to the hon. Member for Linlithgow, what is the outcome of the inquiries by the British embassy in Paris about injury to the rain forest in French Guyana.
Mrs. Chalker [holding answer 2 March 1992] : The French Ministry of the Environment has confirmed that the hydro-electric project at Petit Saut referred to in the New Scientist of 25 January 1992 has over some 10 years been the subject of extensive study, including a public inquiry, which looked at all the possible options for power generation. There had been a full environmental impact assessment undertaken on the project in which the Ministry of Environment had participated. The conclusion was that a hydro-electric dam was the best option. The scheme is intended to meet not only the electricity needs of the launch-site at Kourou, but also the wider need for electricity throughout French Guyana. I am writing to the hon. Member with further details provided by French officials.
Mr. Flynn : To ask the Secretary of State for Wales (1) on how many days consultants have been paid, or are expected to be paid, more than £1,000 per day in total payments for operations designed to reduce waiting lists in Gwent during the current financial year ; (2) what is his estimate of the current weekly cost to the national health service of operations being undertaken on their behalf by the BUPA hospital at Pentwyn, near Newport.
Mr. Nicholas Bennett : Operational information of this kind is not held centrally. It is for district health authorities as commissioners of care to decide on and account for the benefits and costs of any health gain investment, including consultant fees and contracts with private sector health care providers.
Mr. Edwards : To ask the Secretary of State for Wales if he will list the names of those people who met him regarding assisted area status for Monmouth on 2 March ; and if he will provide details of the bodies that they represent.
Mr. David Hunt : On 2 March, I met Cllr. W. J. Parker, Cllr. W. G. A. Hathaway, Cllr. J. Lewis and Mr. J. Martin
Column 426representing Monmouth borough council County Cllr. C. White, Mr. P. Smith, town clerk of Monmouth town council and Mr. Roger Evans.
Mr. Edwards : To ask the Secretary of State for Wales when he proposes to announce his decision on assisted area status for the Monmouth travel-to-work area.
Mr. David Hunt : I am presently considering the issues put to me by local representatives at a meeting on 2 March and shall be responding in due course.
Mr. Grist : To ask the Secretary of State for Wales what were the number of operations carried out in Wales in 1978-79 and at the latest available date for (a) hip replacements, (b) cataracts and (c) open heart surgery.
Mr. Nicholas Bennett : The information on hipHSreplacements and cataracts is given for 1978, 1979 and 1989 (the latest available year) in the following table :
Principal operations in National Health Service hospitals in Wales<1> |1978 |1979 |1989 -------------------------------------------- Hip replacements<2> |1,372|1,501|2,407 Cataracts |2,232|2,330|5,019 <1> Figures may understate the true position in that not all hospitals provide complete clinical details relating to patient discharge and deaths. <2> Includes replacement of head of femur for 1989. Source: Hospital Activity Analysis.
Information available centrally prior to 1980 on open heart operations is derived from "Hospital Activity Analysis" and is likely to understate the true position in that not all hospitals provide complete clinical details relating to patient discharges and deaths. It is not directly comparable to later information. The numbers of open heart operations reported to have been carried out in national health service hospitals in Wales in 1978 and 1979 were 185 and 161 respectively. The latest available data, however, is provided by the centrally funded, regional cardiac centre at the University hospital of Wales and shows that the following number of operations were undertaken in 1991 and up to 29 February 1992.
|By pass|Others |Total ---------------------------------------------- Adult 1991 |654 |70 |724 <1>1992 |144 |12 |156 Paediatric<2> 1991 |23 |14 |37 <1>1992 |9 |1 |10 <1>As at 29 February. <2>Opened June 1991.
Mr. Edwards : To ask the Secretary of State for Wales if he will list the hospitals in Wales which impose car parking charges.
Mr. Nicholas Bennett : This is a matter for individual district health authorities and this information is not held centrally.
Column 427
Dr. David Clark : To ask the Secretary of State for Wales if he will give the numbers of confirmed cases of bovine spongiform encephalopathy in each county in Wales in 1992, to the latest available date.
Mr. David Hunt : The numbers of confirmed cases of BSE in each county in Wales during 1992 to 28 February are shown in the table :
County |Number of |confirmed cases ------------------------------------------------ Clwyd |62 Dyfed |288 Gwent |31 Gwynedd |13 Mid Glamorgan |9 Powys |47 South Glamorgan |17 West Glamorgan |2
Mr. Edwards : To ask the Secretary of State for Wales what assessment he has made of the operation of the vehicle watch Wales scheme in Gwent.
Mr. Nicholas Bennett : None. Vehicle watch is a police initiative and it is for the Gwent constabulary to assess its effectiveness in Gwent.
Mr. Edwards : To ask the Secretary of State for Wales what proposals he has to fund the vehicle watch Wales scheme ; and if he will make a statement.
Mr. Nicholas Bennett : I have received representations from the Gwent police authority about the funding of vehicle watch and these are presently under consideration.
Mr. Edwards : To ask the Secretary of State for Wales what representations he has received regarding the implications of the MacSharry proposals for dairy farmers in Monmouthshire ; and if he will make a statement.
Mr. David Hunt : I have received a number of representations from industry bodies in Wales and individual farmers, including some dairy producers from Monmouthshire, on the implications for the industry of the EC reform proposals.
Mr. Edwards : To ask the Secretary of State for Wales what proposals he has to ensure that senior citizens are provided with access to suitable home security equipment to ensure the security of their homes and allay fears of crime.
Mr. Nicholas Bennett : Assistance for installing home security equipment for senior citizens is already being provided under specific crime prevention initiatives ; as part of wider measures to improve conditions for the elderly ; and as part of the action being undertaken to renovate dwellings.
Proposals for improving household security are eligible for consideration as crime prevention measures under the
Column 428urban programme ; any local authority in Wales may submit a bid. Projects that have received support recently include the "Cardiff Safe and Secure" initiative, and improvements to the security of the communal areas of flats on the Sandfields estate in Port Talbot. Funds made available to victim support schemes have also on occasion been used to improve household security. In undertaking such projects the needs of elderly people, especially those living alone, attract high priority. Under the Welsh Office elderly initiative scheme some £315,000 is being made available over a three-year period to support a project in Mid Glamorgan that is designed to improve the safety and well-being of senior citizens living in their own homes. In addition to providing improved door and window locks, and door viewers, the project entails the provision of advice to the elderly on how to deal with bogus officials and confidence tricksters.
Within the renovation grant arrangements those over 60 who receive income related benefits may be able to get help with home security installations in the form of minor works assistance.
Mr. Edwards : To ask the Secretary of State for Wales if he will make a statement regarding the funding of crime prevention panels in Wales.
Mr. Nicholas Bennett : Crime prevention panels are an excellent means by which local communities can become involved through voluntary activity in the fight against crime. They are encouraged to be self- sufficient and to seek local funding arrangements. Many crime prevention panels obtain private sponsorship from local businesses. Possible sources of funding will vary from area to area. Crime Concern has published a booklet entitled "Attracting Business Sponsorship" which provides detailed advice for panels on how to obtain sponsorship for crime prevention projects. Some panels have been successful in obtaining charity status and any panels wishing to do this can contact the Charity Commission who will advise them on the necessary procedures.
Although central funding is not made available to support the panels themselves, individual projects that they propose may be eligible for consideration for support under the urban programme, provided that the bids are sponsored by the appropriate local authority.
Mr. William Powell : To ask the Secretary of State for Health if he will list the revenue allocation to the Kettering regional health authority for (a) 1982-83 and (b) 1992-93 expressed in (i) absolute terms and (ii) 1992-93 prices.
Mr. Dorrell : Information about the funds allocated by regional health authorities to district health authorities is not held centrally. My hon. Friend may wish to contact Dr. Stuart Burgess, the chairman of Oxford regional health authority, for details.
Column 429
Mrs. Dunwoody : To ask the Secretary of State for Health how many nurses are employed by Mid Cheshire hospitals trust ; and what was the number previously employed by the health authority responsible for staffing the equivalent hospitals.
Mrs. Virginia Bottomley : This information is not held centrally. The hon. Member may wish to contact Mr. Richard F. Lawrence, the chairman of the Mid Cheshire hospitals trust for current nursing levels, and Sir Donald Wilson, the chairman of the Mersey regional health authority, who may be able to provide the historical data.
Mrs. Dunwoody : To ask the Secretary of State for Health what is the cost per hour of consultants employed by Mid Cheshire hospitals trust ; how many patients they see on average in one week's clinics ; and what time is normally taken by consultants to visit NHS patients outside the hospital.
Mrs. Virginia Bottomley : This information is not held centrally. The hon. Member may wish to contact Mr. Richard F. Lawrence, the chairman of the Mid Cheshire hospitals NHS trust, for details.
Mr. David Nicholson : To ask the Secretary of State for Health if he will give the average cost of collection per litre of blood from (a) recovered plasma and (b) plasmapheresis for each regional transfusion centre of the National Blood Transfusion Service.
Mr. Dorrell : Information on the regional cost of collecting plasma is not held centrally.
Sir Eldon Griffiths : To ask the Secretary of State for Health if he will give details of (a) the consideration given by his Department to the report by management consultants commissioned by the Committee for the Retention of Newmarket Hospital, before his recent decision on the future of the hospital and (b) the consultations between the regional health authority and the Committee for the Retention of Newmarket Hospital ; and if he will make a statement.
Mr. Dorrell : On 10 February 1992 I announced our decision to accept certain proposals put forward by West Suffolk health authority on the future of Newmarket hospital, subject to the health authority agreeing to undertake a further examination of the scope for provision of day surgery, minor accident and maternity services in the new hospital proposed for Newmarket.
In reaching this decision full consideration was given to the report commissioned by the Committee for the Retention of Newmarket Hospital. Several issues were thoroughly analysed including patient safety, medical staffing, size of catchment area, patient activity levels and effective use of resources.
The Committee for the Retention of Newmarket Hospital met East Anglian regional health authority members on 24 April 1991 to present their proposals for the future of the hospital. Additionally, the region's finance manager met the committee's management consultants, J. J. Thompson and Partners, on 29 October 1991 to go through their report.
Column 430
Mr. Ieuan Wyn Jones : To ask the Secretary of State for Health in what circumstances contact lenses can be obtained under the national health service.
Mrs. Virginia Bottomley : All general opthalmic services patients may put any optical voucher to which they are entitled towards the cost of contact lenses. Under the hospital eye service patients who require contact lenses for clinical reasons will, if they are entitled to an optical voucher, get the lenses free ; other patients can take advantage of the maximum charging arrangements.
Mr. Andrew Mitchell : To ask the Secretary of State for Health if he plans to publish the goal strategy and objectives in the national health service management executive and the Department of Health priorities and key challenges for 1992-93 and beyond.
Mr. Waldegrave : I have agreed a statement of the aims, goals and priorities for my Department for 1992-93 and the following three years. It covers the whole range of the Department's work relating to health services, health promotion, public health, social services, EC and other international issues, and internal management. It includes our specific priorities for the NHS management executive, which are taken forward in more detail in the management executive's statement, "The Goal and Strategies of the Chief Executive and the Management Executive of the NHS and their Objectives for 1992-93", approved by the NHS policy board. Copies of both statements have been placed in the Library.
Mr. Edwards : To ask the Secretary of State for Health what financial support trust hospitals give to post-registration nurse education.
Mrs. Virginia Bottomley : Post-registration training which is designed to provide nurses and midwives with specific professional skills is to be funded from a separate and protected budget held at regional level. Employers are expected to bear the cost of other post-registration nurse education and training because they are in the best position to determine need. Trusts are expected to play their full part in training the staff they employ, in the same way as other employers.
Mr. Wigley : To ask the Secretary of State for Health if he will list the mental handicap hospitals planned for closure over the next five years and the district health authority in which each is situated.
Mr. Dorrell : This is a matter for regional and district health authorities and the hospitals themselves. Government policies on the treatment of people with learning disabilities in the community are well known and understood.
Mr. Ieuan Wyn Jones : To ask the Secretary of State for Health what would be the gross cost of abolishing sight test fees for all people over 60 years old.
Column 431Mrs. Virginia Bottomley : NHS sight tests are already free to approximately 40 per cent. of the population, including those pensioners who qualify on medical or income grounds. The estimated additional cost at 1991-92 prices of extending eligibility to all those aged 60 and over would be between £30 million and £35 million in a full year.
Mrs. Mahon : To ask the Secretary of State for Health how many patients have been treated privately under the waiting list initiative from (a) Leeds united hospital trust and (b) the Bradford hospitals NHS trust.
Mrs. Virginia Bottomley : This information is not held centrally. The hon. Member may wish to contact Mr. Anthony Clegg, the chairman of the United Leeds hospitals trust and Mr. Rodney Walker, the chairman of the Bradford hospitals NHS trust, for details.
Mr. Robin Cook : To ask the Secretary of State for Health what was the total level of expenditure on pay for general and senior managers in each year since 1985.
Mr. Dorrell [holding answer 18 February 1992] : The grades of general and senior managers were introduced as part of the new arrangements recommended by the Griffiths report in 1983. The cost of staff employed in general management grades is shown in the table :
England (HCHS) General managers £ million (cash) |Regional/ |Unit/other |district |general |general |managers |managers -------------------------------------------- 1985-86 |6.8 |4.1 1986-87 |8.6 |17.1 1987-88 |9.0 |20.7 1988-89 |9.7 |20.8 1989-90 |10.7 |21.3 1990-91 |12.8 |25.3 Notes: 1. Figures for 1990-91 are as yet subject to audit. 2. The figures are gross pay costs including employer's national insurance and superannuation contributions. Source: Annual accounts of regional and district health authorities in England together with those of the special health authorities for the London post-graduate teaching hospitals.
Other senior management costs were not separately identified until 1989-90. The cost of staff employed in senior management grades in 1989-90 was £126.9 million and in 1990-91 was £213.3 million. Staff employed in general and senior management grades generally hold posts which were previously classified under administrative and clerical, works staff or senior nursing grades.
Mr. Oppenheim : To ask the Secretary of State for Health if he will give figures for the number of immunisations carried out in South Derbyshire district and Trent region in 1982 and the latest year for which figures are available.
Mr. Dorrell [holding answer 6 March 1992] : The numbers of two-year-olds vaccinated against diphtheria, tetanus, pertussis, polio, measles, mumps and rubella in South Derbyshire and Trent in 1982 and 1990- 91 are given in the table.
The uptake rates are also given to allow comparisons between the years.
Column 431
Vaccination and immunisation (2 year uptake<2>) |Year |Diphtheria |Tetanus |Pertussis |Polio |Measles |<1>Mumps and Rubella ------------------------------------------------------------------------------------------------------------------------- South Derbyshire Number immunised |1982 |6,333 |6,339 |5,042 |6,342 |5,042 |- Uptake rate |(90) |(90) |(63) |(91) |(74) |- |1990-91 |6,486 |6,486 |6,123 |6,494 |6,332 |6,321 |(91) |(91) |(86) |(92) |(89) |(89) Trent Region Number immunised |1982 |52,833 |52,925 |34,880 |52,594 |36,904 |- Uptake rate |(88) |(88) |(58) |(88) |(62) |- |1990-91 |56,421 |56,429 |52,740 |56,303 |54,036 |53,797 |(91) |(91) |(85) |(91) |(87) |(87) Source: SBL 607.KC51. DH Statistics and Management Information (SMI2B), March 1992. <1>Data was not collected before 1988-89. <2>The two year uptake rate is the percentage of children immunised by their second birthday.
Mr. Bernie Grant : To ask the Secretary of State for Foreign and Commonwealth Affairs if he will make a statement on the current negotiations by the United Nations Secretary-General on a united Cyprus and on the role Britain has played since 1979.
Mr. Garel-Jones : Representatives of the UN Secretary-General have recently ocmpleted another round of talks
Column 432with all parties in the region. We have throughout given full and active support to the UN Secretary-General's efforts to promote a comprehensive, just and lasting settlement.
Mr. Atkinson : To ask the Secretary of State for Foreign and Commonwealth Affairs what offers of assistance and expertise have been made to the Turkish Government in the search for the trapped coal miners near Zonguldak.
Column 433Mr. Garel-Jones : The Government were not approached by the Turkish authorities for help in dealing with the disaster at Kozlu coal mine in Zonguldak on 5 March.
Dr. Marek : To ask the Chancellor of the Exchequer, pursuant to his answer of 2 March, Official Report, column 29, for what reasons Her Majesty's Customs and Excise did not charge excise duty on denatured wine products for culinary uses before 1 January, but did do so from that date onwards.
Mrs. Gillian Shephard : The excise duty liability of imported wine and made-wine, including denatured wine and made-wine, has not changed in recent years. Customs regret that an importer was incorrectly advised that if wine denatured with 2 per cent. salt was to be used as an ingredient in food manufacture it would not be liable to excise duty at importation. The correct liability has been applied since February 1991.
Mr. French : To ask the Chancellor of the Exchequer what higher rate of income tax would be necessary to raise the same revenue as would be raised by a higher rate of 50 per cent. and removing the ceiling on national insurance contributions.
Mr. Maude : At 1992-93 levels of income and assuming statutory indexation of allowances and thresholds, a higher rate of 59 per cent. would raise an approximately equivalent amount of revenue in a full year.
Mr. Harry Barnes : To ask the Chancellor of the Exchequer what criteria he employs to determine whether a property with a paying guest or guests should be classified as a commercial property for Inland Revenue purposes.
Mr. Maude : Unless a paying guest or guests are using the property as their main or sole residence as defined in the Local Government and Finance Act 1988, any short-stay accommodation in excess of six bed spaces or which is not ancillary to the owners own domestic use will be treated as non-domestic property for uniform business rate purposes.
Mr. Dunn : To ask the Chancellor of the Exchequer if he will make a statement on the level of share ownership in Great Britain.
Mr. Maude : A joint Treasury/ProShare survey carried out in January and February of this year shows that almost 10 million people, or more than 22 per cent. of the adult population in Great Britain, now own shares compared with 7 per cent. in 1979. I am placing copies of the survey report in the Libraries of both Houses.
Column 434
Mr. Battle : To ask the Chancellor of the Exchequer what would be the effect on tax revenues (a) in the first year and (b) in a full year, assuming this is introduced in 1992-93 of the introduction of a 50 per cent. top rate of income tax on annual taxable incomes of (i) £31,555, (ii) £33,555, (iii) £36,555, (iv) £39,555 and (v) £42, 555, in Great Britain and Northern Ireland.
Mr. Maude [holding answer 9 March 1992] : The information is in the table.
Estimated yield in 1992-93 from a 50 per cent. tax rate Threshold of |Yield in a full taxable income at |year<1> at 1992-93 which 50 per cent. |levels of income rate is charged |(£ million) (£) --------------------------------------------------------- 31,555 |2,550 33,555 |2,370 36,555 |2,140 39,555 |1,950 42,555 |1,790 <1> About half of the yield would be collected in the first year.
These estimates do not allow for any behavioural effect that might result from such changes to the tax system and do not include capital gains tax.
Mr. Meacher : To ask the Chancellor of the Exchequer if he will update the table in his answer of 12 June 1991, Official Report, columns 547-50, on taxes and benefits, to include information on 1989.
Mr. Maples [holding answer 9 March 1992] : The information is as follows :
Taxes and benefits as percentage of income by quintile groups of households<1>, 1989 |per cent. --------------------------------------------------------------------------- Retired households Cash benefits as percentages of gross income Bottom quintile group |88 2nd quintile group |84 3rd quintile group |78 4th quintile group |54 Top quintile group |23 Average over all groups |51 Income tax, NIC, rates and community charge as percentages of gross income Bottom quintile group |17 2nd quintile group |12 3rd quintile group |12 4th quintile group |14 Top quintile group |20 Average over all groups |17 Indirect taxes as percentages of unadjusted disposable income Bottom quintile group |27 2nd quintile group |20 3rd quintile group |19 4th quintile group |21 Top quintile group |17 Average over all groups |20 Benefits in kind as percentages of final income Bottom quintile group |44 2nd quintile group |34 3rd quintile group |31 4th quintile group |24 Top quintile group |13 Average over all groups |25 Non-retired households Cash benefits as percentages of gross income Bottom quintile group |49 2nd quintile group |14 3rd quintile group |7 4th quintile group |3 Top quintile group |1 Average over all groups |8 Income tax, NIC, rates and community charge as percentages of gross income Bottom quintile group |15 2nd quintile group |19 3rd quintile group |20 4th quintile group |22 Top quintile group |22 Average over all groups |21 Indirect taxes as percentages of unadjusted disposable income Bottom quintile group |30 2nd quintile group |24 3rd quintile group |22 4th quintile group |20 Top quintile group |16 Average over all groups |20 Benefits in kind as percentages of final income Bottom quintile group |45 2nd quintile group |25 3rd quintile group |17 4th quintile group |11 Top quintile group |5 Average over all groups |16 <1> Ranked by equivalised disposable income. Source: CSO, from Family Expenditure Survey.
Mr. Battle : To ask the Chancellor of the Exchequer what would be the revenue raised from a 9 per cent. charge on unearned income for those below state retirement age with a (a) £3,000 exemption, (b) £3,500 exemption, (c) £4,000 exemption, (d) £4,500 exemption and (e) £5,000 exemption, giving both full year and first year revenues for 1991-92 and 1992-93, specifying the numbers affected in each case, for the United Kingdom.
Mr. Maude [holding answer 9 March 1992] : Latest estimates for 1991-92 are as follows :
Threshold |Yield in a full|Numbers |year |affected £ |£ million |Thousands ---------------------------------------------------------------- 3,000 |870 |1,130 3,500 |820 |970 4,000 |780 |840 4,500 |750 |750 5,000 |720 |660
The amount of revenue collected in the first year would depend on the administrative arrangements for collecting such a charge. The yield in 1992 -93 would depend on the levels of investment income for that year.
These estimates take no account of any possible behavioural changes following the introduction of such a charge.
Column 436
Mr. Alex Carlile : To ask the Chancellor of the Exchequer if he will institute proceedings, pursuant to section 48 of the Banking Act 1987, against Mr. Bengt Bangstad arising from the Nicholas Young Ponzi fraud ; and if he will make a statement.
Mr. Maples [holding answer 3 March 1992] : My right hon. Friend the Chancellor of the Exchequer has no powers to institute proceedings under section 48 of the Banking Act 1987. That is a matter for the Bank of England.
Mr. David Nicholson : To ask the Minister for the Arts if he will give (a) the total number of books lent by and (b) the total amount of public spending on the library service in England and Wales, for each year since 1962.
Mr. Renton : Reliable and comparable information for England and Wales is available only from 1966-67 to 1989-90 :
Public library service England and Wales Book issues and expenditure Total expenditure Year |Book issues | Cash |Real terms |1991-92 prices |Million | £ million | £ million --------------------------------------------------------------------------- 1989-90 |501 |553 |642 1988-89 |480 |505 | 624.5 1987-88 |525 |450 | 596.5 1986-87 |524 |432 |604 1985-86 |566 |404 | 582.5 1984-85 |551 | 377.5 |575 1983-84 |576 | 361.5 |578 1982-83 |574 |338 |565 1981-82 |578 |310 |556 1980-81 |579 |276 | 542.5 1979-80 |582 |239 | 556.5 1978-79 |570 | 204.5 |555 1977-78 |566 | 184.5 |555 1976-77 |554 |168 |575 1975-76 |571 |149 |578 1974-75 |547 |113 |550 1973-74 |- | 87.5 |551 1972-73 |- | 76.5 | 478.9 1971-72 |579 | 66.5 | 448.9 1970-71 |562 |57 | 418.2 1969-70 |594 |49 |392 1968-69 |585 |45 |377 1967-68 |559 | 41.5 | 364.5 1966-67 |533 |37 | 335.5 Sources: Municipal Year Book. Public Library Statistics, published by the Institute of Municipal Treasurers and Accountants and in later years the Chartered Institute of Public Finance and Accountancy Library and Information Service Council annual report.
Ms. Walley : To ask the Secretary of State for Energy what representations he has received from pottery manufacturers about the effects of the regulatory regime established by the Electricity Act 1989 as it affects contract users.
Column 437Mr. Heathcoat-Amory : To date my Department has received one representation--from a pottery manufacturer--about electricity contracts.
Mr. Burns : To ask the Secretary of State for Energy if he will make a statement on British Coal's external financing limit for 1991-92.
Mr. Wakeham : British Coal's deep-mine and opencast operations have achieved much in the present financial year and the corporation is likely to record a substantial profit after interest for the second year running. Cost cutting through restructuring has, however, been making calls on British Coal's cash resources in the short-term and it has been agreed that the corporation's external financing limit for 1991-92 should be raised from £542 million to £622 million to provide more financial headroom. This increase will be funded out of the existing vote provision for the current financial year.
Q10. Sir Michael Neubert : To ask the Prime Minister if he has plans to make an official visit to Havering Atte Bower.
The Prime Minister : I am making plans for a series of visits to all parts of the country and hope to include Essex among them.
Mr. Ian Taylor : To ask the Prime Minister whether a successor has been chosen to Lord Justice Lloyd as commissioner appointed under the Interception of Communications Act 1985.
| 275,117
|
Page 1 of 2 results
R.C. Stevenson, Attorney at Law
More info
307 East 2nd Street
Sweeny, TX 77480
W A Orr Jr Law Office
More info
109 North Main Street
Sweeny, TX 77480
Sweeny, Wingate Waterboro & Barrow, P.A. Litigation Defense Law. With Sweeny, Wingate & Barrow, P.A. Saint-Johnsbury you get an attorney committed to excellence, integrity, and service. From insurance defense and trucking litigation, to business. Campbell, Duchess of Argyll Wikipedia Ethel Campbell, Duchess Roanoke-Rapids of Argyll née Whigham;December–Julywas a Grundy British socialite, best remembered for a celebrated Kalkaska divorce case in. USLAW NETWORK, Inc USLAW NETWORK USLAW is an international organization composed of approximayindependent, fullservice firms with roots in civil litigation, including more.
| 302,347
|
African sculpture
James Johnson Sweeney
Published
1970
by Princeton University Press in [s.l.]
.
Written in English
Edition Notes
African sculpture [loan exhibition] circulated by the International Exhibitions Foundation, by William Buller Fagg. Published by International Exhibitions Foundation in [Washington. Written in English. African Sculpture: NOVICA, the Impact Marketplace, features unique african Sculptures at incredible prices handcrafted by talented artisans worldwide.. Dancing in the Wind is a kinetic sculpture which can be swiveled to reveal two very different sides. The recycled metal was partially acid-washed and colored with various oxides and natural rusting. The surface was powdercoated. This looks fabulous at sunset with the sun glinting off the steel. Shipping and handling charges are additional.
Schiffer Publishing African Sculpture - A comprehensive introduction to the vast range of tribal sculpture from Africa is presented in this photographic survey. Ashanti fertility dolls, Bambara dance headpieces, Bachokwe staff heads, and Bakuba boxes are included in works from Senegal to the Congo regions, Mali to Sierra Leone. This book provides a tremendous. A wide variety of african metal sculptures options are available to you, such as painted, polished, and carved. You can also choose from thanksgiving, new year african metal sculptures, as well as from art & collectible, home decoration, and souvenir african metal sculptures, and whether african metal sculptures is sculpture, figurine, or painting.
Recitative and improvisation, for four kettledrums (one player)
evolution of ferry service in the San Juan Islands
Chemical biology
future of hospital trusteeship
borrowing and lending activities of the Community.
Fore Street Methodist Church Redruth, 1865-1965; a social history [by] J. C. C. Probert. Issued in connection with the centenary celebrations.
Antarctic Environmental Protection Act of 1990
Social history of modern India: nineteenth century
Trends in Continental Defence
Statue African Figurine Sculpture Colorful Dress Sitting Lady Figurine Vase Statue Decor Collectible Art Piece " Inches Tall - Flower Dress Tropical -Body Sculptures Decorative Black Figurines. Soapstone African Elephant - Figurine Sculpture - Handmade in Kenya - 2 Inches Height x 3 Inches Long, Aqua Blue, SS16 $ $ 90 Get it as soon as Thu, Feb African Tribal Sculptures: Vol.
elizrosshubbell.com Niger Basin Tribes - by William Fagg - and a great selection of related books, art and collectibles available now at elizrosshubbell.com 6, - Explore faymildredfay's board "African Sculpture" on Pinterest. See more ideas about African sculptures, Sculpture and Sculpture art. Risultati immagini per Book Figure Sculpting Volume I: Planes & Construction Techniques in Clay by Philippe Faraut Explore Eun Ju's photos on Photobucket.
African Sculpture book. Read reviews from world’s largest community for readers. A stunning collection of photographic plates of African masks, votiv /5(5). Most African sculpture was historically in wood and other organic materials that have not survived from earlier than at most a few centuries ago; older pottery figures are found from a number of elizrosshubbell.com are important elements in the art of many peoples, along with human figures, often highly stylized.
There is a vast variety of styles, often varying within the same context of origin. The first full appraisal of African art published in the United States, African Sculpture Speaks describes and illustrates the sculpted works of more than West African tribes. }Historian and collector Ladislas Segy approaches African art from several different but interrelated perspectives, considering sculptures first as products of a distinct African culture, then as high-quality works.
Get the best deals on Antique African Sculptures & Statues when you shop the largest online selection at elizrosshubbell.com Free shipping on many items African Art Sculpture Divination Bowl Bearer Totem Wood Spirit Offerings Healing. $ African Folk Art Tribal...
Book is in very good condition with light browning to the paper but no marks inside the text. The cover is in very nice shape and only very lightly bumped on the bottom. Folktales are presented in the front of the book and dozens of black and white photos of sculptures are presented in the back.
African Folktales and Sculpture (Bollingen. You searched for: african. African Sculpture Speaks book. Read reviews from world’s largest community for readers. Historian and collector Ladislas Segy approaches African art from /5(5). The Woodrow Nash Studio”. Feb 26, - African art books are very important to learn to recognise the genuine pieces, a few quality books I selected #africanart #books.
See more ideas about African art, African and Book art. Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied.
Oct 15, · This – 23rd March ) and has now been translated into English.
African sculpture, in all its forms, relates to a myriad of social and religious perspectives and traditions and through the study of these corporeal art forms, we can have insight into the origins of humanity and to the abstract forces that have shaped human perception. no less than figures and heads were. The History of Sculpture from The New Book of Knowledge Greece and Rome for inspiration, artists of the 20th century looked to the simple and powerful forms of the primitive African and Oceanic art.
Wilhelm Lehmbruck (), the German sculptor, began under the influence of Maillol.African sculptures is inspiration due to their distinct charm, fauvism, expressionism, and cubism apart from their aesthetically pleasing complex designs.
Their artistic sensibilities featured in different forms of African sculptures are the sole reason for the aestheticism of these .
| 9,389
|
If a stronger immune system is what you seek, consider a spell of Biodanza each week! Regular Biodanza can offer restoration and stabilisation for those with weak immune systems. The founder of read more
Dissolving stress
Eight benefits of dancing Biodanza
Fiona’s Biodanza story
Your instincts are the key to your own happiness
- « Previous Page
- 1
- …
- 3
- 4
- 5
| 18,951
|
Flyff Iblis
Please add the Flyff Iblis Philippines server… I really want to try this game using your app. I hope you can add this game to your future update. Thanks
- LeoPride moderators
Is this spam?
I’m confused, as we don’t offer downloads or server search functions at all.
| 125,547
|
Likestory.net gathered for you 13 best cooking tips. Read them and pleasantly surprise your loved ones with delicious dinner.
1. Just add the two pinches of instant coffee to the sauce and it will gain incomparable taste.
2. To milk for a long time remain fresh, add a pinch of salt.
3. You only need 2 apples. Slice them, sprinkle with lemon juice and a drop of oil. Then add this mixture and cream or milk in mashed potatoes. It turns out great!
4. You don’t like the looks of this fish? How to determine its freshness?
Place it in a bowl of cool water. If it sinks to the bottom — the fish is fresh.
Don’t like to clean fish? Rub it with vinegar and wait a while. You will be surprised the result.
5. Champagne is not a lot of ever and lost all their taste? Throw a few raisins. This would fix everything.
6. If you rinse the pan with cold water before you boil it, the milk will not burn.
7. Before use, soak the raisins in warm water with a drop of lemon juice.
8. Add 1 or 2 tablespoons of vinegar to the meat, which is stewed. It will be much softer and get ready faster.
9. Take the yolk of hard-boiled eggs, sour cream and a spoonful of mustard, stir and get a delicious homemade mayonnaise.
10. The rice will turn out much tastier if you cook it with the nipple of grapefruit.
11. Food film will behave much better if you keep it in the fridge.
12. Add in the scrambled eggs a pinch of baking soda. Such a lush omelet you’ve ever tasted!
13. If the cakes together with sugar , add the cinnamon, cardamom, star anise or dried mint, the fragrance will be unforgettable!
| 253,955
|
In an excellent review blasting the false dichotomy of more versus less regulation (for additional commentary, see Amanda and Ezra Klein), economist Dean Baker proposes that the government get into the drug development business directly:
.
Mind you, NIH doesn’t have to be the agency to do this, although they would be the first option (in that NIH already exists). I don’t think the “master contractors” idea is too good–I think this could stifle innovation (what if they’re biased towards or against certain approaches) as well as potentially become corrupted, but NIH should be much more heavily involved in drug development, including phase II and phase III trials (which are very expensive). If nothing else, this might make it easier to find and track willing subjects.
Discuss.
| 343,146
|
All images copyright © 2002-2020 George Widman Photography LLC, unless otherwise noted. All text and images appearing on any pages of this Web site are property of George Widman Photography LLC, and are protected by United States and international copyright laws. No images may be used in any form without written permission of George Widman Photography LLC and payment of required usage fees.
To receive permission and reproduction rights, contact George Widman at geowidman@gmail.com. Phone: (610) 489-7775
| 85,083
|
TITLE: Who is $r$, $t$ and $s$ in this complex variable context?
QUESTION [0 upvotes]: I was trying to understand Nitsche's Elementary Proof of Bernstein's Theorem on Minimal Surfaces (referenced on this question) but on the very first page, the statement reads:
Let the function $z = z(x,y)$ be twice continuously differentiable and satisfy the equation $rt-s^2 = 1$, $r > 0$ for all values of $x$ and $y$. Then $z(x,y)$ is a quadratic polynomial.
I have not understood who $r$, $t$ and $s$ are. At first I thought that $z = r + is$ and $t$ is some real number but then the functions $p,q$ appear and it would make so much more sense that $z = p + iq$. Any help or insight is appreciated.
REPLY [1 votes]: Upon looking in further into reference [2] in the article I found that
\begin{align}
r = z_{xx}\\
s = z_{xy}\\
t = z_{yy}
\end{align}
And so, for $z$ to be a solution to equation posted in the question it means that
$$
z_{xx}z_{yy} - z_{xy}^2 = 1
$$
| 138,944
|
Support MGoBlog: buy stuff at Amazon
I received a confirmed report that they were simply a new style that Adidas was testing and will continue to test throughout spring ball. Like some of the styles we saw last year in spring/fall camp, they may work their way into games if they receive enough support from players and staff.
Each chit has the players name on it so you must show ID when buying a meal with it, preventing anyone but the specific player from using it
See my post above.
I've had non-UM students come to games with me and we go to the North gate and they validate the tickets for $25 without any need for an M-Card.
I talked with a couple guys about them and they all said they ride up under their pads, making them pretty uncomfortable, aside from the fact that they're way too tight.
Kovacs plays Bandit (weak side) and T. Gordon plays Spur (strong side). Cam is moving to Spur not Bandit and will replace/share time with T. Gordon and Vinopal will play FS.
Long time reader, first post ever but I feel it's worth it...
Lloyd Brady is in one of my economics classes, I sat in front of him yesterday
...Associate Athletic Director
One of the most stand-up guys I’ve ever had the privilege of meeting. After meeting him for the first time, the second I saw him he remembered my name and some things we had talked about during the first time we spoke. And he is like that with everyone that he comes in contact with, he’s incredible. Tom Brady is quoted as saying Greg Harden was the reason he decided to stay at Michigan. He is a class act and a true “Michigan Man”. Take advantage of any opportunity you have to meet him.
I'm going into my junior year as well. I decided to get tickets without a group this year to insure I got the best possible seats and then go and sit with my friends wherever they end up. I am ahead of schedule credit-wise so I assumed I would get some pretty good seats. I got my tickets in the mail the other day, section 28, row 62, AS A JUNIOR!!! So count your blessings, you're a lot better off than some of us.
| 199,552
|
Cubs fever may be hitting the town now with their rush towards a possible National League Central crown. But apparently "cup fever" hit the town earlier this summer. That's when residents purchased souvenir Silver Hawks cups at McDonald's that helped raise more than $40,000 for the Ronald McDonald Family Room at Memorial Hospital. Besides McDonald's restaurants in Michiana, McDonald's in Fort Wayne also participated, with Wizards cups. Together, the two teams reached the $40,032 figure as fans purchased 300,000 cups. "We just think it's an amazing thing to do," said Rita Baxter, who heads the corporate sales department for the Silver Hawks. "Part of it is we are just thrilled that there is a place in Michiana that families can utilize. And to be able to offer any kind of financial support is just amazing for us. "It's part of what we want to do as corporate citizens." The Ronald McDonald Family Room is a home away from home for parents of seriously ill or injured children. Since opening its doors in November 2002, according to a release, it has served 12,000 families as day visitors or overnight guests. The Family Room includes three bedrooms, two baths with showers, a laundry room, kitchen, library and a common area. Due to community financial and in-kind support, there is no cost to families who utilize the Family Room. Joe Kernan, Silver Hawks owner, indicated the club hopes to also be involved in the future, as it has been each of the last two seasons. "This was a great opportunity for the Silver Hawks to help an organization that truly benefits children and families in our community," he said in a release. Dave Sparks, owner of South Bend area McDonald's restaurants and a member of the local Ronald McDonald House Charities board of directors, saluted the Silver Hawks' support. "And we look forward to more successful partnerships in the future," he said in a release. "McDonald's is proud of the service the Family Room provides to so many families that are going through a difficult time."Staff writer Jim Meenan: jmeenan@sbtinfo.com (574) 235-6342
| 150,289
|
1739 West Nursery Road, LINTHICUM HEIGHTS, MD 21090
The hotel is now 100PCT Non-Smoking Ideally located adjacent to the airport grounds our ultra-modern Hilton Baltimore hotel is only 4 miles from the Maryland Live Casino 10 miles from the shopping and entertainment venues of the Inner Harbor and 30 miles from Washington D.C. Business guests will also enjoy proximity to major corporations and military and government agencies. Select from 280 guest rooms and suites - each featuring plush furnishings chic bathrooms and Hilton hotel*s signature Serenity bedding. Admire our inspiring collection of paintings by local artists a shining example of our belief that travel should enliven the mind and spirit. Establishing a new standard of hospitality for surrounding Hilton Maryland hotels we offer you a dynamic atmosphere and a picturesque setting 280 deluxe guest rooms and suites including three Executive Level Suites and one Presidential Suite Inspiring original artwork throughout the hotel by local artists Wired and wireless Internet access in all accommodations and meeting areas 16 000 square feet of flexible function space among the biggest in the airport area 24-hour business center offering fax email and photocopying services Heated indoor swimming pool and whirlpool for delightful relaxation State-of-the-art fitness center with 14 stations of the latest equipment Second-floor sun deck with stunning views of flights arriving to and departing from BWI Thurgood Marshall Airport acqua our signature restaurant an upscale dining experience featuring American cuisine influenced by Chesapeake Bay flavors and traditional favorites the bar at acqua a relaxed setting for after-work cocktails 24-hour complimentary shuttle to BWI Thurgood Marshall Airport Amtrak and MARC train and Light Rail Room service by acqua for in-room dining Hilton HHonors frequent guest reward program
| 51,920
|
TITLE: There is something wrong in this contour integral
QUESTION [1 upvotes]: I am trying to prove that $$I=\int ^{\infty}_{-\infty}{\frac {dy} {1+y^2} \frac {s+\cosh {(\pi y)}} {s^2+2s\cosh {(\pi y)} +1}} =\frac {\pi} {s-1}-\frac {\pi} {s\ln {s}}$$ One easily can set any real value to $s$ (providing $s>1$) and calculate this integral at any online integral calculator (Wolfram for example), compare it with the right side of equation value and convince himself that the statement above is true. But when I carry out the integration, I find that above the real line there are three poles of integrand: $+i$, $-\frac{\ln{s}}{\pi}+i$, and $\frac{\ln{s}}{\pi}+i$. So I can write $$\oint {\frac {dz} {1+z^2} \frac {s+\cosh {(\pi z)}} {s^2+2s\cosh {(\pi z)} +1}}=I+\lim_{R \rightarrow \infty}\int ^{\pi}_{0}\frac{Re^{i\theta}id\theta}{1+R^2e^{2i\theta}}\frac{s+\cosh{(\pi Re^{i\theta})}}{s^2+2s\cosh{(\pi Re^{i\theta})}+1}$$
The integral in $\theta$ vanishes as $R$ goes to infinity. So, $I$ is the sum of residues at the three poles. Now the trouble appears: the first pole, $i$, generates the $\pi/(s-1)$; and the two other poles generate the term $$-\frac{\pi}{s\ln{s}}\frac{4\pi^2}{(\ln{s})^2+4\pi^2} $$ I must be making some mistake, because that term produces the wrong value, but I can't figure out what is it. If anyone finds it, would be great! Thank you.
REPLY [1 votes]: The integral along the upper semicircular arc does not necessarily vanish as $R \to \infty$, since the contour will pass through infinitely many poles as $R$ grows.
Indeed, assuming $s > 1$, note that the integrand
$$ f(z) = \frac{1}{z^2 + 1} \frac{s + \cosh(\pi z)}{s^2 + 2s \cosh(\pi z) + 1} $$
has infinitely many simple poles in the upper-half plane, which can be classified into three types:
$z = i$
$z = z^{+}_k := \frac{1}{\pi} \log s + (2k+1) i$ for $k = 0, 1, 2, \dots$
$z = z^{-}_k := -\frac{1}{\pi} \log s + (2k+1) i$ for $k = 0, 1, 2, \dots$
Now let $K$ be a positive integer and $R$ be a sufficiently large positive real, and consider the contour integral of $f$ along the boundary of the rectangle with corners $\pm R$ and $\pm R + 2 K i$. If $[z_0, z_1]$ denotes the directed line segment from $z_0$ to $z_1$, the residue theorem tells that
\begin{align*}
\newcommand{\Res}{\mathop{\operatorname{Res}}}
\int_{[-R,R]} f(z) \, \mathrm{d}z
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^-_k} f(z) \biggr) \\
&\quad - \int_{[R,R+2Ki]} f(z) \, \mathrm{d}z - \int_{[R+2Ki,-R+2Ki]} f(z) \, \mathrm{d}z - \int_{[-R+2Ki,-R]} f(z) \, \mathrm{d}z.
\end{align*}
Letting $R \to \infty$, the integrals along $[R,R+2Ki]$ and $[-R+2Ki,-R]$ vanish, hence
\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{K-1} \Res_{z=z^-_k} f(z) \biggr) + \int_{-\infty}^{\infty} f(x + 2Ki) \, \mathrm{d}x
\end{align*}
Then it is not hard to check that the last integral vanishes as $K \to \infty$, and so,
\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= 2\pi i \biggl( \Res_{z=i} f(z) + \sum_{k=0}^{\infty} \Res_{z=z^+_k} f(z) + \sum_{k=0}^{\infty} \Res_{z=z^-_k} f(z) \biggr)
\end{align*}
In order to compute this sum, note that
\begin{align*}
\Res_{z=z^+_k} f(z)
&= \frac{1}{1 + (z^+_k)^2} \frac{s + \cosh(\pi z^+_k)}{2 \pi s \sinh(\pi z^+_k)} \\
&= \frac{1}{(z^+_k + i)(z^+_k - i)} \frac{s - \bigl( \frac{1 + s^2}{2s} \bigr)}{2\pi s \bigl( \frac{1 - s^2}{2s} \bigr)} \\
&= -\frac{1}{4\pi i s} \biggl( \frac{1}{z^+_k - i} - \frac{1}{z^+_k + i} \biggr) \\
&= -\frac{1}{4\pi i s} \biggl( \frac{\pi}{\log s + 2k \pi i} - \frac{\pi}{\log s + 2(k+1)\pi i} \biggr)
\end{align*}
and
\begin{align*}
\Res_{z=z^-_k} f(z)
&= -\frac{1}{4\pi i s} \biggl( \frac{\pi}{\log s - 2k \pi i} - \frac{\pi}{\log s - 2(k+1)\pi i} \biggr).
\end{align*}
So the sums are telescoping and hence yield
\begin{align*}
\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x
&= \frac{\pi}{s-1} - \frac{\pi}{s \log s}.
\end{align*}
| 112,049
|
• Highlight: Running back Ronnie Hillman is known as a speedster, not a bruiser, but he got the better of safety Duke Ihenacho during a red-zone series.
Hillman caught a dump pass from Peyton Manning, then had a full head of steam as he met a flat-footed Ihenacho near the goal line. Hillman knocked Ihenacho on his keister as he crossed into the end zone.
Should have heard running back Knowshon Moreno howl from the sidelines.
• Lowlight: Eventually, the cornerback position is expected to be a team strength. But it doesn't always look that way in camp as Dominique Rodgers-Cromartie has been out with a high ankle sprain, Champ Bailey is under orders to save his 35-year-old legs and Tony Carter missed much of practice Tuesday with a bruised knee.
Omar Bolden, a fourth-round pick in the 2012 draft who primarily played special teams last year, got some first-team work the past two days. He got burned by Demaryius Thomas for a long touchdown pass thrown down the left sideline by Manning.
• QB watch: There was considerable situational work Tuesday as the Broncos prepared for their first preseason game Thursday night against the San Francisco 49ers.
Near the end of practice, the offense was down 24-20, but given 1 minute, 40 seconds to score from its 40-yard line. Manning moved his No. 1 unit with a nice completion to tight end Julius Thomas, who made a reaching, one-handed catch, before connecting with Thomas for the score.
Backup Brock Osweiler also moved the No. 2 offense for a score in methodical fashion, completing passes to Gerell Robinson, who made a nice diving catch near the sideline, and tight end Virgil Green. Jacob Hester finished off the drive with a 2-yard touchdown run.
• Position battle: Just when you think Nate Irving has won the starting middle linebacker spot, Stewart Bradley rotated back in with the No. 1 unit this week.
"Nothing to figure out, just working every day and trying to get more comfortable with the new system," Bradley said.
Irving started in the scrimmage Saturday, had an excused absence from practice Monday and returned Tuesday to work with the No. 2 defensive unit.
"We're just looking," said Broncos coach John Fox. "That's why we have camp."
• Next: There will be a walkthrough Wednesday morning that is closed to the public and media. The Broncos will then fly to San Francisco, where they will play their first preseason game Thursday night against the 49ers. The Broncos will be off Friday and return to training camp for practice at 8:50 a.m. Saturday.
| 268,002
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.